// Numbas version: finer_feedback_settings {"name": "PHY 2025/26", "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], []], "questions": [{"name": "AC15 Algebraic substitution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Substitute values into an algebraic expression and calculate the result.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate the following expression,

\\[\\simplify{p^{n}+{a}*r*t+{c}},\\]

when $p = \\var{pval}$, $r = \\var{rval}$, and $t = \\var{tval}$.

", "advice": "

In order to evaluate $\\simplify{p^{n}+{a}*r*t+{c}},$ with the given values, $p = \\var{pval}$, $r = \\var{rval}$, and $t = \\var{tval}$, we replace each instance of that letter with its corresponding value and then apply the rules of BIDMAS:

\\[\\var{pval}^\\var{n}+\\var{a}\\times \\var{rval} \\times \\var{tval} + \\var{c}\\]

Which gives the answer $\\var{ans}$.

Follow this link for more help on tackling these kind of questions.

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Rearrange a specific formula. No randomisation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the following equation, to make $y$ the subject:

\\[{cy -b = 3x}\\]

", "advice": "

In order to rearrange the equation so that it is in terms of $y$, we must first add $b$ to both sides, and then divide both sides of the equation by $c$:

\\begin{split} cy-b &= 3x \\\\ cy &= 3x + b \\\\ y &=\\frac{3x+b}{c} \\end{split}

Use this link to find some resources which will help you revise this topic.

$y=$ [[0]]

Simplify (qx+a)/(rx+b) +/- (sx+c)/(tx+d)

x is a randomised variable. a,b,c,d,q,r,s,t are randomised integers. a,b,c,d run from -5 to 5, including 0. q,r,s,t run from -3 to 3, and can be 0 if the constant term is nonzero, but are mostly 1.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express $\\displaystyle{\\var{te[0]}\\var{sgnl}\\var{te[1]}}$ as a single fraction.

", "advice": "

\\[\\begin{align*} \\var{te[0]}\\var{sgnl}\\var{te[1]} &= \\frac{\\var{ndnde[3]}}{\\var{ndnde[3]}}\\times\\var{te[0]}\\var{sgnl}\\frac{\\var{ndnde[1]}}{\\var{ndnde[1]}}\\times\\var{te[1]}\\\\&=\\frac{\\var{cnd[0]}\\var{sgnl}\\var{cnd[1]}}{(\\var{ndnde[1]})(\\var{ndnde[3]})}\\\\&=\\frac{(\\var{cnd[2]})\\var{sgnl}(\\var{cnd[3]})}{(\\var{ndnde[1]})(\\var{ndnde[3]})}\\\\&=\\frac{\\var{cnd[4]}}{(\\var{ndnde[1]})(\\var{ndnde[3]})}\\\\&=\\var{ans} \\end{align*}\\]

There is no benefit in expanding the denominator. In fact, it is best to leave the denominator factorised, because then it is easier to see if the fraction can be simplified.

Use this link to find some resources which will help you revise this topic.

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the variable to use

", "templateType": "anything", "can_override": false}, "xc": {"name": "xc", "group": "Ungrouped variables", "definition": "repeat(weighted_random([ [1,0.7] , [random(1..3),0.3] ])\n ,4)", "description": "

the x-coefficients

", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "map(\nswitch(xc[i]=0, random(2..5),\n xc[i]=1, random(-5..5 except 0),\n xc[i]=2, random(-5..5 except [-4,-2,2,4]),\n random(-5..5 except [-xc[i],xc[i]])\n),i,[0,1,2,3])", "description": "

the constants. Make sure that the constants are coprime with their x-coefficient, and if their x-coefficient is 0, that they are positive. There is a condition in variable testing to ensure that no fraction = 1.

", "templateType": "anything", "can_override": false}, "ndnd": {"name": "ndnd", "group": "Ungrouped variables", "definition": "[\"+\"+c[0],\n xc[1]+\"*\"+v+\"+\"+c[1],\n \"+\"+c[2],\n xc[3]+\"*\"+v+\"+\"+c[3]\n ]", "description": "

numerator 1, denominator 1, numerator 2, denominator 2, as strings.

", "templateType": "anything", "can_override": false}, "sgn": {"name": "sgn", "group": "Ungrouped variables", "definition": "random(\"+\",\"-\")", "description": "", "templateType": "anything", "can_override": false}, "sgnl": {"name": "sgnl", "group": "Ungrouped variables", "definition": "latex(sgn)", "description": "

for display purposes

", "templateType": "anything", "can_override": false}, "ndnde": {"name": "ndnde", "group": "Ungrouped variables", "definition": "map(simplify(expression(x),\"all\"),x,ndnd)", "description": "", "templateType": "anything", "can_override": false}, "cnd": {"name": "cnd", "group": "Ungrouped variables", "definition": "[simplify(expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),\"all\"),\nsimplify(expression(\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),\"all\"),\nsimplify(expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),[\"expandBrackets\",\"all\"]),\nsimplify(expression(\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),[\"expandBrackets\",\"all\"]),\nsimplify(expression(\n string(simplify(\n expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"])\n )+\"+\"+\n string(simplify(\n expression(sgn+\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"])\n )\n),[\"expandBrackets\",\"basic\"]),\n \nsimplify(expression(\n string(simplify(expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"]))+ \"+\" +\n string(simplify(expression(sgn+\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"]))\n ),[\"all\",\"!noLeadingMinus\"])\n ]", "description": "

The combined numerator and denominator terms:

0) numerator term 1, 1) numerator term 2,

2) brackets expanded num t1, 3) brackets expanded num t2

4) numerator, no brackets

5) numerator simplified

", "templateType": "anything", "can_override": false}, "ansnum": {"name": "ansnum", "group": "Ungrouped variables", "definition": "simplify(expression(\n string(simplify(expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"]))+ \"+\" +\n string(simplify(expression(sgn+\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"]))\n ),[\"all\",\"!noLeadingMinus\"])", "description": "", "templateType": "anything", "can_override": false}, "ansden": {"name": "ansden", "group": "Ungrouped variables", "definition": "simplify(expression(\"(\"+ndnd[1]+\")*(\"+ndnd[3]+\")\"),\"all\")", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Ungrouped variables", "definition": "expression(\"(\"+string(ansnum)+\")/(\"+string(ansden)+\")\")", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "(xc[0]<>xc[1] or c[0]<>c[1]) and (xc[2]<>xc[3] or c[2]<>c[3])", "maxRuns": "76"}, "ungrouped_variables": ["v", "xc", "c", "ndnd", "te", "sgn", "sgnl", "ndnde", "cnd", "ansnum", "ansden", "ans"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{ans}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "mustmatchpattern": {"pattern": "?`+/?`+", "partialCredit": 0, "message": "You need to give your answer as just one fraction", "nameToCompare": "", "warningTime": "submission"}, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AC18 Algebraic Fractions - addition (harder)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Simplify the sum of two algebraic fractions where spotting factorising of both numerators and denominators can reduce the work massively.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Write the following as a single fraction $\\frac{\\var{num1}}{\\var{den1}}+\\frac{\\var{num2}}{\\var{den2}}$ simplifying as much as possible. Your answer should be in the form $\\frac{\\alpha\\var{v}+\\beta}{\\delta\\var{v}^2-\\gamma}.$

", "advice": "

To write the following as a single fraction $\\frac{\\var{num1}}{\\var{den1}}+\\frac{\\var{num2}}{\\var{den2}}$ first factorise as much as possible and look for any cancellations:

\\[\\begin{split}
&\\frac{\\var{a}\\times\\var{b}}{\\var{den1fact}} + \\frac{\\var{num2}}{\\var{den2fact}}\\\\
& = \\frac{\\var{b}}{\\var{den1simp}} + \\frac{1}{\\var{f1c}}.
\\end{split}\\]

Then get a common denominator for the two fractions and combine into a single fraction:

\\[\\begin{split}
&\\frac{\\var{b}}{\\var{den1simp}} + \\frac{\\var{f1}}{\\var{den1simp}}\\\\
& = \\frac{\\var{b}+\\var{f1}}{\\var{den1simp}}\\\\
& = \\var{ans}.
\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

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{"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Luke Park", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/826/"}, {"name": "Anna Strzelecka", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2945/"}, {"name": "heike hoffmann", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2960/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": [], "metadata": {"description": "

A question to practice simplifying fractions with the use of factorisation (for binomial and quadratic expressions).

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify the following algebraic expression.

", "advice": "

\\[\\frac{{\\simplify{(n^2+({e1}+{e2})n+{e1}{e2})}}}{{\\simplify{(n^2+({e1}+{e3})n+{e1}{e3})}}}\\]

In this question there is a quadratic expression which needs to be factorised into the products of binomials in both the numerator and denominator.

\\[\\frac{({\\simplify{n+{e1}}})({\\simplify{n+{e2}}})}{({\\simplify{n+{e1}}})({\\simplify{n+{e3}}})}\\]

The repeated binomials in the numerator and denominator cancel, leaving:

\\[\\frac{({\\simplify{n+{e2}}})}{({\\simplify{n+{e3}}})}\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"e2": {"name": "e2", "group": "Ungrouped variables", "definition": "random(-5..5 except 0)", "description": "", "templateType": "anything", "can_override": false}, "e1": {"name": "e1", "group": "Ungrouped variables", "definition": "random(-5..5 except 0)", "description": "", "templateType": "anything", "can_override": false}, "e3": {"name": "e3", "group": "Ungrouped variables", "definition": "random(-5..5 except 0 except e2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["e1", "e2", "e3"], "variable_groups": [], "functions": {"": {"parameters": [], "type": "number", "language": "jme", "definition": ""}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

\\[\\frac{\\simplify{(n^2+({e1}+{e2})n+{e1}{e2})}}{\\simplify{(n^2+({e1}+{e3})n+{e1}{e3})}}\\]

", "answer": "(n+{e2})/(n+{e3})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "notallowed": {"strings": ["^2", "^"], "showStrings": false, "partialCredit": 0, "message": ""}, "valuegenerators": [{"name": "n", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AC20 Multiplication of algebraic fractions 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Lauren Richards", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1589/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": ["adding and subtracting fractions", "adding fractions", "converting between decimals and fractions", "converting integers to fractions", "Fractions", "fractions", "integers", "manipulation of fractions", "subtracting fractions", "taxonomy"], "metadata": {"description": "

Manipulate fractions in order to add and subtract them. The difficulty escalates through the inclusion of a whole integer and a decimal, which both need to be converted into a fraction before the addition/subtraction can take place.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following multiplication, giving the fraction in its simplest form.

\\[\\frac{\\var{a}}{y^\\var{n2}}\\times\\frac{y^\\var{n1}}{\\var{b}}\\]

", "advice": "

To multiply two fractions you just multiply the numerators and multiply the denominators. This means we have,

\\[\\frac{\\var{a}}{y^\\var{n2}}\\times\\frac{y^\\var{n1}}{\\var{b}}=\\frac{\\var{a}\\times{y^\\var{n1}}}{y^\\var{n2}\\times\\var{b}}=\\frac{\\var{a/gcd_ab}\\times{y^\\var{n1-n2}}}{\\var{b/gcd_ab}}\\]

Use this link to find some resources which will help you revise this topic.

[[0]] [[1]]

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Simplifying first is essential in terms of managing expressions that might need factorising.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Expand and simplify $\\displaystyle{\\var{LeftMul}\\times\\var{RightMul}}.$

", "advice": "

Before we multiply the fractions together first lets check if we can do any cancellation. Notice that $\\var{RightMulBottom}$ has a factor of $\\var{Num}$ so we can cancel this straight away.

We also have a factor of $x$ in both $\\var{QuadCoeff[0]}x^2+\\var{QuadCoeff[1]}x$ and $\\var{RightMulTop}$ so we're now left with multiplying

\\[\\frac1{\\var{QuadCoeff[0]}x+\\var{QuadCoeff[1]}}\\times\\frac{\\simplify[all,expandBrackets]{(x+{Lin1Coeff})*({QuadCoeff[0]}x+{QuadCoeff[1]})}}{\\var{Lin2Coeff[0]}x+\\var{Lin2Coeff[1]}}.\\]

We're not necesserily done with cancellation though! To make sure that a fraction with a quadratic is simplified we have to factorise it to make sure there are no linear factors we can cancel. In this case we have
\\[\\simplify[all,expandBrackets]{(x+{Lin1Coeff})*({QuadCoeff[0]}x+{QuadCoeff[1]})}={(x+\\var{Lin1Coeff})(\\var{QuadCoeff[0]}x+\\var{QuadCoeff[1]})}.\\]

This gives us one last factor to cancel and then we can finally multiply whats left of each fraction to give us a final answer of

\\[\\var{ans}.\\]

Use this link to find some resources which will help you revise this topic.

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Rewrite the expression $\\frac{mx^2+nx+k}{(x+a)(x^2+bx+c)}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{Bx+C}{x^2+bx+c}$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rewrite the following expression as partial fractions:

\\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))}. \\]

", "advice": "

To express \\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} \\] as partial fractions, we want to set this equal to the sum of two fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x^2+{b}x+{c}}$. Since we have a linear factor and a quadratic factor, this tells us that the form of the partial fractions will be

\\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\simplify{A/(x+{a}) + (B*x+C)/(x^2+{b}x+{c})},\\]

where $A$, $B$, and $C$ are constants.

To find the values of $A$, $B$, and $C$, we want to first multiply this equation by the denominator of the left-hand side. This gives

\\[ \\simplify{{m}x^2+{n}x+{k}=A(x^2+{b}x+{c})+B*x(x+{a}) + C(x+{a})}.\\]

(Note: To find $A$, $B$, and $C$, we will use a combination of choosing suitable values of $x$ to eliminate terms, and equating coefficients. It can be solved by only equating coefficients, but this is a more efficient process.)

To find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:

\\[ \\simplify{{m*a^2-n*a+k}=A{(a^2-b*a+c)}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]

To find $C$, we can eliminate $B$ by setting $x=0$ and substituting in the result of $A$:

\\[ \\simplify{{k}={c}A+{a}C} \\implies C=\\simplify[all,fractionNumbers]{({k}-{c}A)/{a}}.\\]

Hence,

\\[ C = \\simplify[fractionNumbers]{{Csol}}.\\]

Finally, by equating coefficients of the $x^2$-terms we can find $B$:

\\[ (x^2): \\quad \\var{m} = \\simplify{A+B} \\implies B=\\var{m}-A. \\]

Therefore, \\[ B=\\simplify[fractionNumbers]{{Bsol}}, \\]

and

{check}

Use this link to find some resources which will help you revise this topic.

\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify{{Asol}/(x+{a})+({Bsol}x+{Csol})/(x^2+{b}x+{c})}.\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sol2": {"name": "sol2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify[all,fractionNumbers]{{m*a^2-n*a+k}/({a^2-a*b+c}(x+{a}))+({m*c-m*b*a+n*a-k}x+{k*(a-b)-m*a*c+n*c})/({a^2-a*b+c}(x^2+{b}x+{c}))}.\\\\]

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\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify[all,fractionNumbers]{{m*a^2-n*a+k}/({a^2-a*b+c}(x+{a}))+({(m*c-m*b*a+n*a-k)/simp2}x+{(k*(a-b)-m*a*c+n*c)/simp2})/({(a^2-a*b+c)/simp2}(x^2+{b}x+{c}))}.\\\\]

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[[0]]

Basic calculation from a sum given in Sigma notation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate:

\\[\\displaystyle{\\Sigma_{n=1}^3} \\var{b}n.\\]

", "advice": "

The sigma notation $\\displaystyle\\sum_{n=1}^{3}\\var{b}n$ is asking us to find the sum of the first three terms of the sequence $\\var{b}n$.

\\[\\begin{split}\\Sigma_{n=1}^3 \\var{b}n &\\, = (\\var{b}\\times 1) + (\\var{b}\\times 2) + (\\var{b}\\times 3) \\\\ &\\, = \\var{b1} + \\var{b2} + \\var{b3} \\\\ &\\, = \\var{sum}.\\end{split}\\]

Use this link to find resources to help you revise sigma notation.

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Calculating gradient and finding intercept from a geogebra graph.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

{app}
Find the gradient of the line.

", "advice": "

Firstly draw a right angled 'step' from left to right. This triangle can be anywhere, but it is more helpful for it to have corners on the vertices (whole number points) of the graph and it is easier to calculate with postive numbers.

{app_advice}

Before we start to calculate, notice that the line is {uod}, so the gradient will be {pon} and the line is {sos}, so the absolute value of the number will be {mol}.

Now find the coordinates of the places your triangle meets the line

$(x_1,y_1)=(\\var{ax},\\var{ay})$ and $(x_2,y_2)=(\\var{bx},\\var{by})$

We need to compare the 'rise on the y-axis' to the 'run across the x-axis', we can say that:

$\\text{gradient} = \\frac{\\text{rise}}{\\text{run}}$

This is equivalent to using the formula:

$ m = \\frac{y_2 - y_1}{x_2 - x_1} $

and substitute the coordinates of the vertices of the triangle:

$\\begin{split} &\\, m = \\frac{\\var{by} - \\var{ay}}{\\var{bx} - \\var{ax}} \\\\
&\\, = \\frac{\\var{by-ay}}{\\var{bx-ax}} \\\\
&\\, = \\var[fractionNumbers]{m} \\\\
\\end{split} $

Use this link to find resources to help you revise straight line graphs and how to find the gradient of them.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"app": {"name": "app", "group": "Ungrouped variables", "definition": "geogebra_applet(\n 800,500,\n [\n A: [\n definition: p1,\n label_visible: false,\n visible: false\n ],\n B: [\n definition: p2,\n label_visible: false,\n visible: false \n ],\n line: [\n definition: \"Line(A,B)\",\n label_visible: false\n ]\n ]\n)", "description": "", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "(ay-by)/(ax-bx)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "ay-m*ax", "description": "", "templateType": "anything", "can_override": false}, "P1": {"name": "P1", "group": "Ungrouped variables", "definition": "vector(ax, ay)", "description": "", "templateType": "anything", "can_override": false}, "P2": {"name": "P2", "group": "Ungrouped variables", "definition": "vector(bx,by)", "description": "", "templateType": "anything", "can_override": false}, "uod": {"name": "uod", "group": "Ungrouped variables", "definition": "if(m=0,'horizontal',if(m=abs(m),'going up','going down'))", "description": "

if(m=abs(m),'positive','negative')

", "templateType": "anything", "can_override": false}, "ax": {"name": "ax", "group": "Ungrouped variables", "definition": "random(0,1)", "description": "", "templateType": "anything", "can_override": false}, "ay": {"name": "ay", "group": "Ungrouped variables", "definition": "random(0,1,2,3)", "description": "", "templateType": "anything", "can_override": false}, "bx": {"name": "bx", "group": "Ungrouped variables", "definition": "random(ax+1..3) \n", "description": "", "templateType": "anything", "can_override": false}, "by": {"name": "by", "group": "Ungrouped variables", "definition": "random(0..4 except ay)\n", "description": "", "templateType": "anything", "can_override": false}, "app_advice": {"name": "app_advice", "group": "Ungrouped variables", "definition": "geogebra_applet(\n 800,500,\n [\n A: [\n definition: p1,\n label_visible: false,\n visible: true\n ],\n B: [\n definition: p2,\n label_visible: false,\n visible: true \n ],\n \n C: [\n definition: p3,\n label_visible: false,\n visible: false \n ],\n \n line1: [\n definition: \"Line(A,B)\",\n label_visible: false,\n visible: true\n ],\n \n line2: [\n definition: \"Segment(A,C)\",\n label_visible: false,\n visible: true\n ],\n \n \n \n line3: [\n definition: \"Segment(C,B)\",\n label_visible: false,\n visible: true\n ]\n ]\n)", "description": "", "templateType": "anything", "can_override": false}, "p3": {"name": "p3", "group": "Ungrouped variables", "definition": "vector(bx,ay)", "description": "", "templateType": "anything", "can_override": false}, "pon": {"name": "pon", "group": "Ungrouped variables", "definition": "if(m=0,'zero',if(m=abs(m),'a positive number','a negative number'))", "description": "", "templateType": "anything", "can_override": false}, "sos": {"name": "sos", "group": "Ungrouped variables", "definition": "if(m=0,'horizontal',if(abs(m)<1,'shallow','steep'))", "description": "", "templateType": "anything", "can_override": false}, "mol": {"name": "mol", "group": "Ungrouped variables", "definition": "if(m=0,'zero',if(abs(m)<1,'less than 1','greater than or equal to 1'))", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "m<>1", "maxRuns": 100}, "ungrouped_variables": ["app", "m", "c", "P1", "P2", "uod", "ax", "ay", "bx", "by", "app_advice", "p3", "pon", "sos", "mol"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you have incorrectly rounded this answer. You might want to look at some resources on rounded decimals. You can also leave your answer in fraction form as
$\\var[fractionNumbers]{m}$

", "useAlternativeFeedback": false, "answer": "{m}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.1", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "{m}", "showPreview": true, "checkingType": "dp", "checkingAccuracy": "1", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AF03 Shapes of quadratics", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Multiple choice - select the quadratic graph.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Which of the following is the graph $y=x^2$.

", "advice": "

Use this link to find some resources to help you familiarise yourself with these graphs.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["{geogebra_applet('https://www.geogebra.org/m/tpfzv3w7')}", "{geogebra_applet('https://www.geogebra.org/m/zftpwq64')}", "{geogebra_applet('https://www.geogebra.org/m/we3gngqa')}", "{geogebra_applet('https://www.geogebra.org/m/cadkup6r')}"], "matrix": ["1", 0, 0, 0], "distractors": ["", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AF04 Graphs of trig functions (sin, cos, tan)", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Match the relevant graph (sin, cos, tan) with its equation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

This is about core knowledge of graphs. You should know the shapes of the fundamental trig graphs, if you don't familiarize yourself with them from the resources linked below. In this setting the $x$-axis is given with a scale in radians but you might also find some where it is given in degrees. You should also be aware of the difference between those two different units of angles.

Use this link to find some resources to help you familiarise yourself with these graphs.

Match the graph to its function.

", "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["$\\sin(x)$", "$\\cos(x)$", "$\\tan(x)$"], "matrix": [["1", 0, 0], [0, "1", 0], [0, 0, "1"]], "layout": {"type": "all", "expression": ""}, "answers": ["{geogebra_applet('https://www.geogebra.org/m/ntqvuwqr')}", "{geogebra_applet('https://www.geogebra.org/m/fsqmnhsc')}", "{geogebra_applet('https://www.geogebra.org/m/yg6f9eqz')}"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AF05 Function notation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Evaluating a linear function for a given value of $x$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Given $f(x)=\\simplify{{m}x+{c}}$, find $f(\\var{n})$.

", "advice": "

If $f(x)=\\simplify{{m}x+{c}}$, to find $f(\\var{n})$ we need to evaluate $f(x)$ when $x=\\var{n}$:

\\[ \\begin{split} f(\\var{n}) &\\,= \\simplify[alwaysTimes]{{m}({n})+{c}} \\\\ &\\,= \\simplify[!collectNumbers]{{m*n}+{c}} \\\\ &\\,= \\simplify{{m*n+c}}. \\end{split} \\]

Use this link to find resources to help you revise function notation.

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$f(\\var{n})=$[[0]]

Determining the range of a function of the form $f = m|x| + a$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

The range is the set of values that can be taken by $f(x)$, i.e. the values on the $y$-axis.

{geogebra_applet('https://www.geogebra.org/m/aqcgkurg',[a: a, m: m])}

Therefore, for $f(x)=\\simplify{{m}x^2+{a}}$, the range is $[\\var{a}, \\infty)$.

Use this link to find some resources to help you revise how to find the domain and range of a function.

Given $f(x)=\\simplify{{m}x^2+{a}}$

What is the range of $f(x)$?

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Finding the inverse of a function of the form $f(x)=\\frac{mx+c}{x+a},\\,x\\neq-a$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

If $f(x)=\\simplify{({m}x+{c})/(x+{a})},\\,x\\neq \\simplify{{-a}}$, find the inverse function, $f^{-1}(x)$.

", "advice": "

To find $f^{-1}x$, it can help to first set $f(x)$ to a different variable, which we will call $y$:

\\[ y = f(x) = \\simplify{({m}x+{c})/(x+{a})}\\]

Since the function $f(x)$ takes us from $x$ to $y$, the inverse function $f^{-1}$ will take us from $y$ to $x$. So to obtain $f^{-1}$, we want to rearrange $y=\\simplify{({m}x+{c})/(x+{a})}$ so that it is $x$ as a function of $y$:

\\[ \\begin{split} y &\\,= \\simplify{({m}x+{c})/(x+{a})} \\\\\\\\ \\simplify{(x+{a})y} &\\,= \\simplify{{m}x+{c}} \\\\\\\\ \\simplify{x*y+{a}y} &\\,= \\simplify{{m}x+{c}} \\\\\\\\ \\simplify{x*y - {m}x} &\\,= \\simplify{{c}- {a}y} \\\\ \\\\ \\simplify{x(y-{m})} &\\,= \\simplify{{c}-{a}y} \\\\\\\\ x&\\,= \\simplify{({c}-{a}y)/(y-{m})}. \\end{split} \\]

Hence, $f^{-1}(y) =\\simplify{({c}-{a}y)/(y-{m})}$, and therefore \\[ f^{-1}(x) =\\simplify{({c}-{a}x)/(x-{m})}.\\]

(Note: The inverse is valid for all values of $x$ except $x=\\var{m}$, since this would make the denominator equal to 0.)

Use this link to find resources to help you revise how to find the inverse of functions.

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$f^{-1}(x)=$[[0]]

Finding composite functions of 2 linear functions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

If $f(x)=\\simplify{{m}x+{c}}$ and $g(x)=\\simplify{{n}x+{d}}$, find expressions for $f\\circ g(x)$ and $g \\circ f(x)$.

Recall: $f \\circ g(x) \\equiv f(g(x))$ and $g \\circ f(x) \\equiv g(f(x))$.

", "advice": "

To find the composition $f \\circ g(x)$ we are substituting the expression for $g(x)$ into the function $f(x)$, replacing the $x$-terms with the function $g(x)$. Similarly, to find the composition $g \\circ f(x)$ we are substituting the expression for $f(x)$ into the function $g(x)$, replacing the $x$-terms with the function $f(x)$.

So, for $f(x)=\\simplify{{m}x+{c}}$ and $g(x)=\\simplify{{n}x+{d}}$,

\\[ \\begin{split} f \\circ g(x) \\equiv f(g(x)) &\\,= \\simplify{{m}({n}x+{d})+{c}} \\\\ &\\,=\\simplify[!collectNumbers,unitFactor]{{m*n}x+{m*d}+{c}} \\\\ &\\,=\\simplify{{m*n}x+{m*d+c}}, \\end{split} \\]

and

\\[ \\begin{split} g \\circ f(x) \\equiv g(f(x)) &\\,= \\simplify{{n}({m}x+{c})+{d}} \\\\ &\\,=\\simplify[!collectNumbers,unitFactor]{{m*n}x+{n*c}+{d}} \\\\ &\\,=\\simplify{{m*n}x+{n*c+d}}. \\end{split} \\]

Use this link to find resources to help you revise how to find composite functions.

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$f \\circ g(x)=$[[0]]

$g \\circ f(x)=$[[1]]

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Finding $x$ from a logarithmic equation of the form $\\log_ax = b$, where $a$ and $b$ are positive integers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the value of $x$:

\\[ \\log_\\var{a}x = \\var{n} \\]

", "advice": "

To find the value of $x$, recall that $\\log_a(x)=b$ is equivalent to $x=a^b$.

Therefore, \\[\\log_\\var{a}(x) = \\var{n} \\implies \\simplify[!collectNumbers]{x={a}^{n}}.\\]

Hence, \\[x=\\var{a^n}\\,.\\]

Use this link to find resources to help you revise logarithms.

$x=$ [[0]]

Solving $a\\log(x)+\\log(b)=\\log(c)$ for $x$, where $a$, $b$ and $c$ are positive integers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve for $x$:

\\[ \\var{a}\\log(x)+\\log(\\var{b})=\\log(\\var{c}). \\]

", "advice": "

To solve $\\var{a}\\log(x)+\\log(\\var{b})=\\log(\\var{c})$ for $x$, we want to use the following logarithm rules:

$\\log(a^n) = n\\log(a)$;
$\\log(a)+\\log(b) = \\log(ab)$.

Hence,

\\[ \\begin{split} \\var{a}\\log(x)+\\log(\\var{b}) &\\,=\\log(\\var{c}) \\\\ \\log(x^\\var{a})+\\log(\\var{b}) &\\,= \\log(\\var{c}) \\\\ \\log(\\var{b}x^\\var{a}) &\\,= \\log(\\var{c}). \\end{split} \\]

If $\\log(a)=\\log(b)$ then this implies $a=b$. Therefore,

\\[ \\begin{split} \\var{b}x^\\var{a} &\\,=\\var{c} \\\\ x^\\var{a} &\\,= \\simplify[fractionNumbers]{{c/b}} \\\\ x &\\,= \\simplify[fractionNumbers]{({c/b})^(1/{a})} \\\\ x &\\,= \\var{sol} \\text{ (2 d.p.)}\\end{split} \\]

Use this link to find rsources to help you revise how the rules of logarithms to help you solve logarithmic equations.

$x=$ [[0]] (Give you answer to 2 decimal places where necessary)

Solving an equation of the form $a^x=b$ using logarithms to find $x$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve for $x$:

\\[ \\var{a}^x = \\var{b} \\,. \\]

", "advice": "

To solve $\\var{a}^x = \\var{b}$ for $x$, since $x$ is the exponent we want to make use of the following logarithm rule:

$\\log(a^n) = n \\log(a)$.

By taking the logarithm of each side and applying the above rule:

\\[ \\begin{split}\\var{a}^x &\\,= \\var{b} \\\\ \\log_{10}(\\var{a}^x) & \\,= \\log_{10}(\\var{b})\\\\ x \\log_{10}(\\var{a}) &\\,= \\log_{10}(\\var{b}) \\\\\\\\ x&\\,=\\simplify{log({b})/log({a})} \\\\\\\\ x &\\,= \\var{sol} \\text{ (2 d.p.)}. \\end{split} \\]

Use this link to find resources to help you revise how logarithms.

$x=$ [[0]] (Give you answer to 2 decimal places where necessary)

Solve linear equations with unkowns on both sides. Including brackets and fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

Given $\\simplify{{m}w-{n} = {p}w+{q}}$, we can get all the $w$'s on the left hand side and all the numbers on the right hand side, and then divide both sides by the coefficient of $w$ to get $w$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n



$\\simplify{{m}w+{n}}$	$=$	$\\simplify{{p}w+{q}}$

$\\simplify[!cancelTerms,unitFactor]{{m}w-{n}-{p}w}$	$=$	$\\simplify[!cancelTerms,unitFactor]{{p}w+{q}-{p}w}$

$\\simplify{{m-p}w-{n}}$	$=$	$\\var{q}$

$\\var{m-p}w-\\var{n}+\\var{n}$	$=$	$\\var{q}+\\var{n}$

$\\var{m-p}w$	$=$	$\\var{q+n}$

$\\displaystyle{\\frac{\\var{m-p}w}{\\var{m-p}}}$	$=$	$\\displaystyle{\\frac{\\var{q+n}}{\\var{m-p}}}$

$w$	$=$	$\\displaystyle{\\simplify{{q+n}/{m-p}}} = \\var{precround(ansA,1)} \\text{ to 1 dp}$

Use this link to find resources to help you revise how to solve linear equations

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Solve $\\simplify{({m}w-{n}) = {p}w+{q}}$

$w=$ [[0]]

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Solve linear equations with unkowns on both sides. Including brackets and fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

To solve an equation like

$\\displaystyle{\\frac{x+\\var{num1}}{\\var{num2}}+\\frac{x}{\\var{num3}}=\\var{num4}},$

the first thing to deal with is the denominators of the fractions. In order to do that you multiply both sides of the equation by both denominators $\\var{num2}$ and $\\var{num3}$ (or their lowest common multiple to be slightly more efficient). This will give something equivalent to:

$\\displaystyle{\\var{num3 + num2} x+\\var{num3*num1} = \\var{num2*num3*num4}.}$

Then proceeding by subtracting $\\var{num3*num1} from both sides:

$\\displaystyle{\\var{num3 + num2} x = \\var{num2*num3*num4-num3*num1}.}$

And finally dividing by $\\var{num2+num3}$:

$\\displaystyle{x = \\frac{\\var{num2*num3*num4-num3*num1}}{\\var{num2+num3}}.}$

Use this link to find resources to help you revise how to solve linear equations

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Solve $\\displaystyle{\\frac{x+\\var{num1}}{\\var{num2}}+\\frac{x}{\\var{num3}}=\\var{num4}}$.

$x=$ [[0]]

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Solving a pair of linear simultaneous equations, giving answers as integers or fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve the simultaneous equations for x and y, giving your answers as integers or fractions, but not decimals.

\\[ \\begin{split} \\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\,=\\var{c} \\\\ \\simplify[!noLeadingminus,unitFactor]{{a1}x +{b1}y} &\\,=\\var{c1} \\end{split}\\]

", "advice": "

\\[\\begin{split}\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\,=\\var{c} \\qquad\\qquad&(1)\\\\ \\simplify[!noLeadingminus,unitFactor]{{a1}x +{b1}y} &\\,=\\var{c1} \\qquad\\qquad&(2)\\end{split}\\]

{advice1}

Use this link to find some resources which will help you revise this topic.

For these equations, it is easiest to get a solution for $y$ first, due to the $x$-terms having {eqoroppa} coefficients.

\\n

If we {aorsa} equation (2) {torfa} equation (1) this eliminates the $x$-terms leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({b}+{sgna*(b1)})y} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c}+{sgna*(c1)}}\\\\\\\\ \\\\simplify{{b+sgna*(b1)}y} &\\\\,= \\\\simplify{{c+sgna*(c1)}} \\\\\\\\ y &\\\\,= \\\\simplify[all, fractionNumbers]{{c+sgna*(c1)}/{b+sgna*(b1)}} \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}x + \\\\var{b} \\\\times \\\\simplify[all, fractionNumbers]{{c+sgna*(c1)}/{b+sgna*(b1)}} &\\\\,= \\\\var{c} \\\\\\\\ \\\\var{a}x &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c} - {c*b+b*sgna*(c1)}/{b+sgna*(b1)}} \\\\\\\\ x &\\\\,= \\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "eqoroppb": {"name": "eqoroppb", "group": "Ungrouped variables", "definition": "if(abs(b)*b1=abs(b1)*b,'equal','equal and opposite')", "description": "", "templateType": "anything", "can_override": false}, "eqoroppa": {"name": "eqoroppa", "group": "Ungrouped variables", "definition": "if(abs(a)*a1=abs(a1)*a,'equal','equal and opposite')", "description": "", "templateType": "anything", "can_override": false}, "samey": {"name": "samey", "group": "Ungrouped variables", "definition": "\"

For these equations, it is easiest to get a solution for $x$ first, due to the $y$-terms having {eqoroppb} coefficients.

\\n

If we {aorsb} equation (2) {torfb} equation (1) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({a}+{sgn*(a1)})x} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c}+{sgn*(c1)}}\\\\\\\\ \\\\simplify{{a+sgn*(a1)}x} &\\\\,= \\\\simplify{{c+sgn*(c1)}} \\\\\\\\ x &\\\\,= \\\\simplify[all, fractionNumbers]{{c+sgn*(c1)}/{a+sgn*(a1)}} \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a} \\\\times\\\\simplify[fractionNumbers]{{c+sgn*(c1)}/{a+sgn*(a1)}} + \\\\var{b}y &\\\\,= \\\\var{c} \\\\\\\\ \\\\var{b}y &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c} - {c*a+a*sgn*(c1)}/{a+sgn*(a1)}} \\\\\\\\ y &\\\\,= \\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcmb": {"name": "lcmb", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (2) by $\\\\simplify{{abs(b/b1)}}$ we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\\\,=\\\\var{c} \\\\qquad\\\\qquad&(3)\\\\\\\\ \\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(b/b1)}x +{b1*abs(b/b1)}y} &\\\\,=\\\\var{c1*abs(b/b1)} \\\\qquad\\\\qquad&(4)\\\\end{split}\\\\]

\\n

If we {aorsb} equation (4) {torfb} equation (3) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({a}+{sgn*(a1*abs(b/b1))})x} &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c}+{sgn*(c1*abs(b/b1))}}\\\\\\\\ \\\\simplify{{a+sgn*(a1*abs(b/b1))}x} &\\\\,= \\\\simplify{{c+sgn*(c1*abs(b/b1))}} \\\\\\\\ x &\\\\,= \\\\simplify[all,fractionNumbers]{{c+sgn*(c1*abs(b/b1))}/{a+sgn*(a1*abs(b/b1))}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}\\\\times\\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c+sgn*c1*abs(b/b1)}/{(a)+sgn*a1*abs(b/b1)}) + {b}y} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(a*c)+a*sgn*c1*abs(b/b1)}/{(a)+sgn*a1*abs(b/b1)})} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(a*c+a*sgn*c1*abs(b/b1))/(a+sgn*a1*abs(b/b1))}} \\\\\\\\ y &\\\\,=\\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcmb1": {"name": "lcmb1", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (1) by $\\\\simplify{{abs(b1/b)}}$ we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(b1/b)}x +{b*abs(b1/b)}y} &\\\\,=\\\\var{c*abs(b1/b)} \\\\qquad\\\\qquad&(3) \\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1}x+{b1}y} &\\\\,=\\\\var{c1} \\\\qquad\\\\qquad&(4)\\\\\\\\ \\\\end{split} \\\\]

\\n

If we {aorsb} equation (4) {torfb} equation (3) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({(a*abs(b1/b))}+{sgn*a1})x} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{(c*abs(b1/b))}+{sgn*c1}}\\\\\\\\ \\\\simplify{{(a*abs(b1/b))+sgn*a1}x} &\\\\,= \\\\simplify{{(c*abs(b1/b))+sgn*c1}} \\\\\\\\ x &\\\\,= \\\\simplify[all, fractionNumbers]{{(c*abs(b1/b))+sgn*c1}/{(a*abs(b1/b))+sgn*a1}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}\\\\times\\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({(c*abs(b1/b))+sgn*c1}/{(a*abs(b1/b))+sgn*a1}) + {b}y} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(a*c*abs(b1/b))+a*sgn*c1}/{(a*abs(b1/b))+sgn*a1})} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(a*c*abs(b1/b)+a*sgn*c1)/(a*abs(b1/b)+sgn*a1)}} \\\\\\\\ y &\\\\,=\\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "full": {"name": "full", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (1) by $\\\\var{abs(b1)}$ and equation (2) by $\\\\var{abs(b)}$, we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(b1)}x+{b*abs(b1)}y} &\\\\,=\\\\var{c*abs(b1)} \\\\qquad\\\\qquad&(3)\\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(b)}x +{b1*abs(b)}y} &\\\\,=\\\\var{c1*abs(b)} \\\\qquad\\\\qquad&(4) \\\\end{split}\\\\]

\\n

Now, {aorsb} equation (4) {torfb} equation (3) to eliminate the $y$ terms:

\\n

\\\\[ \\\\begin{split} (\\\\simplify[!collectNumbers]{{a*abs(b1)} +{sgn*a1*abs(b)}}) x &\\\\,= \\\\simplify[!collectNumbers]{{c*abs(b1)}+{sgn*c1*abs(b)}} \\\\\\\\ \\\\simplify{{a*abs(b1)+sgn*a1*abs(b)}} x &\\\\,= \\\\simplify{{c*abs(b1)+sgn*c1*abs(b)}} .\\\\end{split} \\\\]

\\n

So the solution for $x$ is \\\\[ x=\\\\simplify{{c*abs(b1)+sgn*c1*abs(b)}/{a*abs(b1)+sgn*a1*abs(b)}}.\\\\]

\\n

To obtain a solution for $y$ we can substitute this value of $x$ into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[noLeadingminus,fractionNumbers,unitFactor]{{a} {xsimp} + {b}y} &\\\\,=\\\\var{c} \\\\\\\\ \\\\var{b}y &\\\\,= \\\\simplify[!collectNumbers,fractionNumbers]{{c}-{a*xsimp}} \\\\\\\\\\\\var{b}y &\\\\,= \\\\simplify[fractionNumbers]{{c-a*xsimp}} \\\\\\\\y &\\\\,= \\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}} \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "aorsa": {"name": "aorsa", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,'subtract','add')", "description": "", "templateType": "anything", "can_override": false}, "torfa": {"name": "torfa", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,'from','to')", "description": "", "templateType": "anything", "can_override": false}, "sgna": {"name": "sgna", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,-1,1)", "description": "", "templateType": "anything", "can_override": false}, "lcma": {"name": "lcma", "group": "Ungrouped variables", "definition": "\"

To get a solution for $y$, if we multiply equation (2) by $\\\\simplify{{abs(a/a1)}}$ we will have two equations with {eqoroppa} $x$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\\\,=\\\\var{c} \\\\qquad\\\\qquad&(3)\\\\\\\\ \\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(a/a1)}x +{b1*abs(a/a1)}y} &\\\\,=\\\\var{c1*abs(a/a1)} \\\\qquad\\\\qquad&(4)\\\\end{split}\\\\]

\\n

If we {aorsa} equation (4) {torfa} equation (3) this eliminates the $x$-terms, leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({b}+{sgna*(b1*abs(a/a1))})y} &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c}+{sgna*(c1*abs(a/a1))}}\\\\\\\\ \\\\simplify{{b+sgna*(b1*abs(a/a1))}y} &\\\\,= \\\\simplify{{c+sgna*(c1*abs(a/a1))}} \\\\\\\\ y &\\\\,= \\\\simplify[all,fractionNumbers]{{c+sgna*(c1*abs(a/a1))}/{b+sgna*(b1*abs(a/a1))}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{{a}x + {b}}\\\\times \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c+sgna*c1*abs(a/a1)}/{(b)+sgna*b1*abs(a/a1)})} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(b*c)+b*sgna*c1*abs(a/a1)}/{(b)+sgna*b1*abs(a/a1)})} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(b*c+b*sgna*c1*abs(a/a1))/(b+sgna*b1*abs(a/a1))}} \\\\\\\\ x &\\\\,=\\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcma1": {"name": "lcma1", "group": "Ungrouped variables", "definition": "\"

To get a solution for $y$, if we multiply equation (1) by $\\\\simplify{{abs(a1/a)}}$ we will have two equations with {eqoroppa} $x$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(a1/a)}x +{b*abs(a1/a)}y} &\\\\,=\\\\var{c*abs(a1/a)} \\\\qquad\\\\qquad&(3) \\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1}x+{b1}y} &\\\\,=\\\\var{c1} \\\\qquad\\\\qquad&(4) \\\\end{split}\\\\]

\\n

If we {aorsa} equation (4) {torfa} equation (3) this eliminates the $x$-terms, leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({(b*abs(a1/a))}+{sgna*b1})y} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{(c*abs(a1/a))}+{sgna*c1}}\\\\\\\\ \\\\simplify{{(b*abs(a1/a))+sgna*b1}y} &\\\\,= \\\\simplify{{(c*abs(a1/a))+sgna*c1}} \\\\\\\\ y &\\\\,= \\\\simplify[all, fractionNumbers]{{(c*abs(a1/a))+sgna*c1}/{(b*abs(a1/a))+sgna*b1}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{{a}x + {b}}\\\\times \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c*abs(a1/a)+sgna*c1}/{(b*abs(a1/a))+sgna*b1})} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(b*c*abs(a1/a))+b*sgna*c1}/{(b*abs(a1/a))+sgna*b1})} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(b*c*abs(a1/a)+b*sgna*c1)/(b*abs(a1/a)+sgna*b1)}} \\\\\\\\ x &\\\\,=\\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "advice1": {"name": "advice1", "group": "Ungrouped variables", "definition": "if(abs(b)=abs(b1), {samey},if(abs(a)=abs(a1),{samex},if(lcm(abs(b),abs(b1))=abs(b),{lcmb},if(lcm(abs(b),abs(b1))=abs(b1),{lcmb1},if(lcm(abs(a),abs(a1))=abs(a),{lcma},if(lcm(abs(a),abs(a1))=abs(a1),{lcma1},{full}))))))", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "abs(b-b1)>1 and\nabs(a-a1)>1 and\ngcd(a,c)=1 and\ngcd(a1,c1)=1", "maxRuns": 100}, "ungrouped_variables": ["a", "b", "a1", "b1", "c", "c1", "aorsa", "torfa", "aorsb", "torfb", "sgna", "sgn", "xn", "xd", "xsimp", "eqoroppa", "eqoroppb", "advice1", "samey", "samex", "lcmb", "lcmb1", "lcma", "lcma1", "full"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$x=$ [[0]]

$y=$ [[1]]

Solving a pair of simultaneous equations of the form $a_1x+y=c_1$ and $a_2x^2+b_2xy=c_2$ by forming a quadratic equation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve the following simultaneous equations:

\\[ \\begin{split} \\simplify{{a1}x+y} &\\,= \\var{c1} \\\\ \\simplify{{a2}x^2+{b2}x*y} &\\,= \\var{c2} \\end{split} \\]

Give your answers to 2 decimal places where necessary.

", "advice": "

To solve a pair of simultaneous equations of this type we want to rearrange the linear equation such that $y$ is the subject, which we can then substitute into the equation with the quadratic $x$-term. This will result in a quadratic equation in terms of $x$ only.

For the equations

\\[ \\begin{split} \\simplify{{a1}x+y} &\\,= \\var{c1} \\qquad \\qquad &(1) \\\\\\simplify{{a2}x^2+{b2}x*y} &\\,= \\var{c2} \\qquad \\qquad &(2) \\end{split} \\]

we can rearrange equation (1) to make $y$ the subject:

\\[ y = \\simplify{{c1}-{a1}x}. \\qquad\\qquad (3)\\]

Substituting this into equation (2):

\\[ \\begin{split}\\simplify{{a2}x^2+{b2}x({c1}-{a1}x)} &\\,=\\var{c2} \\\\ \\simplify[!cancelTerms,unitFactor]{{a2}x^2+{b2*c1}x-{b2*a1}x^2} &\\,=\\var{c2}. \\end{split} \\]

Collecting similar terms:

\\[ \\simplify{({a2}-{b2*a1})x^2+{b2*c1}x-{c2}} =0. \\qquad\\qquad (4) \\]

Using the quadratic formula, we find two solutions for $x$:

{check}

Therefore, the 2 pairs of solutions for these simultaneous equations are

\\[ (x_1,y_1) = (\\var{x1dp},\\var{y1dp}) \\] and \\[ (x_2,y_2) = (\\var{x2dp},\\var{y2dp}). \\]

Use this link to find some resources which will help you revise this topic.

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\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\, \\\\text{ (2 d.p.)} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "text2": {"name": "text2", "group": "Ungrouped variables", "definition": "\"

\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\text{ (2 d.p.)}\\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\text{(2 d.p.)} \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "text3": {"name": "text3", "group": "Ungrouped variables", "definition": "\"

\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\text{ (2 d.p.)}\\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\text{(2 d.p.)} \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "text": {"name": "text", "group": "Ungrouped variables", "definition": "\"

\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp}\\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}}, "variablesTest": {"condition": "(a2-a1*b2)>0 and (b2^2*c1^2+4(a2-a1*b2)*c2)>0 and gcd(a2,b2)=1", "maxRuns": 100}, "ungrouped_variables": ["a1", "c1", "a2", "b2", "c2", "solx1", "solx2", "soly1", "soly2", "x1dp", "y1dp", "x2dp", "y2dp", "solutions1", "solutions2", "check", "text", "text1", "text2", "text3"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$(x_1,y_1)=$[[0]]

$(x_2,y_2)=$[[1]]

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Solve linear equations with unkowns on both sides. Including brackets and fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

To solve an equation like

$\\displaystyle{\\frac{\\var{a}}{y}=\\frac{\\var{b}}{y+\\var{c}}},$

the first thing to deal with is that the unknown ($y$) that you are trying to find is in the denominator (on the bottom) of the fractions. In order to do that you first times by $y$ on both sides and $(y+\\var{c})$ on both sides leading to

\\[\\var{a}(y+\\var{c}) = \\var{b}y.\\]

From here, multiply out the brackets,

\\[\\var{a}y +\\var{a*c} = \\var{b}y.\\]

Now collect the $y$-terms on one side and the numbers on the other,

\\[\\var{a-b}y=\\var{-a*c}.\\]

Finally divide by the coefficient of $y$,

\\[y=\\frac{\\var{-a*c}}{\\var{a-b}}.\\]

Use this link to find resources to help you revise how to solve linear equations

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Solve $\\displaystyle{\\frac{\\var{a}}{y}=\\frac{\\var{b}}{y+\\var{c}}}$.

$y=$ [[0]] (Give your answer as a fraction)

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "fraction", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "ans1", "maxValue": "ans1", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AS06 Factorising a Quadratic (a=1)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": ["factorisation", "Factorisation", "factorising quadratic equations", "Factorising quadratic equations", "taxonomy"], "metadata": {"description": "

Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Factorise the following quadratic equation.

", "advice": "

Quadratic equations of the form

\\[x^2+bx+c=0\\]

can be factorised to create an equation of the form

\\[(x+m)(x+n)=0\\text{.}\\]

When we expand a factorised quadratic expression we obtain

\\[(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\\]

To factorise an equation of the form $x^2+bx+c$, we need to find two numbers which add together to make $b$, and multiply together to make $c$.

We need to find two values that add together to make $\\var{v3+v4}$ and multiply together to make $\\var{v3*v4}$.

\\[\\begin{align}
\\var{v3} \\times \\var{v4}&=\\var{v3*v4}\\\\
\\var{v3}+\\var{v4}&=\\var{v3+v4}\\\\
\\end{align} \\]

So the factorised form of the equation is

\\[\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\\]

Use this link to find some resources which will help you revise this topic

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$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$

[[0]] $=0$

The answer is a comma-separated list of numbers.

The list is marked correct if each number occurs the same number of times as in the expected answer, and no extra numbers are present.

You can optionally treat the answer as a set, so the number of occurrences doesn't matter, only whether each number is included or not.

", "help_url": "", "input_widget": "string", "input_options": {"correctAnswer": "join(\n if(settings[\"correctAnswerFractions\"],\n map(let([a,b],rational_approximation(x), string(a/b)),x,settings[\"correctAnswer\"])\n ,\n settings[\"correctAnswer\"]\n ),\n settings[\"separator\"] + \" \"\n)", "hint": {"static": false, "value": "if(settings[\"show_input_hint\"],\n \"Enter a list of numbers separated by {settings['separator']}.\",\n \"\"\n)"}, "allowEmpty": {"static": true, "value": true}}, "can_be_gap": true, "can_be_step": true, "marking_script": "bits:\nlet(b,filter(x<>\"\",x,split(studentAnswer,settings[\"separator\"])),\n if(isSet,list(set(b)),b)\n)\n\nexpected_numbers:\nlet(l,settings[\"correctAnswer\"] as \"list\",\n if(isSet,list(set(l)),l)\n)\n\nvalid_numbers:\nif(all(map(not isnan(x),x,interpreted_answer)),\n true,\n let(index,filter(isnan(interpreted_answer[x]),x,0..len(interpreted_answer)-1)[0], wrong, bits[index],\n warn(wrong+\" is not a valid number\");\n fail(wrong+\" is not a valid number.\")\n )\n )\n\nis_sorted:\nassert(sort(interpreted_answer)=interpreted_answer,\n multiply_credit(0.5,\"Not in order\")\n )\n\nincluded:\nmap(\n let(\n num_student,len(filter(x=y,y,interpreted_answer)),\n num_expected,len(filter(x=y,y,expected_numbers)),\n switch(\n num_student=num_expected,\n true,\n num_studentThe separate items in the student's answer

", "definition": "let(b,filter(x<>\"\",x,split(studentAnswer,settings[\"separator\"])),\n if(isSet,list(set(b)),b)\n)"}, {"name": "expected_numbers", "description": "", "definition": "let(l,settings[\"correctAnswer\"] as \"list\",\n if(isSet,list(set(l)),l)\n)"}, {"name": "valid_numbers", "description": "

Is every number in the student's list valid?

", "definition": "if(all(map(not isnan(x),x,interpreted_answer)),\n true,\n let(index,filter(isnan(interpreted_answer[x]),x,0..len(interpreted_answer)-1)[0], wrong, bits[index],\n warn(wrong+\" is not a valid number\");\n fail(wrong+\" is not a valid number.\")\n )\n )"}, {"name": "is_sorted", "description": "

Are the student's answers in ascending order?

", "definition": "assert(sort(interpreted_answer)=interpreted_answer,\n multiply_credit(0.5,\"Not in order\")\n )"}, {"name": "included", "description": "

Is each number in the expected answer present in the student's list the correct number of times?

", "definition": "map(\n let(\n num_student,len(filter(x=y,y,interpreted_answer)),\n num_expected,len(filter(x=y,y,expected_numbers)),\n switch(\n num_student=num_expected,\n true,\n num_studentHas every number been included the right number of times?

", "definition": "all(included)"}, {"name": "no_extras", "description": "

True if the student's list doesn't contain any numbers that aren't in the expected answer.

", "definition": "if(all(map(x in expected_numbers, x, interpreted_answer)),\n true\n ,\n incorrect(\"Your answer contains \"+extra_numbers[0]+\" but should not.\");\n false\n )"}, {"name": "interpreted_answer", "description": "A value representing the student's answer to this part.", "definition": "if(lower(studentAnswer) in [\"empty\",\"\u2205\"],[],\n map(\n if(settings[\"allowFractions\"],parsenumber_or_fraction(x,notationStyles), parsenumber(x,notationStyles))\n ,x\n ,bits\n )\n)"}, {"name": "mark", "description": "This is the main marking note. It should award credit and provide feedback based on the student's answer.", "definition": "if(studentanswer=\"\",fail(\"You have not entered an answer\"),false);\napply(valid_numbers);\napply(included);\napply(no_extras);\ncorrectif(all_included and no_extras)"}, {"name": "notationStyles", "description": "", "definition": "[\"en\"]"}, {"name": "isSet", "description": "

Should the answer be considered as a set, so the number of times an element occurs doesn't matter?

", "definition": "settings[\"isSet\"]"}, {"name": "extra_numbers", "description": "

Numbers included in the student's answer that are not in the expected list.

", "definition": "filter(not (x in expected_numbers),x,interpreted_answer)"}], "settings": [{"name": "correctAnswer", "label": "Correct answer", "help_url": "", "hint": "The list of numbers that the student should enter. The order does not matter.", "input_type": "code", "default_value": "", "evaluate": true}, {"name": "allowFractions", "label": "Allow the student to enter fractions?", "help_url": "", "hint": "", "input_type": "checkbox", "default_value": false}, {"name": "correctAnswerFractions", "label": "Display the correct answers as fractions?", "help_url": "", "hint": "", "input_type": "checkbox", "default_value": false}, {"name": "isSet", "label": "Is the answer a set?", "help_url": "", "hint": "If ticked, the number of times an element occurs doesn't matter, only whether it's included at all.", "input_type": "checkbox", "default_value": false}, {"name": "show_input_hint", "label": "Show the input hint?", "help_url": "", "hint": "", "input_type": "checkbox", "default_value": true}, {"name": "separator", "label": "Separator", "help_url": "", "hint": "The substring that should separate items in the student's list", "input_type": "string", "default_value": ",", "subvars": false}], "public_availability": "always", "published": true, "extensions": []}], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Solving a quadratic equation via factorisation (or otherwise) with the $x^2$-term having a coefficient of 1.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve the following quadratic equation by factorisation or otherwise:

\\[ \\simplify[unitFactor]{x^2+{b}x+{c}=0} \\]

", "advice": "

To solve a quadratic equation of the form \\[ x^2+bx+c=0\\] by factorisation, we want to factorise the equation into the form \\[(x+p)(x+q)=0,\\] where $p+q=b$ and $p \\times q = c$.

Hence, for the equation \\[\\simplify{x^2+{b}x+{c}=0}, \\]

this can be factorised to \\[\\simplify{(x+{p})(x+{q})=0}.\\] This equation is satisfied when either \\[\\simplify{x+{p}=0} \\quad \\text{or} \\quad \\simplify{x+{q}=0}, \\] which implies the solutions to this quadratic equation are \\[ \\simplify{x={-p}} \\quad \\text{and} \\quad \\simplify{x={-q}} .\\]

Use this link to find resources to help you revise how to solve quadratic equations by factorising the expression.

$x= $[[0]]

", "gaps": [{"type": "list-of-numbers", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "settings": {"correctAnswer": "{sol}", "allowFractions": false, "correctAnswerFractions": false, "isSet": false, "show_input_hint": true, "separator": ","}}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AS08 Completing the square", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": ["complete the square", "completing the square", "taxonomy"], "metadata": {"description": "

Rearrange expressions in the form $ax^2+bx+c$ to $a(x+b)^2+c$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

We can rewrite quadratic equations given in the form $ax^2+bx+c$ as a square plus another term - this is called \"completing the square\".

This can be useful when it isn't obvious how to fully factorise a quadratic equation.

Rewrite the following expressions in the form \\[(x+b)^2-c\\]

", "advice": "

Completing the square works by noticing that

\\[ (x+a)^2 = x^2 + 2ax + a^2 \\]

So when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.

Replace $x^2+\\var{evens2}x$ with $(x+\\var{evens2/2})^2 - \\var{evens2/2}^2$. Remember to keep the $\\var{evens2-evens1}$ term on the end!

\\begin{align}
\\simplify[basic]{ x^2 + {evens2}x + {evens2-evens1}} &= \\simplify[basic]{ (x+{evens2/2})^2 - {evens2/2}^2 + {evens2-evens1} } \\\\
&= \\simplify[basic]{ (x+{evens2/2})^2 + {evens2-evens1 - evens2^2/4} }
\\end{align}

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"multiall2": {"name": "multiall2", "group": "Ungrouped variables", "definition": "all2*random(2..10 #2)", "description": "", "templateType": "anything", "can_override": false}, "odds3": {"name": "odds3", "group": "Odds and Evens", "definition": "random(11..30 #2 except odds odds2)", "description": "", "templateType": "anything", "can_override": false}, "evens3": {"name": "evens3", "group": "Odds and Evens", "definition": "random(2..30 #2 except evens1 evens2)", "description": "", "templateType": "anything", "can_override": false}, "evens2": {"name": "evens2", "group": "Odds and Evens", "definition": "random(10..30 #2 except evens1)", "description": "", "templateType": "anything", "can_override": false}, "multiall": {"name": "multiall", "group": "Ungrouped variables", "definition": "all*random(2..10#2)", "description": "", "templateType": "anything", "can_override": false}, "evens1": {"name": "evens1", "group": "Odds and Evens", "definition": "random(10..30 #2)", "description": "", "templateType": "anything", "can_override": false}, "all2": {"name": "all2", "group": "Ungrouped variables", "definition": "random(2..6 except all)", "description": "", "templateType": "anything", "can_override": false}, "odds2": {"name": "odds2", "group": "Odds and Evens", "definition": "random(11..30 #2 except odds)", "description": "", "templateType": "anything", "can_override": false}, "big": {"name": "big", "group": "Ungrouped variables", "definition": "random(30..50)", "description": "", "templateType": "anything", "can_override": false}, "all": {"name": "all", "group": "Ungrouped variables", "definition": "random(2..6)", "description": "", "templateType": "anything", "can_override": false}, "sml": {"name": "sml", "group": "Ungrouped variables", "definition": "random(2..6#2)", "description": "", "templateType": "anything", "can_override": false}, "odds": {"name": "odds", "group": "Odds and Evens", "definition": "random(11..30 #2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["all", "all2", "multiall", "big", "sml", "multiall2"], "variable_groups": [{"name": "Odds and Evens", "variables": ["evens1", "evens2", "evens3", "odds", "odds2", "odds3"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\simplify {x^2+ {evens2}x +{evens2-evens1}} =$ [[0]]

It doesn't look like you've completed the square.

"}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AS09 Quadratics - factorise (a not 1)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": ["coefficient of x^2 greater than 1", "factorisation", "Factorisation", "factorising", "factorising quadratic equations", "Factorising quadratic equations", "factorising quadratic equations with x^2 coefficients greater than 1", "taxonomy"], "metadata": {"description": "

Factorise a quadratic equation where the coefficient of the $x^2$ term is greater than 1 and then write down the roots of the equation

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.

To find $a$ and $c$, we need to consider the factors of $\\var{a*c}$.

You may have to test a a few different options before you find one that works. In this case $a$ and $c$ are $\\var{a}$ and $\\var{c}$.

This means our factorised equation must take the form

\\[(\\var{a}x+b)(\\var{c}x+d)=0\\text{.}\\]

This expands to

\\[ \\simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b} \\]

So we must find two numbers which add together to make $\\var{a*d+b*c}$, and multiply together to make $\\var{b*d}$.

Therefore $b$ and $d$ must satisfy

\\begin{align}
b \\times d &=\\var{b*d}\\\\
\\simplify{{a}d+{c}b} &= \\var{a*d+b*c}\\text{.}
\\end{align}

$b = \\var{b}$ and $d = \\var{d}$ satisfy these equations:

\\begin{align}
\\var{b} \\times \\var{d} &=\\var{b*d}\\\\
\\simplify[]{ {a}*{d} + {b}*{c} } &= \\var{a*d+b*c}
\\end{align}

So the factorised form of the equation is

\\[ \\simplify{({a}x+{b})({c}x+{d}) = 0} \\text{.}\\]

$\\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\\var{a}x+\\var{b} = 0$ or $\\var{c}x+ \\var{d} = 0$.

So the roots of the equation are $\\var[fractionnumbers]{-b/a}$ and $\\var[fractionnumbers]{-d/c}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"b": {"name": "b", "group": "last q", "definition": "random(-5..5 except 0)", "description": "

$b$ in $(ax+b)(cx+d)$

", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "last q", "definition": "random(2..8 except a)", "description": "

$c$ in $(ax+b)(cx+d)$

", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "last q", "definition": "random(2..3)", "description": "

$a$ in $(ax+b)(cx+d)$

", "templateType": "anything", "can_override": false}, "roots": {"name": "roots", "group": "last q", "definition": "sort([-b/a,-d/c])", "description": "

The roots of the equation

", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "last q", "definition": "random(-8..8 except 0)", "description": "

$d$ in $(ax+b)(cx+d)$

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "last q", "variables": ["a", "b", "c", "d", "roots"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Solve the following equation by factorisation to find $x$.

$\\simplify{{a*c}x^2+{a*d+b*c}x+{b*d}=0}\\text{.}$

Input your answers in ascending order.

$x=$ [[0]]

$x=$ [[1]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "roots[0]", "maxValue": "roots[0]", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "roots[1]", "maxValue": "roots[1]", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AS10 Difference of two squares", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Factorising a quadratic expression of the form $a^2x^2-b^2$ to $(ax+b)(ax-b)$, using the difference of two squares formula.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Factorise the following quadratic expression:

\\[ \\simplify[unitFactor]{{a^2}x^2-{c^2}} \\]

", "advice": "

For a quadratic expression of this form we can make use of the Difference of Squares formula, which states that \\[a^2-b^2 = (a+b)(a-b).\\]

Therefore,

\\[ \\simplify[unitFactor]{{a^2}x^2-{c^2} = ({a}x+{c})({a}x-{c})}. \\]

Use this link to find resources to help you revise how to factorise a quadratic equation using the difference of two squares formula.

Find the derivative of a function of the form $y=ax^b$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify{{a}x^{b}}$.

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=kx^n \\] has a derivative \\[ \\frac{df}{dx} = knx^{n-1}. \\]

So, for the function \\[ y=\\simplify{{a}x^{b}}, \\] the derivative is \\begin{split}\\frac{dy}{dx} &= (\\var{a}\\times\\var{b})x^{\\var{b}-1},\\\\ \\\\&= \\simplify{{a*b}x^{{b}-1}}.\\end{split}

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Differentiate a polynomial expression involving coefficients and, negative and fractional indices.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of $y=\\simplify[unitFactor, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}}$.

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=kx^n \\] has a derivative \\[ \\frac{df}{dx} = knx^{n-1}. \\]

Additionally, the derivative of the sum or difference of two or more functions is equal to the sum or difference of the derivatives of each function: \\[ \\frac{d}{dx}(f(x)\\pm g(x)) = \\frac{df}{dx} \\pm \\frac{dg}{dx}.\\]

{advice}

Use this link to find some resources which will help you revise this topic.

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} +(\\\\var[fractionNumbers]{a_2}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} +(\\\\var[fractionNumbers]{a_3}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionb": {"name": "solutionb", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} +(\\\\var[fractionNumbers]{a_3}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionc": {"name": "solutionc", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} +(\\\\var[fractionNumbers]{a_2}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutiond": {"name": "solutiond", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

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$\\frac{dy}{dx}=$[[0]]

Calculating the derivative of a function of the form $a \\ln(bx)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[unitFactor]{{a}*ln({a_1}*x^2+{a_2}*x+{a_3})}.$

", "advice": "

From the Table of Derivatives and the chain rule we see that a function of the form \\[ f(x)=a \\ln(g(x)) \\] has a derivative \\[\\frac{df}{dx}=\\frac{g'(x)}{g(x)}.\\]

In this case $g(x)=\\var{a_1}x^2+\\var{a_2}x+\\var{a_3}$ so

\\[g'(x)=\\var{2*a_1}x+\\var{a_2}\\]

Therefore, the function \\[ \\simplify[unitFactor]{y={a}ln({a_1}*x^2+{a_2}*x+{a_3})}\\] has a derivative \\[(\\var{a*a_1*2}x+\\var{a*a_2})/(\\var{a_1}x^2+\\var{a_2}x+\\var{a_3})\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Calculating the derivative of an exponential function of the form $ae^{bx}$, using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[all]{{a}*e^({b}x)}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a e^{kx} \\] has a derivative \\[ak e^{kx}.\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*e^({b}x)}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})e^{\\simplify[unitFactor]{{b}x}}\\\\ &= \\simplify[unitFactor]{{a*b}e^({b}x)}.\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Find the derivative of a function of the form $y=a \\cos(bx+c)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of $y=\\simplify[unitFactor]{{a}cos({b}x+{c})}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a \\cos(kx+c) \\] has a derivative \\[-ak \\sin (kx+c).\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*cos({b}x+{c})}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=-(\\var{a}\\times \\var{b})\\sin(\\simplify[unitFactor]{{b}x+{c}})\\\\ &= \\simplify[unitFactor]{{-a*b}sin({b}x+{c})}.\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Find the derivative of a function of the form $y=a \\sin(bx+c)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Using the Table of Derivatives, calculate the derivative of $y=\\simplify[unitFactor]{{a}sin({b}x+{c})}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a \\sin(kx+c) \\] has a derivative \\[ak \\cos (kx+c).\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*sin({b}x+{c})}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})\\cos(\\simplify[unitFactor]{{b}x+{c}})\\\\ &= \\simplify[unitFactor]{{a*b}cos({b}x+{c})}.\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Find the derivative of a function of the form $y=a \\tan(bx+c)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Using the Table of Derivatives, calculate the derivative of $y=\\simplify[unitFactor]{{a}tan({b}x+{c})}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a \\tan(kx+c) \\] has a derivative \\[ak \\sec^2(kx+c).\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*tan({b}x+{c})}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})\\sec^2(\\simplify[unitFactor]{{b}x+{c}})\\\\ &= \\simplify[unitFactor]{{a*b}}\\sec^2(\\simplify[unitFactor]{{b}x+{c}}).\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "{a*b}sec^2({b}x+{c})", "answerSimplification": "fractionNumbers, basic, unitFactor", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "sec", "value": ""}, {"name": "x", "value": ""}]}], "answer": "{a*b}sec({b}x+{c})^2", "answerSimplification": "fractionNumbers, basic, unitFactor", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CD08 Finding turning points", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Finding the stationary points of a cubic equation and determining their nature.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Given the function \\[ \\simplify{y={a}x^3+{b}x^2+{c}x+{d}} ,\\] find its stationary points and determine their nature.

", "advice": "

To find the stationary points of the function, we must solve $\\tfrac{dy}{dx}=0$ for $x$. For the function $\\simplify{y={a}x^3+{b}x^2+{c}x+{d}}$,

\\[ \\frac{dy}{dx} = \\simplify{{3a}x^2+{2b}x+{c}}. \\]

Setting $\\frac{dy}{dx}=0$ and solving for $x$:

\\[ \\simplify{{3a}x^2+{2b}x+{c}} =0 \\\\ \\\\ \\implies x=\\var{solx1dp} \\var{x1} \\text{ and } x=\\var{solx2dp} \\var{x2}. \\]

Hence, the function has two stationary points at $x=\\var{solx1dp}$ and $x=\\var{solx2dp}$. To find the corresponding $y$-coordinates, we want to plug these values back into the initial equation.

When $x=\\var{solx1dp}$,

\\[ \\begin{split} y &\\,= \\simplify[unitFactor,!cancelTerms]{{a}*({solx1dp})^3+{b}*({solx1dp})^2+{c}*({solx1dp})+{d}} \\\\ &\\,=\\simplify{{soly1dp}} \\var{y1}. \\end{split} \\]

When $x=\\var{solx2dp}$,

\\[ \\begin{split} y &\\,= \\simplify[unitFactor,!cancelTerms]{{a}*({solx2dp})^3+{b}*({solx2dp})^2+{c}*({solx2dp})+{d}} \\\\ &\\,=\\simplify{{soly2dp}} \\var{y2}. \\end{split} \\]

Therefore, the stationary points of $y=\\simplify{{a}x^3+{b}x^2+{c}x+{d}}$ are

\\[ (\\simplify{{solx1dp}},\\, \\simplify{{soly1dp}}) \\, , \\,(\\simplify{{solx2dp}},\\, \\simplify{{soly2dp}}). \\]

Finally, we need to determine the nature of the stationary points. To do this we want to calculate the second derivative of the initial function and then evaluate it for each $x$-value of the stationary points.

Recall:

If $\\tfrac{d^2y}{dx^2}<0$ the stationary point is a maximum;
If $\\tfrac{d^2y}{dx^2}>0$ the stationary point is a minimum;
If $\\tfrac{d^2y}{dx^2}=0$ the stationary point is a point of inflection.

To calculate $\\tfrac{d^2y}{dx^2}$, we want to differentiate $\\tfrac{dy}{dx}$ again with respect to $x$:

\\[ \\begin{split} &\\frac{dy}{dx} = \\simplify{{3a}x^2+{2b}x+{c}}, \\\\ \\\\\\implies &\\frac{d^2y}{dx^2} = \\simplify{{6a}x+{2b}}. \\end{split}\\]

For $(\\simplify{{solx1dp}},\\, \\simplify{{soly1dp}})$, $\\frac{d^2y}{dx^2} = \\simplify{{check}}$, so it is a minimum.

For $(\\simplify{{solx2dp}},\\, \\simplify{{soly2dp}})$, $\\frac{d^2y}{dx^2} = \\simplify{{check2}}$, so it is a maximum.

Use this link to find some resources which will help you revise this topic.

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There is a minimum point at ([[0]], [[1]]) and a maximum point at ([[2]] , [[3]]).

(Give the coordinates of the stationary points to 2 decimal places where necessary.)

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Calculating the derivative of a function of the form $\\sin(ax^m+bx^n)$ using the chain rule.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[all]{sin({a}*x^{n}+{b}*x^{m})}$.

", "advice": "

If we have a function of the form $y=f(g(x))$, sometimes described as a function of a function, to calculate its derivative we need to use the chain rule:

\\[ \\frac{dy}{dx} = \\frac{du}{dx} \\times \\frac{dy}{du}.\\]

This can be split up into steps:

Let $u=g(x)$;
Rewrite $y$ in terms of $u$, such that $y=f(u)$;
Calculate $\\frac{du}{dx}$ and $\\frac{dy}{du}$;
Write $\\frac{dy}{dx}$ as a product of $\\frac{du}{dx}$ and $\\frac{dy}{du}$;
Make sure $\\frac{dy}{dx}$ is only in terms of $x$. Ensure any $u$ terms have been replaced using the initial substitution.

Following this process, we must first identify $g(x)$. Since the function is of the form $y=f(g(x))$, we are looking for the 'inner' function.

So, for $y=\\simplify[all,fractionNumbers]{sin({a}*x^{n}+{b}*x^{m})}$, \\[g(x)=\\simplify[all, fractionNumbers, unitFactor]{{a}*x^{n}+{b}*x^{m}}.\\]

If we now set $u=g(x)$, we can rewrite $y$ in terms of $u$ such that $y=f(u)$:

\\[y=\\simplify[all, fractionNumbers,unitFactor]{sin(u)}.\\]

Next, we calculate the two derivatives $\\frac{du}{dx}$ and $\\frac{dy}{du}$:

\\[\\frac{du}{dx}=\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}}, \\quad \\frac{dy}{du}=\\simplify[all, fractionNumbers, unitFactor]{cos(u)}.\\]

Plugging these into the chain rule:

\\[ \\begin{split} \\frac{dy}{dx} &= \\frac{du}{dx} \\times \\frac{dy}{du}, \\\\&=(\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}}) \\times\\simplify[all, fractionNumbers, unitFactor]{cos(u)}. \\end{split} \\]

Finally, we need to express $\\frac{dy}{dx}$ only in terms of $x$, so we must replace the $u$ term using the initial substitution $u=\\simplify[all, fractionNumbers, unitFactor]{{a}*x^{n}+{b}*x^{m}}$:

\\[ \\frac{dy}{dx} =(\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}})\\simplify[all, fractionNumbers, unitFactor]{cos({a}*x^{n}+{b}*x^{m})}.\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Calculating the derivative a function of the form $ax^n \\sin(bx)$ using the product rule.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of \\[ \\simplify{y={a}x^{n} sin({b}x)}. \\]

", "advice": "

If we have a function of the form $y=u(x)v(x)$, to calculate its derivative we need to use the product rule:

\\[ \\dfrac{dy}{dx} = u(x) \\times \\dfrac{dv}{dx} + v(x) \\times\\dfrac{du}{dx}.\\]

This can be split up into steps:

Identify the functions $u(x)$ and $v(x)$;
Calculate their derivatives $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$;
Substitute these into the formula for the product rule to obtain an expression for $\\tfrac{dy}{dx}$;
Simplify $\\tfrac{dy}{dx}$ where possible.

Following this process, we must first identify $u(x)$ and $v(x)$.

As \\[ \\simplify{y={a}x^{n} sin({b}x)}, \\]

let \\[ u(x) = \\simplify{{a}x^{n}} \\quad \\text{and} \\quad v(x)=\\simplify{sin({b}x)}.\\]

Next, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:

\\[ \\dfrac{du}{dx} = \\simplify{{a*n}x^{n-1}}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{b}cos({b}x)}.\\]

Substituting these results into the product rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:

\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{du}{dx}\\times v(x) + u(x) \\times\\dfrac{dv}{dx} \\\\ &\\,=\\simplify{{a*n}x^{n-1}} \\times\\simplify{sin({b}x)} +\\simplify{{a}x^{n}} \\times \\simplify{{b}cos({b}x)}. \\end{split}\\]

Simplifying,

\\[\\dfrac{dy}{dx} = \\simplify{{n*a}x^{n-1}sin({b}x) + {a*b}x^{n}cos({b}x)}. \\]

Use this link to find some resources which will help you revise this topic

$\\dfrac{dy}{dx}=$[[0]]

Calculating the derivative of a function of the form $\\frac{ax^n}{bx+c}$ using the quotient rule.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of \\[ \\simplify{y={a}x^{n}/({b}x+{c})}. \\]

", "advice": "

If we have a function of the form $y=\\tfrac{u(x)}{v(x)}$, to calculate its derivative we need to use the quotient rule:

\\[ \\dfrac{dy}{dx} = \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2}\\,.\\]

This can be split up into steps:

Identify the functions $u(x)$ and $v(x)$;
Calculate their derivatives $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$;
Substitute these into the formula for the quotient rule to obtain an expression for $\\tfrac{dy}{dx}$;
Simplify $\\tfrac{dy}{dx}$ where possible.

Following this process, we must first identify $u(x)$ and $v(x)$.

As \\[ \\simplify{y={a}x^{n}/({b}x+{c})}, \\]

let \\[ u(x) = \\simplify{{a}x^{n}} \\quad \\text{and} \\quad v(x)=\\simplify{{b}x+{c}}.\\]

Next, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:

\\[ \\dfrac{du}{dx} = \\simplify{{a*n}x^{n-1}}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{b}}.\\]

Substituting these results into the quotient rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:

\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2} \\\\ \\\\&\\,=\\dfrac{(\\simplify{{b}x+{c}}) \\times\\simplify{{a*n}x^{n-1}} - \\simplify{{a}x^{n}} \\times \\simplify{{b}}}{\\simplify{({b}x+{c})^2}}. \\end{split}\\]

Simplifying,

\\[ \\begin{split} \\dfrac{dy}{dx} &\\,=\\dfrac{(\\simplify{{b}x+{c}})\\simplify{{a*n}x^{n-1}} - \\simplify{{b*a}x^{n}}}{\\simplify{({b}x+{c})^2}} \\\\ \\\\&\\,=\\dfrac{\\simplify[all,!cancelTerms]{{b*a*n}x^{n}+{c*a*n}x^{n-1} - {b*a}x^{n}}}{\\simplify{({b}x+{c})^2}}\\\\ \\\\ &\\,=\\dfrac{\\simplify{{b*a*n}x^{n}+{c*a*n}x^{n-1} - {b*a}x^{n}}}{\\simplify{({b}x+{c})^2}} \\\\ \\\\ &\\,=\\dfrac{\\simplify{{simp}x^{n-1}({(b*a*n-b*a)/simp}x+{c*a*n/simp})}}{\\simplify{({b}x+{c})^2}} \\end{split} \\]

Use this link to find some resources which will help you revise this topic.

$\\dfrac{dy}{dx}=$[[0]]

Calculating the integral of a function of the form $a_1x^{b_1}+a_2x^{b_2}+a_3x^{b_3}$ using a table of integrals.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the integral of $f(x)=\\simplify[unitFactor, unitPower, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}+{a_4}}$.

", "advice": "

From the Table of Integrals we see that a function of the form \\[ f(x)=x^n \\] has the integral \\[ \\int x^n dx = \\frac{x^{n+1}}{n+1}+ c,\\]

and \\[\\int kf(x) dx = k \\int f(x) dx.\\]

Additionally, the integral of the sum or difference of two or more functions is equal to the sum or difference of the integrals of each function: \\[ \\int(f(x)\\pm g(x))\\, dx = \\int f(x)\\, dx \\pm \\int g(x) \\, dx.\\]

So, for the function

\\[f(x)=\\simplify[unitFactor,unitPower]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}+{a_4}},\\]

the integral is

\\[ \\begin{split}\\simplify[unitFactor,unitPower]{int({a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}+{a_4},x)} &\\,= \\simplify{{a_1}int(x^{b_1},x)+{a_2}int(x^{b_2},x)+{a_3}int(x^{b_3},x)+int({a_4},x)} \\\\&\\,= \\simplify[all,fractionNumbers]{({a_1}*x^{b_1+1})/{b_1+1}+({a_2}*x^{b_2+1})/{b_2+1}+({a_3}*x^{b_3+1})/{b_3+1}+{a_4}x}+c.\\end{split} \\]

Note: You only need to put one $c$ term here, you do not need to put a separate constant term for each calculation.

Use this link to find some resources which will help you revise this topic.

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionb": {"name": "solutionb", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} +(\\\\var[fractionNumbers]{a_3}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionc": {"name": "solutionc", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} +(\\\\var[fractionNumbers]{a_2}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutiond": {"name": "solutiond", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

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[[0]]

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

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Calculating the definite integral $\\int_{n_1}^{n_2}a_1x^{b_1}+a_2x^{b_2}+a_3x^{b_3} dx$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate \\[ \\int_{\\var{n_1}}^{\\var{n_2}}\\simplify[unitFactor, unitPower, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}} \\,dx.\\]

", "advice": "

Integrating a function of the form \\[ f(x)=x^n \\] has the integral \\[ \\int_a^b x^n dx = \\left[\\frac{x^{n+1}}{n+1}\\right]_a^b,\\]

and \\[\\int_a^b kf(x) dx = k \\int_a^b f(x) dx.\\]

Therefore,

\\[ \\begin{split}\\simplify[unitFactor,unitPower]{defint({a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3},x,{n_1},{n_2})} &\\,= \\simplify{{a_1}defint(x^{b_1},x,{n_1},{n_2})+{a_2}defint(x^{b_2},x,{n_1},{n_2})+{a_3}defint(x^{b_3},x,{n_1},{n_2})} \\\\ &\\,= \\left[\\simplify[all,fractionNumbers]{{a_1}x^{b_1+1}/{b_1+1}+{a_2}x^{b_2+1}/{b_2+1}+{a_3}x^{b_3+1}/{b_3+1}}\\right]_\\var{n_1}^\\var{n_2} \\\\ &\\,= \\left[\\simplify[all,fractionNumbers,!collectNumbers]{{a_1*n_2^(b_1+1)}/{b_1+1}+{a_2*n_2^(b_2+1)}/{b_2+1}+{a_3*n_2^(b_3+1)}/{b_3+1}}\\right] -\\left[\\simplify[all,fractionNumbers,!collectNumbers]{{a_1*n_1^(b_1+1)}/{b_1+1}+{a_2*n_1^(b_2+1)}/{b_2+1}+{a_3*n_1^(b_3+1)}/{b_3+1}}\\right] \\\\ &\\,= \\simplify[!collectNumbers]{{eval2a}-{eval1a}} \\\\ &\\,=\\var{sol1} \\end{split} \\]

Use this link to find some resources on areas under curves which will help you revise this topic.

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[[0]] (Give answers to 2 decimal places where necessary)

Calculating the integral of a function of the form $\\frac{c}{(x+a)(x+b)}$ using partial fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the integral

\\[ \\simplify{int({c}/((x^2+{aPlusb}x+{ab})),x)} .\\]

", "advice": "

In order to integrate the function \\[ \\simplify{int({c}/((x^2+{aPlusb}x+{ab})),x)}, \\] we want to rewrite it in terms of its partial fractions.

First we need to factorise the denominator so we have

\\[ \\simplify{{c}/((x+{a})(x+{b}))}. \\]

Now to write this as a partial fraction, we want to set the function equal to the sum of 2 fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x+{b}}$. Since these are both distinct linear factors, this tells us that the numerators will be constants, which we will call $A$ and $B$:

\\[ \\simplify{{c}/((x+{a})(x+{b}))} = \\simplify{A/(x+{a}) + B/(x+{b})}.\\]

To find the values of $A$ and $B$, we want to multiply this equation by the denominator of the left-hand side. This gives

\\[ \\simplify{{c}=A(x+{b})+B(x+{a})}.\\]

To find $A$, we can eliminate $B$ by setting $\\simplify{x={-a}}$:

\\[ \\simplify{{c}=A{b-a}} \\implies \\simplify[fractionNumbers]{A={c/(b-a)}}.\\]

Similarly, to find B, we can eliminate $A$ by setting $\\simplify{x={-b}}$:

\\[ \\simplify{{c}=B{a-b}} \\implies \\simplify[fractionNumbers]{B={c/(a-b)}}.\\]

Therefore,

{check1}

and

{check2}

Use this link to find some resources which will help you revise this topic.

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\\\\[ \\\\simplify{{c}/((x+{a})(x+{b}))} = \\\\simplify[all,fractionNumbers]{{Asol}/(x+{a})+{Bsol}/(x+{b})},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "Sol2": {"name": "Sol2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify{{c}/((x+{a})(x+{b}))} = \\\\simplify[all,fractionNumbers]{{c}/(({b-a})(x+{a}))+{c}/(({a-b})(x+{b}))},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "int1": {"name": "int1", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split} \\\\simplify{int({c}/((x+{a})(x+{b})),x)} &\\\\,= \\\\simplify[all,fractionNumbers]{int({Asol}/(x+{a})+{Bsol}/(x+{b}),x)}\\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{Asol} int(1/(x+{a}),x)+{Bsol} int(1/(x+{b}),x)} \\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + C}. \\\\end{split}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "int2": {"name": "int2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split} \\\\simplify{int({c}/((x+{a})(x+{b})),x)} &\\\\,= \\\\simplify[all,fractionNumbers]{int({c}/(({b-a})(x+{a}))+{c}/(({a-b})(x+{b})),x)} \\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[basic,fractionNumbers,zeroFactor,noLeadingMinus]{{Asol} int(1/(x+{a}),x)+{Bsol} int(1/(x+{b}),x)} \\\\\\\\ \\\\\\\\ &\\\\,=\\\\simplify[basic,fractionNumbers,zeroFactor,noLeadingMinus]{{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + C}. \\\\end{split}\\\\]

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[[0]]

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Correct answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": true, "customName": "brackets", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

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Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": true, "answer": "{Asol} ln (x+{a})+{Bsol} ln (x+{b}) + k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Forgotten constant", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b}))", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + c", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CI04 Integration - trig identities", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Using the various versions of $\\cos{2x}$ identity to integrate $\\sin^2{x}$ and $\\cos^2{x}$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Integrate $f(x)=\\var{Func}$.

", "advice": "

We can't integrate $\\var{Coeff}\\sin^2(x)$ directly so first we have to use the double angle formula $\\cos(2x)=1-2\\sin^2(x)$. We re-arrange using the double angle formula to give us,

\\[\\var{Coeff}\\sin^2(x)=\\frac{\\var{Coeff}}2-\\frac{\\var{Coeff}}2\\cos(2x).\\]

From the Table of Integrals we see that a function of the form \\[ f(x)= \\cos(nx) \\] has the integral \\[ \\int \\cos(nx) dx = \\frac{1}{n}\\sin(nx)+c\\]

So, for the function

\\[f(x)=\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}cos(2x)},\\]

the integral is

\\[ \\begin{split} \\int\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}cos(2x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{-Coeff/2}int(cos(2x),x)} &\\,=\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}(1/2 sin({2}x))} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{-Coeff/4} sin(2x)+c}. \\end{split} \\]

The integral of $\\frac{\\var{Coeff}}2$ is

\\[\\int\\frac{\\var{Coeff}}2dx=\\frac{\\var{Coeff}}2x+c,\\]

so combining these our final answer is

\\[\\int\\frac{\\var{Coeff}}2-\\frac{\\var{Coeff}}2\\cos(2x)dx=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}x-{Coeff/4} sin(2x)+c}\\]

We can't integrate $\\var{Coeff}\\cos^2(x)$ directly so first we have to use the double angle formula $\\cos(2x)=2\\cos^2(x)-1$. We re-arrange using the double angle formula to give us,

\\[\\var{Coeff}\\cos^2(x)=\\frac{\\var{Coeff}}2+\\frac{\\var{Coeff}}2\\cos(2x).\\]

From the Table of Integrals we see that a function of the form \\[ f(x)= \\cos(nx) \\] has the integral \\[ \\int \\cos(nx) dx = \\frac{1}{n}\\sin(nx)+c\\]

So, for the function

\\[f(x)=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}cos(2x)},\\]

the integral is

\\[ \\begin{split} \\int\\simplify[unitFactor,fractionNumbers]{{Coeff/2}cos(2x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{Coeff/2}int(cos(2x),x)} &\\,=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}(1/2 sin({2}x))} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{Coeff/4} sin(2x)+c}. \\end{split} \\]

The integral of $\\frac{\\var{Coeff}}2$ is

\\[\\int\\frac{\\var{Coeff}}2dx=\\frac{\\var{Coeff}}2x+c,\\]

so combining these our final answer is

\\[\\int\\frac{\\var{Coeff}}2+\\frac{\\var{Coeff}}2\\cos(2x)dx=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}x+{Coeff/4} sin(2x)+c}\\]

Use this link to find some resources which will help you revise this topic.

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It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "1/2{Coeff}*(x+{OneIfCosMinusOneIfSine}/2*sin(2x))", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "1/2{Coeff}*(x+{OneIfCosMinusOneIfSine}/2*sin(2x))+c", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CI05 Integration - Substitution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculating the integral of a function of the form $\\frac{nx^{n-1}}{x^n+a}$ using integration by substitution.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate \\[ \\simplify[all]{int(({n}x^{n-1})/(x^{n}+{a}),x)}\\]

by using the substitution \\[ \\simplify[all]{u=x^{n}+{a}}.\\]

", "advice": "

Since this integral is of the form \\[ \\int g'(x)f(g(x))\\,dx,\\] we can use the method of substitution to calculate the solution.

Firstly, we must make a change of variables from $x$ to $u$, where $u$ is equal to the 'inner' function $g(x)$.

So, for \\[\\simplify[fractionNumbers]{int(({n}x^{n-1})/((x^{n}+{a})),x)}\\]

let $\\color{red}{u=\\simplify[fractionNumbers]{x^{n}+{a}}}.$

Now, we need to calculate the differential, $du$, where \\[ du = \\left(\\frac{du}{dx}\\right)dx. \\]

Differentiating $u$ with respect to $x$:

\\[ \\frac{du}{dx}= \\simplify[fractionNumbers]{{n}x^{n-1}}.\\]

Therefore, \\[ \\color{blue}{du = \\simplify[fractionNumbers]{{n}x^{n-1}}\\, dx}.\\]

We can now rewrite the original integral in terms of $u$:

\\[ \\int \\frac{\\color{blue}{\\simplify{{n}x^{n-1}}}}{\\color{red}{\\simplify{x^{n}+{a}}}}\\color{blue}{\\text{d}x} = \\int \\frac{1}{\\color{red}{u}}\\color{blue}{\\text{d}u}.\\]

(Note: It is important to see that both the function we are integrating, and the variable we are integrating with respect to, has changed.)

\\[ \\simplify[fractionNumbers]{int(1/u,u) = ln(abs(u)) + c}.\\]

Finally, we must rewrite our solution back in terms of the original variable $x$:

\\[ \\simplify[fractionNumbers]{ln(abs(u)) + c = ln(abs(x^{n}+{a})) + c}.\\]

Use this link to find some resources which will help you revise this topic.

[[0]]

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Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": false, "answer": "ln(x^{n}+{a})+c", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Alternative using \"+k\"", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "ln(abs(x^{n}+{a})) + k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Alternative using brackets and \"+k\"", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": false, "answer": "ln(x^{n}+{a})+k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Forgotten constant", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "ln(abs(x^{n}+{a}))", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "ln(abs(x^{n}+{a}))+c", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CI06 Integration - Parts", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculating the integral of a function of the form $ax^2 \\cos(bx)$ using integration by parts.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the integral \\[ \\simplify{int({a}x^2 cos({b}x),x)}\\]

", "advice": "

If we have a function of $x$ which is the product of two functions of $x$, to integrate such a function it is often necessary to use Integration by Parts. The formula for Integration by Parts is:

\\[ \\int u(x) \\frac{dv}{dx} dx = u(x)v(x) - \\int v(x) \\frac{du}{dx} dx.\\]

Using this method can be broken down into steps:

Identify $u(x)$ and $\\tfrac{dv}{dx}$ (The function you pick for each is important, in general you want $u(x)$ to become simpler when differentiating it, and you must be able to integrate $\\tfrac{dv}{dx}$ to find $v(x)$);
Calculate $\\tfrac{du}{dx}$ and $v(x)$;
Put the functions $u(x)$, $v(x)$, and their derivatives into the Integration by Parts formula;
Calculate the integral $\\int v(x) \\tfrac{du}{dx} dx$ (This may require you to use Integration by Parts again, this is OK!);
Simplify your answer where possible and don't forget to add the constant of integration.

For the integral

\\[ \\simplify{int({a}x^2 cos({b}x),x)},\\]

we must first identify $u(x)$ and $\\tfrac{dv}{dx}$. In this case, let \\[ u(x)=\\simplify{{a}x^2},\\quad \\frac{dv}{dx}= \\simplify{cos({b}x)}. \\]

Next, we need to calculate $\\tfrac{du}{dx}$ and $v(x)$:

\\[ \\begin{split} u(x) = \\var{a}x^2 \\quad &\\implies \\frac{du}{dx} = \\simplify{{2a}x}; \\\\ \\frac{dv}{dx} = \\cos(\\var{b}x) &\\implies v(x) = \\simplify[fractionNumbers]{1/{b} sin({b}x)}. \\end{split} \\]

Plugging these 4 terms into the integration by parts formula:

\\[ \\begin{split} \\simplify{int({a}x^2 cos({b}x),x)} &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) - int({2a/b}x sin({b}x),x)}, \\\\ \\\\ &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}int(x sin({b}x),x)}.\\end{split} \\]

Since the integral on the right-hand side is still the product of two functions of $x$, we need to use integration by parts again.

So, for

\\[ \\simplify{int(x sin({b}x),x)}, \\]

Let $u=x$ and $\\tfrac{dv}{dx} = \\sin(\\var{b}x)$. Therefore, $\\tfrac{du}{dx}=1$ and $v(x)=\\simplify{-1/{b} cos({b}x)}$.

Hence,

\\[ \\begin{split} \\simplify{int(x sin({b}x),x)} &\\,= \\simplify{-1/{b}x cos({b}x)- int(-1/{b} cos({b}x),x)} \\\\ \\\\ &\\,= \\simplify{-1/{b}x cos({b}x)+1/{b^2}sin({b}x)}. \\end{split}\\]

Plugging this back into the original calculation:

\\[ \\begin{split} \\simplify{int({a}x^2 cos({b}x),x)} &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}int(x cos({b}x),x)} \\\\ \\\\ &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}[-1/{b}x cos({b}x)+1/{b^2}sin({b}x)]} \\\\ \\\\ &\\,=\\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) +{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)} + c.\\end{split} \\]

Use this link to find some resources which will help you revise this topic.

[[0]]

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It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "{a/b}x^2 sin({b}x)+{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{a/b}x^2 sin({b}x)+{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)+c", "answerSimplification": "fractionNumbers, basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GA01 Angles in triangles", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Find the missing angle in a triangle using the fact that the angles add to 180 degrees.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the size of the missing angle $C$? All angles are measured in degrees.

{geogebra_applet('https://www.geogebra.org/m/akwnfkfr',[a: a, b: b])}

", "advice": "

Recall that the angles in a triangle add up to $180^{\\circ}$.

We can add together two angles we know and subtract the result from $180$ to find the size of our missing angle,

\\[ \\begin{split} 180 - (\\var{a} + \\var{b}) &\\, = 180 - (\\var{a+b}) \\\\ &\\, = \\var{180-(a+b)}^{\\circ}. \\end{split} \\]

Use this link to find resources to help you revise properties of triangles.

The size of the missing angle is [[0]]$^{\\circ}$.

Draws a triangle based on 3 side lengths and randomises asking for hypotenuse or not.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{statement}

Find $x$.

", "advice": "

Only round your final answer to 1 decimal place.

{advice}

Use this link to find some resources to help you revise how to use pythagoras' theorem.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"setup": {"name": "setup", "group": "Varying q and advice", "definition": "random(1,2)", "description": "", "templateType": "anything", "can_override": false}, "answerside": {"name": "answerside", "group": "Varying q and advice", "definition": "sh1", "description": "", "templateType": "anything", "can_override": false}, "answerhyp": {"name": "answerhyp", "group": "Varying q and advice", "definition": "hyp", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Varying q and advice", "definition": "if(setup=1,answerside,answerhyp)", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "Varying q and advice", "definition": "if(setup=1,advice1,advice2)", "description": "", "templateType": "anything", "can_override": false}, "advice2": {"name": "advice2", "group": "Varying q and advice", "definition": "\"

Avoid using rounded values in calculations and just round for the final answer.

Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:

\\n

\\\\[a^2 + b^2 = c^2\\\\]

\\n

Let\\'s call the unknown value $x$, therefore we can write:

\\n

$a = \\\\var{sh1}$, $b =\\\\var{sh2}$ and $c = x$

\\n

So

\\\\[\\\\var{sh1}^2 + \\\\var{sh2}^2 = x^2\\\\]

\\n

and therefore

\\n

\\\\[x^2 = \\\\var{sh1^2} + \\\\var{sh2^2}\\\\]

\\\\[x = \\\\sqrt{\\\\var{sh1^2} + \\\\var{sh2^2}}\\\\]

\\n

\\\\[x = \\\\sqrt{\\\\var{sh1^2+sh2^2}}\\\\]

$x = \\\\var{hyp}$ to 1 d.p.

\"", "description": "", "templateType": "long string", "can_override": false}, "advice1": {"name": "advice1", "group": "Varying q and advice", "definition": "\"

Avoid using rounded values in calculations and just round for the final answer.

Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:

\\n

\\\\[a^2 + b^2 = c^2\\\\]

\\n

Let\\'s call the unknown value $x$, therefore we can write:

\\n

$a = x$, $b =\\\\var{sh2}$ and $c = \\\\var{hyp}$

\\n

\\\\[x^2 + \\\\var{sh2}^2 = \\\\var{hyp}^2\\\\]

\\n

and therefore

\\n

\\\\[x^2 = \\\\var{hyp^2} - \\\\var{sh2^2}\\\\]

\\n

\\\\[x = \\\\sqrt{\\\\var{hyp^2-sh2^2}}\\\\]

$x = \\\\var{sh1}$ to 1 d.p.

\"", "description": "", "templateType": "long string", "can_override": false}, "statement1": {"name": "statement1", "group": "Varying q and advice", "definition": "{geogebra_applet('https://www.geogebra.org/m/zhu32wjq',[sh1: sh1, sh2: sh2])}", "description": "

", "templateType": "anything", "can_override": false}, "statement2": {"name": "statement2", "group": "Varying q and advice", "definition": "{geogebra_applet('https://www.geogebra.org/m/b8wz2vn9',[sh1: sh1, sh2: sh2])}", "description": "", "templateType": "anything", "can_override": true}, "statement": {"name": "statement", "group": "Varying q and advice", "definition": "if(setup=1,statement1,statement2)", "description": "", "templateType": "anything", "can_override": false}, "hyp": {"name": "hyp", "group": "Unnamed group", "definition": "precround(sqrt(sh1^2+sh2^2),1)", "description": "", "templateType": "anything", "can_override": false}, "sh1_gen": {"name": "sh1_gen", "group": "Unnamed group", "definition": "random(5 .. 10#0.1)", "description": "

one of two shortest sides for calculations.

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$x=$[[0]] to 1 d.p.

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Find the diagonal or one side of a rectangle using Pythagoras' theorem.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the height of the rectangle below (all measurements given in $cm$)? Please give your answer to one decimal place.

{geogebra_applet('https://www.geogebra.org/m/jk3n6sxh',[base: base, hyp: hyp])}

", "advice": "

You can see that the rectangle contains a right-angled triangle. We also have the lengths of the base and the hypoteneuse of the triangle. This means we can use Pythagoras' theorem to calculate the last remaining side of the triangle which is also the height of the rectangle.

\\[ \\begin{split} Height &\\, = \\sqrt{hypoteneuse^2 - base^2} \\\\ &\\, = \\sqrt{\\var{hyp}^2-\\var{base}^2} \\\\ &\\, = \\sqrt{\\var{{hyp}^2}-\\var{{base}^2}} \\\\ &\\, = \\sqrt{\\var{{{hyp}^2}-{{base}^2}}}\\\\ &\\, = \\var{ans}\\\\ &\\, = \\var{ansr} \\text{ to 1 d.p.} \\end{split} \\]

Use this link to find resources to help you revise Pythagoras' theorem.

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[[0]]$cm$ (give your answer to 1 d.p.)

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Draws a triangle based on 3 side lengths. Randomises asking angle or side.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{max_height(25,diagram)}

", "advice": "

Avoid using rounded values in calculations and just round for the final answer.

{advice}

Use this link to find some resources to help you revise how to answer trigonometry questions that ask you to find a missing side.

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"can_override": false}, "d_s_s_1": {"name": "d_s_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle+'\u00b0')\n , b..c label('x cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_1": {"name": "d_c_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , a..c label(ac + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_1": {"name": "d_s_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_1": {"name": "d_t_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..c label(ac + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_2": {"name": "d_t_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle+'\u00b0')\n , b..c label(bc + 'cm')\n , a..c label('x cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_2": {"name": "d_c_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x\u00b0')\n , a..c label(ac + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "SCT": {"name": "SCT", "group": "Unnamed group", "definition": "random('s','c','t')", "description": "", "templateType": "anything", "can_override": false}, "AngORside": {"name": "AngORside", "group": "Unnamed group", "definition": "'side'", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_2": {"name": "d_c_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle+'\u00b0')\n , a..c label('x cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_2": {"name": "d_s_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_2": {"name": "d_s_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle+'\u00b0')\n , b..c label('x cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_2": {"name": "d_t_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0), \n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..c label(ac + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "answer": {"name": "answer", "group": "Unnamed group", "definition": "if(SCT='s',\n if(AngORside='ang',\n angle,\n bc),\n if(SCT='t',\n if(AngORside='ang',\n angle,\n ac),\n if(AngORside='ang',\n angle,ac)))", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "advice", "definition": "if(SCT='s',\n if(AngORside='ang',\n {sin_a},\n {sin_bc}),\n if(SCT='c',\n if(AngORside='ang',\n {cos_a},\n {cos_ac}),\n if(AngORside='ang',\n {tan_a},{tan_ac})))", "description": "", "templateType": "anything", "can_override": false}, "sin_a": {"name": "sin_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:

\\n

\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]

We need to use the \\'inverse $\\\\sin$\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\sin^{-1}(\\\\var{bc}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:

\\n

\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]

We need to use the \\'inverse $\\\\cos$\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\cos^{-1}(\\\\var{ac}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$

We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:

\\n

\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]

We need to use the \\'inverse $\\\\tan$\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\tan^{-1}(\\\\var{bc}/\\\\var{ac})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:

\\n

\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$

We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:

\\n

\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$

We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:

\\n

\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}}, "variablesTest": {"condition": "precround(180*(arcsin(bc/(ab)))/pi,1) = precround(angle,1)", "maxRuns": "6"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "angle", "SCT", "AngORside", "answer"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Given a right angled triangle as shown calculate the value of x.

Angles are given in degrees (make sure you calculator is in the right mode)

Give your answer correct to 2 decimal place.

", "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "2", "precisionPartialCredit": "100", "precisionMessage": "", "strictPrecision": true, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GA05 Trigonometry - missing angle", "extensions": ["eukleides"], "custom_part_types": [], "resources": ["question-resources/Picture1_caMIdF1.png", "question-resources/Picture2_6KE4ZpW.png"], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "David Wishart", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1461/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Draws a triangle based on 3 side lengths. Randomises asking angle or side.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{max_height(25,diagram)}

", "advice": "

Avoid using rounded values in calculations and just round for the final answer.

{advice}

Use this link to find resources to help you revise how to answer trigonometry questions that ask you to find the missing angle.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"ab": {"name": "ab", "group": "Unnamed group", "definition": "precround(sqrt(ac^2+bc^2),1)", "description": "", "templateType": "anything", "can_override": false}, "ac": {"name": "ac", "group": "Unnamed group", "definition": "precround(gen_ac,1)", "description": "", "templateType": "anything", "can_override": false}, "bc": {"name": "bc", "group": "Unnamed group", "definition": "precround(gen_bc,1)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_1": {"name": "d_t_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label(bc)\n , a..c label('x')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_1": {"name": "d_c_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , a..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "diagram": {"name": "diagram", "group": "Unnamed group", "definition": "if(SCT='s',\n if(AngORside='ang',\n random(d_s_a_1,d_s_a_2),\n random(d_s_s_1,d_s_s_2)),\n if(SCT='t',\n if(AngORside='ang',\n random(d_t_a_1,d_t_a_2),\n random(d_t_s_1,d_t_s_2)),\n if(SCT='c',\n if(AngORside='ang',\n random(d_c_a_1,d_c_a_2),\n random(d_c_s_1,d_c_s_2)),'X')))\n ", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_1": {"name": "d_s_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_1": {"name": "d_c_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , a..c label(ac)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_1": {"name": "d_s_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_1": {"name": "d_t_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..c label(ac)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_2": {"name": "d_t_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle)\n , b..c label(bc)\n , a..c label('x')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_2": {"name": "d_c_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x')\n , a..c label(ac)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "SCT": {"name": "SCT", "group": "Unnamed group", "definition": "random('s','c','t')", "description": "", "templateType": "anything", "can_override": false}, "AngORside": {"name": "AngORside", "group": "Unnamed group", "definition": "'ang'", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_2": {"name": "d_c_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle)\n , a..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_2": {"name": "d_s_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_2": {"name": "d_s_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_2": {"name": "d_t_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0), \n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x')\n , b..c label(bc)\n , a..c label(ac)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "answer": {"name": "answer", "group": "Unnamed group", "definition": "precround(angle*180/pi,2)", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "advice", "definition": "if(SCT='s',\n if(AngORside='ang',\n {sin_a},\n {sin_bc}),\n if(SCT='c',\n if(AngORside='ang',\n {cos_a},\n {cos_ac}),\n if(AngORside='ang',\n {tan_a},{tan_ac})))", "description": "", "templateType": "anything", "can_override": false}, "sin_a": {"name": "sin_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:

\\n

\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]

We need to use the \\'inverse sin\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\arcsin(\\\\var{bc}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:

\\n

\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]

We need to use the \\'inverse cos\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\arccos(\\\\var{ac}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$

We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:

\\n

\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]

We need to use the \\'inverse sin\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\arctan(\\\\var{bc}/\\\\var{ac})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:

\\n

\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$

We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:

\\n

\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$

We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:

\\n

\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "angle": {"name": "angle", "group": "Unnamed group", "definition": "If(SCT='c',arccos(ac/ab),if(SCT = 's',arcsin(bc/ab),arctan(bc/ac)))", "description": "", "templateType": "anything", "can_override": false}, "gen_ac": {"name": "gen_ac", "group": "Unnamed group", "definition": "random(3 .. 12#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "gen_bc": {"name": "gen_bc", "group": "Unnamed group", "definition": "random(5 .. 15#0.1)", "description": "", "templateType": "randrange", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "300"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "SCT", "AngORside", "answer", "angle", "gen_ac", "gen_bc"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Given a right angled triangle as shown calculate the value of x.

Give your answer in degrees (make sure you calculator is in the right mode), correct to 2 decimal place.

", "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "2", "precisionPartialCredit": "100", "precisionMessage": "", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GA06 Trigonometry - non-right angled trig", "extensions": ["geogebra"], "custom_part_types": [], "resources": ["question-resources/Picture1_caMIdF1.png", "question-resources/Picture2_6KE4ZpW.png"], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "David Wishart", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1461/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": [], "metadata": {"description": "

Draws a triangle based on 3 side lengths. Randomises asking angle or side.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{diagram}

Find x.

", "advice": "

{Advice}

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"Ruleuse": {"name": "Ruleuse", "group": "Question structure", "definition": "random('s','c','s','c')", "description": "", "templateType": "anything", "can_override": false}, "ANGorSIDE": {"name": "ANGorSIDE", "group": "Question structure", "definition": "random('ang','side')", "description": "", "templateType": "anything", "can_override": false}, "cosSIDEadvice": {"name": "cosSIDEadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that there are the lengths of two sides given and the angle specifically between those two sides. Further to this, the instruction is to find the other missing side. These are the conditions for when to use the $\\\\textit{cosine rule}$.

\\n

The formula for a missing side using the cosine rule is:

\\n

\\\\[ a^2 = b^2 + c^2 - 2bc \\\\cos(A)\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the $a$ (side) and $A$ (angle) labels are applied to the angle given and it\\'s opposite side.

\\n

In this case:

\\n

\\\\[ a=x, \\\\quad b=\\\\var{a}, \\\\quad c=\\\\var{b}, \\\\text{and} \\\\quad A=\\\\var{Cang},\\\\]

\\n

where the choice of which way round $b$ and $c$ are assigned doesn\\'t matter.

\\n

So, we now have:

\\n

\\\\[x^2 = \\\\var{a}^2 +\\\\var{b}^2-2\\\\times\\\\var{a}\\\\times\\\\var{b}\\\\times\\\\cos{(\\\\var{Cang})},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\sqrt{\\\\var{a^2 +b^2-2*a*b*(cos(Cang))}}\\\\]

\\n

\\\\[x=\\\\var{c}\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "

case 1: missing side in the cosine rule.

", "templateType": "long string", "can_override": false}, "cosANGadvice": {"name": "cosANGadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that there are the lengths of all three sides given. Further to this, the instruction is to find the a missing angle. These are the conditions for when to use the $\\\\textit{cosine rule}$ but in its rearranged form to find an angle. You need to identify which side is \\\"$a$\\\" as being the one opposite the angle you are asked to find.

\\n

The formula for a missing angle using the cosine rule is:

\\n

\\\\[ A = \\\\arccos\\\\left(\\\\frac{b^2+c^2-a^2}{2bc}\\\\right)\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the $a$ (side) and $A$ (angle) labels are applied to the side opposite the angle that is asked for and the angle that is asked for.

\\n

In this case:

\\n

\\\\[ a=\\\\var{c_round}, \\\\quad b=\\\\var{a}, \\\\quad c=\\\\var{b}, \\\\text{and} \\\\quad A= x,\\\\]

\\n

where the choice of which way round $b$ and $c$ are assigned doesn\\'t matter.

\\n

So, we now have:

\\n

\\\\[x = \\\\arccos\\\\left(\\\\frac{\\\\var{a}^2+\\\\var{b}^2-\\\\var{c_round}^2}{2\\\\times\\\\var{a}\\\\times\\\\var{b}}\\\\right),\\\\]

\\n

hence,

\\n

\\\\[x=\\\\var{(180/pi)*arccos((a^2 +b^2-c_round^2)/(2*a*b))}\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sinSIDEadvice": {"name": "sinSIDEadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that a single length is provided, along with two angles, crucially including the angle opposite the given side. Further to this, the instruction is to find the a missing angle. These are the conditions for when to use the $\\\\textit{sine rule}$. The sine rule uses the sides and angles in pairs and uses two pairs for any given calculation

\\n

The formula for finding a side using the sine rule can be written as:

\\n

\\\\[ \\\\frac{a}{\\\\sin(A)}=\\\\frac{b}{\\\\sin(B)}\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the side being asked for is in the above notation $a$.

\\n

In this case:

\\n

\\\\[ a=x, \\\\quad b=\\\\var{a}, \\\\quad A=\\\\var{Cang}, \\\\text{and} \\\\quad B= \\\\var{Aang_round}.\\\\]

\\n

So, we now have:

\\n

\\\\[\\\\frac{x}{\\\\sin{(\\\\var{Cang})}}=\\\\frac{\\\\var{a}}{\\\\sin{(\\\\var{Aang_round})}},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\frac{\\\\var{a}}{\\\\sin{(\\\\var{Aang_round})}}\\\\times\\\\sin{(\\\\var{Cang})},\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "

case 3

", "templateType": "long string", "can_override": false}, "sinANGadvice": {"name": "sinANGadvice", "group": "Question structure", "definition": "safe(\"

First recognise that the diagram is a non-right angled triangle and that two lengths are provided, along with an angle, crucially including an angle opposite a given side. Further to this, the instruction is to find the a missing side. These are the conditions for when to use the $\\\\textit{sine rule}$. The sine rule uses the sides and angles in pairs and uses two pairs for any given calculation

\\n

The formula for finding an angle using the sine rule can be written as:

\\n

\\\\[ \\\\frac{\\\\sin(A)}{a}=\\\\frac{\\\\sin(B)}{b}\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the angle being asked for is in the above notation $A$.

\\n

In this case:

\\n

\\\\[ a=\\\\var{c_round}, \\\\quad b=\\\\var{a}, \\\\quad A= x, \\\\text{and} \\\\quad B= \\\\var{Aang_round}.\\\\]

\\n

So, we now have:

\\n

\\\\[\\\\frac{\\\\sin{(x)}}{\\\\var{c_round}}=\\\\frac{\\\\sin{(\\\\var{Aang_round})}}{\\\\var{a}},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\arcsin\\\\left(\\\\var{c_round}\\\\times\\\\frac{\\\\sin{(\\\\var{Aang_round})}}{\\\\var{a}}\\\\right),\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\")", "description": "

case 4

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side length a

", "templateType": "randrange", "can_override": false}, "b": {"name": "b", "group": "Quantities", "definition": "random(5 .. 10#0.1)", "description": "

side length b

", "templateType": "randrange", "can_override": false}, "Cang": {"name": "Cang", "group": "Quantities", "definition": "random(40..140 except 85..95)", "description": "

C angle in degrees

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angle A in degrees

", "templateType": "anything", "can_override": false}, "Bang": {"name": "Bang", "group": "Quantities", "definition": "180-(Aang+Cang)", "description": "", "templateType": "anything", "can_override": false}, "cosSIDEans": {"name": "cosSIDEans", "group": "Quantities", "definition": "c", "description": "", "templateType": "anything", "can_override": false}, "cosANGans": {"name": "cosANGans", "group": "Quantities", "definition": "arccos((a^2+b^2-c_round^2)/(2*a*b))*180/pi", "description": "

Calculated answer for c from rounded values - as these will be seen information by student.

", "templateType": "anything", "can_override": false}, "c_round": {"name": "c_round", "group": "Quantities", "definition": "precround(c,1)", "description": "", "templateType": "anything", "can_override": false}, "Aang_round": {"name": "Aang_round", "group": "Quantities", "definition": "precround(Aang,1)", "description": "", "templateType": "anything", "can_override": false}, "Bang_round": {"name": "Bang_round", "group": "Quantities", "definition": "precround(Bang,1)", "description": "", "templateType": "anything", "can_override": false}, "Cang_roundcos": {"name": "Cang_roundcos", "group": "Quantities", "definition": "Precround((180/pi)*arccos((a^2+b^2-c_round^2)/(2*a*b)),1)", "description": "", "templateType": "anything", "can_override": false}, "sinANGans": {"name": "sinANGans", "group": "Quantities", "definition": "If(Cang<90,arcsin(c_round*(sin(Aang_round*pi/180)/a))*180/pi,180 - arcsin(c_round*(sin(Aang_round*pi/180)/a))*180/pi)", "description": "", "templateType": "anything", "can_override": false}, "sinSIDEans": {"name": "sinSIDEans", "group": "Quantities", "definition": "(a/sin(aang_round*pi/180))*sin(cang*pi/180)", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Quantities", "definition": "precround(If(Ruleuse='c',IF(ANGorSIDE='ang',cosANGans,cosSIDEans),IF(ANGorSIDE='ang',sinANGans,sinSIDEans)),1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "a+b>c and b+c>a and a+c>b", "maxRuns": "200"}, "ungrouped_variables": [], "variable_groups": [{"name": "Question structure", "variables": ["Ruleuse", "ANGorSIDE", "cosSIDEadvice", "cosANGadvice", "sinSIDEadvice", "sinANGadvice", "advice"]}, {"name": "Diagrams", "variables": ["cosSIDEdiagram", "cosANGdiagram", "sinSIDEdiagram", "sinANGdiagram", "diagram"]}, {"name": "Quantities", "variables": ["a", "b", "Cang", "c", "Aang", "Bang", "cosSIDEans", "cosANGans", "sinANGans", "sinSIDEans", "c_round", "Aang_round", "Bang_round", "Cang_roundcos", "ans"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": true, "customName": "Answer", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$x =$[[0]]

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Rewriting a trigonometric expression of the form $A\\sin(\\theta)-B\\cos(\\theta)$ to $R\\sin(\\theta-\\alpha)$ by calculating $R$ and $\\alpha$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

If \\[ \\simplify[unitFactor]{{A}sin(theta)-{B}cos(theta)} = R \\sin (\\theta - \\alpha),\\]

find the values for $R$ and $\\alpha$, given $R>0$ and $0<\\alpha<\\frac{\\pi}{2}$.

", "advice": "

To find $R$ and $\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\sin(a-b)=\\sin(a)\\cos(b)-\\sin(b)\\cos(a)$:

\\[ \\begin{split}\\simplify[unitFactor]{{A}sin(theta)-{B}cos(theta)} &\\,= R \\sin(\\theta-\\alpha) \\\\ &\\,= R(\\sin(\\theta)\\cos(\\alpha) - \\sin(\\alpha)\\cos(\\theta)) \\\\ &\\,= R\\sin(\\theta)\\cos(\\alpha) - R\\sin(\\alpha)\\cos(\\theta). \\end{split} \\]

By comparing the coefficients of $\\sin(\\theta)$ and $\\cos(\\theta)$, we find that

\\[ R\\cos(\\alpha) = \\var{A},\\quad \\text{and} \\quad R\\sin(\\alpha) = \\var{B}. \\]

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\sin^2(\\alpha)+\\cos^2(\\alpha) = 1$:

{Rsol}

Similarly, to find $\\alpha$ we can divide $R\\sin(\\alpha) = \\var{B}$ by $R\\cos(\\alpha) = \\var{A}$, and use the identity $\\tan(\\alpha) = \\frac{\\sin(\\alpha)}{\\cos(\\alpha)}$:

\\[ \\frac{R\\sin(\\alpha)}{R\\cos(\\alpha)} = \\frac{\\var{B}}{\\var{A}} \\implies \\tan(\\alpha) = \\simplify[fractionNumbers]{{B/A}}.\\]

Therefore, \\[ \\begin{split} \\alpha &\\,= \\tan^{-1}\\left(\\simplify[fractionNumbers]{{B/A}}\\right) \\\\ &\\,= \\var{alpharound} \\text{ (2 d.p.)}. \\end{split} \\]

Use this link to find some resources which will help you revise this topic.

\\\\[ \\\\begin{split} R^2\\\\cos^2(\\\\alpha) + R^2 \\\\sin^2(\\\\alpha) &\\\\,= \\\\var{A}^2+\\\\var{B}^2 \\\\\\\\ R^2 (\\\\cos^2(\\\\alpha) +\\\\sin^2(\\\\alpha)) &\\\\,= \\\\var{A^2+B^2} \\\\\\\\ R &\\\\,= \\\\var{R}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "Rsol2": {"name": "Rsol2", "group": "Ungrouped variables", "definition": "\"

\"", "description": "", "templateType": "long string", "can_override": false}, "alpharound": {"name": "alpharound", "group": "Ungrouped variables", "definition": "precround(alpha,2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["A", "B", "R", "Rround", "alpha", "alpharound", "Rsol", "Rsol1", "Rsol2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$R=$[[0]]

$\\alpha=$[[1]]

(Give your answers to 2 decimal places where necessary.)

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{Rround}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{alpharound}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GA08 Trigonometric Identities 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": [], "metadata": {"description": "

Rewriting a trigonometric expression of the form $A\\cos(\\theta)\\pm B\\sin(\\theta)$ to either $R\\sin(\\theta+\\alpha)$ or $R\\cos(\\theta+\\alpha)$ by calculating $R$ and $\\alpha$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

{question}

find the values for $R$ and $\\alpha$, given $R>0$ and $0<\\alpha<\\frac{\\pi}{2}$.

", "advice": "

{answer}

Use this link to find some resources which will help you revise this topic.

\"", "description": "", "templateType": "long string", "can_override": false}, "Rsol2": {"name": "Rsol2", "group": "Ungrouped variables", "definition": "\"

\"", "description": "", "templateType": "long string", "can_override": false}, "alpharound": {"name": "alpharound", "group": "Ungrouped variables", "definition": "precround(alpha,2)", "description": "", "templateType": "anything", "can_override": false}, "question": {"name": "question", "group": "Ungrouped variables", "definition": "if(Q=1,'{q1}','{q2}')", "description": "", "templateType": "anything", "can_override": false}, "Q": {"name": "Q", "group": "Ungrouped variables", "definition": "random(1,2)", "description": "", "templateType": "anything", "can_override": false}, "sign": {"name": "sign", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "templateType": "anything", "can_override": false}, "q1": {"name": "q1", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} = \\\\simplify[unitFactor]{R sin (theta+{sign}*alpha)},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "q2": {"name": "q2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} = \\\\simplify[unitFactor]{R cos (theta+{sign}*alpha)},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "a1": {"name": "a1", "group": "Ungrouped variables", "definition": "\"

To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify[unitFactor]{sin(a+{sign}*b)=sin(a)cos(b)+{sign}*sin(b)cos(a)}$:

\\n

\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} &\\\\,= \\\\simplify{R sin(theta+{sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(sin(theta)cos(alpha) + {sign}*sin(alpha)cos(theta))} \\\\\\\\ &\\\\,= \\\\simplify{Rsin(theta)cos(alpha) + {sign}*R sin(alpha)cos(theta)}. \\\\end{split} \\\\]

\\n

By comparing the coefficients of $\\\\sin(\\\\theta)$ and $\\\\cos(\\\\theta)$, we find that

\\n

\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

\\n

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

\\n

{Rsol}

\\n

Similarly, to find $\\\\alpha$ we can divide $R\\\\sin(\\\\alpha) = \\\\var{B}$ by $R\\\\cos(\\\\alpha) = \\\\var{A}$, and use the identity $\\\\tan(\\\\alpha) = \\\\frac{\\\\sin(\\\\alpha)}{\\\\cos(\\\\alpha)}$:

\\n

\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

\\n

Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "a2": {"name": "a2", "group": "Ungrouped variables", "definition": "\"

To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify{cos(a+{sign}*b)=cos(a)cos(b)-{sign}*sin(a)sin(b)}$:

\\n

\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} &\\\\,= \\\\simplify[unitFactor]{R cos (theta + {sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(cos(theta)cos(alpha) - {sign}*sin(theta)sin(alpha))} \\\\\\\\ &\\\\,= \\\\simplify{Rcos(theta)cos(alpha) - {sign}*R sin(theta)sin(alpha)}. \\\\end{split} \\\\]

\\n

By comparing the coefficients of $\\\\cos(\\\\theta)$ and $\\\\sin(\\\\theta)$, we find that

\\n

\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

\\n

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

\\n

{Rsol}

\\n

\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

\\n

Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "if(Q=1,'{a1}','{a2}')", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["Q", "A", "B", "sign", "R", "Rround", "alpha", "alpharound", "Rsol", "Rsol1", "Rsol2", "question", "q1", "q2", "answer", "a1", "a2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$R=$[[0]]

$\\alpha=$[[1]]

(Give your answers to 2 decimal places where necessary.)

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Match the graphs to the functions. No randomisation. Multiple choice.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

This is about knowledge of graphs. Generally with trigonometric graphs it is best to start with making sure you know and understand the graphs of the functionts $\\sin(x)$, $\\cos(x)$ and $\\tan(x)$. From there you can use knowledge of where they are zero to work out the position of the asymptotes in the graphs of $\\sec(x)$, $\\text{cosec}(x)$ and $\\cot(x)$. However, you still need really to be able to recall the shape of each graph for some purposes and be confident about where the zeros and turning points are.

Use this link to find some resources to help you familiarise yourself with these graphs.

Match the graph to its function.

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Calculate the volume of different 3D shapes, given the units and measurements required. The formulae for the volume of each shape are available as steps if required.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

For a triangular prism, we first need to find the area of one of the faces then multiply this area by the depth of the prism.
In this example the easiest way to calculate the volume is to take the area of the triangular face first with $\\mathrm{base} = \\var{w6}m$ and $\\mathrm{height} = \\var{h6}m\\thinspace$.

\\begin{align}
\\mathrm{Area\\thinspace_\\triangle} &= \\frac{\\mathrm{base} \\times \\mathrm{height}}{2} \\\\
&= \\frac{\\var{w6} \\times \\var{h6}}{2} \\\\
&= \\var{0.5*w6*h6}\\, \\mathrm{m}^2\\,.
\\end{align}

Now that we have the area of the triangular face ($\\mathrm{Area\\thinspace_\\triangle}$) we can multiply this by the $\\mathrm{depth} = \\var{d6}m\\thinspace$.

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\triangle} \\times \\mathrm{depth} \\\\
&= \\var{0.5*w6*h6} \\times \\var{d6} \\\\
&= \\var{0.5*w6*h6*d6}\\, \\mathrm{m}^2\\,.
\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a triangular prism.

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Side of square in cuboid.

", "templateType": "anything", "can_override": false}, "w6": {"name": "w6", "group": "Triangular prism", "definition": "random(5..9#1)", "description": "

Creates base of triangle.

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One side of square base.

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Height of pyramid.

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Depth of cylinder.

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Depth of triangular prism.

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Radius of the cylinder.

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Side of square in cuboid.

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Width of cuboid.

", "templateType": "anything", "can_override": false}, "w8": {"name": "w8", "group": "Square based pyramid", "definition": "random(3..7#1)", "description": "

One side of square base.

", "templateType": "anything", "can_override": false}, "h6": {"name": "h6", "group": "Triangular prism", "definition": "random(2..5#1)", "description": "

Height of traingle.

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Calculate the $\\mathrm{Volume}$ of the following triangular prism.

$\\mathrm{Volume} =$[[0]]$\\mathrm{m}^3$.

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Volume of a triangular prism:

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\triangle} \\times \\mathrm{depth} \\\\
&= \\frac{\\mathrm{base} \\times \\mathrm{height}}{2} \\times \\mathrm{depth}
\\end{align}

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Finding the area of a circle when given the diameter of the circle.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the area of a circle with diameter $\\var{d}$ cm giving your answer to 1 decimal place.

{geogebra_applet('https://www.geogebra.org/m/ngcchpcj',[d: d])}

", "advice": "

To calculate the area of a circle we want to use the formula \\[ A = \\pi r^2, \\]

where $r$ is the radius of the circle.

So, if the diameter, d, is $\\var{d}$ cm, then the radius is, $r=\\frac{d}{2}=\\var{{d}/2}$ cm, then

\\[ \\begin{split} Area &\\,=\\var{{d}/2}^2 \\times \\pi \\text{ cm}^2 \\\\ &\\,= \\simplify[all, fractionNumbers]{{{{d}^2/4}}pi} \\text{ cm}^2 \\\\ &\\,= \\var{precround({d}^2/4*pi,1)} \\text{ cm}^2. \\end{split} \\]

Use this link to find some resources to help you revise how to calculate the area of a circle.

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$Area=$ [[0]] $\\text{ cm}^2$

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Calculate the volume of different 3D shapes, given the units and measurements required. The formulae for the volume of each shape are available as steps if required.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

For a cylinder, we first need to find the area of the circular face then multiply this area by the depth of the cylinder.
In this example the radius of the circular face is $\\mathrm{radius} = \\var{r7}m$ which can be used to calculate the area of the circular face.

\\begin{align}
\\mathrm{Area\\thinspace_\\bigcirc} &= \\pi \\times \\mathrm{radius}^2 \\\\
&= \\pi \\times \\var{r7}^2 \\\\
&= \\var{pi * (r7)^2}\\, \\mathrm{m}^2 \\,.
\\end{align}

Now that we have the area of the circular face ($\\mathrm{Area\\thinspace_\\bigcirc}$) we can multiply this by the $\\mathrm{depth} =\\var{w7}m\\thinspace$.

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\bigcirc} \\times \\mathrm{depth} \\\\
&= \\var{pi*(r7)^2} \\times \\var{w7} \\\\
&= \\var{dpformat(pi*w7*(r7)^2, 5)} \\\\
&= \\var{dpformat(pi*w7*(r7)^2, 1)}\\, \\mathrm{m}^2\\,. \\quad \\text{1 d.p.}
\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a cylinder.

Side of square in cuboid.

", "templateType": "anything", "can_override": false}, "w6": {"name": "w6", "group": "Triangular prism", "definition": "random(5..9#1)", "description": "

Creates base of triangle.

", "templateType": "anything", "can_override": false}, "d8": {"name": "d8", "group": "Square based pyramid", "definition": "random(3..6#0.1)", "description": "

One side of square base.

", "templateType": "anything", "can_override": false}, "h8": {"name": "h8", "group": "Square based pyramid", "definition": "random(3..7#1)", "description": "

Height of pyramid.

", "templateType": "anything", "can_override": false}, "w7": {"name": "w7", "group": "Cylinder", "definition": "random(7..15#0.1)", "description": "

Depth of cylinder.

", "templateType": "anything", "can_override": false}, "d6": {"name": "d6", "group": "Triangular prism", "definition": "random(9..15#0.1)", "description": "

Depth of triangular prism.

", "templateType": "anything", "can_override": false}, "r7": {"name": "r7", "group": "Cylinder", "definition": "random(2..6#1)", "description": "

Radius of the cylinder.

", "templateType": "anything", "can_override": false}, "h4": {"name": "h4", "group": "Cuboid ", "definition": "random(2..5#1 except d4)", "description": "

Side of square in cuboid.

", "templateType": "anything", "can_override": false}, "w4": {"name": "w4", "group": "Cuboid ", "definition": "random(5.5..8#0.1)", "description": "

Width of cuboid.

", "templateType": "anything", "can_override": false}, "w8": {"name": "w8", "group": "Square based pyramid", "definition": "random(3..7#1)", "description": "

One side of square base.

", "templateType": "anything", "can_override": false}, "h6": {"name": "h6", "group": "Triangular prism", "definition": "random(2..5#1)", "description": "

Height of traingle.

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Calculate the $\\mathrm{Volume}$ of the following cylinder.

$\\mathrm{Volume} =$[[0]] $\\mathrm{m}^3$. Round your answer to 1 decimal place.

Volume of a cylinder:

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\bigcirc} \\times \\mathrm{depth} \\\\
&= \\pi \\times \\mathrm{r}^2 \\times \\mathrm{depth}
\\end{align}

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Find the volume of a semicylinder from a diagram.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the volume of this (all lengths are in $cm$):

{geogebra_applet('https://www.geogebra.org/m/vdbvgwkf',[height: height,radius: radius])}

", "advice": "

In order to work out the volume of a prism you need to work out the cross sectional area first. In this question the cross section is a semi-circle. Find the area of a circle and then half it.

The area of a semi-circle is given by:

\\begin{align} \\frac{\\pi\\times r^2}{2} \\end{align}

where $r$ is the radius of the circle.

\\begin{align} \\frac{\\pi\\times\\var{radius}^2}{2} = \\var{precround(semiarea,2)}... \\quad cm^2 \\end{align}

Then to calculate the volume you multiply the cross-sectional area by the length,

\\begin{align} \\frac{\\pi\\times r^2}{2} \\times l \\end{align}

\\begin{align} \\var{precround(semiarea,2)}... \\times \\var{height} = \\var{precround(answer,2)}cm^3.\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a prism.

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[[0]]$cm^3$

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "Volume", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GM05 Volume of a Trapezoidal Prism", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Find the volume of a prism with a trapezium as a cross section from a diagram.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the volume of this (all lengths are in $cm$):

{geogebra_applet('https://www.geogebra.org/m/qvcktek2',[basew: basew, topw: topw, h: h, l: l])}

", "advice": "

In order to work out the volume of a prism you need to work out the cross sectional area first. In this question the cross section is a trapezium. Find the area of a trapezium,

\\begin{align} \\frac{\\var{basew}+\\var{topw}}{2}\\times \\var{h} = \\var{traparea} cm^2 \\end{align}

Then to calculate the volume you times the cross-sectional area by the length,

\\begin{align} \\var{traparea} \\times \\var{l} = \\var{answer}cm^3\\end{align}.

Use this link to find resources to help you revise how to calculate the volume of a prism.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"basew": {"name": "basew", "group": "Ungrouped variables", "definition": "random(12 .. 20#2)", "description": "", "templateType": "randrange", "can_override": false}, "topw": {"name": "topw", "group": "Ungrouped variables", "definition": "random(4 .. 10#2)", "description": "", "templateType": "randrange", "can_override": false}, "h": {"name": "h", "group": "Ungrouped variables", "definition": "random(4 .. 10#1)", "description": "", "templateType": "randrange", "can_override": false}, "l": {"name": "l", "group": "Ungrouped variables", "definition": "random(8 .. 20#1)", "description": "", "templateType": "randrange", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "((topw+basew)/2)*h*l", "description": "", "templateType": "anything", "can_override": false}, "traparea": {"name": "traparea", "group": "Ungrouped variables", "definition": "(basew+topw)/2*h", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["basew", "topw", "h", "l", "answer", "traparea"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

[[0]]$cm^3$

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "Volume", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV01 Adding and subtracting vectors", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Vector Arithmetic

Work through the following questions to ensure you know how to add and subtract vectors in 2D.

For the whole of this question:

$\\bf{a} = \\var{a}$ and $\\bf{b}=\\var{b}$.

", "advice": "

The vectors in this question have two dimensions but the idea of addition and subtraction of vectors works in any number of dimensions (as long as all the vectors being added or subtracted have the same dimensions as each other).

To add two vectors you simply add their corresponding elements. In general:

$$
\\left(\\begin{array}{c}
a \\\\
b \\\\
\\end{array}\\right) +
\\left(\\begin{array}{c}
c \\\\
d \\\\
\\end{array}\\right) =
\\left(\\begin{array}{c}
a+c \\\\
b+d \\\\
\\end{array}\\right).
$$

Subtraction works in the same way so we have:

$$
\\var{a} + \\var{b} = \\var{a+b}.
$$

$$
\\var{a} - \\var{b} = \\var{a-b}.
$$

In order to undertstand the third part of the question you need to know what a \"position vector\" and \"direction vector\" are.

A position vector is defined as a vector that symbolises the location of any given point with respect to the origin. It can be thought of as a coordinate point, but written as a column vector - top entry is the \"x-coordinate\" and the bottome entry is the \"y-coordinate\".

A direction vector is defined as a vector that symbolises a direction and a distance in that direction but with no specified \"starting point\". In 2D it can be summarized as an instruction to go the top element number of units left or right based on the sign of the element and the bottom element number of units up or down based on the sign of the element.

So the direction vector from $A$ to $B$ can be worked out by looking at a route from $A$ to $B$ that travels along the position vectors given. Starting at $A$ we have to go backwards down $\\bf{a}$ to the origin and then forwards along $\\bf{b}$. This corresponds to doing \"minus\" $\\bf{a}$ and \"positive\" $\\bf{b}$:

$$
\\vec{AB} = (-)\\bf{a} + \\bf{b} = \\bf{b}-\\bf{a} = \\var{b}-\\var{a} = \\var{b-a}.
$$

Use this link to find some resources which will help you revise this topic.

Calculate $\\bf{a}+\\bf{b}$.

", "correctAnswer": "answeradd", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "matrix", "useCustomName": true, "customName": "2)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate $\\bf{a} - \\bf{b}$.

Let $\\bf{a}$ be the position vector of point $A$ and $\\bf{b}$ be the position vector of point $B$. Find the direction vector $\\vec{AB}$.

", "correctAnswer": "answerab", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV02 Scalar multiplication of vectors", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Vector Arithmetic

Work through the following questions exploring how to multiply a vector by a scalar.

For the whole of this question:

$\\bf{a} = \\var{a}$ and $\\bf{b}=\\var{b}$.

", "advice": "

The vectors in this question have two dimensions but the ideas herein work in any number of dimensions.

To multiply a vector by a scalar (number) you just multiply each element by that scalar:

$$
k\\left(\\begin{array}{c}
a \\\\
b \\\\
\\end{array}\\right) =
\\left(\\begin{array}{c}
ak \\\\
bk \\\\
\\end{array}\\right).
$$

So we have:

$$
\\var{m}\\var{a} = \\var{m*a}.
$$

The second and third part of this question just combine this idea of multiplying a vector by a scalar and the idea that addition and subtraction work by just calculating element by element (as long as all the vectors involved have the same dimensions).

$$
\\var{p}\\var{a} + \\var{q}\\var{b} =
\\left(\\begin{array}{c}
\\var{p} \\times \\var{a[0]} + \\var{q} \\times \\var{b[0]} \\\\
\\var{p}\\times \\var{a[1]} + \\var{q} \\times \\var{b[1]}\\\\
\\end{array}\\right) = \\var{p*a+q*b}.
$$

$$
\\var{r}\\var{a} - \\var{s}\\var{b} =
\\left(\\begin{array}{c}
\\var{r} \\times \\var{a[0]} - \\var{s} \\times \\var{b[0]} \\\\
\\var{s}\\times \\var{a[1]} - \\var{s} \\times \\var{b[1]}\\\\
\\end{array}\\right) = \\var{r*a-s*b}.
$$

Use this link to find some resources which will help you revise this topic.

Calculate $\\var{m}\\bf{a}$.

", "correctAnswer": "answerma", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "matrix", "useCustomName": true, "customName": "2)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate $\\var{p}\\bf{a}+\\var{q}\\bf{b}$.

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Calculate $\\var{r}\\bf{a} - \\var{s}\\bf{b}$.

", "correctAnswer": "answersub", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV03 Vectors and geometry", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

This question is about understanding the use of vectors to prove that a shape is a trapezium.

The quadrilateral ABCD is shown below.

$$
A = (\\var{a[0]},\\var{a[1]})\\\\
B = (\\var{b[0]},\\var{b[1]})\\\\
C = (\\var{c[0]},\\var{c[1]})\\\\
D = (\\var{d[0]},\\var{d[1]})
$$

{geogebra_applet('https://www.geogebra.org/m/xb9bvcaa',defs)}

", "advice": "

Vectors between points

In order to undertstand how to find the vector between two points it is helpful to know what a \"position vector\" and \"direction vector\" are. This advice should cover parts a), b), e) and f) of this question.

The point $B$ has coordinates $(\\var{B[0]},\\var{B[1]})$ and it has position vector, denoted $\\bf{b}$, given as $\\bf{b} = \\var{b}$.

So the direction vector from $B$ to $C$ can be worked out by looking at a route from $B$ to $C$ that travels along the position vectors given. Starting at $B$ we have to go backwards down $\\bf{b}$ to the origin and then forwards along $\\bf{c}$. This corresponds to doing \"minus\" $\\bf{b}$ and \"positive\" $\\bf{c}$:

$$
\\vec{BC} = (-)\\bf{b} + \\bf{c} = \\bf{c}-\\bf{b} = \\var{c}-\\var{b} = \\var{c-b}.
$$

Parallel vectors

If one vector is a multiple of another then they are vectors that point in the same direction. This means they are parallel. You just need to check what the multplier is between corresponding elements in each vector. If it is the same for both pairs of elements then the vectors are parallel (and if not then they are not).

For example, $\\vec{BC} = \\var{bc}$ and $\\vec{AD} = \\var{k*BC}.$ Since $\\frac{\\var{k*BC[0]}}{\\var{BC[0]}} = \\var{k}$ which gives the same multiplier as $\\frac{\\var{k*BC[1]}}{\\var{BC[1]}} = \\var{k}$ then $\\vec{BC}$ and $\\vec{AD}$ are parallel.

Conclusions about shapes

This question is looking at a trapezium specifically. The key properties of a trapezium are that it is a quadrilateral and there is one pair of parallel sides. This question goes through establishing that one pair of sides are parallel and then does the calculations to show that the other pair is not parallel. At this point we can conclude that $ABCD$ is a trapezium.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"A": {"name": "A", "group": "Ungrouped variables", "definition": "vector(0,0)", "description": "", "templateType": "anything", "can_override": false}, "AB": {"name": "AB", "group": "Ungrouped variables", "definition": "vector(0,random(2..7))", "description": "", "templateType": "anything", "can_override": false}, "B": {"name": "B", "group": "Ungrouped variables", "definition": "A+ab", "description": "", "templateType": "anything", "can_override": false}, "BC": {"name": "BC", "group": "Ungrouped variables", "definition": "vector(repeat(random(2..5),2))", "description": "", "templateType": "anything", "can_override": false}, "C": {"name": "C", "group": "Ungrouped variables", "definition": "B+bc", "description": "", "templateType": "anything", "can_override": false}, "D": {"name": "D", "group": "Ungrouped variables", "definition": "A+k*bc", "description": "", "templateType": "anything", "can_override": false}, "defs": {"name": "defs", "group": "Ungrouped variables", "definition": "[\n ['A',A],\n ['B',B],\n ['C',C],\n ['D',D]\n]", "description": "", "templateType": "anything", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["A", "AB", "B", "BC", "C", "D", "defs", "k"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "matrix", "useCustomName": true, "customName": "a)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find the vector $\\vec{BC}$.

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Find the vector $\\vec{AD}$.

", "correctAnswer": "-a+d", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "numberentry", "useCustomName": true, "customName": "c)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Hence, it is possible to write $\\vec{AD} = k \\times \\vec{BC}$. Find the value of $k$.

", "minValue": "k", "maxValue": "k", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "1_n_2", "useCustomName": true, "customName": "d)", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The fact that $\\vec{AD}$ can be written as a multiplier times by $\\vec{BC}$ means:

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Find the vector $\\vec{AB}$.

", "correctAnswer": "ab", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "matrix", "useCustomName": true, "customName": "f)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find the vector $\\vec{CD}$.

", "correctAnswer": "-c+d", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "m_n_2", "useCustomName": true, "customName": "g)", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Up to this point in the question we have now shown which of the following:

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the scalar (or dot) product of $\\bf{a}$ and $\\bf{b}$.

", "advice": "

It is important to note that for vectors there is more than one type of multiplication. This question is specifically about the scalar (or dot) product.

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2},$ the scalar (or dot) product is defined as

$$
\\mathbf{v \\cdot w} = v_1 \\times w_1 + v_2 \\times w_2.
$$

So for this question:

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b}\\\\
\\bf{a} \\cdot \\bf{b} = \\var{a[0]}\\times\\var{b[0]} + \\var{a[1]}\\times\\var{b[1]} = \\var{adotb}.
$$

Use this link to find some resources which will help you revise this topic.

$\\bf{a} \\cdot \\bf{b} =$ [[0]]

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the scalar (or dot) product of $\\bf{a}$ and $\\bf{b}$.

", "advice": "

It is important to note that for vectors there is more than one type of multiplication. This question is specifically about the scalar (or dot) product.

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2 \\\\ v_3},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2 \\\\ w_3},$ the scalar (or dot) product is defined as

$$
\\mathbf{v \\cdot w} = v_1 \\times w_1 + v_2 \\times w_2 + v_3 \\times w_3.
$$

So for this question:

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b}\\\\
\\bf{a} \\cdot \\bf{b} = \\var{a[0]}\\times\\var{b[0]} + \\var{a[1]}\\times\\var{b[1]} + \\var{a[2]}\\times\\var{b[2]} = \\var{adotb}.
$$

Use this link to find some resources which will help you revise this topic.

$\\bf{a} \\cdot \\bf{b} =$ [[0]]

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "scalar product", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "adotb", "maxValue": "adotb", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV06 Scalar product to find angles between vectors", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Poppy Jeffries", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21275/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the angle between $\\bf{a}$ and $\\bf{b}$.

", "advice": "

To answer these questions, we want to use the equations for the scalar product. Recall:

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2},$

\\[ \\begin{split} \\mathbf{v \\cdot w} &\\,= v_1 \\times w_1 + v_2 \\times w_2 \\\\\\\\ \\mathbf{v \\cdot w} &\\,= |\\mathbf v| |\\mathbf w | \\cos(\\theta), \\end{split} \\]

where $|\\mathbf v|$ and $|\\mathbf w|$ are the magnitudes of the vectors, and $\\theta$ is the angle between the vectors.

For

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b},
$$

we have

$$
\\bf{a} \\cdot \\bf{b} = \\var{adotb},
$$

$$
|\\bf{a}| = \\sqrt{\\var{a[0]^2 +a[1]^2}} \\approx \\var{precround(asize,2)},\\\\
|\\bf{b}| = \\sqrt{\\var{b[0]^2 +b[1]^2}} \\approx \\var{precround(bsize,2)}.
$$

So we have:
$$
\\theta = \\cos^{-1}\\left(\\frac{\\var{adotb}}{\\sqrt{\\var{a[0]^2 +a[1]^2}}\\times\\sqrt{\\var{b[0]^2 +b[1]^2}}}\\right) = \\var{precround(angle,1)}.
$$

Use this link to find some resources which will help you revise this topic.

The angle between $\\bf{a}$ and $\\bf{b}$ is [[0]] degrees.

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the angle between $\\bf{a}$ and $\\bf{b}$.

", "advice": "

To answer these questions, we want to use the equations for the scalar product. Recall:

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2 \\\\ v_3},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2 \\\\ w_3},$

\\[ \\begin{split} \\mathbf{v \\cdot w} &\\,= v_1 \\times w_1 + v_2 \\times w_2 + v_3 \\times w_3 \\\\\\\\ \\mathbf{v \\cdot w} &\\,= |\\mathbf v| |\\mathbf w | \\cos(\\theta), \\end{split} \\]

where $|\\mathbf v|$ and $|\\mathbf w|$ are the magnitudes of the vectors, and $\\theta$ is the angle between the vectors.

For

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b},
$$

we have

$$
\\bf{a} \\cdot \\bf{b} = \\var{adotb},
$$

and

$$
|\\bf{a}| = \\sqrt{\\var{a[0]^2 +a[1]^2 +a[2]^2}} \\approx \\var{precround(asize,2)},\\\\
|\\bf{b}| = \\sqrt{\\var{b[0]^2 +b[1]^2 +b[2]^2}}\\approx \\var{precround(bsize,2)}.
$$

So we have:
$$
\\theta = \\cos^{-1}\\left(\\frac{\\var{adotb}}{\\sqrt{\\var{a[0]^2 +a[1]^2 +a[2]^2}}\\times\\sqrt{\\var{b[0]^2 +b[1]^2 +b[2]^2}}}\\right) = \\var{precround(angle,1)}.
$$

Use this link to find some resources which will help you revise this topic.

The angle between $\\bf{a}$ and $\\bf{b}$ is [[0]] degrees.

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "Angle", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "minValue": "180 - angle", "maxValue": "180 - angle", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "1", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "minValue": "angle", "maxValue": "angle", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "1", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV08 Scalar product to find perpendicular vectors", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\pmatrix{\\var{b1} \\\\ k}$ and that the two vectors are perpendicular.

Find the value of $k$.

", "advice": "

The key thing to understand for this question is that for perpendicular vectors the scalar (or dot) product will give a result of zero.

In this question we have,

\\begin{alignat}{2}
&\\quad
&\\var{a}\\cdot\\pmatrix{\\var{b1} \\\\ k}
& = 0 \\\\
&\\Rightarrow\\quad
&\\var{a[0]}\\times\\var{b1} + \\var{a[1]} \\times k & = 0 \\\\
&\\Rightarrow\\quad
&\\var{a[0]*b1} + \\var{a[1]}k & = 0.
\\end{alignat}

Solving this then gives,

$$
k = \\var{k}.
$$

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "vector(repeat(random(-9..9 except 0),2))", "description": "", "templateType": "anything", "can_override": false}, "asize": {"name": "asize", "group": "Ungrouped variables", "definition": "sqrt(a[0]^2+a[1]^2+a[2]^2)", "description": "", "templateType": "anything", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "-a[0]*b1/a[1]", "description": "", "templateType": "anything", "can_override": false}, "b1": {"name": "b1", "group": "Ungrouped variables", "definition": "random(-5..5)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "asize", "b1", "k"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "numberentry", "useCustomName": true, "customName": "dec", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "minValue": "k", "maxValue": "k", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "sigfig", "precision": "3", "precisionPartialCredit": "100", "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "minValue": "k", "maxValue": "k", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV09 Scalar product to find perpendicular vectors 3D", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\pmatrix{\\var{b1} \\\\ k \\\\ \\var{b3}}$ and that the two vectors are perpendicular.

Find the value of $k$.

", "advice": "

The key thing to understand for this question is that for perpendicular vectors the scalar (or dot) product will give a result of zero.

In this question we have,

\\begin{alignat}{2}
&\\quad
&\\var{a}\\cdot\\pmatrix{\\var{b1} \\\\ k \\\\ \\var{b3}}
& = 0 \\\\
&\\Rightarrow\\quad
&\\var{a[0]}\\times\\var{b1} + \\var{a[1]} \\times k + \\var{a[2]} \\times \\var{b3} & = 0 \\\\
&\\Rightarrow\\quad
&\\var{a[0]*b1 + a[2]*b3} + \\var{a[1]}k & = 0.
\\end{alignat}

Solving this then gives,

$$
k = \\var{k}.
$$

Use this link to find some resources which will help you revise this topic.

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The vectors $\\bf a$ and $\\bf b$ are defined as follows,
$$
{\\bf a} = \\var{a} \\qquad \\text{and} \\qquad \\bf b = \\var{b}.
$$

", "advice": "

The vector product (also referred to as the cross product) of two vectors can be calculated as follows:

$$
\\pmatrix{a_1 \\\\ a_2 \\\\ a_3} \\times \\pmatrix{b_1 \\\\ b_2 \\\\ b_3} = \\pmatrix{a_2b_3-a_3b_2 \\\\ -(a_1b_3-a_3b_1) \\\\ a_1b_2-a_2b_1}.
$$

A way to think of this is as the following determinant:

$$
\\begin{vmatrix}
\\bf i & \\bf j & \\bf k\\\\
a_1 & a_2 & a_3 \\\\
b_1 & b_2 & b_3
\\end{vmatrix},
$$

where $\\bf i$, $\\bf j $, and $\\bf k$ are the standard unit basis vectors.

In this question we therefore have:

$$
\\begin{align*}
\\var{a} \\times \\var{b} & = \\pmatrix{\\var{a[1]} \\times \\var{b[2]} - \\var{a[2]} \\times \\var{b[1]} \\\\ -(\\var{a[0]} \\times \\var{b[2]} - \\var{a[2]} \\times \\var{b[0]}) \\\\ \\var{a[0]} \\times \\var{b[1]} - \\var{a[1]} \\times \\var{b[0]}}\\\\
& = \\var{answer}.
\\end{align*}
$$

Use this link to find some resources which will help you revise this topic.

Find $\\bf a \\times \\bf b$.

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Thank you for completing the Skills Audit for Maths and Stats. Hopefully it has been useful in directing you to resources and services that can support your studies. The Skills Audit for Maths and Stats will remain open to you throughout the academic year and you can always revisit it again later.

For any further information or questions please contact mash@sheffield.ac.uk

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The answer is a comma-separated list of numbers.

The list is marked correct if each number occurs the same number of times as in the expected answer, and no extra numbers are present.

You can optionally treat the answer as a set, so the number of occurrences doesn't matter, only whether each number is included or not.

Is every number in the student's list valid?

Are the student's answers in ascending order?

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Is each number in the expected answer present in the student's list the correct number of times?

", "definition": "all(included)"}, {"name": "no_extras", "description": "

True if the student's list doesn't contain any numbers that aren't in the expected answer.

Should the answer be considered as a set, so the number of times an element occurs doesn't matter?

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Numbers included in the student's answer that are not in the expected list.

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