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COM undergraduate courses

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Simplifying expressions from $\\frac{x^mx^n}{x^p}$ to $x^{m+n-p}$.

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

Simplify the following expression:

\\[x^{\\var{m}}x^{\\var{n}}\\]

", "advice": "

To simplify $x^{\\var{m}}x^{\\var{n}}$, we want to make use of the following rule:

\\[a^n \\times a^m = a^{n+m}\\]

Applying this rule,

\\[\\begin{split}x^{\\var{m}}x^{\\var{n}} &\\,=x^{\\simplify[!collectNumbers]{{m}+{n}}}\\\\ \\\\&\\,=x^{\\var{m+n}}. \\end{split}\\]

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

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Find the missing whole number power in an equation.

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

What is the value of $n$ if

\\[\\frac{x^n}{x^\\var{p}}=x^\\var{m}\\]

", "advice": "

To find $n$ we need to re-write the expression such that we have $x^n$ on the left. We can multiply through by $x^\\var{p}$ to get

\\[x^n=x^\\var{m}{x^\\var{p}}\\]

Then applying the rule $x^p \\times x^q = x^{p+q}$ we get

\\[x^n=x^{\\var{m}+\\var{p}}=x^\\var{m+p}\\]

Hence, $n =\\var{m+p}$

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

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Calculate an answer involving a fractional index.

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

Evaluate the following expression:

\\[\\var{a^n}^{\\frac{1}{\\var{n}}}\\]

", "advice": "

To find $\\var{a^n}^{\\frac{1}{\\var{n}}}$, we want to make use of the fact that a power of $\\frac{1}{n}$ is the same as the $n$th root. Since

\\[\\var{a^n}=\\var{a}^\\var{n},\\]

we have,

\\[ \\var{a^n}^{\\frac{1}{\\var{n}}} =\\left(\\var{a}^\\var{n}\\right)^{\\frac{1}{\\var{n}}}=\\var{a}. \\]

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

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Using indices rules to rewrite an expression from $a^\\frac{m}{n}$ to $b$, for integers $a$, $b$, $m$ and $n$.

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

Evaluate the following expression:

\\[\\var{a^n}^{\\frac{\\var{m}}{\\var{n}}}\\]

", "advice": "

To find $\\var{a^n}^{\\frac{\\var{m}}{\\var{n}}}$, we want to make use of the following rule:

\\[\\left(a^n\\right)^m = a^{n\\times m}\\]

By rewriting the power $\\frac{\\var{m}}{\\var{n}}$ as a product of $\\var{m} \\times \\frac{1}{\\var{n}}$, we can apply this rule:

\\[ \\begin{split} \\var{a^n}^{\\frac{\\var{m}}{\\var{n}}} &\\,= \\var{a^n}^{\\left(\\var{m} \\times \\frac{1}{\\var{n}}\\right)} \\\\ &\\,= \\left(\\var{a^n}^\\frac{1}{\\var{n}}\\right)^\\var{m} \\\\ &\\,= \\var{a}^\\var{m}\\end{split} \\]

Then calculating what is left:

\\[ \\begin{split} \\var{a}^\\var{m} &\\,=\\var{a^(m)} \\end{split} \\]

Therefore,

\\[ \\var{a^n}^{\\frac{\\var{m}}{\\var{n}}} =\\var{a^(m)}. \\]

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

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perform a calculation involving negative indices.

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

Evaluate and simplify the following expression:

\\[\\frac{\\var{x}^{\\var{n}}}{\\var{y}^{\\var{m}}}\\]

", "advice": "

To simplify this expression we use the rule $a^{-n}=\\frac1{a^n}$.

\\[\\frac{\\var{x}^{\\var{n}}}{\\var{y}^{\\var{m}}}=\\frac{\\var{y}^{\\var{-m}}}{\\var{x}^{\\var{-n}}}=\\frac{\\var{y^-m}}{\\var{x^-n}}=\\simplify{{y^-m}/{x^-n}}\\]

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

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Expand two brackets involving powers of $x$.

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

Expand the brackets and simplify

", "advice": "

To expand the brackets $\\simplify{({a[1]}x^{b[1]}+{a[2]}x^{b[2]})({a[3]}x^{b[3]}+{c[1]}x^{b[4]})}$ We first multiply all the terms in the left bracket by all the terms in the right bracket. This gives us

\\[\\var{a[1]}x^\\var{b[1]}\\times\\var{a[3]}x^\\var{b[3]}+\\var{a[1]}x^\\var{b[1]}\\times\\var{c[1]}x^\\var{b[4]}+\\var{a[2]}x^\\var{b[2]}\\times\\var{a[3]}x^\\var{b[3]}+\\var{a[2]}x^\\var{b[2]}\\times\\var{c[1]}x^\\var{b[4]}\\]

We can then simplify to give us the final answer of

$\\simplify{{a[1]*a[3]}*x^{b[1]+b[3]}+{a[1]*c[1]}*x^{b[1]+b[4]}+{a[2]*a[3]}*x^{b[2]+b[3]}+{a[2]*c[1]}*x^{b[2]+b[4]}}.$

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

$\\simplify{({a[1]}x^{b[1]}+{a[2]}x^{b[2]})({a[3]}x^{b[3]}+{c[1]}x^{b[4]})}=$[[0]]

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Fiind the Highest Common Factor of two algebraic expressions involving a coefficient and powers of $x$ and $y$.

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

What is the highest common factor of $\\var{c[0]}x^\\var{xp[0]}y^\\var{yp[0]}$ and $\\var{c[1]}x^\\var{xp[1]}y^\\var{yp[1]}$?

", "advice": "

In order to find the highest common factor of two single term algebraic expressions you can first find the highest common factor of the coefficients.

In this case the Highest common factor of $\\var{c[0]}$ and $\\var{c[1]}$ is $\\var{cans}$.

Then work through each of the variables (letters) in turn and see what powers of each appear. In the first expression there is $x^\\var{xp[0]}$ and the second expression there is $x^\\var{xp[1]}$. So they both have at least $x^\\var{xpans}$ in them. Similarly, the first expression there is $y^\\var{yp[0]}$ and the second expression there is $y^\\var{yp[1]}$. So they both have at least $y^\\var{ypans}$ in them.

Hence, the Highest Common Factor (HCF) of the two expressions is:

\\[\\var{cans}x^\\var{xpans}y^\\var{ypans}.\\]

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

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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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"f1c": {"name": "f1c", "group": "Set up", "definition": "simplify(cf1[0]+\"*\"+v+\"-\"+cf1[1],\"all\")", "description": "", "templateType": "anything", "can_override": false}, "cf2": {"name": "cf2", "group": "Set up", "definition": "repeat(random(2..5),2)", "description": "", "templateType": "anything", "can_override": false}, "f2": {"name": "f2", "group": "Set up", "definition": "simplify(cf2[0]+\"*\"+v+\"+\"+cf2[1],\"all\")", "description": "", "templateType": "anything", "can_override": false}, "den1fact": {"name": "den1fact", "group": "Advice", "definition": "simplify(a+\"*\"+\"(\"+string(f1)+\")*(\"+string(f1c)+\")\",\"all\")", "description": "", "templateType": "anything", "can_override": false}, "num1": {"name": "num1", "group": "Question", "definition": "a*b", "description": "", "templateType": "anything", "can_override": false}, "den2": {"name": "den2", "group": "Question", "definition": "simplify(den2fact,[\"expandBrackets\",\"all\"])", "description": "", "templateType": 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"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": "

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]]

", "gaps": [{"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": "{(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}))", "answerSimplification": "all", "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": "`! (((`+-$n`?*x^2+`+-$n`?*x+`+-$n)/((x+`+-$n)(x^2+`+-$n*x+`+-$n))))", "partialCredit": 0, "message": "", "nameToCompare": "", "warningTime": "submission"}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AF01 Sigma 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": "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": "

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.

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

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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": "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.

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

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]]

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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)

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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)

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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} \\\\]

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

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

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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.

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

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

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

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

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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]]

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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.

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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]]$^{\\circ}$.

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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.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"ab": {"name": "ab", "group": "Unnamed group", "definition": "random(7..12)", "description": "", "templateType": "anything", "can_override": false}, "ac": {"name": "ac", "group": "Unnamed group", "definition": "precround(ab*cos(pi*angle/180),2)", "description": "", "templateType": "anything", "can_override": false}, "bc": {"name": "bc", "group": "Unnamed group", "definition": "precround(ab*sin(pi*angle/180),2)", "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+'\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_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+'\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}, "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}, "angle": {"name": "angle", "group": "Unnamed group", "definition": "random(32..72)", "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+'\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.

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"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

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side length b

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C angle in degrees

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angle A in degrees

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Calculated answer for c from rounded values - as these will be seen information by student.

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$x =$[[0]]

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Finding the area of a circle when given the diameter of the circle.

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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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The following questions are designed to explore the dimensions of matrices and what you can and can't do with matrices of differing dimensions.

", "advice": "

Rows and Columns

The convention in Matrix notation is to give the dimensions of a matrix in the order \"rows\" by \"columns\".

For $\\var{Dimensions}$ there are $\\var{rows[0]}$ rows and $\\var{columns[0]}$ columns. We write this as \"this is a $\\var{rows[0]}$X$\\var{columns[0]}$ matrix\".

When can you add and subtract matrices?

Two Matrices can be added or subtracted if they have the exact same dimensions as each other. For example $\\var{canadd1}$ and $\\var{canadd2}$ are both $\\var{rows[1]}$X$\\var{columns[1]}$ matrices and therefore can be added (or subtracted). However, $\\var{cantaddsub1}$ is a $\\var{rows[3]}$X$\\var{columns[3]}$ matrix and $\\var{cantaddsub2}$ is a $\\var{rows[3]}$X$\\var{columns[3]+1}$ matrix. Since these dimensions are different these matrices cannot be added or subtracted.

Multiplying Dimensions

When you multiply two matrices together the number of columns in the first matrix must match the number of rows in the second matrix. For example in the calculation $\\var{Mult3}$X$\\var{Mult4}$ the first matrix has $3$ columns and the second matrix has $3$ rows so they can be multiplied. In addition to this when multiplying two matrices (that can be multiplied) the result will be a single matrix that has the number of rows of the first matrix and the number of columns of the second matrix. In this example the first matrix has $\\var{rows[0]}$ rows and the second matrix has $\\var{columns[1]}$ columns, so the result of multiplying the two matrices will be a $\\var{rows[0]}$X$\\var{columns[1]}$ matrix.

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

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What are the dimensions of the following matrix?

$\\var{dimensions}$

[[0]]X[[1]]

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Which of the following calculations are defined?

(Indicate ALL possible answers by ticking the corresponding box(es))

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Is this calculation defined?

$\\var{Mult1}$X$\\var{Mult2}$

[[0]]

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What will be the dimensions of the matrix you get when you multiply these two matrices?

$\\var{Mult3}$X$\\var{Mult4}$.

[[0]]X[[1]]

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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.

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Calculate $\\var{m}\\bf{a}$.

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Calculate $\\var{p}\\bf{a}+\\var{q}\\bf{b}$.

", "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": "3)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

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": "NA07 HCF", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}, {"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": "Calculating the LCM and HCF of numbers by using prime factorisation.", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

By considering the prime factorisation of $\\var{x}$ and $\\var{y}$, or otherwise, find the highest common factor (HCF) of $\\var{x}$ and $\\var{y}$.

", "advice": "

We can write $\\var{x}$ and $\\var{y}$ as a product of prime factors as follows:

$\\var{x}=\\var{show_factors(x)}$

$\\var{y}=\\var{show_factors(y)}$

For HCF of $\\var{x}$ and $\\var{y}$ we need to multiply each prime factor the least number of times it occurs in either $\\var{x}$ or $\\var{y}$

i.e. HCF$(x,y) = \\var{show_factors(hcf_xy)}=\\var{hcf_xy}$

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

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By considering the prime factorisation of $\\var{x}$ and $\\var{y}$, or otherwise, find the lowest common multiple (LCM) of $\\var{x}$ and $\\var{y}$.

", "advice": "

We can write $\\var{x}$ and $\\var{y}$ as a product of prime factors as follows:

$\\var{x}=\\var{show_factors(x)}$

$\\var{y}=\\var{show_factors(y)}$.

For LCM of $\\var{x}$ and $\\var{y}$ we need to multiply each factor the greatest number of times it occurs in either $\\var{x}$ or $\\var{y}$.

i.e. LCM$(x,y) = \\var{show_factors(lcm_xy)}=\\var{lcm_xy}$.

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

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State the Upper and lower bound of a distance that has been rounded to either the nearest 10 or 100 miles.

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

The distance between two towns had been rounded to the nearest {x} miles in an aticle in the newspaper. If they reported that the distance was {y} miles, what are the upper and lower bound for the reported number?

", "advice": "

If a number like {y} has been rounded to the nearest {x} then {y} would have been rounded down if it was less than {y+x/2} because {y} is the nearest multiple of {x}.

Similarly {y} would have been rounded up if it was larger than or equal to {y-x/2}. This means the lower bound is {y-x/2} and the upper bound is {y+x/2}.

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

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Upper bound:

[[0]]

Lower bound:

[[1]]

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This question is about correctly interpreting pie charts.

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

{geogebra_applet{\"https://www.geogebra.org/calculator/pmvumdrv\",[C: C,M: M]}}

The Pie Chart above shows the responses to a question asked by someone trying to plan a social event for their workplace. It shows answers given to the question \"Where would you like to go for a staff social?\" with the options \"Meal\", \"Cinema\" and \"Games Cafe\".

", "advice": "

A Pie chart of this type can only be used to make statements about the proportions of data in each category and does not provide information about the actual frequencies.

For more information on Pie Charts follow this link.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"C": {"name": "C", "group": "Ungrouped variables", "definition": "random(60 .. 70#5)", "description": "", "templateType": "randrange", "can_override": false}, "M": {"name": "M", "group": "Ungrouped variables", "definition": "random(5 .. 25#5)", "description": "", "templateType": "randrange", "can_override": false}}, "variablesTest": {"condition": "C<>M", "maxRuns": 100}, "ungrouped_variables": ["C", "M"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "m_n_x", "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": "

From the following comments which can you say are definitely true, definitely false and which do you not have enough information to know?

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Reading a value from a simple bar chart.

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

The bar heights give the values of the spend.

Each company has two bars, the left one for last year (in red) and the right one for this year (in purple).
Isolate last years spend by looking at the the bars on the right side, and choose the tallest bar, corresponding to the highest value.

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

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new y values after the transformation

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the (random) original y values which relate to the x values

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vertical shift

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horizontal shift

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new transformed x values

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original x values

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\n var yo8 = scope.variables.yo8.value;\n var yo9 = scope.variables.yo9.value; \n \n var div = Numbas.extensions.jsxgraph.makeBoard('550px','550px',{boundingBox:[-0.8,82,16,-8], axis:false, grid:true});\n \n question.display.html.querySelector('#dragpoint').append(div);\n \n var board = div.board;\n \n// board.suspendUpdate(); \n\n \n var dataArr = [yo0,yo5,0,yo1,yo6,0,yo2,yo7,0,yo3,yo8,0,yo4,yo9]; \n \n var xaxis = board.create('axis', [[0, 0], [12, 0]], {withLabel: true, name: \"Bank\", label: {offset: [250,-30]}});\n \n xaxis.removeAllTicks(); \n \n board.create('axis', [[0, 0], [0, 10]], {hideTicks:true, withLabel: false, name: \"\", label: {offset: [-110,300]}});\n \n var pop0 = board.create('point', [1.5,0],{name:'Morgan',fixed:true,size:0,color:'black',face:'diamond', label:{offset:[-20,-8]}});\n var pop1 = board.create('point',[4.5,0],{name:'Strome',fixed:true,size:0,color:'black',face:'diamond', label:{offset:[-20,-8]}});\n var pop2 = board.create('point',[7.5,0],{name:'Bentley',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var pop3 = board.create('point',[10.5,0],{name:'Sand',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var pop4 = board.create('point',[13.5,0],{name:'Karchen',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var leg1 = board.create('point',[12,75],{name:'last year',fixed:true,size:6,color:'#DA2228', face:'square', label:{offset:[9,0]}});\n var leg2 = board.create('point',[12,72],{name:'this year',fixed:true,size:6,color:'#6F1B75', face:'square', label:{offset:[9,0]}});\n \n \n// var chart = board.createElement('chart', dataArr, \n // {chartStyle:'bar', fillOpacity:1, width:1,\n // colorArray:['#8E1B77','#8E1B77','Red','Red','blue','red','blue','red','red','blue', 'red','blue','red','red'], shadow:false});\n \n//var chart = board.createElement('chart', dataArr, \n // {chartStyle:'bar', width:1,fillOpacity:1, fillColor:'red', shadow:false}); \n \n \n var a = board.create('chart', [[1,2,3],[yo0,yo5,0]], {chartStyle:'bar',colors:['#DA2228','#6F1B75','#6F1B75'],width:1,fillOpacity:1});\n var b = board.create('chart', [[4,5,6],[yo1,yo6,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var c = board.create('chart', [[7,8,9],[yo2,yo7,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var d = board.create('chart', [[10,11,12],[yo3,yo8,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var e = board.create('chart', [[13,14],[yo4,yo9]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75'],fillOpacity:1});\n \n board.unsuspendUpdate();\n \n var txt1 = board.create('text',[-0.3,30, 'Investment \u00a3(m)'], {fontColor:'black', fontSize:14, rotate:90});\n \n // var txt = board.create('text',[0.5,75, 'Investment (m)'], {fontSize:14, rotate:90});\n \n // var txt1 = board.create('text',[8,76, 'red bars represent 2010'], {fontColor:'red', fontSize:14, rotate:90});\n \n // var txt2 = board.create('text',[8,73, 'blue bars represents 2011'], {fontSize:14, rotate:90});\n\n // var myColors = new Array('red', 'blue', 'white','red', 'blue', 'white','red', 'blue', 'white','red', 'blue', 'white','red', 'blue');\n \n \n \n //board.unsuspendUpdate();\n\n // Rotate text around the lower left corner (-2,-1) by 30 degrees.\n // var tRot = board.create('transform', [90.0*Math.PI/180.0, -1,40], {type:'rotate'}); \n // tRot.bindTo(txt);\n // board.update();\n\n \n//var chart2 = board.createElement('chart', dataArr, {chartStyle:'line,point'});\n//chart2[0].setProperty('strokeColor:black','strokeWidth:2','shadow:true');\n//for(var i=0; i<11;i++) {\n // chart2[1][i].setProperty({strokeColor:'black',fillColor:'white',face:'[]', size:4, strokeWidth:2});\n//}\n//board.unsuspendUpdate(); \n \n //board.unsuspendUpdate();\n\n}\n\nquestion.signals.on('HTMLAttached',function() {\n dragpoint_board();\n});", "css": "table#values th {\n background: none;\n text-align: center;\n}"}, "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, "prompt": "

Banking Sector: IT Infrastructure Spending

What was the maximum spend by a single company last year?

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£{answer} m

", "

£{fakeanswer1} m

", "

£{fakeanswer2} m

", "

£{fakeanswer3} m

", "

£{fakeanswer4} m

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Interpreting the elements of a box plot

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

The diagram below shows a box plot of some data.

{geogebra_applet{\"https://www.geogebra.org/m/aj2hcbhg\",[lv: lv,lq: lq,m: m,uq: uq,hv: hv]}}

", "advice": "

A boxplot (also known as a box-and-whisker diagram or plot) is a convenient way of graphically depicting groups of numerical data through their five-number summaries: the smallest observation (sample minimum), lower quartile (Q1), median (Q2), upper quartile (Q3), and largest observation (sample maximum). A boxplot may also indicate which observations, if any, might be considered outliers.

For more information on box plots follow this link.

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Which of these statements are true and which are false?

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The median of the data is $\\var{m}$.

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Calculate an intersection probability given a two way table.

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

a) Each row and column must sum to the 'total'.

b) Look for the column containing '$\\var{q1a}$' and the row containing '$\\var{q1b}$'. The entry where they intersect, $\\var{q1*total}$, is the value we are interested in.

Since we require a probability, this is $\\var{q1*total}$ out of $\\var{total}$, i.e.

\\[ \\frac{\\var{q1*total}}{\\var{total}} \\]

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

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$\\var{total}$ items are sampled. Complete the table.

\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

	$\\var{nB}$	not $\\var{nB}$	TOTAL
$\\var{nA}$	[[0]]	$\\var{AnB'}$	$\\var{A}$
not $\\var{nA}$	$\\var{notAnB}$	[[1]]	[[2]]
TOTAL	[[3]]	[[4]]	$\\var{total}$

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If one item is picked at random, use the table to calculate the probability that the item is '{q1a}' and '{q1b}'.

Give your answer as a fraction, or a decimal correct to 2dp.

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Reading a value from a histogram.

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

The histogram shows information about distances run on a Sunday by some randomly asked people in a park.

", "advice": "

To calculate the frequencies using the following formula:

frequency = class width $\\times$ frequency density

Hence our table becomes:

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

Distance, $km$	Frequency
$0 < t \\leq 10$	$10 \\times \\var{yo0} = \\var{freq0}$
$10 < t \\leq 15$	$5 \\times \\var{yo1} = \\var{freq1}$
$15 < t \\leq 20$	$5 \\times \\var{yo2} = \\var{freq2}$
$20 < t \\leq 30$	$10 \\times \\var{yo3} = \\var{freq3}$

Hence, to find the total number of people that ran that day:

$\\var{freq0} + \\var{freq1} + \\var{freq2} + \\var{freq3} = \\var{total}.$

The frequency of runners that ran less than $10km$ is found by calculating the frequency from the first bar on the histogram:

class width $\\times$ frequency density $= 10 \\times \\var{yo0} = \\var{freq0}$.

The frequency of runners that ran between $15km$ and $20km$ is found by calculating the frequency from the third bar on the histogram:

class width $\\times$ frequency density $= 5 \\times \\var{yo2} = \\var{freq2}$.

To estimate how many runners ran more than $25km$ we need again need to use the frequency = class width \\times frequency density formula.

Here the class width is $5$ because we are looking for the frequency of runners that ran between $25km$ and $30km$ from the college.

Frequency $= 5 \\times \\var{yo3} = \\var{5*yo3}.$

If this number is a decimal we round up to get $\\var{m25}$.

As in part d),to estimate how many runners ran less than $7km$ we use the frequency = class width \\times frequency density formula.

Here the class width is $7$ because we are looking for the frequency of runners that ran between 0 and 7 km.

Frequency $= 7 \\times \\var{yo0} = \\var{7*yo0}.$

If this number is a decimal we round up to get $\\var{l7}$.

To estimate how many runners ran between $5km$ and $12.5km$ we use a method similar to that in part d) and e) but, this time, we need to use information from both the first and second bar on the histogram.

Let's first calculate how many runners ran between $5km$ and $10km$:

frequency $=$ class width $\\times$ frequency density $= 5 \\times \\var{yo0} = \\var{5*yo0}$.

Now we need to calculate how many runners ran between $10km$ and $12.5km$:

frequency $=$ class width $\\times$ frequency density $= 2.5 \\times \\var{yo1} = \\var{2.5*yo1}$.

Putting this together the number of runners that ran between $5km$ and $12.5km$ is $\\var{5*yo0} + \\var{2.5*yo1} = \\var{5*yo0 + 2.5*yo1}.$

If this number is a decimal we round up to get $\\var{ef}.$

The number of runners that ran between $10km$ and $14km$ is $4 \\times \\var{yo1} = \\var{4*yo1}.$

If this number is a decimal we round up to get $\\var{eh}.$

The number of runners that ran further than $18km$ is:

$2 \\times \\var{yo2} + 10 \\times \\var{yo3} = \\var{2*yo2} + \\var{freq3} = \\var{2*yo2+freq3}.$

If this number is a decimal we round up to get $\\var{ej}.$

Did you get a decimal answer? Were you surprised to see a whole number answer when you got a decimal on one of the estimate questions? Look at the context of the question, you cannot have $0.5$ of a student so we round our answers up to the next whole number!

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

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Frequency Density

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Frequency Density

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Frequency Density

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Frequency Density

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Sum of frequencies

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Frequency

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Frequency

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Frequency

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Frequency

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How many people ran more than 25km

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How many runners ran less than 7km

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Number of people running between 5 & 12.5 km

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Number of runners running between 10 & 14km

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Number of runners running more than 18km

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How many people were asked about the distance they ran that day?

", "minValue": "total", "maxValue": "total", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "b)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

How many runners ran less than 10 km that day?

", "minValue": "freq0", "maxValue": "freq0", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"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": "

How many people ran between 10 and 15 km that day?

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Estimate how many runners ran more than 25 km that day?

", "minValue": "m25", "maxValue": "m25", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "e)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Estimate how many runners ran less than 7 km that day?

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Estimate how many people ran between 5 km and 12.5 km.

", "minValue": "Ef", "maxValue": "Ef", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SD07 Interpreting Line Graphs", "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": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Francesca recorded the number of customers in a supermarket every two hours.

She began at 9 am and finished at 7 pm.

The line graph below shows her results.

{geogebra_applet{\"https://www.geogebra.org/classic/s4w7nmga\",[y1:y1,y3:y3,y4:y4,y6:y6]}}

", "advice": "

a) You want to find the point on the graph with the greatest frequency. You can see that this is at 1pm when there were $\\var{y3}$ cars in the car park.

b) We find 11am on the $x$-axis and look vertically upwards until we find the point. From here we go horiztonally across to the $y$-axis to read the frequency. We can see that at 11am there were $\\var{y2}$ cars in the car park.

c) We want to find $\\var{number}$ on the $y$-axis and then look horiztonally across until we find the point. From here we move down to the $x$-axis to see at which time there were $\\var{number}$ cars in the car park. We can see this occured at $\\var{answerc}\\var{time}$.

d) We must find 6pm on the $x$-axis. This isn't marked on like the other times but we know it sits half way between 5pm and 7pm. From here we look vertically upwards until we meet the red line on our graph. Notice we don't have a point for this time, hence why this is an estimate. From here we move hortizontally across to the $y$-axis to find the frequency. At 6pm we estimate that there were $\\var{y56}$ cars in the car park.

e) We must find the frequency of cars at 1pm and at 3pm using the same steps as in part b. At 3pm there were $\\var{y4}$ cars in the car park and we subtract this from $\\var{y3}$ which is the number of cars in the car park at 1pm. Hence, $\\var{y3}-\\var{y4}=\\var{y3-y4}$.

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Number of customers in supermarket at 7pm.

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Number of customers in supermarket at 11am.

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Number of customers in supermarket at 6pm.

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Associated time to randomly selected frequency variable (number)

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For printing am or pm depending on time selelected - to appear in advice only

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When were the most customers in the supermarker?

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How many customers were in the supermarket at 11 am?

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At what time were there {number} customers in the supermarket?

If your answer was 12pm you would just write 12 in the box.

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Estimate the number of customers in the supermarket at 6pm.

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How many less customers were there in the supermarket at 3pm than 1pm?

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Predicting the probability of an unbiased coin landing on heads based on the results of previous throws.

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When we flip an unbiased coin there are two possible events that we could measure: the coin lands on heads or the coin lands on tails.

Each toss of the coin is independent; if we flip a coin once and it lands on heads then the next time we flip the coin it is still equally likely to land on either heads or tails.

It doesn't matter what the coin landed on previously as this outcome does not affect the outcome of the next flip of the coin.

Even when we flip an unbiased coin $\\var{no_flips}$ times and it lands on heads each time; the next time we flip the coin, it is still equally likely to land on either heads or tails.

So the probability that the coin lands on heads the next time that the coin is flipped is still $\\displaystyle\\frac{1}{2}$.

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

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Number of flips of the coin

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An unbiased coin is flipped $\\var{no_flips}$ times. Given that the coin landed on heads each time, what is the probability of the coin landing on heads the next time it is flipped?

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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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