// Numbas version: exam_results_page_options {"name": "MAS140 Chemical and Biological Engineering draft Skills Audit ", "metadata": {"description": "

Skills Audit for Maths and Stats for MAS140 Chemical and Biological Engineering.

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Metric Unit conversion - division by 1000.

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

Express {liquid} millilitres ($ml$) in litres ($l$). Give your answer to 3 decimal places.

", "advice": "

There are $1000ml$ in $1l$. To work out the conversion: $\\frac{\\var{liquid}}{1000} = \\var{answer}$.

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

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[[0]]$l$

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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 rouned 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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Compound percentage change.

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

The value of a car is initially {StartingPrice}. If the value decreases by {dec}%, and then increases by {inc}%, what is the final value?

Give your answer correct to two decimal places.

", "advice": "

There is a {dec}% decrease in price. This means that price after the decrease will be {100-dec}% of the old price.

\\[\\frac{\\var{100-dec}}{100} \\times \\var{StartingPrice} = \\var{(100-dec)/100*StartingPrice}\\]

Then there is a {inc}% increase in price. This means the final price will be {100+inc}% of the price after the decrease.

\\[\\frac{\\var{100+inc}}{100} \\times \\var{(100-dec)/100*StartingPrice} = £\\var{dpformat((100+inc)/100*(100-dec)/100*StartingPrice,2)}\\]

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

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£[[0]]

Dividing amounts in ratios

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

The ratio of ethanol to water is {a}:{b} for an experiment. If I have {volWater}ml of water, how much ethanol do I need?

", "advice": "

If there is a ratio of {a}:{b} for ethanol:water then that means for every {b}ml of water we need {a}ml of ethanol.

In our experiment there is {volwater}ml of water so to find the amount of ethanol we divide by {b} and then multiply by {a}.

\\[\\var{volwater}\\text{ml}\\times\\frac{\\var{a}}{\\var{b}}=\\var{volwater*a/b}\\text{ml}\\]

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

[[0]]ml

Calculations involving Standard form.

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

To divide two numbers in standard form we can calculate the division of each part of the standard form number separately. In general we have,

\\[\\frac{x\\times10^j}{y\\times10^k}=\\frac xy\\times\\frac{10^j}{10^k}=\\frac xy\\times 10^{j-k}\\]

In this question we therefore have,

\\[\\frac{\\var{a}\\times10^{\\var{n}}}{\\var{b}\\times10^{\\var{m}}}=\\frac{\\var{a}}{\\var{b}}\\times\\frac{10^{\\var{n}}}{10^{\\var{m}}}=\\var{aDivBRound}\\times10^\\var{n-m}.\\]

Since {aDivBRound} is less than 1 then our answer isn't in standard form. In this case we need to reduce the exponent by 1 so the final answer is

\\[\\var{MantAnsRound}\\times10^{\\var{ExponentAns}}.\\]

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

For the equation

\\[\\frac{\\var{a}\\times10^{\\var{n}}}{\\var{b}\\times10^{\\var{m}}}=a\\times10^n\\]

find the values of $a$ and $n$ which keep the answer in standard form.

Give $a$ to two decimal places.

$a=$[[0]]

$n=$[[1]]

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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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Simple exercise in collecting terms in different powers of \$x\$

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

Simplify the following expression by combining \"like\" terms.

", "advice": "

First we expand the minus sign in the bracket.

\\[\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4-({f}x+{e}x^3)}=\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4+{-f}x+{-e}x^3}\\]

The idea is to collect together and combine any terms that are the same kind of term so:

$\\var{b}x$ and $\\var{-f}x$ both have an $x$ term. We can combine them to get $\\var{b-f}x$

We can combine $\\var{a}x^4$ and $\\var{d}x^4$ to get $\\var{a+d}x^4$.

We combine $\\var{c}x^3$ and $\\var{-e}x^3$ to get $\\var{c-e}x^3$. So our answer is:

$\\simplify{{a+d}x^4+{c+e}x^3+{b+f}}$

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

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$\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4-({f}x+{e}x^3)}$

", "answer": "({a}+{d})x^4+({c}-{e})x^3+({b}-{f})x", "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": "`+-$n`?*x^4+`+-$n`?*x^3+`+-$n`?*x", "partialCredit": 0, "message": "", "nameToCompare": ""}, "valuegenerators": [{"name": "x", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AB8 Expand Double Brackets (Hard)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Poppy Jeffries", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21275/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

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]]

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

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

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

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

", "advice": "

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

{advice1}

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

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

\\n

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

\\n

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

\\n

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

\\n

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

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

\\n

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

\\n

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

$x=$ [[0]]

$y=$ [[1]]

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

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

Solve the following simultaneous equations:

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

Give your answers to 2 decimal places where necessary.

", "advice": "

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

For the equations

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

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

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

Substituting this into equation (2):

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

Collecting similar terms:

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

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

{check}

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

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

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

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

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

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

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

", "gaps": [{"type": "matrix", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "matrix", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "correctAnswer": "solutions2", "correctAnswerFractions": false, "numRows": 1, "numColumns": "2", "allowResize": false, "tolerance": 0, "markPerCell": true, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "correctAnswer": "solutions1", "correctAnswerFractions": false, "numRows": 1, "numColumns": "2", "allowResize": false, "tolerance": 0, "markPerCell": true, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "matrix", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "matrix", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "correctAnswer": "solutions1", "correctAnswerFractions": false, "numRows": 1, "numColumns": "2", "allowResize": false, "tolerance": 0, "markPerCell": true, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "correctAnswer": "solutions2", "correctAnswerFractions": false, "numRows": 1, "numColumns": "2", "allowResize": false, "tolerance": 0, "markPerCell": true, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AC6 Rearrange Formulae", "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": "Luigi Pivano", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18182/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Rearrange a specific formula. No randomisation.

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

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

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

", "advice": "

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

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

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

$y=$ [[0]]

The answer is a comma-separated list of numbers.

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

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

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

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

Is every number in the student's list valid?

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

Are the student's answers in ascending order?

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

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

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

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

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

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

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

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

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

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

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

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

Solve the following quadratic equation by factorisation or otherwise:

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

", "advice": "

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

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

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

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

$x= $[[0]]

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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.

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

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

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

"}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AE2 Algebraic Fractions - addition (harder)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

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

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

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

", "advice": "

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

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

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

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

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

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Set up", "definition": "random(2 .. 6#1)", "description": "", "templateType": "randrange", "can_override": false}, "b": {"name": "b", "group": "Set up", "definition": "random(2 .. 5#1)", "description": "", "templateType": "randrange", "can_override": false}, "v": {"name": "v", "group": "Set up", "definition": "random(\"a\",\"b\",\"c\",\"d\",\"f\",\"g\",\"h\",\"k\",\"m\",\"n\",\"p\",\"q\",\"r\",\"s\",\"t\",\"u\",\"v\",\"w\",\"x\",\"y\",\"z\")", "description": "", "templateType": "anything", "can_override": false}, "cf1": {"name": "cf1", "group": "Set up", "definition": "repeat(random(2..4),2)", "description": "", "templateType": "anything", "can_override": false}, "f1": {"name": "f1", "group": "Set up", "definition": "simplify(cf1[0]+\"*\"+v+\"+\"+cf1[1],\"all\")", "description": "", "templateType": "anything", "can_override": false}, "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": "anything", "can_override": false}, "num2": {"name": "num2", "group": "Question", "definition": "simplify(f2,\"all\")", "description": "", "templateType": "anything", "can_override": false}, "den1": {"name": "den1", "group": "Question", "definition": "simplify(den1fact,[\"expandBrackets\",\"all\"])", "description": "", "templateType": "anything", "can_override": false}, "den2fact": {"name": "den2fact", "group": "Advice", "definition": "simplify(expression(\"(\"+string(f1c)+\")*(\"+string(f2)+\")\"),\"all\")", "description": "", "templateType": "anything", "can_override": false}, "ansn": {"name": "ansn", "group": "Question", "definition": "simplify(string(f1) + \"+\" + b,\"all\")", "description": "", "templateType": "anything", "can_override": false}, "ansd": {"name": "ansd", "group": "Question", "definition": "simplify(expression(\"(\"+string(f1)+\")\"+\"*\"+ \"(\"+string(f1c)+\")\"),[\"expandBrackets\",\"all\"])", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Question", "definition": "simplify(expression(\"(\"+string(ansn)+\")\"+\"/\"+\"(\"+string(ansd)+\")\"),\"all\")", "description": "", "templateType": "anything", "can_override": false}, "den1simp": {"name": "den1simp", "group": "Advice", "definition": "simplify(\"(\"+string(f1)+\")*(\"+string(f1c)+\")\",\"all\")", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "f1<>f2 AND f1c<>f2 AND cf1[0]<>cf1[1] AND cf2[0]<>cf2[1]", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "Set up", "variables": ["a", "b", "v", "cf1", "f1", "f1c", "cf2", "f2"]}, {"name": "Question", "variables": ["num1", "den1", "num2", "den2", "ansn", "ansd", "ans"]}, {"name": "Advice", "variables": ["den1fact", "den2fact", "den1simp"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{ans}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "mustmatchpattern": {"pattern": "?`+/?`+", "partialCredit": 0, "message": "", "nameToCompare": ""}, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AE3 - Cancelling algebraic fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Luke Park", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/826/"}, {"name": "Anna Strzelecka", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2945/"}, {"name": "heike hoffmann", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2960/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": [], "metadata": {"description": "

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

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

Simplify the following algebraic expression.

", "advice": "

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

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

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

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

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

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

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

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

", "answer": "(n+{e2})/(n+{e3})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "notallowed": {"strings": ["^2", "^"], "showStrings": false, "partialCredit": 0, "message": ""}, "valuegenerators": [{"name": "n", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AE6 Partial Fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"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": "

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}))}.\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "pairs[index][0]", "description": "", "templateType": "anything", "can_override": false}, "simp1": {"name": "simp1", "group": "Ungrouped variables", "definition": "gcd(k*(a-b)-m*a*c+n*c,m*c-m*b*a+n*a-k)", "description": "", "templateType": "anything", "can_override": false}, "simp2": {"name": "simp2", "group": "Ungrouped variables", "definition": "gcd(simp1,a^2-a*b+c)", "description": "", "templateType": "anything", "can_override": false}, "sol3": {"name": "sol3", "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)/simp2}x+{(k*(a-b)-m*a*c+n*c)/simp2})/({(a^2-a*b+c)/simp2}(x^2+{b}x+{c}))}.\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "random([1,2,3,5,7])", "description": "", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "1", "description": "", "templateType": "anything", "can_override": false}, "pairs": {"name": "pairs", "group": "Ungrouped variables", "definition": "[[1,random(-1,1)*random([1,3,4,5])],[2,random(-1,1)*random([1,2,4,5])],[3,random(-1,1)*random([1,2,3,5])],[5,random(-1,1)*random([1,2,3,4,5,7])],[7,random(-1,1)*random([1,2,3,4,5,7])]]", "description": "", "templateType": "anything", "can_override": false}, "index": {"name": "index", "group": "Ungrouped variables", "definition": "random(0..4)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "a^2-a*b+c>0 or a^2-a*b+c<0", "maxRuns": 100}, "ungrouped_variables": ["a", "pairs", "index", "b", "c", "m", "k", "n", "Asol", "Bsol", "Csol", "check", "sol1", "sol2", "sol3", "simp1", "simp2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "\n

[[0]]

Basic calculation from a sum given in Sigma notation.

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

Calculate:

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

", "advice": "

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

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

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

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Find the volume of a semicylinder from a diagram.

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

Calculate the volume of this (all lengths are in $cm$):

{geogebra_applet('https://www.geogebra.org/m/vdbvgwkf',[height: height,radius: radius])}

", "advice": "

In order to work out the volume of a prism you need to work out the cross sectional area first. In this question the cross section is a semi-circle. Find the area of a circle and then half it.

The area of a semi-circle is given by:

\\begin{align} \\frac{\\pi\\times r^2}{2} \\end{align}

where $r$ is the radius of the circle.

\\begin{align} \\frac{\\pi\\times\\var{radius}^2}{2} = \\var{precround(semiarea,2)}... \\quad cm^2 \\end{align}

Then to calculate the volume you multiply the cross-sectional area by the length,

\\begin{align} \\frac{\\pi\\times r^2}{2} \\times l \\end{align}

\\begin{align} \\var{precround(semiarea,2)}... \\times \\var{height} = \\var{precround(answer,2)}cm^3.\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a prism.

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[[0]]$cm^3$

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "Volume", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GB3 - sec/cosec/cot", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Match the graphs to the functions. No randomisation. Multiple choice.

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

This is about knowledge of graphs. Generally with trigonometric graphs it is best to start with making sure you know and understand the graphs of the functionts $\\sin(x)$, $\\cos(x)$ and $\\tan(x)$. From there you can use knowledge of where they are zero to work out the position of the asymptotes in the graphs of $\\sec(x)$, $\\text{cosec}(x)$ and $\\cot(x)$. However, you still need really to be able to recall the shape of each graph for some purposes and be confident about where the zeros and turning points are.

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

Match the graph to its function.

", "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["$\\sec(x)$", "$\\text{cosec}(x)$", "$\\cot(x)$"], "matrix": [["1", "0", 0], [0, "1", 0], ["0", 0, "1"]], "layout": {"type": "all", "expression": ""}, "answers": ["{geogebra_applet('https://www.geogebra.org/m/h9d8hzna')}", "{geogebra_applet('https://www.geogebra.org/m/kqnrbjzy')}", "{geogebra_applet('https://www.geogebra.org/m/xm44vcwe')}"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GB5 Trigonometric Identities 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": [], "metadata": {"description": "

Rewriting a trigonometric expression of the form $A\\cos(\\theta)\\pm B\\sin(\\theta)$ to either $R\\sin(\\theta+\\alpha)$ or $R\\cos(\\theta+\\alpha)$ by calculating $R$ and $\\alpha$.

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

{question}

find the values for $R$ and $\\alpha$, given $R>0$ and $0<\\alpha<\\frac{\\pi}{2}$.

", "advice": "

{answer}

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

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\\\\[ \\\\begin{split} R^2\\\\cos^2(\\\\alpha) + R^2 \\\\sin^2(\\\\alpha) &\\\\,= \\\\var{A}^2+\\\\var{B}^2 \\\\\\\\ R^2 (\\\\cos^2(\\\\alpha) +\\\\sin^2(\\\\alpha)) &\\\\,= \\\\var{A^2+B^2} \\\\\\\\ R &\\\\,= \\\\var{R}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "Rsol2": {"name": "Rsol2", "group": "Ungrouped variables", "definition": "\"

\"", "description": "", "templateType": "long string", "can_override": false}, "alpharound": {"name": "alpharound", "group": "Ungrouped variables", "definition": "precround(alpha,2)", "description": "", "templateType": "anything", "can_override": false}, "question": {"name": "question", "group": "Ungrouped variables", "definition": "if(Q=1,'{q1}','{q2}')", "description": "", "templateType": "anything", "can_override": false}, "Q": {"name": "Q", "group": "Ungrouped variables", "definition": "random(1,2)", "description": "", "templateType": "anything", "can_override": false}, "sign": {"name": "sign", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "templateType": "anything", "can_override": false}, "q1": {"name": "q1", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} = \\\\simplify[unitFactor]{R sin (theta+{sign}*alpha)},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "q2": {"name": "q2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} = \\\\simplify[unitFactor]{R cos (theta+{sign}*alpha)},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "a1": {"name": "a1", "group": "Ungrouped variables", "definition": "\"

To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify[unitFactor]{sin(a+{sign}*b)=sin(a)cos(b)+{sign}*sin(b)cos(a)}$:

\\n

\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} &\\\\,= \\\\simplify{R sin(theta+{sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(sin(theta)cos(alpha) + {sign}*sin(alpha)cos(theta))} \\\\\\\\ &\\\\,= \\\\simplify{Rsin(theta)cos(alpha) + {sign}*R sin(alpha)cos(theta)}. \\\\end{split} \\\\]

\\n

By comparing the coefficients of $\\\\sin(\\\\theta)$ and $\\\\cos(\\\\theta)$, we find that

\\n

\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

\\n

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

\\n

{Rsol}

\\n

Similarly, to find $\\\\alpha$ we can divide $R\\\\sin(\\\\alpha) = \\\\var{B}$ by $R\\\\cos(\\\\alpha) = \\\\var{A}$, and use the identity $\\\\tan(\\\\alpha) = \\\\frac{\\\\sin(\\\\alpha)}{\\\\cos(\\\\alpha)}$:

\\n

\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

\\n

Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "a2": {"name": "a2", "group": "Ungrouped variables", "definition": "\"

To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify{cos(a+{sign}*b)=cos(a)cos(b)-{sign}*sin(a)sin(b)}$:

\\n

\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} &\\\\,= \\\\simplify[unitFactor]{R cos (theta + {sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(cos(theta)cos(alpha) - {sign}*sin(theta)sin(alpha))} \\\\\\\\ &\\\\,= \\\\simplify{Rcos(theta)cos(alpha) - {sign}*R sin(theta)sin(alpha)}. \\\\end{split} \\\\]

\\n

By comparing the coefficients of $\\\\cos(\\\\theta)$ and $\\\\sin(\\\\theta)$, we find that

\\n

\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

\\n

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

\\n

{Rsol}

\\n

\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

\\n

Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "if(Q=1,'{a1}','{a2}')", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["Q", "A", "B", "sign", "R", "Rround", "alpha", "alpharound", "Rsol", "Rsol1", "Rsol2", "question", "q1", "q2", "answer", "a1", "a2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$R=$[[0]]

$\\alpha=$[[1]]

(Give your answers to 2 decimal places where necessary.)

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{Rround}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{alpharound}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CA3 - Graphs of trig functions (sin, cos, tan)", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

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

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

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

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

Match the graph to its function.

", "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["$\\sin(x)$", "$\\cos(x)$", "$\\tan(x)$"], "matrix": [["1", 0, 0], [0, "1", 0], [0, 0, "1"]], "layout": {"type": "all", "expression": ""}, "answers": ["{geogebra_applet('https://www.geogebra.org/m/ntqvuwqr')}", "{geogebra_applet('https://www.geogebra.org/m/fsqmnhsc')}", "{geogebra_applet('https://www.geogebra.org/m/yg6f9eqz')}"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CA4 Function notation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"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 the inverse of a function of the form $f(x)=\\frac{mx+c}{x+a},\\,x\\neq-a$.

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

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

", "advice": "

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

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

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

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

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

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

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

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

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)

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

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

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

", "advice": "

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

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

{advice}

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

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

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

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

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

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

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

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

\"", "description": "", "templateType": "long string", "can_override": false}, "advice2": {"name": "advice2", "group": "Ungrouped variables", "definition": "if(a_2<0 and a_3>0,'{solutionb}',{advice3})", "description": "", "templateType": "anything", "can_override": false}, "advice3": {"name": "advice3", "group": "Ungrouped variables", "definition": "if(a_2>0 and a_3<0,'{solutionc}','{solutiond}')", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(2..10)", "description": "", "templateType": "anything", "can_override": false}, "b_3": {"name": "b_3", "group": "Ungrouped variables", "definition": "b/c", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "gcd(b,c)=1", "maxRuns": "100"}, "ungrouped_variables": ["a_1", "a_2", "a_3", "b_1", "b_2", "b_3", "b", "c", "advice", "advice2", "advice3", "solutiona", "solutionb", "solutionc", "solutiond"], "variable_groups": [{"name": "Unnamed group", "variables": []}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

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

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

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

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

", "advice": "

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

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

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

Therefore,

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

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

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

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

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

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

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

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

Recall:

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

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

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

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

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

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

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

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

", "advice": "

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

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

This can be split up into steps:

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

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

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

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

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

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

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

Plugging these into the chain rule:

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

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

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

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

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

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

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

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

", "advice": "

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

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

This can be split up into steps:

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

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

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

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

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

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

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

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

Simplifying,

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

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

Differentiating $u$ with respect to $x$:

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

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

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

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

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

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

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

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

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

[[0]]

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

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

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

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

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

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

", "advice": "

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

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

Using this method can be broken down into steps:

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

For the integral

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

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

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

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

Plugging these 4 terms into the integration by parts formula:

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

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

So, for

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

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

Hence,

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

Plugging this back into the original calculation:

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

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

[[0]]

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

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This question provides a list of data to the student. They are asked to find the \"range\".

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

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\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

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

Range is the difference between the highest and the lowest value in the data.

To find this, we subtract the lowest value from the highest value:

\\[ \\var{max(a)} - \\var{min(a)} = \\var{range} \\text{.}\\]

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

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Option 2 for the list. Only used if there is only one mode and option 1 was not used.

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Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

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Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

", "templateType": "anything", "can_override": false}, "modetimes": {"name": "modetimes", "group": "final list", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "description": "

The vector of number of times of each value in the data.

", "templateType": "anything", "can_override": false}, "range": {"name": "range", "group": "final list", "definition": "max(a) - min(a)", "description": "", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "final list", "definition": "if(len(modea1) = 1, a1, if(len(modea2) = 1, a2, a3))", "description": "

The final list.

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Find the range.

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This question asks the student to choose the appropriate measure of average and spread for a data with outliers.

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

Which of the following measures would you choose if you were dealing with data which includes outliers? Select one measure of average and one measure of spread.

", "advice": "

The median is a more appropriate measure of average when your data contains outliers because outliers do not affect the median.

The interquartile range is the best measure of variability for skewed distributions or data sets with outliers. Because it’s based on values that come from the middle half of the distribution, it’s unlikely to be influenced by outliers.

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

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "m_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": "checkbox", "displayColumns": 0, "minAnswers": 0, "maxAnswers": 0, "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["Mean", "Median", "Standard deviation", "P-value", "Range", "Inter-quartile range"], "matrix": [0, "1", 0, 0, 0, "1"], "distractors": ["", "", "", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SA11 Identify measures of spread/location", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Gareth Woods", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/978/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": [], "metadata": {"description": "

Identifying measures of spread or location (average)

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

Match each of the following with what they measure.

", "advice": "

The mean is a measure of location or central tendancy. It is calcuated by summing all of the data values and dividing by the number of values.

The median is a measure of location or central tendancy. It is the middle value of an ordered data set.

The inter-quartile range is a measure of spread. The interquartile range is the difference between upper and lower quartiles.The lower quartile, or first quartile (Q1), is the value under which 25% of data points are found when they are arranged in increasing order. The upper quartile, or third quartile (Q3), is the value under which 75% of data points are found when arranged in increasing order. The inter-quartile range therefore gives us an idea of the middle 50% of the ordered data set.

The standard deviation is a measure of spread. It measures the dispersion of a data set relative to its mean.

The variance is a measure spread because it is the square of the standard deviation.

A p-value the probability that a particular statistical measure, such as the mean or standard deviation, of an assumed probability distribution will be greater than or equal to (or less than or equal to in some instances) observed results. A p-value is used to determine statistical significance, not measures of spread or location.

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

", "rulesets": {"std": ["all", "fractionNumbers"]}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "function dragpoint_board() {\n var scope = question.scope;\n\n JXG.Options.text.display = 'internal';\n \n var yo0 = scope.variables.yo0.value;\n var yo1 = scope.variables.yo1.value;\n var yo2 = scope.variables.yo2.value;\n var yo3 = scope.variables.yo3.value;\n var yo4 = scope.variables.yo4.value;\n var yo5 = scope.variables.yo5.value;\n var yo6 = scope.variables.yo6.value;\n var yo7 = scope.variables.yo7.value; \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).find('#dragpoint').append(div);\n \n var board = div.board;\n \nboard.suspendUpdate(); 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\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": "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, "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["Variance", "Mean", "Median", "Inter-quartile range", "P-value", "Standard deviation"], "matrix": [["1", 0, 0], [0, "1", 0], [0, "1", 0], ["1", 0, 0], [0, 0, "1"], ["1", 0, 0]], "layout": {"type": "all", "expression": ""}, "answers": ["Measure of Spread", "Measure of location (average)", "Neither measure of location nor measure of spread"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SA13 Correlation", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Richard Miles", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/882/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Upuli Wickramaarachchi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23527/"}], "tags": [], "metadata": {"description": "

Tests understanding of scatter plots and related concepts.

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

The scatter plot below shows the relationship between an employee’s height in centimetres and how long it takes them to walk to work in minutes.

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

time (mins)	{drawgraph()}
	height (cm)

", "advice": "

The graph shows that there is a positive correlation between a person's height and how long it takes them to walk to work.

A postive correlation is a relationship between two variables where both variables move in the same diection.

This tells us that as a person's height increases, the time it takes to walk to work increases.

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

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p6y

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Mark the statement that best describes what this scatter plot shows.

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["

In general, there is a positive correlation between a person's height and how long it takes them to walk to work.

", "

In general, there is a negative correlation between a person's height and how long it takes them to walk to work.

", "

In general, there is a no correlation between a person's height and how long it takes them to walk to work.

"], "matrix": ["1", 0, 0], "distractors": ["", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SA14 Probability - \"sample space\"", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}], "tags": [], "metadata": {"description": "

Calculate probability of selecting coloured counters from a bag.

", "licence": "None specified"}, "statement": "

A bag contains:

$\\var{srn}$ small, red tokens,
$\\var{sbn}$ small, blue tokens,
$\\var{brn}$ large, red tokens, and
$\\var{bbn}$ large, blue tokens.

", "advice": "

part a)

A probability is a fraction. You can give your answer as a fraction, decimal or percentage as these are all equivalent.

The formula for probability is:

\\[ P(A) = \\frac{\\text{number of possibilities for A}}{\\text{number of total possible outcomes}} \\]

For this question the total possible outcomes are $\\var{srn}+\\var{sbn}+\\var{brn}+\\var{bbn} = \\var{total}$.

Therefore

\\[ P(\\text{A large red token}) = \\frac{\\var{brn}}{\\var{total}} = \\var[fractionnumbers]{brn/total}\\]

part b)

For this question we need to know the total number of small tokens, i.e. $\\var{srn}+\\var{sbn} = \\var{srn+sbn}$.

Therefore

\\[ P(\\text{A small token}) = \\frac{\\var{srn+sbn}}{\\var{total}} = \\var[fractionnumbers]{(srn+sbn)/total}\\]

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

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"srn": {"name": "srn", "group": "Ungrouped variables", "definition": "random(1..20)", "description": "", "templateType": "anything", "can_override": false}, "brn": {"name": "brn", "group": "Ungrouped variables", "definition": "random(1..20)", "description": "", "templateType": "anything", "can_override": false}, "sbn": {"name": "sbn", "group": "Ungrouped variables", "definition": "random(1..20)", "description": "", "templateType": "anything", "can_override": false}, "bbn": {"name": "bbn", "group": "Ungrouped variables", "definition": "random(1..20)", "description": "", "templateType": "anything", "can_override": false}, "total": {"name": "total", "group": "Ungrouped variables", "definition": "brn+bbn+srn+sbn", "description": "", "templateType": "anything", "can_override": false}, "ans1": {"name": "ans1", "group": "Ungrouped variables", "definition": "precround(brn/total,2)", "description": "", "templateType": "anything", "can_override": false}, "ans2": {"name": "ans2", "group": "Ungrouped variables", "definition": "precround((srn+sbn)/total,2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["srn", "brn", "sbn", "bbn", "total", "ans1", "ans2"], "variable_groups": [{"name": "Unnamed group", "variables": []}], "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": "

You take a token at random.

What is the probability that it is a large, red token?

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

", "alternatives": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "minValue": "ans1", "maxValue": "ans1", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "minValue": "{brn}/{total}", "maxValue": "{brn}/{total}", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

You take a token at random.

What is the probability that it is a small token?

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

", "alternatives": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "minValue": "ans2", "maxValue": "ans2", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "minValue": "{{srn}+{sbn}}/{total}", "maxValue": "{{srn}+{sbn}}/{total}", "correctAnswerFraction": true, "allowFractions": true, "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": "SA15 Intuitive Probability", "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": "Upuli Wickramaarachchi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23527/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": ["taxonomy"], "metadata": {"description": "

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

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

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.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"no_flips": {"name": "no_flips", "group": "Ungrouped variables", "definition": "random(6..9)", "description": "

Number of flips of the coin

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["no_flips"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

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?

", "minValue": "1/2", "maxValue": "1/2", "correctAnswerFraction": true, "allowFractions": true, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}], "allowPrinting": true, "navigation": {"allowregen": true, "reverse": true, "browse": true, "allowsteps": true, "showfrontpage": true, "showresultspage": "oncompletion", "navigatemode": "sequence", "onleave": {"action": "none", "message": ""}, "preventleave": true, "typeendtoleave": true, "startpassword": "", "allowAttemptDownload": true, "downloadEncryptionKey": ""}, "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "feedback": {"showactualmark": true, "showtotalmark": true, "showanswerstate": true, "allowrevealanswer": true, "advicethreshold": 0, "intro": "

This is a tool for you! It is here to help you diagnose whether there are any maths or statistics pre-requisites for your course that you may want to brush up on. If at any point you are struggling with any question you should find a link at the end of the \"reveal answer\" section that will take you to some recommended online resources on that subject area. You can also always contact the Maths and Stats Help team (MaSH) to arrange a one to one appointment or check out our workshop timetable to see if you can access the support you need that way. Find all this information via our website here!

", "end_message": "

Thanks for completing the Skills Audit. You can attempt this as many times as you need. Remember the score is not what matters - this is in no way assessed work - this is simply a tool for working out whether you may need to brush up on anything to ensure that you can access all the material on your course and get off to the best possible start.

Don't forget to look up what support is available to you through our web pages here!

", "reviewshowscore": true, "reviewshowfeedback": true, "reviewshowexpectedanswer": true, "reviewshowadvice": true, "results_options": {"printquestions": true, "printadvice": true}, "feedbackmessages": []}, "diagnostic": {"knowledge_graph": {"topics": [], "learning_objectives": []}, "script": "diagnosys", "customScript": ""}, "contributors": [{"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "extensions": ["geogebra", "jsxgraph", "stats"], "custom_part_types": [{"source": {"pk": 2, "author": {"name": "Christian Lawson-Perfect", "pk": 7}, "edit_page": "/part_type/2/edit"}, "name": "List of numbers", "short_name": "list-of-numbers", "description": "

The answer is a comma-separated list of numbers.

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

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

Is every number in the student's list valid?

Are the student's answers in ascending order?

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

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

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

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

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