// Numbas version: finer_feedback_settings {"name": "Skills Audit for Maths and Stats - Science and Engineering Foundation Year (MAS005) ", "metadata": {"description": "

Skills Audit for Maths and Stats for MAS005 Science and Engineering Foundation Year students who have already studied A-level maths.

", "licence": "Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], []], "questions": [{"name": "NA5 - Convert Units (m/s and km/h)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Unit conversion between two compound units.

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

A Cheetah runs at a speed of {speedkm} kilometres per hour (km/h). What is the Cheetah's speed in metres per second (m/s)?

Give your ansswer to 2 decimal places where appropriate.

", "advice": "

There are a number of ways to work out the conversion. Here are a couple of suggestions.

METHOD 1

Since there are $1000$m in $1$km we first multiply by $1000$ to get the speed in metres per hour:

\\begin{equation} 1000*\\var{speedkm} = \\var{step1}\\end{equation}

Then we divide by $3600$ since that is the number of seconds in an hour to get the speed in metres per second:

\\begin{equation} \\frac{\\var{step1}}{3600} = \\var{speedms} \\end{equation}

Finally we round off to 2 decimal places as required, $\\var{roundanswer}$m/s.

METHOD 2

We can actually do all of the above in one step of working by using a single conversion factor. Since there are $1000$m in a km and $3600$ seconds in an hour, we can calaculate the conversion factor:

\\begin{equation} \\frac{3600}{1000} = 3.6 \\end{equation}

and then simply divide by that conversion factor:

\\begin{equation} \\frac{\\var{speedkm}}{3.6} = \\var{speedms} \\end{equation}

finally rounding off as before, $\\var{roundanswer}$m/s.

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

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[[0]]m/s

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

", "advice": "

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

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

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

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Applying the order of operators.

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

To calculate the following expression you press a sequence of buttons on your calculator.

\\begin{align}\\frac{\\var{num}}{\\var{a}\\times\\var{b}}\\end{align}

Which of the following would give the WRONG answer?

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

This is the standardized order of operations that we carry out and is part of how the calculator is designed to work. The most effective way to use most modern calculators is to use either the fraction button (on scientific calculators) or as is hinted at in this question, use brackets.

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

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

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

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

", "advice": "

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

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

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

Upper bound:

[[0]]

Lower bound:

[[1]]

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Calculations with negative numbers.

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

Calculate $(\\var{x})\\div(\\var{y})$.

", "advice": "

When we divide two numbers the rule is,

plus $\\div$ plus = plus
minus $\\div$ plus = minus
plus $\\div$ minus = minus
minus $\\div$ minus = plus

In this calculation we have

\\[(\\var{x})\\div(\\var{y})=\\var{x/y}.\\]

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": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(-10..10)*y", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(-10..10 except 0)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "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, "minValue": "{x/y}", "maxValue": "{x/y}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "NF4 Reverse percentages", "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": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"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": ["decrease", "percentages", "taxonomy"], "metadata": {"description": "

Find the original price before a discount by dividing the new price by the percentage discount.

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

{name1} and {name2} are friends. {name1} noticed {name2}'s new {item} when he came over to visit her house. He immediately knew he wanted to buy the same model. When he got home, he bought the {item} online for £{newprice}.

", "advice": "

We need to find the original price paid by {name2}. This value represents 100%.

By the time {name1} bought the {item}, the price had decreased by {percentage}%.

{name1} therefore paid {100-percentage}% of the price {name2} paid.

We use the unitary method to find the original price. We know the price paid by {name1}.

\\[\\var{100-percentage}\\text{%} = \\var{newprice} \\text{.}\\]

Divide both sides by {100-percentage} to get

\\[\\begin{align} 1\\text{%} &= \\var{newprice} \\div \\var{100-percentage} \\\\&= \\var{newprice/(100-percentage)} \\text{.} \\end{align}\\]

Multiply both sides by 100 to get

\\[\\begin{align} 100\\text{%} &= \\var{newprice/(100-percentage)} \\times 100 \\\\&= \\var{newprice/(100-percentage)*100} \\\\&= \\var{oldprice}\\text{.} \\end{align}\\]

This is the original price paid by {name2} before the {percentage}% decrease.

We can check our answer with a different method.

\\[\\begin{align} \\var{100-percentage}\\text{% of } \\var{oldprice} &= \\var{(100-percentage)/100} \\times \\var{oldprice} \\\\&= \\var{(100-percentage)/100*oldprice} \\\\&= \\var{precround((100-percentage)/100*oldprice, 2)} \\text{.} \\end{align}\\]

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

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Discount percentage.

", "templateType": "anything", "can_override": false}, "item": {"name": "item", "group": "Ungrouped variables", "definition": "random(\"TV\", \"laptop\", \"smartphone\", \"PC\", \"gaming console\")", "description": "

The bought item.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "precround(precround(oldprice*(100-percentage)/100,2)*100/(100-percentage),2) = oldprice", "maxRuns": "1000"}, "ungrouped_variables": ["item", "name1", "percentage", "name2", "oldprice", "newprice"], "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": "

When {name1} told {name2} how much he had paid for the {item}, {name2} said the price had decreased by {percentage}% since she bought it.

How much did {name2} pay for the {item}?

£ [[0]]

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Given the number of international students enrolled on a course of $n$ students, calculate the percentage of 'home' students.

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

{num_students} of the {class_size} students enrolled on a course are international students. What percentage are 'home' students?

", "advice": "

First work out the number of students who are not international. In this case it is {class_size} - {num_students} = {class_size-num_students} students.

Then write this as a fraction out of {class_size}. $ \\frac{\\var{class_size-num_students}} {\\var{class_size}} $

Then convert this to a percentage. You should put this fraction into your calculator and then multiply by 100:

$ \\frac{\\var{class_size-num_students}} {\\var{class_size}} \\times 100 = \\var{(class_size-num_students)/class_size*100}\\%$

Use this link to find resources to help you revise how to calculate percentages.

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The number of students in the class who do speak a language other than English.

", "templateType": "anything", "can_override": false}, "class_size": {"name": "class_size", "group": "Ungrouped variables", "definition": "random(80..300)", "description": "", "templateType": "anything", "can_override": false}, "per": {"name": "per", "group": "Ungrouped variables", "definition": "random(5..90 except 50)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "num_students = precround(num_students,0) AND (num_students<>class_size/2 AND class_size<>100)", "maxRuns": 100}, "ungrouped_variables": ["num_students", "class_size", "per"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["$\\var{class_size-num_students}\\%$", "$\\var{num_students*100/class_size}\\%$", "$\\var{num_students}\\%$", "$\\var{(class_size-num_students)/class_size*100}\\%$"], "matrix": [0, 0, 0, "1"], "distractors": ["Have you converted this to a percentage? Click on Reveal Answer and scroll down for Advice regarding this question.", "How many students do NOT speak a language other than English at home? Click on Reveal Answer and scroll down for Advice regarding this question.", "How many students do NOT speak a language other than English at home? Then convert this to a percentage. Click on Reveal Answer and scroll down for Advice regarding this question.", "Well done!"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NG8 - Ordering fractions", "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": "

Put fractions in size order.

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

Which of these two fractions is the largest?

", "advice": "

To find which is bigger of $\\frac{\\var{top1}}{\\var{bot1}}$ and $\\frac{\\var{top2}}{\\var{bot2}}$ we need them to have the same denominator. A way to do this is to multiply the top and bottom of $\\frac{\\var{top1}}{\\var{bot1}}$ by $\\var{bot2}$ and multiply the top and bottom of $\\frac{\\var{top2}}{\\var{bot2}}$ by {bot1}. This doesn't change the the value of the fractions as this is just like multiplying by one.

\\[\\frac{\\var{top1}}{\\var{bot1}}=\\frac{\\var{top1*bot2}}{\\var{bot1*bot2}},\\quad \\frac{\\var{top2}}{\\var{bot2}}=\\frac{\\var{top2*bot1}}{\\var{bot1*bot2}}\\]

Now we can easily see which is bigger by comparing the numerator.

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

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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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Calculate an answer involving a fractional index.

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

Evaluate the following expression:

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

", "advice": "

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

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

we have,

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

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

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

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

Evaluate the following expression:

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

", "advice": "

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

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

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

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

Then calculating what is left:

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

Therefore,

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

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

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perform a calculation involving negative indices.

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

Evaluate and simplify the following expression:

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

", "advice": "

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

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

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

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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.

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

", "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", "type": "question"}, {"name": "AB4 multiply terms", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Filling in the blanks from the answer to a simplified expression involving indices.

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

Find the missing factor in the following statement:

\\[ \\var{3*a}x^\\var{b}y^\\var{c} = \\var{a}x^\\var{d}(?)\\]

", "advice": "

We can divide the left handside of the expression by the factor given on the right hand side of the expression to work out the missing factor:

\\[\\begin{split}
\\var{3*a}x^\\var{b}y^{\\var{c}}&=\\var{a}x^\\var{d}(?)\\\\
\\Rightarrow \\frac{\\var{3*a}x^\\var{b}y^{\\var{c}}}{\\var{a}x^\\var{d}}&=(?)\\\\
\\Rightarrow 3x^{\\var{b}-\\var{d}}y^\\var{c} &=(?),
\\end{split}\\]

which after simplifying gives the answer:

\\[(?) = 3x^{\\var{b-d}}y^{\\var{c}}\\]

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

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Expand two brackets involving powers of $x$.

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

Expand the brackets and simplify

", "advice": "

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

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

We can then simplify to give us the final answer of

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

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

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"c": {"name": "c", "group": "Ungrouped variables", "definition": "repeat(2*random(6..25),2)", "description": "", "templateType": "anything", "can_override": false}, "xp": {"name": "xp", "group": "Ungrouped variables", "definition": "repeat(random(1..8),2)", "description": "", "templateType": "anything", "can_override": false}, "yp": {"name": "yp", "group": "Ungrouped variables", "definition": "repeat(random(1..8 except [xp[0],xp[1]]),2)", "description": "", "templateType": "anything", "can_override": false}, "cans": {"name": "cans", "group": "Ungrouped variables", "definition": "GCD(c[0],c[1])", "description": "", "templateType": "anything", "can_override": false}, "xpans": {"name": "xpans", "group": "Ungrouped variables", "definition": "min(xp[0],xp[1])", "description": "", "templateType": "anything", "can_override": false}, "ypans": {"name": "ypans", "group": "Ungrouped variables", "definition": "min(yp[0],yp[1])", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "xp[0]<>xp[1] AND c[0]<>c[1]", "maxRuns": 100}, "ungrouped_variables": ["c", "xp", "yp", "cans", "xpans", "ypans"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{cans}*x^{xpans}*y^{ypans}", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": true, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}, {"name": "y", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "AC3 Algebraic substitution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Substitute values into an algebraic expression and calculate the result.

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

Evaluate the following expression,

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

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

", "advice": "

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

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

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

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

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

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

Solve the following simultaneous equations:

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

Give your answers to 2 decimal places where necessary.

", "advice": "

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

For the equations

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

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

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

Substituting this into equation (2):

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

Collecting similar terms:

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

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

{check}

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

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

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

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

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

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

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Rearrange 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]]

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

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

To solve an equation like

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

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

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

From here, multiply out the brackets,

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

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

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

Finally divide by the coefficient of $y$,

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

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

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

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

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

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

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

Factorise the following quadratic equation.

", "advice": "

Quadratic equations of the form

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

can be factorised to create an equation of the form

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

When we expand a factorised quadratic expression we obtain

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

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

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

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

So the factorised form of the equation is

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

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

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

[[0]] $=0$

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

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

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

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

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

", "advice": "

Completing the square works by noticing that

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

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

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

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

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

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

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

"}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "AD5 Difference of two squares", "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": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

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

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

Factorise the following quadratic expression:

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

", "advice": "

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

Therefore,

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

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

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

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

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

", "advice": "

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

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

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

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

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

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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", "type": "question"}, {"name": "AE5 - Multiplication of algebraic fractions 2", "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": "

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

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

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

", "advice": "

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

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

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

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

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

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

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

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

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

Rewrite the following expression as partial fractions:

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

", "advice": "

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

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

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

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

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

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

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

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

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

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

Hence,

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

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

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

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

and

{check}

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

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

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

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

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

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[[0]]

Calculate the volume of different 3D shapes, given the units and measurements required. The formulae for the volume of each shape are available as steps if required.

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

For a triangular prism, we first need to find the area of one of the faces then multiply this area by the depth of the prism.
In this example the easiest way to calculate the volume is to take the area of the triangular face first with $\\mathrm{base} = \\var{w6}m$ and $\\mathrm{height} = \\var{h6}m\\thinspace$.

\\begin{align}
\\mathrm{Area\\thinspace_\\triangle} &= \\frac{\\mathrm{base} \\times \\mathrm{height}}{2} \\\\
&= \\frac{\\var{w6} \\times \\var{h6}}{2} \\\\
&= \\var{0.5*w6*h6}\\, \\mathrm{m}^2\\,.
\\end{align}

Now that we have the area of the triangular face ($\\mathrm{Area\\thinspace_\\triangle}$) we can multiply this by the $\\mathrm{depth} = \\var{d6}m\\thinspace$.

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\triangle} \\times \\mathrm{depth} \\\\
&= \\var{0.5*w6*h6} \\times \\var{d6} \\\\
&= \\var{0.5*w6*h6*d6}\\, \\mathrm{m}^2\\,.
\\end{align}

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

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Side of square in cuboid.

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Creates base of triangle.

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One side of square base.

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Height of pyramid.

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Depth of cylinder.

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Depth of triangular prism.

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Radius of the cylinder.

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Side of square in cuboid.

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Width of cuboid.

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One side of square base.

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Height of traingle.

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Calculate the $\\mathrm{Volume}$ of the following triangular prism.

$\\mathrm{Volume} =$[[0]]$\\mathrm{m}^3$.

", "stepsPenalty": "1", "steps": [{"type": "information", "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": "

Volume of a triangular prism:

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\triangle} \\times \\mathrm{depth} \\\\
&= \\frac{\\mathrm{base} \\times \\mathrm{height}}{2} \\times \\mathrm{depth}
\\end{align}

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Find the volume of a prism with a trapezium as a cross section from a diagram.

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

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

{geogebra_applet('https://www.geogebra.org/m/qvcktek2',[basew: basew, topw: topw, h: h, l: l])}

", "advice": "

In order to work out the volume of a prism you need to work out the cross sectional area first. In this question the cross section is a trapezium. Find the area of a trapezium,

\\begin{align} \\frac{\\var{basew}+\\var{topw}}{2}\\times \\var{h} = \\var{traparea} cm^2 \\end{align}

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

\\begin{align} \\var{traparea} \\times \\var{l} = \\var{answer}cm^3\\end{align}.

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

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[[0]]$cm^3$

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

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

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

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

", "advice": "

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

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

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

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"hyp": {"name": "hyp", "group": "Ungrouped variables", "definition": "random(10 .. 15#1)", "description": "", "templateType": "randrange", "can_override": false}, "base": {"name": "base", "group": "Ungrouped variables", "definition": "random(3 .. 8#1)", "description": "", "templateType": "randrange", "can_override": false}, "ans": {"name": "ans", "group": "Ungrouped variables", "definition": "sqrt(hyp^2-base^2)", "description": "", "templateType": "anything", "can_override": false}, "ansr": {"name": "ansr", "group": "Ungrouped variables", "definition": "precround(ans,1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["hyp", "base", "ans", "ansr"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": true, "customName": "Answer", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

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

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "ans", "maxValue": "ans", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "1", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "GB1 Trigonometry - missing side", "extensions": ["eukleides"], "custom_part_types": [], "resources": [["question-resources/Picture1_caMIdF1.png", "/srv/numbas/media/question-resources/Picture1_caMIdF1.png"], ["question-resources/Picture2_6KE4ZpW.png", "/srv/numbas/media/question-resources/Picture2_6KE4ZpW.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "David Wishart", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1461/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

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

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

{max_height(25,diagram)}

", "advice": "

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

{advice}

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

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"ab": {"name": "ab", "group": "Unnamed group", "definition": "random(7..12)", "description": "", "templateType": "anything", "can_override": false}, "ac": {"name": "ac", "group": "Unnamed group", "definition": "precround(ab*cos(pi*angle/180),2)", "description": "", "templateType": "anything", "can_override": false}, "bc": {"name": "bc", "group": "Unnamed group", "definition": "precround(ab*sin(pi*angle/180),2)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_1": {"name": "d_t_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle+'\u00b0')\n , b..c label(bc + 'cm')\n , a..c label('x cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", 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c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_1": {"name": "d_t_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..c label(ac + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_2": {"name": "d_t_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle+'\u00b0')\n , b..c label(bc + 'cm')\n , a..c label('x cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_2": {"name": "d_c_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x\u00b0')\n , a..c label(ac + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "SCT": {"name": "SCT", "group": "Unnamed group", "definition": "random('s','c','t')", "description": "", "templateType": "anything", "can_override": false}, "AngORside": {"name": "AngORside", "group": "Unnamed group", "definition": "'side'", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_2": {"name": "d_c_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0),\n\n 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"description": "", "templateType": "anything", "can_override": false}, "sin_a": {"name": "sin_a", "group": "advice", "definition": "\"

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

\\n

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

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

\\n

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

\\n

Now we subsitute the values we have in this particular question

\\n

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

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

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

\\n

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

\\n

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

\\n

Round as required:

\\n

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

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

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

\\n

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

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

\\n

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

\\n

Now we subsitute the values we have in this particular question

\\n

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

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

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

\\n

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

\\n

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

\\n

Round as required:

\\n

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

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

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

\\n

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

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

\\n

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

\\n

Now we subsitute the values we have in this particular question

\\n

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

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

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

\\n

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

\\n

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

\\n

Round as required:

\\n

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

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

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

\\n

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

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

\\n

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

\\n

Now we subsitute the values we have in this particular question

\\n

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

and rearrange to give:

\\n

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

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

\\n

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

\\n

Round as required:

\\n

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

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

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

\\n

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

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

\\n

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

\\n

Now we subsitute the values we have in this particular question

\\n

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

and rearrange to give:

\\n

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

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

\\n

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

\\n

Round as required:

\\n

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

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

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

\\n

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

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

\\n

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

\\n

Now we subsitute the values we have in this particular question

\\n

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

and rearrange to give:

\\n

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

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

\\n

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

\\n

Round as required:

\\n

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

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

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

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

Give your answer correct to 2 decimal place.

", "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "2", "precisionPartialCredit": "100", "precisionMessage": "", "strictPrecision": true, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "GB4 Trigonometric Identities 1", "extensions": [], "custom_part_types": [], "resources": [["question-resources/image_jPzIKnS.png", "/srv/numbas/media/question-resources/image_jPzIKnS.png"], ["question-resources/image_kkFBamd.png", "/srv/numbas/media/question-resources/image_kkFBamd.png"], ["question-resources/image_K3sk7RI.png", "/srv/numbas/media/question-resources/image_K3sk7RI.png"], ["question-resources/image_SC1KBmT.png", "/srv/numbas/media/question-resources/image_SC1KBmT.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": [], "metadata": {"description": "

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

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

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

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

", "advice": "

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

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

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

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

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

{Rsol}

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

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

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

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

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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}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["A", "B", "R", "Rround", "alpha", "alpharound", "Rsol", "Rsol1", "Rsol2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$R=$[[0]]

$\\alpha=$[[1]]

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

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

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

{diagram}

Find x.

", "advice": "

{Advice}

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

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"Ruleuse": {"name": "Ruleuse", "group": "Question structure", "definition": "random('s','c','s','c')", "description": "", "templateType": "anything", "can_override": false}, "ANGorSIDE": {"name": "ANGorSIDE", "group": "Question structure", "definition": "random('ang','side')", "description": "", "templateType": "anything", "can_override": false}, "cosSIDEadvice": {"name": "cosSIDEadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that there are the lengths of two sides given and the angle specifically between those two sides. Further to this, the instruction is to find the other missing side. These are the conditions for when to use the $\\\\textit{cosine rule}$.

\\n

The formula for a missing side using the cosine rule is:

\\n

\\\\[ a^2 = b^2 + c^2 - 2bc \\\\cos(A)\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the $a$ (side) and $A$ (angle) labels are applied to the angle given and it\\'s opposite side.

\\n

In this case:

\\n

\\\\[ a=x, \\\\quad b=\\\\var{a}, \\\\quad c=\\\\var{b}, \\\\text{and} \\\\quad A=\\\\var{Cang},\\\\]

\\n

where the choice of which way round $b$ and $c$ are assigned doesn\\'t matter.

\\n

So, we now have:

\\n

\\\\[x^2 = \\\\var{a}^2 +\\\\var{b}^2-2\\\\times\\\\var{a}\\\\times\\\\var{b}\\\\times\\\\cos{(\\\\var{Cang})},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\sqrt{\\\\var{a^2 +b^2-2*a*b*(cos(Cang))}}\\\\]

\\n

\\\\[x=\\\\var{c}\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "

case 1: missing side in the cosine rule.

", "templateType": "long string", "can_override": false}, "cosANGadvice": {"name": "cosANGadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that there are the lengths of all three sides given. Further to this, the instruction is to find the a missing angle. These are the conditions for when to use the $\\\\textit{cosine rule}$ but in its rearranged form to find an angle. You need to identify which side is \\\"$a$\\\" as being the one opposite the angle you are asked to find.

\\n

The formula for a missing angle using the cosine rule is:

\\n

\\\\[ A = \\\\arccos\\\\left(\\\\frac{b^2+c^2-a^2}{2bc}\\\\right)\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the $a$ (side) and $A$ (angle) labels are applied to the side opposite the angle that is asked for and the angle that is asked for.

\\n

In this case:

\\n

\\\\[ a=\\\\var{c_round}, \\\\quad b=\\\\var{a}, \\\\quad c=\\\\var{b}, \\\\text{and} \\\\quad A= x,\\\\]

\\n

where the choice of which way round $b$ and $c$ are assigned doesn\\'t matter.

\\n

So, we now have:

\\n

\\\\[x = \\\\arccos\\\\left(\\\\frac{\\\\var{a}^2+\\\\var{b}^2-\\\\var{c_round}^2}{2\\\\times\\\\var{a}\\\\times\\\\var{b}}\\\\right),\\\\]

\\n

hence,

\\n

\\\\[x=\\\\var{(180/pi)*arccos((a^2 +b^2-c_round^2)/(2*a*b))}\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sinSIDEadvice": {"name": "sinSIDEadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that a single length is provided, along with two angles, crucially including the angle opposite the given side. Further to this, the instruction is to find the a missing angle. These are the conditions for when to use the $\\\\textit{sine rule}$. The sine rule uses the sides and angles in pairs and uses two pairs for any given calculation

\\n

The formula for finding a side using the sine rule can be written as:

\\n

\\\\[ \\\\frac{a}{\\\\sin(A)}=\\\\frac{b}{\\\\sin(B)}\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the side being asked for is in the above notation $a$.

\\n

In this case:

\\n

\\\\[ a=x, \\\\quad b=\\\\var{a}, \\\\quad A=\\\\var{Cang}, \\\\text{and} \\\\quad B= \\\\var{Aang_round}.\\\\]

\\n

So, we now have:

\\n

\\\\[\\\\frac{x}{\\\\sin{(\\\\var{Cang})}}=\\\\frac{\\\\var{a}}{\\\\sin{(\\\\var{Aang_round})}},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\frac{\\\\var{a}}{\\\\sin{(\\\\var{Aang_round})}}\\\\times\\\\sin{(\\\\var{Cang})},\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "

case 3

", "templateType": "long string", "can_override": false}, "sinANGadvice": {"name": "sinANGadvice", "group": "Question structure", "definition": "safe(\"

First recognise that the diagram is a non-right angled triangle and that two lengths are provided, along with an angle, crucially including an angle opposite a given side. Further to this, the instruction is to find the a missing side. These are the conditions for when to use the $\\\\textit{sine rule}$. The sine rule uses the sides and angles in pairs and uses two pairs for any given calculation

\\n

The formula for finding an angle using the sine rule can be written as:

\\n

\\\\[ \\\\frac{\\\\sin(A)}{a}=\\\\frac{\\\\sin(B)}{b}\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the angle being asked for is in the above notation $A$.

\\n

In this case:

\\n

\\\\[ a=\\\\var{c_round}, \\\\quad b=\\\\var{a}, \\\\quad A= x, \\\\text{and} \\\\quad B= \\\\var{Aang_round}.\\\\]

\\n

So, we now have:

\\n

\\\\[\\\\frac{\\\\sin{(x)}}{\\\\var{c_round}}=\\\\frac{\\\\sin{(\\\\var{Aang_round})}}{\\\\var{a}},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\arcsin\\\\left(\\\\var{c_round}\\\\times\\\\frac{\\\\sin{(\\\\var{Aang_round})}}{\\\\var{a}}\\\\right),\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\")", "description": "

case 4

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side length a

", "templateType": "randrange", "can_override": false}, "b": {"name": "b", "group": "Quantities", "definition": "random(5 .. 10#0.1)", "description": "

side length b

", "templateType": "randrange", "can_override": false}, "Cang": {"name": "Cang", "group": "Quantities", "definition": "random(40..140 except 85..95)", "description": "

C angle in degrees

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angle A in degrees

", "templateType": "anything", "can_override": false}, "Bang": {"name": "Bang", "group": "Quantities", "definition": "180-(Aang+Cang)", "description": "", "templateType": "anything", "can_override": false}, "cosSIDEans": {"name": "cosSIDEans", "group": "Quantities", "definition": "c", "description": "", "templateType": "anything", "can_override": false}, "cosANGans": {"name": "cosANGans", "group": "Quantities", "definition": "arccos((a^2+b^2-c_round^2)/(2*a*b))*180/pi", "description": "

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

", "templateType": "anything", "can_override": false}, "c_round": {"name": "c_round", "group": "Quantities", "definition": "precround(c,1)", "description": "", "templateType": "anything", "can_override": false}, "Aang_round": {"name": "Aang_round", "group": "Quantities", "definition": "precround(Aang,1)", "description": "", "templateType": "anything", "can_override": false}, "Bang_round": {"name": "Bang_round", "group": "Quantities", "definition": "precround(Bang,1)", "description": "", "templateType": "anything", "can_override": false}, "Cang_roundcos": {"name": "Cang_roundcos", "group": "Quantities", "definition": "Precround((180/pi)*arccos((a^2+b^2-c_round^2)/(2*a*b)),1)", "description": "", "templateType": "anything", "can_override": false}, "sinANGans": {"name": "sinANGans", "group": "Quantities", "definition": "If(Cang<90,arcsin(c_round*(sin(Aang_round*pi/180)/a))*180/pi,180 - arcsin(c_round*(sin(Aang_round*pi/180)/a))*180/pi)", "description": "", "templateType": "anything", "can_override": false}, "sinSIDEans": {"name": "sinSIDEans", "group": "Quantities", "definition": "(a/sin(aang_round*pi/180))*sin(cang*pi/180)", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Quantities", "definition": "precround(If(Ruleuse='c',IF(ANGorSIDE='ang',cosANGans,cosSIDEans),IF(ANGorSIDE='ang',sinANGans,sinSIDEans)),1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "a+b>c and b+c>a and a+c>b", "maxRuns": "200"}, "ungrouped_variables": [], "variable_groups": [{"name": "Question structure", "variables": ["Ruleuse", "ANGorSIDE", "cosSIDEadvice", "cosANGadvice", "sinSIDEadvice", "sinANGadvice", "advice"]}, {"name": "Diagrams", "variables": ["cosSIDEdiagram", "cosANGdiagram", "sinSIDEdiagram", "sinANGdiagram", "diagram"]}, {"name": "Quantities", "variables": ["a", "b", "Cang", "c", "Aang", "Bang", "cosSIDEans", "cosANGans", "sinANGans", "sinSIDEans", "c_round", "Aang_round", "Bang_round", "Cang_roundcos", "ans"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": true, "customName": "Answer", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$x =$[[0]]

Calculating gradient and finding intercept from a geogebra graph.

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

{app}
Find the gradient of the line.

", "advice": "

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

{app_advice}

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

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

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

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

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

This is equivalent to using the formula:

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

and substitute the coordinates of the vertices of the triangle:

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

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

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

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

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

", "useAlternativeFeedback": false, "answer": "{m}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.1", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "{m}", "showPreview": true, "checkingType": "dp", "checkingAccuracy": "1", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "CA2 - Shapes of quadratics", "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": "

Multiple choice - select the quadratic graph.

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

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

", "advice": "

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

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["{geogebra_applet('https://www.geogebra.org/m/tpfzv3w7')}", "{geogebra_applet('https://www.geogebra.org/m/zftpwq64')}", "{geogebra_applet('https://www.geogebra.org/m/we3gngqa')}", "{geogebra_applet('https://www.geogebra.org/m/cadkup6r')}"], "matrix": ["1", 0, 0, 0], "distractors": ["", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"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.

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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]]

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

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

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

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

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

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

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

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

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

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

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

Find the value of $x$:

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

", "advice": "

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

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

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

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

$x=$ [[0]]

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

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

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

", "advice": "

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

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

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

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

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

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

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

", "advice": "

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

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

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

So, for the function

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

the integral is

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

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

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

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}, "a_4": {"name": "a_4", "group": "Ungrouped variables", "definition": "random(-10..10 except 0)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "b_1>b_2 and b_2>b_3", "maxRuns": "100"}, "ungrouped_variables": ["a_1", "a_2", "a_3", "b_1", "b_2", "b_3", "advice", "advice2", "advice3", "solutiona", "solutionb", "solutionc", "solutiond", "a_4"], "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": "

[[0]]

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

", "useAlternativeFeedback": false, "answer": "{a_1}x^{{b_1}+1}/{b_1+1}+{a_2}x^{{b_2}+1}/{b_2+1}+{a_3}x^{{b_3}+1}/{b_3+1}+{a_4}x", "answerSimplification": "all, fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{a_1}x^{{b_1}+1}/{b_1+1}+{a_2}x^{{b_2}+1}/{b_2+1}+{a_3}x^{{b_3}+1}/{b_3+1}+{a_4}x+c", "answerSimplification": "all, fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "CB4 Definite integration", "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": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": [], "metadata": {"description": "

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

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

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

", "advice": "

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

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

Therefore,

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

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

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

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

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

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

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

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

Recall:

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

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

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

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

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

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

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

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

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

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

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

", "advice": "

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

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

This can be split up into steps:

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

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

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

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

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

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

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

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

Simplifying,

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

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

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

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

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

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

", "advice": "

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

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

This can be split up into steps:

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

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

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

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

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

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

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

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

Simplifying,

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

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

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

Decide whether each of the described sets of data is drawn from a discrete or continuous distribution.

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

Decide whether the following data sets are discrete or continuous.

", "advice": "

Data can either be discrete or continuous.

Discrete data consists of integers, units of measurement cannot be split. Only certain values have a meaning.
Continuous data can be shown on a number line, all points on this line have a meaning. In between two measurements there's always a valid \"middle\" measurement. So for example, between 156.3cm and 156.4cm there's 156.35cm.

a)

Height is a continuous variable. For example, 180.3cm and 180.4cm have a valid midpoint 180.35cm.Weight is a continuous variable. For example, 54.5kg and 54.6kg have a valid midpoint 54.55kg.Time is a continuous variable. For example, 54.2s and 54.3s have a valid midpoint 54.25s.Temperature is a continuous variable, it can take any value between -273.15°C (absolute zero) and positive infinity. For example, 25°C and 26°C have a valid midpoint 25.5°C. Hence, this data is continuous.

b)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not.The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy a half of a bar. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5. The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

c)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not..The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy a half of a bar. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5.The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

d)

e)

When we round continuous variables to the nearest integer, this data becomes discrete, as there are no valid middle measurements between the integers. Therefore, the weight of a dog to the nearest kgthe height of Olympic medalists to the nearest cmthe time taken to run 10km to the nearest min is discrete and not continuous.

f)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not.The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy half of a bar of chocolate. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5.The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

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

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{cont[ranc]}

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Discrete

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Continuous

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{disc[rand]}

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Discrete

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"], "matrix": ["1", 0], "distractors": ["", ""]}, {"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

{disc[rand2]}

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Discrete

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Continuous

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Discrete

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Continuous

{disc[rand3]}

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Discrete

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This question is about identifying what types of charts or visual representations of data you can use for different data sets.

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This question is about recognising what types of charts or visual representations of data you can use with what types of data sets.

", "advice": "

There are many different types of visual representations of data and sometimes there will be a choice of what you use.

Start by looking at these resources to build up your understanding of data display options and methods.

The table shows different names of charts on the left hand side and different descriptions of data sets along the top.

Pair up each description with the chart that would be most suitable.

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Interpreting the elements of a box plot

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

The diagram below shows a box plot of some data.

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

", "advice": "

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

For more information on box plots follow this link.

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Which of these statements are true and which are false?

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

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

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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.

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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.

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

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

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

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

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

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

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

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Number of flips of the coin

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

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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!

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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!

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