// Numbas version: finer_feedback_settings {"name": "AER 2025/26", "metadata": {"description": "

Skills Audit for Students on Aerospace programmes 2025/26

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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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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.

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\\\\[ \\\\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})}.\\\\]

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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}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]]

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

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

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

", "advice": "

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

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

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

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

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

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

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

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

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

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

Find the value of $x$:

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

", "advice": "

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

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

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

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

$x=$ [[0]]

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

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

Solve for $x$:

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

", "advice": "

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

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

Hence,

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

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

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

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

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

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

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

Solve for $x$:

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

", "advice": "

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

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

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

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

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

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

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

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

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

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

", "advice": "

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

{advice1}

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

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

\\n

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

\\n

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

\\n

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

\\n

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

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

\\n

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

\\n

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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

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

\\n

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

\\n

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

\\n

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

\\n

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

\\n

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

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$x=$ [[0]]

$y=$ [[1]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}", "answerSimplification": "fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"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": "{(c-a*xsimp)/b}", "answerSimplification": "fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AS07 Quadratics - solve", "extensions": [], "custom_part_types": [{"source": {"pk": 2, "author": {"name": "Christian Lawson-Perfect", "pk": 7}, "edit_page": "/part_type/2/edit"}, "name": "List of numbers", "short_name": "list-of-numbers", "description": "

The answer is a comma-separated list of numbers.

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

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

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

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

Is every number in the student's list valid?

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

Are the student's answers in ascending order?

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

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

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

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

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

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

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

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

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

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

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

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

Solve the following quadratic equation by factorisation or otherwise:

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

", "advice": "

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

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

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

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

$x= $[[0]]

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

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

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

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

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

", "advice": "

Completing the square works by noticing that

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

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

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

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

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

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

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

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

"}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CD01 Differentiate Polynomials 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}], "tags": [], "metadata": {"description": "

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]]

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

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

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

", "advice": "

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

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

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

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

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

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

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

", "advice": "

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

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

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

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

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

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

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

", "advice": "

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

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

This can be split up into steps:

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

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

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

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

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

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

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

Plugging these into the chain rule:

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

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

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

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

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

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "({a*n}*x^{n-1}+{b*m}*x^{m-1})*cos({a}x^{n}+{b}x^{m})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CD10 Product Rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}], "tags": [], "metadata": {"description": "

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]]

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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]]

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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]]

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

", "useAlternativeFeedback": false, "answer": "{a_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"}, {"name": "CI02 Definite integration", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "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)

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

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

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

", "advice": "

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

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

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

So, for the function

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

the integral is

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

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

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

so combining these our final answer is

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

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

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

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

So, for the function

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

the integral is

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

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

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

so combining these our final answer is

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

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

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

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

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

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

", "advice": "

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

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

Using this method can be broken down into steps:

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

For the integral

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

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

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

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

Plugging these 4 terms into the integration by parts formula:

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

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

So, for

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

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

Hence,

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

Plugging this back into the original calculation:

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

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

[[0]]

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

", "useAlternativeFeedback": false, "answer": "{a/b}x^2 sin({b}x)+{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{a/b}x^2 sin({b}x)+{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)+c", "answerSimplification": "fractionNumbers, basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GM04 Volume of a semi-cylinder", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Find the volume of a semicylinder from a diagram.

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

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

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

", "advice": "

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

The area of a semi-circle is given by:

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

where $r$ is the radius of the circle.

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

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

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

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

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

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[[0]]$cm^3$

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Generates two sets of matrices that can be added and asks the student to add them. Will ensure that the dimensions in part a) are different to the dimensions in part b) to offer some variety.

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

Adding Matrices

Calculate the following:

", "advice": "

Matrix addition is fairly straightforward in that each entry is calculated by adding the corresponding entries. For example the top left entry of the answer to part a) can be calculated as $\\var{A[0][0]} + \\var{Apair[0][0]} = \\var{A[0][0] + Apair[0][0]}$. So for this question:

a) $\\var{A}+\\var{Apair} = \\var{A+Apair}$

b) $\\var{B}+\\var{Bpair} = \\var{B+Bpair}$

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

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$\\var{A} + \\var{Apair}=$

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$\\var{B} +\\var{Bpair}$

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Multiplying Matrices

Calculate the following:

", "advice": "

When you are multiplying two matrices together you need to know what elements of each matrix multiply with each other to produce the answer. In order to look at this let's define the two matrices:

$A=\\var{MatrixA1}$

and

$B=\\var{MatrixA2}.$

These two matrices match the dimensions of the two given in is part a) of this question. When they are multiplied together as we have seen the answer will be a $\\var{row[0]}$ by $\\var{col[0]}$ matrix. So we can denote this matrix in the following way:

$\\var{MatrixA1}$X$\\var{MatrixA2}=\\var{Answermatrixa}$

The way in which this is calculated is as follows in terms of a formula:

$\\var{Answermatrixa}=\\var{matrixmultcalc}$

If you are trying to calculate $c_{11}$ for example one way to think of this is to imagine picking up the first column of the second matrix and tipping it on its side and laying it on top of the top row of the first matrix so, for example, $a_{11}$ gets paired with $b_{11}$ and $a_{12}$ gets paired with $b_{21}$ etc. As we can see in the above formula.

a) $\\var{A}$X$\\var{Apair} = \\var{A*Apair}$

b) $\\var{B}$X$\\var{Bpair} = \\var{B*Bpair}$

b) $\\var{C}$X$\\var{Cpair} = \\var{C*Cpair}$

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

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\"Latexcodebits\" is a JME function written in the Extensions and scripts section. It is a useful function for creating a list of symbols such as \"a_11\" to populate a matrix for the explanation section.

", "templateType": "anything", "can_override": false}, "row": {"name": "row", "group": "Ungrouped variables", "definition": "repeat(random(2..4),3)", "description": "", "templateType": "anything", "can_override": false}, "answermatrixa": {"name": "answermatrixa", "group": "Advice", "definition": "latex('\\\\begin{pmatrix}'+ stringify(rawanswermatrixa) + '\\\\end{'+'pmatrix}')", "description": "

\"stringify\" is a javascript written in the Extensions and scripts section - this converts a list of things into a string suitable for latex to read as a matrix.

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$\\var{A}$X$\\var{Apair}=$

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$\\var{B}$X$\\var{Bpair}$

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$\\var{C}$X$\\var{Cpair}$

", "correctAnswer": "{C}*{Cpair}", "correctAnswerFractions": false, "numRows": "{row[2]}", "numColumns": "{col[2]}", "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LM05 Determinant of a 2x2 matrix", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The determinant of a 2x2 matrix:

Calculate the determinant of the following 2X2 matrices.

", "advice": "

Determinant

The detrminant of a 2x2 matrix can be calculated using the following process:

$$
\\begin{vmatrix}
a & b \\\\
c & d \\\\
\\end{vmatrix} = ad-bc.
$$

Example

For $A = \\var{nonzerodet2},$

we have that,

$$
\\begin{vmatrix}
\\var{nonzerodet2[0][0]} & \\var{nonzerodet2[0][1]}\\\\
\\var{nonzerodet2[1][0]} & \\var{nonzerodet2[1][1]}\\\\
\\end{vmatrix}=(\\var{nonzerodet2[0][0]})(\\var{nonzerodet2[1][1]}) - (\\var{nonzerodet2[0][1]})(\\var{nonzerodet2[1][0]}) =\\var{detnonzerodet2}
$$

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

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$$
A=\\var{nonzerodet2}
$$

$
\\text{What is the value of}\\begin{vmatrix}
A
\\end{vmatrix}?
$

answer:[[0]]

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$$
m=\\var{zerodet1}
$$

$
\\text{What is the value of}\\begin{vmatrix}
M
\\end{vmatrix}?
$

answer:[[0]]

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The inverse of a 2x2 matrix

", "advice": "

Can the inverse of a 2X2 Matrix be found?

The inverse of a matrix can be found if the matrix is a square matrix and has a non-zero determinant. In the first part of this question all the matruces are 2X2 square matrices. The detrminant can be calculated by doing:

$$
\\begin{vmatrix}
a & b \\\\
c & d \\\\
\\end{vmatrix} = ad-bc
$$

If this comes out to be non-zero then the determinant can be found.

Finding the Inverse of a 2x2 Matrix

If there are two 2x2 matrices $A$ and $B$ such that:

$$
AB=BA=
\\left( \\begin{matrix}
1 & 0 \\\\
0 & 1 \\\\
\\end{matrix}\\right)
$$

then we can say that $A$ and $B$ are inverses of each other. The notation for this is that the inverse of a matrix $C$ is written as $C^{-1}$.

If,

$$
C=
\\left(\\begin{matrix}
a & b \\\\
c & d \\\\
\\end{matrix}\\right),
$$

then, the inverse of $C$ is given by the formula:

$$
C^{-1}=\\frac{1}{ad-bc}
\\left(\\begin{matrix}
d & -b \\\\
-c & a \\\\
\\end{matrix}\\right),
$$

So for part 2) of this question let's call the given matrix $D$:

$D=\\var{nonzerodet1}.$

The inverse of $D$ is calculated as follows:

$$
D^{-1} = \\frac{1}{(\\var{nonzerodet1[0][0]})(\\var{nonzerodet1[1][1]})-(\\var{nonzerodet1[1][0]})(\\var{nonzerodet1[0][1]})}\\left(\\begin{matrix}
\\var{nonzerodet1[1][1]} & \\var{-nonzerodet1[0][1]} \\\\
\\var{-nonzerodet1[1][0]} & \\var{nonzerodet1[0][0]} \\\\
\\end{matrix}\\right)
= \\var{inversenonzerodet1}
$$

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

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For which of the following matrices can the inverse be calculated?

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What is the inverses of the following matrix?

(Enter your answers as fractions or decimals to 3 significant figures).

$\\var{nonzerodet1}$

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Find the transpose of the following matrices:

", "advice": "

Transposing a matrix

To transpose a matrix you have to swap over the rows and columns. So for example a 3 by 2 matrix will have a transpose that is a 2 by 3 matrix. The elements of the transpose of a matrix also swap places following the same rule.

Example:

From part b) of the question you have the matrix $M$ given by:

$M=\\var{Transpose1}.$

Swapping the rows with the columns (for example meaning the first row becomes the first column in the transpose), gives the answer:

$M^{T}=\\var{answer}.$

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

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$\\var{squaretranspose}$

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$\\var{Transpose1}$

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Vector Arithmetic

Work through the following questions to ensure you know how to add and subtract vectors in 2D.

For the whole of this question:

$\\bf{a} = \\var{a}$ and $\\bf{b}=\\var{b}$.

", "advice": "

The vectors in this question have two dimensions but the idea of addition and subtraction of vectors works in any number of dimensions (as long as all the vectors being added or subtracted have the same dimensions as each other).

To add two vectors you simply add their corresponding elements. In general:

$$
\\left(\\begin{array}{c}
a \\\\
b \\\\
\\end{array}\\right) +
\\left(\\begin{array}{c}
c \\\\
d \\\\
\\end{array}\\right) =
\\left(\\begin{array}{c}
a+c \\\\
b+d \\\\
\\end{array}\\right).
$$

Subtraction works in the same way so we have:

$$
\\var{a} + \\var{b} = \\var{a+b}.
$$

$$
\\var{a} - \\var{b} = \\var{a-b}.
$$

In order to undertstand the third part of the question you need to know what a \"position vector\" and \"direction vector\" are.

A position vector is defined as a vector that symbolises the location of any given point with respect to the origin. It can be thought of as a coordinate point, but written as a column vector - top entry is the \"x-coordinate\" and the bottome entry is the \"y-coordinate\".

A direction vector is defined as a vector that symbolises a direction and a distance in that direction but with no specified \"starting point\". In 2D it can be summarized as an instruction to go the top element number of units left or right based on the sign of the element and the bottom element number of units up or down based on the sign of the element.

So the direction vector from $A$ to $B$ can be worked out by looking at a route from $A$ to $B$ that travels along the position vectors given. Starting at $A$ we have to go backwards down $\\bf{a}$ to the origin and then forwards along $\\bf{b}$. This corresponds to doing \"minus\" $\\bf{a}$ and \"positive\" $\\bf{b}$:

$$
\\vec{AB} = (-)\\bf{a} + \\bf{b} = \\bf{b}-\\bf{a} = \\var{b}-\\var{a} = \\var{b-a}.
$$

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

Calculate $\\bf{a}+\\bf{b}$.

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Calculate $\\bf{a} - \\bf{b}$.

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Let $\\bf{a}$ be the position vector of point $A$ and $\\bf{b}$ be the position vector of point $B$. Find the direction vector $\\vec{AB}$.

", "correctAnswer": "answerab", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV03 Vectors and geometry", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

This question is about understanding the use of vectors to prove that a shape is a trapezium.

The quadrilateral ABCD is shown below.

$$
A = (\\var{a[0]},\\var{a[1]})\\\\
B = (\\var{b[0]},\\var{b[1]})\\\\
C = (\\var{c[0]},\\var{c[1]})\\\\
D = (\\var{d[0]},\\var{d[1]})
$$

{geogebra_applet('https://www.geogebra.org/m/xb9bvcaa',defs)}

", "advice": "

Vectors between points

In order to undertstand how to find the vector between two points it is helpful to know what a \"position vector\" and \"direction vector\" are. This advice should cover parts a), b), e) and f) of this question.

The point $B$ has coordinates $(\\var{B[0]},\\var{B[1]})$ and it has position vector, denoted $\\bf{b}$, given as $\\bf{b} = \\var{b}$.

So the direction vector from $B$ to $C$ can be worked out by looking at a route from $B$ to $C$ that travels along the position vectors given. Starting at $B$ we have to go backwards down $\\bf{b}$ to the origin and then forwards along $\\bf{c}$. This corresponds to doing \"minus\" $\\bf{b}$ and \"positive\" $\\bf{c}$:

$$
\\vec{BC} = (-)\\bf{b} + \\bf{c} = \\bf{c}-\\bf{b} = \\var{c}-\\var{b} = \\var{c-b}.
$$

Parallel vectors

If one vector is a multiple of another then they are vectors that point in the same direction. This means they are parallel. You just need to check what the multplier is between corresponding elements in each vector. If it is the same for both pairs of elements then the vectors are parallel (and if not then they are not).

For example, $\\vec{BC} = \\var{bc}$ and $\\vec{AD} = \\var{k*BC}.$ Since $\\frac{\\var{k*BC[0]}}{\\var{BC[0]}} = \\var{k}$ which gives the same multiplier as $\\frac{\\var{k*BC[1]}}{\\var{BC[1]}} = \\var{k}$ then $\\vec{BC}$ and $\\vec{AD}$ are parallel.

Conclusions about shapes

This question is looking at a trapezium specifically. The key properties of a trapezium are that it is a quadrilateral and there is one pair of parallel sides. This question goes through establishing that one pair of sides are parallel and then does the calculations to show that the other pair is not parallel. At this point we can conclude that $ABCD$ is a trapezium.

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

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Find the vector $\\vec{BC}$.

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Find the vector $\\vec{AD}$.

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Hence, it is possible to write $\\vec{AD} = k \\times \\vec{BC}$. Find the value of $k$.

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The fact that $\\vec{AD}$ can be written as a multiplier times by $\\vec{BC}$ means:

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Find the vector $\\vec{AB}$.

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Find the vector $\\vec{CD}$.

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Up to this point in the question we have now shown which of the following:

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the scalar (or dot) product of $\\bf{a}$ and $\\bf{b}$.

", "advice": "

It is important to note that for vectors there is more than one type of multiplication. This question is specifically about the scalar (or dot) product.

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2},$ the scalar (or dot) product is defined as

$$
\\mathbf{v \\cdot w} = v_1 \\times w_1 + v_2 \\times w_2.
$$

So for this question:

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b}\\\\
\\bf{a} \\cdot \\bf{b} = \\var{a[0]}\\times\\var{b[0]} + \\var{a[1]}\\times\\var{b[1]} = \\var{adotb}.
$$

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

$\\bf{a} \\cdot \\bf{b} =$ [[0]]

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The vectors $\\bf a$ and $\\bf b$ are defined as follows,
$$
{\\bf a} = \\var{a} \\qquad \\text{and} \\qquad \\bf b = \\var{b}.
$$

", "advice": "

The vector product (also referred to as the cross product) of two vectors can be calculated as follows:

$$
\\pmatrix{a_1 \\\\ a_2 \\\\ a_3} \\times \\pmatrix{b_1 \\\\ b_2 \\\\ b_3} = \\pmatrix{a_2b_3-a_3b_2 \\\\ -(a_1b_3-a_3b_1) \\\\ a_1b_2-a_2b_1}.
$$

A way to think of this is as the following determinant:

$$
\\begin{vmatrix}
\\bf i & \\bf j & \\bf k\\\\
a_1 & a_2 & a_3 \\\\
b_1 & b_2 & b_3
\\end{vmatrix},
$$

where $\\bf i$, $\\bf j $, and $\\bf k$ are the standard unit basis vectors.

In this question we therefore have:

$$
\\begin{align*}
\\var{a} \\times \\var{b} & = \\pmatrix{\\var{a[1]} \\times \\var{b[2]} - \\var{a[2]} \\times \\var{b[1]} \\\\ -(\\var{a[0]} \\times \\var{b[2]} - \\var{a[2]} \\times \\var{b[0]}) \\\\ \\var{a[0]} \\times \\var{b[1]} - \\var{a[1]} \\times \\var{b[0]}}\\\\
& = \\var{answer}.
\\end{align*}
$$

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

Find $\\bf a \\times \\bf b$.

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Find a percentage of an amount.

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

What is {x}% of {y}?

", "advice": "

Taking {x}% of {y} is calculated by multiplying,

\\[\\frac{\\var{x}}{100}\\times\\var{y}.\\]

For this question we can calculate this by noticing that 10% of {y} is {y*0.1} and then since $\\var{x}\\%=\\var{x/10}\\times10\\%$ we can calculate {x}% of {y} as,
\\[\\var{x/10}\\times10\\%\\times \\var{y}=\\var{x/10}\\times\\var{y*0.1}=\\var{x/100*y}.\\]

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

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This question tests the student's ability to identify equivalent fractions through spotting a fraction which is not equivalent amongst a list of otherwise equivalent fractions. It also tests the students ability to convert mixed numbers into their equivalent improper fractions. It then does the reverse and tests their ability to convert an improper fraction into an equivalent mixed number.

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

A mixed number is a number consisting of an integer and a proper fraction, i.e. a number in the form $ a \\displaystyle \\frac{b}{c}$ where $a$ is an integer and $\\displaystyle\\frac{b}{c}$ is a proper fraction: $b$ is smaller than $c$.

An improper fraction is a fraction where the numerator is larger than the denominator, i.e. a number of the form $\\displaystyle\\frac{d}{e}$ where the numerator, $d$, is greater than the denominator, $e$.

To convert an improper fraction into a mixed number, find out how many times the denominator \\var{h_coprime/gcdb} goes into the numerator \\var{num/gcdb}. You can do this by dividing the numerator by the denominator and taking the whole number part or you can just add the denominator to itself until one more addition would make it bigger. This gives us a whole number part of our mixed fraction of \\var{f}.

The numerator of our mixed fraction is what is left from dividing out the whole number. For this question that is $\\var{num/gcdb}-\\var{f*h_coprime}.

Finally the denominator of our mixed fraction is just the denominator of the improper fraction.

\\[
\\frac{\\var{num/gcdb}}{\\var{h_coprime/gcdb}} = {\\var{f}\\frac{\\var{g_coprime}}{\\var{h_coprime}}}\\text{.}
\\]

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

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Random number between 1 and 15

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PART C

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PART C

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numerator for the improper fraction c(i)

", "templateType": "anything", "can_override": false}, "h": {"name": "h", "group": "Ungrouped variables", "definition": "random(10 .. 24#1)", "description": "

Random number between 1 and 24

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PART C

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gcd of num and h

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Random number between 1 and 5

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Write the improper fraction as a mixed number and reduce it down to its simplest form.

$\\displaystyle{\\frac{\\var{num/gcdb}}{\\var{h_coprime/gcdb}}} = $ [[2]] [[0]] [[1]] .

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Convert a percentage to a fraction.

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

Convert $\\var{perc}$% into its equivalent fraction expressed in its simplest form.

", "advice": "

Percentages can be converted to fractions by treating them as fractions out of $100$:

\\[\\frac{\\var{perc}}{100},\\]

and then simplifying. In this case giving:

\\[\\frac{\\var{ansn}}{\\var{ansd}}\\]

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

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[[0]] numerator

[[1]] denominator

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Convert numbers greater than 1 into standard form/scientific notation.

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

Write the following numbers in scientific notation.

", "advice": "

Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. The decimal point is currently to the right of the last digit in $\\var{q2}$ and needs to be between the first and second digits, i.e $\\var{dec2}$. Count the places that the digits must move and you get $\\var{pow2}$ places. That is,

\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]

We have a positive $\\var{pow2}$ as the power because we need to make the number $\\var{dec2}$ bigger to get to $\\var{q2}$.

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

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$\\var{q2} =$ [[0]]$\\times 10$ ^[[1]]

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

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

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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Reading a value from a histogram.

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

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

", "advice": "

To calculate the frequencies using the following formula:

frequency = class width $\\times$ frequency density

Hence our table becomes:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Frequency Density

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Frequency Density

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Frequency Density

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Frequency Density

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Frequency

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Frequency

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Frequency

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Frequency

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How many people ran more than 25km

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How many runners ran less than 7km

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Number of people running between 5 & 12.5 km

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Number of runners running between 10 & 14km

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Number of runners running more than 18km

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

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How many runners ran less than 10 km that day?

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

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

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

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

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

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

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

", "advice": "

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

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

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

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This frequence table stores data about the lengths of some plants:

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

Length ($x$ cm)	Frequency
10 < $x$ $\\leq$ 20	{f1}
20 < $x$ $\\leq$ 30	{f2}
30 < $x$ $\\leq$ 40	{f3}
40 < $x$ $\\leq$ 50	{f4}

", "advice": "

To find the median we must find the middle value. To do this we sum the frequencies and divide the result by 2:

$\\frac{\\var{f1}+\\var{f2}+\\var{f3}+\\var{f4}}{2} = \\frac{\\var{total}}{2} = \\var{middle1}$.

Hence we must find value number $\\var{middle1}$, we can do this by totalling the cumulative frequency in the table below.

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

\n Length $x$ cm \n	Frequency	Cumulative Frequency
1. 10 < $x$ $\\leq$ 20	{f1}	$\\var{f1}$
2. 20 < $x$ $\\leq$ 30	{f2}	$\\var{f1}+\\var{f2}=\\var{cf2}$
3. 30 < $x$ $\\leq$ 40	{f3}	$\\var{f1}+\\var{f2} +\\var{f3} =\\var{cf3}$
4. 40 < $x$ $\\leq$ 50	{f4}	$\\var{f1}+\\var{f2} +\\var{f3} + \\var{f4} =\\var{total}$

You can see value number $\\var{middle1}$ lies in interval number $\\var{interval1}$ hence this is the median class interval for the time taken.

We can estimate the median by using interpolation:

$$
\\text{Estimate of median} = \\text{class start value} + \\frac{\\text{position in class}}{\\text{frequency in class}}\\times \\text{class width}
$$

In this case this is $\\var{lower1}+\\frac{\\var{middle1}-\\var{cf11}}{\\var{fint1}}\\times 10 =\\var{lower1}+\\var{interpolation1}=\\var{median1}.$

(Note: that sometimes it is convenient to use $(\\var{total} + 1)/2 = \\var{middle2}$ as the way to work out the median position. The distinction does not really matter as this is an estimated value and often the decision is made based on which is the most convenient to calculate). This question has been written so that both answers will be marked correctly.

To calculate the $i^{th}$ quartile ($Q_i$) we must use the following formula:

$Q_i = l + \\frac{\\frac{iN}{4}-F}{f}\\times h,$

where:

$l$ = lower limit of the interval in which $Q_i$ lies;

$N$ = Total number of observations;

$F$ = Cumulative frequency of class previous to the $i^{th}$ quartile class;

$f$ = Frequency of $i^{th}$ quartile class.

Since we want to calculate the lower quartile, $i=1$.

If we calculate $\\frac{iN}{4} = \\frac{\\var{total}}{4} = \\var{quarter}$, we can see in the table from part a) that the lower quartile will be in interval number $\\var{intervallq}$. This means that $F = \\var{fprev}$ and $f = \\var{frequency}$.

Hence,

$Q_1 = \\var{lowerlq} + \\frac{\\var{quarter} - \\var{Fprev}}{\\var{frequency}} \\times 10 = \\var{lowerlq} + \\frac{\\var{quarter-Fprev}}{\\var{frequency}} \\times 10 = \\var{lowerlq} + \\var{((quarter-Fprev)/frequency)*10} = \\var{lq}.$

So, The lower quartile is $\\var{lqround}$ to two decimal places.

To calculate the upper quartile we use the same formula as in part b) but this time $i=3$. So, $\\frac{iN}{4} = \\frac{\\var{3*total}}{4} = \\var{3*quarter}$, we can see in the table from part a) that the lower quartile will be in interval number $\\var{intervaluq}$. This means that $F = \\var{ufprev}$ and $f = \\var{ufrequency}$.

Hence,

$Q_3 = \\var{upperlq} + \\frac{\\var{3*quarter} - \\var{uFprev}}{\\var{ufrequency}} \\times 10 = \\var{upperlq} + \\frac{\\var{3*quarter-uFprev}}{\\var{ufrequency}} \\times 10 = \\var{upperlq} + \\var{((3*quarter-uFprev)/ufrequency)*10} = \\var{uq}.$

So, The lower quartile is $\\var{uqround}$ to two decimal places.

To calculate the interquartile range we subtract the lower quartile from the upper quartile.

Hence,

$IQR = Q_3 - Q_1 = \\var{uq}-\\var{lq}=\\var{iqr}$.

So the interquartile range rounded to two decimal places is $\\var{iqrround}$.

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

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Frequency 1

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Frequency 2

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Frequency 3

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Frequency 4

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Sum of frequencies

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Middle Value

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Cumulative Frequency - first and second frequencies

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Lower bound of interval which median sits

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Formula to find lower quartile

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Interval which lower quartile sits in

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Quarter value

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Cumulative frequency of class previous to one which lower quartile sits in

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Frequency of class which lower quartile sits in

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Lower quartile rounded to 2 decimal places

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Upper quartile rounded to 2 decimal places

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Upper quartile

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cumulative frequency of class previous to the one which the upper quartile sits in

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Frequency of class upper quartile sits in

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class which upper quartile sits in

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Interquartile range

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width of class which lower quartile sits in

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width of class upper quartile sits in

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Round interquartle range to two decimal places

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Calculate the lower quartile, give your answer to 2 decimal places.

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Calculate an estimate for the upper quartile using linear interpolation, give your answer to two decimal places.

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Estimate the interquartile range, give your answer to two decimal places.

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Tests understanding of scatter plots and related concepts.

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

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

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

", "advice": "

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

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

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

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

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p6y

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

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

", "

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

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

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age

Thank you for completing the Skills Audit for Maths and Stats. Hopefully it has been useful in directing you to resources and services that can support your studies. The Skills Audit for Maths and Stats will remain open to you throughout the academic year and you can always revisit it again later.

For any further information or questions please contact mash@sheffield.ac.uk

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The answer is a comma-separated list of numbers.

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Is every number in the student's list valid?

Are the student's answers in ascending order?

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