Formula Rearrange (Std 2-X)

Algebra, STD2 A1 2015 HSC 28d v1

The formula \(C=\dfrac{5}{9}(F-32)\) is used to convert temperatures between degrees Fahrenheit \((F)\) and degrees Celsius \((C)\).

Convert 18°C to the equivalent temperature in Fahrenheit. (2 marks)

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\(64.4\ \text{degrees}\ F\)

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\(C\)	\(=\dfrac{5}{9}(F-32)\)
\(F-32\)	\(=\dfrac{9}{5}C\)
\(F\)	\(=\dfrac{9}{5}C+32\)

\(\text{When}\ \ C = 18,\)

\(F\)	\(=\dfrac{9}{5}\times 18+32\)
	\(=64.4\ \text{degrees}\ F\)

Algebra, STD2 A1 2013 HSC 21 MC v1

Which equation correctly shows \(n\) as the subject of \(V=600(1-n)\)?

\(n=\dfrac{V-600}{600}\)
\(n=\dfrac{600-V}{600}\)
\(n=V-600\)
\(n=600-V\)

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\(B\)

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♦♦♦ Mean mark 27%

\(V\)	\(=600(1-n)\)
\(1-n\)	\(=\dfrac{V}{600}\)
\(n\)	\(=1-\dfrac{V}{600}\)
	\(=\dfrac{600-V}{600}\)

\(\Rightarrow B\)

Algebra, STD2 A1 2012 HSC 21 MC v1

Which of the following correctly expresses \(r\) as the subject of \(V=\pi r^2+x\) ?

\(r=\pm\sqrt{\dfrac{V}{\pi}}-x\)
\(r=\pm\sqrt{\dfrac{V}{\pi}-x}\)
\(r=\pm\sqrt{\dfrac{V-x}{\pi}}\)
\(r=\pm\dfrac{\sqrt{V-x}}{\pi}\)

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\(C\)

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\(V\)	\(=\pi r^2+x\)
\(\pi r^2\)	\(=V-x\)
\(r^2\)	\(=\dfrac{V-x}{\pi}\)
\(\therefore\ r\)	\(=\pm\sqrt{\dfrac{V-x}{\pi}}\)

\(\Rightarrow C\)

Algebra, STD2 A1 2011 HSC 18 MC v1

Which of the following correctly expresses \(b\) as the subject of \(y= ax+\dfrac{1}{4}bx^2\)?

\(b=\dfrac{4y-ax}{x^2}\)
\(b=\dfrac{4(y-ax)}{x^2}\)
\(b=\dfrac{\dfrac{1}{4}y-ax}{x^2}\)
\(b=\dfrac{\dfrac{1}{4}(y-ax)}{x^2}\)

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\(B\)

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\(y\)	\(= ax+\dfrac{1}{4}bx^2\)
\(\dfrac{1}{4}bx^2\)	\(=y-ax\)
\(bx^2\)	\(=4(y-ax)\)
\(b\)	\(=\dfrac{4(y-ax)}{x^2}\)

\(\Rightarrow B\)

Algebra, STD2 A1 2010 HSC 18 MC v1

Which of the following correctly express \(h\) as the subject of \(A=\dfrac{bh}{2}\) ?

\(h=\dfrac{A-2}{b}\)
\(h=2A-b\)
\(h=\dfrac{2A}{b}\)
\(h=\dfrac{Ab}{2}\)

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\(C\)

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\(A\)	\(=\dfrac{bh}{2}\)
\(bh\)	\(=2A\)
\(\therefore\ h\)	\(=\dfrac{2A}{b}\)

\(\Rightarrow C\)

Algebra, STD2 A1 2004 HSC 11 MC v1

If \(m = 8n^2\), what is a possible value of \(n\) when \(m=7200\)?

\(0.03\)
\(30\)
\(240\)
\(900\)

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\(B\)

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\(m\)	\(=8n^2\)
\(n^2\)	\(=\dfrac{m}{8}\)
\(n\)	\(=\pm\sqrt{\dfrac{m}{8}}\)

\(\text{When}\ m=7200:\)

\(n\)	\(=\pm\sqrt{\dfrac{7200}{8}}\)
	\(=\pm 30\)

\(\Rightarrow B\)

Algebra, STD2 A1 2006 HSC 18 MC v1

What is the formula for \(g\) as the subject of \(7d=8e+5g^2\)?

\(g =\pm\sqrt{\dfrac{8e-7d}{5}}\)
\(g =\pm\sqrt{\dfrac{7d-8e}{5}}\)
\(g =\pm\dfrac{\sqrt{7d+8e}}{5}\)
\(g =\pm\dfrac{\sqrt{8e-7d}}{5}\)

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\(B\)

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\(7d\)	\(=8e+5g^2\)
\(5g^2\)	\(=7d-8e\)
\(g^2\)	\(=\dfrac{7d-8e}{5}\)
\(g\)	\(=\pm\sqrt{\dfrac{7d-8e}{5}}\)

\(\Rightarrow B\)

Algebra, STD2 A1 2007 HSC 19 MC v1

Which of the following correctly expresses \(X\) as the subject of \(Y=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)?

\(X=\dfrac{Y}{\pi}-L\)
\(X=\dfrac{Y}{\pi}-4L\)
\(X=4L-\dfrac{Y}{2\pi}\)
\(X=\dfrac{Y}{8\pi}-\dfrac{L}{4}\)

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\(B\)

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\(Y\)	\(=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)
\(\dfrac{Y}{4\pi}\)	\(=\dfrac{X}{4}+L\)
\(\dfrac{X}{4}\)	\(=\dfrac{Y}{4\pi}-L\)
\(X\)	\(=4\Bigg(\dfrac{Y}{4\pi}-L\Bigg)\)
\(X\)	\(=\dfrac{Y}{\pi}-4L\)

\(\Rightarrow B\)

Algebra, STD2 A1 2005 HSC 24c v1

Make \(r\) the subject of the equation \(V=4\pi r^2\). (2 marks)

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\(r=\pm\sqrt{\dfrac{V}{4\pi}}\)

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\(V\)	\(=4\pi r^2\)
\(r^2\)	\(=\dfrac{V}{4\pi}\)
\(\therefore\ r\)	\(=\pm\sqrt{\dfrac{V}{4\pi}}\)

Algebra, STD2 A1 2016 HSC 24 MC v1

Which of the following correctly expresses \(M\) as the subject of \(y=\dfrac{M}{V}+cX\)?

\(M=Vy-VcX\)
\(M=Vy+VcX\)
\(M=\dfrac{y-cX}{V}\)
\(M=\dfrac{y+cX}{V}\)

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\(A\)

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\(y\)	\(=\dfrac{M}{V}+cX\)
\(\dfrac{M}{V}\)	\(=y-cX\)
\(\therefore\ M\)	\(=V(y-cX)\)
	\(=Vy-VcX\)

\(\Rightarrow A\)

Algebra, STD2 A1 2017 HSC 28d v1

Make \(b\) the subject of the equation \(a=\sqrt{bc-4}\). (2 marks)

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\(b=\dfrac{a^2+4}{c}\)

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♦ Mean mark 46%.

\(a\)	\(=\sqrt{bc-4}\)
\(a^2\)	\(=bc-4\)
\(bc\)	\(=a^2+4\)
\(\therefore\ b\)	\(=\dfrac{a^2+4}{c}\)

Algebra, STD2 A1 EO-Bank 6

Make \(r\) the subject of the equation \(u=\dfrac{5}{4}r+25\). (2 marks)

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\(r=\dfrac{4}{5}u-20\)

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\(u\)	\(=\dfrac{5}{4}r+25\)
\(\dfrac{5}{4}r\)	\(=u-25\)
\(r\)	\(=\dfrac{4}{5}(u-25)\)
\(r\)	\(=\dfrac{4}{5}u-20\)

Algebra, STD2 A1 2019 HSC 11 MC v1

Which of the following correctly expresses \(y\) as the subject of the formula \(5x-2y-9=0\)?

\(y=\dfrac{5}{2}x-9\)
\(y=\dfrac{5}{2}x+9\)
\(y=\dfrac{5x+9}{2}\)
\(y=\dfrac{5x-9}{2}\)

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\(D\)

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♦ Mean mark 50%.

\(5x-2y-9\)	\(=0\)
\(2y\)	\(=5x-9\)
\(\therefore\ y\)	\(=\dfrac{5x-9}{2}\)

\(\Rightarrow D\)

Algebra, STD2 A1 EO-Bank 11

Make \(V\) the subject of the equation \(E=\dfrac{3}{2}mV^3\). (3 marks)

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\(v=\sqrt[3]{\dfrac{2E}{3m}}\)

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\(E\)	\(=\dfrac{3}{2}mV^3\)
\(2E\)	\(=3mV^3\)
\(\dfrac{2E}{3}\)	\(=mV^3\)
\(V^3\)	\(=\dfrac{2E}{3m}\)
\(V\)	\(=\sqrt[3]{\dfrac{2E}{3m}}\)

Algebra, STD2 A1 EO-Bank 12

Make \(x\) the subject of the equation \(y=\dfrac{2}{7}(x-25)\). (2 marks)

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\(x=\dfrac{7y}{2}+25\)

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\(y\)	\(=\dfrac{2}{7}(x-25)\)
\(7y\)	\(=2(x-25)\)
\(\dfrac{7y}{2}\)	\(=x-25\)
\(\therefore\ x\)	\(=\dfrac{7y}{2}+25\)

Algebra, STD1 A1 2019 HSC 34 v1

Given the formula \(D=\dfrac{B(x+1)}{18}\), calculate the value of \(x\) when \(D=90\) and \(B=400\). (3 marks)

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\(3.05\)

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\(\text{Make}\ x\ \text{the subject:}\)

\(D\)	\(=\dfrac{B(x+1)}{18}\)
\(18D\)	\(=B(x+1)\)
\(x+1\)	\(=\dfrac{18D}{B}\)
\(x\)	\(=\dfrac{18D}{B}-1\)
\(\text{When }\)	\(D=90, B=400\)
\(\therefore\ x\)	\(=\dfrac{18\times 90}{400}-1=3.05\)

Algebra, STD2 A2 2022 HSC 14 MC v1

Which of the following correctly expresses \(x\) as the subject of \(y=\dfrac{mx-c}{3}\) ?

\(x=\dfrac{3y}{m}+c\)
\(x=\dfrac{y}{3m}+c\)
\(x=\dfrac{y+c}{3m}\)
\(x=\dfrac{3y+c}{m}\)

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\(D\)

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\(y\)	\(=\dfrac{mx-c}{3}\)
\(3y\)	\(=mx-c\)
\(mx\)	\(=3y+c\)
\(\therefore\ x\)	\(=\dfrac{3y+c}{m}\)

\(\Rightarrow D\)

Algebra, STD2 A1 EO-Bank 9

The volume of a sphere is given by \(V=\dfrac{4}{3}\pi r^3\) where \(r\) is the radius of the sphere.

If the volume of a sphere is \(385\ \text{cm}^3\), find the radius, to 1 decimal place. (3 marks)

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\(4.5\ \text{cm (to 1 d.p.)}\)

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\(V\)	\(=\dfrac{4}{3}\pi r^3\)
\(3V\)	\(= 4\pi r^3\)
\(r^3\)	\(=\dfrac{3V}{4\pi}\)

\(\text{When}\ \ V =385\)

\(r^3\)	\(=\dfrac{3\times 385}{4\pi}\)
	\(=91.911\dots\)
\(\therefore\ r\)	\(=\sqrt[3]{91.911\dots}\)
	\(=4.512\dots\ \ \text{(by calc)}\)
	\(=4.5\ \text{cm (to 1 d.p.)}\)