smc-5232-10-Linear

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

\(\Rightarrow D\)