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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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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\(64.4\ \text{degrees}\ F\)
\(C\) | \(=\dfrac{5}{9}(F-32)\) |
\(F-32\) | \(=\dfrac{9}{5}C\) |
\(F\) | \(=\dfrac{9}{5}C+32\) |
\(\text{When}\ \ C = 18,\)
\(F\) | \(=\dfrac{9}{5}\times 18+32\) |
\(=64.4\ \text{degrees}\ F\) |
Which equation correctly shows \(n\) as the subject of \(V=600(1-n)\)?
\(B\)
\(V\) | \(=600(1-n)\) |
\(1-n\) | \(=\dfrac{V}{600}\) |
\(n\) | \(=1-\dfrac{V}{600}\) |
\(=\dfrac{600-V}{600}\) |
\(\Rightarrow B\)
Which of the following correctly express \(h\) as the subject of \(A=\dfrac{bh}{2}\) ?
\(C\)
\(A\) | \(=\dfrac{bh}{2}\) |
\(bh\) | \(=2A\) |
\(\therefore\ h\) | \(=\dfrac{2A}{b}\) |
\(\Rightarrow C\)
Which of the following correctly expresses \(X\) as the subject of \(Y=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\)?
\(B\)
\(Y\) | \(=4\pi\Bigg(\dfrac{X}{4}+L\Bigg)\) |
\(\dfrac{Y}{4\pi}\) | \(=\dfrac{X}{4}+L\) |
\(\dfrac{X}{4}\) | \(=\dfrac{Y}{4\pi}-L\) |
\(X\) | \(=4\Bigg(\dfrac{Y}{4\pi}-L\Bigg)\) |
\(X\) | \(=\dfrac{Y}{\pi}-4L\) |
\(\Rightarrow B\)
Which of the following correctly expresses \(M\) as the subject of \(y=\dfrac{M}{V}+cX\)?
\(A\)
\(y\) | \(=\dfrac{M}{V}+cX\) |
\(\dfrac{M}{V}\) | \(=y-cX\) |
\(\therefore\ M\) | \(=V(y-cX)\) |
\(=Vy-VcX\) |
\(\Rightarrow A\)
Make \(r\) the subject of the equation \(u=\dfrac{5}{4}r+25\). (2 marks)
\(r=\dfrac{4}{5}u-20\)
\(u\) | \(=\dfrac{5}{4}r+25\) |
\(\dfrac{5}{4}r\) | \(=u-25\) |
\(r\) | \(=\dfrac{4}{5}(u-25)\) |
\(r\) | \(=\dfrac{4}{5}u-20\) |
Which of the following correctly expresses \(y\) as the subject of the formula \(5x-2y-9=0\)?
\(D\)
\(5x-2y-9\) | \(=0\) |
\(2y\) | \(=5x-9\) |
\(\therefore\ y\) | \(=\dfrac{5x-9}{2}\) |
\(\Rightarrow D\)
Make \(x\) the subject of the equation \(y=\dfrac{2}{7}(x-25)\). (2 marks)
\(x=\dfrac{7y}{2}+25\)
\(y\) | \(=\dfrac{2}{7}(x-25)\) |
\(7y\) | \(=2(x-25)\) |
\(\dfrac{7y}{2}\) | \(=x-25\) |
\(\therefore\ x\) | \(=\dfrac{7y}{2}+25\) |
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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\(3.05\)
\(\text{Make}\ x\ \text{the subject:}\)
\(D\) | \(=\dfrac{B(x+1)}{18}\) |
\(18D\) | \(=B(x+1)\) |
\(x+1\) | \(=\dfrac{18D}{B}\) |
\(x\) | \(=\dfrac{18D}{B}-1\) |
\(\text{When }\) | \(D=90, B=400\) |
\(\therefore\ x\) | \(=\dfrac{18\times 90}{400}-1=3.05\) |
Which of the following correctly expresses \(x\) as the subject of \(y=\dfrac{mx-c}{3}\) ?
\(D\)
\(y\) | \(=\dfrac{mx-c}{3}\) |
\(3y\) | \(=mx-c\) |
\(mx\) | \(=3y+c\) |
\(\therefore\ x\) | \(=\dfrac{3y+c}{m}\) |
\(\Rightarrow D\)