Aussie Maths & Science Teachers: Save your time with SmarterEd
The distance in kilometres (\(D\)) of an observer from the centre of a thunderstorm can be estimated by counting the number of seconds (\(t\)) between seeing the lightning and first hearing the thunder.
Use the formula \(D=\dfrac{t}{3}\) to estimate the number of seconds between seeing the lightning and hearing the thunder if the storm is 2.1 km away. (1 mark)
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\(6.3\ \text{seconds}\)
\(D=\dfrac{t}{3}\)
\(\text{When}\ \ D = 2.1,\)
| \(\dfrac{t}{3}\) | \(=2.1\) |
| \(t\) | \(=6.3\ \text{seconds}\) |
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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\(64.4\ \text{degrees}\ F\)
| \(C\) | \(=\dfrac{5}{9}(F-32)\) |
| \(F-32\) | \(=\dfrac{9}{5}C\) |
| \(F\) | \(=\dfrac{9}{5}C+32\) |
\(\text{When}\ \ C = 18,\)
| \(F\) | \(=\dfrac{9}{5}\times 18+32\) |
| \(=64.4\ \text{degrees}\ F\) |
For adults (18 years and older), the Body Mass Index is given by:
\(B=\dfrac{m}{h^2}\), where \(m=\) mass in kilograms and \(h=\) height in metres.
The medically accepted healthy range for \(B\) is \(21\leq B\leq 25\).
What is the minimum weight for a 172 cm adult female to be considered healthy, correct to 1 decimal place? (2 marks)
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\(62.1\ \text{kg}\)
\(B=\dfrac{m}{h^2}\)
\(h=172\ \text{cm} =1.72\ \text{m}\)
\(\text{Given}\ \ 21\leq B\leq 25,\)
\(\rightarrow\ B = 21\ \text{for minimum healthy weight.}\)
| \(21\) | \(=\dfrac{m}{1.72^2}\) |
| \(\therefore\ m\) | \(=21\times 1.72^2\) |
| \(=62.1264\) | |
| \(=62.1\ \text{kg}\ \text{(1 d.p.)}\) |
The distance, \(d\) metres, travelled by a car slowing down from \(u\) km/h to \(v\) km/h can be obtained using the formula
\(v^2=u^2-100 d\)
What distance does a car travel while slowing down from 100 km/h to 70 km/h? (2 marks)
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\(51\ \text{metres}\)
\(u=100 \ , \ v=70\)
| \(v^2\) | \(=u^2-100d\) |
| \(70^2\) | \(=100^2-100d\) |
| \(100d\) | \(=100^2-70^2\) |
| \(\therefore\ d\) | \(=\dfrac{100^2-70^2}{100}\) |
| \(=51\ \text{metres}\) |
If \(m = 8n^2\), what is a possible value of \(n\) when \(m=7200\)?
\(B\)
| \(m\) | \(=8n^2\) |
| \(n^2\) | \(=\dfrac{m}{8}\) |
| \(n\) | \(=\pm\sqrt{\dfrac{m}{8}}\) |
\(\text{When}\ m=7200:\)
| \(n\) | \(=\pm\sqrt{\dfrac{7200}{8}}\) |
| \(=\pm 30\) |
\(\Rightarrow B\)
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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\(3.05\)
\(\text{Make}\ x\ \text{the subject:}\)
| \(D\) | \(=\dfrac{B(x+1)}{18}\) |
| \(18D\) | \(=B(x+1)\) |
| \(x+1\) | \(=\dfrac{18D}{B}\) |
| \(x\) | \(=\dfrac{18D}{B}-1\) |
| \(\text{When }\) | \(D=90, B=400\) |
| \(\therefore\ x\) | \(=\dfrac{18\times 90}{400}-1=3.05\) |
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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\(4.5\ \text{cm (to 1 d.p.)}\)
| \(V\) | \(=\dfrac{4}{3}\pi r^3\) |
| \(3V\) | \(= 4\pi r^3\) |
| \(r^3\) | \(=\dfrac{3V}{4\pi}\) |
\(\text{When}\ \ V =385\)
| \(r^3\) | \(=\dfrac{3\times 385}{4\pi}\) |
| \(=91.911\dots\) | |
| \(\therefore\ r\) | \(=\sqrt[3]{91.911\dots}\) |
| \(=4.512\dots\ \ \text{(by calc)}\) | |
| \(=4.5\ \text{cm (to 1 d.p.)}\) |