Aussie Maths & Science Teachers: Save your time with SmarterEd
The length \((L)\) of a shark, in centimetres, can be modelled by the formula
\(L = 7.3a + 38\),
where \(a\) is the age of the shark, in years.
At what age should the shark reach a length of 156 cm, correct to the nearest year? (2 marks)
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\(16\ \text{years}\)
| \(L\) | \(=7.3a+38\) |
| \(156\) | \(=7.3a+38\) |
| \(7.3a\) | \(=156-38\) |
| \(a\) | \(=\dfrac{118}{7.3}\) |
| \(=16.16\dots\) | |
| \(=16\ \text{years (nearest year)}\) |
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 70 km/h to 40 km/h? (2 marks)
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`33 \ text{metres}`
`u = 70 \ , \ v = 40`
| `v^2` | `= u^2-100d` |
| `40^2` | `= 70^2-100d` |
| `100d` | `= 70^2-40^2` |
| `:. d` | `= frac{70^2-40^2}{100}` |
| `= 33 \ text{metres}` |
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 70 km/h to 40 km/h? (2 marks)
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`33 \ text{metres}`
`u = 70 \ , \ v = 40`
| `v^2` | `= u^2-100d` |
| `40^2` | `= 70^2-100d` |
| `100d` | `= 70^2-40^2` |
| `:. d` | `= frac{70^2-40^2}{100}` |
| `= 33 \ text{metres}` |
Given the formula `C = (A(y + 1))/24`, calculate the value of `y` when `C = 120` and `A = 500`. (3 marks)
`4.76`
`text(Make)\ \ y\ \ text(the subject:)`♦♦ Mean mark 32% (Std1).
| `C` | `= (A(y + 1))/24` |
| `24C` | `= A(y + 1)` |
| `y + 1` | `= (24C)/A` |
| `y` | `= (24C)/A-1` |
| `= (24 xx 120)/500-1` | |
| `= 4.76` |
For adults (18 years and older), the Body Mass Index is given by
`B = m/h^2` where `m = text(mass)` in kilograms and `h = text(height)` in metres.
The medically accepted healthy range for `B` is `21 <= B <= 25`.
What is the minimum weight for a 163 cm adult female to be considered healthy? (2 marks)
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`55.8\ text(kg)`
`B = m/h^2`
`h = 163\ text(cm) = 1.63\ text(m)`
`text(Given)\ \ 21 <= B <= 25,`
`=>B = 21\ \ text(for minimum healthy weight.)`
| `21` | `= m/1.63^2` |
| `:. m` | `= 21 xx 1.63^2= 55.794…= 55.8\ text(kg)\ text{(1 d.p.)}` |
The volume of a sphere is given by `V = 4/3 pi r^3` where `r` is the radius of the sphere.
If the volume of a sphere is `\text{220 cm}^3`, find the radius, to 1 decimal place. (3 marks)
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`3.7\ \ text{cm (to 1 d.p.)}`
| `V` | `= 4/3 pi r^3` |
| `3V` | `= 4 pi r^3` |
| `r^3` | `= (3V)/(4 pi)` |
`text(When)\ \ V = 220`
| `r^3` | `= (3 xx 220)/(4 pi)` |
| `= 52.521…` | |
| `:. r` | `=root3 (52.521…)` |
| `= 3.744…\ \ \ text{(by calc)}` | |
| `= 3.7\ \ text{cm (to 1 d.p.)}` |
The formula `C = 5/9 (F-32)` is used to convert temperatures between degrees Fahrenheit `(F)` and degrees Celsius `(C)`.
Convert 3°C to the equivalent temperature in Fahrenheit. (2 marks)
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`37.4`
| `C` | `= 5/9(F-32)` |
| `F-32` | `= 9/5C` |
| `F` | `= 9/5C + 32` |
`text(When)\ \ C = 3,`
`F= (9/5 xx 3) + 32= 37.4`
If `d = 6t^2`, what is a possible value of `t` when `d = 2400`?
`B`
| `d` | `= 6t^2` |
| `t^2` | `= d/6` |
| `t` | `= +- sqrt(d/6)` |
`text(When)\ \ d = 2400:`
`t= +- sqrt(2400/6)= +- 20`
`=> B`
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 = t/3` to estimate the number of seconds between seeing the lightning and hearing the thunder if the storm is 1.2 km away. (1 mark)
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`3.6\ text(seconds)`
`D = t/3`
`text(When)\ \ D = 1.2,`
| `t/3` | `= 1.2` |
| `t` | `= 3.6\ text(seconds)` |