If `d = 6t^2`, what is a possible value of `t` when `d = 2400`?
- `0.05`
- `20`
- `120`
- `400`
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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 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 `220\ text(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 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:)`
`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 `220\ text(cm)^3`, find the radius, to 1 decimal place. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`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\ text(degrees)\ F`
`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\ text(degrees)\ F` |
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)` |