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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.)}` |
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`