Question

Please help me on these two questions

Answer 1

1. FIRST QUESTION.

M is directly proportional to
`r^3`

Suppose you have two variables, say,
`x \ and \ y`. We say that
`y` varies directly as
`x` or, in other words,
`y` is directly proportional to
`x` if and only if:

`y=kx`

Where
`k` is a nonzero constant called the constant of proportionality.

Knowing this, we can solve this problem. Here
`M` is directly proportional to
`r^3`. So:

`M=kr^(3)`

We know that when
`r=4, \ M=160`, hence the constant of proportionality can be found as follows:

`M=kr^(3) \\ \\ 160=k(4^3) \\ \\ 160=k(64) \\ \\ Solving \ for \ k: \\ \\ k=(160)/(64) \therefore k=(5)/(2)`

So:

`M=(5)/(2)r^3`

PART A)

When
`r=2`, then:

`M=(5)/(2)(2^3) \\ \\ M=(5)/(2)(8) \\ \\ \boxed{M=20}`

PART B)

When
`M=540`, then:

`540=(5)/(2)(r^3) \\ \\ r^3=(2(540))/(5) \\ \\ r^3=216 \\ \\ r=\sqrt[3]{216} \\ \\ \boxed{r=6}`

2. SECOND QUESTION.

M is directly proportional to
`r^2`

We know that when
`r=2, \ M=14`, hence the constant of proportionality is:

`M=kr^(2) \\ \\ 14=k(2^2) \\ \\ 14=k(4) \\ \\ Solving \ for \ k: \\ \\ k=(14)/(4) \therefore k=(7)/(2)`

So:

`M=(7)/(2)r^2`

PART A)

When
`r=12`, then:

`M=(7)/(2)(12^2) \\ \\ M=(7)/(2)(144) \\ \\ \boxed{M=504}`

PART B)

When
`M=224`, then:

`224=(7)/(2)(r^2) \\ \\ r^2=(224(2))/(7) \\ \\ r^2=64 \\ \\ r=√(64) \\ \\ \boxed{r=8}`