1. FIRST QUESTION.
M is directly proportional to [tex]r^3[/tex]
Suppose you have two variables, say, [tex]x \ and \ y[/tex]. We say that [tex]y[/tex] varies directly as [tex]x[/tex] or, in other words, [tex]y[/tex] is directly proportional to [tex]x[/tex] if and only if:
[tex]y=kx[/tex]
Where [tex]k[/tex] is a nonzero constant called the constant of proportionality.
Knowing this, we can solve this problem. Here [tex]M[/tex] is directly proportional to [tex]r^3[/tex]. So:
[tex]M=kr^{3}[/tex]
We know that when [tex]r=4, \ M=160[/tex], hence the constant of proportionality can be found as follows:
[tex]M=kr^{3} \\ \\ 160=k(4^3) \\ \\ 160=k(64) \\ \\ Solving \ for \ k: \\ \\ k=\frac{160}{64} \therefore k=\frac{5}{2}[/tex]
So:
[tex]M=\frac{5}{2}r^3[/tex]
PART A)
When [tex]r=2[/tex], then:
[tex]M=\frac{5}{2}(2^3) \\ \\ M=\frac{5}{2}(8) \\ \\ \boxed{M=20}[/tex]
PART B)
When [tex]M=540[/tex], then:
[tex]540=\frac{5}{2}(r^3) \\ \\ r^3=\frac{2(540)}{5} \\ \\ r^3=216 \\ \\ r=\sqrt[3]{216} \\ \\ \boxed{r=6}[/tex]
2. SECOND QUESTION.
M is directly proportional to [tex]r^2[/tex]
We know that when [tex]r=2, \ M=14[/tex], hence the constant of proportionality is:
[tex]M=kr^{2} \\ \\ 14=k(2^2) \\ \\ 14=k(4) \\ \\ Solving \ for \ k: \\ \\ k=\frac{14}{4} \therefore k=\frac{7}{2}[/tex]
So:
[tex]M=\frac{7}{2}r^2[/tex]
PART A)
When [tex]r=12[/tex], then:
[tex]M=\frac{7}{2}(12^2) \\ \\ M=\frac{7}{2}(144) \\ \\ \boxed{M=504}[/tex]
PART B)
When [tex]M=224[/tex], then:
[tex]224=\frac{7}{2}(r^2) \\ \\ r^2=\frac{224(2)}{7} \\ \\ r^2=64 \\ \\ r=\sqrt{64} \\ \\ \boxed{r=8}[/tex]
