Help me please...............

Answer:
a) [tex]M = 20[/tex]
b) [tex]r=6[/tex]
Step-by-step explanation:
a)
Given that
[tex]M[/tex] ∝ [tex]r^3[/tex]
Say
[tex]M=kr^3[/tex] --------------(A)
Where is k is constant of proportionality
We are also given that
When r = 4 , M = 160
Hence
[tex]160 = k \times 4^3[/tex]
[tex]160=k \times 64[/tex]
[tex]k = \frac{160}{64}[/tex]
[tex]k = \frac{5}{2}[/tex]
Now we are asked to determine
a) M when r = 2
Putting the values of k and r in equation (A)
[tex]M= \frac{5}{2} \times 2^3[/tex]
[tex]M= \frac{5}{2} \times 8[/tex]
[tex]M= \frac{5 \times 8}{2}[/tex]
[tex]M= 5 \times 4[/tex]
[tex]M = 20[/tex]
b) r , when M = 540
Putting the value of M and k in equation A again
[tex]540 = \frac{5}{2} \times r^3[/tex]
[tex]r^3=540 \times \frac{2}{5}[/tex]
[tex]r^3 = \frac{2 \times 540}{5}[/tex]
[tex]r^3 = 2 \times 108[/tex]
[tex]r^3=216[/tex]
[tex]r=\sqrt[3]{216}[/tex]
[tex]r=6[/tex]