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

Help me please...............-example-1
User Psanford
by
8.3k points

1 Answer

4 votes

Answer:

a)
M = 20

b)
r=6

Explanation:

a)

Given that


M
r^3

Say


M=kr^3 --------------(A)

Where is k is constant of proportionality

We are also given that

When r = 4 , M = 160

Hence


160 = k * 4^3


160=k * 64


k = (160)/(64)


k = (5)/(2)

Now we are asked to determine

a) M when r = 2

Putting the values of k and r in equation (A)


M= (5)/(2) * 2^3


M= (5)/(2) * 8


M= (5 * 8)/(2)


M= 5 * 4


M = 20

b) r , when M = 540

Putting the value of M and k in equation A again


540 = (5)/(2) * r^3


r^3=540 * (2)/(5)


r^3 = (2 * 540)/(5)


r^3 = 2 * 108


r^3=216


r=\sqrt[3]{216}


r=6

User Galaxy
by
8.2k points

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