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Please help me on these two questions

Please help me on these two questions-example-1
Please help me on these two questions-example-1
Please help me on these two questions-example-2

1 Answer

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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}

User Prashant Jajal
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