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frez [133]
3 years ago
15

Approximate the change in the volume of a sphere when its radius changes from r= 5ft to r= 5.1ft (v(r)= 3/4 pi r^3)?

Mathematics
1 answer:
leva [86]3 years ago
4 0

Answer:

18.03 cubic feet

Step-by-step explanation:

Hello,

Step 1

find the volume of the sphere when radius= 5 ft

v(r)= \frac{3}{4} \pi r^3\\v(5)= \frac{3}{4}\pi (5\ ft)^3\\v(5)= \frac{3}{4}\pi *125\ ft^{3} \\v(5)=294.52\ cubic\ feet\\

Step 2

find the volume of the sphere when radius= 5.1 ft

v(r)= \frac{3}{4}\pi r^3\\v(5.1)= \frac{3}{4} \pi (5.1\ ft)^3\\v(5.1)= \frac{3}{4}\pi *132.651\ ft^{3} \\v(5.1)=312.551\ cubic\ feet\\\\

Step 3

Compare the Volumes to find the change

\frac{v(r_{2})}{v(r_{1})} =\frac{312.551}{294.52} =1.06

the volumen of the sphere with radius = 5.1 is 1.06 times bigger than the first one(r=5)

Now, find the change

change= {v(r_{2})-{v(r_{1})}} \\change=312.551\ cubic\ feet\ -294.52 cubic feet \\ change=18.031\ cubic\ feet

change=18.03 cubic feet

Have a great day.

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Group terms that contain the same variable, and move the constant to the opposite side of the equation

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therefore

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Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}+2x=4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}+2x+1=4+1

x^{2}+2x+1=5

Rewrite as perfect squares

(x+1)^{2}=5

(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}

therefore

case c) is not the solution of the problem

<u>case d)</u> x^{2}-2x-4=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x^{2}-2x=4

Complete the square. Remember to balance the equation by adding the same constants to each side.

x^{2}-2x+1=4+1

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