Exercise 43 Page 496

Practice makes perfect

a Let's recall the formula for the volume of a cube.

V=l ^3 Here l is the side length. In our case l= 4s. Let's substitute it into the formula and simplify.

V=l ^3

Substitute

l= 4s

V=( 4s)^3

PowProdII

(a b)^m=a^m b^m

V=4^3s^3

CalcPow

Calculate power

V=64s^3

The formula for the volume of the cube in terms of s is V=64s^3.

b Let's use the given formula for the volume of a cylinder.

V=π r^2h Here r is the radius of the base and h is the height of the cylinder. In our case r= s and h= 48. Let's substitute these expressions into the formula and simplify.

V=π r^2h

SubstituteII

r= s, h= 48

V=π s^2( 48)

Multiply

V=48(π)s^2

The formula for the volume of the cylinder in terms of s is V=48(π)s^2.

c The volume V of the metal left after the cylinder has been removed is equal to the volume of the cube minus the volume of the cylinder. We have found these volumes in Part A and Part B, respectively.

V=64s^3-48(π)s^2 The formula in terms of s for the volume of the metal left after the cylinder has been removed is V=64s^3-48(π)s^2.

d We will factor the formula from Part C.

V=64s^3-48(π)s^2 To do so, we need to find the Greatest Common Factor (GCF) of the two terms and factor it out. Let's factor each term first. 64s^3 & = 2* 2* 2* 2* 2* 2* s* s* s 48(π)s^2 & = 2* 2* 2* 2* 3* π* s* s Now we will identify the factors common to both terms. 64s^3 & = 2* 2* 2* 2* 2* 2* s* s* s 48(π)s^2 & = 2* 2* 2* 2* 3* π* s* s The GCF is 2*2*2*2*s*s, or 16s^2. Finally, we will factor out the GCF.

V=64s^2-48(π)s^2

Rewrite

Rewrite 64s^2 as 16s^2* 4s

V=16s^2* 4s-48(π)s^2

Rewrite

Rewrite 48(π)s^2 as 16s^2* 3π

V=16s^2* 4s-16s^2* 3π

FactorOut

Factor out 16s^2

V=16s^2(4s-3π)

e In Part C we have found the following formula.

V=64s^3-48(π)s^2We have to evaluate V for s= 15in. Let's substitute this value into the formula and simplify.