Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
2. Multiplying Powers With the Same Base
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Exercise 58 Page 431

Practice makes perfect
a

Since the bottom of the box is square and the circular mirror fits inside of the box, the length of one side of the square bottom should be as long as the diameter of the circular mirror. For this reason, the length of one side of the square bottom is 2r.

Since the shape of the bottom of the box is a square, each side of it is of length 2r. The area of a square is length times width. l* w=A ⇒ 2r* 2r=4r^2 Hence, the expression for the area of the bottom of the box is 4r^2.

b

Using the expression that we found in Part A, we can find the area of the bottom of the box when the radius is 4. We will do this by substituting 4 for r in the expression 4r^2.

4r^2
4( 4)^2
4(16)
64

The area of the bottom of the box is 64 inches^2.

c

Since each box is produced with the same dimensions, the area of the bottom of the second box can be expressed using the same formula we found in Part A. If the area of the box is 196 inches^2, we can find the radius of the largest possible mirror by setting this equal to expression 4r^2.

4r^2 =196
â–¼
Solve for r
r^2 =49
r =7

We found the radius. We want to know the diameter, which is twice the length of the radius. Therefore, the diameter of the largest possible mirror is 14inches.