The school athletic director has a storage closet that 4 1/2 feet long, 2 2/3 feet deep, and 6 feet tall. a. She wants to put a carpet in the closet. How much carpeting will she need? b. The athletic director wants to store cube boxers that are 1/2 foot wide. How many boxers will the storage hold?

Answer and Explanation:

The closet is 4 1/2 feet long, 2 2/3 feet deep, and 6 feet tall

Since we are not given width of the closet, we assume the base of the closet has equal width and length = 4 1/2 feet(closet base is a square)

Closet is 2 2/3 feet deep and 6 feet tall, therefore closet height internal = 8 2/3 feet

a. How much carpeting will she need?

The carpeting needed for the closet = area of base of the closet(square base)= 4 1/2* 4 1/2 = 4.5*4.5= 20.25 square feet

b. The athletic director wants to store cube boxers that are 1/2 foot wide. How many boxers will the storage hold?

Area of base = 20.25

Area of base of each cube on the floor = 0.5*0.5= 0.25

So we know that the base of the closet will hold

20.25/0.25= 81 boxer cubes

Or we could straight to the point

Find volume of the closet( the closet is clearly a cuboid) and divide by the volume of one cube to find number of cubes that will fit into the closet

Volume of the closet/cuboid = l*w*h =4.5*4.5*8.67/0.25

=702.27

Therefore number of boxer cubes to fit in closet= 702

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

answer: the distance

step-by-step explanation:

to calculate the distance d of a line segment with endpoints (x1, y1) and (x2, y2) use the formula d (x2 x1)2 (y2 y1)2. to calculate the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) use the formula