2 years ago

Can someone help this is so herd.

Mathematics

1 answer:

Finger [1]2 years ago

7 0

The answer is C
.........................

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You have an office building that is 600 square feet in area. What are two possible dimensions how your office could be construct

stepan [7]

Answer:

$20\ ft$ by $30\ ft$

$15\ ft$ by $40\ ft$

Step-by-step explanation:

In this problem I'm assuming the office is rectangular.

The area of rectangle is equal to

$A=LW$

where

L is the length of the rectangle

W is the width of the rectangle

In this problem we have

$A=600\ ft^{2}$

$600=LW$ ------> equation A

Find two possible dimensions of the office

case A) Assume a length side L and find the value of W in the equation A

For $L=30\ ft$

substitute in the equation and solve for W

$600=(30)W$

$W=600/30=20\ ft$

The dimensions are $20\ ft$ by $30\ ft$

case B) Assume a length side L and find the value of W in the equation A

For $L=40\ ft$

substitute in the equation and solve for W

$600=(40)W$

$W=600/40=15\ ft$

The dimensions are $15\ ft$ by $40\ ft$

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