Rewriting the left-hand side as follows,

Answer:
HI
Step-by-step explanation:
Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:
In rectangle ABCD, AB = 6, BC = 8, and DE = DF.
ΔDEF is one-fourth the area of rectangle ABCD.
We want to determine the length of EF.
First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:

The area of the triangle is 1/4 of this. Therefore:

The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:

Since DE = DF:

Thus:

Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:

Therefore:

Square:

Add:

And finally, we can take the square root of both sides:

Answer:
19.15
Step-by-step explanation:
so find the circumference of the circle using c=2*pi*r it's like 18.849. then find the perimeter of the box so p=2l+2w and it's like 38. so subtract the circle circumference from 38 so 38-18.849=19.15