Answer:
π x³ /12 cubic units.
hope this helps you and have a nice day
Answer:
5 ft
Step-by-step explanation:
Since we have a rectangle, we know for sure that perimeter = double the width and double the height. Algebraically, that looks like:
P = 2W + 2H
Let's sub in the given values, P and W:
18 = 2(4) + 2H
Now, let's solve for the height, H:
18 = 8 + 2H
10 = 2H
10/2 = H
<u>5 = H</u>
I hope this helps!
He earns 200 dollars per day.
He earns 75 dollars per room.
He can paint about 16 rooms in 6 days.
Answer: $2.40
Step-by-step explanation:
$60.00 x .04 = $2.40
Answer:

Step-by-step explanation:
Given

Required
Find

Calculate 
Using tan rule

So:






'
'So:


