SOLUTION: A square with an area of 49 in. To the second power had a perimeter of________.
I tried solving it i don't know if its correct.
49 in. To the second power =2401/4= 600.25·4= 240
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-> SOLUTION: A square with an area of 49 in. To the second power had a perimeter of________.
I tried solving it i don't know if its correct.
49 in. To the second power =2401/4= 600.25·4= 240
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Question 759570: A square with an area of 49 in. To the second power had a perimeter of________.
I tried solving it i don't know if its correct.
49 in. To the second power =2401/4= 600.25·4= 2401 so the perimeter is 2401. Is that correct? Answer by Cromlix(4381) (Show Source):
You can put this solution on YOUR website! A square with an area of 49 ins^2
has a perimeter of 28 ins.
Area of square = length^2
Length^2 = 49
Therefore length = square root of 49
length = 7 ins
Perimeter = 2*length + 2* width
(With a square, length = width)
= 2*7 + 2*7
= 28 ins.
I think you are confusing the second power.
The second power is associated with inches
i.e. 49 ins^2 was the area.
Hope this helps,.
:-)
