SOLUTION: The perimeter of a rectangle is 52 inches squared. Its area is 153 inches squared. What is its length and width?
What I've done so far y=26-x
-x^2+ 26x -52 = 0
I get stuck
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-> SOLUTION: The perimeter of a rectangle is 52 inches squared. Its area is 153 inches squared. What is its length and width?
What I've done so far y=26-x
-x^2+ 26x -52 = 0
I get stuck
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Question 370119: The perimeter of a rectangle is 52 inches squared. Its area is 153 inches squared. What is its length and width?
What I've done so far y=26-x
-x^2+ 26x -52 = 0
I get stuck while using the quadratic formula to use numbers to substitute into the next equation... Your help would greatly be appreciated. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! The perimeter of a rectangle is 52 inches squared. Its area is 153 inches squared. What is its length and width?
What I've done so far y=26-x
-x^2+ 26x -52 = 0
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x + y = 26
x*y = 153
x*(26 - x) = 153 (you had 52)
x^2 - 26x + 153 = 0
(x-9)*(x-17) = 0
x = 9, 17
--> 9 by 17 inches
