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hey, guys, I

need help ASAP I can't figure this out...

 

 

find the dimensions for a rectangle with the area of...

A=32x^2 + 8x

 

help me, please

 Dec 19, 2018

Best Answer 

 #1
avatar+5795 
+2

\(A=32x^2+8x = \ell \cdot w\\ A = 8x(4x+1)\\ \text{Given the info you've provided there's really no further we can go}\\ \ell = 8x \\ w = (4x+1)\ \text{all you can really say about }x \text{ is }\\ x>0 \text{ since length must be positive}\)

.
 Dec 19, 2018
 #1
avatar+5795 
+2
Best Answer

\(A=32x^2+8x = \ell \cdot w\\ A = 8x(4x+1)\\ \text{Given the info you've provided there's really no further we can go}\\ \ell = 8x \\ w = (4x+1)\ \text{all you can really say about }x \text{ is }\\ x>0 \text{ since length must be positive}\)

Rom Dec 19, 2018

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