Can someone please help with this?: The length of a rectangle is 2 inches longer than the width. If the area is 320 square inches, find the rectangle's dimensions. Round your answers to the nearest tenth of an inch.

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Can someone please help with this?: The length of a rectangle is 2 inches longer than the width. If the area is 320 square inches, find the rectangle's dimensions. Round your answers to the nearest tenth of an inch.

Mathematics
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how do we determine that area of any rectangle?
you have a problem with two unknown variables - x and y one being width and other being length - and two conditions 1) length being 2 inches longer than width and 2) area is 320.
@amistre64 Length x Width

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Other answers:

correct :) using that and the given information, we can determine the rest. Area = Length x Width ; Area = 320, and it says the Length is 2 times the Width 320 = (2xWidth)xWidth 320 = 2xW^2
i read 2 times larger for some reason ....
same concept tho; just a correct reading :) Area = Length x Width ; Area = 320, and it says the Length is (Width + 2) 320 = (Width+2)xWidth 320 = W^2 + 2W
would it be (W)(W+2)=320 ?
yes
then would you subtract 320 from both sides?
yes, and expand the left side; this turns it into a usual looking quadratic as a result
so it would be: W^2+2W-320=0?
correct, and then you just use whatever method you like for the determining "w" complete the square, or quadratic formula
since the "c" part is relatively overpowering; id opt complete the square if i had to do by hand w^2 + 2w +1 -1 -320 = 0 (w+1)^2 -321 = 0 (w+1)^2 = 321 w+1 = +- sqrt(321) w = -1 +- sqrt(321)
then use a calculator to work out the decimal :) and ignore the negative since negative length is not defined
yea complete the square seems better in this situation. I always seem to miss a step with the quadratic formula. So W=16.91 and L=18.91?
nearest tenth; but yes
THANKS FOR YOUR HELP!!!!

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