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seattle12345
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.
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
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 ?
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!!!!