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seattle12345
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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.
 one year ago
 one year ago
seattle12345 Group Title
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.
 one year ago
 one year ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.1
how do we determine that area of any rectangle?
 one year ago

msayer3 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

seattle12345 Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 Length x Width
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i read 2 times larger for some reason ....
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

seattle12345 Group TitleBest ResponseYou've already chosen the best response.0
would it be (W)(W+2)=320 ?
 one year ago

seattle12345 Group TitleBest ResponseYou've already chosen the best response.0
then would you subtract 320 from both sides?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
yes, and expand the left side; this turns it into a usual looking quadratic as a result
 one year ago

seattle12345 Group TitleBest ResponseYou've already chosen the best response.0
so it would be: W^2+2W320=0?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
correct, and then you just use whatever method you like for the determining "w" complete the square, or quadratic formula
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
then use a calculator to work out the decimal :) and ignore the negative since negative length is not defined
 one year ago

seattle12345 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
nearest tenth; but yes
 one year ago

seattle12345 Group TitleBest ResponseYou've already chosen the best response.0
THANKS FOR YOUR HELP!!!!
 one year ago
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