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 2 years ago
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.
 2 years ago
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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amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1how do we determine that area of any rectangle?

msayer3
 2 years ago
Best ResponseYou've already chosen the best response.0you 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.

seattle12345
 2 years ago
Best ResponseYou've already chosen the best response.0@amistre64 Length x Width

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1correct :) 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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1i read 2 times larger for some reason ....

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1same 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

seattle12345
 2 years ago
Best ResponseYou've already chosen the best response.0would it be (W)(W+2)=320 ?

seattle12345
 2 years ago
Best ResponseYou've already chosen the best response.0then would you subtract 320 from both sides?

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

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

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1since 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)

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

seattle12345
 2 years ago
Best ResponseYou've already chosen the best response.0yea 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?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.1nearest tenth; but yes

seattle12345
 2 years ago
Best ResponseYou've already chosen the best response.0THANKS FOR YOUR HELP!!!!
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