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 one year ago
check my answer!
Which ratio represents the area of the smaller rectangle compared to the area of the larger rectangle? (Figure not drawn to scale)
 one year ago
check my answer! Which ratio represents the area of the smaller rectangle compared to the area of the larger rectangle? (Figure not drawn to scale)

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Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0You sure it's not 4/x+2?

allie_bear22
 one year ago
Best ResponseYou've already chosen the best response.1hmm thats not an option but 4x(x+2) is

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x }{ x+5 }=\frac{ 4*(x) }{ (x+2)*(x+5) }\]

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Hm, actually nevermind sorry.

allie_bear22
 one year ago
Best ResponseYou've already chosen the best response.1lol okay so the other guy was right?

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0I'm not 100% sure but I'll have to take his word for it

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1Area of the smaller rectangle = \(x(x+5)\) Area of the larger rectangle = \(4x(x^2+7x+10) = 4x(x+2)(x+5)\) Ratio of areas (smaller to larger) = \[\frac{x(x+5)}{4x(x+2)(x+5)} = \frac{x}{4x(x+2)} = \frac{1}{4(x+2)}\]

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1@Luigi0210 was taking the ratio of the sides of the small rectangle and comparing it to the ratio of the sides of the large rectangle, which is not what problem asks....that would be the right setup for determining if the rectangles were proportional in shape, however.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1Gotta watch out, multiplechoice questions often have wrong answers that represent results you could get if you set up the problem incorrectly. Don't assume the answer is right just because it is one of the choices :)

Luigi0210
 one year ago
Best ResponseYou've already chosen the best response.0Yup, I realized that right when I lef
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