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anonymous
 4 years ago
I need to finish my homework ASAP! someone please help!! :(
A square has sides of length 3x cm. A rectangle is 2x cm by (x+7)cm. The area of the square is twice that of the rectangle. Find the dimensions of each figure.
anonymous
 4 years ago
I need to finish my homework ASAP! someone please help!! :( A square has sides of length 3x cm. A rectangle is 2x cm by (x+7)cm. The area of the square is twice that of the rectangle. Find the dimensions of each figure.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0area of the square is 9 , areas of the rectangle is \[2x(x+3)\] and twice that number is 9 so solve \[4x(x+3)=9\] \[4x^2+12x9=0\]via the quadratic formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the area of the rectangle is (2x)(x+7) = A and the area of the square is twice that of the rectangle. Therefore area of the square = 2A . so we know (2x)(x+7) = A and (3x)^2 = 2A . so (3x)^2 = 2[ (2x)(x+7)] ====> 9x^2 = 4x^2 + 28x ====> 5x^2 = 28x ====> x = 28/5 . Now verify if my solution is correct by plugging in the value of x and finding the area of the rectangle and square

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i was wrong, sorry. it says side is 3x, not 3. so malcolm is right and i am wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[4x(x+7)=9x^2\] \[4x^2+28x=9x^2\] \[5x^228x=0\] \[x(5x28)=0\] \[x=\frac{28}{5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Satelite is this how I shall write the whole thingie?
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