Question 675962
If a = lw, then w = a/l
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{{{(2x^2-11x+15)/(2x-5)}}}
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(2x-5) goes into {{{2x^2-11x}}}: x
to get
{{{2x^2 - 5x}}}
subtracting
2x^2 - 2x^2 - 11x - -5x
yielding
-6x
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(2x-5) goes into -6x + 15: -3
to get
-6x + 15
subtracting
-6x - -6x + 15 - 15
yielding
0
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{{{(2x^2-11x+15)/(2x-5) = (x-3)}}}
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Width is {{{highlight_green(x-3)}}}
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