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Solve for x. AB DE x = 19 x = 12 x = 9 x = 16
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I guess that there is a diagram for this. am I right?
YEA THEIR IS THIS IS IT
The ratios of sides \(\triangle DCE\) and \(\triangle ACB\) are equal since similar triangles. That is \(\large\frac{12}{12+x-3}=\large\frac{3}{3+4}\) \(\large\frac{12}{9+x}=\large\frac{3}{7}\) cross multiply and solve for x. Can u do?
I WILL NEED SOME HELP IN THIS
\(\LARGE\frac{12}{9+x}=\LARGE\frac{3}{7}\) Cross multiply. Then u get \(\large7*12=3(9+x)\) \(\large84=27+3x\) Subtract 27 bot sides. Then divide by 3 both sides. Can u do nw?
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SO X =19
ya.
THANKS
yw
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