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3 years ago

D is between C and E, CD = x 2 , CE = 32-2x, and DE = 12x. Find CD, DE, and CE.

Mathematics

1 answer:

Gemiola [76]3 years ago

3 0

Answer:

CD = 4
DE = 24
CE = 28

Step-by-step explanation:

The segment addition theorem tells you ...

CD +DE = CE

x^2 +12x = 32 -2x

Subtract the right side to put this in standard form.

x^2 +14x -32 = 0

(x +16)(x -2) = 0

x = -16 or 2

In order for DE to have a positive length, we must have x > 0. So ...

CD = x^2 = 2^2 = 4

DE = 12x = 12(2) = 24

CE = 32 -2x = 32 -2(2) = 28

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