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puteri [66]
3 years ago
15

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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sergeinik [125]
<span>x^2 = 81
x = </span>±√81
x = ±9<span>

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x^2 = 20
x = </span>±√20
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</span>3x^2 = 125 \\ &#10;x^2 =   \frac{125}{3} \\ x=б \sqrt{ \frac{125}{3} }<span>
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Ulleksa [173]

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(-5,-2)

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Step-by-step explanation:

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In problems 1-3, a and b are legs and c is the hypotenuse. Find the missing side length of the
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Answer:

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Step-by-step explanation:

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