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Vera_Pavlovna [14]
3 years ago
13

Ach equation: - 5(– 2x - 4) + 5x - 4 =- 29

Mathematics
1 answer:
Zinaida [17]3 years ago
3 0

- Multiply (-2x-4) by -5:

[(-5)(-2x) + (-5)(-4)] +5x - 4 = -29

= 10x+20+5x-4=-29

- Combine Like Terms:

(10x+5x) + (20-4) = -29

15x+16=-29

- Subtract 16 from each side

15x+16 -16 = -29 -16

15x = -45

x = -3

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Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.
OverLord2011 [107]

When a line is bisected, the line is divided into equal halves.

See below for the proof of \mathbf{KE \cong FD}

The given parameters are:

  • <em>AC bisects CD</em>
  • <em>IJ bisects CE</em>
  • <em>BH bisects ED</em>

<em />

By definition of segment bisection, we have:

  • \mathbf{CK \cong KE}
  • \mathbf{EF \cong FD}
  • \mathbf{CE \cong ED}

By definition of congruent segments, the above congruence equations become:

  • \mathbf{CK = KE}
  • \mathbf{EF = FD}
  • \mathbf{CE = ED}

By segment addition postulate, we have:

  • \mathbf{CE = CK + KE}
  • \mathbf{ED = EF + FD}

Substitute \mathbf{ED = EF + FD} in \mathbf{CE = ED}

\mathbf{CK + KE = EF + FD}

Substitute \mathbf{CK = KE} and \mathbf{EF = FD}

\mathbf{KE + KE = FD + FD}

Simplify

\mathbf{2KE = 2FD}

Apply division property of equality

\mathbf{KE = FD}

By definition of congruent segments

\mathbf{KE \cong FD}

Read more about proofs of congruent segments at:

brainly.com/question/11494126

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