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Dovator [93]
3 years ago
14

Select the image below

Mathematics
1 answer:
Georgia [21]3 years ago
4 0

Pretend to (or actually) cut out the given image, stick a pin in point P so it cannot move, and rotate the cutout up and to the left around point P about 1/8 of the way around the circle. The result of doing that is the picture you're looking for.

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If m∠OZQ = 125 and m∠OZP = 62, what is m∠PZQ
const2013 [10]

Answer:

<h2>               63° or  173°</h2>

Step-by-step explanation:

You didn't specify where the point P is so there are two posibilites:

1.

P₁Z is dividing angle OZQ

then:

      m∠P₁ZQ = m∠OZQ - m∠OZP₁ = 125° - 62° = 63°

2.

P₂Z is outside angle OZQ

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5 0
3 years ago
Rhombus ADEF is inscribed in △ABC such that the vertices D, E, and F lie on the sides AB , BC , and AC respectively. Find the si
Oksanka [162]
<h3>Answer:</h3>

3 11/15 cm

<h3>Step-by-step explanation:</h3>

AE is the angle bisector of ∠A, so divides the sides of the triangle into a proportion:

... BE:CE = BA:CA = 7:8

Then ...

... BE:BC = 7 : (7+8) = 7:15

ΔDBE ~ ΔABC, so DE = 7/15 × AC

... DE = 7/15 × 8 cm = (56/15) cm

... DE = 3 11/15 cm

3 0
3 years ago
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