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Nookie1986 [14]
3 years ago
14

Two right triangles are graphed on a coordinate plane. One triangle has a vertical side of 4 and a horizontal side of 10. The ot

her triangle has a vertical side of 12 and a horizontal side of 30.
Could the hypotenuses of these two triangles lie along the same line?

No, because they are not similar triangles
No, because one is larger than the other
Yes, because they are similar triangles
Yes, because they are both right triangles
Mathematics
1 answer:
miskamm [114]3 years ago
8 0
No, <span>hypotenuses of these two triangles won't lie along the same line </span><span>because one is larger than the other
</span><span>
In short, Your Answer would be Option B

Hope this helps!</span>
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Answer:

it can not go any more

Step-by-step explanation:

7/45 is the simplified terms

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The fastest a human has ever run is 27 miles per hours. How many miles per minute did the human run?
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Answer all for Brainliest (easy math) 5 questions for 30 points
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7 0
3 years ago
Suppose a circle has a radius of 13 inches. How far would a 24 inch chord be from the center of the circle? (Hint: Draw a diagra
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Answer:

Step-by-step explanation:

Directions

  • Draw a circle
  • Dear a chord with a length of 24 inside the circle. You just have to label it as 24
  • Draw a radius that is perpendicular and a bisector through the chord
  • Draw a radius that is from the center of the circle to one end of the chord.
  • Label where the perpendicular radius to the chord intersect. Call it E.
  • You should get something that looks like  the diagram below. The only thing you have to do is put in the point E which is the midpoint of CB.

Givens

AC = 13 inches                   Given

CB = 24 inches                  Given

CE = 12 inches                    Construction and property of a midpoint.

So what we have now is a right triangle (ACE) with the right angle at E.

What we seek is AE

Formula

AC^2 = CE^2 + AE^2

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169 - 144 = 144-144 + AE^2     Combine

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√25 = √AE^2

5 = AE

Answer

The 24 inch chord is 5 inches from the center of the circle.

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A=πr^2
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