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11Alexandr11 [23.1K]
3 years ago
10

In AFGH, m_F= 65°, m_G = 55°, and m_H = 60° Which side of AFGH is

Mathematics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

  D.  FH

Step-by-step explanation:

Angle G is the smallest, so the side of the triangle opposite that angle will be the shortest. That is the side that does not contain G in its segment name:

  FH is the shortest side . . . . choice D

_____

The attachment shows a triangle with these angle measures drawn to scale. You will note that FH is the shortest side.

__

Please be aware that different instances of this question may have different angle measures, or may have answer choices in a different order. The thing to remember is that <em>the lengths of sides have the same ordering (least to greatest) that the angle measures do</em>.

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A: 40

B:400

C:4000

D:40000
4 0
2 years ago
What is the value of x?? please help
DaniilM [7]

Answer:

x=64

Step-by-step explanation:

Both the semicircle that says 180, and the right angle sign in the image serve to support the idea that all parts of the figure equal 90

Therefore, it is safe to say that you need to subtract 26 from 90 in order to solve for x.

90-26 = 64

8 0
3 years ago
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What is the value of m in the equation 5m − 7 = 6m 11? 18 1 −18 −1.
denpristay [2]

Answer:

-7 = m

Step-by-step explanation:

5m − 7 = 6m

Subtract 5m from each side

5m-5m − 7 = 6m-5m

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6 0
2 years ago
What are the endpoint coordinates for the midsegment of △PQR that is parallel to PQ¯¯¯¯¯?
andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5)  and

N (x₅ ,y₅) = ( -1 , -0.5 )

Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)  

To Find:

M(x₄ ,y₄) = ?  and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})

As M is the mid point of PR and N is the mid point of RQ so we will have

Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})

Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})

Substituting the given value in above equation we get

Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)

Similarly,

Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)

∴ M(x₄ ,y₄) = (-3.5 , 0.5)  and

  N (x₅ ,y₅) = ( -1 , -0.5 )

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3 years ago
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Ymorist [56]

Answer: B) 2(x + 3) - 1

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