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For this item, any answers that are not whole numbers should be entered as a decimal, rounded to the tenths place. In the figure below, line AB, line CD, and line EF intersect at point Q. Line AB is perpendicular to line CD. Complete the following equations.

x =
m∠CQF =
m∠AQE =

For this item any answers that are not whole numbers should be entered as a decimal rounded to the tenths place In the figure below line AB line CD and line EF class=

Respuesta :

akposevictor

Answer:

[tex] x = 10 [/tex]

m<CQF = 32°

m<AQE = 32°

Step-by-step explanation:

m<CQB = m<CQA = 90° (right angle)

m<CQB = m<CQF + m<FQB

m<CQF = 3x + 2

m<FQB = 58°

Therefore,

[tex] 90 = 3x + 2 + 58 [/tex]

Solve for x:

[tex] 90 = 3x + 60 [/tex]

[tex] 90 - 60 = 3x + 60 - 60 [/tex]

[tex] 30 = 3x [/tex]

[tex] \frac{30}{3} = \frac{3x}{3} [/tex]

[tex] 10 = x [/tex]

[tex] x = 10 [/tex]

m<CQF = 3x + 2

Plug in the value of x to find m<CQF

m<CQF = 3(10) + 2 = 30 + 2

m<CQF = 32°

m<CQF and m<AQE are vertical opposite angles, therefore, they are congruent.

Thus,

m<AQE = 32°

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