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Integrated math II

I need help ASAP Integrated math II-example-1

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Answer:


\boxed{ \bold{ \sf{x = 16}}}


\boxed{ \bold{ \sf{m < AFD = 82}}}

Explanation:


\sf{5x + 18 = 3x + 50}

( Being vertically opposite angles)

Vertically opposite angles are always equal.

Move variable to L.H.S and change it's sign

Similarly, Move constant to R.H.S and change it's sign


\sf{5x - 3x = 50 - 18}

Collect like terms


\sf{2x = 50 - 18}

Subtract 18 from 50


\sf{2x = 32}

Divide both sides of the equation by 2


\sf{ (2x)/(2) = (32)/(2) }

Calculate


\sf{x = 16}

The value of x is 16

Now, let's find value of m<AFD :


\sf{m < afd + 5x + 18 = 180} ( sum of angle in straight line )

plug the value of x


\sf{m < AFD + 5 * 16 + 18 = 180}

Multiply the numbers


\sf{m < AFD + 80 + 18 = 180}

Add the numbers


\sf{m < AFD + 98 = 180}

Move constant to R.H.S and change it's sign


\sf{m < AFD = 180 - 98}

Subtract 98 from 180


\sf{m < AFD = 82}

Value of m<AFD = 82

Hope I helped!

Best regards!!

User Darlene
by
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