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soldier1979 [14.2K]
3 years ago
7

HELP IS MUCH NEEDED :)

Mathematics
2 answers:
ElenaW [278]3 years ago
5 0
∠ADC = 112°
∠ADC =6x + 7 + 12x - 3

Equate both:
6x + 7 + 12x - 3 = 112

Solve for x :
6x + 7 + 12x - 3 = 112

Combine like terms:
18x + 4= 112

Take away 4 from both sides:
18x = 108

Divide both sides by 18:
x = 6

Find ∠ADB:
6x + 7 = 6(6) + 7 = 43


Answer: ∠ADB = 43°

Elena L [17]3 years ago
4 0
If m∠ABC = 112°, then m∠ADB + m∠BDC must equal 112°.

Make m∠ADB + m∠BDC = 112.

6x + 7 + 12x - 3 = 112
18x + 4 = 112
18x = 108
x = 6

Now plug x into 6x + 7 to find m∠ADB.

6(6) + 7 = 43

m∠ADB = 43°
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3 years ago
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k:6x-14y=-28\ \ \ \ |subtract\ 6x\ from\ both\ sides\\\\-14y=-6x-28\ \ \ \ \ \ |divide\ both\ sides\ by\ (-14)\\\\y=\frac{-6}{-14}x-\frac{28}{-14}\\\\y=\frac{3}{7}x+2\\----------------------------\\l:3y-7x=-14\ \ \ \ \ |add\ 7x\ to\ both\ sides\\\\3y=7x-14\ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\\y=\frac{7}{3}x-\frac{14}{3}

Two\ lines\ are\ perpendicular\ if\ product\ of\ the\ slopes\ is\ equal\ -1.\\\\k:y=\frac{3}{7}x+2\to the\ skolpe\ m_k=\frac{3}{7}\\\\l:y=\frac{7}{3}x-\frac{14}{3}\to the\ slope\ m_l=\frac{7}{3}\\\\m_k\times m_l=\frac{3}{7}\times\frac{7}{3}=1\neq-1\\\\conclusion:the\ lines\ are\ not\ perpendicular\\\\Two\ lines\ are\ parallel\ if\ the\ slopes\ are\ equal.\\\\m_k=\frac{3}{7};\ m_l=\frac{7}{3}\to m_k\neq m_l\\\\conclusion:the\ lines\ are\ not\ parallel


Answer:\boxed{C-The\ lines\ intersect\ but\ are\ not\ perpendicular.}
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3 years ago
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