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DaniilM [7]
2 years ago
12

Can someone help me find AB and BC

Mathematics
1 answer:
Bond [772]2 years ago
7 0

  • 3x + 2x + 5 + 5x + 15 = 180 [angles on a line add to 180 degrees]
  • 10x + 20 = 180 [combine like terms]
  • 10x = 160 [subtract 20 from both sides]
  • x = 16 [divide both sides by 10]

Arc AB measures 3(16) = 48 degrees.

Arc BC measures 2(16) + 5 = 41 degrees.

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What is the equation of the line perpendicular to 3x+y=6 that passes through the point (0,8)​
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3x + y = 6

y =  - 3x + 6

The coefficient of x is the slope of the line , So the slope of the above equation is -3 .

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Remember from now on ,

Product of multiplying the slopes of two lines which are perpendicular to each other , is -1 .

Thus ;

( - 3) \times (m) =  - 1

m is the slope of the line which we want.

- 3m =  - 1

Negatives simplifies

3m = 1

Divide sides by 3

\frac{3m}{3}  =  \frac{1}{3}  \\

m =  \frac{1}{3}  \\

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We have following equation to find the point-slope form of the linear functions :

y - y(0) = m(x - x(0))

x(0) and y(0) are the coordinates of the point which the line passed through.

y  - 8 =  \frac{1}{3} (x - 0) \\

y - 8 =  \frac{1}{3} x \\

Add sides 8

y =  \frac{1}{3} x + 8 \\

Done....

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