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3 years ago

12

NEED HELP PLEASE don't know how to solve

1 answer:

amid [387]3 years ago

3 0

The angle formed by the intersection of tangent and secant outside the circle equals half the difference of the <span>intercepted arcs.

</span> $\frac{10x+20-80}{2}=2x+15 \\ 10x-60=2(2x+15) \\ 10x-60= 4x+30 \\ 10x-4x=30+60 \\ 6x=90 \\ x=90/6 \\ x=15$ <span>
</span>

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Find the slope of the line through each pair of points. (11, 1) (-7, 18)

Sonbull [250]

$\bf (\stackrel{x_1}{11}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{-7}~,~\stackrel{y_2}{18}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{18-1}{-7-11}\implies \cfrac{17}{-18}$

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xeze [42]

Answer:

D

Step-by-step explanation:

The formula for C is

C=2πr (using 3.14 as pi)

so C=2(3.14)(45)

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kkurt [141]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

The shown pair of angles are Co interior angle pair, therefore the sum of those angles is 180°

$\qquad \sf \dashrightarrow \: 130 \degree + 7x + 1 \degree = 180 \degree$

$\qquad \sf \dashrightarrow \: 131 \degree + 7x = 180 \degree$

$\qquad \sf \dashrightarrow \: 7x= 180 \degree - 131 \degree$

$\qquad \sf \dashrightarrow \: 7x= 49 \degree$

$\qquad \sf \dashrightarrow \: x=49 \degree \div 7$

$\qquad \sf \dashrightarrow \: x=7 \degree$

The required value of x is 7°

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ryzh [129]

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please give me brainlest

Step-by-step explanation:

16 is the answers for the question

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