Answer:
Step-by-step explanation:
<u>The line with points (1,2) and (-1,-8). Work out its equation.</u>
<u>The slope is:</u>
- m = (-8 - 2)/(-1 - 1) = -10/-2 = 5
<u>To find the y intercept, substitute x and y-xoordinates of point (1,2):</u>
- 2 = 5(1) + b
- b = 2 - 5
- b = -3
<u>The line is:</u>
<u>Point (x, 17), substitute y-coordinate and solve for x</u>
- 17 = 5x - 3
- 5x = 17 + 3
- 5x = 20
- x = 20/5
- x = 4
Answer:
The other person is correct, my bad I got mixed up with the numbers
Wow...we have lots of numbers to go through. we know that a 6 sided figure is a hexagon and the interior angles add up to be 720°. so.....
∠A + ∠B + ∠ C + ∠D + ∠E + ∠F = 720°
(x - 60) + (x - 40) + 130 + 120 + 110 + (x - 20) = 720
3x + 240 = 720 (combined all like terms)
3x = 480 (subtracted 240 from both sides)
x = 160 (divided both sides by 3)
put the value of x into ∠A (x - 60) = 160 - 60 = 100
∠A = 100°
Answer:
B.
Step-by-step explanation:
so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant
![\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx%7D%7B9%7D~~%2C~~%5Cstackrel%7By%7D%7B-3%7D%29%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B9%5E2%2B%28-3%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B81%2B9%7D%5Cimplies%20c%3D%5Csqrt%7B90%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
