Mathematics, 14.09.2020 23:01 iicekingmann

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GEOMETRY QUESTIONS, PLEASE HELP!

Answers: 2

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14.
Angles 4 and 6 are supplementary, because they are on the same line. Supplementary angles add up to 180 degrees, and a line must be 180 degrees.

15.
Angles 1 and 8 are congruent, because they are alternate exterior angles

16.
m = y2 - y1 / x2 - x1
m = 7 - 2 / 4 - 5
m = 5 / -1
m = -5

17.
m = 3 - 3 / 7 - (-5)
m = 0 / 12
m = 0

18.
m = 1 - (-2) / 5 - (-4)
m = 3 / 9
m = 1/3

19.
A = (0, 3) - B = (3,0)
m = 0 - 3 / 3 - 0
m = -3 / 3
m = -1
C = (0, -2) - D = (4, 2)
m = 2 - (-2) / 4 - 0
m = 4 / 4
m = 1
Perpendicular, because the slopes are opposite reciprocals.

20.
E = (1, 2) - F = (0, 0)
m = 0 - 2 / 0 - 1
m = -2 / -1
m = 2
G = (1, -3) - H = (3, 0)
m = 0 - (-3) / 3 - 1
m = 3 / 2
Neither, because the slopes are different.

21.
I = (0, 1) - J = (2, -4)
m = -4 - 1 / 2 - 0
m = -5/2
K = (-1, -2) - L = (4, 0)
m = 0 - (-2) / 4 - (-1)
m = 2/5
Perpendicular, because the slopes are opposite reciprocals.

22.
M = (-2, 2) - N = (2, 2)
Horizontal line
m = 0
O = (3, 0) - P = (-3, 0)
Horizontal line
m = 0
Parallel, because the slopes are the same.

23.
Angle 2 is congruent to angle 1 because of the alternate exterior angle theorem.
Angle 1 is congruent to angle 3 because of the vertical angle theorem.
Angle 2 is congruent to angle 3 because of substitution.
Line l is parallel to line m because the corresponding angles are congruent.

Answer from: wes94

14. Supplementary
15. Congruent

16. $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 2}{4 - 5} = \frac{5}{-1} = -5 \\\\y - y_1 = m(x - x_1) \\y - 2 = -5(x - 5) \\y - 2 = -5(x)+ 5(5) \\y - 2 = -5x + 25 \\y = -5x + 27$

17. $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 3}{-5 - 7} = \frac{0}{-12} = 0 \\\\y - y_1 = m(x - x_1) \\y - 3 = 0(x - 7) \\y - 3 = 0(x) - 0(7) \\y - 3 = 0x - 0 \\y = 0x + 3$

18. $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-2)}{5 - (-4)} = \frac{1 + 2}{5 + 4} = \frac{3}{9} = \frac{1}{3} \\\\y - y_1 = m(x - x_1) \\y - (-2) = \frac{1}{3}(x - (-4)) \\y + 2 = \frac{1}{3}(x + 4) \\y + 2 = \frac{1}{3}(x) + \frac{1}{3}(4) \\y + 2 = \frac{1}{3}x + 1\frac{1}{3} \\y = \frac{1}{3}x - \frac{2}{3}$

19. Perpendicular
20. Neither
21. Perpendicular
22. Parallel

23. Given: Transversal r cuts lines
   l and m: <2 = <1
      Prove: l || m

Statements | Reasons
1. <2 ≡ <1 | 1. They are congruent.
2. <1 ≡ <3 | 2. They are congruent.
3. <2 ≡ <3 | 3. They are parallel.
4. l || m    | 4. They are parallel.

Answer from: Quest

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒ $y=\frac{+1}{+1}x+0$

or $y=\frac{-1}{-1}x+0$

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.

Answer from: Quest

Ithink the answer is c just trying to others