Help me fam I need help

Answers

Answer 1

Answer: 1). The 2 angles add up to 180 degrees. 2). And 3). The 2 angles are equal.

Related Questions

A pink crayon is made with 12 mL of red wax for every 5 mL of white wax.Which of the following wax mixtures will create the same shade of pink?(A) 10 mL of red wax mixed wit 24 mL of white wax(B) 48 mL of red mixed with 12 of white wax(C) 13 mL of red mixed with 6 mL of white wax(D) 36 mL of red mixed with 15 mL of white wax(E) 60 mL of red mixed with 24 mL of white wax
If a borrower's monthly interest payment on an interest-only loan at an annual interest rate of 7.3% is $877, how much was the loan amount (rounded to the nearest hundred)?A. $120,100B. $100,300C. $144,200D. $134,200
NEED HELP ASAPSolve 4 – x = –8.
H(x) = –x + 5. Which could you change to move the graph down 3 units
What is the slope of the line which passes through (−2, 0) and (0, 4)? (5 points) 2 Undefined −2 0

Find the slope of the line that contains these two points.
(2, -5) and (7, -10)

Answers

$\sf Slope= (y_2-y_1)/(x_2-x_1)$

Where $\sf (x_1,y_1), (x_2,y_2)$ are the two points.

So

$\sf Slope = ( -10-(-5))/(7-2)=(-10+5)/(7-2)=(-5)/(5)=-1$

So the slope of (2,-5) and (7,-10) was $\boxed{ -1}$

m= $(y1-y2)/(x1-x2)$ = $(-5-(-10))/(2-7)$ = $(5)/(-5) =-1$
Answer: slope is -1

What is the surface area of the prism? A. 180 in2 B. 204 in2 C. 120 in2 D. 192 in2

Answers

I believe from what i have learned from over the years that your answer should be c

Calculate the perimeter of the quadrilateral 2 cm 3 cm 2 cm 5 cm is it A. 10 cm B. 12 cm C. 24 cm D. 60 cm

Answers

P = 2 + 3 + 2 + 5 , where P is the perimeter;
P = 12 cm;
The correct answer is B.

Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas. He starts by assigning coordinates as given.A right triangle is graphed on a coordinate plane. The horizontal x-axis and y-axis are solid, and the grid is hidden. The vertices are labeled as M, K, and L. The vertex labeled as M lies on begin ordered pair 0 comma 0 end ordered pair. The vertex labeled as K lies on begin ordered pair 0 comma 2 b end ordered pair. The vertex labeled as L lies on begin ordered pair 2a comma 0 end ordered pair. A bisector is drawn from point M to the line KL. The intersection point on line KL is labeled as N.

Enter the answers to complete the coordinate proof.
N is the midpoint of KL¯¯¯¯¯KL¯ . Therefore, the coordinates of N are (a,
).

To find the area of △KNM△KNM , the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM△KNM is
.

To find the area of △MNL△MNL , the length of the base ML is
, and the length of the height is
. So an expression for the area of △MNL△MNL is ab.

Comparing the expressions for the areas shows that the areas of the triangles are equal.

Answers

The coordinates of N is (a,b) using the midpoint formula.
The area for △KNM is (1/2)(a)(2b) = ab
The area of △MNL is ab.
Since the area of △KNM = △MNL and the area of △KML is 2ab, then we have proved that a segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.

1. N is a midpoint of the segment KL, then N has coordinates

$\left((x_K+x_L)/(2),(y_K+y_L)/(2) \right) =\left((0+2a)/(2),(2b+0)/(2) \right) =(a,b).$

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

$A_(KMN)=(1)/(2)\cdot \text{base}\cdot \text{height}=(1)/(2)\cdot 2b\cdot a=ab.$

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

$A_(MNL)=(1)/(2)\cdot \text{base}\cdot \text{height}=(1)/(2)\cdot 2a\cdot b=ab.$

4. Comparing the expressions for the areas you have that the area $A_(KMN)$ is equal to the area $A_(MNL)$ . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.