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3 years ago

∠1 and ∠2 are supplementary angles. ∠ 1 = x - 19 ∠ 2 = x - 17

1 answer:

4 0

∠1 and ∠2 are supplementary angles, or they equal 180° (supplementary = 180°)

∠1 = x - 19
∠2 = x - 17

Plug in the corresponding expressions with the angles.

(x - 19) + (x- 17) = 180

Combine like terms

x + x = 2x
-19 - 17 = -36

2x - 36 = 180

Isolate the x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

2x - 36 (+36) = 180 (+36)

2x = 180 + 36

2x = 216

Divide 2 from both sides to isolate the x

2x/2 = 216/2

x = 216/2

x = 108

Plug in 108 for x

∠1 = (108) - 19
∠1 = 89°
∠2 = (108) - 17
∠2 = 91°

hope this helps

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Find n. Use the pictures for more info

SSSSS [86.1K]

9514 1404 393

Answer:

A. 5

Step-by-step explanation:

The base AB is twice the length of the midsegment CD.

8n +10 = 2(5n)

8n +10 = 10n . . . . . eliminate parentheses

10 = 2n . . . . . . . . . subtract 8n

n = 5 . . . . . . . . . . . divide by 2

8 0

3 years ago

What is 88% of 600 shirts

vlabodo [156]

Hack:
x% of 100=x
88% of 100=88
times both sides by 6
88% of 600=6 times 88
88% of 600=528

answer is 528

8 0

3 years ago

Read 2 more answers

Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.

Gelneren [198K]

Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

Step 1

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

Step 2

Using the distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given E(-2,3), F(1,6)

$|EF|=\sqrt{(6-3)^2+(1-(-2))^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given F(1,6), G(4,3)

$|FG|=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given G(4,3), H(1,0)

$|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}$

Given E (−2, 3), H (1, 0)

$|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}$

Step 3

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|= $\frac{6-3}{1+2} =\frac{3}{3}=1$

Segment GH, G (4, 3), H (1, 0)

Slope of |GH| $= \frac{0-3}{1-4} =\frac{-3}{-3}=1$

Step 4

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH| $= \frac{0-3}{1+2} =\frac{-3}{3}=-1$

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| $=\frac{3-6}{4-1} =\frac{-3}{3}=-1$

Step 5

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

Step 6

Segment EF, segment FG, segment GH, and segment EH are congruent.

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

Step 7

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

4 0

3 years ago

Read 2 more answers

Scale factor word problem Someone please help me!

natima [27]

1. Niles is making a map of his neighborhood. He uses a scale in which 1 cm = 5 km. The distance between his house and the school is 8.5 km.

How many centimeters will there be between his house and school on the map?

1 cm = 5km

8.5 ÷ 5 = 1.7

There will be 1.7 centimeters between his house and school.

3 0

3 years ago

Highschool level trigonometry!!! Need help knowing step by step how to find answer!

Goshia [24]

Answer:

$\theta \approx 82.34^\circ$

Step-by-step explanation:

Please refer to the attached diagram. (Apologies if the shading or resolution is a bit off.)

So, we want to find θ.

Since we formed a right triangle, we can use right triangle ratios.

We know the measure of the adjacent side to θ and the hypotenuse.

Therefore, we will use the cosine ratio:

$\displaystyle \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$

The adjacent side is 2 and the hypotenuse is 15. By substitution:

$\displaystyle \cos(\theta)=\frac{2}{15}$

Now, we will take the inverse cosine of both sides. So:

$\displaystyle \theta=\cos^{-1}\Big(\frac{2}{15}\Big)$

Use a calculator. Hence:

$\theta \approx 82.34^\circ$

The angle between the ground and the ladder is about 82.34°.

5 0

3 years ago