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2 years ago

If p∥q, m∠2=5b+1, and m∠12=6b−8, find m∠1 and m∠3.

Mathematics

1 answer:

lesya [120]2 years ago

7 0

Answer:

Step-by-step explanation:

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Help yall pls! !!!!!

Anika [276]

Answer:

20 stickers

Step-by-step explanation:

sine she gave 20 to core one and wants to give the same amount to core two, she'd just give both core one and two 20 and then be left with 10

7 0

2 years ago

What is the correct answer..

solong [7]

Answer:

64 feet

Step-by-step explanation:

To find the height of the kite we use sin trigonometric ratio.

Let the height of the kite from the ground be x feet.

$\therefore \: \sin 40 \degree = \frac{x}{100} \\ \\ \therefore \: 0.6427876097= \frac{x}{100} \\ \\ \therefore \: 0.6427876097 \times 100= x \\ \\ \therefore \:x = 64.29 \\ \\ \therefore \:x = 64 \: feet$

3 0

3 years ago

A homeowner wants to include a semicircular patio at the end of a square sandbox. She knows that the area of the semicircular pa

stepladder [879]

Answer:

8cm

Step-by-step explanation:

First you must know that the diameter of the semi-circular patio is equal to the length of one side of the square

Area of a semi circle = πr²/2

25.12 = πr²

25.12 = 3.14r²/2

r² = 25.12(2)/3.14

r² = 16

r = √16

r = 4

Get the diameter

d = 2r

d =2(4)

d = 8cm

Hence the length of the side of the square is 8cm

6 0

2 years ago

Can someone help me whats x-9>2

skelet666 [1.2K]

Answer:

x>11

11,∞

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

How to solve this :') please help

blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

$\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\$

Use $\cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1$

$\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}$

7 0

2 years ago