All categories

3 years ago

If you good in 7th grade math pls help

Mathematics

1 answer:

Furkat [3]3 years ago

7 0

Answer:

y = 37

Step-by-step explanation:

Since we know all the terms will equal 180°, we can set up the following equation to solve for y:

y + 29 + 40 + 2y = 180

3y + 69 = 180

3y = 111

y = 37

You might be interested in

Brainliest to best answer (:

grigory [225]

Its C
A positive intercept of 4
A solid line because of the equal to part
And shading down because of the sign direction

3 0

3 years ago

Read 2 more answers

2х + Зу = 16.9 5х = y + 7.4

Maurinko [17]

Answer:

(2.3, 4.1)

Step-by-step explanation:

I don't exactly know what you want answer wise

6 0

3 years ago

Find the slope of the line through each pair of points (2, -11), (7, -20)

baherus [9]

7 - 2 = 5
-20 + 11 = -9
slope is 5/-9 or -5/9
(5 right / 9 down)

5 0

2 years ago

What is the volume of the rectangular prism?

Korvikt [17]

Answer:

7 yd^3

hope this helps

have a good day :)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Please help me to prove this!

Sophie [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

Proof LHS → RHS:

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$

$\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C$

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C $\checkmark$

5 0

3 years ago