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3 years ago

Which graph shows a proportional relationship?

Mathematics

2 answers:

dybincka [34]3 years ago

8 0

It’s both graphs c and d

n200080 [17]3 years ago

5 0

Answer:

I think it's graph C and graph D

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O'Farrell offers 2 different after school clubs. The Ukulele club has 30 students and attendance is increasing by 6 students per

Bond [772]

Month 6, work down below

months / students
0 30 45
1 36 48
2 42 51
3 45 54
4 51 57
5 57 60
6 63 63

8 0

3 years ago

Read 2 more answers

Question: The stem-and-leaf plot shows the test scores of a science class. Use the plot to answer the question.

Nikitich [7]

The mode of the data is 8

7 0

3 years ago

Read 2 more answers

NEED HELP !!!!!!!!!!!!!!

chubhunter [2.5K]

Answer:

$31.55

Step-by-step explanation:

She earns $25.75 one week and $37.80 another week that means she earned $25.75 + $37.80 = $63.55 in total

She spent $32 therefore she has $63.55 - $32 = $31.55 left

8 0

3 years ago

Please help me to prove this!

Sophie [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

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)

<u>Proof LHS → RHS:</u>

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

For every touchdown scored by the Timberwolves, the mascot does 333 back flips and the cheerleaders set off 666 confetti cannons

e-lub [12.9K]

Answer:

273 Touchdowns

Step-by-step explanation:

You divide the amount of confetti cannons set off by the amount of confetti cannons set off every time there is a touchdown scored. So in this case 181818 ÷ 666

5 0

3 years ago