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2 years ago

Solve for c C^2-25-11

Mathematics

1 answer:

olya-2409 [2.1K]2 years ago

4 0

C3-36

the 3 is “c3” is an exponent btw!

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The temperature of a city changed by −15 degrees Celsius over a 5-week period. The temperature changed by the same amount each w

tankabanditka [31]

Do you just wan't us to answer it or give you an explanation?

5 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

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

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

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

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago

What is the sales tax of a $249 cell phone if the tax rate is 8%

raketka [301]

$19.92 is the answer

7 0

3 years ago

Read 2 more answers

Reagan will use a random number generator 1,200 times. Each result will be a digit from 1 to 6. Which statement best predicts ho

Stolb23 [73]

The answer is B.

The only reason it will be 200 is because the numbers 1,2,3,4,5,and 6 are all an option so the probability of 5 is 1/6 and if you multiply 1/6 by 1,200 you will get 200. Hope this helped!

5 0

2 years ago

The 1985 Mexico City earthquake measured a magnitude 8.0 on the Richter scale. The Richter scale uses the function M (I) = log (

Travka [436]

Answer:

The correct option is C: I = 10⁸I₀.

Step-by-step explanation:

The function of the Richter scale is:

$M_{(I)} = log(\frac{I}{I_{0}})$

Since the magnitude of the earthquake is 8.0, the relation between the intensity, I, with the intensity of the threshold quake, Io, is:

$8.0 = log(\frac{I}{I_{0}})$

$10^{8} = \frac{I}{I_{0}}$

$I = 10^{8}I_{0}$

Therefore, the correct option is C: I = 10⁸I₀.

I hope it helps you!

7 0

3 years ago