Answer:
Step-by-step explanation:
We are to show that 
<u>Proof:</u>
From trigonometry identity;


From trigonometry, 2sinAcosA = Sin2A

Also note that sin(B-C) = sinBcosC - cosBsinC
sin420cos140 - cos420sin140 = sin(420-140)
The resulting equation becomes;

= 
Answer:
g(-2) = -6
g(0) = 0
g(5) = 15
Step-by-step explanation:
For each of the evaluations, you have to plug in the number everywhere where there is an x
Answer:
The correct answer is (5 - 7)^2
Step-by-step explanation:
To find this, start with the outermost equation g(x).
g(x) = x^2
Now put the next equation (h(x)) in for x in this equation.
(G ▪ H)(x) = (x - 7)^2
Now, put the number in for x in the equation given.
(G ▪ H)(x) = (5 - 7)^2
We have an equation with parentheses. To make our lives easier, first get rid of the parentheses. We do this by using the distributive property.
It is used like this: a(b + c) = ab + ac.
Use the distributive property on the left-hand side.
3(x - 1) = 6
3x - 3 = 6
Now we have an equation that is easier.
The x variable is being multiplied by 3 and added to -3.
Reverse all of these operations with their inverse operation.
3x - 3 = 6
3x = 9 <--- I got rid of the -3 term by using the inverse of subtraction. Addition.
And I did the same for both sides to keep the equation true.
x = 3 <--- The inverse of multiplication is division.
So, I divided both sides by 3.
So, x is equal to 3.

A linear function has 1 as the highest power of the variable.
A. f(x) = 2 - 7x
here, the highest power of the variable x is 1.
Hence it is a LINEAR function.
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B. f(x) = 2 + x + x^2
here, the highest power of the variable x is 2.
Hence it is NOT a linear function.
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
here, the highest power of the variable x is 1/2.
Hence it is NOT a linear function.
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