Y = 5x + 4
Gradient = slope = 5
(0,4) = y - intercept
The tan(-x) is the same thing as -tan(x). The tangent function is also the same thing as sin(x)/cos(x), right? So let's rewrite that tan in terms of sin and cos:
![[cos(x)][tan(-x)]](https://tex.z-dn.net/?f=%5Bcos%28x%29%5D%5Btan%28-x%29%5D)
is the same as
![[cos(x)][ -\frac{sin(x)}{cos(x)}]](https://tex.z-dn.net/?f=%5Bcos%28x%29%5D%5B%20-%5Cfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%5D%20)
We can now cancel out the cos(x), which leaves us only with -sin(x) remaining. So your answer is A.
Answer:
x=6,y=7
Step-by-step explanation:
In a coordinate pair it is assumed that the first number is x, and the 2nd is y.
Answer:
there whould be 1176 botone on all the pairs combined
Step-by-step explanation:
hope this will help
Remark
First of all you have to declare the meaning of g(f(x)) After you have done that, you have to make the correct substitution.
Givens
f(x) = 4x^2 + x + 1
g(x) = x^2 - 2
Discussion
What the given condition g(f(x)) means is that you begin with g(x). Write down g(x) = x^2 - 2
Wherever you see an x on either the left or right side of the equation, you put fix)
Then wherever you see f(x) on the right you put in what f(x) is equal to.
Solution
g(x) = x^2 - 2
g(f(x)) = (f(x))^2 - 2
g(f(x)) = [4x^2 + x + 1]^2 - 2
f(x)^2 =
4x^2 + x + 1
<u>4x^2 + x + 1</u>
16x^4 + 4x^3 + 4x^2
4x^3 + x^2 + x
<u> 4x^2 + x + 1</u>
16x^4 + 8x^3 + 9x^2 + 2x + 1
Answer
g(f(x)) = 16x^4 + 8x^3 + 9x^2 + 2x + 1 - 2
g(f(x)) = 16x^4 + 8x^3 + 9x^2 + 2x - 1
