You got it right because you solved it correctly
Answer:
The statement is false
Step-by-step explanation:
we know that
The tangent function will be positive when the sine function and the cosine function have the same sign
so
In the first quadrant the tangent function is positive
In the third quadrant the tangent function is positive
so
The statement is false
So, we are looking for F of G of 2. So, we need to work backwards and find G(x) first.
So we know that g(-2) = -6, now we take thatt -6 and plug in into f(x)
68 is the answer
Answer:
f + g)(x) = f (x) + g(x)
= [3x + 2] + [4 – 5x]
= 3x + 2 + 4 – 5x
= 3x – 5x + 2 + 4
= –2x + 6
(f – g)(x) = f (x) – g(x)
= [3x + 2] – [4 – 5x]
= 3x + 2 – 4 + 5x
= 3x + 5x + 2 – 4
= 8x – 2
(f × g)(x) = [f (x)][g(x)]
= (3x + 2)(4 – 5x)
= 12x + 8 – 15x2 – 10x
= –15x2 + 2x + 8
\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(
g
f
)(x)=
g(x)
f(x)
= \small{\dfrac{3x+2}{4-5x}}=
4−5x
3x+2
My answer is the neat listing of each of my results, clearly labelled as to which is which.
( f + g ) (x) = –2x + 6
( f – g ) (x) = 8x – 2
( f × g ) (x) = –15x2 + 2x + 8
\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}
Ni because if you divide or mulriply the denominator you have to multiply or divide the numerator by the same amount. It is not possible dince you can divide the denominator by 2 to get 3 but you cant divude 5 by 2 and get an evn number
