Hiii! To work it out, you would have to do 11 ÷ 3. Which would equal 3.6 (Recurring). If it's to one decimal place, it would be 4 if not, it would just be 3.6 ^^
Hope it helps!! Good luck~
Given that f(x) = x/(x - 3) and g(x) = 1/x and the application of <em>function</em> operators, f ° g (x) = 1/(1 - 3 · x) and the domain of the <em>resulting</em> function is any <em>real</em> number except x = 1/3.
<h3>How to analyze a composed function</h3>
Let be f and g functions. Composition is a <em>binary function</em> operation where the <em>variable</em> of the <em>former</em> function (f) is substituted by the <em>latter</em> function (g). If we know that f(x) = x/(x - 3) and g(x) = 1/x, then the <em>composed</em> function is:



The domain of the function is the set of x-values such that f ° g (x) exists. In the case of <em>rational</em> functions of the form p(x)/q(x), the domain is the set of x-values such that q(x) ≠ 0. Thus, the domain of f ° g (x) is
.
To learn more on composed functions: brainly.com/question/12158468
#SPJ1
Answer:
8
Step-by-step explanation:
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}}} }}
<u>ANSWER</u>
True
<u>EXPLANATION</u>
The given trigonometric equation is

Recall the double angle identity:

We apply this identity to obtain:

We maintain the LHS and simplify the RHS to see whether they are equal.
Expand the parenthesis


Recall that:


We apply these identities to get:




Also



Therefore the correct answer is True