Problem 1
<h3>Answer: False</h3>
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Explanation:
The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.
So,
f(x) = x+1
f( g(x) ) = g(x) + 1 .... replace every x with g(x)
f( g(x) ) = 6x+1 ... plug in g(x) = 6x
(f o g)(x) = 6x+1
Now let's flip things around
g(x) = 6x
g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)
g( f(x) ) = 6(x+1) .... plug in f(x) = x+1
g( f(x) ) = 6x+6
(g o f)(x) = 6x+6
This shows that (f o g)(x) = (g o f)(x) is a false equation for the given f(x) and g(x) functions.
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Problem 2
<h3>Answer: True</h3>
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Explanation:
Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.
For example, let
f(x) = 1/(x+2)
g(x) = -2
The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.
So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).
D is the answer hope this helps
12h + 30w represents the amount paid to each employee.....where h is the number of hrs and w is the number of wagons sold.
working 6 hrs and selling 3 wagons.....so h = 6 and w = 3
12h + 30w = total pay
12(6) + 30(3) =
72 + 90 =
162 = total pay <===
Answer:
50
Step-by-step explanation:
answer is 50
Answer:
After simplifying, the exponent of x is 33 and the exponent of y is 0
Step-by-step explanation:
For us to simplify, we bring the x base together and the y base together
Recall from the law of indices;
x^a/x^b = x^(a-b)
and x^a * x^b = x^(a + b)
We shall be applying these here
Let us bring the x terms together
we have these as;
x^8/x^14 * 1/x^-39 = x^( 8 - 14 -(-39))
= x^(8 + 39-14) = x^33
for y, we have it that;
y^-26/y^-5 * 1/y^-21
= y^(-26 -(-5) - (-21)
= y^-26 + 5 + 21 = y^0
