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).
Divide the previous value by 5.
<span>arccos (cos pi/2)=?
</span>cos pi/2=0, arccos (0)=?, we know that cos pi/2=0, so cos^-1 (o)= Pi/2, but cos^-1 (o)=arccos (0), so arccos (cos pi/2)=Pi/2
Answer:
I think its 27
Step-by-step explanation:
GCF = 3a^2b
Solution: A, u are correct :)
