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).
Initial velocity (u) = 0m/s
Final velocity (v) = 20m/s
Time (t) = 10 s
Acceleration (a)
= (v - u)/t
= [(20m/s) - (0m/s)]/10s
= (20m/s)/10s
= (20m/s²)/10
=> 2m/s²
1) in one minute for first wiil be 1/20
the other 1/30, it means that A paint in one minute 1/20 and B 1/30 then A+B in one minute 1/20 + 1/30 = 3+2/60 = 5 / 60, 5/60 = 1/12
if in one min they make 1/12 then in 12 min they paint all the wall
Answer:
4 (3 x + 2)
Step-by-step explanation: