Answer:
You could write -5.16 as -5 4/25
(2, -5) is the mid point
(3, 6) is (2+1, -5+11) is an end point
therefor the other end point should be (2-1, -5-11) = (1, -16)
You could visualise this better if you draw it
Hope this helps, I don't really know a better way to explain it, sorry =P
I'm assuming that you mean
because if you meant
then u would simplify and you couldn't make it the subject.
Under this assumption, we start with
We multiply both sides by 
We expand the left hand side:
We move all terms involving u to the left and all terms not involving u to the right:
We factor u on the left hand side:
We divide both sides by 
If you add all of them together you will get 32
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.
===============================================
Problem 2
<h3>Answer: True</h3>
---------------------------------
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).
