For this type of problem you would have to move the decimal as many places to the right as needed to make the divisor a whole number the you would move the dividend the same amount of times even if it is a whole number . Then you use what you have to divide normally. Hope this helps!
Know the order of operations. However, one of the trickiest things about solving an algebra<span> equation as a beginner is knowing where to start. Luckily, there's a specific order for solving these problems: first </span>do<span> any math operations in parentheses, then </span>do<span> exponents, then multiply, then divide, then add, and finally subtract</span>
Oh my, ummm if u go on slander and search that book name then type that page number it will give u all the answers, and the answers are probably at the back of the book <3
Problem 1
<h3>Answer: False</h3>
---------------------------------
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).