Let x be the number of hours ling work on monday.
We know that she worked three more hours on tuesday that in monday, this can be express as :
We also know that in wednesday she worked on more hour than twice the number on mondays, this can be expressed as:
The total number of hours she worked this three days in two more than five the number of hours she worked on monday, this can be express as :
Now , once we have all the expressions we add the expressions of the days and equate them to the total
Now we solve the equation
Therefore , she worked 2 hours on monday.
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It's "commutative property," which says that (for addition/multiplication) order of the operator doesn't matter. For eg, 3 * 5 = 5 * 3.
Associative property (again, of multiplication and addition) means that it doesn't matter how you solve an expression if the same operand is used and some numbers are grouped. For eg. 3 * (5 * 4) = (3 * 5) * 4.
If you are 16 years old then you are a teenager Question 14
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).
