The statement that -6 is in the domain of f(g(x)) is true
<h3>Complete question</h3>If f(x) = -2x + 8 and g(x) = , which statement is true?
We have:
f(x) = -2x + 8
Start by calculating the function f(g(x)) using:
f(g(x)) = -2g(x) + 8
Substitute
Set the radicand to at least 0
Subtract 9 from both sides
This means that the domain of f(g(x)) are real numbers greater than or equal to -9. i.e. -9, -8, -7, -6, ...........
Hence, the statement that -6 is in the domain of f(g(x)) is true
Read more about domain at:
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Initial:
Changed to:
As you can see the change is: from -1 to +6.
In a equation as follow:
The k is a transformation that moves up or down the graph of the function. If k is changed to a less value the graph moves down, If k is changed to a greather number the graph moves up.
In this case the graph moves up
To find the number of units the graph moves find the difference betwwen values of k:
Answer:
(1)Equivalence Relation
(2)Not Transitive, (0,3) is missing
(3)Equivalence Relation
(4)Not symmetric and Not Transitive, (2,1) is not in the set
Step-by-step explanation:
A set is said to be an equivalence relation if it satisfies the following conditions:
(1) {(0,0), (1,1), (2,2), (3,3)}
(3) {(0,0), (1,1), (1,2), (2,1), (2,2), (3,3)}
The relations in 1 and 3 are Reflexive, Symmetric and Transitive. Therefore (1) and (3) are equivalence relation.
(2) {(0,0), (0,2), (2,0), (2,2), (2,3), (3,2), (3,3)}
In (2), (0,2) and (2,3) are in the set but (0,3) is not in the set.
Therefore, It is not transitive.
As a result, the set (2) is not an equivalence relation.
(4) {(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,2), (3,3)}
(1,2) is in the set but (2,1) is not in the set, therefore it is not symmetric
Also, (2,0) and (0,1) is in the set, but (2,1) is not, rendering the condition for transitivity invalid.
Answer:
Perimeter of quadrilateral will be
Step-by-step explanation:
Given quadrilateral is a reflection of quadrilateral
over the line
.
So, all the corresponding sides of both the quadrilateral should be equal.
Also, we can see from the diagram that
As both image are reflection of same scale we can add corresponding sides.
So, the perimeter of will be
And we have
Now, perimeter of will be
Perimeter of quadrilateral is