The two real values that are not in the domain of the composition are x = 2 and x = -2.
<h3>
What two numbers are not in the domain of f°g?</h3>
Here we have:
f(x) = 1/x
g(x) = x^2 - 4
The composition is:
f°g = f(g(x)) = 1/(x^2 - 4)
The two values that are not in the domain are the values of x such that g(x) = 0, because we can't divide by zero.
g(x) = 0 = x^2 - 4
4 = x^2
±√4 = x
±2 = x
So g(x) = 0 when x = 2 or x = -2, so these are the two real values that are not in the domain of f°g.
If you want to learn more about domains:
brainly.com/question/1770447
#SPJ1
Answer:
140
Step-by-step explanation:
7×20=140 & 7×4=28 that is why
The question did not show the original vertices of A, B, and C.
Generally, for a triangle ABC that is dilated by a scale factor of four to form triangle A'B'C', the coordinates of vertices A', B', C' are:
For A(x, y)
The coordinate of vertex A' will be A' (4x, 4y)
For B(x, y)
The coordinate of vertex B' will be B' (4x, 4y)
For C(x, y)
The coordinate of vertec C' will be C' (4x, 4y)
Simplify the following:
(6^5/7^3)^2
7^3 = 7×7^2:
(6^5/(7×7^2))^2
7^2 = 49:
(6^5/(7×49))^2
7×49 = 343:
(6^5/343)^2
6^5 = 6×6^4 = 6 (6^2)^2:
((6 (6^2)^2)/343)^2
6^2 = 36:
((6×36^2)/343)^2
| | 3 | 6
× | | 3 | 6
| 2 | 1 | 6
1 | 0 | 8 | 0
1 | 2 | 9 | 6:
((6×1296)/343)^2
6×1296 = 7776:
(7776/343)^2
(7776/343)^2 = 7776^2/343^2:
7776^2/343^2
| | | | 7 | 7 | 7 | 6
× | | | | 7 | 7 | 7 | 6
| | | 4 | 6 | 6 | 5 | 6
| | 5 | 4 | 4 | 3 | 2 | 0
| 5 | 4 | 4 | 3 | 2 | 0 | 0
5 | 4 | 4 | 3 | 2 | 0 | 0 | 0
6 | 0 | 4 | 6 | 6 | 1 | 7 | 6:
60466176/343^2
| | | 3 | 4 | 3
× | | | 3 | 4 | 3
| | 1 | 0 | 2 | 9
| 1 | 3 | 7 | 2 | 0
1 | 0 | 2 | 9 | 0 | 0
1 | 1 | 7 | 6 | 4 | 9:
Answer: 60466176/117649
