The solutions to f(x) = 64 is x = 7 and x = –7.
Solution:
Given data:
– – – – (1)
– – – – (2)
To find the solutions to f(x) = 64.
Equate equation (1) and (2), we get

Subtract 15 from both sides of the equation.



Taking square root on both sides of the equation, we get
x = ±7
The solutions to f(x) = 64 is x = 7 and x = –7.
The answer to this question is FALSE.
Domain is the set of all the numbers that we can input to the function or that can be used in place of x. The numbers which make the function undefined are excluded from the domain.
In the given exponential function, there is no any value of x which will make the function undefined, so the domain of the function if set of All real numbers. In general, domain of exponential functions is Set of All real numbers.
Answer:

Step-by-step explanation:
,
,
, Subtract 23 and 3x from both sides and simplify:
, Divide both sides by 2 and simplify: 
<em>Hope this helps!!!</em>
Step-by-step explanation:
Find the Center and Radius (x-4)^2+y^2=4
(
x
−
4
)
2
+
y
2
=
4
This is the form of a circle. Use this form to determine the center and radius of the circle.
(
x
−
h
)
2
+
(
y
−
k
)
2
=
r
2
Match the values in this circle to those of the standard form. The variable
r
represents the radius of the circle,
h
represents the x-offset from the origin, and
k
represents the y-offset from origin.
r
=
2
h
=
4
k
=
0
The center of the circle is found at
(
h
,
k
)
.
Center:
(
4