.7 repeating would make it 7/9
We have been given a quadratic function 
and we need to restrict the domain such that it becomes a one to one function.
We know that vertex of this quadratic function occurs at (5,2).
Further, we know that range of this function is 
.
If we restrict the domain of this function to either ![(-\infty,5]](https://tex.z-dn.net/?f=%28-%5Cinfty%2C5%5D)
or 
, it will become one to one function.
Let us know find its inverse.
Upon interchanging x and y, we get:
Let us now solve this function for y.
Hence, the inverse function would be 
if we restrict the domain of original function to and the inverse function would be 
if we restrict the domain to .
Answer:
Algebraically, f is even if and only if f(-x) = f(x) for all x in the domain of f. A function f is odd if the graph of f is symmetric with respect to the origin. Algebraically, f is odd if and only if f(-x) = -f(x) for all x in the domain of f.
Step-by-step explanation:
Answer:
1/8
Step-by-step explanation:
Answer:
The 6th term of the sequence is 25
Step-by-step explanation:
first term = a = 5
common difference = d = 4
The nth term of a sequence is
T_n = a + (n - 1)d
Therefore, the 6th term of the sequence is
T_6 = a + (6 - 1)d
T_6 = a + 5d
T_6 = 5 + 5(4)
T_6 = 5 + 20
T_6 = 25
