Given:
The given function is:

The graph of the function is given.
To find:
The end behavior of the given function.
Solution:
We have,

From the given graph it is clear that the function approaches to -4 at x approaches negative infinite and the function approaches to negative infinite at x approaches infinite.
as 
as 
Therefore, the end behaviors of the given function are:
as 
as 
Answer:
itd x=68 and y=5 I apolgize if its wrong
Answer:
The 13th term is 81<em>x</em> + 59.
Step-by-step explanation:
We are given the arithmetic sequence:

And we want to find the 13th term.
Recall that for an arithmetic sequence, each subsequent term only differ by a common difference <em>d</em>. In other words:

Find the common difference by subtracting the first term from the second:

Distribute:

Combine like terms. Hence:

The common difference is (7<em>x</em> + 5).
To find the 13th term, we can write a direct formula. The direct formula for an arithmetic sequence has the form:

Where <em>a</em> is the initial term and <em>d</em> is the common difference.
The initial term is (-3<em>x</em> - 1) and the common difference is (7<em>x</em> + 5). Hence:

To find the 13th term, let <em>n</em> = 13. Hence:

Simplify:

The 13th term is 81<em>x</em> + 59.
Answer:
B is correct.
Step-by-step explanation:
Quadratic Expression (Standard Form) is:

To say it simply, the expression must have 2 as the highest degree.
This means that if there are any higher or lower degrees than 2 then they are not quadratic expression.
A choice is not quadratic expression because it has 1 as the highest degree.
B choice is correct because 2 is the highest degree.
C choice is wrong because 3 is the highest degree.
D choice is wrong because it is not a polynomial.
Therefore, B is correct.
The function "choose k from n", nCk, is defined as
nCk = n!/(k!*(n-k)!) . . . . . where "!" indicates the factorial
a) No position sensitivity.
The number of possibilities is the number of ways you can choose 5 players from a roster of 12.
12C5 = 12*11*10*9*8/(5*4*3*2*1) = 792
You can put 792 different teams on the floor.
b) 1 of 2 centers, 2 of 5 guards, 2 of 5 forwards.
The number of possibilities is the product of the number of ways, for each position, you can choose the required number of players from those capable of playing the position.
(2C1)*(5C2)*(5C2) = 2*10*10 = 200
You can put 200 different teams on the floor.