Answer: 
<u>Do Keep Change Flip (KCF)</u>
Keep: 
Change: ÷ into ×
Flip:
into 
<u>Multiply</u>

<u>Divide</u>

From the given table, we have that the lateral limits of f(x) as x -> 3 are different, hence the limit of f(x) does not exist at x = 3.
<h3>What is a limit?</h3>
A limit is given by the value of function f(x) as x tends to a value. For the limit to exist, the lateral limits have to be the same, as follows:

In this problem, we have that:
- To the left of x = 3, that is, for values that are less than x = 3, f(x) - > -3.
- To the right of x = 3, that is, for values that are greater than x = 3, f(x) -> 4.
Hence the lateral limits are given as follows:
Since the lateral limits are different, the limit does not exist.
More can be learned about lateral limits at brainly.com/question/26270080
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Answer:
i. terms = 5xy(z)^2 , -3zy
coefficient = 5 and -3
ii. terms = 1 , X ,x^2
coefficient = 1 , 1
iv. terms = 3 , -pq , qr, -rp
coefficient = -1 ,1 , -1
v. terms = X/2 , y/2 , -xy
coefficient= 1/2 ,1/2 and -1
The GCF of 6, 42, and 12 is:
6.
Pretty sure I can't give you test answers