Answer:
the answer is 6
Step-by-step explanation:
When you compare two functions f(x) and g(x), you're looking for a special input
such that

Since you have the table with some possible candidates for
, you simply have to choose the row that gives values for f(x) and g(x) that are as close as possible (the exact solution would give the same value for f(x) and g(x), so the approximate solution will give values for f(x) and g(x) that are close to each other).
In your table, the values for f(x) and g(x) are closer when x=-0.75
F=72
g=6
------------

Therefore:


But what is e?
E=76
G=32
g=6
And:

Which means that:

If you take this value into account, you will discover that f is...

So I would have to say that the answer is approximately (c).
#7) -7,-2,0,3,4
#8) -9,-6,-5,1,8
Answer:
Answer is n^4
Step-by-step explanation:
You add the exponent when they are multiplying and have the same base