Answer:
Step-by-step explanation:
nth term = (n-1)th term + common difference
d = -10
a₁ = 3
a₂ = a₁ + d = 3 + (-10) = -7
a₃ = a₂ + d = -7 + (-10) = -17
a₄ = a₃ + d = -17 + (-10) = -27
a₅ =a₄ + d = -27 + (-10) = -37
a₆ = a₅ + d = -37 + (-10) = -47
First six terms: 3 , -7 , -17, -27, -37 , -47
To solve this problem follow the steps below.
Step 1:List out all of the factors for -36
-36: 1,2,3,4,6,9,12,18,36
-1,-2,-3,-4,-6,-9,-12,-18,-36
Step 2: Start going through a process of guessing and checking (basically take any two factors and add them together to see if they equal 13)
1 plus 12=13
9 plus 4=13
Step 3:Check your calculations and make sure you didn't make any errors
13-12=1 and 13-1=12
13-9=4 and 13-4=9
Answer: 1 and 12
4 and 9
I hope this helps and if any false information was given I apologize in advance.
Answer:
the question is about the table you have to make a table it is very easy
The slope of the tangent line to
at
is given by the derivative of
at that point:

Factorize the numerator:

We have
approaching -1; in particular, this means
, so that

Then

and the tangent line's equation is
