The limit is equivalent to the value of the derivative of

at

. (See definition of derivative)
Answer:
5,4,6,-7
Step-by-step explanation:
The y's are range
Answer:
Hope this helps!
Step-by-step explanation:
Answer:
∠B and ∠F
∠F and ∠E . . . already listed in the problem statement, so may not be considered an "other pair"
Step-by-step explanation:
∠C and ∠F are vertical angles, so are congruent. Then any angle complementary to one of those will also be complementary to the other.
Likewise, ∠B and ∠E are vertical angles and congruent. Any angle complementary to one of them will also be complementary to the other. Here, ∠E and ∠F are listed as complementary, so we know ∠B and ∠F will be also.
Given a series, the ratio test implies finding the following limit:

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

Then the limit is:

We can simplify the expressions inside the absolute value:

Since none of the terms inside the absolute value can be negative we can write this with out it:

Now let's re-writte n/(n+1):

Then the limit we have to find is:

Note that the limit of 1/n when n tends to infinite is 0 so we get:

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.