Determine whether the series [infinity] cos(n) n2 n = 1 = cos(1) 12 + cos(2) 22 + cos(3) 32 + ⋯ is convergent or divergent. SOLUTION The series has both positive and negative terms, but it is not alternating. (The first term is positive, the next three are negative, and the following three are positive: The signs change irregularly.) We can apply the Comparison Test to the series of absolute values [infinity] cos(n) n2 n = 1 = [infinity] |cos(n)| n2 n = 1 . Since |cos(n)| ≤ 1 Correct: Your answer is correct. for all n, we have |cos(n)| n2 ≤ 1 n2 . We know that Σ 1/n2 is convergent p−series with p = and therefore Σ |cos(n)|/n2 is convergent by the Comparison Test. Thus the given series Σ (cos(n))/n2 is absolutely convergent and therefore convergent by this theorem.

both equations and proportions are better.

the red lines denote the population of the bacteria after 35 hours. that would be 2,409.24. the y-intercept is the blue line which is 800 because that is the starting point. the equation used to make the graph is 800(1.032)^h, where h is the number of hours.

step-by-step explanation:

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation: