Use the reduction formula:
∫sin^n(x) dx = − 1/n cos(x) sin^n − 1(x) + n − 1/n ∫sin^n − 2(x) dx, where n > _ 2 is an integer to determine which of the following reduction formulas is correct.
∫π/2 0 sin^n (x) dx = n+2 / n ∫π/2 0 sin^ n + 1 (x) dx
∫π/2 0 sin^n (x) dx = n+1 / n ∫π/2 0 sin^ n + 2 (x) dx
∫π/2 0 sin^n (x) dx = n-1 / n ∫π/2 0 sin^ n - 2 (x) dx
∫π/2 0 sin^n (x) dx = n-2 / n ∫π/2 0 sin^ n - 1 (x) dx

$650 per month plus utilities

step-by-step explanation:

just based on the info you gave, it would seem more cost effective to rent at $650 per month plus utilities.   if utilities are, in fact, $150 per month, your monthly living costs would be $800 as opposed to the other option of $825.

answer: list the statements and i will follow up!

step-by-step explanation:

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.