Write the given third order linear equation as an equivalent system of first order equations with initial values. 3y′′′+(t3−t4)y′+ycos(t)=3t4 withy(−1)=−1, y′(−1)=0, y′′(−1)=2 Use x1=y, x2=y′, and x3=y′′.

e) none of the above because the measure of angle cba is 101°

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.