36pi = 4/3 * pi * r^3
27 = r^3
r = 3 in
A^2-5a = 6
a^2-5a-6 = 0
Discriminant = (-5)^2-4(1)(-6) = 49
Discriminant is positive so there are two solutions
Answer:
19
Step-by-step explanation:
thats what the calculator said sheesh sheesh it was really hard
• First way to solve:
We'll manipulate the expression of the equation:

If we have y=0:

Then, the function has one real zero (x=3) and two imaginary zeros (4i and -4i).
Answer: B
• Second way to solve:
The degree of the function is 3. So, the function has 3 complex zeros.
Since the coefficients of the function are reals, the imaginary roots are in a even number (a imaginary number and its conjugated)
The function "has only one non-repeated x-intercept", then there is only one real zero.
The number of zeros is 3 and there is 1 real zero. So, there are 2 imaginary zeros.
Answer: B.