If 2 + 5i is a zero, then by the complex conjugate root theorem, we must have its conjugate as a zero to have a polynomial containing real coefficients. Therefore, the zeros are -3, 2 + 5i, and 2 - 5i. We have three zeros so this is a degree 3 polynomial (n = 3).
f(x) has the equation
f(x) = (x+3)(x - (2 + 5i))(x - (2 - 5i))
If we expand this polynomial out, we get the simplest standard form
f(x) = x^3-x^2+17x+87
Therefore the answer to this question is f(x) = x^3-x^2+17x+87
8. $298 4. 250 9. 28 6. 8.7 5. 465
Answer: 13% or 0.1314
Step-by-step explanation: Binomial PDF on the calculator or by hand its
22 nCr 15 (0.60)^15(1-0.6)^7 = 0.131378
Answer:
12
Step-by-step explanation:
Answer:
Step-by-step explanation:
To write this equation with h isolated, we cancel everything else on the right hand side of the equals sign. This means we cancel π and r².
Since both of these are attached to h by multiplication, we divide by both to cancel. We can divide by both at the same time:
