The focus is above the directrix, so the parabola opens upward—its vertical scale factor is positive (+1/12). The line of symmetry is x=-5, so the vertex form of the equation will have the factor (x -(-5))² = (x+5)². The choice that meets both these requirements is
D. f(x) = (1/12)(x + 5)² + 2
A polynomial function of least degree with integral coefficients that has the
given zeros 
Given
Given zeros are 3i, -1 and 0
complex zeros occurs in pairs. 3i is one of the zero
-3i is the other zero
So zeros are 3i, -3i, 0 and -1
Now we write the zeros in factor form
If 'a' is a zero then (x-a) is a factor
the factor form of given zeros

Now we multiply it to get the polynomial

polynomial function of least degree with integral coefficients that has the
given zeros 
Learn more : brainly.com/question/7619478
Answer:
-25.44
Step-by-step explanation:
All three series converge, so the answer is D.
The common ratios for each sequence are (I) -1/9, (II) -1/10, and (III) -1/3.
Consider a geometric sequence with the first term <em>a</em> and common ratio |<em>r</em>| < 1. Then the <em>n</em>-th partial sum (the sum of the first <em>n</em> terms) of the sequence is

Multiply both sides by <em>r</em> :

Subtract the latter sum from the first, which eliminates all but the first and last terms:

Solve for
:

Then as gets arbitrarily large, the term
will converge to 0, leaving us with

So the given series converge to
(I) -243/(1 + 1/9) = -2187/10
(II) -1.1/(1 + 1/10) = -1
(III) 27/(1 + 1/3) = 18