Consider the system of inequalities and its graph. y ≥

y ≤ x – 3

On a coordinate plane, 2 solid straight lines are shown. The first line has a positive slope and goes through (negative 8, negative 2) and (0, 0). Everything above the line is shaded. The second line has a positive slope and goes through (0, negative 3) and (3, 0). Everything to the right of the line is shaded. The section in quadrant 1 is labeled 3, the section in quadrant 2 is labeled 2, the section in quadrant 3 is labeled 1, and the section in quadrant 4 is labeled 4.
In which section of the graph does the actual solution to the system lie?

1
2
3
4

i do not know what this is

Step-by-step explanation:

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt