Find the area of the parallelogram with vertices ps1, 0, 2d, qs3, 3, 3d, rs7, 5, 8d, and ss5, 2, 7d.

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

hello umm i don't understand your question can you explain it more pls

step-by-step explanation:

2*2=3

two times two is four

i dont understand what you mean so pls comment

3x^3-2x^2+4x-3= -3

x^2+3x+3= 3

-3÷3= -1