Combine the like terms 4x+ 17x + 5y + 19 - 28b + 48y +87a - 24x

-3x + 53y + 19 + (-28b) + 87a

First off, you need to circle every single thing on this expression, including the math symbols. So in order to do that, circle the symbols with the number behind:

19 - 28b -> (19) (-28b)

All of your numbers should end up like this:

(4x) (17x) (5y) (19) (-28b) (48y) (87a) (-24x)

Next, place an addition symbol in between all of them.

4x + 17x + 5y + 19 + (-28b) + 48y + 87a + (-24x)

Then, rearrange the numbers with the same variable (ex. a in “87a” is the variable), and the numbers without a variable as-well.

(4x + 17x + (-24x)) + (5y + 48y) + (19) + (-28b) + 87a

Finally, add all of the variables with the same letter with each other. You also need to add the numbers without one but since 19 is the only number without a variable in this expression you don’t add it to anything.

Your answer should come out to this:

-3x + 53y + 19 + (-28b) + 87a

The answer of the above question is

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

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