Carlos has \dfrac45 5 4 start fraction, 4, divided by, 5, end fraction of a tank of fuel in his car. he uses \dfrac{1}{10} 10 1 start fraction, 1, divided by, 10, end fraction of a tank per day.

8 days.

Step-by-step explanation:

Consider the question: Carlos has 4/5 of a tank of fuel in his car. He uses 1/10 of a tank per day. How many days will his fuel last?

To find the number of days Carlos' fuel last, we need to divide amount of fuel in tank by amount of fuel used per day as:

Convert into multiplication problem by flipping the second fraction as:

Therefore, Carlos's fuel will last for 8 days.

the workers who do no drive has no reasonable reason to buy gasoline due to they don't have anything to use it for

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