The denarius was a unit of currency in ancient rome. suppose it costs the roman government 1 0 1010 denarius per day to support 3 33 legionaries and 3 .33 archers. it only costs 3.33 denarius per day to support one legionary and one archer. use a system of linear equations in two variables. can we solve for a unique cost for each soldier?

assuming you copied and pasted this from khan academy and the original numbers are 10, 3, 3 and 3, make two equations:

x+y=3 and 3x+3y=10

x+y=3 -> solve for y -> y=-x+3

plug into 3x+3y=10

3x+3(-x+3)=10

3x-3x+9=10 -> the 3x's cancel out

9=10

however 9 does not equal 10 and therefore, there is no solution

answer: its c because when it has a magnet next to it it will move away from the mixture. hope this

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