15 points!

daniel bought hamburgers and fries for his classmates. the hamburgers costed 1.49 each and a bag of fries cost 1.19. ben spent a total of 50.93. the number of hamburgers was five less than twice the amount of fries, how many hamburgers and fries did he buy

Number of hamburgers bought is 23 and number of bags of fries bought is 14

Let "h" be the number of hamburgers bought

Let "f" be the number of bags of fries bought

Cost of one hamburger = $ 1.49

Cost of one bag of fries = $ 1.19

Given that Ben spent a total of 50.93

Thus we can frame a equation as:

number of hamburgers bought x Cost of one hamburger + number of bags of fries bought x Cost of one bag of fries = $ 50.93

1.49h + 1.19f = 50.93 ------ eqn 1

Also given that the number of hamburgers was five less than twice the amount of fries

number of hamburgers bought = 2(number of fries) - 5

h = 2f - 5  ---------- eqn 2

Let us solve eqn 1 and eqn 2 to find values of "h" and "f"

Substitute eqn 2 in eqn 1

1.49(2f - 5) + 1.19f = 50.93

2.98f - 7.45 + 1.19f = 50.93

4.17f = 50.93 + 7.45

4.17f = 58.38

Substitute f = 14 in eqn 2

h = 2(14) - 5

h = 28 - 5 = 23

Summarizing the results:

number of hamburgers bought = h = 23

number of bags of fries bought = f = 14

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)