Heather can complete two problems in ten minutes, and when she started she had three problems done. joel can complete three problems in fifteen minutes, and when he started he had two problems done. if you include the number of problems that were complete when they started, when will they have completed between 30 and 50 problems? which inequality would you use to solve this problem, if x stands for the number of hours that have passed?

a. 30 ≤ 24x + 5 ≤ 50

b. 30 ≤ 5x + 5 ≤ 50

c. 30 < 24x + 5 < 50

d. 30 < 10x + 5 < 50

Question

Mathematics

Jazmine Schmidt

21 October, 04:23

Heather can complete two problems in ten minutes, and when she started she had three problems done. joel can complete three problems in fifteen minutes, and when he started he had two problems done. if you include the number of problems that were complete when they started, when will they have completed between 30 and 50 problems? which inequality would you use to solve this problem, if x stands for the number of hours that have passed?

a. 30 ≤ 24x + 5 ≤ 50

b. 30 ≤ 5x + 5 ≤ 50

c. 30 < 24x + 5 < 50

d. 30 < 10x + 5 < 50

+4

Answers (1)

Know the Answer?

Kayden Mclean · Answer 1

heather can do 2 problems in 10 minutes ... 60 minutes in an hr

2/10 = x / 60 ... 2 problems to 10 min = x problems to 60 min

10x = 120

x = 120/10

x = 12 ... so she can do 12 problems in 1 hr

Joel can complete 3 problems in 15 minutes ...

3/15 = x / 60

15x = 180

x = 180/15

x = 12 ... so he can do 12 problems in 1 hr

so 24 problems can be done in 1 hr

Heather started with 3 problems done and Joel started with 2 problems done ... added = 5 problems already done

and it is asking between 30 and 50 hrs ... so there is no equal sign in ur problem

ur answer is : 30 < 24x + 5 < 50 <==