Mike can walk 3.9 mph and Leslie can walk 1.8 mph. If they started 12 miles apart and they walked toward each other, about how long before they would meet? SHOW WORK!
a) 2 hrs.
b) 3 hrs.
c) 4 hrs.
d) 5 hrs.
Thanks SO much!
-Andrea
2007-01-25 11:12:21 · 6 answers · asked by Andrea 1 in Education & Reference ➔ Homework Help
Let x = distance Mike walked, then 12-x = distance Leslie walked.
Remember that d = rt
For Mike
x = 3.9t
For Leslie
12-x = 1.8t
You need to solve the above two equations.
Substitution can be used.
Replace x wtih 3.9t in the second equation.
12 - 3.9t = 1.8t
Add 3.9t to each side
12 = 5.7t
2.105 = t
So about 2 hours
2007-01-25 11:19:26 · answer #1 · answered by MsMath 7 · 2⤊ 0⤋
oh k, i have never been so great at this but here goes
um i think that each person has to walk 6 miles each to meet at the 12 miles
3.9 rounds to 4
4divideby =1.5
1.8rounds to 2 6 divide by 2 =3
total time 3 hours.
but dont take my word for it lol
i am sorry
2007-01-25 11:25:05 · answer #2 · answered by Anonymous · 0⤊ 0⤋
3.9x +1.8x = 12
where x = time in hours
5.7x = 12 miles
2.105 = x
so the answer would be A. (closest possible answer)
2007-01-25 11:18:22 · answer #3 · answered by westdyk1 2 · 0⤊ 0⤋
add up their speeds together since they are walking towards each other.
3.9 + 1.8 = 5.7
12/5.7 = just over 2 hours
2007-01-25 11:18:26 · answer #4 · answered by David W 3 · 1⤊ 0⤋
3.9x +1.8x = 5.7
x = time in hours
5.7x/12 miles
2.105 = x
so... A: 2 hours
2007-01-25 11:22:21 · answer #5 · answered by Jessica T 2 · 0⤊ 0⤋
Since they are heading toward each other, you can think of it as a combined speed.
3.9x + 1.8x = 12
5.7x = 12
x = 2.1(ish)
2007-01-25 11:18:28 · answer #6 · answered by mirramai 3 · 1⤊ 0⤋