Answer:
Step-by-step explanation:
<em>Set up equations (L= Lisa; M = Maria; A = April; J = Jim)</em>
L = 12M
A = 2J
<em>Substitute x for Maria and 6x for Jim</em>
L = 12x
A = 2(6x) = 12x
<em>So, you get:</em>
Maria = x
Jim = 6x
Lisa = 12x
April = 12x
Dale drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 7 hours. when dale drove home, there was no traffic and the trip only took 5 hours. if his average rate was 18 miles per hour faster on the trip home, how far away does dale live from the mountains? do not do any rounding.
Answer:
Dale live 315 miles from the mountains
Step-by-step explanation:
Let y be the speed of Dale to the mountains
Time taken by Dale to the mountains=7 hrs
Therefore distance covered by dale to the mountain = speed × time = 7y ......eqn 1
Time taken by Dale back home = 5hours
Since it speed increased by 18 miles per hour back home it speed = y+18
So distance traveled home =speed × time = (y+18)5 ...... eqn 2
Since distance cover is same in both the eqn 1 and eqn 2.
Eqn 1 = eqn 2
7y = (y+18)5
7y = 5y + 90
7y - 5y = 90 (collection like terms)
2y = 90
Y = 45
Substitute for y in eqn 1 to get distance away from mountain
= 7y eqn 1
= 7×45
= 315 miles.
∴ Dale leave 315 miles from the mountains
Answer:
I don't know what's I don't know I'm really sorry very very very very very very very very very very very very sorry very sorry
Answer:
The value of this investment at the end of the 5 years is of $662.5.
Step-by-step explanation:
Compound interest:
The compound interest formula is given by:
Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.
Dina invests $600 for 5 years at a rate of 2% per year compound interest.
This means that 
. Thus
Calculate the value of this investment at the end of the 5 years.
This is A(5). So
The value of this investment at the end of the 5 years is of $662.5.
