X = jane, y = jasmine, z = jocelyn
x + y + z = 56
x = 3z
y = 2z + 2
3z + 2z + 2 + z = 56
6z + 2 = 56
6z = 56 - 2
6z = 54
z = 54/6
z = 9 <==== jocelyn
x = 3z
x = 3(9)
x = 27 <=== jane
y = 2z + 2
y = 2(9) + 2
y = 18 + 2
y = 20 <=== jasmine
Answer:
<em>1500(1.02)^x + 600x</em> is how much he has in savings at the end of x years where it be in the bank or elsewhere
Step-by-step explanation:
x is in years
Let's just think about the investment of 1500 in an account earning 2% per year.
Before the years even start, you are at 1500 ( present value).
The next year (year 1), it would be 1500*.02+1500=(1500)(1.02).
The next year (year 2), it would be 1500(1.02)(.02)+1500(1.02)=1500(1.02)(1.02).
We keep multiplying factors of (1.02) each time.
So for year x, you would have saved 1500(1.02)^x.
Now we are saving 50 cash per month. Per year this would be 12(50) since there are 12 months in a year. 12(50)=600.
So the first year you would have 600.
The second year you would have 600(2) or 1200.
The third year you would have 600(3) or 1800.
Let's put this together:
1500(1.02)^x + 600x
Answer:
1 just solid one i hope this helps you!!!
