<span>Martin deposits $200
in a savings account that earns 5% annual interest.
year interest balance
1 200 * 5% 200(1.05)
2 200(1.05) * 5% 200(1.05)^2
3 200(1.05)^2*5% 200(1.05)^3
y 200(1.05)^y
=> m = 200 (1.05)^y
four years later,
cary deposits $200 in an account earning the same interest.
</span>
<span><span>year interest balance
5 200 * 5% 200(1.05)
6 200(1.05) * 5% 200(1.05)^2
7 200(1.05)^2*5% 200(1.05)^3
y 200(1.05)^(y-4)
=> c = 200(1.05)^ (y-4)
</span>
Answer:
Martin: 200(1.05)^y
Cary: 200(1.05)^(y–4)</span>
Answer:
5
Step-by-step explanation:
f(1) means what is the value when x=1. y is at 5.
The answer is 19.11 hope this helps
Number of tickets: T.
Number of customers: c
Initially the number of tickets is T0=150, when the group hasn't sold any tickets (c=0). Then the graph must begin with c=0 and T=150. Point=(0,150). Possible options: Graph above to the right and graph below to the left.
They sell the tickets in pack of three tickets per customer c, then each time they sell a pack of three tickets to a customer, the number of tickets is reduced by 3 (-3c). Then the number of tickets, T, the group has left after selling tickets to c customers is:
T=150-3c→T=-3c+150
For T=0→-3c+150=0→150=3c→150/3=c→c=50. The graph must finish with c=50, T=0. Final point=(c,T)=(50,0)
Answer:
The correct graph is above to the right, beginning on vertical axis with T=150 and finishing on horizontal axis with c=50.
The correct equation is T=-3c+150
Answer:
55 days
Step-by-step explanation:
Given
Jim ran 15 miles in 5 days
no. of miles ran in 5 days = 15 miles
dividing LHS and RHS by 5
no. of miles ran in 5/5(=1) days = 15/5 miles = 3 miles
no. of miles ran in 1 day = 3 miles
let the no. of days taken to run 165 miles be x ----A
No of miles ran in x days = x*no. of miles ran in 1 day = 3x miles
thus, From A
3x = 165
x = 165/3 = 55
Thus, it took 55 days for JIM to run 165 miles
