Answer:
Yes
Step-by-step explanation:
6 + 8y - 3+ 4y
Add like terms
(8y+4y)+(6-3)
12y+3
Answer:
20/ 11 and if u want it in decimal it's 1.818182
Answer:
1st problem: b) 
2nd problem: c) 
Step-by-step explanation:
1st problem:
The formula/equation you want to use is:

where
t=number of years
A=amount he will owe in t years
P=principal (initial amount)
r=rate
n=number of times the interest is compounded per year t.
We are given:
P=2500
r=12%=.12
n=12 (since there are 12 months in a year and the interest is being compounded per month)

Time to clean up the inside of the ( ).


----------------------------------------------------
2nd Problem:
Compounded continuously problems use base as e.

P is still the principal
r is still the rate
t is still the number of years
A is still the amount.
You are given:
P=2500
r=12%=.12
Let's plug that information in:
.
200,000,000 that your answer