Set up a system of equations.
0.10d + 0.25q = 39.25
d + q = 250
Where 'd' represents the number of dimes, and 'q' represents the number of quarters.
d + q = 250
Subtract 'q' to both sides:
d = -q + 250
Plug in '-q + 250' for 'd' in the 1st equation:
0.10(-q + 250) + 0.25q = 39.25
Distribute 0.10:
-0.10q + 25 + 0.25q = 39.25
Combine like terms:
0.15q + 25 = 39.25
Subtract 25 to both sides:
0.15q = 14.25
Divide 0.15 to both sides:
q = 95
Now plug this into any of the two equations to find 'd':
d + q = 250
d + 95 = 250
Subtract 95 to both sides:
d = 155
So there are 95 quarters and 155 dimes.
Answer:
B, C, D and E. A and F are incorrect
Step-by-step explanation:
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 ( ).


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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:
.
Answer:
what it is language
Step-by-step explanation: