ince the problem is only asking for 4 years, we can just calculated it out year by year. Recall the formula for compounding interest: A = P(1+r)n, where A is the total amount, P is the principle (amount you start with), r is the interest rate per period of time, and n is the number of periods (in this case, r is annual interest rate, so n is number of years). At the beginning (Year 0), Lou starts off with 10000: A = 10000 At the end of Year 1, Lou earned interest on that amount, plus he has deposited another 5000: A = 10000(1.08) + 5000 End of Year 2, Lou's interest from the year 0 amount has compounded, he has started earning interest on the amount deposited last year, and he deposits another 5000: A = 10000(1.08)2 + 5000(1.08) + 5000 End of Year 3, same idea. Lou has earned compounding interest on all existing deposits, and deposits another 5000: A = 10000(1.08)3 + 5000(1.08)2 + 5000(1.08) + 5000 End of Year 4, same idea: A = 10000(1.08)4 + 5000(1.08)3 + 5000(1.08)2 + 5000(1.08) + 5000 = 36135.45
Answer:
A
Step-by-step explanation:
Using the cosine ratio in the right triangle
cosA =
=
=
, thus
A =
(
) ≈ 48° → A
Answer:
The answer to your question is letter C
Step-by-step explanation:
This is a problem with the inverse rule of three because we expect that more machines can complete the same work in fewer days.
6 machines ------------------ 12 days
x machines ------------------- 8 days
x machines = 

x = 4 machines
The answer is 345 students.
I did reverse math(i don't remember the name)
138/.4=345
Now check it:
345*.4=138
I hope I helped somehow!