Answer:
140°
Step-by-step explanation:
155°-15° =140°
You would do an arithmetic sequence
use A1 +(n-1)d to calculate number of seats in the last row
An = 4+(24-1)5
An = 4 + (23)(5)
An = 4 + 115
An = 119 seats in the last row
now use a sum sequence to find total number of seats
Sn = n(a1 + an)/2
Sn = 24(4+119)/2
Sn = 24(123) /2
Sn = 1476 total seats
The correct answer is 10.
In order to evaluate any composite function, you need to first put the value in for the inside function. In this case f(x) is on the inside along with the number 3. So, we input 3 in for x in f(x).
f(x) = 2x + 1
f(3) = 2(3) + 1
f(3) = 6 + 1
f(3) = 7
Now that we have the value of f(3), we can stick the answer in for the outside function, which is g(x).
g(x) = (3x - 1)/2
g(7) = (3(7) - 1)/ 2
g(7) = (21 - 1)/2
g(7) = 20/2
g(f(3)) = 10
Answer:
There will be $5624.32 in the account after 3 years if the interest is compounded annually.
There will be $5630.812 in the account after 3 years if the interest is compounded semi-annually.
There will be $5634.125 in the account after 3 years if the interest is compounded quarterly.
There will be $5636.359 in the account after 3 years if the interest is compounded monthly
Step-by-step explanation:
Tamira invests $5,000 in an account
Rate of interest = 4%
Time = 3 years
Case 1:
Principal = 5000
Rate of interest = 4%
Time = 3 years
No. of compounds per year = 1
Formula :

A=5624.32
There will be $5624.32 in the account after 3 years if the interest is compounded annually.
Case 2:
Principal = 5000
Rate of interest = 4%
Time = 3 years
No. of compounds per year = 2
Formula : 

A=5630.812
There will be $5630.812 in the account after 3 years if the interest is compounded semi-annually.
Case 3:
Principal = 5000
Rate of interest = 4%
Time = 3 years
No. of compounds per year = 4
Formula : 

A=5634.125
There will be $5634.125 in the account after 3 years if the interest is compounded quarterly.
Case 4:
Principal = 5000
Rate of interest = 4%
Time = 3 years
No. of compounds per year = 4
Formula :

A=5636.359
There will be $5636.359 in the account after 3 years if the interest is compounded monthly
3a- 2b + 8b - 2 + 6a + 3c
3a + 6a -2b +8b + 3c - 2
9a + 6b + 3c - 2