First we will find the 11th term
an = a1 + (n-1) * d
a11 = 12 + (11 - 1) * 5
a11 = 12 + 10 * 5
a11 = 12 + 50
a11 = 62
now we use the sum formula...
Sn = (n (a1 + an)) / 2
S11 = (11 (12 + 62)) / 2
S11 = (11 (74)) / 2
S11 = 814/2
S11 = 407
Answer:
14
Step-by-step explanation:
add the stuff together
I use a bit of a different looking formula.
A(t)=P(1+r/n)^nt
P=amount of money. (500)
r= rate (in decimal. 4%=0.04)
n=number of times per year (1 in this problem)
t=amount of time. (5 years)
Plugged in it looks like this:
A(t)=500 (1+ 0.04/1)^1x5
Then I put it into my calculator like this:
0.04/1+ 0.04
Then add one to the above answer:
0.04+1=1.04
Then raise the above answer to the 1x5:
1.04^5=1.2166......
Then multiply the above answer by 500:
1.2166.... x 500=608.3264512
She has $608 after 5 years.
Hope this helps, let me know if you have any questions.
Answer:
157
Step-by-step explanation:
formula- 2 x 3.14 x 5 squared
6.28 x 5 squared
6.28 x 25
157
2,500 subtracted by 372 equals 2148. 2148 dollars is how much savings he has left.
