Sn=A1(1-r^n)/(1-r)
Sum of a finite geometric series formula
We know the sum, number of terms and rate, so you plug those in.
-255=A1(1-(-4)^4))/(1-(-4)
-255=A1(1-256)/5
-255=A1(-255)/5
-1275=A1(-255)
5=A1
Hope this helped! Let me know if you have any questions.
Using the Fundamental Counting Theorem, it is found that:
The 2 people can arrange themselves in 40 ways.
<h3>What is the Fundamental Counting Theorem?</h3>
It is a theorem that states that if there are n things, each with
ways to be done, each thing independent of the other, the number of ways they can be done is:

With one people in the aisle and one in the normal seats, the parameters are:
n1 = 4, n2 = 7
With both in the aisle, the parameters is:
n1 = 4, n2 = 3
Hence the number of ways is:
N = 4 x 7 + 4 x 3 = 28 + 12 = 40.
More can be learned about the Fundamental Counting Theorem at brainly.com/question/24314866
#SPJ1
13 x 6 = 78 it cant be multiplied anymore so it can go into 80 about 6 times.
![\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases} 343=7\cdot 7\cdot 7\\ \qquad 7^3\\ 36=6\cdot 6\\ \qquad 6^2\\ 256=4\cdot 4\cdot 4\cdot 4\\ \qquad 4^4 \end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}} \\\\\\ \sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4} \\\\\\ 7^2+6-4^3\implies 49+6-64\implies -9](https://tex.z-dn.net/?f=%5Cbf%20343%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B36%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-256%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cbegin%7Bcases%7D%0A343%3D7%5Ccdot%207%5Ccdot%207%5C%5C%0A%5Cqquad%207%5E3%5C%5C%0A36%3D6%5Ccdot%206%5C%5C%0A%5Cqquad%206%5E2%5C%5C%0A256%3D4%5Ccdot%204%5Ccdot%204%5Ccdot%204%5C%5C%0A%5Cqquad%204%5E4%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%20%287%5E3%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B%286%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-%284%5E4%29%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B%287%5E3%29%5E2%7D%2B%5Csqrt%5B2%5D%7B%286%5E2%29%5E1%7D-%5Csqrt%5B4%5D%7B%284%5E4%29%5E3%7D%5Cimplies%20%5Csqrt%5B3%5D%7B%287%5E2%29%5E3%7D%2B%5Csqrt%5B2%5D%7B%286%5E1%29%5E2%7D-%5Csqrt%5B4%5D%7B%284%5E3%29%5E4%7D%0A%5C%5C%5C%5C%5C%5C%0A7%5E2%2B6-4%5E3%5Cimplies%2049%2B6-64%5Cimplies%20-9)
to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.