If it's just distribution, not solving for s (I mean I could but do you need to?):
5(2s² - 3s + 4) Distribute/multiply 5 into (2s² - 3s + 4)
(5)2s² - (5)3s + (5)4
10s² - 15s + 20
22 i think idk sorry if i’m wrong
His first payment is $100, thus a₁ = 100.
the next "term", month will be 1.1 times more than the one before, namely r = 1.1, the common ratio.
![\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=100\\ r=1.1\\ n=20 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bsum%20of%20a%20finite%20geometric%20sequence%7D%0A%5C%5C%5C%5C%0AS_n%3D%5Csum%5Climits_%7Bi%3D1%7D%5E%7Bn%7D%5C%20a_1%5Ccdot%20r%5E%7Bi-1%7D%5Cimplies%20S_n%3Da_1%5Cleft%28%20%5Ccfrac%7B1-r%5En%7D%7B1-r%7D%20%5Cright%29%5Cquad%20%0A%5Cbegin%7Bcases%7D%0An%3Dn%5E%7Bth%7D%5C%20term%5C%5C%0Aa_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%0Ar%3D%5Ctextit%7Bcommon%20ratio%7D%5C%5C%0A----------%5C%5C%0Aa_1%3D100%5C%5C%0Ar%3D1.1%5C%5C%0An%3D20%0A%5Cend%7Bcases%7D)
![\bf \sum\limits_{i=1}^{20}~100(1.1)^{i-1}\qquad \qquad\qquad \qquad S_{20}=100\left( \cfrac{1-1.1^{20}}{1-1.1} \right) \\\\\\ S_{20}=100\left( \cfrac{1-\stackrel{\approx}{6.727499949}}{-0.1} \right)\implies S_{20}\approx 100(57.27499949) \\\\\\ S_{20}\approx 5727.4999493256](https://tex.z-dn.net/?f=%5Cbf%20%5Csum%5Climits_%7Bi%3D1%7D%5E%7B20%7D~100%281.1%29%5E%7Bi-1%7D%5Cqquad%20%5Cqquad%5Cqquad%20%20%5Cqquad%20S_%7B20%7D%3D100%5Cleft%28%20%5Ccfrac%7B1-1.1%5E%7B20%7D%7D%7B1-1.1%7D%20%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0AS_%7B20%7D%3D100%5Cleft%28%20%5Ccfrac%7B1-%5Cstackrel%7B%5Capprox%7D%7B6.727499949%7D%7D%7B-0.1%7D%20%5Cright%29%5Cimplies%20S_%7B20%7D%5Capprox%20100%2857.27499949%29%0A%5C%5C%5C%5C%5C%5C%0AS_%7B20%7D%5Capprox%205727.4999493256)
is the serie divergent or convergent?
well, to make it short, when the common ratio is 0 < | r | < 1, namely a fraction between 0 and 1, only then the serie is convergent, namely it reaches a fixed value, now in this case, 1.1 is a value larger than anything between 0 and 1, so no dice.
Answer:
10
Step-by-step explanation:
f(x) =3x-x=2x,
f(7-2)=f(5)= 2*5=10
The equation is a circle centered at the origin with radius 8 (sqrt(64))
Therefore, the bounded region is just a quarter circle in the first quadrant.
Riemann Sum: ∑⁸ₓ₋₋₀(y²)Δx=∑⁸ₓ₋₋₀(64-x²)Δx
Definite Integral: ∫₀⁸(y²)dx=∫₀⁸(64-x²)dx