The answer to this question is 10.6
The graph continues up off the graph ( indicated by the red arrows.)
This means the domain is all real numbers written as (-∞,∞).
The lowest part of the graph is on Y=1 and the lines curve upwards, this makes the range [1,∞)
The answer is negative 1. You divide the data set into 2 parts. Find the median of each and subtract the left side from the right
The first step is to simplify the absolute value. 0.8 - 3/5 is equal to 0.8 - 0.6 which is 0.2. Now we do order of operations. Since the absolute value of 0.2 is 0.2, we can now do PEMDAS and solve. First is multiplication. 0.2 * 10 is 2. Now we have to do 5 - 3 which gives us our answer of 3, or C.
![\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k+1-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{(k+1)^2-1}}{k(k+1)}-\dfrac{\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\ ](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D%7D%7Bk%28k%2B1%29%7D-%5Cdfrac%7B%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%2B1-1%7D%7D%7Bk%28k%2B1%29%7D-%5Cdfrac%7B%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7B%28k%2B1%29%5E2-1%7D%7D%7Bk%28k%2B1%29%7D-%5Cdfrac%7B%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%0A)
![=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\dfrac{\sqrt{(k+1)^2-1}}{\sqrt{(k+1)^2}}-\dfrac{1}{k+1}\cdot\dfrac{\sqrt{k^2-1}}{\sqrt{k^2}}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{\dfrac{(k+1)&^2-1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{\dfrac{k^2-1}{k^2}}\right]=\\\\\\= \sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\dfrac{1}{(k+1)^2}}-\dfrac{1}{k+1}\cdot\sqrt{1-\dfrac{1}{k^2}}\right]=\\\\\\ ](https://tex.z-dn.net/?f=%3D%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B1%7D%7Bk%7D%5Ccdot%5Cdfrac%7B%5Csqrt%7B%28k%2B1%29%5E2-1%7D%7D%7B%5Csqrt%7B%28k%2B1%29%5E2%7D%7D-%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Ccdot%5Cdfrac%7B%5Csqrt%7Bk%5E2-1%7D%7D%7B%5Csqrt%7Bk%5E2%7D%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B1%7D%7Bk%7D%5Ccdot%5Csqrt%7B%5Cdfrac%7B%28k%2B1%29%26%5E2-1%7D%7B%28k%2B1%29%5E2%7D%7D-%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Ccdot%5Csqrt%7B%5Cdfrac%7Bk%5E2-1%7D%7Bk%5E2%7D%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B1%7D%7Bk%7D%5Ccdot%5Csqrt%7B1-%5Cdfrac%7B1%7D%7B%28k%2B1%29%5E2%7D%7D-%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Ccdot%5Csqrt%7B1-%5Cdfrac%7B1%7D%7Bk%5E2%7D%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%0A)
![=\sum\limits_{k=1}^n\arcsin\left[\dfrac{1}{k}\cdot\sqrt{1-\left(\dfrac{1}{k+1}\right)^2}-\dfrac{1}{k+1}\cdot\sqrt{1-\left(\dfrac{1}{k}\right)^2}\right]=\\\\\\= \sum\limits_{k=1}^n\left[\arcsin\left(\dfrac{1}{k}\right)-\arcsin\left(\dfrac{1}{k+1}\right)\right]=\\\\\\](https://tex.z-dn.net/?f=%3D%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B1%7D%7Bk%7D%5Ccdot%5Csqrt%7B1-%5Cleft%28%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Cright%29%5E2%7D-%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Ccdot%5Csqrt%7B1-%5Cleft%28%5Cdfrac%7B1%7D%7Bk%7D%5Cright%29%5E2%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D1%7D%5En%5Cleft%5B%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bk%7D%5Cright%29-%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bk%2B1%7D%5Cright%29%5Cright%5D%3D%5C%5C%5C%5C%5C%5C)
![=\bigg[\arcsin(1)-\arcsin\left(\frac{1}{2}\right)\bigg]+\bigg[\arcsin\left(\frac{1}{2}\right)-\arcsin\left(\frac{1}{3}\right)\bigg]+\\\\\\+ \bigg[\arcsin\left(\frac{1}{3}\right)-\arcsin\left(\frac{1}{4}\right)\bigg]+\dots+ \bigg[\arcsin\left(\frac{1}{n}\right)-\arcsin\left(\frac{1}{n+1}\right)\bigg]=\\\\\\](https://tex.z-dn.net/?f=%3D%5Cbigg%5B%5Carcsin%281%29-%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Cbigg%5D%2B%5Cbigg%5B%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29-%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B3%7D%5Cright%29%5Cbigg%5D%2B%5C%5C%5C%5C%5C%5C%2B%0A%5Cbigg%5B%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B3%7D%5Cright%29-%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7B4%7D%5Cright%29%5Cbigg%5D%2B%5Cdots%2B%0A%5Cbigg%5B%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7Bn%7D%5Cright%29-%5Carcsin%5Cleft%28%5Cfrac%7B1%7D%7Bn%2B1%7D%5Cright%29%5Cbigg%5D%3D%5C%5C%5C%5C%5C%5C)
![\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)}](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bn%2B1%7D%5Cright%29%7D)
![\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \lim\limits_{n\to\infty}\sum\limits_{k=1}^n\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\\\\\\= \lim\limits_{n\to\infty}\Bigg(\dfrac{\pi}{2}-\arcsin\left(\dfrac{1}{n+1}\right)\Bigg)=\dfrac{\pi}{2}-\lim\limits_{n\to\infty}\arcsin\left(\dfrac{1}{n+1}\right)=\\\\\\= \Bigg\{\dfrac{1}{n+1}\xrightarrow{n\to\infty}0\Bigg\}=\dfrac{\pi}{2}-\arcsin(0)=\dfrac{\pi}{2}-0=\dfrac{\pi}{2}](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bk%3D1%7D%5E%5Cinfty%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5Csum%5Climits_%7Bk%3D1%7D%5En%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5CBigg%28%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bn%2B1%7D%5Cright%29%5CBigg%29%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Clim%5Climits_%7Bn%5Cto%5Cinfty%7D%5Carcsin%5Cleft%28%5Cdfrac%7B1%7D%7Bn%2B1%7D%5Cright%29%3D%5C%5C%5C%5C%5C%5C%3D%0A%5CBigg%5C%7B%5Cdfrac%7B1%7D%7Bn%2B1%7D%5Cxrightarrow%7Bn%5Cto%5Cinfty%7D0%5CBigg%5C%7D%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Carcsin%280%29%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D-0%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D)
![\sum\limits_{k=1}^\infty\arcsin\left[\dfrac{\sqrt{k^2+2k}-\sqrt{k^2-1}}{k(k+1)}\right]=\dfrac{\pi}{2}](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bk%3D1%7D%5E%5Cinfty%5Carcsin%5Cleft%5B%5Cdfrac%7B%5Csqrt%7Bk%5E2%2B2k%7D-%5Csqrt%7Bk%5E2-1%7D%7D%7Bk%28k%2B1%29%7D%5Cright%5D%3D%5Cdfrac%7B%5Cpi%7D%7B2%7D)
