Sarah- German
Guadalupe- French
Faith- Spanish
I figures this out by seeing the fact that they gave which was that Sarah’s best friend takes French and it can’t be Faith because they learn a language that doesn’t start with the same letter as their name. If Sarah’s friend takes French then it can’t be Faith. What’s left is Spanish and German and Sarah and Faith. Faith can’t take French and Sarah can’t take Spanish so what’s left is Spanish for Faith and German for Sarah.
Answer: 4*p=8
Step-by-step explanation:
<em>You can use inverse operation to answer this equation.</em>
- <em>write out you problem: 4 x p =8</em>
- <em>The inverse of multiplication is division</em>
- <em>you do 4/4 which gives you one but the 4 will cancel itself out</em>
- <em>Do 8/4 which gives you 2</em>
- <em>under the equation write p = 2</em>
- <em>And be sure to line the p up with the p, the equal sign with the equal sign, and the 2 stays where it is ( which should be already lined up with the 8)</em>
Your answer would be 20 minutes.
12/3=4 grapefruits. 4*5=20
Answer:
blue line : y = 3
black horizontal line : y = 0
black vertical line : x = 0
Step-by-step explanation:
it is really simple.
blue line : no matter what value we pick for x, the functional value for y is always 3. and that defines the equation y = 3 + 0*x or simply y = 3.
the same with the black horizontal line. the difference is that y is now always 0. so, y = 0.
![\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)