Answer:
16
Step-by-step explanation:
50 - 16 = 34
34 / 2 = 17
34 + 17 = 51, which is too much; so we take 1 souvenir out which would be 16.
Quick check:
34 + 16 = 50
Hope this helps!
We know that 3 point can creating always a plane
so from this result the right choice is A. so is true
keeping in mind that when the logarithm base is omitted, the base 10 is assumed.
![\textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \log(x)=2\implies \log_{10}(x)=2\implies 10^2=x\implies 100=x](https://tex.z-dn.net/?f=%5Ctextit%7Bexponential%20form%20of%20a%20logarithm%7D%20%5C%5C%5C%5C%20%5Clog_a%28b%29%3Dy%20%5Cqquad%20%5Cimplies%20%5Cqquad%20a%5Ey%3D%20b%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Clog%28x%29%3D2%5Cimplies%20%5Clog_%7B10%7D%28x%29%3D2%5Cimplies%2010%5E2%3Dx%5Cimplies%20100%3Dx)
The volume of the large cylinder will be twice the volume of the small cylinder.
<u>Step-by-step explanation:</u>
Let the radius of the cylinder be 'r'
Let the height of the cylinder be'h'
Volume of the small cylinder=π r² h
Volume of the large cylinder= π r² (2h)
Volume of large cylinder/volume of small cylinder= 2/1
The large cylinder volume is twice the volume of the small cylinder.
Answer:
The correct option is;
![4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )](https://tex.z-dn.net/?f=4%20%5Cleft%20%28%5Cdfrac%7B50%20%2850%2B1%29%20%282%5Ctimes%2050%2B1%29%7D%7B6%7D%20%5Cright%20%29%20%2B3%20%20%5Cleft%20%28%5Cdfrac%7B50%2851%29%20%7D%7B2%7D%20%5Cright%20%29)
Step-by-step explanation:
The given expression is presented as follows;
![\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )](https://tex.z-dn.net/?f=%5Csum%5Climits%20_%7Bn%20%3D%201%7D%5E%7B50%7Dn%5Ctimes%20%5Cleft%20%284%5Ccdot%20n%20%2B%203%20%20%5Cright%20%29)
Which can be expanded into the following form;
![\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n](https://tex.z-dn.net/?f=%5Csum%5Climits%20_%7Bn%20%3D%201%7D%5E%7B50%7D%20%5Cleft%20%284%5Ccdot%20n%5E2%20%2B%203%20%20%5Ccdot%20n%5Cright%20%29%20%3D%204%20%5Ctimes%20%5Csum%5Climits%20_%7Bn%20%3D%201%7D%5E%7B50%7D%20%5Cleft%20%20n%5E2%20%2B%203%20%20%5Ctimes%5Csum%5Climits%20_%7Bn%20%3D%201%7D%5E%7B50%7D%20%20n)
From which we have;
![\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}](https://tex.z-dn.net/?f=%5Csum%5Climits%20_%7Bk%20%3D%201%7D%5E%7Bn%7D%20%5Cleft%20%20k%5E2%20%3D%20%5Cdfrac%7Bn%20%5Ctimes%20%28n%2B1%29%20%5Ctimes%282n%2B1%29%7D%7B6%7D)
![\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}](https://tex.z-dn.net/?f=%5Csum%5Climits%20_%7Bk%20%3D%201%7D%5E%7Bn%7D%20%5Cleft%20%20k%20%3D%20%5Cdfrac%7Bn%20%5Ctimes%20%28n%2B1%29%20%7D%7B2%7D)
Therefore, substituting the value of n = 50 we have;
![\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}](https://tex.z-dn.net/?f=%5Csum%5Climits%20_%7Bn%20%3D%201%7D%5E%7B50%7D%20%5Cleft%20%20k%5E2%20%3D%20%5Cdfrac%7B50%20%5Ctimes%20%2850%2B1%29%20%5Ctimes%282%5Ccdot%2050%2B1%29%7D%7B6%7D)
![\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}](https://tex.z-dn.net/?f=%5Csum%5Climits%20_%7Bk%20%3D%201%7D%5E%7B50%7D%20%5Cleft%20%20k%20%3D%20%5Cdfrac%7B50%20%5Ctimes%20%2850%2B1%29%20%7D%7B2%7D)
Which gives;
![4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}](https://tex.z-dn.net/?f=4%20%5Ctimes%20%5Csum%5Climits%20_%7Bn%20%3D%201%7D%5E%7B50%7D%20%5Cleft%20%20n%5E2%20%3D%20%204%20%5Ctimes%20%5Cdfrac%7Bn%20%5Ctimes%20%28n%2B1%29%20%5Ctimes%282n%2B1%29%7D%7B6%7D%20%3D%204%20%5Ctimes%20%5Cdfrac%7B50%20%5Ctimes%20%2850%2B1%29%20%5Ctimes%282%20%5Ctimes%2050%2B1%29%7D%7B6%7D)
![3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}](https://tex.z-dn.net/?f=3%20%20%5Ctimes%5Csum%5Climits%20_%7Bn%20%3D%201%7D%5E%7B50%7D%20%20n%20%3D%203%20%20%5Ctimes%20%5Cdfrac%7Bn%20%5Ctimes%20%28n%2B1%29%20%7D%7B2%7D%20%3D%203%20%20%5Ctimes%20%5Cdfrac%7B50%20%5Ctimes%20%2851%29%20%7D%7B2%7D)
![\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}](https://tex.z-dn.net/?f=%5Csum%5Climits%20_%7Bn%20%3D%201%7D%5E%7B50%7Dn%5Ctimes%20%5Cleft%20%284%5Ccdot%20n%20%2B%203%20%20%5Cright%20%29%20%3D%204%20%5Ctimes%20%5Cdfrac%7B50%20%5Ctimes%20%2850%2B1%29%20%5Ctimes%282%5Ctimes%2050%2B1%29%7D%7B6%7D%20%2B3%20%20%5Ctimes%20%5Cdfrac%7B50%20%5Ctimes%20%2851%29%20%7D%7B2%7D)
Therefore, we have;
.