Distribute the 1/4:
1/4(16x) + 1/4(20) - 7x
Do the multiplication:
1/4(16x) = 16x/4 = 4x
1/4 (20) = 20/4 = 5
Now you have:
4x + 5 - 7x = 0
Combine like terms (4x and -7x):
-3x + 5 = 0
Subtract 5 from both sides:
-3x = -5
Divide by -3:
x = -5/-3 = 5/3
Answer:
x= 5/3
Answer:
It means
also converges.
Step-by-step explanation:
The actual Series is::
![\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%5Cinf%7D%20%3D%20%5Cfrac%7B7n%5E2-4n%2B3%7D%7B12%2B2n%5E6%7D)
The method we are going to use is comparison method:
According to comparison method, we have:
![\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Binf%7Da_n%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5Csum_%7Bn%3D1%7D%5E%7Binf%7Db_n)
If series one converges, the second converges and if second diverges series, one diverges
Now Simplify the given series:
Taking"n^2"common from numerator and "n^6"from denominator.
![=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bn%5E2%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7Bn%5E6%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7Bn%5E4%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D)
![\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Binf%7Da_n%3D%5Csum_%7Bn%3D1%7D%5E%7Binf%7D%5Cfrac%7B%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7B%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5Csum_%7Bn%3D1%7D%5E%7Binf%7Db_n%3D%5Csum_%7Bn%3D1%7D%5E%7Binf%7D%20%5Cfrac%7B1%7D%7Bn%5E4%7D)
Now:
![\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Binf%7Da_n%3D%5Csum_%7Bn%3D1%7D%5E%7Binf%7D%5Cfrac%7B%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7B%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D%5C%5C%20%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%5B7-%5Cfrac%7B4%7D%7Bn%7D%2B%5Cfrac%7B3%7D%7Bn%5E2%7D%5D%7D%7B%5B%5Cfrac%7B12%7D%7Bn%5E6%7D%2B2%5D%7D%5C%5C%3D%5Cfrac%7B7-%5Cfrac%7B4%7D%7Binf%7D%2B%5Cfrac%7B3%7D%7Binf%7D%7D%7B%5Cfrac%7B12%7D%7Binf%7D%2B2%7D%5C%5C%5C%5C%3D%5Cfrac%7B7%7D%7B2%7D)
So a_n is finite, so it converges.
Similarly b_n converges according to p-test.
P-test:
General form:
![\sum_{n=1}^{inf}\frac{1}{n^p}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Binf%7D%5Cfrac%7B1%7D%7Bn%5Ep%7D)
if p>1 then series converges. In oue case we have:
![\sum_{n=1}^{inf}b_n=\frac{1}{n^4}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7Binf%7Db_n%3D%5Cfrac%7B1%7D%7Bn%5E4%7D)
p=4 >1, so b_n also converges.
According to comparison test if both series converges, the final series also converges.
It means
also converges.
-4(-5-b)=1/3(b+16) Multiply both sides by 3 to get rid of the fraction
-12(-5-b)=b+16 distribute the -12 to get rid of the parenthesis
60+12b=b+16 get the b on the left side and non b values to the right side
11b=-44 solve for b
b=-44/11 simplify the fraction
b=-4
3/5(t+18)=-3(2-t) multiply both sides by 5/3 to get rid of thefraction
t+18=-5(2-t) distribute the -5 to get rid of the parenthises
t+18=-10+5t get the t to the left side and non t values to the right
-4t=-28 solve for t
t=7
Answer:
A,D,C would be your answers
Step-by-step explanation:
Answer:
6N - 5
Step-by-step explanation:
five less than the product of six and N: <em>6N</em>
five less than the product of six and N: <em>6N </em><em>- 5</em>
Answer: 6N - 5