I believe is C and D because
C. The original rectangle width is 2m and the length is 4m. And to make the other rectangle they are adding .5 m to both the width and length
D. The original rectangle has a width of 2m and length of 4m. To make the other rectangle they are substracting 1 m from the width and length
Answer:
It means 
also converges.
Step-by-step explanation:
The actual Series is::
The method we are going to use is comparison method:
According to comparison method, we have:
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:
if p>1 then series converges. In oue case we have:
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.
Answer:
Set up an equation
-4+x=4 Add 4 to each side
x=8
So
-4 plus eight is 4
Well, you can't really determine the price of 1 book, but you can find out the average cost.
280 - 16 = 8x (x = cost of 1 book)
264 = 8x
264/8 = 8x/8
33 = x
So the average cost of one book is 33$( A little expensive if I do say so myself)
Hope this helped!
C=2πr
r=C/(2π), now the surface area is:
A=4πr^2, using r found above we have:
A=4πC^2/(4π^2)
A=C^2/π, we are given that C=28 so:
A=28^2/π in^2
A=784/π in^2
A≈250 in^2 (to nearest square inch)
