Answer:
17 inches.
Step-by-step explanation:
If the area of a rectangle is the product of its length and width, then;
Area of a rectangle = Length × Width
Given that the width of the poster is greater than 10 inches and is prime, this means and W>10
Area of the rectangle = 204in²
On substituting the values in the formula;
A = LW
204 = LW
Since W is greater than 10 and is prime, it can be between the prime numbers 11, 13 and 17. Note that L must be a whole number as well for any number to be the right answer we seek.
Let's test each values of the prime width that will give a length that is a whole number.
If W = 11
204 = 11×L
L = 204/11
L = 18.54
Since the length didn't give us a whole number, this means our width is not 11.
If W = 13
204 = L × 13
L = 204/13
L = 15.69
Also, we can see that the length is not also a whole number for the value of 13 as the prime width.
If W = 17
204 = L × 17
L = 204/17
L = 12
It can be seen that the length of the rectangle gave us a whole number when we used the prime width of 17, hence the width of the poster that is greater than 10 inches and is prime that makes both length and width to be a whole number is 17 inches.
Answer:
niceeee
Step-by-step explanation:
Answer:
3048 x 1.04 ^3 = 3428.585474
Answer:
For company A, y = 24x + 42
For company B, y = 28x + 25
Step-by-step explanation:
x is the number of containers
y is the total cost
For company A, y = 24x + 42
For company B, y = 28x + 25
For the cost of both companies to be the same, then
24x + 42 = 28x + 25
28x - 24x = 42 - 25
4x = 17
x = 4.25
Mr Lycan would have to order about 4 containers so the cost would be the same from each company
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.
