Answer:
Composite numbers have more than 2 factors
4|-7-2| = 4|-9| = 4x9 = 36
Answer:
inches
Step-by-step explanation:
The "x" represents the number of buttons.
There are 2 sizes longer than 1 inch:
in and 
in
There are NO 1 1/8 inch buttons
THere are two 1 1/4 inch buttons
Hence the combined length would be:
inches
Answer:
For a function f(x), a vertical stretch of scale factor K is written as:
g(x) = k*f(x)
And for a function f(x), a vertical translation of N units is written as:
g(x) = f(x) + N
If N is positive, the translation is up
if N is negative, the translation is down.
1) We have the function:
y = f(x) = 5^(x - 2)
A vertical stretch of factor 3 will be:
y = g(x) = 3*f(x) = 3*(5)^(x - 2)
2) We have the function:
y = f(x) = 7.2^x
A translation up of 8 units will be written as:
y = g(x) = f(x) + 8 = 7.2^x + 8
Answer:
Integration of I= 
=![[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7D%20x%5E%7B%28n%2B1%29%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7Dx%5E%7B%28n%2B1%29%7D%5D)
Step-by-step explanation:
Given integral is I=
Take logx=t
I= 
I= 
Using integration by part,
![I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]](https://tex.z-dn.net/?f=I%3D%20%28t%29%5Cint%20%5Be%5E%7B%28n%2B1%29t%7D%5D%5C%2C%20dt-%5Cint%5B%5Cfrac%7Bd%7D%7Bdt%7D%7Bt%7D%5Ctimes%5Cint%20%28e%5E%7B%28n%2B1%29t%7D%29%5D%5C%5C%5C%5CI%3D%20%28t%29%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5Cint%5B1%5Ctimes%5Cfrac%7B1%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D%5C%2Cdt%5C%5C%5C%5CI%3D%5B%5Cfrac%7Bt%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29t%7D%5D)
Writing in terms of x
I=![[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bt%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29t%7D%5D)
I=![[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29logx%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29logx%7D%5D)
I=![[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7De%5E%7Blogx%5E%7B%28n%2B1%29%7D%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7Blogx%5E%7B%28n%2B1%29%7D%7D%5D)
I=
Thus,
Integration of I= =
