A,D and E are <em><u>not</u></em> polynomials.
Step-by-step explanation:
Well you see my friend, that's how things work.
A polynomial <u>does</u><u> </u><u>not</u> have the variable in the <em>denominator</em>.
First you would subtract 7 from each side to get -18. Then you would change the negative to a positive to make the x positive. X equals 18
Answer:
The value of k is greater than or equal to 0, i.e. k≥7.
Step-by-step explanation:
The given equation is
![k!+48=48((k+1)^m)](https://tex.z-dn.net/?f=k%21%2B48%3D48%28%28k%2B1%29%5Em%29)
The value of k must be a positive integer because k! is defined for k≥0, where k∈Z.
Subtract 48 from both the sides.
![k!=48((k+1)^m)-48](https://tex.z-dn.net/?f=k%21%3D48%28%28k%2B1%29%5Em%29-48)
![k!=48((k+1)^m-1)](https://tex.z-dn.net/?f=k%21%3D48%28%28k%2B1%29%5Em-1%29)
![k!=48(k+1-1)(\frac{(k+1)^m-1}{(k+1)-1})](https://tex.z-dn.net/?f=k%21%3D48%28k%2B1-1%29%28%5Cfrac%7B%28k%2B1%29%5Em-1%7D%7B%28k%2B1%29-1%7D%29)
Using
, we get
![k!=48k((k+1)^{m-1}+(k+1)^{m-2}+...+1)](https://tex.z-dn.net/?f=k%21%3D48k%28%28k%2B1%29%5E%7Bm-1%7D%2B%28k%2B1%29%5E%7Bm-2%7D%2B...%2B1%29)
Divide both sides by 48k.
![\frac{k!}{48k}=(k+1)^{m-1}+(k+1)^{m-2}+...+1](https://tex.z-dn.net/?f=%5Cfrac%7Bk%21%7D%7B48k%7D%3D%28k%2B1%29%5E%7Bm-1%7D%2B%28k%2B1%29%5E%7Bm-2%7D%2B...%2B1)
![\frac{k(k-1)!}{48k}=(k+1)^{m-1}+(k+1)^{m-2}+...+1](https://tex.z-dn.net/?f=%5Cfrac%7Bk%28k-1%29%21%7D%7B48k%7D%3D%28k%2B1%29%5E%7Bm-1%7D%2B%28k%2B1%29%5E%7Bm-2%7D%2B...%2B1)
![\frac{(k-1)!}{48}=(k+1)^{m-1}+(k+1)^{m-2}+...+1](https://tex.z-dn.net/?f=%5Cfrac%7B%28k-1%29%21%7D%7B48%7D%3D%28k%2B1%29%5E%7Bm-1%7D%2B%28k%2B1%29%5E%7Bm-2%7D%2B...%2B1)
Note: The value of m can be 0 or 1.
The value of k is positive integer, so the right hand side of the above equation must be a positive integer.
Since RHS of the equation is positive integer, therefore (k-1)! is completely divisible by 48.
![k-1\geq 6](https://tex.z-dn.net/?f=k-1%5Cgeq%206)
Add 1 on both sides.
![k\geq 6+1](https://tex.z-dn.net/?f=k%5Cgeq%206%2B1)
![k\geq 7](https://tex.z-dn.net/?f=k%5Cgeq%207)
Therefore the value of k is greater than or equal to 0.
Get n by itself in both inequalities
(This program doesn't allow for regular less than or greater than signs, so for the purposes of this problem I will be using less than or equal to (≤) and greater than or equal to (≥) signs)
Inequality 1:
![n-4 \geq 5 \\ n \geq 9](https://tex.z-dn.net/?f=n-4%20%20%20%5Cgeq%205%20%5C%5C%20n%20%5Cgeq%209)
Inequality 2:
![2 \geq n+3 \\ -1 \geq n](https://tex.z-dn.net/?f=2%20%5Cgeq%20n%2B3%20%5C%5C%20-1%20%5Cgeq%20n)
Now that you have both of the n's by themselves, you can put the two inequalities together, like this
![-1 \geq n \geq 9](https://tex.z-dn.net/?f=%20-1%20%5Cgeq%20n%20%5Cgeq%209)
Just keep in mind that these are supposed to be regular less than signs