+0  
 
0
91
3
avatar

is there a formula to divde x by itself y times.

 

eg.

x=3

y=4

 

3/3/3/3

 

P.S. This is not my homework, it's for a complex dice roller program

Guest Jun 23, 2017

Best Answer 

 #1
avatar+4172 
+3

Maybe this:     \(\frac{x}{x^y}\qquad\text{which is equal to}\qquad x^{1-y}\)

 

Here,   " 1 "   is the number of    x's    that you want in the numerator, and  " y "   is the number of    x's    that you want in the denominator.

 

So for example, if  y = 2   it becomes   x/(x*x)  =   ((x / x) / x)   =  x / x / x

 

..........

This doesn't work on your example though.

If you want your example to work, just replace   " y "  with  " (y - 1) "

hectictar  Jun 23, 2017
Sort: 

3+0 Answers

 #1
avatar+4172 
+3
Best Answer

Maybe this:     \(\frac{x}{x^y}\qquad\text{which is equal to}\qquad x^{1-y}\)

 

Here,   " 1 "   is the number of    x's    that you want in the numerator, and  " y "   is the number of    x's    that you want in the denominator.

 

So for example, if  y = 2   it becomes   x/(x*x)  =   ((x / x) / x)   =  x / x / x

 

..........

This doesn't work on your example though.

If you want your example to work, just replace   " y "  with  " (y - 1) "

hectictar  Jun 23, 2017
 #2
avatar
0

I solved my question !!!!!

 

I thought it over in my head cuz' 3/3/3/3 = 1/9

Then I realized it's either 1/x^(sqrt(y)) or 1/x^(y/2) or 1/x^(y-2)

The 3rd one is correct. The calculator simplified it to x^(2-y)

Guest Jun 27, 2017
 #3
avatar+4172 
+1

Also......1/9   =  1 / (32)

 

And.....x^(2 - y) is what you get if you replace  " y "  with  " y - 1 "  in:     \(x^{1-y}\)

 

\(x^{1-(y-1)}=x^{1-y+1}=x^{2-y}\)          smiley

hectictar  Jun 27, 2017

12 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details