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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+7266 
+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
 #1
avatar+7266 
+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+7266 
+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

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