How many 3 digit pass codes can be made from the digits 0 to 9 if the first number is not allowed to be a 0?

total 900 pass codes can be made.

Step-by-step explanation:

Given situation is 3 digit pass codes are required made from the digits 0 to 9.

But the first number is not allowed to be a 0.

So for the third place we have 10 choices ( 0,1,2,3,4,5,6,7,8,9 )

For the second place we have 10 choices ( 0,1,2,3,4,5,6,7,8,9 )

But for the first place we have 9 choices ( 1,2,3,4,5,6,7,8,9 )

( 9 × 10 × 10 ) = 900

Therefore, total 900 pass codes can be made from the digits 0 to 9.

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: