The access code to a house’s security system consists of eight digits. How many different codes are available if each digit can be repeated? Please show work!

Consider one of eight positions separately from all others. There are 10 possible ways to choose

a digit to place on this position since there are ten digits from 0 to 9, and this choice is

independent of all other positions since each digit can be repeated and there is no restriction on

how to choose digits.

So, we have 8 positions and each one is filled by a digit in 10 ways independently from each other,

therefore the total number of possible codes equals to ways it can be done for each independent

position multiplied altogether. This is 10 × 10 × … × 10 = 108

(100 million).

108 different codes.

Step-by-step explanation:

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363