In a different plan for area codes, the first digit could be any number from 0 throughnbsp 5, the second digit was either 6, 7, 8, and the third digit could be any number except bsp 3nbsp bsp or 4. With this plan, how many different area codes are possible?

There are 144 different area codes possible

Step-by-step explanation:

To solve this problem, we need to know how many possibilities are for each number, and them multiply them:

For the first digit, there are 6 possibilities (0,1,2,3,4,5);

For the second digit, there are 3 possibilities (6,7,8);

For the third digit, there are 8 possibilities (0,1,2,5,6,7,8,9).

So, the number of different area codes are 6*3*8 = 144

Hey!

Alright, let's take a look at this problem.
The first step would be to rearrange the equation. To do that we will group the like elements.

Original Equation :
7x - 8 - 3x - 4

New Equation {Changed by Grouping Like Elements} :
7x - 3x - 4 - 8

Next, we'll add both the 7x and 3x since they both contain variables.

Old Equation :
7x - 3x - 4 - 8

New Equation {Changed by Adding Similar Elements} :
4x - 4 - 8

Again, we'll add the other similar elements in the equation.

Old Equation :
4x - 4 - 8

New Equation {Changed by Adding Similar Elements} :
4x - 12

Our equation can no longer be simplified any farther, so we have our final answer.

The equation 7x - 8 - 3x - 4 simplified is  4x - 12.

Hope this helps!

- Lindsey Frazier ♥