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

144 codes are possible

Step-by-step explanation:

Okay for the first digit, we shall be selecting one out of 3,4,5,6.

Meaning we are selecting one out of four choices

The number of ways this can be done is 4C1 ways = 4 ways

For the second digit, we have 5,6,7 or 8, we are still selecting 1 out of 4 selections and the number of ways we can do this is also 4 ways

And lastly , we can choose any digit for the last number expect 5 , so from 0 to 9, we are removing 1 which means we are left with 9 choices

So the number of different area codes possible are ; 9 * 4 * 4 = 144 codes

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1

c) yes, because of asa or aas