Using the fundamental counting theorem, we have that:
- 648 different area codes are possible with this rule.
- There are 6,480,000,000 possible 10-digit phone numbers.
- The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.
The fundamental counting principle states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are ways to do both things.
For the area code:
- 8 options for the first digit.
- 9 options for the second and third.
Thus:

648 different area codes are possible with this rule.
For the number of 10-digit phone numbers:
- 7 digits, each with 10 options.
- 648 different area codes.
Then

There are 6,480,000,000 possible 10-digit phone numbers.
The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.
A similar problem is given at brainly.com/question/24067651
If f(1) =5-2(1)
F(2)=5-2(2)
B but I'm not as sure though
Answer:
C
Step-by-step explanation:
We have:

Where a<0.
We would like to solve for a. So, let’s divide both sides by a.
Notice that a<0. Therefore, a is negative. Hence, we must flip the sign since we are dividing by a negative. This yields:

Now, we will subtract b from both sides. Therefore, our inequality is:

Hence, our answer is C.
-3 + 3n = -6 - 6n. Expand the brackets
3n + 6n = -6 + 3. Collect like terms
9n = -3
n = -3/9 = -1/3