Answer:
No
Step-by-step explanation:
3(r-6)+2=4(r+2)-21
3r-18+2=4r+8-21
3r-16=4r-13
-3=r
r=-3
15 + 2x - 4 = 9x + 11 - 7x
2x + 11 = 2x + 11
always true
2x + 3(4x - 1) = 2(5x + 3) + 4x
2x + 12x - 3 = 10x + 6 + 4x
14x - 3 = 14x + 6
never true
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