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
The answer to the problem is going to be 3
Answer:
1.a=2
2. C x=2 and x=-3
Step-by-step explanation:
The standard form for the quadratic function is
ax^2 +bx+c
so we need to rewrite the function to be in this form
2x^2 -10 = 7x
Subtract 7x from each side
2x^2 -7x-10 = 7x-7x
2x^2 -7x-10 = 0
a =2, b= -7 c=-10
2. The quadratic formula is
-b ± sqrt(b^2 -4ac)
----------------------------
2a
2x^2 + 2x=12
Lest get the equation in proper form
2x^2 + 2x-12 = 12-12
2x^2 +2x-12 =0
a=2 b=2 c=-12
Lets substitute what we know
-2 ± sqrt(2^2 -4(2)(-12))
----------------------------
2(2)
-2 ± sqrt(4+96)
----------------------------
2(2)
-2 ± sqrt(100)
----------------------------
4
-2 ± 10
----------------------------
4
-2 + 10 -2-10
----------- and --------------
4 4
8/4 and -12/4
2 and -3
T / 3/4 ; t = 9 3/4
9 3/4 / 3/4
9 3/4 = 39/4
39/4 / 3/4
39/4 x 4/3
39 x 4 = 156
4 x 3 = 12
156/12 / 4/4 = 39/3 = 13