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xxMikexx [17]
2 years ago
15

How many phone numbers (without area codes) are possible? the first digit cannot be 0 or 1

Mathematics
1 answer:
masha68 [24]2 years ago
4 0

Answer:

Each three-digit area code may contain up to 7,919,900 unique phone numbers:

NXX may begin only with the digits [2–9], providing a base of 8 million numbers: ( 8 x 100 x 10000 ) .

However, the last two digits of NXX cannot both be 1, to avoid confusion with the N11 codes (subtract 80,000).

Despite the widespread usage of NXX "555" for fictional telephone numbers — see 555 (telephone number) — today, the only such numbers specifically reserved for fictional use are "555-0100" through "555-0199", with the remaining "555" numbers released for actual assignment as information numbers (subtract 100).

In individual geographic area codes, several other NXX prefixes are generally not assigned: the home area code(s), adjacent domestic area codes and overlays, area codes reserved for future relief nearby, industry testing codes (generally NXX 958 and 959) and special service codes (such as NXX 950 and 976). Subtract for 911 411etc emergency and informational numbers

Step-by-step explanation:

Hope this helps! Have a nice day!

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16 hours

Step-by-step explanation:

Multiply the 4/5 by 4 to get the denominator as 20. So you do 4 x 4 which equals 16 and 5 x 4 which equals 20 so you get 16/20 miles or 16 hours.

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Question Progress<br> Homework Progress<br> 4/<br> Work out the size of angle x.<br> 105<br> 140
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Answer:

Step-by-step explanation:

Sum of angles of a triangle = 180

x + (180 - 105) + (180 -140) = 180

x + 75 + 40 = 180

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Answer:

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Please show your work and explain it.
Maru [420]

Answer:

f(x)=\dfrac{x+2}{2(x-2)}

Step-by-step explanation:

Remember when you divide fractions, you need to get the reciprocal of the divisor and multiply. So your first simplification would be:

\dfrac{x^2+4x+4}{x^2-6x+8}\div\dfrac{6x+12}{3x-12}\\\\=\dfrac{x^2+4x+4}{x^2-6x+8}\times\dfrac{3x-12}{6x+12}\\\\=\dfrac{(x^2+4x+4)(3x-12)}{(x^2-6x+8)(6x+12)}

Next we factor what we can so we can further simplify the rest of the equation:

=\dfrac{(x^2+4x+4)(3x-12)}{(x^2-6x+8)(6x+12)}\\\\=\dfrac{(x+2)(x+2)(3x-12)}{(x^2-6x+8)(6(x+2))}\\\\

We can now cancel out (x+2)

=\dfrac{(x+2)(3x-12)}{(x^2-6x+8)(6)}

Next we factor out even more:

=\dfrac{(x+2)(3)(x-4)}{(x-2)(x-4)(6)}

We cancel out x-4 and reduce the 3 and 6 into simpler terms:

=\dfrac{(x+2)(1)}{(x-2)(2)}

And we can now simplify it to:

=\dfrac{x+2}{2(x-2)}

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