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san4es73 [151]
3 years ago
13

ABCD is a quadilatoral, the prove AB+BC+CD+DA is grater than AC+BD? ​

Mathematics
1 answer:
konstantin123 [22]3 years ago
4 0

Answer:

yes it's greater then anything

Step-by-step explanation:

really just look it up or ask someone

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Group together the like terms (plus 2x with 2x) (subtract 10 by 6)

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Move plus four to the other side as a negative

4x=32

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Step-by-step explanation:

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How many area codes (abc) would be possible if all three digits could be any vaiue 1-9
kifflom [539]

Answer:

<u>The total number of area codes that would be possible if all three digits could be any value from 1 to 9 is 729.</u>

Step-by-step explanation:

1. Let's start our reasoning to get this answer:

From 000 to 999, we have 1,000 possible area codes.

But for this question, we should find out the answer without considering the digit 0, therefore we must exclude:

A. From 000 to 099, that are 100 possible area codes.

B. We also have to exclude, from 100 to 110, 120, 130, 140, 150, 160, 170, 180 and 190. Those are 19 area codes more.

C. We also have to exclude these same options of code areas, for 200 - 210 and 2x0, and 19 more for each subset that starts with 3, 4, 5, 6, 7, 8 and 9. In total, 19 multiplied by 8, 152 numbers more.

<u>Therefore, the total number of area codes that would be possible if all three digits could be any value from 1 to 9 is:</u>

<u>1,000 - 100 - 19 - 152 = 729</u>

<u>The simplest way to find it, is:</u>

<u>For 000 to 999 = 1,000 or 10³</u>

<u>Excluding the digit zero = 729 or 9³</u>

3 0
3 years ago
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