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zmey [24]
3 years ago
15

Sorry guys but I don't understand please help me

Mathematics
2 answers:
son4ous [18]3 years ago
8 0
SO 5 GOES IN 15 3 TIMES AND SINCE THEY = ITS THE SAME WITH 3
Elan Coil [88]3 years ago
6 0
Look at the denominators. The second fraction's denominator is 3 times the first fraction's denominator. Since the fractions are equal, it means you have to multiply the first fraction's numerator by 3 too. So, b=9. Good luck!
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What is the area of a triangle with a base of five units and a height of 15 units
Leviafan [203]

Answer:

37.5

Step-by-step explanation:

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4 0
3 years ago
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>

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3 years ago
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If we're supposed to put them in order then:
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4 years ago
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I am Lyosha [343]

Answer: b

Step-by-step explanation: step  step

5 0
3 years ago
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