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oooo Math...wait, Math? Oh No >.<


Bloodrun

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Pretty the title says it all..

 

I have some questions for you math wizards out thar..

First question:

How many children are there in a family wherein each girl has as many brothers as sisters, but each boy has twice as many sisters as brothers?

 

o.0" toughy? lets see.

 

next question:

 

Which terms must be be removed from 1/2+1/4+1/6+1/8+1/10+1/12 if the sum of the remaining terms is to equal 1?

 

o.0"" toughy? lets move on.

 

next question:

 

use inductive reaoning, to find the next term that should follow this sequence of numbers ;)

1, 8, 27, 64, 125....

 

o.0""" toughy? reguires thought, lol

also, dont ask me for the answers, i dont know em >.<

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' pid='1018447' dateline='1219592048']

1st: Five. Three girls and two boys.

 

2nd: 1/8 and 1/10 should be removed.

 

3rd:

1=1^4

8=2^3

27=3^3

64=4^4

125=5^3

196=6^3

So' date=' my guess is for 196.

[/quote']

 

wow, nice, but no cigar on the last one.

1, should be 1^3, 64, should be 4^3, and 6^3, is 216, so the answer is 216 =)

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First question is a simultaneous equation. The first statement is b = g-1, where b = boys, and g = girls. The second statement is g = 2b-1.

 

[statement 1]b = g-1

[statement 2]g = 2b-1

 

[-b] 0 = g-b-1

[x2] 0 = 2g-2b-2

[+ statement 2] g = 2g-3

[-g] 0 = g-3

[+3] 3 = g

 

So, there are three girls. Now we can adapt statement 1 accordingly.

 

[statement 1]b = g-1

b = 3-1

[=]b = 2

 

There are three girls and two boys.

 

Next question.

 

1/2+1/4+1/6+1/8+1/10+1/12

 

Convert the base terms, to make it:

 

60/120 + 30/120 + 20/120 + 15/120 + 12/120 + 10/120

 

The total of that is:

 

147/120. We need 120/120, so we need to lose 27/120. 15/120 + 12/120 = 27/120, so 15/120 and 12/120 need to be removed, and they are 1/8 and 1/10.

 

Next question.

 

1, 8, 27, 64, 125

 

--7-19-37-61

1, 8, 27, 64, 125

 

---12-18-24

--7-19-37-61

1, 8, 27, 64, 125

 

-----6-6-6-

---12-18-24

--7-19-37-61

1, 8, 27, 64, 125

 

Sub in to the formula. 6 isn't in the baseline, so you halve it, leaving 3. There are no other numeral changes, making the term n^3. As an example.

 

1^3=1

2^3=8

3^3=27

4^3=64

5^3=125

 

Which fits perfectly. There, the 6th term in a n^3 equation is 6^3 = 216. 216 is the 6th term of n.

 

That was easy.

 

EDIT: For bonus prizes, Umbra, 64=4^4 is incorrect. 4^4=256. Also, Mariojo2, the square root of Pi is an impossible question - since Pi never ends, or at least, we don't know what it's end is, we can't discover it's square root. However, to the nearest 5 decimal places, the sqrt of Pi is 1.77245.

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i thought there was a rule agenced fail pics

 

I think a rule might have been made after fail pics became a fad...but as they are no longer a fad' date=' one fail pic isn't going to destroy the board.

 

First question is a simultaneous equation. The first statement is b = g-1, where b = boys, and g = girls. The second statement is g = 2b-1.

 

[statement 1]b = g-1

[statement 2]g = 2b-1

 

[-b] 0 = g-b-1

[x2] 0 = 2g-2b-2

[+ statement 2] g = 2g-3

[-g] 0 = g-3

[+3] 3 = g

 

So, there are three girls. Now we can adapt statement 1 accordingly.

 

[statement 1]b = g-1

b = 3-1

[=]b = 2

 

There are three girls and two boys.

 

Wrong.

 

Statement 2 says g = 2(b-1), not g = 2b-1.

 

There are actually three boys and four girls.

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