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Math Quiz: 9 is number one

when
a = b
multiply both sides by 8a
8a^2 = 8ab
add (a^2  9ab) to both sides
8a^2 + (a^2  9ab) = 8ab + (a^2  9ab)
simplify them to
9(a^2  ab) = (a^2  ab)
divide both sides by (a^2  ab)
9 = 1

There's a a^2  ab ≠ 0 missing somewhere, I think
(by the way, I suck at math)

bureX said:
There's a a^2  ab ≠ 0 missing somewhere, I think
(by the way, I suck at math)
That's right.
9(a^2  ab) = (a^2  ab) → 0 = 0

algorith said:bureX said:*snip*
That's right.
9(a^2  ab) = (a^2  ab) → 0 = 0
OK!
Another one,lim x>0 [{sin(9x)}/x] = ?

bureX said:
There's a a^2  ab ≠ 0 missing somewhere, I think
(by the way, I suck at math)
if a == b, then a^2 == ab and a^2ab==0. Division by zero is not permitted here You can get any answer you want if you divide by zero, just as you can "prove" any proposition if you accept a contradiction, just as any statement about members of the empty set is logically true!

brianbec said:bureX said:*snip*
if a == b, then a^2 == ab and a^2ab==0. Division by zero is not permitted here You can get any answer you want if you divide by zero, just as you can "prove" any proposition if you accept a contradiction, just as any statement about members of the empty set is logically true!
This statement is false.

iStation said:algorith said:*snip*
OK!
Another one,lim x>0 [{sin(9x)}/x] = ?
Nice one. Here's another
Find the first positive real value of t for which the following is zero
256 cos^9(pi t / 162)  576 cos^7(pi t / 162) + 432 cos^5(pi t / 162) 120 cos^3(pi t/162) + 9 cos(pi t / 162)

brianbec said:iStation said:*snip*
Nice one. Here's another
Find the first positive real value of t for which the following is zero
256 cos^9(pi t / 162)  576 cos^7(pi t / 162) + 432 cos^5(pi t / 162) 120 cos^3(pi t/162) + 9 cos(pi t / 162)
Wow!
9! Right? [27, 45, 63, 81, 99....]
How about this one?
+ ∞
∫ [{sin(9x)}/x]dx = ?
− ∞ 
@iStation:
You are dividing by 0!
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