Math Problems

[Katt_knome_hobbit]

ARG! I love my calculous homework!

It's all cos and sin and tan and sec and csc and cot.

The thing that bugs me is csc=(1/sin) while sec=(1/cos) it should be the other way around.

Better yet, make sec the word for cos and cos the word for csc and csc the word for sec.

The world would be a lot simpler.

[Count Comfect]

Therefore x^n + y^n = x^2 z^(n-2) + y^2 z^(2-n)
And I think... therefore z^(n-2) must = x and must = y

Not quite. Because it's a sum - no rule says that the "x^2 z^(n-2)" must equal x^n and the "y^2 z^(n-2)" must equal the y^n. Their sums just have to be the same. Like 1+4 =2+3. 1 doesn't equal 2 or 3=4, but the sums are still the same.

[Janny]

Yeah, I knew it was somewhere.
Talk about a post to kill athread tho!
Um... problems... problems... anyone?

[Count Comfect]

Hmmm...
A warrior has a 1/5 probability of winning a large battle and a 1/3 probability of winning a small battle. If he has a choice of fighting either several small battles (campaign A) or 3 larges and a small (campaign B), how many smalls would need to be in A to make B worth it?

[Janny]

Ok.
Campaign B:
1/5 * 1 /5 * 1 /5 * 1/3
= um.. 25, 125... 375
= 1 / 375

Campaign A
375 / 3
= 125 / 3
= 41 and 2 / 3 *finds calc*
= 13.8rec
= 4.something
=1.5
0.5

Six battles? According to my calculator anyway.

[Count Comfect]

Correct. Your calculator can go next, if it wants (says my calculator).

[Janny]

Ok...

The Truel.
Mr Black in a truel with Mr White and Mr Grey.
Mr White is a perfect shot.
Mr Grey hits only twice in every three shots.
Regardless of Mr Blacks own skill, what action should he take?

[Count Comfect]

The Truel.
Mr Black in a truel with Mr White and Mr Grey.
Mr White is a perfect shot.
Mr Grey hits only twice in every three shots.
Regardless of Mr Blacks own skill, what action should he take?

Shoot Mr. White? Because he's dead in 2 if he doesn't. The other option I've heard is to miss entirely.

[Janny]

Damn you. Yes. Shoot into the air, thus making yourself the least dangerous opponent. Whoever wins out of Mr White and Mr Grey, Mr. Black will get the first shot at. From the first shot in a Truel to the first shot in a duel.

[Count Comfect]

'course, if Mr. White were not a perfect shot, no such optimal strategy presents itself, as he cannot ensure it will be a duel.
Or one of them could shoot him. If he is a perfect shot, say, then White will kill him first.

James is twice as old as John. John is 5 years older than 1/3 of how old James was 5 years ago. How old is John?

[Janny]

James = n
John = m
2m = n
Errrm... i can never do the algebra on these ones.
*looong pause*
20 and 10

[Count Comfect]

*Looong pause* Yes.
I actually did that after asking and was glad it worked...

[mithrand1r]

Originally posted by Count Comfect
'course, if Mr. White were not a perfect shot, no such optimal strategy presents itself, as he cannot ensure it will be a duel.
Or one of them could shoot him. If he is a perfect shot, say, then White will kill him first.

James is twice as old as John. John is 5 years older than 1/3 of how old James was 5 years ago. How old is John?

Today James is 4 and John is 2. (of course the answer does not make sense when applied to the real world )

James Age today = 2x
John Age today = x

(1/3[x-5])=2x-5
(1/3)x-(5/3)=2x-5 === add 5 to both sides; sub. (1/3x from both sides)
(10/3)=(5/3)x === div. both sides by (5/3)
10/3 ÷ 5/3 =x
(10/3) (3/5) = x
2=x

The only way that the answer could be James=20 and John=10 is if

James is twice as old as John. John is 5 years older than 1/3 of how old James was 5 years ago.

is changed to read

James is twice as old as John. John is 5 years older than 3 times of how old James was 5 years ago.

[Janny]

I believe I read what I wanted to in that what I did was easier, and produced higher numbers.
but good going on setting questions you hadn't worked out, County! That's what got Edexcel into trouble!

[Count Comfect]

James is twice as old as John. John is 5 years older than 1/3 of how old James was 5 years ago.

I think your algebra setup is wrong, mithrandir.
James = 2x
John = x

(2x-5) x 1/3 = x-5
2x -5 = 3x -15
10 = x
20 = 2x

you set is up as if James were half as old as John (1/3 times x-5) instead of 1/3 times (2x-5) as it should be.
Which is how 10 and 20 work.

Sorry if I was confusing

[mithrand1r]

Originally posted by Count Comfect
I think your algebra setup is wrong, mithrandir.
James = 2x
John = x

(2x-5) x 1/3 = x-5
2x -5 = 3x -15
10 = x
20 = 2x

you set is up as if James were half as old as John (1/3 times x-5) instead of 1/3 times (2x-5) as it should be.
Which is how 10 and 20 work.

Sorry if I was confusing

You are correct. I just re-read the problem and see where I made my error.

Ironically, I was able to make the problem workout with using the reciprocal of 1/3. Sometimes with enough wrong turns you still can wind up where you should.

[Count Comfect]

's ok. Very weird that it worked with the reciprocal... odd, in fact.

[Janny]

Why is that particularly weird? I bet quite a few recipricols wopuld work. 1 over 1 and 1 over 1, for example!

[Count Comfect]

The reason it is weird is that mithrand1r did two things that don't appear similar (exchanging an x and a 2x and switching 1/3 to 3) - so it seems odd that a reciprocal and a doubling/halving would work out to the same answer as the original.

[Janny]

Yeah I didn't really follow the algebra anyway, it's too hard to read on computers!