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Offline Chazz

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Re: Math Questions
« Reply #30 on: May 10, 2015, 11:43:19 pm »
Here it is
before we go on though, could you explain how we do this, since I tried a few methods, but all failed
Gladly ^^

It's been so long I can't find my original solution typed up in LaTeX ;_;

Anyway, utilizing the fact that 1/cos=sec, every term now has a sec factored in. So you have
Integral -1 to 1 of x2011*sec(x) + sec(x)tan(x) + sec(x)

Using even/odd identities, that an odd function times an even function results in an odd function, and seeing that x2011 is odd while sec(x) is even, we know that their product is odd. Thus the integral from -a to a of this product will always be zero.

The second part simply evaluates to sec(x)

The third part is ln(sec(x)+tan(x)) (Not gonna prove it here, just look up integral of sec(x))

And finally evaluating this result at the bounds, we get
sec(1) + ln(sec(1)+tan(1)) - sec(-1) - ln(sec(-1)+tan(-1))

Again, even/odd tells us for even (sec), f(-x)=f(x) and for odd(tan) f(-x)=-f(x)
So this is equal to

sec(1)+ln(sec(1)+tan(1)) - sec(1) - ln(sec(1)-tan(1))

Secants cancel and you're left with the logs. Using log rules, log(a) - log(b)= log(a/b), combine the two and you end up with

ln((sec2(1)-tan2(1))/(sec-tan)2)

Top goes to -1. Flip the 2 power to -2 and bring it up. Use log rules to bring down exponent as coefficient. Combine sec(1)-tan(1) using only cos and sin into -(cos(1))/(sin(1)-1). Negate the negative with the -1 we got earlier. Using the negative from the coefficient of the log, flip the inside fraction of the log (log rules), and you get Option B

I guess it was a little more work than I promised.. haha...
« Last Edit: May 14, 2015, 08:40:24 am by Chazz »


Archive: http://pastebin.com/LTEXJMgE
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Offline Chazz

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Re: Math Questions
« Reply #31 on: May 11, 2015, 05:48:30 am »
New problemo

Since u guys r boring and don't like challenges, no big prize on this one q_q



lets make it 2 for the helluvit

« Last Edit: May 11, 2015, 06:12:12 am by Chazz »


Archive: http://pastebin.com/LTEXJMgE
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Offline Vaughn147523

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Re: Math Questions
« Reply #32 on: May 13, 2015, 07:33:23 pm »
Any attempt yet? Already got the answers :D.

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Offline Yaz

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Re: Math Questions
« Reply #33 on: May 13, 2015, 07:35:52 pm »
Any attempt yet? Already got the answers :D.
Those questions are some next level ****

Easier pls :s




Bubblebeam2 [11|Sep 01:08 am]:   Yaz is love, Yaz is Life
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Offline Range 4 Meee

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Re: Math Questions
« Reply #34 on: May 13, 2015, 07:50:38 pm »
Any attempt yet? Already got the answers :D.
Those questions are some next level ****

Easier pls :s

#Old, I know

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Offline Yaz

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Re: Math Questions
« Reply #35 on: May 13, 2015, 07:52:05 pm »
Any attempt yet? Already got the answers :D.
Those questions are some next level ****

Easier pls :s

#Old, I know
old for u. new to me 19 hhhh




Bubblebeam2 [11|Sep 01:08 am]:   Yaz is love, Yaz is Life
Drugs [12|Sep 05:08 pm]:   just.saw.yazan.on.esl.dubai
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Offline Someone12116

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Re: Math Questions
« Reply #36 on: May 13, 2015, 08:02:44 pm »
x=x

I win!

Offline Chazz

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Re: Math Questions
« Reply #37 on: May 13, 2015, 08:45:14 pm »
Any attempt yet? Already got the answers :D.
Those questions are some next level ****

Easier pls :s
hint: try to get both equalities to the form a^(something)=a^(somethingelse), and that the log base a of both sides. This should leave you with a system of equations that can easily be solved.
Another hint: this can be done by changing for example 27 to 33, and subsequently using power rules to factor the 3 into the current exponent.


Archive: http://pastebin.com/LTEXJMgE
IGN: Rican soul15

Offline Suryoyo K0

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Re: Math Questions
« Reply #38 on: May 14, 2015, 12:37:28 am »
Any attempt yet? Already got the answers :D.
Those questions are some next level ****

Easier pls :s
hint: try to get both equalities to the form a^(something)=a^(somethingelse), and that the log base a of both sides. This should leave you with a system of equations that can easily be solved.
Another hint: this can be done by changing for example 27 to 33, and subsequently using power rules to factor the 3 into the current exponent.
because that totally made it easier for him :kappa:
« Last Edit: May 14, 2015, 12:40:29 am by Suryoyo K0 »
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Offline Nany499

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Re: Math Questions
« Reply #39 on: May 14, 2015, 08:32:50 am »
Any attempt yet? Already got the answers :D.
Those questions are some next level ****

Easier pls :s
hint: try to get both equalities to the form a^(something)=a^(somethingelse), and that the log base a of both sides. This should leave you with a system of equations that can easily be solved.
Another hint: this can be done by changing for example 27 to 33, and subsequently using power rules to factor the 3 into the current exponent.

You haven't added my previous answer to the archive btw  >:( >:( >:( >:( >:(
('ll try to solve the latest questions soon-ish)
(also since I do math in german I'll have to read a bit on what constant terms of expansion are cuz languages)
edit:this took longer than expected because I am a moron who forgot how exponents work, but I hope I made no mistakes
this time:



« Last Edit: May 14, 2015, 09:03:16 am by Nany499 »
“He who has a why to live for can bear almost any how.”
― Friedrich Nietzsche
“the voice of beauty speaks softly; it creeps only into the most fully awakened souls”
― Friedrich Nietzsche

Offline Chazz

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Re: Math Questions
« Reply #40 on: May 14, 2015, 08:12:40 pm »
Any attempt yet? Already got the answers :D.
Those questions are some next level ****

Easier pls :s
hint: try to get both equalities to the form a^(something)=a^(somethingelse), and that the log base a of both sides. This should leave you with a system of equations that can easily be solved.
Another hint: this can be done by changing for example 27 to 33, and subsequently using power rules to factor the 3 into the current exponent.

You haven't added my previous answer to the archive btw  >:( >:( >:( >:( >:(
('ll try to solve the latest questions soon-ish)
(also since I do math in german I'll have to read a bit on what constant terms of expansion are cuz languages)
edit:this took longer than expected because I am a moron who forgot how exponents work, but I hope I made no mistakes
this time:

pic

Close. you error is in calculating x after you find y.


Archive: http://pastebin.com/LTEXJMgE
IGN: Rican soul15

Offline Nany499

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Re: Math Questions
« Reply #41 on: May 15, 2015, 08:44:58 am »
Yep, I always do mistakes like that, even in exams, is the answer 24? since if y=6 and x=3y+3 then x=3*6+3=21 and then 2*21-3*6=24

edit: I even fixed it in my text book



« Last Edit: May 15, 2015, 09:00:08 am by Nany499 »
“He who has a why to live for can bear almost any how.”
― Friedrich Nietzsche
“the voice of beauty speaks softly; it creeps only into the most fully awakened souls”
― Friedrich Nietzsche

Offline Pure Ranged1

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Re: Math Questions
« Reply #42 on: May 15, 2015, 10:53:05 am »
I have learned this subject 1 year ago, it took me sometime to remember how to do such stuff ;p
Nany has beaten me, but I had fun solving it anyway :)


(if there's anything not clear, ask me about it)
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Offline Chazz

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Re: Math Questions
« Reply #43 on: May 15, 2015, 04:27:27 pm »
Yep, I always do mistakes like that, even in exams, is the answer 24? since if y=6 and x=3y+3 then x=3*6+3=21 and then 2*21-3*6=24

edit: I even fixed it in my text book

pic

Congratz.
Now for the second part :^)



Archive: http://pastebin.com/LTEXJMgE
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