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March 15, 2026, 01:00:32 am
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Author Topic: Andiio needs help.  (Read 3060 times)  Share 

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98.40_for_sure

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Re: Andiio needs help.
« Reply #15 on: December 01, 2010, 12:18:41 am »
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Quote from: Andiio on December 01, 2010, 12:14:21 am
Quote from: 99.95_for_sure on December 01, 2010, 12:02:48 am
Uhh... this is the answer i got?



OH you were supposed to add AX + XC like vectors? I got up to AX and CY and didn't know why the answer was combining the two.. lol thanks!!! :D

Btw what is the answer from back of book? Mine looks wrong...
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mandy

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Re: Andiio needs help.
« Reply #16 on: December 01, 2010, 12:22:31 am »
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Quote from: 99.95_for_sure on December 01, 2010, 12:17:37 am
Not vectors, they are just lengths. By definition, vectors have direction and magnitude. Haven't you seen despicable me? :P

Don't talk about Despicable Me. It's math time LOL.
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98.40_for_sure

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Re: Andiio needs help.
« Reply #17 on: December 01, 2010, 12:26:17 am »
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Quote from: mandy on December 01, 2010, 12:22:31 am
Quote from: 99.95_for_sure on December 01, 2010, 12:17:37 am
Not vectors, they are just lengths. By definition, vectors have direction and magnitude. Haven't you seen despicable me? :P

Don't talk about Despicable Me. It's math time LOL.

The villain is called 'vector' and he has a line saying he possesses direction and magnitude you gronk. If you saw it you would understand... clueless as always.
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mandy

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Re: Andiio needs help.
« Reply #18 on: December 01, 2010, 12:31:50 am »
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Quote from: 99.95_for_sure on December 01, 2010, 12:26:17 am
Quote from: mandy on December 01, 2010, 12:22:31 am
Quote from: 99.95_for_sure on December 01, 2010, 12:17:37 am
Not vectors, they are just lengths. By definition, vectors have direction and magnitude. Haven't you seen despicable me? :P

Don't talk about Despicable Me. It's math time LOL.

The villain is called 'vector' and he has a line saying he possesses direction and magnitude you gronk. If you saw it you would understand... clueless as always.

This movie has never looked so appealing. And sif "as always".. this would be the first!
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Andiio

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Re: Andiio needs help.
« Reply #19 on: December 01, 2010, 12:48:33 am »
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You get what I mean :P

That is the answer from the back of the book, but I assume it isn't +/- as lengths can only be +ve! :D
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98.40_for_sure

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Re: Andiio needs help.
« Reply #20 on: December 01, 2010, 01:09:47 am »
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Ooh sweet, the working out was long and tedious so i thought i was doing it the hard way or something
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Andiio

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Re: Andiio needs help.
« Reply #21 on: December 01, 2010, 01:32:21 am »
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Quote from: 99.95_for_sure on December 01, 2010, 01:09:47 am
Ooh sweet, the working out was long and tedious so i thought i was doing it the hard way or something

LOL yeah, I hate those kinda questions where it's all long and complex haha
Pretty interesting question though!
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dptjandra

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Re: Andiio needs help.
« Reply #22 on: December 01, 2010, 01:32:11 pm »
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Quote from: Andiio on December 01, 2010, 12:48:33 am
You get what I mean :P

That is the answer from the back of the book, but I assume it isn't +/- as lengths can only be +ve! :D

I haven't actually read the question (skipped to the second page because my attention span's so short :P)

But it seems -rt(x^2 + 9) + x + 3 is always positive, in fact.

Consider that x + 3 = rt((x+3)^2) = rt(x^2 + 9 + 6x)
rt((x^2 + 9 + 6x)) > rt(x^2 + 9) [due to the additional 6x term...I'm assuming that geometrically, x > 0]
Then -rt(x^2 + 9) + x + 3 > 0
Consequently, r > 0

I hope I haven't done something wrong!
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