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November 01, 2025, 03:47:01 pm

Author Topic: My 1/2 question thread  (Read 3934 times)  Share 

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WhoTookMyUsername

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My 1/2 question thread
« on: June 19, 2011, 10:14:20 am »
0
can anyone help me out with interpreting this question? I just can't get my head around it :(

A man has to travel 50km in 4 hours. he does it by walking the first 7 km at km/h, cycling the next 7 km at 4x km/h and motoring the remainder at km/h.

Show that

Thanks!

EDIT: Solved :|

It was very easy =.=
« Last Edit: June 19, 2011, 12:05:58 pm by Bazza16 »

WhoTookMyUsername

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Re: My 1/2 question thread
« Reply #1 on: June 19, 2011, 12:08:17 pm »
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Find the value of


why...? (question over post)


its a sum to infinity sequence , of the rule

my first thought was to times the whole thing by 9, in order to get a nice
and then divide by 9 later on

this brought the answer

But why does treating as

and dividing by not get the same answer?

which way is right?

EDIT - how the *%ck do i use latex?
« Last Edit: June 19, 2011, 01:11:36 pm by Bazza16 »

b^3

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Re: My 1/2 question thread
« Reply #2 on: June 19, 2011, 12:32:55 pm »
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EDIT - how the *%ck do i use latex?
LaTeX help thread -> http://vce.atarnotes.com/forum/index.php?topic=3137.0
LaTeX generator - >http://www.codecogs.com/latex/eqneditor.php the link was on the thread but anyway, credit to Stroodle
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WhoTookMyUsername

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Re: My 1/2 question thread
« Reply #3 on: June 19, 2011, 12:48:31 pm »
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Ah the beauty!

WhoTookMyUsername

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Re: My 1/2 question thread
« Reply #4 on: June 19, 2011, 01:22:08 pm »
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edited second question, is now completely different


tearing hair out, please prevent me going bald by responding ASAP :D
« Last Edit: June 19, 2011, 01:24:00 pm by Bazza16 »

Truck

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Re: My 1/2 question thread
« Reply #5 on: June 19, 2011, 01:25:47 pm »
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edited second question, is now completely different


tearing hair out, please prevent me going bald by responding ASAP :D
9^1/3 x 9^1/9 x 9^1/27
=9^(1/3 + 1/9 + 1/27)
Sum to infinity of powers; a=1/3, r=1/3.
1/3 / 1 - 1/3 = 1/2 (skipped working out here).
=> 9^1/2 = 3 = answer.
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gossamer

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Re: My 1/2 question thread
« Reply #6 on: June 19, 2011, 01:29:56 pm »
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Find the value of


why...? (question over post)


its a sum to infinity sequence , of the rule

my first thought was to times the whole thing by 9, in order to get a nice
and then divide by 9 later on

this brought the answer

But why does treating as

and dividing by not get the same answer?

which way is right?

EDIT - how the *%ck do i use latex?

and now you use the rule (since it is sum, not product). Did it not work out?

WhoTookMyUsername

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Re: My 1/2 question thread
« Reply #7 on: June 19, 2011, 01:42:38 pm »
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i mean, if i times the whole thing by 9, then sub it in and divide the answer by 9, why is this incorrect? (gives 1.5 which is too small)

EDIT: i have a very small brain capacity

scrap everything
« Last Edit: June 19, 2011, 01:46:19 pm by Bazza16 »

Mao

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Re: My 1/2 question thread
« Reply #8 on: June 19, 2011, 11:16:53 pm »
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It is not a geometric series. If it was, it would look like

This series can be represented by

If this series is infinite (as you have written it), then it is divergent (to positive infinity). We know this because for any positive n. Thus , the latter sum is divergent (to positive infinity), therefore the first one must be divergent.
« Last Edit: June 19, 2011, 11:23:38 pm by Mao »
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WhoTookMyUsername

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Re: My 1/2 question thread
« Reply #9 on: July 31, 2011, 01:56:56 pm »
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can someone prove that
i just can't do it :'(

secx - sinx = 

also
how to i copy paste an image or upload a word document?

pi

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Re: My 1/2 question thread
« Reply #10 on: July 31, 2011, 02:23:02 pm »
+2
I'll do it a really cheap way, but I like it  ;)

So we have the identities:






No, we look at the LHS of your equation:


   (using the above mix of identities)



QED.
« Last Edit: July 31, 2011, 02:30:00 pm by Rohitpi »

pi

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Re: My 1/2 question thread
« Reply #11 on: July 31, 2011, 02:33:27 pm »
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I know uploads haven't come though yet, but this sheet I typed up last year is gold for this stuff (attached)

WhoTookMyUsername

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Re: My 1/2 question thread
« Reply #12 on: July 31, 2011, 02:38:25 pm »
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I'll do it a really cheap way, but I like it  ;)

So we have the identities:






No, we look at the LHS of your equation:


   (using the above mix of identities)



QED.

thanks a lot rohitpi :D

any chance someone could do it the longisher way so i can improve my basic techniuqe (i'm a noob)?:D + how did you even see that :O
I figure i should have been able to do it just with the things i've been taught, more about developing technique, and i'm lost as f^!$

« Last Edit: July 31, 2011, 03:23:58 pm by Bazza16 »

pi

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Re: My 1/2 question thread
« Reply #13 on: July 31, 2011, 04:07:13 pm »
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thanks a lot rohitpi :D

...how did you even see that :O

As for this question, I'm used to instantly going for pythagorean trig identities whenever I see a in a "show" or "prove" question. So I saw where the top part was coming from. Secondly, another good technique to try is if there is a reciprocal of the original function present in it's simplified form, multiply both sides by that reciprocal (in the form of 1) straight away, it means that I don't have to worry about the denominator.

As for a "long" way, I'm not really sure :(

xZero

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Re: My 1/2 question thread
« Reply #14 on: July 31, 2011, 04:15:35 pm »
+2
firstly

     











LHS=RHS

as required

Edit:wtf i can tweet, share, google+ and like my AN post? :D
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