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December 30, 2025, 04:06:11 pm

Author Topic: Silly Questions Thread  (Read 33104 times)  Share 

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Aqualim

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Re: Silly Questions Thread
« Reply #60 on: February 08, 2010, 11:00:21 pm »
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did you long divide it properly heh xD you shouldn't get another polynomial as the numerator.
When I put both equations into the calculator they are exactly the same, unless i'm supposed to simplify it further? if thats what your saying, haha sorry bit slow

TrueTears

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Re: Silly Questions Thread
« Reply #61 on: February 08, 2010, 11:02:11 pm »
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well yeah haha, you need to long divide them
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the.watchman

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Re: Silly Questions Thread
« Reply #62 on: February 09, 2010, 05:50:53 am »
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There is a shortcut! :o

For horizontal asymptotes, it's the top co-efficient of x over the bottom co-efficient of x (in this case, )

For vertical asymptotes, let the denominator equal 0, ,

So to divide, pull the out as 'c', and work out what is left as the numerator ('a').

There shouldn't be any x in the numerator


Moderator Action: fixed LaTeX
« Last Edit: February 09, 2010, 08:59:03 am by Mao »
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Aqualim

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Re: Silly Questions Thread
« Reply #63 on: February 23, 2010, 09:27:33 pm »
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Given that when and where , show that is not defined.

How would I go about showing the working out for that equation?

I already know by looking at it, that in order for to be undefined, as the square root cannot be a negative number..

superflya

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Re: Silly Questions Thread
« Reply #64 on: February 23, 2010, 09:29:46 pm »
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for f(g(x)) to be defined
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Re: Silly Questions Thread
« Reply #65 on: February 23, 2010, 09:33:26 pm »
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Well for to be defined







is not defined.
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Aqualim

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Re: Silly Questions Thread
« Reply #66 on: February 23, 2010, 09:44:36 pm »
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Cheers guys :)

Aqualim

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Re: Silly Questions Thread
« Reply #67 on: February 26, 2010, 05:22:18 pm »
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Let
Show that is divisible by for all k

Would that just be as simple as;



Meaning that is divisible??
« Last Edit: February 26, 2010, 05:38:57 pm by Aqualim »

Syncness

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Re: Silly Questions Thread
« Reply #68 on: February 26, 2010, 05:35:26 pm »
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Well for to be defined







is not defined.

Isn't it different when they ask you to define something, and prove if it exists?

Defining is just writing it, regardless if it exists?

Proving it exists is matching the range and domains.

Aqualim

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Re: Silly Questions Thread
« Reply #69 on: February 28, 2010, 02:21:04 pm »
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Another question;

The graph of the cubic function cuts the x-axis at the point with coordinates and touches the x-axis when . It has a y-axis of . Find the equation of the cubic function?

Ok, so I know that it has to be a repeated factor, since it touches one x-intercept whereas it cuts another, but how would I get ?

TrueTears

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Re: Silly Questions Thread
« Reply #70 on: February 28, 2010, 02:23:24 pm »
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sub in x = 0 and y = a^3b^3 to work out the other constant (q), there is another constant q because of the fundamental theorem of algebra.

y = q(x-r_1)(x-r_2)^2

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the.watchman

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Re: Silly Questions Thread
« Reply #71 on: February 28, 2010, 02:24:16 pm »
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Ok:

Let

Sub into above





So the equation is

EDIT: Beaten yet again :D
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Aqualim

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Re: Silly Questions Thread
« Reply #72 on: February 28, 2010, 02:28:40 pm »
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Ok:

Let

Sub into above





So the equation is

EDIT: Beaten yet again :D

hmm it really wasn't hard at all... how sad

the.watchman

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Re: Silly Questions Thread
« Reply #73 on: February 28, 2010, 02:29:58 pm »
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hmm it really wasn't hard at all... how sad

There's no need to say that, now you know how to do these types of questions! :)
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Aqualim

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Re: Silly Questions Thread
« Reply #74 on: February 28, 2010, 02:31:17 pm »
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hmm it really wasn't hard at all... how sad

There's no need to say that, now you know how to do these types of questions! :)
Very true, I suppose the letters just through me off, if they were numbers it would have clicked.

Thanks for your help guys :)