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March 17, 2026, 12:07:37 am

Author Topic: what does this mean...  (Read 1966 times)  Share 

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bar0029

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what does this mean...
« on: October 30, 2010, 06:14:27 pm »
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q1 mc from exam 2 2010 of tssm

what does the highlighted sentence mean ...

basically the question was asking to solve a system of linear equationd










 
« Last Edit: October 30, 2010, 06:32:42 pm by bar0029 »
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kenhung123

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Re: what does this mean...
« Reply #1 on: October 30, 2010, 06:37:01 pm »
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Its telling you how to interpret the calculator. The calculator may give you another variable, but the question requires in term of p, so just change the variable to p

TrueTears

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Re: what does this mean...
« Reply #2 on: October 30, 2010, 06:40:21 pm »
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A parameter is another "variable" which describes your variables. So here, you would choose some values of p, maybe pick integer values, say 1, 2, 3... etc
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bar0029

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Re: what does this mean...
« Reply #3 on: October 30, 2010, 06:41:05 pm »
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it doesnt give me an answer as a variable, it gives me exact constants ....

and also, if i have a question says, draw this graph for a pdf, and the variable for the pdf is t, for time, then when i draw it, along the x-axis do i draw y=0 along the entire axis or greater or equal to zero... because it is time even though it says f(t)=0 elsewhere???

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TrueTears

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Re: what does this mean...
« Reply #4 on: October 30, 2010, 06:47:11 pm »
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when it says elsewhere it means ANYWHERE besides the domain for f(t). So sketch y = 0 for the part of the complement of the domain :)
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TrueTears

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Re: what does this mean...
« Reply #5 on: October 30, 2010, 06:47:55 pm »
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it doesnt give me an answer as a variable, it gives me exact constants ....

and also, if i have a question says, draw this graph for a pdf, and the variable for the pdf is t, for time, then when i draw it, along the x-axis do i draw y=0 along the entire axis or greater or equal to zero... because it is time even though it says f(t)=0 elsewhere???


No the parameter p IS a variable, after substitution for some values of p obviously yields exact values...
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bar0029

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Re: what does this mean...
« Reply #6 on: October 30, 2010, 06:49:17 pm »
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noooo noo noo, there are NO values given in terms of a parameter, p, i get exact values, like 13/5 .. there is no p in my answer.....

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TrueTears

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Re: what does this mean...
« Reply #7 on: October 30, 2010, 06:51:21 pm »
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lol i know... p is a variable that describes YOUR variables x, y, z (that's the definition of a parameter) after you substitute some values in for p you will get exact values.

limme give you an example

say we got x = 2-p y = p z = 2/p

now if we set p = 1 we get x = 1, y = 1, z = 2.

See the set of solutions (1,1,2) is described by the parameter p = 1 :)



You may ask "what's the point of having p in there? Why not just state x, y, z" this is because in some higher levels of maths, often parameters can describe curves which normal functions cannot (or will be ridiculously hard).
« Last Edit: October 30, 2010, 06:53:12 pm by TrueTears »
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bar0029

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Re: what does this mean...
« Reply #8 on: October 30, 2010, 06:57:38 pm »
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oh m gee, i am so confused..

okay when i type it into my cas, to solve, am i supposed to get values in terms of p or no :S

im so sorry, im a drainer lol
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kenhung123

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Re: what does this mean...
« Reply #9 on: October 30, 2010, 07:21:09 pm »
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Are you using ti89? You should get something with n, that's the parameter in calc. The question is in terms of p. p=n.

bar0029

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Re: what does this mean...
« Reply #10 on: October 30, 2010, 07:59:56 pm »
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im using ti-inspire.. and i dont get anything i terms of p or n ...
is it like when you solve for a gen solution for a trig function...
cause im not getting that

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itolduso

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Re: what does this mean...
« Reply #11 on: October 30, 2010, 08:10:49 pm »
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A parameter is not a variable. It is a constant in the considered case.

bar0029

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Re: what does this mean...
« Reply #12 on: October 30, 2010, 08:12:43 pm »
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...
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itolduso

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Re: what does this mean...
« Reply #13 on: October 30, 2010, 08:23:10 pm »
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look up a dictionary

bar0029

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Re: what does this mean...
« Reply #14 on: October 30, 2010, 08:27:28 pm »
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lol whatever, i'm getting nowhere with this, catch
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