
subiedude
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Stupid math question
#11232685  10/12/09 01:52 PM (8 years, 11 days ago) 


real quick.. what is:
y^{2}+5y ?
can it be broken down any?
Edited by subiedude (10/12/09 01:53 PM)

Tedwilto
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Re: Stupid math question [Re: subiedude]
#11232699  10/12/09 01:53 PM (8 years, 11 days ago) 


y=0 or y=5
thats if y^{2}+5y=0

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Re: Stupid math question [Re: subiedude]
#11232701  10/12/09 01:54 PM (8 years, 11 days ago) 


it depends, what does it equal?
does it equal x or 5, or what?
there's way too many variables for the equation, what are the instructions? are you trying to get an answer or a factor or what?
the only thing i can think of is y(y+5)
Edited by PMasterPoo (10/12/09 01:55 PM)

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Re: Stupid math question [Re: Tedwilto]
#11232703  10/12/09 01:54 PM (8 years, 11 days ago) 


Quote:
Tedwilto said: y=0 or y=5
thats if y^{2}+5y=0
exactly.

learningtofly
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Re: Stupid math question [Re: Tedwilto]
#11232705  10/12/09 01:54 PM (8 years, 11 days ago) 


the derivative is 2y+5. does that help?


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Re: Stupid math question [Re: subiedude]
#11232708  10/12/09 01:55 PM (8 years, 11 days ago) 


That's already simplified. if you factor it its y(y+5). But I can't think of any answer to your question unless you're plugging in a value.

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DragonChaser
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Quote:
PMasterPoo said: the only thing i can think of is y(y+5)
You got it, and from that, y=0 or 5.
Unless you're in complex variables, then there are several more answers in the imaginary plane.
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learningtofly
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i mean if you really wanted to you could put i^4 everywhere. but.. thats pointless


subiedude
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solving rational equations
i gotta find the lcd with the denomonators being:
y^{2}+5y
y+5
and..
y.
so what would the lcd be
Edited by subiedude (10/12/09 02:00 PM)

learningtofly
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Re: Stupid math question [Re: subiedude]
#11232734  10/12/09 02:01 PM (8 years, 11 days ago) 


lowest common denominator?
I would leave it as y(y+5)
Thats about as far as you can go.


subiedude
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Re: Stupid math question [Re: subiedude]
#11232735  10/12/09 02:02 PM (8 years, 11 days ago) 


so lcd would be:
y(y+5), right?

subiedude
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Re: Stupid math question [Re: subiedude]
#11232737  10/12/09 02:03 PM (8 years, 11 days ago) 


ok thanks guys
i love the shroomery

learningtofly
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Re: Stupid math question [Re: subiedude]
#11232742  10/12/09 02:04 PM (8 years, 11 days ago) 


ya shroomery > math tutor. lol
i do the same thing though.


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Re: Stupid math question [Re: subiedude]
#11232774  10/12/09 02:08 PM (8 years, 11 days ago) 


Quote:
subiedude said: solving rational equations
i gotta find the lcd with the denomonators being:
y^{2}+5y
y+5
and..
y.
so what would the lcd be
wait, you mean there's three different denominators? that doesn't make sense...
then what's the numerator?
you must not be giving us all the information...

DragonChaser
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Quote:
learningtofly said: i mean if you really wanted to you could put i^4 everywhere. but.. thats pointless
Not quite that simple.
Have you taken a class on complex variables? I'm not talking about when they introduce the topic of sqrt(1) in Algebra 2, I'm talking about a college course on the complex domain.
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Re: Stupid math question [Re: subiedude]
#11232820  10/12/09 02:14 PM (8 years, 11 days ago) 


Quote:
subiedude said: solving rational equations
i gotta find the lcd with the denomonators being:
y^{2}+5y
y+5
and..
y.
so what would the lcd be
if you have all those in the denominator, you would multiply each by each other to get the LCD.
so your LCD would be (y^{2}+5y)(y+5)(y)

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Re: Stupid math question [Re: I AM SWIM]
#11232855  10/12/09 02:21 PM (8 years, 11 days ago) 


Why doesn't OP just post the whole fucking problem?
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subiedude
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3y+5 over y^{2}+5y + y+4 over y+5 = y+1 over y
solve the rational equation
Note: i dont know how to make fractions on here so i just used the word "over" as the fraction bar

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Re: Stupid math question [Re: subiedude]
#11232931  10/12/09 02:34 PM (8 years, 11 days ago) 


(mathy shit)/(more mathy shit) is how I normally do fractions.

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Re: Stupid math question [Re: subiedude]
#11233165  10/12/09 03:19 PM (8 years, 11 days ago) 


Quote:
subiedude said: 3y+5 over y^{2}+5y + y+4 over y+5 = y+1 over y
solve the rational equation
Note: i dont know how to make fractions on here so i just used the word "over" as the fraction bar
your LCD would be
(y^2+5y)(y+5)(y)
the '^ sign' implies that the y is raised to a power.
You multiply the LCD by each fraction and do the appropriate cancellation.
Eventually you will get
(3y+5)(y^2+5y) + (y+4)(y^3+5y) = (y+1)(y^3+10y^2+25y)
=
(3y^3+20y^2+25y) + (y^4+5y^2+4y^3+20y) = (y+1)(y^3+10y^2+25y)
then
y^4 + 7y^3 + 25y^2 + 45y = (y+1)(y^3+10y^2+25y)
then
(y^4 + 11y^3 +35y^2 +25y) = (y^4 + 7y^3 + 25y^2 + 45y)
so the statement is not true.
unless i did some thang wrong.
or you could move the right side to the left side and set it equal to zero.
or
(4y^3+10y^2) = 20
y = 10  y^2 ALL over 4
I could be wrong. how about you give it a try and see what you come up with.

