Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: Kamikaze on July 03, 2003, 01:42:21 am
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As I'm going into High School soon I was interested in what kind of math you all took in your HS years, I'm taking some "advanced algebra" course (hopefully) but I have no idea what that's supposed to be (I was thinking it might be linear algebra with matrices and vector spaces but that seems too advanced). Also I was going to study ahead so maybe I can just get into the analysis class with a placement test :p
Also did any others of you think basic Euclidean Geometry as a total bore? :p
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:wtf: how old are you?
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offhand, i'd say around 13-14 years...
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Originally posted by Knight Templar
:wtf: how old are you?
e^9! and a half, duh :rolleyes:
*mumbles about kids these days*
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Annoyingly, Kam is one of the youngest on the board, while being one of the most mature.
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You go to high-school at 13, get to choose the math you take?
oh, the weirdness of the American school system.
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Originally posted by Stunaep
You go to high-school at 13, get to choose the math you take?
oh, the weirdness of the American school system.
hmm...well sorta...
if you did good in math in middle school often times you were bumped up to the next level of math, i.e. instead of taking 8th grade math you'd take intro to algebra instead.
Now that being said, there are basically 3 variants on the math courses...retarded, normal, and advanced. So yes, if you're ahead of the game in 8th grade (in algebra or geometry), you can take the next level math in 9th grade, and pick one of the flavors I was typing about earlier.
On reflection, it is rather weird, isn't it?
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wait a minute.... from what grade does High-school begin?
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Originally posted by Stunaep
wait a minute.... from what grade does High-school begin?
9th.
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I'm taking 10th grade math in my 9th grade year (hopefully, I have to pass this test coming up).
Can anyone help me with these problems?
5x^2+2x-1=0
-c^2=3c-3
y=-14x^2-(5/9)x+2
Please?
Pretty, pretty, pretty please?
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I'm about to take Pre-Calc in 11th Grade.
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Originally posted by Unknown Target
I'm taking 10th grade math in my 9th grade year (hopefully, I have to pass this test coming up).
Can anyone help me with these problems?
5x^2+2x-1=0
-c^2=3c-3
y=-14x^2-(5/9)x+2
Please?
Pretty, pretty, pretty please?
Erm.... where's the problem? First equation is the second simplest form of square function
x=(-b�}�ãb^2-4ac)/2a
Where a is the number before x^2, b is number before x, and c is the free number, or whatever you call it in english.
Same thing for b and c. get the equation into a ax^2+bx+c=0 state, and use the the thingy above.
Back to the the school system... odd certainly. At what age do you go to school. I mean, i'm about to enter high-school (10th grade here), and I'm 15. But I skipped 1st grade, meaning most of my classmates are 16, one or two even 17.
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oh boy! I love it when you guys have problems I know how to do! :D
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Isn't that just a derivative of the quadratic equation?
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Originally posted by Unknown Target
I'm taking 10th grade math in my 9th grade year (hopefully, I have to pass this test coming up).
Can anyone help me with these problems?
Use the Quadratic formula to solve the equation
1) 5x^2+2x-1=0
2) -c^2=3c-3
Tell whether it opens up or down, find axis of symetry+graph
y=-14x^2-(5/9)x+2
Please?
Pretty, pretty, pretty please?
Srry, forgot what to say to do.
Srry if these are really stupid, but first person to answer gets a part in the next section of the HLP movie :D
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To solve quadratic equations you need to know the solution formula.
Arrange the equation like this
a,b,c are parameters for your equations.
a*x^2+b*x+c=0 - Pull out a
a*(x^2+(b/a)*x+(c/a))=0
Convert the x^2+(b/a)*x to a complete square:
(x^2+(b/a)*x+(b^2/4a^2))=(x+b/2a)^2
So x^2+(b/a)*x can be turned into the line above like this:
(x^2+(b/a)*x+(b/2a)^2) - (b/2a)^2
so it's gona be:
(x+(b/a))^2 - (b/2a)^2
(x+(b/a))^2- (b^2/4a^2)
I hope you understand what (a+b)^2 is :D
Let's put that back into the original equation:
a*[(x+(b/a))^2 - (b^2/4a^2) + (c/a)] = 0
-(b^2/4a^2)+(c/a) can be simplified:
-[(b^2-4ac)/4a^2]
Put that back into the original:
a*[(x+(b/2a))^2 - ((b^2-4ac)/4a^2)] = 0
If (b^2-4ac)<0 then we multply a positive number with a so the result can't be 0.
So (b^2-4ac)>=0 is the only case when there can be a solution to the equation.
This is also called as the second degree equation's discriminant - this determines how many solutions the equation has.
If (b^2-4ac) is bigger than 0 we can have its square root:
a*[x^2-SQR{(b^2-4ac)}/2a]
What's nice about this?
It corresponds with the a^2-b^2=(a+b)*(a-b) formula:
a*[x+(b/2a)+SQR{b^2-4ac}/2a]*[x+(b/2a)-SQR{b^2-4ac}/2a]
The result will be 0 if either of the parts are equal to 0 since its a multyplication.
a can't be 0 since then it would be a first degree equation.
So we have 2 solutions:
x1+b/2a+SQR{b^2-4ac}/2a=0
and
x2+b/2a-SQR{b^2-4ac}/2a=0
So x would be:
x1=(-b+SQR{b^2-4ac})/2a
x2=(-b-SQR{b^2-4ac})/2a
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okay, number three:
Very simple. If the x^2 is positive, the graph opens up, if negative then down. My part please.
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expanding on that:
Tell whether it opens up or down, find axis of symetry+graph
y=-14x^2-(5/9)x+2
1: Tell whether it opens up or down.
Since the square member, in this case -14x^2 is a negative value (remember, whether x is positive or negative doesn't matter, because no square can be negative), then the parabole will open down.
2. Graph: Replace x with random numbers, usually -3 to 3, calculate Y accordingly, make graph.
3.Axis of symmetry: +2. Because the free member is +2. Because a +2 always added to the equation, It doesn't depend on the value of X. So when X is 0. Y will still be 2. Or in other words, the axis of symmetry.
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Originally posted by Flaser
To solve quadratic equations you need to know the solution formula.
You know, I really need to get on ICQ with you math-knowing people. I'd love to learn the english terminology for maths.
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Originally posted by Unknown Target
Srry, forgot what to say to do.
Srry if these are really stupid, but first person to answer gets a part in the next section of the HLP movie :D
even if you're already dead? :D
*puzzles with #2*
EDIT: Duh...feel stupid now...
-c^2 = 3c - 3
0 = c^2 + 3c - 3
and this corresponds to the ax^2+bx+c, solve as per flaser's instructions :D
(does this mean I'm a zombie now?)
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The quadratic forumla we learned was -b + The square root of B^2-4AC, all divided by 2.
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that's the one.
Then there's the simplified one:
-p/2+-sqr.rt.of (p/2)Ž2-q
Which is used when the equation is in the form of
xŽ2+px+q=0
That can also be solved using the Viete theorem.
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Originally posted by Unknown Target
....
5x^2+2x-1=0
-c^2=3c-3
y=-14x^2-(5/9)x+2
These are the most basic things in the Hungarian Education System, we are learning all we have no use for in the life.
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Originally posted by TopAce
These are the most basic things in the Hungarian Education System, we are learning all we have no use for in the life.
That's why i'm proud to be an American :D :nod:
BTW, even tho I couldn't understand them Flaser + Stunaep get kudos. Thnx guys, enjoy your parts ;)
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Originally posted by Unknown Target
The quadratic forumla we learned was -b + The square root of B^2-4AC, all divided by 2.
Wrong. All divided by 2a.
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Don't make me start calculating the volume of revolution for an exponential curve containing complex polynomials between a and b!
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It's really hard to write down anything mathematical in vbulletin that's visally easy to understand as well.
I'll try to make a rtf file for you.
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I made it:
http://users.freestart.hu/szandtner/equation.rtf
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Originally posted by Levyathan
Wrong. All divided by 2a.
Oops, you're right, I thought I put that A in the post. Oh well :doubt:
My brain hurts.
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have any of you seen the analogous formulas for cubic and quartic equations? Those are some of the messiest things I have ever seen, even edging out the euler-maclaurin formula. :D
I have also been stuck on a math problem for most of today actually; anyone here good with limits?
lims®n BinC(s+k-1,s-1) [ z[/size]'(s+k) + ( y(s+k) - y(s) ) z(s+k) ]
(n is a negative integer, k is an integer between 0 and s and BinC is the binomial coef)
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Y'know, it's times like these that I feel better about the fact that I chose alcohol over study.
Also, on an unrelated note: Muwhaha. I'm gonna be relatively rich for doing sweet ****-all. Mwuahaha. All thanks to the Cyprian mafia and the dodgyness of my uncle.
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To me, Hell is a big hairy math school.
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one more thing a bout all the quadratic formulas, pretty basic, but it might be of some help.
if you can write the thing as (x+-number)(x+-number) like you can write x^2+x-12 as (x+4)(+-3), you can solve it without any formula, in this case, x is either -4 or +3.
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Originally posted by kasperl
one more thing a bout all the quadratic formulas, pretty basic, but it might be of some help.
if you can write the thing as (x+-number)(x+-number) like you can write x^2+x-12 as (x+4)(+-3), you can solve it without any formula, in this case, x is either -4 or +3.
kasperl, those are the thanks to the Viet formula:
a*x^2+b*x+c=a*(x-x1)*(x-x2)=0
Beside the original are:
x1+x2=-b/a
x1*x2=c/a
You can get both by simply doing the procedures with the quadratic equation solution formula.
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Originally posted by CP5670
lims®n BinC(s+k-1,s-1) [ z[/size]'(s+k) + ( y(s+k) - y(s) ) z(s+k) ]
(n is a negative integer, k is an integer between 0 and s and BinC is the binomial coef)
Ahh, limits scare me.. I've been trying to learn infinitesimal calculus for the past bit and I can understand them enough to find derivatives, but my mind still doesn't want to accept their existence.... :p
and the only formula related to cubic equations I could find in my math encyclopedia was Cardano's Formula... hmmm.
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Originally posted by CP5670
I have also been stuck on a math problem for most of today actually; anyone here good with limits?
lims®n BinC(s+k-1,s-1) [ z[/size]'(s+k) + ( y(s+k) - y(s) ) z(s+k) ]
(n is a negative integer, k is an integer between 0 and s and BinC is the binomial coef)
Hmm... [Deciphering] I need a better picture of this... thing. See if you can get a graphic out of MathType or Equation editor for this equation. Say, is that zeta-prime (deriv. of zeta)?
By the way, I'll give you the link to one of the greatest mathematics references around: Click Here (http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=200&Page=0) Put it in your "Favorites" menu.
To answer Kamikaze, my son will be a senior taking AP Calculus BC at his high school. He initiated a taste for mathematics back in eighth grade, similar to your case. When he entered 9th grade, he asked the math department head for a calculus textbook. He basically mastered calculus (with my help) in several months. Say, does your high school offer AP Physics C?
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Originally posted by Kamikaze
As I'm going into High School soon I was interested in what kind of math you all took in your HS years,
How good are you in math?
i'm not real good with it so i took the IAG series ( intergrated algebra/geometry). This was a three year/credit course and they teach you what you learn in Algebra and Geometry. The difference is it's three years instead the usual two. I took the first cource in summer school, the second course in my sophmore year, and a full geometry course in the following summer school also in replacement of the third IAG course.
i'll be a senior this year and i'm already finished with math so....
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the only math I remember from High School (get it? High? heh... hehehe... *cough*) was trig... mainly because it was simple.... the rest I forgot 10 minutes after writing the exam.. I'll probably never use any of it again, so why bother? trig I actually use when working with lightwave, I can actually do some of it in my head no problem (which is amazing for me, I'm terrible at doing math in my head), the rest of it went up in a cloud of smoke....
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Yeah, that is a true problem that we had been suffering with a part of science for years, and we forget everything a few month after we made the exam. These are the things you don't find interesting, and you won't use it in your own life. I absolutely hate maths, simply has no use.
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I love math, so I think I'll be taking some advanced route in high school... I'm one of those people who don't particularly care about the practical application of math. I just enjoy the profound elegance of the system.
The thing is, obviously the math they teach in school sucks... it's basically like those use-and-throw-away-cameras, you learn it, you use it (exams), you throw it in your mental trash can. Horribly wasteful... (both the cameras and school math)
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Originally posted by CP5670
have any of you seen the analogous formulas for cubic and quartic equations? Those are some of the messiest things I have ever seen, even edging out the euler-maclaurin formula. :D
I have also been stuck on a math problem for most of today actually; anyone here good with limits?
lims®n BinC(s+k-1,s-1) [ z[/size]'(s+k) + ( y(s+k) - y(s) ) z(s+k) ]
(n is a negative integer, k is an integer between 0 and s and BinC is the binomial coef)
I think I've come up with some plugged-in values that may help, although, I had to assume that k = floor(s/2) since you mentioned k is "an integer between 0 and s"
-5, 8.31616406969419E-03
-4, 1.39440426165502E-03
-3, -0.002358750031144
-2, 7.98380171900115E-03
-1, -3.04483748327652E-02
These are approximate values.
*Goes back to programming algorithms for the construction of antidifferentiable polynomials able to calculate the definite integral of existing functions through non-singular matrices* ;7