Need Math Help?

[Druggedfox]

Lol this is a kewl thread. There are many kewl things like 1= 0 And i havent read the whole thread but if this hasnt been mentioned then:

1/9 = .111111
2/9 = .222222
3/9 = 1/3 = .333333
etc. etc... etc....
So y does :

9/9 = 1 not .99999?!?!?!
ITs a math mindgame lol. And this is a cool thread because when school starts high schoolers can come to this thread (including me though i have a 100 in math XP)

 

[AltF4]

You Are Ugly:

Did your math teacher make these problems up by hand? Hah, because your work looks fine, but you're right that it doesn't really make sense in terms of the problem. Good catch on that.

What Doraki said is true, but not usually an issue they make Geometry students worry about. Basically he said that any angle A is coterminal with angle A + 360. (Or minus 360) Makes sense right? If you spin around in a full circle, you end up where you started.

It should be noted that you can't say that angle A is equal to angle A+360 or equivalent, or any such terms that imply that they are the same. They are not. They only happen to be coterminal, meaning that they wind up in the same place. In many cases, this difference is not important and you can just treat them like equals. But there certainly is a difference between turning a bottlecap 0 degrees and turning it 360 degrees, right!

DruggedFox:

There's a whole thread dedicated to that. Short story 1 does equal .9999~.

Also, any number divided by 9 ius equal to itself repeating. But sometimes it gets crowded. 11/9 looks like this.

Put the number 11 in the "tenths" place. A one carries over to the "ones" place, the other one stays put. The next 11 goes in the "hundredths" place, a one carries over to the "tenths" making it a two, then the one stays put in the "hundredths". This process repeats and you get:

1.222222~

You can do it for any number, There's lots of odd number things.

In case anyone doesn't know, there's an easy way to tell if a number is divisible by three. Just add all the digits in the number together until you get a single digit number. If it's a 3 6 or 9, it's divisible by three!

Example: 26,184. Is this number divisible by three?

2+6+1+8+4 = 21
2+1 = 3 <----Good! It is divisible by three.

 

[plasmawisp6633]

Lol this is a kewl thread. There are many kewl things like 1= 0 And i havent read the whole thread but if this hasnt been mentioned then:

1/9 = .111111
2/9 = .222222
3/9 = 1/3 = .333333
etc. etc... etc....
So y does :

9/9 = 1 not .99999?!?!?!
ITs a math mindgame lol. And this is a cool thread because when school starts high schoolers can come to this thread (including me though i have a 100 in math XP)

ALTF4 explained that very well. Just keep in mind, that numbers can be divided infinitely into infinitely smaller pieces. If we didn't round to the next number when we had to, then we would have never even got to the next number in the first place.

Numbers are a B***h

 

[xyouxarexuglyx2]

Yeah. Me and the kid next to me had the same question. The teacher said he made them up by hand and it didn't matter if the answers didn't make sense, and that the point of the assignment was just to make sure that we knew the postulates and how to set up the equations.

btw, AltF4, do you know any good place or source I can go to in order to find summer classes for Algebra 3-4/Algebra 2 so I can take Pre Calc in Sophomore year?

 

[plasmawisp6633]

Yeah. Me and the kid next to me had the same question. The teacher said he made them up by hand and it didn't matter if the answers didn't make sense, and that the point of the assignment was just to make sure that we knew the postulates and how to set up the equations.

btw, AltF4, do you know any good place or source I can go to in order to find summer classes for Algebra 3-4/Algebra 2 so I can take Pre Calc in Sophomore year?

Dude, you got high goals. I'm only going to be a Junior and I'm taking pre-calc. If you took it your junior year, you can take calc in your senior year. Why would you want to finish math a year early?

 

[psicicle]

oh cool a math thread. My math background: high school sophmore going to be junior, finished AP calculus but never took geometry or a formal algebra 2 course.

For some reason I was placed into pre-call as a freshman but I was getting Bs because I didn't learn the stuff like logs and trigonometry before. I ended up in AB precal because of this so I need (well want) to learn stuff about matrices and vectors which I know I missed.

So, I taught myself how to add and multiply matrices. How do you divide and subtract?

I know how to add vectors. Im not so sure how to subtract, and I kind of know how to multiply but not really. Also, what notation is used for them?

Next: Can you give a proof for the divisible by three thing? WHat about being divisible by nine and so on?

Next: Can you explain the proof of the millenium math thing that was solved? Well actually probably nobody here can do that since it is a pretty hard problem.

Also, can anybody explain some of the practical uses of complex numbers and where they arise in real life? Maybe a little bit of a refresher on their properties?

I am going into a multivariable calc course next year, has anyone taken that? How is it?

 

[Druggedfox]

Lol i was just bringing about something i thought was kewl. I hate Southern states (im in GA) because there math programs suck. My middle school considered it "advanced" to do Algebra I in 8th grade. because of them i wasted a year or two of my life, and know in my sophmore year im stuck taking Geometry, then in my Junior year trig... It sux for me and because i enjoy math a lot so it ruins so much. I considered trying to test out of Geometry (no ive never taken it but no enough to probably pass the test lol) so what do u think... should i have tested out of it?

Peace

 

[xyouxarexuglyx2]

Dude, you got high goals. I'm only going to be a Junior and I'm taking pre-calc. If you took it your junior year, you can take calc in your senior year. Why would you want to finish math a year early?

idk. ****s and giggles?

 

[Doraki]

oh cool a math thread. My math background: high school sophmore going to be junior, finished AP calculus but never took geometry or a formal algebra 2 course.

For some reason I was placed into pre-call as a freshman but I was getting Bs because I didn't learn the stuff like logs and trigonometry before. I ended up in AB precal because of this so I need (well want) to learn stuff about matrices and vectors which I know I missed.

So, I taught myself how to add and multiply matrices. How do you divide and subtract?

substracting is always adding the opposite.
The opposite of a vector and of a matrix is simply when you replace every number by its opposite. Substracting is as easy as adding there.

As for division, you can only divide with some square matrixes.
(division wouldn't make sense with non square matrixes).
it's multiplying by the inverse of the matrix.
So if your matrixe is A, you have to find some B matrix so that A * B = I
(I = diagonal matrix with 1s on the diagonal, also neutral for the multiplication)
If you can solve linear systems, you should be able to figure out a way to compute that B, when it is possible.
After all, solving a linear system is solving A*X = Y, thus finding X = A ^ -1 * Y.
(A = square matrix ; X,Y = vectors).
Of course, there are matrixes that are non-invertible.

Next: Can you give a proof for the divisible by three thing? WHat about being divisible by nine and so on?

have you seen congruence ?
why that proof works is because 10 = 1 + 9 so 10 = 1 mod 9.
A number x is dividible by n if x = 0 mod n.
When you add the digits of number, you're only computing another value of that number mod 9 :
take 234

234 = 100*2 + 10*3 + 1*4
= 99*2 + 2 + 9*3 + 3 + 4
= 2+3+4 + 9*(11*2+3)
= 2+3+4 mod 9
= 9 mod 9
= 0 mod 9.

It's the same with 3 because 10 = 3*3+1 = 1 mod 3.

Also, can anybody explain some of the practical uses of complex numbers and where they arise in real life? Maybe a little bit of a refresher on their properties?

They are incredibly useful in electronics. Otherwise I don't think I know practical uses

 

[AltF4]

I'll write up a good description on how to do matrix and vector operations. (A vector is a maxtrix by the way) It's kinda late right now....

Complex Number Usages:

Complex Numbers are often taught in algebra only as an introduction to vectors. Truth be told, you won't see them again in any non-contrived way until much further down the line. Complex numbers are used widely sciences like fluid dynamics and modern physics.

 

[psicicle]

Well, Im planning to go into modern physics later on in life, so yeah, I'm wondering where in nature they appear.

I kind of understand the proof for being divisible by three, but how would you work out being divisible by 11? I've seen methods to do that although I forgot them.

About subtracting matrices, what do you mean by the opposite of the number?

About dividing matrices, is it basically trying to get the inverse of the matrix?

And I'm also curious about how matrices and vectors are the same.

thanks math guys!

 

[AltF4]

-A Vector looks like this < x, y> Which is essentially just a matrix with 1 row and 2 columns right?

-Subtracting Matrices: It works just like you think it would...
[ -1 4] [3 2]
[2 -5] - [-2 4]

=

[-4 2
4 -9]

You just subtract each element rather than adding them!

-Matrix Division: There is no such thing as matrix division! So then, how can you solve for Y in this equation? (y and c are real numbers, A is a matrix. x is the cross product symbol, not a variable. I is the identity matrix)

y * A = c

You have to "divide by A" right? Well, instead, you just do the cross product of A inverse. It's kind of like how dividing by two is the same thing as multiplying by one half.

(A^-1) x (y * A) = c * (A^-1)
y * (A^-1) X A = c * (A^-1)
y * I = c * (A^-1)
y = c * (A^-1)

 

[psicicle]

oh I see, so you find the inverse and multiply both sides by it and then the matrix becomes 1 and the other side is whatever, thanks.

Also I'm not familiar with vector notation

 

[Doraki]

I kind of understand the proof for being divisible by three, but how would you work out being divisible by 11? I've seen methods to do that although I forgot them.

It also works with congruence
with 3, you have
10 = 1 mod 3 ; 100 = 1 mod 3 ; etc etc, so you can replace 10,100,1000 etc by 1, thus adding the digits work

with 11, you have
10 = (-1) mod 11 ; 100 = 10^2 = (-1)^2 = 1 mod 11 ; and so on.
so basically, 10^n = (-1)^n
Then you use that in your number, for example

23456 = 2*1 + 3*(-1) + 4*1 + 5*(-1) + 6*1 mod 11
= (2+4+6) - (3+5) mod 11
= 4 mod 11

So for divisibility by 11, you have to compare the sum of the digits in odd places with the sum of the digits in even places.

The same kind of things exist for every number, of course.
But it won't be as simple as this.
With 3 and 9 you always have 10^n = 1
With 11 you have 10^n = (-1)^n
But for say, 17, you'll get complicated weights all over the place.

You have to "divide by A" right? Well, instead, you just do the cross product of A inverse. It's kind of like how dividing by two is the same thing as multiplying by one half.

(A^-1) x (y * A) = c * (A^-1)
y * (A^-1) X A = c * (A^-1)
y * I = c * (A^-1)
y = c * (A^-1)

Multiplication isn't commutative, the 1st line is wrong.
And why do you use x and * to note the product ? that's confusing.

psicicle, how did you learn to multiply matrixes ?

 

[AltF4]

Matrix multiplication isn't commutative, but the y is a scalar. You can move a scalar around. If y were a Matrix, it would be wrong.

 

[Doraki]

so you just put a horrible equation where a scalar times a matrix is supposed to equal another scalar ??
c and y are supposed to be vectors there or else it doesn't make sense.

edit : isn't cross product an operation between (n-1) vectors of size n, that results in an nth vector or size n ?? why are you mentioning it ? =/

 

[psicicle]

wait, since it isn't commutative, when you want to multiply both sides of an equation by a matrix how would you go about doing that?

I learned it from my parents in prep for the math SAT 2, but I didn't have much time so Im revising over it

 

[Doraki]

you just multiply it to the right way
when you solve a linear equations system (n equations with n variables), it's exactly like finding a solution vector X so that you have an equation
A X = Y ; where A is the square matrix of all the coefficients in the equations and Y is the vector of the right-hand terms of the equations.

If you find the inverse of the matrix A, you multipliy with it to the left, and you get
A^-1 A X = A^-1 Y, thus X = A^-1 Y.
So you get the solution X from multiplying the inverse of A with the Y vector.

If you have to solve the system
2x+y=5
3x-2y=2

then the matrix A is

[2 1]
[3 (-2)]

Y is the vector

[5]
[2]

and X is simply

[x]
[y]

You can check that finding a vector X so that A X = Y is the same as finding x and y so that those equations hold.

In my example it just happens that A * A = 7 * I (you can check this)
So you get 7*X = A * Y = [12 11]
and then you find 7x = 12 and 7y = 11 (and you can check that the solution is right).

 

[FauxShaux]

Hey guys.

I'm in Algebra 2, and since its just the beginning of the year, this should be easy stuff for you guys, considering all the stuff I've seen you talking about previous to this post. I normally get this stuff really quick, but my teacher is worthless this year, and expects us to figure out everything on our own.

Anyways, my question is how do I figure out the domain and range of an equation like (x-4)/(x+2)

How do I know the lines in which each of the curves never touches?

And a stupid question that I just forgot the answer to: does the parent parabola function have unlimited domain?

 

[GamerGuitarist7]

Dan are you good at physics too? It's pretty math oriented. but i have this physics extra credit assignment due on tuesday and i thought maybe you'd be able to help me with it. please PM me if you read this becuase i don't check this thread too often. see you at the Brawl tourney in Dec (unless tucson has another tourney before then)

 

[AltF4]

Faux:

Think of a function as a black box. You put stuff into the box, it does something to what you put in, and you get something out.

Now, you can't go and just put ANYTHING into every function. There are restrictions sometimes. The list of stuff that you're allowed to put into the function is called the "Domain".

Similarly, not every function can give you ANYTHING back out. Some can only give you back a certain list of things. The list of stuff that a function could potentially give you is called the "Range".

So for the function: f(x) = (x-4)/(x+2) Try to answer the questions: "What am I allowed to replace x with?" and "For every number I'm allowed to use, what numbers will I get back?"

The first question (finding the domain) is usually pretty easy. Just look for things you know you're not allowed to do in math. Like dividing by zero. You're not allowed to divide by zero, which means you can't replace x with -2. So that means your domain is all real numbers, except negative two.

Then the range is a little trickier. I think at the Algebra 2 level nowadays, they just have you graph it and just sort of look and see. And when you do, you'll notice that the function never reaches 1. So it winds up being all reals except 1.

 

[shiva39]

Hey guys.

I'm in Algebra 2, and since its just the beginning of the year, this should be easy stuff for you guys, considering all the stuff I've seen you talking about previous to this post. I normally get this stuff really quick, but my teacher is worthless this year, and expects us to figure out everything on our own.

Anyways, my question is how do I figure out the domain and range of an equation like (x-4)/(x+2)

How do I know the lines in which each of the curves never touches?

And a stupid question that I just forgot the answer to: does the parent parabola function have unlimited domain?

EDIT. Hmm. I guess that turned out slower and later than I had thought. GG with the much quicker explanation...er, AltF4.

All right, I'll randomly interject.

f(x)=(x-4)/(x+2) is a rational function. Now, the "lines the curve never touches", the asymptotes, can be determined by finding the zeros of the denominator--the values of x that make the denominator zero. In this case,

x+2=0
x+2-2=0-2
x=-2

So, at x=-2, the denominator is zero, and a vertical asymptote exists there.

In some cases, there can be multiple vertical asymptotes. Take, for example,
Here, you must factor the denominator:

x^2-9=0

(x+3)(x-3)=0 At this point, we can ignore one binomial when zeroing the other, since 0*anything =0.
So, x+3=0
x-3=0
x=3, -3.

At both x equals 3 and negative 3, there is a vertical asymptote.

Horizontal asymptotes can be determined by comparing the leading terms of the numerator and denominator--the terms with the highest exponent power. In the above example, a power of 1 (1^1) is compared to 2 (x^2). Since the denominator is growing faster than the numerator, the function will approach 0 in the long run.

In the first example, with x as the highest power in both the numerator and denominator, the ratio is 1--both are growing equal quickly. Thus, the asymptote is y=1. If the numerator was 2x, then the asymptote would be y=2, since the numerator would be growing twice as fast as the denominator (essentially).

Well, there's a rough explanation. Someone should really clean up, though, as this is really messy. As for the pretty print:

http://prettyprint.free.fr/

French. I'm sure everyone can manage.
​

 

[psicicle]

Hey guys.

I'm in Algebra 2, and since its just the beginning of the year, this should be easy stuff for you guys, considering all the stuff I've seen you talking about previous to this post. I normally get this stuff really quick, but my teacher is worthless this year, and expects us to figure out everything on our own.

Anyways, my question is how do I figure out the domain and range of an equation like (x-4)/(x+2)

How do I know the lines in which each of the curves never touches?

And a stupid question that I just forgot the answer to: does the parent parabola function have unlimited domain?

TO figure out the domain and range of (x-4)/(x+2):

Domain is determined by the bottom part which in this case is (x+2). You know that x can't equal -2 because then the bottom is 0 and you can't divide by 0. For all other values of x there seems to be a real number, so in this case the domain would be x != -2.

The range would be trickier here. You know that as x approaches -2 from the right the whole thing goes to minus infinity because values like -1 give you -5/4 and as you get closer it becomes more negative. From the left, the whole thing goes to minus infinity as well because you can substitute values in and it will get more and more negative. Then... well I dont know how to see the maximum without calculus... I don't think you'll be asked for the range of those sorts of functions. (unless there is a way that I'm forgetting... or you can use a calculator)

for the question "How do I know the lines in which each of the curves never touches?"

I don't really know what the question is talking about as in what lines. I guess it would be the y= some constant above the range lines for this function.


"And a stupid question that I just forgot the answer to: does the parent parabola function have unlimited domain"

Not sure what the parent parabola is, but for parabolas of the form ax^2+bx+c, yeah it does have unlimited domain, that is, the domain is all reals.

EDIT: ooh yeah I am the slowest

 

[Rici]

x^2-9=0
(x+3)(x-3)=0 At this point, we can ignore one binomial when zeroing the other, since 0*anything =0.
So, x+3=0
x-3=0
x=3, -3.

Edit: oops, **** **** **** **** master.
It's been a while ok? I just had a 2 month vacation without any math at all.

I feel so ashamed ;_;


I didn't say anything. Don't look at the post.
(my last teacher just smashed in our heads with that (x+3)^2 is not x^2+9, so when I saw x^2+9 I automaticly thought it can't be (x+3)^2, which is true, but I didn't notice the -3 even though I read it).

Srry ^_^''

 

[GoldShadow]

(x+3)(x-3) is not x^2+6x-9 Riciardos!

 

[Doraki]

(x-4)/(x+2) = (x+2-6)/(x+2) = 1 - 6 / (x - 2).

Written this way, it's easier to see the range of this function.

 

[WaterTails]

(x+3)(x-3) is not x^2+6x-9 Riciardos!

Yeah, Even I know that it's x^2-9

 

[Rici]

Please, let the pain stop.

 

[vericz]

Alt4-Says you're a physics tutor in your profile, I just started Physics with Calc 1 so you may see a few questions from me soon, like tomorrow

 

[cF=)]

Oh god, my math teacher is showing us matrix.

Kill me.

PS: It's so boring I want to die.

 

[Keku]

I might aswell contribute to the thread aswell. We a have a bit different courses here in Finland, but they are on the same basis anyway. Math's always been quite easy for me, aswell as the logical thinking it requires. We haven't yet had calculus, but I could try to help with any questions concerning geometry, analytic geometry, everyday maths, vectors, polynomic equations and so on. Ask away!

 

[vericz]

Yooo, Just want to make sure I got this problem correct:
A vehicle starts from rest and accelerates at a rate of 2.0 m/s^2 in a straight line until it reaches a speed of 20m/s. The vehicle then slows down at a constant rate of 1.0m/s^2 until it stops. How much time elapses from start to stop? How far does the vehicle travel from start to stop?

For the first question I used V=Vo+at. Since initial velocity is 0 solve for time i get t=V/a.

 

[MVPaintballer]

walk me through #4 ???? I don't get it @ all. Its the only one I haven't done


EDIT ITs just c i need

 

[Keku]

Yooo, Just want to make sure I got this problem correct:
A vehicle starts from rest and accelerates at a rate of 2.0 m/s^2 in a straight line until it reaches a speed of 20m/s. The vehicle then slows down at a constant rate of 1.0m/s^2 until it stops. How much time elapses from start to stop? How far does the vehicle travel from start to stop?

For the first question I used V=Vo+at. Since initial velocity is 0 solve for time i get t=V/a.

Just forget all the formulas, and use your reason. Here, you have a car.

CAR.

The car is in rest.

CAR.

The car accelerates 2.0m/s^2, this means that each second it's speed is increasing by 2,0 m/s.

CAR -->--->-ACCELERATION--->--->---> CAR (Speed 20m/s, and with 2 m/s increase per second, this takes 10 seconds.

Allright, so, the car starts to slow down. The acceleration is now -1,0m/s^2.

CAR ----<----<----<----<-----<----<---< CAR (Speed 0 m/s. Because it's speed was 20 m/s, and it slowed down 1 m/s^2, it took 20 seconds for the car to stop.

The duration is therefore 10 seconds plus 20 seconds = 30 seconds total.

We still need to calculate how far the car went. Well, during the first 10 seconds, the average speed of the car was 10 m/s. Because it accelerated in the same, constant rate, we can say that it moves the same amount that it would move if it went for 10 seconds at the speed of 10 m/s. Because it goes 10m/s, and we have 10 seconds, we get 100m in total.

The second part takes 20 seconds. The object's average speed is 10m/s in this case, too. We get 10 * 20 = 200. 100m + 200m is 300m total.

The final answer would be 30 seconds and 300 metres. Sometimes it's just good to forget all about formulas and just think about it with pure reason and from a realistic view. It really helps to understand how the formulas are formed and thus, how they are to be used, especially in easy every-day cases such as this.

 

[Varuna]

Am I doing this right?

I'm supposed to be getting the derivatives.

 

[AltF4]

Varuna:

27) The last term says 7 / t right? Which you correctly rewrite as 7 * (t^-1). But when you take the derivative of that, you should subtract one from the exponent of t and multiply the old one down. You did the multiplying down right, but then where did the exponent of 6 come from with t? It should be -7 (t^2).

29) Correct. But what's with the mixed numbers? Simple fractions are easier to deal with.

MVP:

That question might be a little ambiguous. It says the woman walks the block and it is 100m long. Then it says how long would it be if it weren't slanted... well... it would still be 100m, just level. But I don't think that's the answer they're looking for.

I think they want to ask you "what is the horizontal component of a right triangle with a hypotenuse of 100m and with an interior angle of (what you computer earlier)"

It's trying to ask you the how far horizontally she walked when she walked diagonally 100m.

 

[Varuna]

thanks, I corrected 27 and how would you suggest I write 29's answer?

 

[Doraki]

Woah I can't read your handwriting at all.

it's all bars and bars... not cute.. angles everywhere.. argh..
how come x transforms into +

And I find things like 2 1/2 for 5/2 totally ugly.

You could transform things like ... let's say | (1/2)x+7 | into 1 1/2 ++71, thus 3/2++71.. totally different yay.

Now I don't want to be a math teacher in the us.

 

[Puffinator]

today it took me forever to do the problem

prove that the perpendicular bisector of a circles chord goes through the center

i had to do it algebraically; no geometry....took forever to figure it out x_x

 

[Varuna]

How about these three, any takers? Thanks in advance.