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[Feep]

My brother brought home some prep work on mathematical induction, and I could do most of it fine (it's been awhile), but the last one is giving me fits. Can anyone take a look at number 3 and tell me what they think? I'm not sure how the hint is relevant.

https://docs.google.com/viewer?a=v&pid=gmail&attid=0.1&thid=12707bc1b5d384f9&mt=application%2Fpdf&url=https%3A%2F%2Fmail.google.com%2Fmail%2F%3Fui%3D2%26ik%3D8b7ea1a01c%26view%3Datt%26th%3D12707bc1b5d384f9%26attid%3D0.1%26disp%3Dattd%26realattid%3Df_g649gpop0%26zw&sig=AHIEtbSrPMUNErcMTOQy7BVTHhmQTurvmw

Edit: Solved it. Here is the explanation for anyone who is interested.

First, we prove the base cases, n = 0 and n = 1. Easy. (0 nCr 0) = 1, which is less than or equal to 1, and (3 nCr 1) = 3, which is less than or equal to 27/4.

Now, assuming our bound is true, we need to prove that (3(n + 1) nCr (n + 1)) is less than or equal to (27/4)^(n+1). Expanding out both sides with the nCr formula, we get:

(3n + 3)! / ((n + 1)! (2n + 2)!) <= (27/4)^n * (27/4)

Expanding further, we get

(3n + 3)(3n + 2)(3n + 1)(3n)! / ((n + 1) n! (2n + 2) (2n + 1) (2n)!) <= (27/4)^n * (27/4)

Bringing out all those extra binomials on the left side, we get

((3n + 3)(3n + 2)(3n + 1)) / ((n + 1)(2n + 2)(2n + 1)) * (3n + 3)! / ((n + 1)! (2n + 2)!) <= (27/4)^n * (27/4)

And we notice that without those extra binomials, it's the old (3n nCr n) formula, so

((3n + 3)(3n + 2)(3n + 1)) / ((n + 1)(2n + 2)(2n + 1)) * (3n nCr n) <= (27/4)^n * (27/4)

Now, we're assuming (3n nCr n) < (27/4)^n, so if we can prove that under all circumstances, ((3n + 3)(3n + 2)(3n + 1)) / ((n + 1)(2n + 2)(2n + 1)) is less than or equal to (27/4), our proof by induction is complete. First, let's simplify a bit; we can cancel two of the binomials after factoring out a three:

(3 (3n + 2)(3n + 1)) / ((2n + 2)(2n + 1)) <= 27/4

Divide by 3:

((3n + 2)(3n + 1)) / ((2n + 2)(2n + 1)) <= 27/12

Expand the binomials out:

(9n^2 + 9n + 2) / (4n^2 + 6n + 2) <= 27/12

If we can prove this is true, we're golden. Here's where his hint comes in. Translating it into English, if every coefficient in the numerator is less than or equal to the corresponding coefficient in the denominator times gamma (in this case, 27/12), then the fraction as a whole must be less than or equal to gamma. Well,

9 <= (4 * 27/12) is true,
9 <= (6 * 27/12) is true, and
2 <= (2 * 27/12) is true.

This shows the lemma true, and therefore, that fraction is less than or equal to 27/12, proving the inductive step, and, along with the two base cases demonstrated earlier, proves the upper bound correct under all circumstances.

I didn't really need to type all of this out to GAF, but I needed to for my brother anyway, so I thought I might as well go ahead and post it here. = D

 

[Tater Tot]

Need some knowledge thrown my way.

5/4(4)+(-2)-1=0

Fraction keep throwing me off how do I get that thing out of there?
/=fraction

Its a substitution problem.

Original

5/4x+y-1=0

x=4
y=-2

Edit: Nevermind, I know how to solve it. I just had a horrible brain malfunction or something. :lol

 

[hamchan]

So, apparently I'm not very good at set theory . Any help is appreciated.

Prove the following statements if they are true and give a counter-example if they are false:

a)For all sets A,B and C, if A &#8745; C &#8838; B &#8745; C and A&#8746;C &#8838; B&#8746;C then A&#8838;B.

 

[Demokrit]

So, apparently I'm not very good at set theory . Any help is appreciated.

Prove the following statements if they are true and give a counter-example if they are false:

a)For all sets A,B and C, if A &#8745; C &#8838; B &#8745; C and A&#8746;C &#8838; B&#8746;C then A&#8838;B.

To get you started:

[;A \cap C \subseteq B \cap C \wedge A \cup C \subseteq B \cup C\\
\Leftrightarrow\\
\forall x : x \in A \cap C \Rightarrow x\in B \cap C \wedge \forall x : x \in A \cup C \Rightarrow x\in B \cup C\\
\Leftrightarrow\\
\forall x : (x \in A \wedge x \in C \Rightarrow x\in B \wedge x \in C) \wedge (x \in A \vee x \in C \Rightarrow x\in B \vee x \in C) ;]
etc.

You basically just have to translate the set theory into logic and the proof comes together on its own - this always works. (If it doesn't, you should start looking for a counter-example.)

 

[Father_Mike]

Can any one help with these?

5. Use vectors to prove that the line joining the midpoints of two sides of any triangle in R3 is parallel to the other side and half as long as the other side.

6. Let P =(x0,y0,z0) be any point in R3, and let a,b,c be any non-zero real numbers. Prove that the line though P in any direction d = (a,b,c) consists of all points (x,y,z) in R3 satisfying the equations

(x-x0)/a = (y-y0)/b = (z-z0)/c

 

[peakish]

Can any one help with these?

5. Use vectors to prove that the line joining the midpoints of two sides of any triangle in R3 is parallel to the other side and half as long as the other side.

Don't have time for the second one, but:

Write the ends of the triangle as coordinates (x0,y0), (x1,y1) and (x2,y2). Construct the vectors between the points: (x2-x0,y2-y0), (x2-x1,y2-y1) and (x1-x0,y1-y0). Then you get the midpoints between the points by dividing these in half.

Finally construct the vectors between two of these midpoints in the same way and ta-da!

 

[close to the edge]

Okay, weird problem here. I'm trying to convert a complex number from Polar form to Cartesian.
z = [2, pi/2]
a = 2 * cos(pi/2)
b = 2 * sin(pi/2)
Now, when I type 2*cos(pi/2) into my calculator (TI 30X) it says 1.9992...something (~ 2) and for 2*sin(pi/2) it says 0.054... For Google and WolframAlpha 2*cos(pi/2) = 0 and 2*sin(pi/2) = 2 however. (http://www.wolframalpha.com/input/?i=2+cos(pi/2) and http://www.google.com/search?q=2*co...s=org.mozilla:en-US:official&client=firefox-a) What the hell? Is my calculator broken?

 

[torontoml]

^^^ you are using degrees and the other things are using radians, you can switch between them by pushing the mode button on your calculator.

 

[close to the edge]

Yup, that did it. Thanks.

 

[Scalemail Ted]

Quick stupid question...

I'm doing some series problems with limit comparison and ran into this derivative.

I shamefully had to use wolfram alpha and still cannot recall the rule for taking the derivative when your variable is the exponent.

So

when ever you have a constant with your variable as its exponent then its derivative is original value * log(constant), right?

and

wouldn't

3^n * ln(3) == 1?

or am i thinking of

e^3 * ln(3)?

 

[tsef]

3^n = exp(n*ln(3))

The derivative of exp(x) is exp (x)
The derivative oh n*ln(3) is ln(3)

the derivative of f(g(x)) is g'(x)*f'(g(x)) (chain rule if I'm not mistaken)

So the derivative of 3^n is ln(3)*exp(n*ln(3)) = ln(3)*3^n

Edit :

3^0 = 1

3^n is not a constant so ln(3)*3^n is not either.

 

[Scalemail Ted]

3^n = exp(n*ln(3))

The derivative of exp(x) is exp (x)
The derivative oh n*ln(3) is ln(3)

the derivative of f(g(x)) is g'(x)*f'(g(x)) (chain rule if I'm not mistaken)

So the derivative of 3^n is ln(3)*exp(n*ln(3)) = ln(3)*3^n

Edit :

3^0 = 1

3^n is not a constant so ln(3)*3^n is not either.

Ok, thanks. That makes sense now. just like riding a bicycle.

Also disregard the later question, of course its not a constant.. I just missed something incredibly stupid in the answer. (i.e. to take the limit while n approaches infinity: "1 -(1/(3^n*ln(3)))" of course that's 1-0 leaving me that 1. >_<

 

[unconcerned]

Gaf i need some help, I'm totally lost on this problem.

The Maclaurin series for f(x) is given by:

(1/2) - (x^2/(2^3 * 3!)) + (x^4/(2^5 * 5!)) - (x^6/(2^7 * 7!)) +...+ ((-1)^n x^(2n)/(2^(2n+1) * (2n+1)!)

a. Find f'(0) and f''(0) and explain why f(x) has a local maximum, local minimum or neither at x = 0.

b. Let g(x) = 2xf(2x). Give the first three nonzero terms and the general term for g(x).

 

[liquidspeed]

does any one here know how to do power series solutions to differential equations? I kinda get it, but in some ways I'm lost..... If somone responds, I'll throw a hw problem out there

 

[hemtae]

Gaf i need some help, I'm totally lost on this problem.

The Maclaurin series for f(x) is given by:

(1/2) - (x^2/(2^3 * 3!)) + (x^4/(2^5 * 5!)) - (x^6/(2^7 * 7!)) +...+ ((-1)^n x^(2n)/(2^(2n+1) * (2n+1)!)

a. Find f'(0) and f''(0) and explain why f(x) has a local maximum, local minimum or neither at x = 0.

b. Let g(x) = 2xf(2x). Give the first three nonzero terms and the general term for g(x).

Don't know about part b, but in a Maclaurin series, the second term is given by f'(0)x, since there is no term resembling that f'(0) = 0. The third term is given by (f"(0)/2!)x^2 so therefore f"(0) = -(1/24). Since the first derivative is 0 and the second derivative is a negative number, it has a local minimum at x = 0

That looks right to me anyways

 

[unconcerned]

Don't know about part b, but in a Maclaurin series, the second term is given by f'(0)x, since there is no term resembling that f'(0) = 0. The third term is given by (f"(0)/2!)x^2 so therefore f"(0) = -(1/24). Since the first derivative is 0 and the second derivative is a negative number, it has a local minimum at x = 0

That looks right to me anyways

i think i'm starting to get it now but why is f''(0) = -(1/24) and not -(1/48) seeing that (2^3 * 3!) = 48?

 

[Therion]

i think i'm starting to get it now but why is f''(0) = -(1/24) and not -(1/48) seeing that (2^3 * 3!) = 48?

(f"(0)/2!)x^2 = -1/48, so you need to multiply by two to get f',

 

[Rich Uncle Skeleton]

So, apparently I'm not very good at set theory . Any help is appreciated.

Prove the following statements if they are true and give a counter-example if they are false:

a)For all sets A,B and C, if A &#8745; C &#8838; B &#8745; C and A&#8746;C &#8838; B&#8746;C then A&#8838;B.

Probably you figured this out ages ago, but: You want to show that everything that's in A is also in B. Well, if x is in A and C, use one of your assumptions to show that x is also in B; and if x is in A but not in C use the other assumption to show that again x must be in B.

 

[Feep]

Here's a quiz with some interesting math problems, for you math nerds out there. I got 9/10, made a stupid mistake, blegh. = D

http://www.funtrivia.com/flashquiz/index.cfm?qid=97961

 

[Lebron]

Need some help with this problem for a friend of mine, but my Trig sucks. Any help would be appreciated. Think he said the answer was plus or minus (square root 5/5), but only the plus ends up working.

arcsin2x + arcsinx=(pie/2)
solve for x, obviously.

 

[Rich Uncle Skeleton]

Need some help with this problem for a friend of mine, but my Trig sucks. Any help would be appreciated. Think he said the answer was plus or minus (square root 5/5), but only the plus ends up working.

arcsin2x + arcsinx=(pie/2)
solve for x, obviously.

Let y_1 = arcsin(2x) and y_2 = arcsin x.
Then, since y_1 + y_2 = pi/2, we can draw a right triangle with angles pi/2, y_1, and y_2.
Now sin(y_1) = 2x and sin(y_2) = x, so if the hypotenuse has length 1 then the other two sides have length x and 2x. Pythagorean theorem then tells you that x^2 must be 1/5, hence the two answers.

Edit: No, you're right. Only the positive one works. Note that arcsine takes values between -pi/2 and pi/2, so you see from the beginning that y_1 and y_2 must be nonnegative, since they sum to pi/2. This is what allows me to say that they're the angles of some triangle, and since x must then also be nonnegative, we can say (as I did) that x and 2x are the lengths of the sides. Pythagorus then gives you the answer (but only the positive one).

 

[innervision961]

As you guys get deeper in math school, or after university, do you feel like you forget a lot of what you learned early on, are you constantly needing a refresher? Can someone put down a problem in front of you and you can usually just tackle it pretty straight forrward.

 

[Lebron]

Let y_1 = arcsin(2x) and y_2 = arcsin x.
Then, since y_1 + y_2 = pi/2, we can draw a right triangle with angles pi, y_1, and y_2.
Now sin(y_1) = 2x and sin(y_2) = x, so if the hypotenuse has length 1 then the other two sides have length x and 2x. Pythagorean theorem then tells you that x^2 must be 1/5, hence the two answers.

Edit: No, you're right. Only the positive one works. Note that arcsine takes values between -pi/2 and pi/2, so you see from the beginning that y_1 and y_2 must be nonnegative, since they sum to pi/2. This is what allows me to say that they're the angles of some triangle, and since x must then also be nonnegative, we can say (as I did) that x and 2x are the lengths of the sides. Pythagorus then gives you the answer (but only the positive one).

Yeah, you're right. I took your advice and played around with it, and found the tan of the triangle, which was 1/2. Then simply found the hypo by doing Pythagorean, which was obviously square root 5. Then just solved for sin. Seems so simple now, but for whatever reason I couldn't see it. His stupid book had him doing like a page of work, or some BS. Going to play around with similar problems so I can explain it to him better. Thanks for the help.

 

[ili0926]

As you guys get deeper in math school, or after university, do you feel like you forget a lot of what you learned early on, are you constantly needing a refresher? Can someone put down a problem in front of you and you can usually just tackle it pretty straight forrward.

Definitely need a refresher. I can do multivariate calculus now, but I honestly couldn't for the life of me do basic trig problems from high school if you asked me to.

 

[Rich Uncle Skeleton]

Yeah, you're right. I took your advice and played around with it, and found the tan of the triangle, which was 1/2. Then simply found the hypo by doing Pythagorean, which was obviously square root 5. Then just solved for sin. Seems so simple now, but for whatever reason I couldn't see it. His stupid book had him doing like a page of work, or some BS. Going to play around with similar problems so I can explain it to him better. Thanks for the help.

No problem. By the way, I fixed another typo in my answer but you probably already caught it. (90 degrees is pi/2, not pi. Woops.)

As you guys get deeper in math school, or after university, do you feel like you forget a lot of what you learned early on, are you constantly needing a refresher? Can someone put down a problem in front of you and you can usually just tackle it pretty straight forrward.

I forget a lot of stuff from earlier courses. But usually if I spend a few minutes refreshing my memory I'm good. It's interesting returning to material you learned earlier with new eyes and having a deeper appreciation for it.

 

[jepense]

As you guys get deeper in math school, or after university, do you feel like you forget a lot of what you learned early on, are you constantly needing a refresher? Can someone put down a problem in front of you and you can usually just tackle it pretty straight forrward.

I think everyone remembers something and forgets a lot. It of course depends on how much you've had to use the stuff. And sometimes you just remember some things for no particular reason.

Personally, I remember my basic maths quite well, and of the advanced stuff functional analysis and measure theory have stuck ok. For stochastics and statistics I only remember the basic stuff and would need to refresh if I were to return to them. I still read the secondary level maths final exam (or whatever it should be called in English) from the newspaper every spring and autumn, and can consistently solve all the problems there.

 

[Minamu]

Geometry pisses me off :'(

Can someone please explain this to me? I'm supposed to calculate exactly how big that square inside the triangle is. all I know is that A to B is 12 centimeters & A to C is 18 centimeters. Thanks to Pythagora's something something, B to C should be about 21.6 centimeter. The book doesn't say but the smaller triangles look like they are in proportion to the bigger one but I can't get anything but conflicting numbers :S The picture is not to scale & it's not that way in the book either so it shouldn't matter. I assume I should be using a/(a+b)=c/(c+d)=e/f in some way but other examples give me at least a few more numbers to work with :lol

The square is not supposed to be a rectangle, that's just ms paint being ms paint.

 

[torontoml]

and it doesn't tell you anything about where D, E, and F are?

 

[Chris R]

Geometry pisses me off :'(

Can someone please explain this to me? I'm supposed to calculate exactly how big that square inside the triangle is. all I know is that A to B is 12 centimeters & A to C is 18 centimeters. Thanks to Pythagora's something something, B to C should be about 21.6 centimeter. The book doesn't say but the smaller triangles look like they are in proportion to the bigger one but I can't get anything but conflicting numbers :S The picture is not to scale & it's not that way in the book either so it shouldn't matter. I assume I should be using a/(a+b)=c/(c+d)=e/f in some way but other examples give me at least a few more numbers to work with :lol

The square is not supposed to be a rectangle, that's just ms paint being ms paint.

Never assume ANYTHING from ANY picture. And are you sure that is all the information that you are given? Nothing about midpoints, or bisection or anything else?

edit: Nevermind, I'm an idiot, and totally missed the point where the shape inside the triangle is pointed out as a square :lol I just thought it was a random rectangle :lol

 

[DogWelder]

Geometry pisses me off :'(

Can someone please explain this to me? I'm supposed to calculate exactly how big that square inside the triangle is. all I know is that A to B is 12 centimeters & A to C is 18 centimeters. Thanks to Pythagora's something something, B to C should be about 21.6 centimeter. The book doesn't say but the smaller triangles look like they are in proportion to the bigger one but I can't get anything but conflicting numbers :S The picture is not to scale & it's not that way in the book either so it shouldn't matter. I assume I should be using a/(a+b)=c/(c+d)=e/f in some way but other examples give me at least a few more numbers to work with :lol

The square is not supposed to be a rectangle, that's just ms paint being ms paint.

I'm gonna redefine a few things:

DE = EF = x
BE = h1
EC = h2

[Getting rid of this stuff so as not to confuse you. Look at my post below.]

 

[Minamu]

esc: How did you come up with 6/9x & 9/6x exactly? :S I haven't seen numbers written that way before :lol I only assume they are proportionate because that's what the last few pages of the chapter have been about. The answer is supposed to be that one side of the square is 7.2 centimeter (basically AB minus BD for example). I got BC by square rooting 12^2+18^2=21.63^2. Your formula doesn't say but shouldn't I do the same to get h1 & h2? If I know what (6/9x)^2 means in centimeters, figuring out what 2x^2 is shouldn't be so hard, right? But if I know that, then I would already have the answer since all sides are the same. Ugh, wrapping my head around this is hard enough as it is & English being my second language sure doesn't help

Never assume ANYTHING from ANY picture. And are you sure that is all the information that you are given? Nothing about midpoints, or bisection or anything else?

Yep, that's all the info I was given. Two sets of numbers, go have fun

 

[DogWelder]

esc: How did you come up with 6/9x & 9/6x exactly? :S I haven't seen numbers written that way before :lol I only assume they are proportionate because that's what the last few pages of the chapter have been about. The answer is supposed to be that one side of the square is 7.2 centimeter (basically AB minus BD for example). I got BC by square rooting 12^2+18^2=21.63^2. Your formula doesn't say but shouldn't I do the same to get h1 & h2? If I know what (6/9x)^2 means in centimeters, figuring out what 2x^2 is shouldn't be so hard, right? But if I know that, then I would already have the answer since all sides are the same. Ugh, wrapping my head around this is hard enough as it is & English being my second language sure doesn't help


Yep, that's all the info I was given. Two sets of numbers, go have fun

6/9 is just 12/18 reduced. Maybe I should've left it as 12/18. It's the ratio of one length of the triangle to the other. Because the large one has 12/18 proportions and the inset triangles have the same proportions, if DE is defined as "x" then BD must be (12/18)x. Understand?

Sorry, I made bit of a mistake in writing out the equations, I forgot to square h1 and h2 (keep in mind what I defined everything as). Let's try this again:

h1^2 = x^2 + (6/9 x)^2
h1^2 = 13/9 x^2
h1 = &#8730;(13/9) x

And for the other triangle:

h2^2 = x^2 + (9/6 x)^2
h2^2 = 13/4 x^2
h2 = &#8730;(13/4) x

So now you have values for h1 and h2. Remember h1 and h2 are the hypotenuse of the smaller triangles and that if you add them together you get the hypotenuse of the large triangle. Therefore:

h1 + h2 = 21.63 = &#8730;(13/9) x + &#8730;(13/4) x
21.63 = 3 x
x = 7.2

 

[Minamu]

This is so damn abstract xD I don't really get where you got 13/9 & 13/4 from. How would I enter this on my calculator though? 12 18ths of x doesn't help me much when x is undefined (though my TI-83 Plus says that x = 5 by default for some reason). Previous examples have been kind enough to give a number for at least one of the shorter sides & then it's real easy to decipher the other sizes thanks to this formula: a/(a+b)=c/(c+d)=e/f. All examples have revolved around this formula. You say I have solid values for h1 & h2 but "13/9 x^2" doesn't tell me much, I'm afraid.

Edit: You know what? I just had an epiphany :lol I just figured that if two sides are proportionate to their bigger counterparts, then the third one, h1 or h2, surely must be as well... So if either one is 12 18ths, or rather 2 3rds, then the other one must be 6 18ths or 1 3rd. And simply removing 2 3rds from 21.63 gives me 7.21. Screw complex equations, a 12 year old can figure out what 21/3 is :lol I'm not entirely sure why it gives me the correct answer since no hypotenuse actually is 7.21 centimeter though, but hey it works.

 

[Subliminal]

This is so damn abstract xD I don't really get where you got 13/9 & 13/4 from. How would I enter this on my calculator though? 12 18ths of x doesn't help me much when x is undefined (though my TI-83 Plus says that x = 5 by default for some reason). Previous examples have been kind enough to give a number for at least one of the shorter sides & then it's real easy to decipher the other sizes thanks to this formula: a/(a+b)=c/(c+d)=e/f. All examples have revolved around this formula. You say I have solid values for h1 & h2 but "13/9 x^2" doesn't tell me much, I'm afraid.

> Pythagoras: a^2+b^2=c^2

>Don't. Use Paper. Algebra works so much better when you have it written down

 

[SDMz]

Edit: You know what? I just had an epiphany :lol I just figured that if two sides are proportionate to their bigger counterparts, then the third one, h1 or h2, surely must be as well... So if either one is 12 18ths, or rather 2 3rds, then the other one must be 6 18ths or 1 3rd. And simply removing 2 3rds from 21.63 gives me 7.21. Screw complex equations, a 12 year old can figure out what 21/3 is :lol I'm not entirely sure why it gives me the correct answer since no hypotenuse actually is 7.21 centimeter though, but hey it works.

You actually don't need h1 and h2 to solve this. You already know that AF = EF (it's a square) and that EF : FC = 2:3 (proportionate sides). Since AC = AF + FC, dividing AC in 5 equal parts gives you a nice partition: the first two of these parts are AF and the last 3 are FC. Therefore AC = 2* AF/ 5 = 2* 18/5 (and this is easily done with just pen and paper )

 

[Minamu]

What a wonderful first post on gaf :lol Welcome, and thanks.

I'm doing an entry exam sort of thing in 10 days where math is a big part so I'm trying to brush up on it before I hopefully get into a university. Luckily, I think this kind of math is a bit more advanced than what will be on said test. It's just that they use the most convoluted & abstract diagrams etc they can find just to fuck with you :lol

 

[SDMz]

Thanks and you're welcome

If you think such diagrams are convoluted and abstract, you have to write down everything you already know about the things in the diagram. E.g. you know that AE, DE, EF and AF form a square so AE=DE=EF=AF. And you know that AE + FC = AC. You know that AB :AC = EF = FC, etc. If the problem was well chosen (by the teacher or the one that made the test) you will recognise something eventually, provided you have enough experience and have learnt your math well enough.

Anyway good luck with your entry test!

 

[Minamu]

Nah, that problem wasn't that abstract now that I kinda got it, and it was the very last problem in that chapter so it was bound to be a bit tougher than the rest. I actually meant the diagrams in said entry test. They usually look like they're drawn by 3-year-olds or contain a chunk of wall of text. Paper to write notes on is not allowed And no calculator or anything more advanced than a standard ruler.

 

[eosos]

Since everyone here seems to be really good at math I'd thought that I would sure you some extra credit problems that I'm doing! It's from this calender with a problem written under each date. You guys ready?

1. What is the value of 1-2+3-4+...+99-100

2. A used car salesman sold two cars and received $840 for each car. One of these transactions yielded a 40 percent profit for the dealer(compared to his purchase price), whereas the other amounted to a 25% loss. What is the dealers net profit(or loss) on the two transactions?

3. Johnny was ill and had to take a test a day late. His 96 raised the class average from 71 to 72. How many students, including Johnny, took the test?

4. Joan holds a 26-inch piece of light weight string with a heavy bead on it. One end is is in her right hand; the other is in her left hand. Initially, her hands are together. How many inches apart must she pull her hands, keeping them at the same height, so the bead moves upwards by eight inches.

5. A bowl contains 50 colored balls: 13 green, 10 red, 9 blue, 8 yellow, 6 black, and 4 white. If you are blindfolded, what is the smallest number of balls you must pick to guarantee that you have at least 7 balls of the same color?

I'll add more later. Good luck!

 

[ananag]

http://www.wolframalpha.com/

I wish I had this back in the day. Its an excellent tool to help solve tons of different math and science problems.

 

[jepense]

Since everyone here seems to be really good at math I'd thought that I would sure you some extra credit problems that I'm doing! It's from this calender with a problem written under each date. You guys ready?

1. What is the value of 1-2+3-4+...+99-100

2. A used car salesman sold two cars and received $840 for each car. One of these transactions yielded a 40 percent profit for the dealer(compared to his purchase price), whereas the other amounted to a 25% loss. What is the dealers net profit(or loss) on the two transactions?

3. Johnny was ill and had to take a test a day late. His 96 raised the class average from 71 to 72. How many students, including Johnny, took the test?

4. Joan holds a 26-inch piece of light weight string with a heavy bead on it. One end is is in her right hand; the other is in her left hand. Initially, her hands are together. How many inches apart must she pull her hands, keeping them at the same height, so the bead moves upwards by eight inches.

5. A bowl contains 50 colored balls: 13 green, 10 red, 9 blue, 8 yellow, 6 black, and 4 white. If you are blindfolded, what is the smallest number of balls you must pick to guarantee that you have at least 7 balls of the same color?

I'll add more later. Good luck!

1. Think (1-2) + (3-4) + ...
2. He got 840, you know the profit/loss percentages, what did he originally pay for the cars?
3. Class has n students. Others scored x combined: x = (n-1)*71 and (x+96) = n*72
4. Initially, the beads are 13 inches below the hands. In the end, the string forms a triangle where the beads are 5 inches below her hands. Trigonometry.
5. To guarantee 7 of same color, in the worst case you have to take 4 whites, 6 of every other color and 7 of one color. (It's the pigeonhole principle)

 

[Dogenzaka]

I need help with:
- Identifying graphs of logarithmic equations.
- Wtf is a logarithm in relationship to a function/inverse?

Any good math video sources?

 

[narcosis219]

If I have a system of differential equations...

Y' = AY

where A has the eigenvalues (I'll use L for lambda)

L1 = -1 + i
L2 = -1 -i

with the corresponding eigenvectors
V1 = [1 + i, 1 - i]' (' means transpose)
V2 = [1 - i, 1 + i]'

how would i compute the matrix A given this information? I'm asked to write the MATLAB commands that would compute the first column of A, but I can figure that out if I know how to find A from the above info

 

[Parham]

Anyone who is having difficulty with Calculus should check out this website. It clears up just about any conceptual issues you may have.

I need help with:
- Identifying graphs of logarithmic equations.
- Wtf is a logarithm in relationship to a function/inverse?

Any good math video sources?

Does this video help?

 

[jepense]

If I have a system of differential equations...

Y' = AY

where A has the eigenvalues (I'll use L for lambda)

L1 = -1 + i
L2 = -1 -i

with the corresponding eigenvectors
V1 = [1 + i, 1 - i]' (' means transpose)
V2 = [1 - i, 1 + i]'

how would i compute the matrix A given this information? I'm asked to write the MATLAB commands that would compute the first column of A, but I can figure that out if I know how to find A from the above info

The diff. equation is irrelevant. You have A's eigendecomposition.

 

[Aeris130]

Need some dice-related probability help. Basically, I want to compute the probability of scoring a certain dice combination involving x dies, using y dies (where y >= x). I've found tons of dice-examples on the net, but all of them seem to fall short of describing this particular scenario.

I assume that it's (# of beneficial combinations) / (total # of combinations), but I'm unsure of how the first one is calculated when there's multiple copies of one value ([1,1,1] should be more difficult to score than [1,2,3], since in the latter example, the first dice rolled has a 50% chance of getting the correct value, right?).

 

[Minamu]

I have a new question. It's not really from a test, just something I think most would have an economic interest in.

So, is there some way of figuring out how to most efficiently repay a student loan? Let's assume one does not have enough money to repay it all on day 1. Or don't. Thing is, what's left of the debt gets deleted when you retire at around 65 years old. I'm kinda curious as to see what's wiser from an economic stand point. Just pay it off as fast as possible to avoid interest or just let it die after working for 40+ years Surely there's must be a turning point somewhere where one or the other is smarter? I mean, if the amount of money one would've repayed in 40 years is less than the original debt due to interests, just letting it get nullified would obviously save you a few dollars It's purely theoretical so I don't really have any numbers to put in an equation.

 

[Khalifa Jayy]

Solve the proportion.

I've never really seen a fraction written like that. wtf?

 

[mcrae]

(2/5)/y is the same as (2/5)/(y/1), which you can change into (2/5)*(1/y), then multiply across to get 2/5y ... is that easier?

or you could look at it like 0.4/y ... but i always got yelled at when i mixed decimals and fractions.

 

[mcrae]

I have a new question. It's not really from a test, just something I think most would have an economic interest in.

So, is there some way of figuring out how to most efficiently repay a student loan? Let's assume one does not have enough money to repay it all on day 1. Or don't. Thing is, what's left of the debt gets deleted when you retire at around 65 years old. I'm kinda curious as to see what's wiser from an economic stand point. Just pay it off as fast as possible to avoid interest or just let it die after working for 40+ years Surely there's must be a turning point somewhere where one or the other is smarter? I mean, if the amount of money one would've repayed in 40 years is less than the original debt due to interests, just letting it get nullified would obviously save you a few dollars It's purely theoretical so I don't really have any numbers to put in an equation.

wait, if your debt ceases to exist when you turn 65, why would you ever repay any of it?
(i obviously dont understand what you wrote)

Need some dice-related probability help. Basically, I want to compute the probability of scoring a certain dice combination involving x dies, using y dies (where y >= x). I've found tons of dice-examples on the net, but all of them seem to fall short of describing this particular scenario.

I assume that it's (# of beneficial combinations) / (total # of combinations), but I'm unsure of how the first one is calculated when there's multiple copies of one value ([1,1,1] should be more difficult to score than [1,2,3], since in the latter example, the first dice rolled has a 50% chance of getting the correct value, right?).

can you explain more? the chances of 3 dice rolling 3 1's is 1/6*1/6*1/6= 1/216, and the chances of 3 dices rolling a 1, a 2, and a 3, are 3/6*2/6*1/6 = 1/36, or 6 times as likely