Need Math Help?

[POKE40]

That's not what my teacher wrote, I typed it exactly as it was on my paper =\ and my teacher got a diff answer than you. I'll post what her answer was one sec.

Edit: Original problem:
99xy - 12a^5 - 63b^4 + 20c^7
11xy^3 - 4a^3 + 7b^2 - 10c^4

The answer my teacher got:
9y^-2 + 3a^2 - 9b^2 - 2c^3

How did my teacher get that answer?

Your teacher is wrong...the original problem is already irreducible anyways in that form. lol
There needs to be space b/w each line. Try this:

 

[zeldspazz]

Your teacher is wrong...the original problem is already irreducible anyways in that form. lol
There needs to be space b/w each line. Try this:

Wow, I see what you mean now, that's interesting I should point that out to my teacher, maybe that's why I don't understand it =P
Edit: BTW how do you post the "white board" that you wrote on? thats cool

 

[POKE40]

Wow, I see what you mean now, that's interesting I should point that out to my teacher, maybe that's why I don't understand it =P
Edit: BTW how do you post the "white board" that you wrote on? thats cool

I used paint
then inserted it as a picture

I wish there was a white board so I don't have to go through all the hassle through imageshack...
your welcome anyways

 

[zeldspazz]

Ah yes thank you Poke, I never did say thank you =)

Unfortunatly I got sick last night, so I was unable to go to my teacher today =P
I'll get to it though.

 

[Corpsecreate]

Posted a question that I have already figured out myself. WTB Method of deleting my own posts

 

[Binx]

I need help ^_^

144^3 = 1/3*Pi*R^2*4 + Pi*R^2*6

Obviously I'm solving for R I just don't know where to even start.

I tried cubing and then dividing and I get lost and I am pretty sure its not right.

I figured it out, NVM ^_^

 

[Death]

I've got my math exam in 5 days time and I'm really struggling on identities... Help??

Prove:

sin^2 x + tan^2 x = sec^2 x - cos^2 x

 

[Corpsecreate]

If you have the mathematics formulae and statistics tables book then you should see these identities:

Sin^2x + Cos^2x = 1 (You should commit this one to memory as it is used quite a bit)
1 + Tan^2x = Sec^2x

That particular proof you have there is a simple one :D

Sin^2x + Tan^2x = Sec^2x - Cos^2x
(1 - Cos^2x) + Tan^2x = Sec^2x - Cos^2x
1 + Tan^2x - Cos^2x = Sec^2x - Cos^2x
Sec^2x - Cos^2x = Sec^2x - Cos^2x

I've got a question:

The mean typing speed of a class is 175 Words Per Minute (WPM) and the Standard Deviation of the class is 20 WPM. A student that is within the top 3% of the class is awarded with a prize. What is the minimum number of WPM required to be awarded the prize?

I got an answer of 222 WPM which looks correct but im not sure if it is, would be cool if someone else could have a try at it and let me know what they get :D

NOTE: Sorry about the double post but answering a question and then asking a question in the same post is just....weird.
NOTE2: Mod merged my posts together. Oh well

 

[1048576]

I think it should be under 215, since above that is where the top 2.5% lie. +-2 is 95%, so there's 2.5% left each way.

 

[Corpsecreate]

How do you know where the top 3% are? Thats really what the question is, the rest is easy.

 

[GoldShadow]

Well I can't remember the equations or methods that may help to figure this out (maybe I'll look for them later) so I just ran it in MiniTab and got 212.616 WPM (that is the 97th percentile). So you should be looking for a method that gives an answer in that vicinity. Although it is worth noting that sometimes statistical software uses slightly different algorithms and may give you a different answer than if you calculate by hand, so take it with a grain of salt.

edit: If you have a normal distribution table, you should just be able to look at it and find approximately how many standard deviations the top 3% is, and use that to find the answer.

 

[Corpsecreate]

If you have a normal distribution table, you should just be able to look at it and find approximately how many standard deviations the top 3% is, and use that to find the answer.

This is how I solved it and I got 222 WPM.

 

[Shaya]

Within the realms of Complex Numbers...

I'm confused on how to properly get the argument (this is for any complex number)

Take
z = -1 - I

which I would believe
Arg(z) =
sin -1 / 2^(1/2)
= -pi/4
However the argument is said to be -3Pi/4

I'm confused with the relationships between the certain quadrants of 2Pi and the minus or additions of Pi to the answers I feel I should be getting...

 

[solitonmedic]

I might keep this in mind.

 

[GoldShadow]

This is how I solved it and I got 222 WPM.

Okay, the normal dist table I just looked at from my old textbook says top 3% is approximately 1.88 standard deviations (or 37.6 WPM). 175+37.6 = 212.6 WPM.

 

[AltF4]

Oh, wow. I forgot about this thread. I'll try to pop in here more often again. Is everyone just taking summer classes? Or do you live in the southern hemisphere, lol.

...I just ran it in MiniTab and got 212.616 WPM ...

O M G. I remember Minitab. I can't believe you still have a copy of that around! Do you actually use it for Sciencey stuff?

Within the realms of Complex Numbers...

I'm confused on how to properly get the argument (this is for any complex number)

Take
z = -1 - I

which I would believe
Arg(z) =
sin -1 / 2^(1/2)
= -pi/4
However the argument is said to be -3Pi/4

I'm confused with the relationships between the certain quadrants of 2Pi and the minus or additions of Pi to the answers I feel I should be getting...

The "Argument" (poor choice of terminology if you ask me) of the complex number is just the angle towards that point when mapped in Cartesian space.

Maybe it would help you to make graph?

Your complex number is "-1 - i", so it is essentially the point (-1, -1), right?

Right off the bat, you can see why the angle is -3/4 Pi. The angle would start at the "East" position, go to the "South" position (using up Pi / 2) then go halfway to "West" to get 3/4 Pi.

Or just do: arctan ( y / x ) to find the "Principal" of the angle. (Pi / 4 in this case) Then determine which quadrant you're in. (3rd quadrant in this case). So now you know you have to subtract Pi from the principal to get -3/4 Pi.


A good description is here:

http://scholar.hw.ac.uk/site/maths/topic11.html

Specifically, look at the animated tutorial second from the bottom of the page.

 

[GoldShadow]

O M G. I remember Minitab. I can't believe you still have a copy of that around! Do you actually use it for Sciencey stuff?

As a matter of fact, I do! Ever since I took stats I've been using MiniTab for all my lab reports and stuff. I used to use Excel but MT knows a lot more tricks.

 

[Corpsecreate]

I re-checked using the normal distribution and I also got 212. Maybe when I did it in the exam I accidently added it wrong and got 222. sux!

Another question!

What is the sum of 1/(X² + 2X) from X = 1 to X = Infinity?

 

[GOD!]

What is the sum of 1/(X² + 2X) from X = 1 to X = Infinity?

I broke in my calculus brain tonight after my reprieve from school:

integral from 1 to infinity of 1/(x^2 + 2x) can be written as (through integration by parts, I'm assuming you know how to do that, if not I'll try and explain on request):

--> integral from 1 to infinity ( .5 * (1/x - 1/ (x+2))) dx. (use common denominator to check this if you want to)

using u-substitution with the x+2 term (and separating the integral), you get:

--> .5 * [ integral from 1 to infinity ( 1/x) - integral from 3 to infinity 1/x ] dx

This becomes:

--> .5* [integral from 1 to 3 (1/x) dx

Which equals ln(3)/2 .

I hope this is right, almost positive.

 

[Corpsecreate]

I know how to do integration by parts. Unfortunately your answer of ln(3)/2 is not correct. From running a summation calculation on my computer I get an answer very close to 3/4. ln(3)/2 is 0.549...... which is quite a bit off.

You can see this just from the 1st few sums.

1/3 + 1/8 + 1/15 + 1/24 = 0.5666666

And I just realised how to solve it. The answer is in fact 3/4.

1 / (X² + 2x) = 0.5 * [1 / x - 1 / (x+2)]

look at the first few terms and ignore the 0.5 for the moment.

(1/1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + (1/5 - 1/7)...

as X --> infinity, everything cancels itself out except for the 1/1 and the 1/2 = 3/2. 3/2 * 0.5 = 3/4 which is the answer

 

[GOD!]

I know how to do integration by parts. Unfortunately your answer of ln(3)/2 is not correct. From running a summation calculation on my computer I get an answer very close to 3/4. ln(3)/2 is 0.549...... which is quite a bit off.

You can see this just from the 1st few sums.

1/3 + 1/8 + 1/15 + 1/24 = 0.5666666

And I just realised how to solve it. The answer is in fact 3/4.

1 / (X² + 2x) = 0.5 * [1 / x - 1 / (x+2)]

look at the first few terms and ignore the 0.5 for the moment.

(1/1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + (1/5 - 1/7)...

as X --> infinity, everything cancels itself out except for the 1/1 and the 1/2 = 3/2. 3/2 * 0.5 = 3/4 which is the answer

Ouch.. three weeks and I don't know math anymore.. My bad. Can someone show me where my problem went wrong? Like, in which step?

 

[Corpsecreate]

Ouch.. three weeks and I don't know math anymore.. My bad. Can someone show me where my problem went wrong? Like, in which step?

--> .5 * [ integral from 1 to infinity ( 1/x) - integral from 3 to infinity 1/x ] dx which is 0.5lnx - 0.5ln(x+3)

that was a correct integration except it should be the integral from 1 to infinity for both segments and not 3 to infinity for one of them.

Although, the sum of an integral of a function is not the same as the sum of the original function so integrating it in the 1st place is useless and thats where your mistake was For example. the sum of 1/X² from 1 to infinity converges to a number but the sum of -1/X from 1 to infinity does not.

 

[Corpsecreate]

This question is driving me insane. Paramtetric equations are evil!

x(t) = 3t^2
y(t) = t^3-3t

Find the Arc length of the curve from t = 0 to t = sqrt3

Arc Length = Integral of Sqrt [ (dx/dt)^2 + (dy/dt)^2 ].

The problem with this is that its impossibly difficult to integrate sqrt [ (6t)^2 + (3t^2-3)^2 ]

EDIT: I managed to solve it

 

[Don Kirby]

I don´t need help!

 

[Elder Sister]

om goodness
how old are ya all?
how comes i dun understand anything?

 

[M.K]

PLEASE PLEASE PLEASE HELP ME

This problem set is for a grade equal in weight to a quiz tomorrow and I am HONESTLY completely stuck on these problems.
I am in Pre-Calculus:

1)

csc^2 x - 4 cot x = -2

2)

csc^2 x - .5 cot x = 5

PLEASE PLEASE PLEASE AND THANK YOU THANK YOU THANK YOU!!!

(show work please, so I can basically grasp it without looking like a total fool)

 

[Corpsecreate]

I would just graph it on a calculator and find the values for x. I tried doing the first question manually but could only get as far as:

Sin²x -2SinxCosx = -1/2

 

[Proud_Smash_N00b]

I have some unusual/hard problems for a Calc BC summer project. Idk why such Algebra 2 problems are needed for a Calc BC class but whatever.

1) If the remainders of divison of f(x) by x-1 and x-4 are 4 and 1 respectively, what is the remainder of division of f(x) by (x-1)(x-4)?

For this problem do I find f(x)? If so, what is f(x)

-----

2) If tan (x+y)= 1/2 and tan (x-y)= 1/3, then what is the value of tan 2x?

What identities do I use and how do I cancel out the y's if I have to use substitution? Do I solve for x and y? Do I even have to use identites?

 

[★Malik★™]

are most of these problems college problems? i looked at some of these and i'm like "uuuuhh". well i'm in 9th, just started, so most of the math my teacher is teaching is last year stuff. now in like 4 years, i'll be back here.

 

[nux23]

well, I guess the only question I have is...what are algorithms?

also, I'm a huge fan of xkcd too

 

[Proud_Smash_N00b]

Well, I'm gonna be a senior in high school who's gonna take AP Calc BC.
My teacher has a knack for giving such hard problems, idk why

 

[bleyva]

om goodness
how old are ya all?
how comes i dun understand anything?

maybe you shouldnt have skipped math class all those times
but on to more serious matters......................................

1)how many times can 5 billion go into 74 gazillion?

2)prove that 1 puppy = 0 puppies

 

[1048576]

I have some unusual/hard problems for a Calc BC summer project. Idk why such Algebra 2 problems are needed for a Calc BC class but whatever.

1) If the remainders of divison of f(x) by x-1 and x-4 are 4 and 1 respectively, what is the remainder of division of f(x) by (x-1)(x-4)?

For this problem do I find f(x)? If so, what is f(x)

-----

2) If tan (x+y)= 1/2 and tan (x-y)= 1/3, then what is the value of tan 2x?

What identities do I use and how do I cancel out the y's if I have to use substitution? Do I solve for x and y? Do I even have to use identites?

Well, number 2 seems fairly straightforward. x+y = arctan(1/2) = 0.46365
x-y = arctan(1/3) = 0.32175
Using linear combinations: 2x = 0.78540
tan(2x) = 1.00000

Alternatively tan(2x) = tan(arctan(1/2)+arctan(1/3)) which is apparantly 1. Unfortunately, my trig is pretty rusty.

 

[AltF4]

Oh, wow. I was totally about to bump this thread just now since school started up again. But you guys beat me to it!

I'll check in on this thread semi-regularly. Hopefully some of our other math-inclined members as well.

 

[UTDZac]

Ditto


<3 Math

 

[Zero Beat]

Well, I'm gonna be a senior in high school who's gonna take AP Calc BC.
My teacher has a knack for giving such hard problems, idk why

Wow, BC in high school? You're in for some pain buddy. I'll be seeing you in this thread a lot:-p.

 

[Proud_Smash_N00b]

Wow, BC in high school? You're in for some pain buddy. I'll be seeing you in this thread a lot:-p.

this thread was why I came back to Smashboards lol

 

[[oni]LoKo]

I <3 this thread
J/k, I love math, although I already opted out of Engineering because of, and other reasons >.>;

 

[wool]

Dude I'm taking BC junior year (this year) in highschool... I have a feeling that I'll be here a lot >.>

 

[Corpsecreate]

Evil Question:

Integrate SQRT(x^2 + 4 + 4/x^2)

EDIT: Solved it myself