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Lyde Lyde
09-24-2005, 02:39 AM
Here you can challenge other members with math problems or give links to quiz sites here visit this one, I rather enjoy it. This is fun.
http://users.adelphia.net/~sismondo/VisCalc.html and here's the first question
127x16 divided by 4 is what % of 1000?
A)25 B)58 C)54 D)52 E)none of these
Be careful choose wisely.

Doomie
09-24-2005, 02:42 AM
E) None of the above

radyk05
09-24-2005, 02:43 AM
E)none

60.4%

Shlup
09-24-2005, 02:45 AM
I hate math soooo much. :mad2:

radyk05
09-24-2005, 02:59 AM
Picture this:

Two cars are placed on a ceratin point in a highway facing oposite directions (they are on the same point). Car A drives at 50 miles/hr and car B drives at 60 miles/hr for HALF an hour. They both make a right turn. Car A then drives at 10 miles/hr while car B drives at 20 miles/hr for another half hour before they both stop. How far away are the cars from each other?

(A) 15 miles (B) 55 miles (C) 70 miles (D) 57 miles (E) 63 miles

Mo-Nercy
09-24-2005, 03:06 AM
(D)

...approximately.

radyk05
09-24-2005, 03:07 AM
*golden star*

Mo-Nercy
09-24-2005, 03:14 AM
Yay! Banzai!

Here's a question from last years HSC (this is 2U only, so it's a cinch):

---

Show that cos x . tan x = sin x
And hence solve 8sinx . cos x . tan x = cosec x (for values between zero and 2pi)

Doomie
09-24-2005, 03:15 AM
E)none

60.4%

I got 50.8 :O

radyk05
09-24-2005, 03:18 AM
tan x = sin x/ cos x
so
cos x * sin x/ cos x = sin x (cancel the cosines)

8*sin x * cos x * tan x = 8*(sin x)^2

del mordor: oops....

Doomie
09-24-2005, 03:19 AM
I think the 8 was a mistake.

Mo-Nercy
09-24-2005, 03:21 AM
8*sin x * cos x * tan x = 8*(sin x)^2

You haven't solved for x, nor proved that it's equal to cosec x.

And yes, the 8's supposed to be there.

radyk05
09-24-2005, 03:21 AM
no matter

sin x * cos x * tan x = (sin x)^2

Mo-Nercy
09-24-2005, 03:24 AM
no matter

sin x * cos x * tan x = (sin x)^2
sinē x does not equal cosec x.

And there IS an 8.

Doomie
09-24-2005, 03:24 AM
cosec x doesn't equal sin^2. cosec x = 1/sin x

radyk05
09-24-2005, 03:25 AM
yes, i know. what is the problem?

Doomie
09-24-2005, 03:26 AM
yes, i know. what is the problem?

All you did was change the right side of the equation from cosec x to sinx^2. That's the problem. xD

Mo-Nercy
09-24-2005, 03:26 AM
Show that cos x . tan x = sin x
And hence solve 8sinx . cos x . tan x = cosec x (for values between zero and 2pi)
You get marks for part one but you haven't proven 8sinx . cos x . tan x is equal to cosec x. You also haven't solved for x.

loza
09-24-2005, 03:28 AM
I hate math soooo much. :mad2:

now that is the truth my friends! :p

radyk05
09-24-2005, 03:31 AM
yes, i know. what is the problem?

All you did was change the right side of the equation from cosec x to sinx^2. That's the problem. xD

didn't i showed that 8* sin x * cos x * tan x is not equal to cosec x?

8 * sin x * (cos x * tan x) = cosec x
8 * sin x * (sin x) = cosec x
8 * (sin x)^2 = cosec x
8* (sin x)^2 = (sin x)^(-1) ---> false

;)

Doomie
09-24-2005, 03:33 AM
yes, i know. what is the problem?

All you did was change the right side of the equation from cosec x to sinx^2. That's the problem. xD

didn't i showed that 8* sin x * cos x * tan x is not equal to cosec x?

8 * sin x * (cos x * tan x) = cosec x
8 * sin x * (sin x) = cosec x
8 * (sin x)^2 = cosec x
8* (sin x)^2 = (sin x)^(-1) ---> false

;)

I'm doing it on paper and got the EXACT same thing. But apparently there's a way to solve this.

-N-
09-24-2005, 03:41 AM
Taking what's done...

8*(sin x)^2 = (sin x)^-1
8*(sin x)^3 = 1
2 sin x = 1
x = 30 degrees

radyk05
09-24-2005, 03:43 AM
i fail to recognize that as a proof.

edit: nevermind. the question is poorly writen, tho.
ahem: for which values of x is 8 * sin x * cos x * tan x = cosec x?

Doomie
09-24-2005, 03:54 AM
I say we give it to Neel even if the question isn't completely answered. I hate proving trig identities + anything to do with radians. Not that it's hard, just that I dislike it. The thing with the generators and such. Bleh.

eestlinc
09-24-2005, 03:54 AM
sin<sup>2</sup>x looks better than sin^2 x

-N-
09-24-2005, 03:54 AM
Mo never asked for a proof, simply a solution.

Here's a proof-type question for you.

Say I start walking at the Cartesian coordinate (0,0) and I need to go to (1,1). I can walk at right angles, going to (0,1) then (1,1) for a distance of 2, but it would be shorter to go at 45 degrees straight to (1,1), for a distance of 1.414. But say I move infinitesimally north, to (0,delta), then infinitesimally east, to (epsilon,delta), then repeat this in a staircase pattern. As epsilon and delta go to zero, this path should represent the 45 degree path from (0,0) to (1,1). However, as long as epsilon and delta are finite, I will have to walk a distance of 2. Is it a contradiction that my right-angle approximation of the 45 degree path still results in a distance of 2? Is the 45 degree path really shorter?

Doomie
09-24-2005, 03:56 AM
sin<sup>2</sup>x looks better than sin^2 x

Teach me how do do a squared thingie!

radyk05
09-24-2005, 03:58 AM
he did answer the question because

8 * sin (30) * cos (30) * tan (30) = cosec (30) ---> true

if anyone cares for an easy-but-hard projectile problem i will be more than happy to bring it to the thread (even tho its a physics problem but lots of math are involved).

eestlinc
09-24-2005, 04:00 AM
sin<sup>2</sup>x looks better than sin^2 x

Teach me how do do a squared thingie!
sin< sup >2< /sup >x

-N-
09-24-2005, 04:00 AM
Wow, posts go boom. I posted a thing to solve up there.

The Summoner of Leviathan
09-24-2005, 04:00 AM
if anyone cares for an easy-but-hard projectile problem i will be more than happy to bring it to the thread (even tho its a physics problem but lots of math are involved).

It has been a while but I can give it a shot.

Doomie
09-24-2005, 04:01 AM
Mo never asked for a proof, simply a solution.

Here's a proof-type question for you.

Say I start walking at the Cartesian coordinate (0,0) and I need to go to (1,1). I can walk at right angles, going to (0,1) then (1,1) for a distance of 2, but it would be shorter to go at 45 degrees straight to (1,1), for a distance of 1.414. But say I move infinitesimally north, to (0,delta), then infinitesimally east, to (epsilon,delta), then repeat this in a staircase pattern. As epsilon and delta go to zero, this path should represent the 45 degree path from (0,0) to (1,1). However, as long as epsilon and delta are finite, I will have to walk a distance of 2. Is it a contradiction that my right-angle approximation of the 45 degree path still results in a distance of 2? Is the 45 degree path really shorter?

...yes.

Mo-Nercy
09-24-2005, 04:02 AM
Mo never asked for a proof, simply a solution.

Yeah. Sorry guys. Bad wording on my part. Somewhere along the line I asked for a proof when all that was required was a solution.

*blush*

So embarassed. So much for being good at maths.

eestlinc
09-24-2005, 04:06 AM
Mo never asked for a proof, simply a solution.

Here's a proof-type question for you.

Say I start walking at the Cartesian coordinate (0,0) and I need to go to (1,1). I can walk at right angles, going to (0,1) then (1,1) for a distance of 2, but it would be shorter to go at 45 degrees straight to (1,1), for a distance of 1.414. But say I move infinitesimally north, to (0,delta), then infinitesimally east, to (epsilon,delta), then repeat this in a staircase pattern. As epsilon and delta go to zero, this path should represent the 45 degree path from (0,0) to (1,1). However, as long as epsilon and delta are finite, I will have to walk a distance of 2. Is it a contradiction that my right-angle approximation of the 45 degree path still results in a distance of 2? Is the 45 degree path really shorter?
Yes the 45 degree path is shorter. As long as your minute stairsteps have a definite length, each individual staircase is an exact replica of the full coordinate square between (0,0) and (1,1) and each stairstep can be more quickly traversed by taking the hypotenuse rather than both sides of the stairstep. The smaller the size of each stairstep, the more stairsteps are necessary to cover the full distance, and although each tiny stairstep is closer in length to the actual hypotenuse, say perahps a difference of .00001, you will have to travel 100000 of these steps to arrive at the point (1,1).

Or do you want a more formal proof?

X = horizontal distance traveled = 1
Y = vertical distance traveled = 1
X+Y = 2
k = number of steps (let's say for convenience that each step is the same size, although it doesn't matter whether they are or not)

distance of each step horizontal = D<sub>h</sub>
distance of each step vertical = D<sub>v</sub>

D<sub>h</sub> = X/k
D<sub>v</sub> = Y/k

D<sub>total per step</sub> = X/k + Y/k

D<sub>interval</sub> = D<sub>total per step</sub> * k

D<sub>interval</sub> = k(X/k + Y/k) = k(X/k) + k(Y/k) = x+y = 2

this is true regardless of the value of K and is thus independent of the value of K (the number of steps)

therefore:

lim<sub>k-oo</sub> D<sub>interval</sub> = lim<sub>k-oo</sub> (X+Y) = lim<sub>k-oo</sub> (2) = 2.

radyk05
09-24-2005, 04:19 AM
a projetile that is fired with an initial velocity v at an angle θ will pass through two points at hight h. show that, if the gun is set for maximun reach, the distance between the two points is

d = v/g * (v^2 - 4*g*h)^(1/2)

where g stands for the gravitational acceleration.
i'm going to make things a bit easier for those who haven't taken a physics or calculus course:
remember that we're dealing in two dimenssions so,
velocity in y, vy = sin θ * v
velocity in x, vx = cos θ * v

the equation for movement are:
x = xinitial + vx,initial * t
y = yinitial + vy,initial + (1/2) * g * t^2

where t stands for time.

that is all that you need.

have fun, i sure did.

Del Murder
09-24-2005, 04:20 AM
Math looks so ugly typed out.

radyk05
09-24-2005, 04:24 AM
Math looks so ugly typed out.yes it does.

Endless
09-24-2005, 10:13 AM
y = yinitial + vy,initial + (1/2) * g * t^2

I sense a typo.
Edit:
Try y = y<sub>0</sub> + v<sub>y,0</sub> * t - (1/2) * g * tē

Edit2:
While I'm at it:

v<sub>x</sub> = dx/dt = v<sub>x,0</sub>
v<sub>y</sub> = dy/dt = v<sub>y,0</sub> - g * t

At the max height reached, v<sub>y</sub> = 0, therefore t<sub>max height</sub> = v<sub>y,0</sub>/g
At that t<sub>max height</sub>, half the horizontal distance is covered, so x<sub>max height</sub> = x<sub>0</sub> + v<sub>x,0</sub> * v<sub>y,0</sub>/g = sin θ * v * cos θ * v /g, which is max for θ = pi/4.
Now we can clean the mess, and rewrite:
x = x<sub>0</sub> + sqrt(2)/2 * v * t
y = y<sub>0</sub> + sqrt(2)/2 * v * t - (1/2) * g * tē

Now, let's take a point y<sub>h</sub>.
y<sub>h</sub> = y<sub>0</sub> + sqrt(2)/2 * v * t<sub>h</sub> - (1/2) * g * t<sub>h</sub>ē
h = y<sub>h</sub> - y<sub>0</sub>
h = sqrt(2)/2 * v * t<sub>h</sub> - (1/2) * g * t<sub>h</sub>ē

We solve this for t<sub>h</sub>, we get:
t<sub>h,2</sub> = sqrt(2)/2g *(v + sqrt(vē - 4*h*g))
and
t<sub>h,1</sub> = sqrt(2)/2g *(v - sqrt(vē - 4*h*g))

d = x<sub>2</sub> - x<sub>1</sub> = sqrt(2)/2 * v * (t<sub>h,2</sub> - t<sub>h,1</sub>)
d = sqrt(2)/2 * v * (sqrt(2)/2g *(v + sqrt(vē - 4*h*g)) - sqrt(2)/2g *(v - sqrt(vē - 4*h*g)))
d = v/g * sqrt(vē -4*h*g)

theundeadhero
09-24-2005, 05:34 PM
Prove 2.9999999~ doesn't = 3 :p

Odaisé Gaelach
09-24-2005, 05:38 PM
Pi is exactly 3!

Destai
09-24-2005, 05:40 PM
Math? *vomits*

("x=none of the above" is for wusses)

radyk05
09-24-2005, 06:52 PM
*two golden stars for endless* (one is for fixing the equation)

-N-
09-25-2005, 01:07 AM
Yes, eestlinc. Moreover, the "paradox" illustrates the fact that linear paths don't need to be approximated, and illustrates how problems arise as a result.

eestlinc
09-25-2005, 01:09 AM
Prove 2.9999999~ doesn't = 3 :p
or rather, prove that it does.

Nick Schovitz
09-25-2005, 03:53 PM
Oh my God man my brain hurts! The college professor had to start this. I took that quiz and made a 42. Gosh nerds man!!!! I'm staying out of this thread now and beware that first question was just a warm up from Lyde and do you think that Quistis looks more like a librarian or a math teacher than a military teacher? Yes I do.

Alive-Cat
09-25-2005, 04:19 PM
What??? Maths? This is some plot to confuse me...

Lyde Lyde
09-26-2005, 06:18 PM
Actually the answer to mu first question was 50.8% so that person was right now keep it up, I'm in a rush now, so no time for another question, good bye.