Math Forum game Yay!

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.

E) None of the above

E)none

60.4%

I hate math soooo much. :mad2:

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

(D)

...approximately.

*golden star*

Yay! Banzai!

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

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

Quote:

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Originally Posted by radyk05

E)none

60.4%

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I got 50.8 :O

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....

I think the 8 was a mistake.

Quote:

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Originally Posted by radyk05

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

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

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

no matter

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

Quote:

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Originally Posted by radyk05

no matter

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

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sin² x does not equal cosec x.

And there IS an 8.

cosec x doesn't equal sin^2. cosec x = 1/sin x

yes, i know. what is the problem?

Quote:

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Originally Posted by radyk05

yes, i know. what is the problem?

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

Quote:

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Originally Posted by Mo-Nercy

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

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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.

Quote:

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Originally Posted by ShlupQuack

I hate math soooo much. :mad2:

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now that is the truth my friends! :p

Quote:

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Originally Posted by Del Mordor

Quote:

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Originally Posted by radyk05

yes, i know. what is the problem?

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

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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

;)

Quote:

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Originally Posted by radyk05

Quote:

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Originally Posted by Del Mordor

Quote:

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Originally Posted by radyk05

yes, i know. what is the problem?

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

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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

;)

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

Taking what's done...

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

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?

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.

sin²x looks better than sin^2 x

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?

Quote:

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Originally Posted by eestlinc

sin²x looks better than sin^2 x

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Teach me how do do a squared thingie!

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).

Quote:

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Originally Posted by Del Mordor

Quote:

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Originally Posted by eestlinc

sin²x looks better than sin^2 x

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Teach me how do do a squared thingie!

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sin< sup >2< /sup >x

Wow, posts go boom. I posted a thing to solve up there.

Quote:

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Originally Posted by radyk05

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).

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

Quote:

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Originally Posted by Neel With A Hat

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?

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...yes.

Quote:

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Originally Posted by Neel With A Hat

Mo never asked for a proof, simply a solution.

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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.

Quote:

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Originally Posted by Neel With A Hat

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?

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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_h
distance of each step vertical = D_v

D_h = X/k
D_v = Y/k

D_{total per step} = X/k + Y/k

D_interval = D_{total per step} * k

D_interval = 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_k-oo D_interval = lim_k-oo (X+Y) = lim_k-oo (2) = 2.

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.

Math looks so ugly typed out.

Quote:

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Originally Posted by Del Murder

Math looks so ugly typed out.

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yes it does.

Quote:

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Originally Posted by radyk05

y = yinitial + vy,initial + (1/2) * g * t^2

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I sense a typo.
Edit:
Try y = y₀ + v_y,0 * t - (1/2) * g * t²

Edit2:
While I'm at it:

v_x = dx/dt = v_x,0
v_y = dy/dt = v_y,0 - g * t

At the max height reached, v_y = 0, therefore t_{max height} = v_y,0/g
At that t_{max height}, half the horizontal distance is covered, so x_{max height} = x₀ + v_x,0 * v_y,0/g = sin θ * v * cos θ * v /g, which is max for θ = pi/4.
Now we can clean the mess, and rewrite:
x = x₀ + sqrt(2)/2 * v * t
y = y₀ + sqrt(2)/2 * v * t - (1/2) * g * t²

Now, let's take a point y_h.
y_h = y₀ + sqrt(2)/2 * v * t_h - (1/2) * g * t_h²
h = y_h - y₀
h = sqrt(2)/2 * v * t_h - (1/2) * g * t_h²

We solve this for t_h, we get:
t_h,2 = sqrt(2)/2g *(v + sqrt(v² - 4*h*g))
and
t_h,1 = sqrt(2)/2g *(v - sqrt(v² - 4*h*g))

d = x₂ - x₁ = sqrt(2)/2 * v * (t_h,2 - t_h,1)
d = sqrt(2)/2 * v * (sqrt(2)/2g *(v + sqrt(v² - 4*h*g)) - sqrt(2)/2g *(v - sqrt(v² - 4*h*g)))
d = v/g * sqrt(v² -4*h*g)

Prove 2.9999999~ doesn't = 3 :p

Pi is exactly 3!

Math? *vomits*

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

*two golden stars for endless* (one is for fixing the equation)

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.

Quote:

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Originally Posted by theundeadmonkey

Prove 2.9999999~ doesn't = 3 :p

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or rather, prove that it does.

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.

What??? Maths? This is some plot to confuse me...

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.