TODAY'S PI DAY!!!

In honor of today, I will post math problems! If someone dares to answer a problem and answers correctly, I will post an even more difficult problem.

Your rewards through this process:
- new mathematical knowledge
- a newfound appreciation of math
- miscellaneous unknown things

FUN! FUN! FUN! GO!



Q1:
Solve.

1+1

An extreme math challenge, eh? How edumacational! Okay, I'll solve this one:

1+1=2

Ha! That was too easy (and I really suck at math)!

Hard mode guys.

4x - 1 = 7

WHAT IS X?

4x - 1 = 7
4x = 8 by the Subtraction Property of Equality
x = 2 by the Division Property of Equality

Therefore x = 2

:D!

Increase difficulty!

Solve for x:

|2x-4| = 12

2x-4 = 12
2x = 16
x = 8

2x-4 = -12
2x = -8
x = -4

sooo…
x equals -4 or 8.

:D!

Increase difficulty!

Solve for x:

x^2 + 24 = 10x

x^2 + 24 = 10x

x^2 -10x + 24 = 0

(x-6)(x-4) = 0

x = 6 or 4


Increase difficulty!

In a right angle triangle, the two shorter sides are 3 and 4 inches. What is the length of the longest side?

Since it's a Pythagorean triple, 5!

:3

Increase difficulty!

What is the sum of the coefficients in this expre20ssion:

20x^3 + 3x^2 - 7x + 13

The coefficients of
20x^3 + 3x^2 - 7x + 13
are 20, 3, -7 and 13.

Therefore the sum is 20 + 3 + (-7) + 13 = 29.


Increase difficulty!

Evaluate d/dx(x^2 + 2x + 4)

Decrease Dificulty

3x4+56-1223+10000=

Baconator's answer:
2x + 2

cool guy's answer:
8845


Increase difficulty!


Solve y(t) in terms of t:

y'(t) - y(t) = 0

I'm not quite sure about that one, but the way I see it…

y'(t) - y(t) = 0
y'(t) = y(t)

And for the derivative of a function y to be exactly equal to that function y, the only thing I could think is that..

y(t) = e^t
since d/dt(e^t) = e^t

But I'm not sure that I can exactly assume that…nor am I sure if that's the kind of thing you're looking for.

—————————–

Decrease difficulty!

If vectors u = (2x, 3(y^2), 1/z) and v = (1/x, -(y^-2), z), then what is the dot product of u and v?

WTF?

…

Shoop Da Hoop?

Hah I get where this is going.

ok Guys listen up!

1 + 1 =


1 = ln (e)
1 also = Sin^2(theta) + Cos^2(theta)

Where Cos^2 (theta) = (Cos2*(theta)+1)/2
Where Sin^2(theta) = (-Cos2*(theta)+1)/2


Thus giving you

ln(e) + ((Cos2*(theta)+1)/2 ) + ((-Cos2*(theta)+1)/2 ) = 1 + 1

Simple as Pi

Did I make a mistake?

Q2 was easy, like the guy up top said, e^y will do it :) since F' = F

To Baconator's last question:

Dot product equals 0.


*heh heh*
WARNING: EXTREME DIFFICULTY INCREASE!!!


Find the general solution of

y"(t) + y(t) = 2y'(t)

oh bugger, you said general… oh well, its a general ODE anyways, no forcing function. I don't wanna write it out…

The answer is y(t) = c1te^t + c2e^t in which c1 and c2 are distinct constants. Simple, is it not?

Reset difficulty.

Simplify: 2+2

4

INCREASE DIFFICULTY!

Solve:
6x9

6x9 = 54


—

INCREASE DIFFICULTY!

What is the slope of the line described by the following equation:

y^2 = 16(x^2)

Do you mean the gradient?

Nope.

Hint: the equation I gave is a line defined in 2-space…so compare it to the general equation of a line (y=mx + b)

It's a really simple question actually =P

Actually, it looks like a trick question since y^2 = 16(x^2) consists of two lines with slopes of 4 and -4 intersecting at the origin.

=) Good call

*when Bacon realizes he did something smart by accident and plays it off as if it were intentional*

XD

Okay, time for a word problem:

You are in a room with three doors and a guard. The guard tells you that behind one door is an exit and behind the other two are rabid tigers. He also, says that he knows which door the exit is behind. You choose a random door, but before you open it, the guard opens a different door which contains a tiger and then immediately closes it. He, then, gives you the option of choosing the other unopened door.

The question: Which door would you choose?


Let me clarify:

- You chose door #1, a tiger is behind door #2, you have the option of choosing door #3

- This ISN'T a lateral thinking puzzle in the sense that you can't stay in the room until the tigers die or make the tigers attack the guards while you escape or do some abstract stunt that “magically” gets you out.

- This question DOES involve math (mainly probability), so to get full credit, SHOW YOUR WORK (AKA Don't guess, think)!!!

so the chances of a door not having a tiger are 1/3
but he eliminates one of the doors which does have a tiger so the chances are 1/2

but the chances of a door having a tiger are 2/3
with the revealed tiger that makes it… 1/3?

so the chances of your chosen door having a tiger are 1/3 still… i think…

I'd stay with my chosen door

I hate showing work… but I'd have to say door number three.

The Psycho… Guard I mean… shows the middle door, so, the two tigers would be in adjacent rooms. By doing this, I'd feel MUCH safer with the next door… on the chance that he showed me the SECOND tiger.

X*X=Y
Y+X=Z
Z/X=B
B-X=1

This is a little problem I figured out by accident in math once: All of the above problems leads to the next. Solve for X.

humorman
Okay, time for a word problem:

You are in a room with three doors and a guard. The guard tells you that behind one door is an exit and behind the other two are rabid tigers. He also, says that he knows which door the exit is behind. You choose a random door, but before you open it, the guard opens a different door which contains a tiger and then immediately closes it. He, then, gives you the option of choosing the other unopened door.

The question: Which door would you choose?

Well, if you initially pick the door with the exit, and the guard shows you a door with a tiger, then you have the choice between your exit and a tiger door, that's 1/2 chance each.

But, if you choose a door with a tiger initially, then, since the guard reveals the other door with a tiger, by switching, you'd end up at the exit.

Since there are two doors with tigers, this switching to the exit can occur twice, while switching to a tiger can only occur once. Thus, the probability of switching doors to the exit is 2/3, and switching to a door is 1/3.

And since there are 2 doors with a tiger, and one with the exit, by keeping the door you originally chose, that's 2/3 chance of failure, and 1/3 chance of success.

Thus it is best to switch doors.

BloodTh
X*X=Y
Y+X=Z
Z/X=B
B-X=1

This is a little problem I figured out by accident in math once: All of the above problems leads to the next. Solve for X.

B-X=1
B=1+X

Z/X=B
Z/X=1+X
Z=X+X^2

Y+X=Z
Y+X=X+X^2
Y=X^2

X*X=y
X*X=X^2
X^2=X^2

Err, yeah there is no real “solving” for x here.

———–

Here's a cute and simple one:

How do you make 1000 by the addition of numbers whose digits are only 8's and where the sum of the number of digits of all numbers used is 8?

(incase you don't understand that second part of the question, I mean say you add 34 and 233, then the sum of the number of their digits is 5)