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
Start publishing on
DD Comics!
Extreme Math Challenge... Squared!
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?
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
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