It's inconsistent.

Eh, gives multiple problems to do. I've actually been trying to find out that puzzle for a long time now. Very hard.

No, there is no solution to the problem.

Which would explain why it's vexed you for a while.

Inverse sine or arcsine really just makes you ask yourself "At what angle is sine ___?"
If you had the sin^-1(1); the answer would be "at ∏/2 on the unit circle.
Are you familiar at all with the unit circle?


If two chords on a circle are the same length, then the angle that makes the two segments are also congruent.

[Mod-Edit: No need to double post.]

Ok so I'm having trouble with this one.

What is the solution to the equation 1 over the square root of 8 = 4^(m + 2) ?

Now, I could be horribly wrong. But this is how I'd do it:

If you can see that

Ok, I think you might be right, thanks.

Is this proof correct?

(cos(x))(sec(x)) = 1
(cos(x))(1/cos(x)) = 1
1 = 1

Yes, it's correct. However, it makes more sense to set it out in the 'lhs/rhs' format. Makes it more coherent.

And don't forget domain! x can't equal pi/2, 3pi/2, etc.

Would a physics question go here too?

Yeah, anything related to Math can go here.

>:3c Good for me...

A 3.7 g bullet leaves the muzzle of a rifle with
a speed of 317 m/s

What total constant force is exerted on the
bullet while it is traveling down the 0.78 m
long barrel of the rifle?
Answer in units of N

Can we assume u=0? In that case, just use v²=u²+2as, to find a. Then F=ma.

I have no idea what u is

It would be the coefficient of friction. But with none given, I'd assume you're working in a frictionless environment.

Well I have it all figured out and I'm not sure why I couldn't before...

You use the velocity and distance to calculate the time it takes to leave the barrel with this formula
d = 1/2 ( Vf + Vi ) × t
You then use the time to calculate the acceleration using this formula
a = ( Vf - Vi ) / t
Now that you have acceleration, you can use
F = m a
just remember that the mass of the bullet has been given in grams and thus needs to be converted to kilograms
When you are calculating acceleration, you WILL get a big number
but it gets more normal when you calculate force because the mass of the bullet is so small.

Note: This was just copypasta'd from my facebook chat explaining to a friend.

Just remember this, and everything should be fine when it comes to basic kinematics.

I use u=initial velocity (i.e. u=V_i)

That makes more sense, as I use V_o for initial.

f([0, ln2))

Is anyone familiar with this notation? I am told to find its value for f(x) = e^x for a homework assignment, but I'm not sure what it means. The assignment gives the hint "use some calculus", so I'm wondering if it's shorthand for integration over the interval, or something like that.

On a related note, does anyone know how to prove that a function which is not one-to-one is nonetheless onto? The only method I've learned for demonstrating a function is onto involves finding its inverse and showing that each element in the range maps to an element in the domain via the inverse function, but of course I can't use that method if the function is not one-to-one.

By the way, this is for a Classical Analysis course.

I've never seen that notation myself, but it looks like it would be some sort of integral.

I had to prove this for subsets of real numbers for a combinatorics course I took. I think you can take the inverse of the function and show.that every element of the range produces some, not necessarily unique, value of the domain. It will be also one to one if each range produces a unique domain value. So if the inverse is defined for all the range, bit doesn't pass the vertical line test it still can be onto.

It's late and my class was a year ago, but I hope it helps.

The problem is that because the function is not one-to-one, its inverse function does not exist.

Thank you for your time.