The Forum > Math & Science > Math Problem Thread
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 Hydrogen777 Administrator5990 Posts11 Ineo 9:0 - 9.15.74940.954 days ago It's inconsistent. [Quote] [Link]
 Eh, gives multiple problems to do. I've actually been trying to find out that puzzle for a long time now. Very hard. [Quote] [Link]
 No, there is no solution to the problem. [Quote] [Link]
 Hydrogen777 Administrator5990 Posts11 Ineo 9:0 - 11.79.31940.822 days ago Which would explain why it's vexed you for a while. [Quote] [Link]
 SporeInsanity said:Sophrosyne said:Arcsine undoes Sine. The information you gather from a triangle is the Sine of ABC, which is not useful in finding the angle measurement. We can undo Sine though with Arcsine, just like how the sqrt(x2) == x. At least, I think that's how it is explained. ... but how does putting sine under a sign that means "basically" to give the reciprocal... how does that worK? 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? Never said:It's a Given. The two little dashy lines *tell* you that they're equal. 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.] [Quote] [Link]
 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) ? [Quote] [Link]
 Now, I could be horribly wrong. But this is how I'd do it: If you can see that [Quote] [Link]
 Ok, I think you might be right, thanks. [Quote] [Link]
 Is this proof correct? (cos(x))(sec(x)) = 1 (cos(x))(1/cos(x)) = 1 1 = 1 [Quote] [Link]
 Yes, it's correct. However, it makes more sense to set it out in the 'lhs/rhs' format. Makes it more coherent. [Quote] [Link]
 And don't forget domain! x can't equal pi/2, 3pi/2, etc. [Quote] [Link]
 Would a physics question go here too? [Quote] [Link]
 Yeah, anything related to Math can go here. [Quote] [Link]
 >:3c Good for me... [Quote] [Link]
 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 [Quote] [Link]
 Can we assume u=0? In that case, just use v2=u2+2as, to find a. Then F=ma. [Quote] [Link]
 I have no idea what u is [Quote] [Link]
 It would be the coefficient of friction. But with none given, I'd assume you're working in a frictionless environment. [Quote] [Link]
 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. [Quote] [Link]
 Just remember this, and everything should be fine when it comes to basic kinematics. [Quote] [Link]
 I use u=initial velocity (i.e. u=Vi) [Quote] [Link]
 That makes more sense, as I use Vo for initial. [Quote] [Link]
 Hydrogen777 Administrator5990 Posts11 Cresco 13:0 - 15.65.56826.629 days ago f([0, ln2)) Is anyone familiar with this notation? I am told to find its value for f(x) = ex 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. [Quote] [Link]
 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. [Quote] [Link]
 Hydrogen777 Administrator5990 Posts11 Cresco 13:0 - 16.77.13826.573 days ago The problem is that because the function is not one-to-one, its inverse function does not exist. Thank you for your time. [Quote] [Link]
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The Forum > Math & Science > Math Problem Thread