Recent content by mathie.girl

If you're writing these in cycle notation, then be careful that you permute any cycles bigger than two. When you chose two numbers for a transposition, say 2 and 5, then order didn't matter because (2,5) = 2 -> 5, 5 -> 2 = (5,2). But in a three-(or bigger)-cycle, (2,3,4) \neq (2,4,3)...

I assume you're asking how many elements are in the conjugacy class of (1,2)(3,4)(5,6)(7,8)(9,10)(11,12) in S_12. In that case, the answer that you posted said that you first chose two elements for the first transposition, then two for the second and so on. You divide by 6! because right now...

Great! Good luck with the rest of your work.

For associativity, you should look at both sides of the equation you listed ((x*y)*z = x*(y*z)) and use your rule to expand. For your associativity condition, z is any third element of R, as both x and y must be. So if you're asked to do y*z, then you fill in y and z in the place of x and y in...

I haven't used Maple 13, but it seems very unlikely to me that it can't plot. So you should be able to plot a graph in it, even if the language isn't the same as in v11. Did you try it in v13 with round brackets () instead of square []?

For (a) I entered the command: plot((x^3-1)/(sqrt(x)-1), x = 0 .. 10) into Maple (version 11) and it plotted okay. Are you inputting "f" or the actual function? Did you define f with "f:="? When I entered either function into Maple with the square brackets it didn't work. It did work...

Factoring was a correct way to do it (though that doesn't imply to me that it needs to be shown, especially in a calculus course). Alternatively, you could have just as easily have done: \frac{-32xh - 16h^2 + 20h}{h} = \frac{-32xh}{h} - \frac{16h^2}{h} + \frac{20h}{h} = -32x - 16h + 20...

I honestly couldn't imagine anyone taking off points for that, unless you were told specifically to include that step (in which case I don't think you'd have much of an argument). I encourage students to factor and then cancel because so many write something like \frac{xh + 6}{3h} =...

No problem. Good luck with the course!

That looks good. Since you're looking for the rate of change, you sub it into the derivative. And yup, -15/7 looks right. Sorry about doubting you before - I must have been having a weak moment! :wink:

When I did it, I got: \frac{dD}{dt} = 10(2)[2-\frac{t}{14}]*(-\frac{1}{14}) = -\frac{40}{14} + \frac{20t}{196} = \frac{5t}{49} - \frac{20}{7} which is exactly what you got. So I think it's likely right. For future reference, if you put the numbers between tags like...

Is this the equation you have? D=10(2-\frac{t}{14})^2 If so, then wouldn't you use the chain rule to get: \frac{dD}{dt} = 10(2)[2-\frac{t}{14}]*(-\frac{1}{14}) If you were avoiding the chain rule, and expanding instead, then you should have a -4 instead of a -2 here: And you...

Actually, looking at this particular problem, there's an easier way to do it. The last two terms you have are both \frac{1}{2^{12}}, so when you add them together, what do you get? What about when that's simplified? Will that work again? ((I'm leaving my other comment about the exponents...

If you were going to factor the expression x^5 + x^4 + x^2, what would you factor out? The x^2, right? That's because it has the smallest exponent. So for your expression, you want to factor out the power with the smallest exponent, which is -12. You could also factor out 2^{-10}, just like...

If you plugged 5 into f(x), then you calculated f(5) = f(g(x)), since g(x) = 5. You want to find g(f(x)), which means plug f(x) into g(x). If you're having trouble, maybe ask yourself, what's g(1), g(10), g(1000)? Once you see a pattern, then g(f(x)) shouldn't be much harder.