Recent content by InfinityZero

You are right, but checking it with a calculator doesn't work because the screen of the calculator can't display all the digits of the product, if you use the calculator on your computer it should display the full number.

Well there is a quick way of finding the inverse of a 2x2 matrix A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} A^{-1}=\frac{1}{det(A)}*adj(A) and the adjugate of a 2x2 matrix is just \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} the determinant is ad-bc which...

But since 3-x is always positive we can eliminate one of the solutions to the quadratic and be left with a unique inverse, right? This is of course only if the domain only includes real numbers.

Well that would be right if it weren't for the pressure thing. Basically what you would do is 3r_1=r_2 because 12/4=3 V_1=4/3 \pi r_1^3V_2=4/3 \pi r_2^3=4/3 \pi (3r_1)^3 so you will get V_2=27V_1 then do the same for pressure, then try to express n_2~ in~terms ~of ~n_1 and remember that...

Well by from the gas law PV=nRT where P is pressure, V is volume, n is the number of mols gas involved(or for this question we can just say it is the number of molecules), R is the gas constant, and T is temperature. You will want to find a ratio between the number of molecules in the 8cm...

Did you account for the fact that gasses under higher pressure occupy a smaller volume when under pressure? That seems to be vital to the question.

Does the question state where the hole is compared to the fox and rabbit. Your result is correct if the rabbit is directly between the fox and the hole. but what if the hole was between the fox and rabbit?

This is right, note that there are two terms on the right integral. Umm, I don't think that's right remember that a^{b}*a^{c}=a^{b+c} I also don't know where your dx terms are coming from there. You seem to have multiplied your exponents correctly, but forgot that there is another...

The limit is of indeterminate form \frac{0}{0} so you can apply L'Hopital's rule to it to get the limit \lim_{x\rightarrow0}\frac{af'(ax)-af'(bx)}{1} which since we know f'(0)=2 should be easy to evaluate.