Recent content by LearninDaMath

If I want to complete the square with (-x^{2}-4x+3) I would write (-x^{2}-4x+(...) +3 - (...)) = (-x^{2}-4x+4+3-4) = (-x^{2}-4x+4-1) = (x^{2}+4x-4) - 1Why does adding the parentheses to separate the -1 change all the signs. I understand it has something to do with factoring out a...

∫[(2)/(x-3)(√(x+10))]dx U-Substitution → du/dx or dx/du, & why? Okay, here I solve \int\frac{2}{(x+3)\sqrt{x+10}}dx in two ways. The problem I'm having however is during the U substitution. The first method, I take \frac{du}{dx}= (function in terms of x) For the second method, I set...

Thanks, from there, I have tried to work it out, however it just doesn't feel correct for some reason. These are all the steps I made, is it correct?4\int\frac{1}{u^{2}-7} = \frac{A}{u+\sqrt{7}}+\frac{B}{u-\sqrt{7}} A(u - \sqrt 7) + B(u + \sqrt 7) = (A + B)u + (B - A)\sqrt 7 = 1 thus A + B =...

Cool, thanks, so in proceeding into partial fractions, I am a little hesitant in how to set it up. I have two ways in mind. ________________________________________ Attempt 1: 4\int\frac{1}{u^{2}-7} = \frac{Au+B}{u^{2}-7} [1 = Au + B], ? Maybe B = 1, but then what does A equal? Would A=0...

\int\frac{2}{(x+3)\sqrt{x+10}}dx _____________________________________ First thing would be u-substitution, finding what I can replace in terms of u: let u=\sqrt{x+10} \frac{du}{dx}=\frac{1}{2}(x+10)^{\frac{1}{2}-\frac{2}{2}}(x+10)' du=\frac{1}{2\sqrt{x+10}}dx → dx=2\sqrt{x+10}du...

I wonder if this question makes any sense or not? If I have 10 blue marbles and drop them into a Liter of water, the concentration will be 10 blue marbles/1L If I have 10 red marbles and drop them into a liter of water, the concentration will be 10 red marbles/1L With 10 blue marbles...

Say I have 10(H2O) that ionizes into H and OH. The reaction would be 10(H2O)→10(H) +10(OH) Because if you have 10 of something, and then then split it into 2 pieces, you have 10 of each piece. If I have 1 piece of paper and I tear it into 2 pieces, I have 1 piece in my right hand and 1...

------------------------- Thanks just did that. Got 1 out of every 5.58x10^{9} H2O molecules will dissociate into H^{+} and OH^{-} I also did \frac{55.5molH2O}{10^{-7}}= 5.5x10^{9} Since both values are the same, I take it that for the first result, I can say 1 molecule out of every 5.5E9...

This is where I saw that particular 1 out of 10E7 value: I stopped watching about a minute into the video when she said 10^7 moles of H2O on one side of the equation and then on the other side of the equation she had 10^{-7) or something to that affect. I couldn't see how those numbers...

Firstly, I have just read that at 25degC, 1 out of every 10,000,000 H2O molecules breakes into H^{+} and OH^{-}. If this true, then would this be correct: Assuming no ionization occurs yet.. If 1 liter of H2O = 1000grams, and 1mol H2O = 18.015g, then by Dimensional Analysis...

Thanks SammyS, Also, You show that the derivative of arctanx is: \displaystyle \frac{d}{dx}\tan^{-1}\left(\frac{x}{a}\right)=\frac{1}{a}\frac{1}{(x/a)^2+1} But if I am interested in deriving the function \frac{1}{a}tan^{-1}(\frac{x}{a}) for this particular formula, does the extra fraction...

sin^{-1}\frac{x}{a}=y \frac{d}{dx}sin^{-1}=\frac{d}{dx}\frac{x}{a} cos(y)y'= \frac{1}{a} y'=\frac{1}{acos(y)} y'=\frac{1}{acos(sin^{-1}(\frac{x}{a})^{2})} Trig Identity: sin^{2}y+cos^{2}y=1 cosy=\sqrt{1-(\frac{x}{a})^2} y'=\frac{1}{acos(y)}=\frac{1}{a\sqrt{1-(\frac{x}{a})^{2}}}So when I...

I didn't mean proof, sorry about that. I guess I just meant formula. Thanks, I'll give it a try.

Hi Hallsofivy, really appreciate it. However, of the links you kindly provided, there is only one instance of a formula of which I'm describing. One for arcsin. But there seems to be no resource on the web that displays all or even 2 or 3 of these particular formulas. I've searched every...

Yes, they are talking about degrees defined in terms of something similar with smaller degrees when mentioning "reduction formulas." Thanks for clarifying and confirming they are different words meaning the same thing in this case.