Search results

For chess.com, membership is required. Lichess can be played without an account.

Yeah, meant OTB ratings. What are the main reasons for these inaccuracies? Is it because people are less serious in online play? (And probably because some people don't play that much or often online.)

By the way, how accurate are the ratings on lichess? Are they like decently accurate at the upper end but not so much towards the lower end?

Yes, this is the one.

$\noindent 1. For $x\geq 0$, it just looks like the graph of $y=x^3+1$. For $x<0$, reflect the $x\geq 0$ part of the graph about the $y$-axis. So it looks like this:$...

$\noindent We just write that as $5 + \frac{7}{x} - \frac{6}{x^2}$, and then it's a standard integration problem.$

What is? The questions here, or LaTeX?

I used this thing called LaTeX, which you can use on this forum to type mathematical expressions. There's a short guide to how to use it here on BOS: http://community.boredofstudies.org/13/mathematics-extension-1/234261/short-guide-latex.html

$\underline{QUESTION 4}$ $\noindent I think this was a typo too, since the answer written here doesn't agree with the question (just try some values for $a$ and $b$ to see this; also, the right bracket of the answer would have had $3a^2$ rather than $a^2 + 2a^2$, so I suppose there's a typo...

$\underline{QUESTION 3}$ $\noindent I think the expression for Question 3 is meant to be $a^4 + a^2b^2 + b^4$, rather than what you typed (I think you may have done a typo; there's no really nice way to factorise the one you typed). Assuming it's $a^4 + a^2b^2 + b^4$, we have$...

$\underline{QUESTION 2}$ $\noindent We have$ $$\begin{align*}\left (ax+by \right )^2+\left ( ay-bx \right )^2+c^2 \left ( x^2 + y^2 \right ) &= a^2 x^2 \color{red}{+ 2axby}\color{black} + b^2y^2 + a^2 y^2 \color{red}- 2aybx \color{black}+ b^2 x^2 + c^2x^2 + c^2 y^2 \quad (\text{via...

$\underline{QUESTION 1}$ $We have$ $$\begin{align*}\left(a^2-b^2-c^2\right)^2 - 4b^2c^2 &=\left(a^2-b^2-c^2\right)^2 - \left(2bc\right)^2 \\ &= \left(a^2-b^2-c^2 + 2bc \right)\left(a^2-b^2-c^2 - 2bc \right) \quad (\text{difference of two squares}) \\ &= \left ( a^2-\left ( b^2-2bc+c^2...

Oh yeah, sorry, that was just a typo by me. It doesn't affect the rest of the working. :)

I wouldn't say it's cheating. According to Google, ''cheat'' means ''act dishonestly or unfairly in order to gain an advantage''. It's not like the students acted unfairly or anything, since the questions are publicly available. If the questions were online in a secure place though (and needed...

How can so many people copy off one person in a Yearly exam? Wouldn't you need to be sitting right next to that person to copy? Wouldn't they make the class sit apart from each other during exams to prevent copying? And wouldn't it be obvious to the exam invigilators if so many people were...

$\noindent So here's how to do it. We have$ $$\begin{align*}4x^4 - 2x^3 y -3xy^3 - 9y^4 &= \left(4x^4 - 9y^4\right) - \left(2x^3y+3xy^3 \right) \\ &= \left(2x^2 + 3y^2 \right) \left(2x^2 - 3y^2 \right) - xy \left(2x^2 + 3y^2 \right) \quad (\text{since }4x^4 - 9y^4= \left(2x^2 + 3y^2 \right)...

$\noindent Your answer is correct except that you should have a bracket around the $-xy$ instead. I.e. it should be $\left( 2x^2+3y^2\right)\left(2x^2-3y^2 -xy\right)$, so maybe you just made a silly mistake somewhere? Once you have this, you just need to factorise the $\left(2x^2-3y^2...

What is the question?

Maybe check out this thread: http://community.boredofstudies.org/263/space/334473/g-force.html

$\noindent What do you mean? You find the actual value of $b$ by solving the equations using reversible steps (since there turns out to be exactly one solution, $b$ is unique). Taking logs of negative numbers though would be non-elementary so you'd maybe want to justify that $b$ is positive...