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Jun
28
comment Integration with substitution
The answer is correct. There are limits on the $x$ for which this makes sense.
Jun
28
comment Issue with modelling word problems as cubic equations
The perimeter of the original rectangle is $20$. So length plus width is $10$. One side (call it the length) was divided into $4$ equal parts of size $x$, so the length is $4x$. The width therefore is $10-4x$.
Jun
28
comment A and B play until one has 2 more points than the other…
MSE has an enormous number of such problems, just hard to find. I have heard good things about the Khan Academy.
Jun
28
comment A and B play until one has 2 more points than the other…
Here it is with lots of symbols. Let $W$ be the event A wins, let $I$ be the event "instant win" ($2$ games) and let $T$ be the event "tie after $2$." Then $W=I\cup (T\cap W)$. So $\Pr(W)=\Pr(I)+\Pr(T\cap W)$. And $\Pr(T\cap W)=\Pr(T)\Pr(W|T)$. But $\Pr(W|T)=\Pr(W)=a$. But the point is that if there is a tie after $2$, A has just the same chance of ultimately winning as she had at the beginning.
Jun
28
revised A and B play until one has 2 more points than the other…
added 595 characters in body
Jun
28
comment A and B play until one has 2 more points than the other…
Player A wins if she wins the first two games. Clear. She also wins if the first two games end AB or BA, but A ultimately manages to win. Given they are tied after the first two games, A's probability of winning is exactly the same as her initial probability of winning, hence the result. I can make an explicit conditional probability calculation if you wish. I will also add a remark giving a different way of solving the A wins problem.
Jun
28
revised A and B play until one has 2 more points than the other…
added 22 characters in body
Jun
28
comment A and B play until one has 2 more points than the other…
I will be around for a while, I have made the argument somewhat compact.
Jun
28
revised A and B play until one has 2 more points than the other…
added 370 characters in body
Jun
28
answered A and B play until one has 2 more points than the other…
Jun
28
comment Did I Solve This Composite Function Correctly or Not?
There is a technical point that the official solution probably slides over. I would not be surprised if they gave $\frac{x^2+x+1}{x^2+2x+1}$ as the answer. This is simpler than the more correct answer that has been given below. However, the simplified version is well-defined at $x=0$, while $f(x)$ is not.
Jun
28
answered Prove that $\dfrac{0.5x^2 + x + 1}{x^2 + x + 1}$ is a strictly decreasing function.
Jun
28
comment Is some thing wrong with the epsilon-delta definition of limit??
Downvoting seems inappropriate. OP has thought about the matter and happens to have reached an incorrect conclusion.
Jun
28
revised When are there no critical points?
added 266 characters in body
Jun
28
answered When are there no critical points?
Jun
28
comment Help with improper integral
This has been done a bunch of times on MSE, for example here. You can probably find others.
Jun
28
comment How to calculate number of different combinations along a line?
You are looking for compositions. (The parameter is different by $1$.)
Jun
28
revised Show that all real roots of the polynomial $P (x) = x^5 − 10x + 35$ are negative.
added 350 characters in body
Jun
28
answered Show that all real roots of the polynomial $P (x) = x^5 − 10x + 35$ are negative.
Jun
28
awarded  Nice Answer