Recent content by Olinguito

Well, note that $G$ is a group whereas $RG$ is a ring, so not every theorem about $G$ may be applicable to $RG$. For example, $G$ may be a cyclic group, but there is no such thing as a cyclic ring, so a theorem about cyclic groups may not make sense when applied to rings. What you can say is...

What do you mean by “ring groups”?

Hello gazparkin. You’re on the right line, but when you move the $20$ from the RHS to the LHS, you should have $-20$, not $+20$. The answer is not just -16. The answer involves $-16$, the variable $x$, and an inequality sign in between. It’s important to get the inequality sign right, or you...

Let $a\in A$. Then $(a,a)\in A\times A$. Since we’re assuming $A\times A=B\times B$, this means $(a,a)\in B\times B$ and thus $a\in B$. Therefore $A\subseteq B$. The same argument with $A$ and $B$ interchanged shows that $B\subseteq A$. Hence $A=B$.

A neat trick to do is to differentiate each of the multiple-choice answers in turn until you get the expression to be integrated.

Substitute $x=-1$ and $x=2i$ in each of the given expressions. Do you get $0$? If the same expression gives $0$ for both these two values of $x$, then the equation is the one you’re looking for; otherwise, it isn’t. Right, let’s do it one at a time. Start with $x=-1$. Substitute this in each of...

Hint: Rewrite the equation of the parabola as $$16x\ =\ 68-12y-y^2\ =\ 104-(y+6)^2$$ whence $$(y+6)^2\ =\ 104-16x\ =\ 4(-4)\left(x-\frac{13}2\right)$$ in the form you have been using.

What do you mean by “highest order coefficient”? Do you mean the leading coefficient or the constant term?

Hi Yankel. As Opalg pointed out, there is a typo with the co-ordinates of the point A; it should be $(6,3)$ rather than $(3,6)$. Let B with co-ordinates $(u,v)$ be the point opposite A on the circle. Then the line segment BA is perpendicular to the tangent line $y=\frac12x$ and so has gradient...

Let $x\in[a,\,b]$. Then $x\in U$ for some $U\in\mathscr U$. As $U$ is open, there is an open interval $I_x$ such that $x\in I_x\subseteq U$. So $I_x\in\mathscr I$ and $[a,\,b]\subseteq\bigcup_xI_x$, i.e. $\mathscr I$ is a cover for $[a,\,b]$.

If the Cartesian equation of the second line (curve) is $y=f(x)$, change it to $y=100\cdot f(x)$.

$r$ is the minimum of all the $r_j$ and so $r\leqslant r_j$ for all $j$; hence $B_r(a)\subseteq B_{r_j}(a)\subset U_j$ for all $j=1,\ldots,n$.

Hi Avro. You can also use this formula for any sets $A$, $B$, $C$: $$|A\cup B\cup C|\ =\ |A|+|B|+|C|-|A\cap B|-|B\cap C|-|C\cap A|+|A\cap B\cap C|.$$ So, in this problem, $A$ might be the set of girls playing fall sports, $B$ the set of those playing winter sports, and $C$ the set of those...

Hi Peter. $\cal O^{\prime\prime}$ is only a finite subcover of $\cal O^\prime$. In order to prove $K$ compact, we need to find a finite subcover of $\cal O$. That’s what’s going on.