Recent content by Olinguito

[FONT=Times New Roman]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...

[FONT=Times New Roman]What do you mean by “ring groups”?

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

[FONT=Times New Roman]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 $A=B$.

[FONT=Times New Roman]A neat trick to do is to differentiate each of the multiple-choice answers in turn until you get the expression to be integrated.

[FONT=Times New Roman]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 $x=-1$...

[FONT=Times New Roman]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.

[FONT=Times New Roman]What do you mean by “highest order coefficient”? Do you mean the leading coefficient or the constant term?

[FONT=Times New Roman]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...

[FONT=Times New Roman]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 $[a,\,b]$.

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

[FONT=Times New Roman]$r$ 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 $j=1,\ldots,n$.

[FONT=Times New Roman]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 $C$...

[FONT=Times New Roman]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.

[FONT=Times New Roman]