shyvera

Treating the left-hand expression as a quadratic in a, verify that the determinant is never positive.

If the determinant is negative (treating the LHS as a quadratic in a), then the inequality holds for all real values of a. Otherwise it holds for all a such that [math]a<\alpha_1[/math] or [math]a>\alpha_2[/math] where [math]\alpha_1,\ \alpha_2[/math] are the real roots of the equation [math]x^2+(b-4)x+3=0[/math] with [math]\alpha_1\leqslant\alpha_2.[/math]

Use the following rules: [math]a\Leftrightarrow b\ \equiv\ (a\wedge b)\vee(\neg a\wedge \neg b)[/math] [math]\neg(a\Leftrightarrow b)\ \equiv\ (a\wedge \neg b)\vee(\neg a\wedge b)[/math] Thus: [math]\neg[(p\Leftrightarrow q)\Rightarrow r][/math] [math]\implies\ (p\Leftrightarrow q)\wedge\neg r[/math] [math]\implies\ [(p\wedge q)\vee(\neg p\wedge\neg q)]\wedge\neg r[/math] [math]\implies\ [(p\wedge q)\wedge\neg r]\vee[(\neg p\wedge\neg q)\wedge\neg r][/math] [math]\implies\ [p\wedge(q\wedge\neg r)]\vee[\neg p\wedge(\neg q\wedge\neg r)][/math] [math]\implies\ \{[p\wedge(q\wedge\neg r)]\vee[\neg p\wedge(\neg q\wedge\neg r)]\}\vee\{[p\wedge(\neg q\wedge r)]\vee[\neg p\wedge(q\wedge r)]\}[/math] [math]\implies\ \{[p\wedge(q\wedge\neg r)]\vee[p\wedge(\neg q\wedge r)]\}\vee\{[\neg p\wedge(q\wedge r)]\vee[\neg p\wedge(\neg q\wedge\neg r)]\}[/math] [math]\implies\ \{p\wedge[(q\wedge\neg r)\vee(\neg q\wedge r)]\}\wedge\{\neg p\wedge[(q\wedge r)\vee(\neg q\wedge\neg r)]\}[/math] [math]\implies\ [p\wedge\neg(q\Leftrightarrow r)]\wedge[\neg p\wedge(q\Leftrightarrow r)][/math] [math]\implies\ \neg[p\Leftrightarrow(q\Leftrightarrow r)][/math]

From BBC News: http://news.bbc.co.uk/1/hi/science_and_environment/10139906.stm http://z8.invisionfree.com/DYK/index.php?showtopic=908 From Wikipedia: http://en.wikipedia.org/wiki/Biodiversity

No, the odds do not “shoot up” the second time. The fact that he has already won the lottery before does not affect his chances of winning the lottery again in any way whatsoever. You are probably confused by two totally different questions: (i) What is the probability of winning the lottery twice (rather than once)? (ii) What is the probability of winning the lottery a second time (given that he has already won it once before)? If the probability of winning the lottery in any one week is p, then the answer to (i) is p2 whereas the answer to (ii) is just p. It is important to be clear about the distinction between (i) and (ii).

Mind you, not everyone uses oil to heat their home (which is what you seem to be implying). In the UK, for example, many (if not most) homes use gas-based central heating.

One thing that hasn’t been mentioned in this thread so far is formatting colors in LaTeX. [math]\color{red}{\rm red}[/math] [math]\color{blue}{\rm blue}[/math] [math]\color{green}{\rm green}[/math] [math]\color{yellow}{\rm yellow}[/math] [math]\color{black}{\rm black}[/math] [math]\color{white}{\rm white}[/math]

So there are only three cases, not four.

The question is all about minimizing [math]a+b+c[/math] since [math]x[/math] is independent of [math]a,\ b[/math] and [math]c.[/math] From the first equation, [math]b=\tfrac32a[/math] and [math]c=2a.[/math] Hence [math]a+b+c\ =\ a+\frac32a+2a\ =\ \frac92a[/math] The minimum value of this is [math]18[/math] corresponding to [math]a=4.[/math] Moreover [math]a=4\ \implies\ b=6[/math] and [math]c=8,[/math] i.e. choosing [math]a=4[/math] also conveniently makes [math]b[/math] and [math]c[/math] composite. Hence the minimum value of [math]\frac{a+b+c+x}2[/math] is [math]\frac{18+4}2=11.[/math] Hopefully this proves that the tree is right.

The visible part of the electromagnetic spectrum is in the frequency range from about 4×1014 Hz (red) to about 8×1014 Hz (violet), if that’s what you mean.

You know that [math]a\cdot\left(\frac1a\right)=1[/math] and [math]1>0.[/math] If [math]a<0[/math] then, since [math]\frac1a>0,[/math] the LHS would be negative. As 1 is positive, and [math]a\ne0,[/math] you must have [math]a>0.[/math]