Melody

Wed 10/12/14

1) Happy's tribute to many on the forum.  Thanks Happy   

$${\frac{{\mathtt{100}}}{{\mathtt{12}}}} = {\frac{{\mathtt{25}}}{{\mathtt{3}}}} = {\mathtt{8.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

8 hours and 20 minutes.

Wouldn't a sled be better than a Bicycle?  I am sure that it would     LOL     

250 *   3/5 miles

It is not a fraction - It is a FUNCTION.

How do you evaluate the FUNCTION when x=-2,0 and 3? g(x)=-3*x+5

g(-2) = -3*-2+5 = +6+5 = 11

g(0) = -3*0+5    = 0+5   = 5

Perhaps you can do    g(3)    by yourself ?

For

x+(1/x) = -1

Alan and Heureka are both telling you that there are no real solutions.

Alan has said x cannot equal 0 because you cannot divide by 0

Alan has rearranaged the equation to give a quadratic.

$$x^2+x+1=0$$

For any quadratic you can determine the nature of the roots by examining the discriminate

$$\\\triangle =b^2-4ac\\
\triangle =1-4=-3\\$$

Since the discriminant (which is under a square root) is negative, there are no real roots.

Alan's answer is an approximation.

$$\frac{1}{2^6}=\frac{1}{64}$$

$$$\int \frac{1}{\left(\frac{\sqrt{3}}{2}v\right)^2+\frac{3}{4}}\frac{\sqrt{3}}{2}$$$

I assume that this is meant to be

$$\frac{\sqrt{3}}{2}\int \frac{1}{\left(\frac{\sqrt{3}}{2}v\right)^2+\frac{3}{4}}dv\\\\
=\frac{\sqrt{3}}{2}\int \frac{1}{\frac{3}{4}v^2+\frac{3}{4}}\;dv\\\\
=\frac{\sqrt{3}}{2}\int \frac{1}{\frac{3}{4}(v^2+1)}\;dv\\\\
=\frac{\sqrt{3}\times 4}{2\times 3}\int \frac{1}{(v^2+1)}\;dv\\\\
=\frac{2}{\sqrt 3}\int \frac{1}{1+v^2}\;dv\\\\
=\frac{2}{\sqrt 3}\;tan^{-1}\;v +c$$

I think that is correct.