Alan

I think Ramanujan's number is 1729 Bertie.  It's the smallest integer that can be expressed as a sum of two cubes in two different ways:

1729 = 1³ + 12³

1729 = 9³ + 10³

$${\mathtt{ExpectedValue}} = {\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{0.962\: \!3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{0.035\: \!7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,000}}{\mathtt{\,\times\,}}{\mathtt{0.000\: \!036\: \!9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,000\,000}}{\mathtt{\,\times\,}}{\mathtt{0.000\: \!000\: \!143}} \Rightarrow {\mathtt{ExpectedValue}} = {\mathtt{0.287}}$$

Expected value = 0.287 dollars = 28.7cents.  

To the nearest whole number of cents: Expected value = 29

Let x = log₇(4)

This means that

7^x = 4

Take the natural log of both sides (ln means log to base e)

ln(7^x) = ln(4)

By a property of logarithms (see the Formulary at the top of this page) we have

x*ln(7) = ln(4)

so x = ln(4)/ln(7) or, substituting for x from the first line above):

log₇(4) = ln(4)/ln(7)

\frac{-1-x^2}{(1+x*y)^2+(x-y)^2} $$\\ Expand the denominator$$ (1+x*y)^2+(x-y)^2=1+2x*y+x^2*y^2+x^2-2x*y+y^2 $$\\ Collect terms in the denominator$$ 1+x^2+y^2+x^2*y^2 = 1+x^2+y^2*(1+x^2)=(1+x^2)*(1+y^2) $$\\ The original numerator can be written as$$ -(1+x^2) $$\\ Putting the numerator and denominator together again we have$$ \frac{-(1+x^2)}{(1+x^2)*(1+y^2)}= \frac{-1}{1+y^2}$$

Rate = 0.150 M/s

[OCl-] = 0.500 M

[I-] = 0.050 M

0.150 = k*0.5*0.05

Divide both sides by 0.5*0.05

$${\mathtt{k}} = {\frac{{\mathtt{0.15}}}{\left({\mathtt{0.5}}{\mathtt{\,\times\,}}{\mathtt{0.05}}\right)}} \Rightarrow {\mathtt{k}} = {\mathtt{6}}$$

k = 6 M^-1s^-1

If we have a general point (x, y) in Cartesian coordinates, the polar form is (R, Θ) where R = √(x² + y²) and

Θ = tan^-1(y/x). 

And if we are in the complex plane, the point z = x + iy is z = Re^iΘ in polar coordinates.

Your point is z = -3 + 4i  so, in polar form it is z = Re^iΘ where R = 5 and Θ = tan^-1(4/-3) (note that this is an angle in the 2nd quadrant, because the x-value is negative while the y-value is positive). Because you can add any multiple of 2pi (360°) to Θ without changing the value of e^iΘ, we can also write z = re^i(Θ+2pi*k) where k is an integer.

To find the fourth root of a complex number in polar form we simply take the fourth root of r and divide the angle by 4.  That is: z^1/4 = r^1/4e^{i(θ+2pi*k)/4}.  So  z^1/4 = 5^1/4e^{i(tan-1(4/-3)+2pi*k}). 

We can let k = 1, 2, 3 and 4 to get the four roots.

We can write this as z1/4 = re^iθ,  where r = 5^1/4 and θ = tan^-1(4/-3) + 2pi*k.

The result of doing this is summarised in the image below:

There are 20 options in total, 5 with holly, 4 with dragon scale.  However, there is one with both holly and dragon scale, so, to avoid double-counting that one, we have 5 + 4 - 1 = 8 possibilities.  So the overall probability is 8/20 = 2/5 = 0.4.

1. p = 1*(1/6)*(1/6) = 1/36

2. p = (5/6)*(5/6)*(5/6) = 125/216

3. p = (1/2)*(1/3)*(1/6) = 1/36

This is the sum of a converging infinite geometric series, so it can be written as a/(1-r) where a = 6 (the first term)  and r = -1/2 (the common ratio).  So the sum is 6/(3/2) = 4. 

Ok.  If I have f = (a*x+b)^c and want to differentiate say  with respect to x, I can first set z = a*x+b and I have f = z^c.  I can differentiate this with respect to z to get df/dz = c*z^c-1.  But I want it differentiated with respect to x not z, so I must use the chain rule, which says that df/dx = df/dz*dz/dx.

Now, we have found df/dz, so we need dz/dx.  Since z = a*x+b then dz/dx = a.  Hence overall we have df/dx = df/dz*dz/dx = c*z^c-1*a = c*(a*x+b)^c-1*a, where I've replaced z by its definition in terms of x.

In your case we have a = 5;  b = -4; and c = 1/2, so in your case df/dx = (1/2)*(5*x-4)^-1/2*5.

I did this all in one step. I'm so used to doing this I don't need to break it down into separate steps, but clearly I should have done here!

If this still isn't clear, let me know.