reinout-g

I think you either forgot to mention some of the details, or the variables (other than i off course) have an implicit meaning which I'm unaware of. 

Anyway, I can make it a little easier for you by doing the following. 

$$\begin{array}{lcl}
\alpha - i\beta &= \frac{1}{a-ib}\\
\alpha - i\beta &= \frac{1}{a-ib}\frac{a+ib}{a+ib}\\
\alpha - i\beta &= \frac{a+ib}{a^2+ aib - aib -i^2b^2}\\
\alpha - i\beta &= \frac{a+ib}{a^2+b^2}\\
(\alpha - i\beta)(\alpha + i\beta) &= \frac{(a+ib)(\alpha + i\beta)}{a^2+b^2}\\
\alpha^2 +i\alpha \beta - i\alpha \beta - i^2 \beta^2 &= \frac{(a+ib)(\alpha + i\beta)}{a^2+b^2}\\
\alpha^2 + \beta^2 &= \frac{(a+ib)(\alpha + i\beta)}{a^2+b^2}\\
(\alpha^2 + \beta^2)(a^2+b^2)&= (a+ib)(\alpha + i\beta)
\end{array}$$

 Now if you can prove that

$$(a+ib)(\alpha + i\beta) = 1$$

You're there.

Reinout

This problem is an altered form of an easier problem, which I will explain first before I'll move on to your problem.

Suppose we want to add the numbers 1 to 10. 

Then we could go about this by simply calculating

$$\begin{array}{lcl}
1+1 = 2\\
2+3 = 5\\
5+4 = 9\\
9+5 = 14\\
\mbox{etc.}
\end{array}$$

Since this takes quite a long time, we can also try to find a more logical way to solve this.

Let's try to find pairs which make it easier for us to solve this summation.

$$\begin{array}{lcl}
0+10 = 10\\
1+9 = 10\\
2+8 = 10\\
3+7 = 10\\
4+6 = 10\\
\ \ \ \ \ \ 5 = 5\\
\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
1+...+10 = 55\\
\mbox{Hence the sum of all the numbers between 1 and 10 is 55}\\
\end{array}$$

$$\begin{array}{lcl}
\mbox{Note that we could also write this as } 10\times \frac{10}{2}+\frac{10}{2}\\
\mbox{This again can be rewritten to} \frac{10 \times 10 + 10}{2}\\
\mbox{and } \frac{10 \times 10 + 10}{2} = \frac{10 \times (10+1)}{2}
\end{array}$$

Try to see for yourself that this idea doesn't just hold for the sum of the numbers 1 to 10, but also for numbers 1 to n. where n is any number you want to choose. Therefore

$$1+2+...+n = \frac{n(n+1)}{2}$$

Now find the sum of 1 to 201, the sum of 1 to 100 and then substract the second from the first :D

Then you can find the sum of 1 to 201.

Good luck!

Have a look at the definition of a inversely propotional relationship.

Hence

$$m = \frac{k}{g} \Leftrightarrow k = mg \Leftrightarrow k = 90 \times 40 = 3600$$

Now if

$$m = g$$

we can write

$$m^2 = k = 3600 \Leftrightarrow m = \sqrt{3600} = 60$$

Therefore

$$m = g = 60$$

(11-11-2014 edit)

or

$$m = g = -60$$

 dangerous

$$\begin{array}{lcl}
\mbox{For } f(x)=x^2-1 \mbox{ and } g(x)=x-3 \mbox{ , calculate } g(f(2)).\\
\mbox{Note that } g(f(2)) \mbox{ means that we calculate } f(2) \mbox{ and we fill in this answer in function } g(x)\\
\mbox{Then}\\
f(2) = 2^2-1 = 4-1 = 3\\
\mbox{Hence } g(f(2)) = g(3)\\
g(3) = 3-3 = 0\\
\mbox{ So } g(f(2)) = 0


\end{array}$$

So you have this function, which for your convenience I'm going to rewrite as $$y = 4^x$$.

Now make a plot of the x- and y-axis. 

Now calculate some values for x such as;

x=-2, x=-1, x=0, x=1 and x=2 which give you respectively, y = 1/16, y = 1/4, y = 1, y = 4 and y = 16.

Now mark these points in your graph and then you probably have an idea of what the graph looks like. 

If you want to check your answer, visit this website;

$$(x*0.07)+x = 113.04\\
0.07x + 1x = 113.04\\
(1+0.07)x = 113.04\\
1.07x = 113.04\\
x = 113.04/1.07 = 105.64$$

according to the density for each 5.53 grams of wood the wood will take up 1 cm3 in space. Therefore the volume of the wood is given by 33.3/5.53 = 6.02 cm3 wood

if you travel 500m/s over 45m it'll take you 45/500= 0.09 seconds to get there

See for yourself;