Bertie

Here's an alternative method.

Fold the 49+48+...+2+1 underneath the 1+2+...+48+49 and you have

$$\begin{array}{l}\quad 1 \;+ \;2+...+48+49+50\\+49+48+...+2 \;+1\end{array}$$

Now add 'columnwise' 1+49, 2+48, and so on, (the 50 on the top line is left by itself),  and you can see that you finish up with 50 50's giving a total of 2500.

You should ask your teacher if each piece of work carried the same weighting.

It could be that one or two of the assignments, (if that's what they were), were considered to be minor compared with the other (s) and therefore contributed less to your final grade.

Take logs (any base) and then use the rule

 $$\log (a^{n})=n\log(a)$$.

In the USA where they put the month before the day, March 14th is known as pi-day.

Here in the UK, sadly, we don't celebrate pi-day. 

The iteration in the previous post converges to the square root of 2.

For convergence to the square root of N, the formula should be

$$x_{i+1}=\frac{1}{2}\left(x_{i}+\frac{N}{x_{i}}\right)$$.

As a decimal 8/108 = 0.074 repeating, as in the first post.

It's not possible to answer this question unless you are told the relative elevations of Rowena and the top or bottom of the pyramid. That is, is the ground totally flat so that Rowena's feet are on the same level as the base of the pyramid or are they on the same level as the top of the pyramid, or at some other height ?

Write your integrand as $$1.(\log(x))^{n}$$ and integrate by parts, integrating the 1 and differentiating the log.

That gets you a reduction formula which you have to use twice, once for P10 and then again for P9.

There are usually two methods of solution for his type of problem, a method based on counting, as used by Melody and CPhill, and a basic probability method.

The counting method has already been discussed, so here is the probability method.

Suppose that the six smarties fall in the sequence BBBBOO, where B represents a blue smartie and O one of the other colours.

The probability of that sequence happening will be

(6/19)*(5/18)*(4/17)*(3/16)*(13/15)*(12/14).

If you consider any other sequence consisting of four blues and two others, the individual fractions will be different but the combined top and bottom lines will be the same.

So, for example, if the sequence happened to be OBBOBB, the probability would be (13/19)*(6/18)*(5/17)*(12/16)*(4/15)*(3/14).

Different fractions, but the combined top line is still 13*12*6*5*4*3 and the bottom line 19*18*17*16*15*14.

So, the overall probability will be this probability multiplied by the number of possible sequences consisting of four blues and two others, that is,

(6/19)*(5/18)*(4/17)*(3/16)*(13/15)*(12/14)*6C4 = 0.0028748*15 = 0.0431(4dp).

mymathquestions:

If I have an angle whose measure is 10 degrees more than twice it's complement, what is the measure of that angle?