Melody

The area of the circle is 10

So   
\(10=\pi*r^2\\ r^2=\frac{10}{\pi }\)

And the area of the sector is 

\(=60/360  * \pi * r^2\\ =\frac{  \pi * \frac{10}{\pi }}{6}\\ =\frac{ 10}{6}\\ =1.\dot6\)

Maybe you should think about the rest yourself  :)

I am sure you are grateful to both oRom and Nickolas for their time and effort,

but I am equally sure they would both like to hear that from you.

Fractions are very difficult to learn. Much more difficult than most mathematicians realize and it is difficult to teach it on the internet.

But i will give it a go.  

\(19\frac{3}{4}\times 1\frac{2}{5}\)

You cannot multiply mixed numerals.  You have to turn them into improper fractions first.

So how many quarters are there in 19+3/4

there are 4 quarters in 1

4*2 = 8 quarters in 2

4*3= 12 quarters in 3   etc

so 

there are 4*19 = 76 quarters in 19   and the other 3 and you have  79 quarters in 19 and 3/4

so

\(19\frac{3}{4}= \frac{4*19+3}{4}=\frac{79}{4}\)

Using the same idea

\(1\frac{2}{5}=\frac{5*1+2}{5}=\frac{7}{5}\)

so

\(19\frac{3}{4}\times 1\frac{2}{5}\\=\frac{79}{4}\times \frac{7}{5}\\ =\frac{79\times 7}{4\times 5}\\ =\frac{553}{20}\\\)

I used a calculator to help me get 553 but you could do it on paper.

Now how many whole numbers is that and how much is left over?

there are 5 twenties in 100 

so there are 5*5=25    twenties in 500

then 2 more in 40

so that is  27  lots of 20 in 540

\(\frac{553}{20}=\frac{540}{20}+\frac{13}{20}=27+\frac{13}{20} = 27\frac{13}{20}\)

Think about it and then you can let me know if that helps    

Hi Tertre,

I've only learnt this recently and I do not do it quite like that.  My technique is pretty bad.

But your way looks really good.

Most times I do not think it would work. I mean i doubt that questions like this can be simplified all that often.

If the questions says 'simplify' it is a reasonable to assume that it can be done in that instance.

You are very welcome :)

Do your own homework !!

Yes you were  

SECOND WAY (probably easier for you)

Use you calculator to get values for     cot(0.5*3pi),    cot(0.5*0),    cot (0.5*2pi),  Cot(0.5*-2pi)   and cot(0.5*pi/2)

If you get a an error then that vaue of x is an asymptote.

If your calc does not have cot then use   1/tan( )

Lets look at this anyway. You should not rely on desmos, that is just for checking.

FIRST WAY  (The most formal way)

\(y=3cot(0.5x)-4\)

The 4 only pushed the graph up or down so it has nothing to do with aspymptotes.

The 3 just exaggerates the graph in a up down direction so that has nothing to do with it either.

so that asymptotes will be the same as for

\(y=cot(0.5x)\)

now

\(y=cot(0.5x)=\frac{cos(0.5x)}{sin(0.5x)}\\ \text{You cannot divide by 0 so the asymptotes will be where sin(0.5x)=0}\\ 0.5x\ne n\pi \qquad where \;n\;is\;an\;integer\\ so\\ x\ne 2n\pi \)

so x canot be 0 or  pi or -pi    they are three of the asymptotes.

OK :) 

It is the second and the third one.

The first and the forth BOTH cross the graph, so they are not asymptotes.