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Like so:

.

\(\textrm{Let}\quad\overline{BC}=a,\quad\overline{CA}=b,\quad\overline{AB}=c\\ \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R\quad (\textrm{R=radius of the circumscribed circle of triangle ABC})\\ \therefore sinA=\frac{a}{2R},\quad sinB=\frac{b}{2R},\quad sinC=\frac{c}{2R}\\ cosA=\frac{b^2+c^2-a^2}{2bc},\quad cosB=\frac{c^2+a^2-b^2}{2ca},\quad cosC=\frac{a^2+b^2-c^2}{2ab}\\ \therefore \frac{c}{2R}=\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}}\\ \quad c=\frac{2abc(a+b)}{a(b^2+c^2-a^2)+b(c^2+a^2-b^2)}=\frac{2abc(a+b)}{ab^2+c^2a-a^3+bc^2+a^2b-b^3}=\frac{2abc(a+b)}{c^2(a+b)-(a-b)^2(a+b)}=\frac{2abc}{c^2-(a-b)^2}\\ \therefore c^2-(a-b)^2=2ab\\ \quad c^2=2ab+(a-b)^2=a^2+b^2\\ \therefore \textrm{Triangle ABC is a right triangle that has angle C as the right angle.}\)

OK, thanks, so that makes it easy than.

BM = QB

AN=AQ

AB=BM+AN

thanks again

Let the crossing point between AQ and PM B.

PA+AB+PB=PA+AQ+QB+PB

              =(PA+AQ)+(PB+BQ)

              =(PA+AN)+(PB+BM)

              =PN+PM

              =22cm

remove the segment twice

or (10-1.6)2=16.8

I know this much but I dont know where to start on the problem

Hey Melody,

I did 10+10 and then subtracted the 1.6666... and got 18.333333... which simplifies to 18 1/3. This answer is wrong. Did I make a mistake? Thanks

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    

Nickolas, I could understand your answer.  Rom's... well, I'm sure it's accurate, but I have no idea where those numbers came from. 

\((x+8)^2 = 44\\ (x+8)=\pm \sqrt{44} = \pm 2\sqrt{11}\\ x = -8 \pm 2\sqrt{11}\)

\(\text{Can always apply the quadratic formula}\\ x = \dfrac{9\pm \sqrt{81-4(5)(8)}}{10} = \\ \dfrac{9\pm\sqrt{81-160}}{10} = \\ \dfrac{9\pm i \sqrt{79}}{10}\)

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.

\(\text{Couple ways to do this}\\ 1)~\text{convert both to proper fractions and multiply}\\ 19\dfrac 3 4 = \dfrac{79}{4}\\ 1\dfrac 2 5 = \dfrac 7 5\\ \dfrac{79}{4}\dfrac{7}{5} = \dfrac{553}{20}\)

\(2) \text{distribute }\\ \left(19+\dfrac 3 4\right)\left(1+\dfrac 2 5\right) = \\ 19 + \dfrac 3 4 + \dfrac{38}{5} + \dfrac{6}{20} = \dfrac{553}{20}\)

Pretty trivial with the distance formula!

1. I'll just try this problem...I don't know if I'll get it !

Let \(\sqrt{49-20\sqrt{6}}=a-b\).

We can square both sides to get, \(49-20\sqrt{6}=(a-b)^2\).

Expanding this, gives us \(49-20\sqrt{6}=a^2+b^2-2ab\).

Now, by a bit of matching, we can see that \(a^2-b^2=49\) and \(-2ab=-20\sqrt{6}\) (Same thing as \(2ab=20\sqrt{6}\))

This might be a bit of tedious work, but by a bit of inspection \(a=5\) and \(b=2\sqrt{6}\) work.

Thus, the answer is \(a-b=\boxed{5-2\sqrt{6}}.\)

You are very welcome :)

Awesome... Thank you so much!! 

Do your own homework !!

Yes you were