JovenlyCosmo

Sorry for the late posting (understatement).

Anyways, you can imagine this question as a group of 5 people and 2 other people - that group of 5 people will always sit next to each other, such that Pierre/Rosa/Thomas will be in the first, third, and fifth position of this group of 5 people. Now, you have 2 positions left, and since there are 4 people left, you can calculate this as ${4 \choose 2}=6$. Also, their are $3!=6$ ways to have the 3 people moved. Multiplying writing these numbers, you get $36$. Therefor, you have 36 ways to arrange the group of 5 (whom you can now treat as a single person!). On the other side of the round table, you have 2 people that can be arranged in $2!=2$ ways, so you have $36*2=72$, and then multiplying by 3 gives us $72*3!=72*6=432$. However, since this is a round table, you must divide by 3 since you can rotate the table as 

A

B           C

or

B

C            A

or

C

A          B,

Giving us a total of 3 ways you overcounted the original $432$, so you must divide by 3 to get $\frac{432}{3}=\boxed{144}$.

(However, you divide by 3 because this question only asks for rotations, but other questions ask for reflections/flips, in which case divide by an additional 2.)

I finally managed to solve it!!

(Thanks to my dad)

The question I was answering was:

$(\sqrt5+\sqrt3-\sqrt2)(3 \sqrt a -\sqrt b+2 \sqrt c)=12$

Find the ordered pair $(a,b,c)$.

I am planning on putting the solution on another post, though.

However, the clue is that $5=3+2$, which makes this possible.

I wil post the solution on this question If I can't post it on the other post.

Hint:

Expanded, you an get 

$x^2+6x>49-5x$ which simplfies into

$x^2+11x-49>0$

Which can be factored - You can then draw a graph on a numberline where the 2 roots of the quadratic are points on a parabola, and find the ranges on the numberline for which the Y value of the parabola is positive.

Sorry, I don't know how to draw the graph on web2.0calc.

Hint 1: Use the quadratic formula, most notably, the discrimant.

Hint 2: Part a) is a single number, the rest are inequalities.

Here's the answer for part (a):

If you put the 2 equations together, you will get $x^2-9x+c=8x^2-13x$, from which you will get $7x^2-4x-c=0$. In order for their to be only 1 solution, the discriminant must be equal to $0$. The discrimant of any quadratic is $b^2-4ac$ (It is the part in the square root of the quadratic formula, which is the formula that pops up whenever you press the LaTeX button on your screen.)

$b^2-4ac = 0$, so

$16+28c=0$, from which you can solve

$c=-\frac{16}{28}=-\frac{4}{7}$

Sorry I meant $z^3+1=0$, and $z^2-z=-1$.

 Sorry if I confused anyone.

Here is the processed version.

Given $30-60-90$ triangle,

Hypotenuse $=$ $q\sqrt3$

The short side that is opposite of the $30$ degree angle is $p\sqrt2$

And the long side that is opposite to the $60$ degree angle is $p\sqrt6$

Find the values of $p$ and $q.$

Anyways, $2p\sqrt2 = q\sqrt3$, techincally, there is 1 equation and 2 variables, making this problem more difficult to solve.

I'm going to have this problem solved by someone else (Because unfortunately, I must do my homework). 

Yeah, I managed to solve it 30 minutes later. (Sorry, the correct answer was -1.)

Anyways, thanks for the answer. What I did was take

$z + \frac{1}{z}=1$ and multiply it by $z^2$, getting $z^3-z^2+z=0$. However, when you multiply $z+ \frac{1}{z}=1$ by $z$, you will get

$z^2-z=1$

From which you can replace the $-z^2+z$ with $-1$ (The negative of the equation above) to get

$z^3-1=0$

From which you can easily get $z^3=-1$

What side is $\frac{p}{\sqrt2}$? What side is $\frac{p}{\sqrt6}?$