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 \(\frac{p}{\sqrt6}?\)