Rom

treat couples as units

there are 3!=6 ways to seat the 3 couples

there are 2^3 = 8 ways to choose the order each couple sits in.

This is 6*8 = 48 ways to seat the 3 couples as couples.

There are a total of 6! = 720 ways the 6 individuals can sit.

P[they sit as couples] = 48/720 = 1/15

don't use all caps, it's rude.

I assume you mean the other 9 marbles are some other color than red, green, and blue.

It matters if there are colors that are repeated.  I'll assume not.

I also assume that the order of the 4 chosen marbles doesn't matter.

We have to pick a red, green, or blue marble first.  There are 3 ways to do that.

We then have 3 marbles to choose for the 9 other marbles.  There are 9 choose 3 = 84 ways to choose these.

So there are 3 * 84 = 252 ways to choose marbles as specified

Again this one is easier if we consider the distributions where 2 boys are not next to each other and subtract that off of the total number.

The only distribution that has no boys together is B G B G B G B

The only degrees of freedom we have is in selecting which girls and which boys get which seat.

There are 4! ways to seat the boys, and 3! ways to seat the girls that gets us N=4! 3! = 144

There are a total of 7!=5040 possible ways to seat everyone

Thus the number of seatings where at least 2 boys are adjacent is 5040-144 = 4896

\(2a^2 + 4a - 48 \\ 2(a^2 + 2a - 24) \\ 2(a-4)(a+6)\)

let that expression equal x, then

\(\sqrt{12+x}=x \\ 12 + x = x^2 \\ x^2 -x -12 = 0 \\ (x-4)(x+3) = 0 \\ x = 4,~-3 \\ \text{however as x is a square root x > 0, thus }x=4\)

The key here is that if two lines, l1 and l2, with slopes, m1 and m2, are perpendicular then

m2 = -1/m1

1) is correct

2) don't guess

(-3 - i) + (5+6i) = (-3+5) + (-1+6)i = 2+5i

3) your guesses.... are not good

(8-3i)-(6+i) = (8-6) + (-3-1)i = 2 - 4i

-i is correct though it would be better if you understood why.

\(\forall n \in \mathbb{Z} \\ i^{4n} = 1 \\ i^{4n+1}=i \\ i^{4n+2}=-1 \\ i^{4n+3}=-i\)

for the first problem we have

\(2\sqrt{-27} = 2\sqrt{27}\sqrt{-1} = 6\sqrt3 i = 6i \sqrt{3}\)

the 3rd choice

The answer is x=2

\(a b^4 = 12 \Rightarrow a^5 b^{20} = 248832 \\ b^{15}=\dfrac{a^5 b^{20}}{a^5 b^5} = \dfrac{248832}{7776} = 32 \\ b = 32^{1/15} = 2^{1/3} \\ a b^4 = 12 \\ a \left(2^{1/3}\right)^4 = 12 \\ a 2^{4/3} = 2^2(3) \\ a = 3\cdot 2^{2/3} \)