Melody

I got 36,  it might not be right.

There are only 3 ways to pair the siblings so that no two siblings are together.

But then either person can be in front of the pair so Ill double it to 6

The first pair has a choice of 3 seats in the row, the second pair has a choice of 2 and the last has the choice of 1

So that is    3*2*3*2*1 = 36

Where is the diagram?

3!*4! = 6*24= 144 ways

$3.05

I have answered on the original question.

the second one is very similar 

You just have to know that 

     \(a^3+b^3=(a+b)(a^2-ab+b^2)\\ \text{rearranging I get}\\ a^3+b^3=(a+b)(a^2+2ab+b^2-3ab)\\ a^3+b^3=(a+b)([a+b]^2-3ab)\\ \text{now check what I have done and do the appropriate substitutions to finish the question.}\)

1)    

\(7x^2+x-5=-4x-9\\ 7x^2+5x+4=0\\\)

The sum of the roots          a+b= -5/7

The product of the roots       ab=4/7

\((a-4)(b-4)\\ =ab-4a-4b+16\\ =ab-4(a+b)+16\\~\\ \text{Substitute and you get}\\ =\frac{4}{7}-4*\frac{-5}{7}+16\\ =\frac{4}{7}+\frac{20}{7}+16\\ =\frac{24}{7}+16\\ =19\frac{3}{7}\)

Hi Juriemagic,

A smart little kid could use trial and error.

Your question is incomplete.

The big hand is on the 11    

The little hand is   11/12 of the way from 8 to 9

So they are both on the left hand side of the clock.

from 9 to 11 is   2/12 *360 = 60 degrees

1/12 of 1/12 * 360 = 2.5  degrees

So I get  62.5 degrees

You need to check.