Melody

Hi Kitty,

I used Coordinate geometry as shown in the diagram and then solved the bottom circle equation (centred at A) with x=0 and then y=0 respectively to find the points I and H.

Then just find the distance between them.

I will leave the rest to you,

You could have 5 ones and a 6.  The six can be any of the tosses so that is 6 ways   (for 6 tosses.)

Or you could roll a 2, a 3 and four 1's  in any order.

you have 1 1 1 1 there are 5 places where you can slot in a 2 

Now you have     X X X X X    where one of the x's is a 2 and the others are all 1s.

There are 6 ways to now slot in the 3.

Than makes    5*6 = 30 ways to have a 2, a 3 and four ones.

30+6  - 36 possibilities.

I think that is correct.

Hi EP,

Your annser is right but you confused me by accidentaly using the power ^ notation.

\(f(a)=a* 5+2\\ f(a)=5a+2\\~\\ f(3)=5*3+2=17\\~\\ f(10)=5*10+2=52\)

Rather than using brute force Bosco you could have used this division fact

Divisibility rule of 11: This particular rule states that the given number can only be completely divided by 11 if the difference of the sum of digits at odd position and sum of digits at even position in a number is 0 or 11.

mmm

\(\frac{x}{x - 5} = \frac{4}{2x - 4} + \frac{2}{x + 3}\)

for starters x cannot equal  5, 2 or  -3

\(\frac{x}{x - 5} = \frac{4}{2x - 4} + \frac{2}{x + 3}\\ \frac{x}{x - 5} = \frac{2}{x - 2} + \frac{2}{x + 3}\\ x(x-2)(x+3)=2(x-5)(x+3)+2(x-5)(x-2)\)

Now expand out and see where it goes.

It did not fall out for me.

Thanks Pythagorean.

I think I will try this one too.

I'll use a method called stars and bars.

Put the 12 stickers in a row (they are the stars)

Now we need to divide them into 4 groups.  We can do this using 3 bars. 

eg  X X | X X X X | | X X X X X X        represents the outcome of   2, 4, 0, 6

The question now becomes:  How many positions can the 3 bars be placed in

The answer is    15C3 = 455 ways

Our answers are close Pythagoras but I think you have double counted some. ..

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.