Melody

Why has just made a careless mistake and it you were interested in more than just plugging a numberinto a box you would easily have found it.

Hence you are just here to cheat.

Please whyamIdoingthis, do not change your answer, you logic is good and it is good to out cheaters.

In my view your answer is better than if you had not made this trivial error.

there are 10 ways that 3 consecutive people can be chosen.

There are 10* 6 ways that 2 (but not 3) consecustive people can be chosen

So that is  70 ways that at least 2 consecutive people are chosen.

There are 10C3 = 120 ways that any 3 people can be chosen

so maybe  70/120 ways.

That is my best guess.

A parabola with equation y=ax^2+bx+c has a vertical line of symmetry at x=2 and goes through the two points (1, 1) and (4, -7). The quadratic ax^2+bx+c has two real roots. The greater root is sqrt(n)+2. What is n?

The axis of symetry is just   -b/2a 

so   -b/2a = 2

-b = 4a

b=-2a

So the equation so far becomes       y=ax^2 -2ax  +c

Sub in your two points and solve simulataneously and what do you get for a and c

Can you take it from here?

4/5 meters in 8 minutes

is

8 minutes for  4/5 of a meter

8*5/4  minutes for  4/5  *  5/4 metres.

you can finish it.

Why?

Hi Atlas,

I don't know the answer, I am just playing with the question.

We have     \(t_n=\frac{n(n+1)}{2}\)

so              \(4t_n=2n(n+1)\)

the questions says what is the greatest possible value of the GCD of     2n(n+1)    and   n-1   

If n=3 then the greatest divisor is 2

If n=5 then the greatest divisor is 4

I don't know what the biggest one is though.  Maybe 4 ?????

split the bath into 5 parts.

2/5 will be cold water and 3/5 will be hot water.

Hot  23*3/5 = 13.8 minutes       

Cold 17*2/5 = 6.8 minutes

13.8-6.8 =7 minutes

I don't think there is a trick as such.  It is just getting your to think about numbers, which is what you have done,

The apothem is not 3.6 it is approximately 3.44  

So the question is nonsense.

I suppose you can ignore this and take it at face value.

Have you done anything yourself towards answering this question?  What?

Hi Heureka,

It is really nice to see you here again :)