Melody

Thanks Heureka,

You mean that there is a maximum of 8 steps if the element exists in the array, but 9 steps if the element does not exist?

Thanks Heureka but can you explain to me why you have to add the 1 ?

Hi EMP,

I did one of these for you the other day :)

Just assume that it is the very last one that is the one you are looking for and see how lon it would take.

193 in the array.

Let the first on be in postion 0 and the last one be in position 193

1.      (0+193)/2 =  96     rounded down

2.      (0+96)/2   =  48

3.      (0+48)/2   =  24

4.      (0+24)/2   = 12

5.      (0+12)/2=    6

6.      6/2=3

7.      3/2=1

8.       1/2=0

That is 8

So I think that 9 will do it.

Thanks   ♪ ♫

Hi Piperson:)

(12-5)^2+(8/2)=?

=7^2 + 4

=49+4

=53

Well, if you put any eggs in it it won't be empty any more SO

I guess if you put them in one at a time then you can only put one in BUT

If you put them in all at once then I guess you can put any number of eggs in it - it just won't be empty any more.  :)

Maths exists so that children, small and large, can be tortured  !!

let 

\(T_0=2,\;\;\;T_1=3 \qquad \dots \qquad T_{14}=47,\;\;\;T_{15}=53, \qquad ... T_{63}=311 \)

About 7 

\(T_{( \lfloor 0.5(0+63)\rfloor)}=T_{31}=too\;big\\ T_{( \lfloor 0.5(0+31)\rfloor)}=T_{15}=53=too\;big\\ T_{( \lfloor 0.5(0+15)\rfloor)}=T_{7}=19=too\;small\\ T_{( \lfloor 0.5(7+15)\rfloor)}=T_{11}=37=too\;small\\ T_{( \lfloor 0.5(11+15)\rfloor)}=T_{13}=43=too\;small\\ T_{( \lfloor 0.5(13+15)\rfloor)}=T_{14}=47=too\;small\\ T_{( \lfloor 0.5(14+15)\rfloor)}=T_{14}=47=umm\; already\; seen\;that\;one.\\ \text{Conclusions: 52 is not a prime number} \)

[-10u ^ 5 * y ^ 5 * l ^ 3 (g-13 * o) / 20 (13 * 0-g) * u ^ 3 * y ^ 4 * l ^ 4] = (u ^ 2 * y) / (2 * l)

(9 * p ^ 2-9 * V ^ 2) / (9 * ^ v 2-18pv + 9 * p ^ 2) = (p + v) / (VP)

(32t * u + 8 * u-16 * t-8) / (16 * t * u ^ 2 + 8 * u ^ 2-16 * tu-8 * u) = (2 * t) / u

[(2x + 2 * t) / (m * f) / (4 * x * t + 4) / (3 * m-3 * f)] = 8/3

These are your questions (more or less anyway)   Are they really what you want?   Question 2 doesn't make sense.

\(1) \qquad \frac{-10u ^ 5 y ^ 5 l ^ 3 (g-13 o) }{ 20 (13 0-g) u ^ 3 y ^ 4 l ^ 4 }=\frac{ (u ^ 2 y) }{(2 l)}\\ \)

\(2) \qquad \frac{(9 p ^ 2-9 V ^ 2)}{ (9 * ^ v 2-18pv + 9 p ^ 2)} = \frac{(p + v) }{ (VP)}\\ \)

\(3) \qquad \frac{(32tu + 8u-16t-8) }{ (16 t u ^ 2 + 8 u ^ 2-16 tu-8 u) }=\frac{ (2 t)}{u}\\ \)

\(4) \qquad \frac{(2x + 2 t)} { (mf) (4 x t + 4) (3m-3f)} =\frac{ 8}{3}\)

Hi rarinstraw,

Gino is really good at these he may have a better way to do it.

I set up these four equations

\(cost = ax^2+bx+c\\~\\ 550=a*225+b*15+c\\ 700=a*289+b*17+c\\ 1500=a*529+b*23+c\\ 2500=a*900+b*30+c\)

Then I solved them simultaneously.

I got 

\(a\approx 3.23\\ b\approx -28.40\\ c\approx 249\\ so\\ cost\approx 3.23x^2-28.40x+249\)

Hi TandCforlife,

Are you T or are you C ?  Anyway I wish you both a happy life journey :)

What is the answer pls ? |3−6|−(12÷6+1)^2 I am trying to figure it out 

\( |3−6|−(12÷6+1)^2 \\ = |-3|−(2+1)^2 \\ = 3−1*(3)^2 \\ = 3−1*9 \\ = 3−9 \\ = -6\\ \)