The GCF of 18 and 20 is 2, so the maximum size square tile is 2m x 2m.

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Thanks Hectictar, I love those colours :)

Thanks Max, I will have some of you wonderful cookies with a cup of coffee right now.  I think I will start  with a smartie cookie.  I love smarties :)    (Maybe you call them m&m's)

And Yomyhomies, your balloons and your funky writing are really great, they continue, or maybe finish, the thread with great pizazz.  

I am ever so pleased.  Thank you everyone.

Thanks Heureka,

I have answered this before I saw your answer.  I'd like to upload mine even though it is essentially the same as yours.

\(s(t)=122+45t-16t^2\)

s is height (feet),

which is a funtion of t which is time (seconds)

To derive the given answer, h is a difference in time.  Using h for this seems really confusing to me, and I suspect that this is part of your problem.

As you already know height at time t is given by

\(s(t)=122+45t-16t^2\)

h seconds later the height is given by

\(s(t+h)\\ =122+45(t+h)-16(t+h)^2\\ =122+45t+45h-16(t^2+h^2+2th)\\ =122+45t+45h-16t^2-16h^2-32th\\\)

Now the difference quotient is the  gradient of the secant joining those two elevations with regards to time.

That is

\(difference\;\; quotient \\ =\frac{difference \;in\; height}{difference \;in\;time}\\ =\frac{s(t+h)-s(t)}{(t+h)-t}\\ =\frac{(122+45t+45h-16t^2-16h^2-32th)-(122+45t-16t^2)}{h}\\ =\frac{45h-16h^2-32th}{h}\\ =45-16h-32t\\\)

So the difference quotient over the time interval   [2.1,8]  is

\(t=2.1\\ h=8-2.1=5.9\\ DQ=45-16*5.9-32*2.1 = -116.6\)

Well  Veteren, it took a while but now you have 2 solid answers  :))

I didnt understand the guest's way, so ill solve it my way:

P(k)=the number of times the digit k appears between 0 and 999,999. P(0)*0+P(1)*1+P(2)*2+....+P(9)*9+1= the sum we need to find (i replaced 1,000,000 with 0, all we need to do is add 1 to the final solution.

all the numbers between 0 and 999,999 are also every possible combination of 6 digits (instead of 1 we write 0000001, instead of 2 we write 000002. Yes, a large number of zeroes is added, but that doesnt change the solution because x+n*0=x+0=x)

Here's a link to a similar question someone asked about the number of times a digit appears between 1 and 10ⁿ(n is a positive integer):

https://web2.0calc.com/questions/how-many-3s#r21

. i told him that between 0 and 10ⁿ-1 ,every digit (if we write it the way i write the numbers) appears n*10^n-1 times, meaning every digit except for 0 appears n*10^n-1 (but we dont care about the zeroes). what we want to find is the number of times every digit (except for 0) appears between 0 and 999,999=10⁶-1. so, the answer is 6*10^6-1=6*10⁵.

so, P(k)=6*10⁵ for every k that is between 1 and 9.

P(0)*0+P(1)*1+P(2)*2+....+P(9)*9+1=P(1)*1+P(2)*2+....+P(9)*9+1=6*10⁵(1+2+3+4....+9)+1=6*10⁵*45+1=27,000,000+1=

27,000,001

In the following h is the time increment:

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Here is my take on this one. Will do a twist on Gauss's method as follows:

0, 1, 2, 3..............999,996, 999,997,999,998, 999,999

Will add first and last=0 +999,999 =9 x 6 =54

Will add second to second last =1 + 999,998 =999,999 =9 x 6=54........and so on.

Since we have: 1,000,000/2 =500,000 pairs, each equaling 54, we therefore have:

500,000 x 54 =27,000,000 !!.

The only number left is the 1,000,000 itself, and since it has a value of 1, we therefore have:

27,000,000 + 1 =27,000,001 !!. And that is that, I THINK !!.

One possible solution:

Individual elephant weights are w:

so the lightest is 100kg.

Calculate the digital root:  

6*(9+7+5+3+1) = 150

1+5+0 = 6  

6 is divisible by 3 hence the original number is divisible by 3, so it is composite, not prime.

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Perhaps these graphs will help:

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Dont come on often but luckily i saw this awesome post so...

Happy birthday Melody, 

Cookies are great, hope you enjoy: 

Notice something else:

If we use modular arithmetic.....it's easy to evaluate i to any positive integer power

Exponent mod 4  = 1    →   i

Exponent mod 4  = 2    → -1

Exponent mod 4  = 3   →   -i

Exponent mod 4  = 0   → 1

So

9 mod 4  = 1   =  i

100 mod 4  = 0  = 1

2017 mod 4  = 1  = i

This is how I think of it:

For example:

i^7 = i i i i i i i

Then circle each pair of i's and replace each pair with -1.

i^7 = i i i i i i i = (i i)(i i)(i i)i = (-1)(-1)(-1)i

Then circle each pair of -1's and replace each pair with 1.

i^7 = i i i i i i i = (i i)(i i)(i i)i = (-1)(-1)(-1)i = [ (-1)(-1) ] (-1)i = [1](-1)i = -i

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For i^2017 , we could do it the same way,

...but I don't feel like writing out all those i's!

But if I WERE to write them all out, and circle each pair of i's,

how many pairs would I get?

I would get 2017/2 = 1008 pairs of i's with a remainder of 1 i's.

Then replace every pair with a negative one and then put one i at the end.

Then if I circled each pair of negative ones, how many pairs would I get?

I would get 1008/2 = exactly 504 pairs of negative ones.

(There are no negative ones left unpaired.)

Then I would replace each pair of negative ones with a positive one.

And then I would "erase" every " 1 " that I wrote because those are not necessary to write. Whatever is left = i^2017

i^2017 = i

I hope this made sense!! 

Very impressive, hectictar....!!!!

Notice that the selection of point "A"  is irrelevent......the fact that BA  and CA are perpendicular to l and m means that  angle BAC always = 152°.......!!!!

Hmmmmm! This is an interesting one...

Here is a picture:

There is a quadrilateral formed in the middle by:

point O, the intersection point of AB and l, point A, and the intersection point of AC and m.

The sum of the interior angles in every quadrilateral = 360º

Look! We know all 3 angles in the quadrilateral except angle BAC.

28º + 90º + m∠BAC + 90º = 360º

m∠BAC = 360º - 90º - 90º - 28º

m∠BAC = 152º

700500000000000

Tada!

We have the following two equations

(x - 2)^2 + (y + 1)^2   = 16

(x - 2)^2 + (y - 5)^2  =  16

Setting the two equations equal and subtracting  (x-2)^2  from both sides. we have that

(y + 1)^2  =  (y - 5)^2

y^2 + 2y + 1  = y^2 - 10y + 25   rearrange

12y  = 24

y = 2

And subbing this into either of the original equation we have that

(x - 2)^2 + 3^2  = 16

(x - 2)^2  = 7        take both roots

x - 2  =   ±√7       add 2 to both sides

x = ±√7 + 2

So....the intersection points  are A, B =    (√7 + 2, 2)   and ( -√7 + 2, 2)

My question, tertre, is how to interpret (AB)^2.......are we supposed to take the dot product of these points and square that???

If so..... [A (dot) B]^2  =   [ 4 - 7 + 4]^2 =  1^2   = 1

When it hits the ground, h(t)  will = 0  .....so.....

0  =  -16t^2 + 48t + 4     divide through by -4

0 = 4t^2 -12t -1

4t^2 -12t - 1  = 0        

Using the quadratic formula....the ball hits the ground at t ≈  3.08 sec

18.  tan ( x - pi/4)

This is the graph of the tangent shifted to the right by pi/4 units

The normal tangent graph would have an asymptote at pi/2......so this graph will have an asymptote, for example, at

x = (pi/4 + pi/2)   =  3pi/4

See the graph here : https://www.desmos.com/calculator/ufvhykjqvu

19.  f(x) = 3cos(2pi *x)

The period, P, can be found as

P(2pi) = 2pi    →  P  = 1 radian

Here's the graph :   https://www.desmos.com/calculator/a4od0m3gco

Wow !!!

Happy Birthday ! ! ! ! ! ! ! ! ! ! !

Thanks Chris, you are a great friend. :)

I'll have my cake along with MJ's icecream for morning tea.

Yum :))

1.

a.

 6x + 2 - 8x + 5   =

6x - 8x + 2 + 5  =

-2x  + 7

b.

{8x + 12 ] -  [5x - 17 ] =

8x  - 5x  + 12  - (-17)  =

3x + 29

c.  (7x) (x - 9)  =  7x * 4x  - 7x * 9    =   28x^2  - 63x

2.

2/3 (3x + 14) = 7x + 6     multiply through by 3

2(3x + 14)  = 21x + 18    simplify

6x + 28  =  21x + 18       subtract 6x, 18 from both sides

10 = 15x                         divide both sides by 15 and reduce the fraction

10/15  = x  = 2/3

Using the Law of Sines, we have

sin P / 47.6  =  sin 73 / 68     multiply both sides by 47.6

sin P  = ( 47.6 sin (73) / 68 )

Take the sine inverse to find  P

arcsin  ( 47.6 sin (73) / 68 )  = P  ≈   42.02°

Using the Law of Cosines

d =  sqrt [ 21^2 + 27^2 - 2(21)(27)cos(25) ]  ≈  11.93

The field is halved after every round......so.......after 6 rounds we have

256 / 2^6  =   256 / 64  =  4 players left

BRAVO, CPhill !!!.