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Matrix system:

\(\begin{pmatrix} 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} w \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ 19 \\ 12 \end{pmatrix} \)

Inverting the matrix leads to w = -1, x = -9, y = 6, z = 12, so wx + yz = 63.

You can track the probabilities using different case, i.e. Anty gets to 0 after 1 steps, 2 steps, 3 steps, and so on.  This gives us e = 1/2*1 + 1/4*2 + 1/8*3 + 1/16*4 + ... By arithmetico-geometric series, e = 2.  Similarly, f = 1/2*2 + 1/4*3 + 1/8*4 + 1/16*5 + ... = 3, so (e,f) = (2,3).

For the first break, the probability that you get a piece that is 1 or longer is 4/5.  Then for the second break, the probability that you get a piece that is 1 or longer is 3/5.  Therefore, the probability is 4/5*3/5 = 12/25.

The smallest fourth lucky number = 38.

Check: [5+7+1+3+7+2+3+8] = 1/5*[57+13+72+38] = 36 = 36

thanks, all of you!

The smallest n =408, so that: 2550 * 408 =1,040,400 whose square root =1020.

A particular fortune cookie lists four two-digit positive integers as your lucky numbers. The first three are 57, 13, and 72, but you got sauce on the last one and can't read it. If the sum of the digits of all four numbers equals 1/5 of the sum of all four numbers, what is the smallest possibility for the fourth lucky number?

My computer says there are: 181,440 eight-digit numbers between 10,236,789 and 98,763,210 that are divisible by 9. Somebody will confirm this using combinatorics such as 9! / 2! or 10 nPr 8 / 10 or 9 nPr 8 / 2!...etc. 

Oh... I didn't read the question right...

Wouldn't just two colors do it?

Equation of the line is     y = x+1

the shaded area is everything GREATER than this line but NOT including the line (because it is dashed)

so

y > x+1

cross multipy the R side to get   [ a(x+1)  + b(x-2) ] / [(x-2)(x+1) ]

Now you can equate the numerators of both sides of the equation

4x+13 = ax+a   + bx-2b        and now you can equate   the 'x' components coefficient and the constant

4 = a+b

13 = a-2b 

solve the simple system of red equations to find      a = 7 and   b = -3

Total Height of the rectangle (which represents the property) is   21 + x + x = 21 + 2x

Width = 39 + x + x = 39 + 2x      (or   2x + 39)

Area = Height * Width    =

(2x+21) * (2x+39)   = 4x^2 +120x + 819

d =  #dimes                     value = 10d

q = quarters = (50-d)       value = 25(50-d)            summed  = 800

10d    +   25(50-d)    = 800           solve for 'd' the number of dimes      then # quarters = 50 - d

Equation of line: x + y + 1 = 0

Inequality: x + y + 1 < 0 (check point (0,0))

From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B.  Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.

Then the line is y = -x/3 - 19/3.  Then 3y = -x - 19, so 3y + x = -19.  We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3.  Therefore, A + B = -9/19 - 3/19 = -12/19.

1/(9L)   = 4     L = 1/36

1/(6s)   = 8     s =  1/48

1/ (  (4/36) + (8/48) )  = 1/t

1 / ( 16/144 + 24/144) = 1/t

1/ (40/144) = 1/t

t = 3.6 hrs

I think Alan made a typo. You can't solve for x in the last equation! 

Cos(25/65) = 65 /(x + 25) - Now you can solve for x

169 =x + 25

x =169 - 25 = 144

From 1st eq     b =   (1-a)/2

From 3rd eq     c = 7-2a            sub these into second equation

(1-a)/2   + 14-4a = 3

1-a        + 28-8a = 6

29 -9a = 6

a = 23/9

b= -14/18 = -7/9

c= 17/9

Hi, I don't know you but welcome back! :)

The most frequent digit between 1 and 100 inclusive is the digit "1" as you can see in this breakdown of ALL 10 digits (0 to 9): (11, 21, 20, 20, 20, 20, 20, 20, 20, 20) >>>Total Number = 192

As follows:

Solve for x.

In the small right triangle on the RHS with sides of 65 and 25, wil find the altitude by Pythagoras's Theorem:
65^2 =25^2 + A^2
A  =sqrt(3600) =60
The measure of the altitude drawn from the vertex of the right angle of a right triangle to its hypotenuse is the geometric mean of the measures of the two segments of the hypotenuse.
Sqrt(25x) =60
5sqrt(x) =60      Divide both sides by 5
sqrt(x) =12         Square both sides
x = 144

we have (360-59-77-66-62)/12=8 so x=y since all the external angles add up to 360 degrees

we can use the pythagorenas theorem to find the altitude is 60 and we have te equation 60^2+x^2=y^2 and (x+25)^2-65^2=y^2 where y is the base