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Using Catmg's combinations that do not appear to be in question

1, 1, 4, 6       

1, 2, 2, 6

1, 2, 3, 4

2, 2, 2, 3

Assume all the dice are different in colour.  (That will make it much easier)

Also assume that the dice are rolled one at a time.  It is probably easier to act as if order counts.

That means they are all unique and a green 1 is different from a blue 1

The total number of combinations is 6^4=1296,   just as catmg found.

Each of those 4 combinations above can be formed in 4! ways.  But I do need to allow for the double counting here

1, 1, 4, 6       4!/2 = 12

1, 2, 2, 6        4!/2 =12

1, 2, 3, 4           4!=24

2, 2, 2, 3           4

total  52

So we have a prob of    52/1296 =  13/324

I am not 100% sure that this is correct .......      But it is the same as guest found so maybe it is ok.

The sum of all a is 8.

b - a = 27.

b/a = 1 - i and c/a = 1 + i.

There are 104 pairs of integers (a,b).

[ABEF] = 2[(1/2) / tan(22.5º)] = 1/2 [ABCDEFGH]

@amygdaleon305, nice solution! Next time, consider using the \begin{align*} environment.

Your solution in that environment(you can view the code by double-clicking the TeX and clicking view TeX commands):

\(\begin{align*} x^2+13x+30 &= x^2+3x+10x+30 \\ &= x(x+3) + 10(x+3) \\ &= (x+3)(x+10) \end{align*}\)

$\Rightarrow$ $a = 3$ and $b = 10$

\(\begin{align*} x^2+5x-50 &= x^2+10x-5x-50 \\ &= x(x+10) - 5(x+10) \\ &= (x+10)(x-5) \end{align*}\)

$\Rightarrow c = 5$

$a + b + c = 3 + 10 + 5 = \boxed{18}$

$x^2 + 13x + 30 = (x+10)(x+3)$

$x^2 + 5x - 50 = (x+10)(x-5)$

$10 + 3 + 5 = \boxed{18}$

i'm not CPhill, but here is my take on this problem:

first, let's expand and simplify the expression - 

pq + 3q + 4p + 12 - (pq + q + 2p + 2) = 44

pq + 3q + 4p + 12 - pq - q - 2p - 2 = 44

2q + 2p + 10 = 44

2q + 2p = 34

and when we divide both sides of the equation by 2, we get p + q = 17.

hope this helped! please let me know if you are confused about anything i did 

Find the number of paths from P to Q, where each step is to the right or up.

\(3y=2x^2+16x+28\)

      \(=2(x^2+8x+16)-2\)

\(y={2\over 3}(x+4)^2-{2\over 3}\)

This equation is similar to  \(y=a(x-h)^2+k\)  in which (h,k) is the vertex 

⇒ (m,n) = (-4, -2/3)

 \(m+n=-4-{2\over 3} = -{14\over 3}\) 

\(x^2+13x+30 = x^2+3x+10x+30\)

                           \(=x(x+3)+10(x+3)\)

                           \(=(x+3)(x+10)\)

⇒ a = 3   and  b = 10 

\(x^2+5x-50 = x^2+10x-5x-50\)

                         \(=x(x+10)- 5(x+10)\)

                         \(=(x+10)(x-5)\)

⇒ c = 5

\(a+b+c = 3+10+5 = 18\)

Smart. :)

Surprisingly, cbrt(9999) isn't that big, I assumed there were a lot of cases. 

=^._.^=

(2/4)x =  140 + 20  -  68

(1/2)x  =  92

"Half of John's  money   equals $92.    How  much does  he have? "

I brought x dollars. 

I bough an 140 dollar cat, then my cat and I went to get a massage for 20 dollars. Finally, I got 68 dollars from the ATM. 

I only had 1/2 of the money I originally had. 

=^._.^=

Let, Cayden had $x and Jayden had             $y.

According to the question,

                  x + y = 784          ...............(1)

and,         x - \({x \over 6}\frac{}{}\) = y - 124

                5x - 6y = -744        ................(2)

Solving equation (1) and (2)

eq(1) *6 + eq(2)

              6x + 6y = 4704

              5x - 6y = -744

         ...............................................

                   11x = 3960

                       x = 360

Therefore, Cayden had $360 at first.

Area =  2s^2  ( 1 +  sqrt 2)               where  s is the side length

1  =  2s^2  ( 1 + sqrt 2)

s^2   =      1 /  [ 2 (1  + sqrt 2)  ] 

Connect  HC   and draw a perpendicular  BI from B to HC

Triangle   BIC is a 45-45-90  right triangle

IC  =  s / sqrt 2   

HC  =   s  +  2s / sqrt (2)   =     s  +  s *  sqrt (2)  =   s ( 1  +  sqrt (2) )

[ ABEF  ]  = [ HCDG  ]  =       HC *  CD   =    

s ( 1 + sqrt (2) )  *  s =

s^2  ( 1 + sqrt (2) )     =  

1 /  [ 2 (1 + sqrt (2) )  ]   *   ( 1 + sqrt (2) )    = 

1 /  2

Let's give some distances    100 up    100 flat   100 down      meters

    then return                        100 up    100 flat   100 down

time required   (distance/rate = time  )    200/100  + 200/120  + 200 / 150    = 5 hr

              distance 600

distance / time = average speed  =       600/5 = 120 m/s   

Need a little more information.....    are you looking for the initial deposit?   Or a series of payments ?

If you are looking for initial deposit   we will need compounding period    (monthly ?  Weekly?  Quarterly ? yearly? )

               n= 62      periods ....but what is the period?   

See my answer here  :

https://web2.0calc.com/questions/logarithms_34

Hard to tell exactly, but it  looks as  though

f (-1/4)   = -1

f(2 + 1/4)  =  f( 9/4)  =    -1

So

c =  -1/4    and  9/4

Just  a little  trial and error, catng

21^3   = 9261      ⇒  sum =  18        

20^3  = 8000       ⇒  sum = 8

19^3  = 6859       ⇒  sum   = 28

18^3  =  5832     ⇒  sum  = 18

Nice answer, but how did you solve it?

I tried putting it in this form, but I don't know where to go from there. 

1000a + 100b + 10c + d = (a+b+c+d)^3

=^._.^=

2.

5832  =   

5 + 3 + 8 +  2   =   18

18^3  =  5832