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I counted them and got

6 of 1 and 0 of the other      (6,0)  (0,6)

5 of 1 and 1 to the other      (6,2)  (2,6)

4 of 1 and 2 of the other      (6,3)  (3,6)                  also (6,2)  (2,6) again

                                            (6,4)  (4,6)

3 each                                  (5,5)

                                             (6,5)  (5,6)

                                             (3,3)

So I counted    12 possibilities

Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

If I understand your question, these are all the sequences possible on a roll of 1 die 5 times:

24556 , 24565 , 24655 , 25456 , 25465 , 25546 , 25564 , 25645 , 25654 , 26455 , 26545 , 26554 , 34455 , 34545 , 34554 , 35445 , 35454 , 35544 , 42556 , 42565 , 42655 , 43455 , 43545 , 43554 , 44355 , 44535 , 44553 , 45256 , 45265 , 45345 , 45354 , 45435 , 45453 , 45526 , 45534 , 45543 , 45562 , 45625 , 45652 , 46255 , 46525 , 46552 , 52456 , 52465 , 52546 , 52564 , 52645 , 52654 , 53445 , 53454 , 53544 , 54256 , 54265 , 54345 , 54354 , 54435 , 54453 , 54526 , 54534 , 54543 , 54562 , 54625 , 54652 , 55246 , 55264 , 55344 , 55426 , 55434 , 55443 , 55462 , 55624 , 55642 , 56245 , 56254 , 56425 , 56452 , 56524 , 56542 , 62455 , 62545 , 62554 , 64255 , 64525 , 64552 , 65245 , 65254 , 65425 , 65452 , 65524 , 65542 , Total =  90

1/(2L+s)   =4     and

1/(L+3s)=4         equating the denominators leads to L = 2s    subbing into one of the equations finds

                                       s = 1/20     so L = 2/20

now for FOUR of them

1 /( 4(2/20) + 4 (1/20) )    =     1/ (12/20) = 20 /12 = 1 2/3 hr   for 4 of each pump to fill the pool

mmm that answers wrong

1200  factors  as      2^4 * 3  * 5^2 =   2 * 2 * 2 * 2 * 3 * 5 * 5  =  4 * 4 * 3 * 5 * 5

So....the number of different identifiable roll sequenes  =

        5!                        120      

    _______  =            ____    =     30

     2!  * 2!                      4

Hi Guest! This is the harmonic mean problem. The answer is \(\frac{2}{\frac{1}{30} + \frac{1}{20}}=24\)

Approach (closing speed) speed  is   50+x

in 6 seconds at this speed the trains must cover 100 m + 120 m = 220 m

   (it is like holding one completely still..... when the engines meet the moving train must move 100 m to get the 'caboose' to the nose of the non-moving train....then an ADDITIONAL 120 m to get that caboose the end of the non-moving train....do you see?)

220 m /1000 m/km   /   (6 sec / 3600 sec/hr)    = 50 +x    km/hr

x = 82 km/hr

Bottom side is    6 x 10

other two rectangular sides    two x    5x10

trianglular ends  (two of them)    area =   1/2 (6) (4)  x two        add all of this together.........

It is a square so the side lengths are equal

area =  s x s =

                    s^2 = 225 ft^2

                        so the side length = 15 ft    she needs three side lengths of fencing =........

Slope of the line you are given    = 3           perpindicular   is   -1/3

  so far you will have    y = -1/3 x + b       put in the point given for x and y   to compute 'b'

                                    8 = -1/3 (6) + b            so b = 10

the line you are looking for becomes   y = -1/3 x + 10

We can start by plugging in values.

If we let x = 1 and y = 2, then 9x + 5y = 19 and x + 9y = 19.  We can add multiple of 9 to another number which is the remainder when the number is divided by 9 to get a multiple of 19.

We can also try other values, like x = 2 and y = 1.  Then 9x + 5y = 23 and x + 9y = 11, so these numbers are not divisible by 19.  We add a multiple of 9 to another number which is not the remainder when the number is divided by 9, which is why the numbers are not divisible by 19.

We can create congruences to account for the conditions in the problem.  We want both 9x + 5y and x + 9y to be divisible by 19, so 9x + 5y == 0 (mod 19) and x + 9y == 0 (mod 19).  Therefore, 9x + 5y == x + 9y (mod 19).  This simplifies to 8x == 4y (mod 19).  Then y == 2x (mod 19).

This tells us that the only cases where both 9x + 5y and x + 9y are divisible by 19 are the cases where y = 2x, like the case x = 1 and y = 2 we looked at above.  Every other solution will be a multiplicative scale of this solution (like x = 2 and y = 4, or x = 3 and y = 6).  This proves the result.

p + q = -6 + 8.5 = 2.5.

Thanks for the help!! I believe a-2b should equal 7, though, not -7.

1/5   /   4/1      =   1/5  *  1/4 = 1/20

Thank you!

[ sqrt( x + sqrt( x + sqrt( x = ... ]  =  5

Square both sides:

x + [ sqrt( x + sqrt( x + sqrt( x = ... ]   =  25

Since  [ sqrt( x + sqrt( x + sqrt( x = ... ]  =  5:

x + 5  =  25

      x  =  20

Sure....think about a kite....

I understand now thank you for your help!

-2(3x+2)>-8x+6     expand L side

-6x-4  > - 8x + 6    add 8x to both sides

2x-4 > 6

2x>10

x > 5

oops sorry

thank you!