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Sorry! I missed quite a few, but I think that is all there is.

https://web2.0calc.com/questions/quadratic_21219#r1

I also realized its not 64 maybe there was a repeated digit somehwere?

How many ways can you roll a sum of 18 with 3 dice?

How many ways can you roll the 3 dice? 

Thanks for listing them out but do you know any way to do it without listing them out? Because that could take a while

(105, 120, 125, 130, 135, 145, 150, 165, 170, 175, 185, 195, 205, 210, 215, 230, 235, 240, 245, 250, 260, 265, 270, 275, 280, 285, 290, 295, 305, 310, 315, 320, 325, 340, 345, 350, 360, 365, 370, 375, 380, 385, 390, 395, 405, 415, 420, 425, 430, 435, 450, 465, 470, 475, 485, 495, 510, 520, 530, 540, 560, 570, 580, 590, 605, 615, 620, 625, 630, 635, 645, 650, 670, 675, 685, 695, 705, 710, 715, 720, 725, 730, 735, 740, 745, 750, 760, 765, 780, 785, 790, 795, 805, 815, 820, 825, 830, 835, 845, 850, 865, 870, 875, 895, 905, 915, 920, 925, 930, 935, 945, 950, 965, 970, 975, 985)==116 such 3-digit integers

4x  -6y   = - 2    →   mult through by   9   →     36x - 54y  =  -18      

kx - 9y    =  3   →    mult through  by  6   →      6kx - 54y =    18

This will have no solution when

6k =  36

k  = 6

DUH!!!

Nice work figuring out the arithmetic series  !!!

b₁₉₉₉ =  2(1999) + 1  =  3999

b ₂₀₀₀ =  2(2000)   +1  =  4001

b_n  =      1  / [ (a_n )^2  - 1] 

Note that we can write this as partal fractions as follows :

b_n  =   1 / [ ( a_n - 1) (a_n + 1)]     =   

1/ [ 2 (a_n -1) ]  - 1/ [2(a_n + 1) ]  =

 (1/2) [ 1/(a_n - 1) - 1/(a_n + 1)]

So  we can write

(1/2)  [  ( 1/2 - 1/4 ) +  (1/4  - 1/6) + ( 1/6 - 1/8) + ( 1/8 - 1/10)  +  ....... + ( 1/3998 - 1/4000) + (1/ 4000 - 1/4002 )

Notice that most of the terms  "cancel"   and we are left with

(1/2) ( 1/2  - 1/4002)  =

(1/2)  [ 4002 - 2 ]  / ( 8004)    =

(1/2) (4000)  / 8004  =

2000 / 8004

500/ 2001    =  p/q

p - q    =   500 - 2001   =  -1501

There are \(900 - 5 \times 6 \times 6 = \color{brown}\boxed{720}\) numbers with at least 1 prime digit. 

a= 32     b = 81   c   = 64    d  = 25

You should be able to finish this now....

3x^2 + 4x + 12  =  2x^2  - 7x  + 1        rearrange as

x^2 + 11x + 11  =  0

r + s  =  -11     square both sides     

r^2 + 2rs + s^2    =121

And  rs =  11    so .....  2rs  =  22

So

r^2 + 22 + s^2  =  121

r^2 + s^2  + 22  =  121

r^2 + s^2   = 99

Here's my best attempt  (I'm not great at these counting problems !!! )

Non-prime digits = (0, 1, 4, 6, 8, 9 )

Number of three digit numbers not having any prime digits

Hundreds place  = (Choose  any  one of the digits 1,4,6,8,9) =  C(5,1) = 5

Tens place  = ( Choose any one of the digits 0,1,4,6,8,9) = C(6,1)  = 6

Ones place = (Choose any one  of the digits 0,1,4,6,8,9) = C(6,1)  = 6

So   5 * 6 * 6 =    180 have no prime digits

So

900  three digit numbers -  180 with no prime digits  =    

720  have at least one prime digit

f (f(x)) = 3(3x - 2) - 2 =  9x - 8

f (f(f(x)))  = f (9x  -8)  =  3(9x - 8) - 2  =  27x - 26

f(f(f(f(x)))) =  f(27x - 26)  =  3 (27x  -26)  - 2 =   81x - 80  =  g(x)

g ( 1)  =   81(1) - 80  =  1

g(2) =  81(2)  -  80  =   82

Domain  of g  =  [ 1 , 82 ] 

Solve this

x^2  = 8

x = sqrt (8)  = 2sqrt (2)     or   x = -sqrt(8)  =  -2sqrt (2)

The width of the interval   is     2sqrt (2)  -   - 2sqrt (2)  =   4sqrt(2)

THX, BuilderBoi.....I didn't see the "+" sign  !!!

We have the form

y = a ( x -h)^2  + k          where (h,k)  = the vertex  and  (x,y) is a point on the parabola

So

175 = a ( 1 - - 4)^2 +  0

175 =  a (5)^2 

175  = 25a

175 / 25  = a =   7

Note that  x^2  - 4x - 5 =  ( x - 5) ( x + 1)

Multiply through by these two factors and we  get

12x + 22  =   A( x + 1)  + B ( x -5)       simplify

12x + 22 =  (A + B)x  +  ( A - 5B)        equating terms we have

A + B  = 12     →   B = 12 -  A     (1)

A - 5B  =  22       (2)

Sub (1) into (2)

A - 5 ( 12 - A) = 22

A - 60 + 5A  = 22

6A  = 82

A =  82/6  =  41/3

And   

B   = 12  - 41/3  =   36/3 - 41/3   =  - 5 / 3

A + B  =   36/3  =  12

Three sides of the rectangle (a square) will be fenced  = 3*35  =  105ft

The semi-circle will have a radius of   (35/2)  =17.5 ft

And the additional fencing required = (1/2) the perimeter of the circle  = pi (17.5)  = 17.5 pi ft

So.....the total fencing required =  105 + 17.5 pi ≈  160 ft

Let his salary ==S

S  -  0.20S  -  0.40S ==1200

0.40S == 1200

S ==1200 / 0.40 ==$3,000 - his monthly salary.

The radius^2  =   (PQ)^2  =  ( 9 - - 3)^2  +  ( 4 - -3)^2  =  12^2 + 7^2  =  193

The area =     pi * 193  =

193 pi

 Number of terms in the series =  (104 - 2)  / 2  + 1  = 51 + 1  =    52

Sum of the series  =   

(2 + 104) * (52)  / 2  =

53 * 52  =

2756

2756 mod 7  =   5

sumfor(n, 2/2, 104/2, 2*n)==2756 mod 7 == 5 - the remainder 

f(-2) =  (-2)^2 + 3 =   7

t( f(-2)) =   t (7)  =  9 + 2f (-7) =  9 + 2 (7^2 + 3) =  9 + 2( 52)  =  113