CPhill

Assuming  that letters can be repeated......we have 7 letters to hose from in each case....so.....

7  * 7  * 7   =   343  possibilities

The common ratio, r,  can be determined  as  the ratio of the terms

a_n+1  / a_n    so we have....  (1/8)  / (1/4)   =  4/8  =  1/2

And we have that

255/512  =  1/4  [  1 -  (1/2)^n ] /  [  1  - 1/2  ]

255/512 =   1/4   [ 1 - (1/2)^n ]  /  (1/2)

255/512  =  (1/2)  [ 1 - (1/2)^n ]     multiply both sdes by  2

255* 2 / 512   =  1 - (1/2)^n  rearrange  as

(1/2)^n  =  1  -  510/512

(1/2)^n   =  [ 512  - 510 ] / 512    =  2 / 512  =   1/256

Take the log of both sides

log (1/2)^n  =  log (1/256)     and we can write

n  =  log (1/256) / log (1/2)

n  =   8    ⇒  8 terms

Each girl  will play all five of the boys....so....

5 *  5    =  25 games

2x^2 -5x  - 4  = 0     divide through by 2

x^2  - 5/2 x  -  2   =  0      add 2 to both sides

x^2  -  5/2 x   =  2        take 1/2 of 5/2  = 5/4.....square it =  25/16...add to both sides

x^2  -5/2 x  +  25/16   =  2 +  25/16      simplify both sides

(x  - 5/4)^2  =  57/16     take both rots

x - 5/4   = ±√57 / 4       add  5/4 to both sides

x  =  [  ±√57   +  5  ]  / 4

So...n  = 57

f(x)  =  -3 /  [5x^3]

Note that  x  cannot be 0  because it will make the denominator 0

Also.....the range willl be the same....all y except  y = 0

To see this....set the left side  = 0

0  = -3 / [ 5x^3]         multiply  both sides by 5x^3

0 * 5x^3  =  - 3

0  =  -3         this says that there is no x value that makes y  = 0

So...the domain and range is  (- inf, 0 ) U (0, inf)

Here's a graph :  https://www.desmos.com/calculator/1vbys9muqn

The second graph....if we let  x  = 0  the y intercept  is

1 / [ 0 - 1  ]  - 2  =

-1   -  2  =

-3

And the second graph is the only one with a y intercept of -3

The third graph.....one way of telling.....let  x  =  0  and we have the y intercept

2 /  [ 0 - 1]  + 3  =

-2  +  3  =

1

And the third graph is the only one that has a y intercept  =  1

Assuming we are talking about real zeroes....

Let one function be    2( x - 2) ^2   = 2 x^2  - 8 x  + 8    whch  has one zero  at  x  = 2

Let the second be   -x^2  -  17       which has no real zeroes

So....their sum  is

x^2  -  8x  -  9         factor

(x - 9) (x + 1)   

The zeroes  are   x  = 9  and x  =  -1

See the graphs here :  https://www.desmos.com/calculator/fre2biof7f

If  -8  +  k  is the y intercept of one line  and  8 is the y intercept of the other....and we want the y intercepts to be the same, we have that

-8  +  k  =  8      add 8 to both sides

k   =  16

Given that f(x) = 4x + k , and g(x) = x - 2 , find the possible value of m and k if  f[g(x)] = mx + 8

I'm assuming that these represent linear equations  with m = slope and k  = y intercept

f (g (x) )  =   4 (x - 2)  +  k  

So.....

4(x - 2) +  k  =  mx +  8     

4x  -  8  + k  =  mx  +  8            let k  =  16

4x  + 8  =  mx  + 8         ⇒      m   =  4

So

m  =  4     and  k   =  16