CPhill

20)     x - 2y  = 18   ⇒  x = 18 + 2y

Sub the into the second equation for x    and we have

-6(18 + 2y) - 4y = 4      simplify

-108 - 12y - 4y = 4      

-108 - 16y = 4            add 108 to both sides

-16y = 112            divide both sides by -16

y  =  -7 

And

x  =  18 + 2(-7)   =   18 - 14    =    4

 sin2x / [1-cos2x ]  =  cotx           work with the left side

sin 2x  /  [ 1 - (1 - 2sin ^2 x ) ]  

sin 2x / [ 2 sin^2 x ] [ 2sin x cosx] / [ 2 sin ^2 x ]

sin x cos x  / [ sin ^2 x]

sin x cos x / [ sin x * sin x ] =

cos x / sin x      =

cot x

We have the following set of moves

(right, right, right, right, right, up, up, up ,up)

So.....we have 9 possible moves  and we want all the ordered sets where any  4 of them are  "up" or any 5 of them are "right"

So....the total pathways are  = C(9,4) = C(9,5)   = 126 paths

Answered earlier today here :

https://web2.0calc.com/questions/need-help-plzzz

f(x) = -2x^2 + 24x + 1280

This is a downward turning parabola........we can find the x coodinate of the vertex as

-24/ [ (2 (-2) ]  =    -24 / -4   =  6

This means that  the number of people that maximizes the revenue is  32 - 6   =  26

This is the answer to (2)

And for (1)....the max revenue is   

-2(6)^2 + 24(6) + 1280 =

-2(36) + 144 + 1280 =  $1352

For   f(x) =   1 /  [x - 3]

y intercept.....let x = 0     and we have   1 / [ 0 - 3]  =   -1/3

x intercept...let   y = 0 and we have that      0 = 1 / [ x - 3]    multiply both sides by x - 3   and we get that

0 = 1......so.....there is no x intercept

Vertical asymptote.....set  x - 3 =  0

Add 3 to both sides

x = 3   ⇒  the vertical asymptote

For  g(x)

y intercept =   0

x intercept = 0

Vertical asymptote ⇒  x = 4

LOL!!!

Yes....it was  "2"  complicated.....!!!!

Thanks, EP......I think either RainbowPanda or jennylove posted this before, but I couldn't find the question        

Thanks, heureka.....that is a nice comprehensive answer !!!

Just raise this to the 6th power and we have

( 2 * 2^(1/3) )^6   =

2^6 *  (2^(1/3))^6 =     { remember  that  (a^m)^n  =  a^(m*n)  }

64  *  2^2 = 

64 * 4 =

256