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L = 13/2.

Sorry

We can solve this using geometry. :))

Drawing a 1 by 1 square, let one side be x, from 0 to 1. Let the other be y, from 0 to 1. 

Then shade in the area where y <= 2x, the area is 3/4. :))

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5 = (x^3 - x^2 - 8x)/(x + 2)    multiply  both sides by  x + 2

5 (x + 2)  =  x^3  -x^2  -  8x

5x + 10  = x^3  - x^2  -8x        rearrange  as

x^3 - x^2  - 13x  -  10  =   0

We  have the form    ax^3  + bx^2  + cx + d

By  Vieta....the  sum of  the roots   =   -b /a  =  -(-1) / 1   =    1  

AX = 50/3.

We can't see the problem. 

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We  can use  the "points"    (5,  3.85)   and  ( 10 , 5.7)   =   (  t . h)

The slope of the line is    ( 5.7 -  3.85)  / ( 10 - 5)  = 1.85  /  5    =   37 / 100

The  equation  of  the line is

h =  (37/100) ( t  -10)  + 5.7       simplify

h  = (37/100)t  - 3.7 +  5.7

h =  .37t  +  2       where h is the height after  t  seconds

If     p*p*p  =  27....rhen   p^3  =  27  ⇒  p = 3

And

p * b * p  =   18

3 * b * 3  =   18

9 b  =  18

b = 18 / 9  =  2

And  

b* b * b   =  t

2 * 2 * 2  =  8  = t 

So

t * t * t  =  8 * 8 * 8  =   256

Note  x^2  - 1  = (x -1) (x + 1)

( x + 3)  / ( x^2 - 1)  =  A / ( x - 1)  +  B / ( x + 1)        multiply  through  by   ( x - 1) ( x + 1)

x + 3  =  A(x + 1)  + B  (x - 1)

x + 3 =  (A + B)x  + (A - B)

Equating terms .......this implies that  A - B =   3

The initial  height   =  16.4  ft

So.....the  hright  at landing is    16.4 - 3.2 =  13.2  ft

We can solve this

13.2   =  -16t^2  +  24t  +  16.4       rearrange as

16t^2  - 24t  - 3.2  =    0

Using  the  quad  formula we  get

t  =   [ 24  +  sqrt ( 24^2  - 4 (15)(-3.2) )  ]  /  ( 2 * 16)      ≈ 1.6  seconds

x + 4

_____  =    4       multiply  through  by  x + 3

x + 3

x + 4  =  4 ( x + 3)

x + 4  =   4x  +  12

x - 4x  =  12 -  4

-3x  =  8

x =   -8/3

So

1 / ( x + 3)  =  1 / (  -8/3 + 3)  =  1 / ( 1/3)  =    3

The  function  is

y = .46  + .20 * ceiling (x - 1)

This is definitely  a  step function ....see the  graph here  :  https://www.desmos.com/calculator/xahfr9lse3

Arc  length  =  r  * (theta  in radians)

12  = 8   * (theta  in rads )

12 / 8    =    3 / 2 =  1.5  rads

\(\frac{1}{3} t - 5 < t - 2 \le -3t + 7\\ \frac{1}{3} t - 5 < t - 2\qquad and \qquad t - 2 \le -3t + 7\\ \frac{-2}{3} t < 3\qquad \qquad and \qquad \qquad 4 t \le 9\\ -2 t < 9\qquad \qquad and \qquad \qquad t \le 2.25\\ t > -4.5\qquad \qquad and \qquad \qquad t \le 2.25\\ -4.5\)

The Latex is not working so here is a pic of the preview:

Why do you think we should give priority to your question.

Are you more important to us than other question askers?

Skjn...:

If you think it is not correct then state why.   

I partially agree with catmg.  There are 2 different equally correct answers to this.

Do not vote a member down for trying to help you.  Instead, thank them for their time.

Catmg is a highly valued answerer here and when you insult her/him you insult all of us.

Have you drawn it?

Have you done anything to help yourself answer this?

ABD = x

CBD = 2x

AD = y

DC = 2y

BD = h

Note that BAD * 2 = BCD. 

2*5*sine(x)*h = 6*h*sine(2x)

10*sine(x) = 6*sine(2x)

5/3sine(x) = sine(2x)

5/3sine(x) = 2*cos(x)*sine(x)

5/6 = cos(x)

x = 33.5573098

3x = 100.6719294

AC^2 = 5^2 + 6^2 - 5*6*cos(100.6719294 degrees)

AC = 8.15815883728

That problem took way too much effort, please tell me if it's correct and if there's a better solution. :))

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The largest 4 digit palindrome in base 3 is 2222. (only numbers from 0 - 2)

2222 from base 3 to base 3 = 2222 

2222 from base 3 to base 4 = 1100

2222 from base e to base 5 = 310

2222 from base 3 to base 6 = 212   # This works

Our answer is 6. :)))

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f(f(-2)) = -4/3.

Sorry, its wrong

It works out that b = 5.

The point is (18,12), so the sum of the coordinates is 18 + 12 = 30.

You should start by squaring sides.  This will get rid of the square roots.

THX!!!