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83-__=11     subtract 83 from both sides

-___  = - 72

So ___  =  72

19.   Domain  =  (-4, 6]

        Range  =  [-4, 4]

20. Domain  =  [-3, 4 ]   I'm assuming that the function becomes a vertical line when x = 3, but it's hard to tell from the picture.....

Range  =  ( -infinity,   4  ] 

slope = y value

           -----------

            x- value

Your welcome!

At 6.25 seconds

(6.25,625)

*Hint: usually a square root of a square are the x and y values

vf = vo + a t

60 = 50 + (5) t

2=t

Thanks, qazwsxedc123,  for those good answers  !!!!!

h=200t-16t^2

1. Just sub in  "4"  fot t   and we have 544 ft

2. at t = 10 , we have  400 ft

3. Falling......if it were rising, the height at 10 seconds would br greater than at 4 seconds

4.   Table values

5 = 600 ft

6 = 624 ft

7 = 616 ft

8 = 576 ft

9 = 504 ft

5.   Here's the graph......I'll let you fill in the details  - you can "trace" the graph by holding down your mouse button :  https://www.desmos.com/calculator/1hlpee4759

6.  The shell reaches max height at 6.25 seconds  [ see the graph]

7. Here's the graph of this function : https://www.desmos.com/calculator/yce2kwfm01

No I can't graph it my computer won't load.

Do I have to graph it. I can if you want?

The shell is falling at 10 seconds. This is because the height of the shell decreased by 144 feet.

I am sorry I meant 400 feet and 544 feet

Here are the answers:

h = 200t-16t^2

h = 200(4) -16(4)^2

h = 800 - 256

h = 544

h = 200t-16t^2

h = 200(10) - 16(10)^2

h = 2000 - 1600

h = 400

The whole pg.?

I estimate that each pizza costs about $ 10 each    5 pizzas x 10 = $ 50 total       20% would be $10      so $ 5 would be a small tip of 10%

19 or 21

depends

All we need to do is to compare the cubes of their radiuses.....so we have

[ 9.5 / 2]^3   /  [ 2.5 / 2]^3   =  [9.5]^3 / [2.5]^3  = about 54.872  times as great

You left off the velocity the rocket has acheived.....I assume it is 50 m/s again

Velocity changed from 0   to 50 m/s     in   t   time

v= vo + at

50 = 0 + at

50/t = a             (Equation 1)

Now we know   x = xo + vo t + 1/2 a t^2   for the thrust portion of the rocket flight   (xo and vo are both ZERO) so we are left with

x= 1/2 a t^2          From Equation 1 :  50/t = a      Substitute this value for a

105 = 1/2 (50/t) t^2

 105 =  25 t

t = 4.2 seconds

oh wow it just came to me, i'm sorry. I'm just being an idiot...

thanks for the help!

Which following points I don't see anything?

IDK ;D