ElectricPavlov

The shaded area is everything ABOVE and INCLUDING the line...

Calculate the equation of the line to be   y = - x + 1

    so the shaded area      y >= -x+1

c = 6      v = 22       

h = - 16t^2 + 22t + 6

10 = - 16t^2 + 22t  + 6

-16t^2 + 22t -4 = 0       Use quadratic formula to find t = .215    and     1.15     sec    

                                          so yes, it will reach 10 ft as we have times for it on the way up and down

sin 55 = opp/ hyp = height/22

sin 55 = height / 22

22 sin 55   = height           Get out your calculator to finish !   ( make sure it is in 'degree mode'  )

Common d = -  3/4

20 + (-.75)n < 0

.75n > 20

n> 26.66             The  27th

Opposite sides of a parallelogram are =

7 = 8y-1    solve for y

9 = 2x+3    solve for x

Whatever value of 'd' makes the lines' slopes the same would be what you are looking for .....same slope lines are parallel and do not intersect.

Place each line into y = mx + b   form     m is the slope

15y = 10x - 17

y = 2/3 x - 17/15          slope = 2/3      so  we need a value of d to make slope of second line = 2/3

dx + 12y = 8

12 y = - dx + 8

   y =  - d/12  + 2/3             we need    -d/12 = 2/3     so d = -8

Nice work MPS      but that should be    x = 11   or   - 4

3x^2-24x+72  

3 ( x^2 - 8x )    + 72      complete the square for x

 3 ( x-4)^2  - 48 + 72

3 ( x-4)^2 + 24           You can finish from here !  

Difference between day 1 and day 2 was 5 student tix and 25 dollars      so each student ticket is 5 dollars

    Day 1 was 1 adult and two student for 18 dollars    so an adult costs  8 dollars .

   x     -   2/5 x    - 28   = 44

  3/5 x =  72

x = 120 dollars to start