ElectricPavlov

1 st patient at 1200    in lobby until 12 35

2nd patient at 12 05    in lobby until 12 40

3rd patient   at 12 10   in lobby until 12 45

4th patient   at 12 15    in lobby until 12 50

5th patien   at 12 20      getting a shot at 12 34

6th patient   at 12 25     in lobby until 12 35

7th patient   at 12 30     in lobby until 12 40                                  6 patients are in the lobby at 12 34

Hour hand starts at 5 *30 = 150 ^o    moves 1/2  degree per minute     150 + 1/2 x 51 = 175.5⁰

minute hand moves 6^o per min     6 x 51 = 306^o

306 - 175.5 = 130.5^o

(5x+2)(x-3)

This question is rather vague

The car travels a total   130 m in 20seconds     = 6.5 m/s   OVERALL

As far as it's DISPLACEMENT is concerned , it is   90 - 40 meters from the starting point = 50 meters   in 20 s = 2.5 m/s  NET

You could just superimpose them to see it is not true:

****  deleted ****    Misread the question....Sorry !     ..

Here is one way:

x = y+4

xy = 2

(y+4)(y)= 2

y^2 + 4y-2 = 0     Quadratic Formula shows    y = .44949      and  -4.44949

                                                        then x =          4.44949     and   -.44949

|x+y| = 4.89898

Put D = 150 into your equation      4 (150) - 250 = .......gpm

Max value will occur at   - b/2a  = - p/2

Put this into the equation to find the min value =3

(-p/2)² + p (-p/2) + 4 = 3

p^2 /4   - p^2 /2 +1 = 0

- p^2 /4 + 1 = 0

-p^2 = -4

p = 2   

This question is a bit unclear....

  I will assume the travel distance is the hypotenuse of a right triangle

  the opposite leg of the angle of climb is 250

sin(angle) = 250/1000

arcsin (250/1000) = angle = 14.48 degrees

(if the 1000 is the ground distance travel   then    tan (angle ) = 250/1000    angle = 14.04 degrees)