A golfer strikes a golf ball into the air from the ground. The height, h metres, of the ball can be modelled by h = 32t – 4t2, where t is the time in seconds after it leaves the
ground.

i. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the t -axis
and 1 cm to represent 5 units on the h -axis, draw the graph of h = 32t – 4t2
ii Use your graph to find he maximum height of the ball above the ground and the time

at which it occurs.
u dont have to draw the graph can u just find the values for me thanku

64 m, at time t = 4 sec

Step-by-step explanation:

We must maximize the function h = 32t – 4t^2.  One way in which to do this is to find the axis of symmetry, t = -b / [2a], and from that calculate the height:

a = -4, b = 32, c = 0

                                                                          -32

and therefore the axis of symmetry is t = = 4

                                                                          2(-4)

The height, h, at time t = 4, is h(4) = 32(4) - 4(4)^2 = 64 m

The maximum height of the ball is 64 m, at time t = 4 sec

answer:

the position of equilibrium will shift to the left.

step-by-step explanation:

the general equation for the dissociation of a weak base is

b + h₂o   ⇌   bh⁺ + oh⁻

according to le châtelier's principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

if you apply a stress by adding more bh⁺ or oh⁻,   the system will respond by moving in a direction that gets rid of the added ions.

the position of equilibrium will shift to the left, forming more of the undissociated base.

c: lkllmn.

hope it's you