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

the answer is x equals 50, this is because the two angles are the same measurements because they are vertical angles. in corresponding angles, vertical angles are two angles with one diagonally above another. those two angles will be the same measurements.

110 - 10 = 100

100 / 2 = 50

x   = 50

110 = 2(50) + 10 = 110

110 = 110

can u give brainliest if i'm correct?

for the points

step-by-step explanation: