The height y (in feet) of a ball thrown by a child is y=-1/14 x^2 + 2x +3 where a is the horizontal distance in feet from the point at which the ball is thrown. (a) how high is the ball when it leaves the child's hand? preview feet (b) what is the maximum height of the ball? preview feet (c) how far from the child does the ball strike the ground? preview feet get : video

Part 1) Height of ball at launch = 3 feet.

Part 2) Maximum height achieved by ball = 17 feet

Part 3) Point at which ball strikes the ground = 29.43 feet from point of launch.

Step-by-step explanation:

The height as a function of position of launch is given as

Part 1)

Since at the time the ball leaves the hand of the child the distance of the ball from point of throwing is zero thus putting x = 0 in the above equation we get

Height of throw

Thus the ball is at 3 feet high when it leaves the child's hand.

Part 2)

The maximum height the ball achieves can be found by maximizing the given function

Thus differentiating equation 'i' with respect to 'x' and subsequently equating to zero we get

Thus the maximum value of y(x) occurs at x = 14 feet thus

Part 3)

At the moment the ball strikes the ground it's height from the ground should be = 0

Thus equating y(x) = 0 we get

Taking the positive value we find the distance at which the ball strikes the ground equals 29.43 feet.

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.