67. Maximizing the Height of an Object If an object is thrown upward with an initial velocity of 32 ft/second, 32 , ft slash , second , comma then its height after t seconds is given by h(t)=32t−16t2. h , open t close , equals 32 t minus 16 , t squared , . (a) Find the maximum height attained by the object. (b) Find the number of seconds it takes the object to hit the ground.

(a)Therefore the maximum height of the object is 16 feet.

(b)Therefore the object hits the ground after 2 seconds.

Step-by-step explanation:

(a)

Given function of height is

h(t)= 32t-16t²

where t is time in second

[The unit of h(t) is not given, so consider the unit of h(t) in feet]

We know that,

If a function y = ax²+bx+c.

The maximum value of the y is at

Here a= -16 and b =32.

The object attains its maximum height when

Its means after 1 s , the object attains its maximum height.

Therefore the maximum height of the object is

h(1) = 32.1-16(1)²    [putting t= 1]

     =32-16

    =16 feet

(b)

When the object teaches the ground then height of the object will be zero.

i.e h(t) = 0

Therefore,

32t-16t² =0

⇒16t(2-t)=0

⇒t=0 or, t=2

Therefore the object hits the ground after 2 seconds.

walter needs to make at least 5 more cookies but no more than 40. 5 ≤ x ≤ 40

step-by-step explanation:

20 ≤ x + 15 ≤ 55

subtract 15 from all parts

20-15 ≤ x + 15-15 ≤ 55-15

5 ≤ x ≤ 40

this means how many more cookies he has to make

walter needs to make at least 5 more cookies, but no more than 40 cookies; 5 ≤ x ≤ 40