Abeach ball rolls off a cliff and onto the beach. the height, in feet, of the beach ball can be modeled by the function h(t)=64–16t^2, where t represents time, in seconds. what is the average rate of change in the height, in feet per second, during the first 1.25 seconds that the beach ball is in the air?

You are given h(t) = 64–16t^2.

You need to find h(1.25).

Set up:

h(1.25) = 64 - 16(1.25)^2

Take it from here.

Step-by-step explanation:

Given the height of a beach ball rolling off a cliff as a function of time

h = 64 — 16t²

We want to find the average change in height when t = 1.25s

The rate of change of height is the differentiation of the height I.e

∆h/∆t → dh/dt

Therefore

h = 64 — 16t².

Therefore

Differential of a constant is zero

dh/dt = —32t

At t = 1.25s

dh/dt = —32(1.25)

dh/dt = —40 ft/s

So, the height is reducing at a rate of 40ft in every seconds.

do you have the measurements?

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841