Question

The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the function h(t)=300-16t squared which expression could be used to determine the average rate at which the object falls during the first 3 seconds of its fall

Answer 1

Answer:

EDGE 2020

Step-by-step explanation:

D) h(3)-h(0)/3

trust me

Answer 2

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby \begin{array}{llll} average\ rate\\ of\ change \end{array}\\\\ -------------------------------\\\\ f(x)= 300-16t \qquad \begin{cases} x_1=0\\ x_2=3 \end{cases}\implies \cfrac{f(3)-f(0)}{3-0} \\\\\\ \cfrac{[300-16(3)]~~-~~[300-16(0)]}{3}\implies \cfrac{[300-48]~-~[300]}{3} \\\\\\ \cfrac{252-300}{3}\implies -\cfrac{48}{3}\implies -\cfrac{16}{1}\implies -16$

is a negative rate, because the object is falling down, at 16 feet every passing second.