How many groups does it take to equally distribute 2 5/8?

If a football is thrown upward, its height above the

dybincka [34]

Answer:

a) s = -16 $t^{2}$ + 192t + 96

b) Height of the football after 2 seconds is 448 feet.

ii. The football would not hit the ground after 2 seconds.

Step-by-step explanation:

Given:

s = -16 $t^{2}$ + $V_{o}$ t + $s_{o}$

where: s is the height of the football above the ground, t is the time, $V_{o}$ is the initial velocity and $s_{o}$ is the initial height.

The height of the building = $V_{o}$ = 192 feet and the initial speed of the ball = $s_{o}$ = 96 feet per second, then;

s = -16 $t^{2}$ + 192t + 96

a) The polynomial is given as:

s = -16 $t^{2}$ + 192t + 96

b) The height of the football when t = 2 seconds is;

s = -16(2) + 192(2) + 96

= -32 + 384 + 96

= -32 + 480

s = 448

The height of the football when t = 2 seconds is 448 feet.

ii. The football would not hit the ground after 2 seconds.

7 0

2 years ago

1. Find the area of the region bounded above by y = e^x, bounded below by y = x, and bounded on the sides

salantis [7]

The area of the region bounded above by y= eˣ bounded by y = x, and bounded on the sides; x =0; and x = 1 is given as e¹ - 1.5.

<h3>What is the significance of "Area under the curve"?</h3>

This is the condition in which one process increases a quantity at a certain rate and another process decreases the same quantity at the same rate, and the "area" (actually the integral of the difference between those two rates integrated over a given period of time) is the accumulated effect of those two processes.

<h3>What is the justification for the above answer?</h3>

Area = $\int\limits^1_0 {(e^x-x)} \, dx$