Answer:
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
Step-by-step explanation:
If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -
x = ( 30 cos 20° )( time ),
y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2
To determine " ( 30 cos 20° )( time ) " you would do the following calculations -
( x = 30 * 0.93... = ( About ) 28.01t
This represents our horizontal distance, respectively the vertical distance should be the following -
y = 30 * 0.34 - 4.9t^2,
( y = ( About ) 10.26t - 4.9t^2 + 2
In other words, our solution should be,
x = 28.01t,
y = 10.26t - 4.9t^2 + 2
<u><em>These are are parametric equations</em></u>
The <em>proposed</em> design of the atrium (<em>V < V'</em>) is possible since its volume is less than the <em>maximum possible</em> atrium.
<h3>Can this atrium be built in the rectangular plot of land?</h3>
The atrium with the <em>maximum allowable</em> radius (<em>R</em>), in feet, is represented in the image attached. The <em>real</em> atrium is possible if and only if the <em>real</em> radius (<em>r</em>) is less than the maximum allowable radius and therefore, the <em>real</em> volume (<em>V</em>), in cubic feet, must be less than than <em>maximum possible</em> volume (<em>V'</em>), in cubic feet.
First, we calculate the volume occupied by the maximum allowable radius:
<em>V' =</em> (8 · π / 3) · (45 ft)³
<em>V' ≈</em> 763407.015 ft³
The <em>proposed</em> design of the atrium (<em>V < V'</em>) is possible since its volume is less than the <em>maximum possible</em> atrium. 
To learn more on volumes, we kindly invite to check this verified question: brainly.com/question/13338592
Average rate of change from x = -3 to x = 1 is (f(1) - f(-3))/(1 - (-3)) = (5 - (-3))/(1 + 3) = (5 + 3)/4 = 8/4 = 2
So basically what you have to do is