The diagonal of a rectangle = sqrt(w^2 + l^2)
w = width
l = length
In this problem,
The diagonal = 20 in
w = x
l = 2x + 8
Let's plug our numbers into the formula above.
20in = sqrt((x)^2 + (2x + 8)^2)
Let's simplify the inside of the sqrt
20 in = sqrt(5x^2 + 32x + 64)
Now, let's square both sides.
400 = 5x^2 + 32x + 64
Subtract 400 from both sides.
0 = 5x^2 + 32x - 336
Factor
0 = (5x - 28)(x + 12)
Set both terms equal to zero and solve.
x + 12 = 0
Subtract 12 from both sides.
x = -12
5x - 28 = 0
Add 28 to both sides.
5x = 28
Divide both sides by 5
x = 28/5
The width cant be a negative number so now we know that the only real solution is 28/5
Let's plug 28/5 into our length equation.
Length = 2(28/5) + 8 = 56/5 + 8 = 96/5
In conclusion,
Length = 96/5 inches
Width = 28/5
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
D(x)=1375-110x
when it reaches 0 it has reached destination
0=1375-110x
add 110x to both sides
110x=1375
divide both sides by 110
x=12.5
answer is 12.5 hour
we know that
If the point has an nonzero abscissa (x coordinate) then it can be located in any of the four quadrants
If the point has an negative ordinate (y coordinate) then it can be located in Quadrant III or Quadrant IV
therefore
the answer is the option
Quadrant III or Quadrant IV