<h3><u>Given Information :</u></h3>
- Length of parallel sides = 60 ft and 40 ft
- Height of the trapezoid = 30 ft
<h3><u>To calculate :</u></h3>
<h3><u>Calculation :</u></h3>
As we know that,

- a and b are length of parallel sides.
- h denotes height.
<em>S</em><em>u</em><em>b</em><em>s</em><em>t</em><em>i</em><em>t</em><em>u</em><em>t</em><em>i</em><em>n</em><em>g</em><em> </em><em>valu</em><em>es</em><em>,</em><em> </em><em>we</em><em> </em><em>get</em><em> </em>:
Area =
× ( 60 + 40 ) × 30 ft
Area =
× 100 × 30 ft
Area = 1 × 100 × 15 ft
Area = 100 × 15 ft
<u>Area = 1500 ft</u>
Therefore,
- Area of the trapezoid is <u>1500 ft
</u>
Neagtive means go backwards or oposite
it takes away that many units
Distance Formula: 
Apply the points: 
Solve: 
Since the square root of 218 cannot be simplified, the answer is

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Answer:
Time = 3secs
Max height = 56m
Step-by-step explanation:
Given the height reached by the rocket modeled by the equation;
h(x) = -5(x - 3)² + 56 where;
x is in seconds
The rocket velocity at its maximum height is zero.
Hence dh/dx = 0
dh/dx = 2(-5)(x-3)
dh/dx = -10(x-3)
Since dh/dx = 0
0 = -10(x-3)
0 = -10x + 30
10x = 30
x = 3secs
<em>Hence it takes 3 secs to reach the maximum height.</em>
Get the maximum height reached by the rocket. Substitute x = 3 into the equation given;
Recall that:
h(x) = -5(x - 3)² + 56
If x = 3
h(3) = -5(3 - 3)² + 56
h(3) = 0 + 56
h(3) = 56
<em>Hence the maximum height that the rocket reaches is 56m</em>