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a. Length of the fence around the field = perimeter of quarter circle = 892.7 ft.
b. The area of the outfield is about 39,584 sq. ft..
<h3>What is the Perimeter of a Quarter Circle?</h3>Perimeter of circle = 2πr
Perimeter of a quarter circle = 2r + 1/4(2πr).
a. The length of the fence around the field = perimeter of the quarter circle fence
= 2r + 1/4(2πr).
r = 250 ft
Plug in the value
The length of the fence around the field = 2(250) + 1/4(2 × π × 250)
= 892.7 ft.
b. Size of the outfield = area of the full field (quarter circle) - area of the infield (cicle)
= 1/4(πR²) - πr²
R = radius of the full field = 250 ft
r = radius of the infield = 110/2 = 55 ft
Plug in the values
Size of the outfield = 1/4(π × 250²) - π × 55²
= 49,087 - 9,503
= 39,584 sq. ft.
Learn more about perimeter of quarter circle on:
Answer:
Step-by-step explanation:
Q1)
Use Phythogoras theorem:
Q2)
Apply phythogoras theorem:
Q3)
Apply phythogoras theorem again:
I have an attached an image for Question 2 for better understanding the length of DE in question equals 15 - 6 = 9
Answer:
Graph B
No
Inside
Step-by-step explanation:
Let, the number of apples = x and number of peaches = y.
It is given that the farmer can afford maximum of 54 acres.
Thus, x + y ≤ 54.
Also, apples require 3000 gallons of water and peaches require 800 gallons of water each day.
Since, maximum amount of water is 80,000 gallons per day,
We get, 3000x + 800y ≤ 80,000.
It is required to maximize the profit given by z = 3400x + 1600y.
Thus, the system of equations becomes,
z = 3400x + 1600y
3000x + 800y ≤ 80,000.
x + y ≤ 54.
Plotting these equations gives the following graph.
We can see that the graph obtained is equivalent to Graph B.
Further, we know that the maximum value of the objective function z = 3400x + 800y is obtained at the boundary point of the solution region.
Thus, any point inside the solution region will give a feasible solution not maximum solution.
Hence, ( 30,20 ) will not maximize the profit as it lies inside the solution region.
