pi is the ratio of circumference to the diameter of same circle. The length of the metal frame of trampoline considered is: Option D: 37. 7 feet.
<h3>How to find the length of the outer line of a circle?</h3>
Length of the outer line of a circle is called its circumference.
If the circle has got radius of r units length, then
Circumference of that circle = 2π r units = πd units where d is diameter of that circle.
This is why, we get the ratio of circumference to diameter as:
It is given that the trampoline in consideration has got diameter of 12 feet.
It is said that π = 3.14 to be considered (its not its exact value though).
Therefore, we get
Hence, the length of the metal frame of trampoline considered is obtained as: Option D: 37. 7 feet.
Learn more about circumference of circle here:
brainly.com/question/15211210
The optimum shape of such a box is half a cube. The corresponding cube will have a volume of 2×256 ft³ = 512 ft³ = (8 ft)³. Such a box has a square base that is 8 ft on a side. If the height is half that of the cube, it will be 4 ft.
The dimensions of your box will be 8 ft square by 4 ft high.
_____
If the base dimension is x ft, the area (quantity of material) is
... a = x² + 4x(256/x²)
... a = x² + 1024x⁻¹
Then the derivative of area with respect to x is
... a' = 2x -1024x⁻²
Setting this derivative to zero and solving for x gives the value of x for minimum area.
... 0 = 2x -1024/x²
... 512 = x³
... x = 8 . . . . . . . . same as above.
Answer:
12
Step-by-step explanation:
you have to add everything
Answer:
x = 7
y = 3
z (max) = 4950/3 = 1650
Step-by-step explanation:
Let call
x numbers of church goup and
y numbers of Union Local
Then
First contraint
2*x + 2*y ≤ 20
Second one
1*x + 3*y ≤ 16
Objective Function
z = 150*x + 200*y
Then the system is
z = 150*x + 200*y To maximize
Subject to:
2*x + 2*y ≤ 20
1*x + 3*y ≤ 16
x ≥ 0 y ≥ 0
We will solve by using the Simplex method
z - 150 *x - 200*y = 0
2*x + 2*y + s₁ = 20
1*x + 3*y + 0s₁ + s₂ = 16
First Table
z x y s₁ s₂ Cte
1 -150 -200 0 0 = 0
0 2 2 1 0 = 20
0 1 3 0 1 = 16
First iteration:
Column pivot ( y column ) row pivot (third row) pivot 3
Second table
z x y s₁ s₂ Cte
1 -250/3 0 0 200/3 = 3200/3
0 - 4/3 0 -1 2/3 = -20/3
0 1/3 1 0 1/3 = -20/3
Second iteration:
Column pivot ( x column ) row pivot (second row) pivot -4/3
Third table
z x y s₁ s₂ Cte
1 0 0 750/12 700/6 = 4950/3
0 1 0 3/4 -1/2 = 7
0 0 1 -1/4 1/2 = 9/3
