A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Four hu
1 answer:
Answer:
h = 120 ft; w = 80 ft
A = 9600 ft^2
Explanation:
Let h and w be the dimensions of the playground. The area is given by:
A = h*w (eq1)
The total amount of fence used is:
L = 2*h + 2*w + w (eq2) (an extra distance w beacuse of the division)
Solving for w:
w = L - 2/3*h = 480 - 2/3*h (eq3) Replacing this into the area eq:
We derive this and equal zero to find its maximum:
Solving for h:
h = 120 ft. Replacing this into eq3:
w = 80ft
Therefore the maximum area is:
A = 9600 ft^2
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Answer:
<h2>a) 50°</h2><h2>b) 40°</h2>Explanation:
Check the complete diagram n the attachment below
a) The angle of incidence on a plane surface is the angle between the incidence ray and the normal ray acting on a plane surface. The normal ray is the ray perpendicular to the surface while the incidence ray is the ray striking a plane surface.
According to the diagram, the angle of reflection r₂ on M₂ is 90°-g where g is the angle of glance.
Given angle of glance on M₂ to be 40°, r₂ = 90-40 = 50°
According the second law of reflection, the angle of incidence = angle of reflection, therefore i₂ = r₂ = 50° (on M₂)
Also ∠OO₂O₁ = ∠OO₁O₂ = 40° (angle of glance on M₁){alternate angle}
The angle of incidence on M₁ = 90° - 40° = 50°
b) The angle of incidence to the surface of M₁(∠PO₁A)will be the angle of glance on M₁ which is equivalent to 40°
