Let
x-------> the width of the rectangular area
y------> the length of the rectangular area
we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2
substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft
see the attached figure
the solution is<span> the shaded area</span>
Volume of sphere: V(s) = 4/3*pi*R^3 = (4/3)*pi*(D/2)^3 = (1/6) * pi * D^3
Volume of cube: V(c) = s^3
Volume of them is the same, I'm assuming you actually want to know the length of the cubic vertice
So s^3 = (1/6)*pi*D^3 -> s = (1/6 * pi)^1/3 * 6 = (36pi)^1/3
Answer:
AB = 10
Step-by-step explanation:
Use the distance formula: √(12 - 18)^2 + (1 - 9)^2
√36 + 64
10
Calculate the hypotenuse( h ) using Pythagoras' identity
h =√(4² + 7²) = √(16 + 49) = √65
sinΘ = 4 / √65 ⇒ cscθ = √65 / 4
cosθ = 7/ √65 ⇒ secθ = √65 / 7
tanθ =
⇒ cotθ = 