1. ## problem set

1.) A vertical cylindrical storage tank, with a diameter 20 ft, is filled with oil to a depth of 40 ft. Sometime later, the oil is drained, decreasing the depth at a rate of 8 in per hour. Write an equation for the volume of oil (ft^3) remaining in the tank t hours later as a function of t, V(t). Draw a geometrically correct sketch supporting your solution.

2.) The area of a series of isosceles triangles is modeled by a triangle that has its base along the positive x-axis, with one base angle at the origin and its vertex in the first quadrant, positioned on the graph of the function f(x)=6-x^2. Write the area of the triangle as a function of x, A(x). Make a suitable sized graphical sketch of the problem, showing relevant details supporting your solution.

thank you so much with the help :]

2. 1.) A vertical cylindrical storage tank, with a diameter 20 ft, is filled with oil to a depth of 40 ft. Sometime later, the oil is drained, decreasing the depth at a rate of 8 in per hour. Write an equation for the volume of oil (ft^3[SIZE=2]) remaining in the tank t hours later as a function of t, V(t). Draw a geometrically correct sketch supporting your solution.

That means the volume is being diminished per hour by the volume of a circular disk whose radius is 10 ft and whose thickness is 1 inch.
In t hours, the volume is diminished by pi(10^2)(1/12) *t
That is (25pi/3)t cu.ft.

The original volume in the tank is pi(10^2)(40)= 4000pi cu.ft.

So, the remaining volume in the tank after t hours is

If you like,
V(t) = 25pi(160 -(t/3)) ------answer also.

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2.) The area of a series of isosceles triangles is modeled by a triangle that has its base along the positive x-axis, with one base angle at the origin and its vertex in the first quadrant, positioned on the graph of the function f(x)=6-x^2. Write the area of the triangle as a function of x, A(x). Make a suitable sized graphical sketch of the problem, showing relevant details supporting your solution.

I don't know how to draw sketches or graphs in computers, so you have to do the sketches yourself.

Anyway, here, the apex of the isosceles triangle is on the first-quadrant branch of the vertical parabola, opening downward, y = 6 -x^2.

The coordinates of that apex is (x,y).
Since you want the area as a function of x, then you have to express the y in terms of x.
So the apex is (x,(6 -x^2))

The area of the isosceles triangle then is
A = (1/2)(base)(altitude)
A(x) = (1/2)(2x)(y)
A(x) = (1/2)(2x)(6 -x^2)