A square is inscribed in a circle. Find the ratio of the area of the square to the area of the circle.
We can refer to the radius of the circle as r. Since r is equal to half of the diagonal of the square, we can use Pythagorean theorem to solve for the side length of the square, which we will refer to as s. r^2 + r^2 = s^2 => 2r^2 = s^2 Therefore, s = r x sqrt(2) The area of the circle is equal to πr^2. The area of the square is equal to s^2, which is equal to [ r x sqrt(2) ] ^2 = (r^2) x 2 = 2r^2 Therefore, the ratio of the area of the square to the area of the circle is 2r^2 : πr^2, which is equal to 2 : π after simplifying.
A cup is in the shape of a cone, with the vertex pointed downward. The radius of the cone is 3 inches, and the height is 9 in. The cup is being filled with water at a constant rate of 10 cubic inches per minute. What is the rate at which the depth is changing when the water is 6 inches deep?
Volume of the cone = (1/3) x π x r^2 x h The depth of the water and the radius of the water-filled section of the cone are always proportional, so h = 3r, and r = (1/3)h We can substitute this into the volume equation, resulting in: V = (1/3) x π x [(1/3)h]^2 x h = (1/27) x π x h^3 Taking the derivative of the volume, we get that the rate of change of the volume is: dV/dt = (1/27) x π x 3 x h^2 x dh/dt = (1/9) x π x h^2 x dh/dt We know from the problem that dV/dt = 10 cubic inches per minute. We want to find the change in depth when h = 6 inches. Plugging in all the information, we get the equation: 10 = (1/9) x π x (6)^2 x dh/dt = 4π x dh/dt Solving for dh/dt (which is the change in depth), we get: dh/dt = 10/(4π) = 5/(2π) The change in depth is 5/(2π) inches per minute when the water is 6 inches deep.
Solve for x: x^2 + 3x + 4 = 7x + 16
Question: x^2 + 3x + 4 = 7x + 16 Step 1: Subtract 7x from both sides. x^2 + 3x + 4 - 7x = 16 Step 2: Subtract 16 from both sides. x^2 + 3x + 4 - 7x - 16 = 0 Step 3: Combine like terms x^2 - 4x - 12 = 0 Step 4: Factor the quadratic (x - 6)(x + 2) = 0 Step 5: Solve for x. x = 6, or x = -2