Given:
Volume of the cone = π r² h/3
radius of the cone is half the radius of the cylinder ⇒ r/2
height of the cone is equal to the radius of the cylinder. ⇒ r
V = 3.14 * (r/2)² * r/3
V = 3.14 * r²/4 * r/3
V = 3.14r³ / 12
In order to do this you can <span>factorise it to get
(x+1)^2 + (y-5)^2 = 16 = 4^2
hence centre is (-1,5) and radius is 4
</span>I hope this is what you are looking for
Answer:
The answer is below
Step-by-step explanation:
Find the dimensions of the rectangle of maximum area with sides parallel to the coordinate axes that can be inscribed in the ellipse 4x² + 16y² = 16
Solution:
Given that the ellipse has the equation: 4x² + 16y² = 16
let us make x the subject of the formula, hence:
4x² + 16y² = 16
4x² = 16 - 16y²
Dividing through by 4:
x² = (16 - 16y²)/4
x² = 4 - 4y²
Taking square root of both sides:

The points of the rectangle vertices is at (x,y), (-x,y), (x,-y), (-x,-y). Hence the rectangle has length and width of 2x and 2y.
The area of a rectangle inscribed inside an ellipse is given by:
Area (A) = 4xy
A = 4xy

Therefore the length = 2x = 2√2, the width = 2y = 2/√2
Answer:
x= 
Step-by-step explanation: