What is the extracts volume of the largest cone that can fit into a cube with edges of 10 inches

1 answer:

4 0

Volume of a cone = π r² h/3

I envisioned a cone inside a cube. I only identified the necessary value to compute for the exact volume of the largest cone.

Edges of the cube is 10 inches. When the circle part of the cone is placed in the circle, the diameter would be 10 inches also. I divided it by 2 to get the value of the radius, resulting to 5 inches.

The height of the cone is also the height of the cube. Thus, it is 10 inches.

Volume of the cone = 3.14 * (5in)² * 10in/3
V = 3.14 * 25in² * 3.33 in
V = 261.405 in³ or 261.41 in³ VOLUME OF THE LARGEST CONE THAT CAN FIT IN A 10 INCH CUBE

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Consider the half of the sphere of radius 2 centered at the origin which lies above the xy-plane (a hemisphere). Suppose the den

Harlamova29_29 [7]

Answer:

25.133 units

Step-by-step explanation:

Since the density ρ = r, our mass is

m = ∫∫∫r³sinθdΦdrdθ. We integrate from θ = 0 to π (since it is a hemisphere), Φ = 0 to 2π and r = 0 to 2 and the maximum values of r = 2 in those directions. So

m =∫∫[∫r³sinθdΦ]drdθ

m = ∫[∫2πr³sinθdθ]dr ∫dФ = 2π

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3 years ago

Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract

anyanavicka [17]

Answer:

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8$

Step-by-step explanation:

The sum you are trying to understand is this.

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}$

Remember that in general when you have a geometric series

$\sum\limits_{n = 0}^{\infty} a*r^n$ you have that

$\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}$ and that equality is true as long as $|r| < 1$ .

Therefore here we have

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n$ and $\big|\frac{-1}{15} \big| = \frac{1}{15} < 1$