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4 years ago

A volcanic cone has a diameter of 300 meters and a height of 150 meters. What is the volume of the cone?

Mathematics

1 answer:

Slav-nsk [51]4 years ago

4 0

I think it's 3532500 because h= 150 / r= 150 which equals the answer.

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Find the slope(6,8) and (9,10)

USPshnik [31]

name the points

a=(x1,y1) b=(x2,y2)

a=(6,8) b=(9,10)

use the slope formula

$m=\frac{y2-y1}{x2-x1}$

replace

$\begin{gathered} m=\frac{10-8}{9-6} \\ m=\frac{2}{3} \end{gathered}$

answer= The slope is equal to 2/3

a=(9,10) b=(6,8)

using the formula

$\begin{gathered} m=\frac{y2-y1}{x2-x1} \\ m=\frac{8-10}{6-9} \\ m=\frac{-2}{-3}=\frac{2}{3} \end{gathered}$

slope will also be 2/3

5 0

1 year ago

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andriy [413]

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6 0

3 years ago

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Yuliya22 [10]

The correct answer is the 2nd one

3 0

3 years ago

Sum of the ages of a father and the son is 40 years. If father's age is 3 times that of his son, then father's age is ________ *

nikklg [1K]

Answer:

Step-by-step explanation:

Sum of ages= 40

Let the sons age be x

Fathers age = 3 times x or ,3x

So 3x+x=40

4x= 40

x= 40/4

= 10

Therefore the fathers age = 3x

= 3*10

= 30

8 0

4 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago