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3 years ago

What's two equivalent from of3/5

Mathematics

1 answer:

jolli1 [7]3 years ago

4 0

6/10
300/500
30/50
60/100
12/20

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What’s the distance between (-9,-3) and y=x-6

jeyben [28]

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3 years ago

What are the powers of three that are in the range 3 through 1000

kirill115 [55]

3^1
3^2
3^3
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3 years ago

Balu and Pumba shared 2/3 of a cake. Balu got to eat three times as much cake as Pumba. What fraction of the whole cake did Balu

MatroZZZ [7]

Answer:

In fraction, Balu ate 1/2 of the whole cake

Step-by-step explanation:

Balu and Pumba shared 2/3 of a cake.

Balu eats three times as much cake as Pumba.

So let's take the 2/3 they shared as a whole.

Let's Balu share be x

And pumbs share be y

X = 3y

But x + 3y = 2/3

Since x = 3y

Y = x/3

x + x/3 = 2/3

4x/3 = 2/3

X = (2*3)/(4*3)

X = 2/4

X = 1/2

Balu ate half of the whole cake

In fraction, Balu ate 1/2 of the whole cake

6 0

2 years ago

Julia has 14 pounds of nuts. There are 16 ounces in 1 pound. How many ounces of nuts does she have?

AlladinOne [14]

She has 224 ounces of nuts

5 0

3 years ago

Read 2 more answers

Use spherical coordinates to find the volume of the region that lies outside the cone z = p x 2 + y 2 but inside the sphere x 2

Harman [31]

I assume the cone has equation $z=\sqrt{x^2+y^2}$ (i.e. the upper half of the infinite cone given by $z^2=x^2+y^2$ ). Take

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

The volume of the described region (call it $R$ ) is

$\displaystyle\iiint_R\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\sqrt2}\int_{\pi/4}^\pi\rho^2\sin\varphi\,\mathrm d\varphi\,\mathrm d\rho\,\mathrm d\theta$

The limits on $\theta$ and $\rho$ should be obvious. The lower limit on $\varphi$ is obtained by first determining the intersection of the cone and sphere lies in the cylinder $x^2+y^2=1$ . The distance between the central axis of the cone and this intersection is 1. The sphere has radius $\sqrt2$ . Then $\varphi$ satisfies

$\sin\varphi=\dfrac1{\sqrt2}\implies\varphi=\dfrac\pi4$

(I've added a picture to better demonstrate this)

Computing the integral is trivial. We have

$\displaystyle2\pi\left(\int_0^{\sqrt2}\rho^2\,\mathrm d\rho\right)\left(\int_{\pi/4}^\pi\sin\varphi\,\mathrm d\varphi\right)=\boxed{\frac43(1+\sqrt2)\pi}$

4 0

2 years ago