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kvasek [131]
3 years ago
12

1 Point

Mathematics
1 answer:
I am Lyosha [343]3 years ago
7 0

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

\text { Volume of a cylinder }=\pi r^{2} h

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e  

\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}

<em><u>Ratio for the volumes of two cylinders</u></em>

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}

Cancelling the common terms, we get

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)

Substituting we get,

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}

Hence, the ratio of volume of two cylinders is 8 : 27

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One more question on this test guys. please answer as fast as possible. If you do,
GaryK [48]

1. The given rectangular equation is x=2.

We substitute x=r\cos \theta.

r\cos \theta=2

Divide through by \cos \theta

r=\frac{2}{\cos \theta}

r=2}\sec \theta

\boxed{x=2\to r=2\sec \theta}

2. The given rectangular equation is:

x^2+y^2=36

This is the same as:

x^2+y^2=6^2

We use the relation r^2=x^2+y^2

This implies that:

r^2=6^2

\therefore r=6

\boxed{x^2+y^2=36\to r=6}

3. The given rectangular equation is:

x^2+y^2=2y

This is the same as:

We use the relation r^2=x^2+y^2 and y=r\sin \theta

This implies that:

r^2=2r\sin \theta

Divide through by r

r=2\sin \theta

\boxed{x^2+y^2=2y\to r=2\sin \theta}

4. We have x=\sqrt{3}y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta\sqrt{3}

This implies that;

\tan \theta=\frac{\sqrt{3}}{3}

\theta=\frac{\pi}{6}

\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}

5. We have x=y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta

This implies that;

\tan \theta=1

\theta=\frac{\pi}{4}

\boxed{x=y\to \theta=\frac{\pi}{4}}

6 0
3 years ago
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miskamm [114]
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8 0
3 years ago
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Alecsey [184]

Answer:

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Step-by-step explanation:

7 0
3 years ago
Help me please i’m really confused on it :) thankyou
Nata [24]

A is 6:55

B is 37 Minutes

C is 15:12

Step-by-step explanation:

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5 0
3 years ago
Solve the system of equations.
guajiro [1.7K]

Answer:

Option 2: (1, 0) and (0, -5)

Step-by-step explanation:

Let's solve this system of equations using the elimination method.

Start by labelling the two equations.

5x -y= 5 -----(1)

5x² -y= 5 -----(2)

(2) -(1):

5x² -y -(5x -y)= 5 -5

Expand:

5x² -y -5x +y= 0

5x² -5x= 0

Factorise:

5x(x -1)= 0

5x= 0 or x -1= 0

x= 0 or x= 1

Now that we have found the x values, we can substitute them into either equations to solve for y.

Substitute into (1):

5(0) -y= 5 or 5(1) -y= 5

0 -y= 5 or -y= 5 -5

y= -5 or -y= 0

y= 0

Thus, the solutions are (0, -5) and (1, 0).

3 0
2 years ago
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