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3 years ago

Hey! pls help! will mark brainiest!

Mathematics

1 answer:

jeka943 years ago

4 0

Answer:

This relation is not a function since two inputs i.e. -3 and 10 gives the same output i.e. 6.

Step-by-step explanation:

The domain is first number in the ordered pair.

Domain: {-2, -3, 4, 10}

Best Regards!

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Explain how the digit 7 can have different values

kakasveta [241]

It could be 7 groups of 1 3 groups of 2 with one left over or 2 groups of 3 with 1 left over

7 0

3 years ago

Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6

Anton [14]

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder, $x^{2}+y^{2} =1$ and the ellipsoid, $4x^{2}+ 4y^{2} + z^{2} =64$

In polar coordinates, we know that

$x^{2}+y^{2} =r^{2}$

So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

#SPJ4

3 0

2 years ago

Can someone please help me with my homework <br> Ignore the bottom question

pav-90 [236]

1. x = 1 2. x = -30 3. x = -2 4. x = 15 5. x = 7, y = 11 6. x = 4, y = 1

7. x = 1/2, y = 25/5 8. x = 1/4, y = 11/2

hope this helps :3

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3 years ago

How to write 0.12 repeating as a fractiom

frosja888 [35]

Answer:

4/33

Step-by-step explanation:

do the normal steps to conversion

5 0

3 years ago

What is the answer 8s + 8 + 3st + 4s

Illusion [34]

Answer:

12s + 8 + 3st

Step-by-step explanation:

Combine like terms:

8s + 4s = 12s

3sr

6 0

3 years ago