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2 years ago

2.

Mathematics

1 answer:

spin [16.1K]2 years ago

3 0

Answer:

D, (4,-4) satisfies both equations.

Step-by-step explanation:

The answer to a system of equations is where the lines intersect, and these lines intersect at (4,-4)

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HELP!!!!:(<br>2.) WHAT'S THE RESULT?

Juli2301 [7.4K]

-1+5x

You combine like terms:

Here there are two different ones.

the number without the variable (the 1s) and the one with the variable (5x)

All you have to do is combine them:

1+(-1)+(-1) = 1-1-1 = -1

5x = 5x

So, the answer would be -1+5x

Hope this helps :D

6 0

3 years ago

The jogger can run at an average speed of 5.5 miles per hour up the slope and 6.5 miles per hour going down the slope. The jogge

Bond [772]

<span>DOC]<span>Venn Diagram Task – Differentiation - Summer Summit

paste that into google and the last page is the answer sheet</span></span>

7 0

3 years ago

I really need some help so pls someone help me

Gemiola [76]

Answer:

its b ez

Step-by-step explanation:

5 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

What is the sum of -1/2 and 8/6? Please help! I need to pass this test! 15 points!

notsponge [240]

Answer: C) 5/6

Step-by-step explanation:

-1/2= -3/6 + 8/6 is 5/6

3 0

3 years ago