Sheri's freezer is 2 feet wide 6feet long and 2 feet deep what's the volume of her freeser

Which of the following have at least two congruent parallel bases?

koban [17]

Answer: B and C
Hope this help :)

3 0

3 years ago

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Sin^-1(1/2)<br> Please help I don't know how to do inverse or arcs<br> Thanks

Anit [1.1K]

Answer:

pi/6

Step-by-step explanation:

An inverse sin is asking which point on the unit circle has that value as its sin

Which in this case, is pi/6

$(\frac{\sqrt{3} }{2} ,\frac{1}{2})$

Hope this helps

6 0

2 years ago

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Help please!!! Which graph represents a system of equations with NO SOLUTION. Explain how you know.

ELEN [110]

Answer:graph B

Step-by-step explanation:because if a line is parallel they have no solutions with a certain slope

If two lines intersect then they must have at least a solution

7 0

2 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

1 year ago

HELP PLEASE MARK BRAINLIEST ANSWER PLEASE HELP ME THE GRAPH IS IN THE PICTURE THANK YOU AND PLEASE REALLY HELP ME SOMEONE THAT I

kipiarov [429]

Answer:

The answer would be G.

Step-by-step explanation.

To get your answer, you have to pick two points. Ex: (0,4) (4,1)

Put them into the slop equation: slope= y2-y1/x2-x1

slope:1-4/4-0= -3/4

5 0

2 years ago