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Assoli18 [71]
3 years ago
14

A sphere of radius 3 inches is sliced with two parallel planes: one passes through the equator and the other is H inches above t

he first plane. The resulting portion of the sphere between the two planes is called a spherical segment; see the picture: In Math 125, you will show that the volume V of the spherical segment is given by this formula (which we will assume): V = (π/3)H(27 – H2). Give EXACT ANSWERS to the questions below. (a) Find the volume of the spherical segment if H=1: (b) Find the rate of change of the volume with respect to H of the spherical segment at H=1: (c) Use the tangent line approximation at H=1 to estimate the value of H that will yield a spherical segment having volume 25 cubic inches:

Mathematics
1 answer:
Leni [432]3 years ago
6 0

Answer:

V = (π/3)H(27 – H2) : Please see attachment

a. Volume =26π/3

b.rate of change of the volume with respect to H = 8π

c.H=0.0114 inch

Step-by-step explanation:

Please see attachment

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Can someone help me with these questions please?
kotykmax [81]

Step-by-step explanation:

You can solve systems of equations using either substitution or elimination.  For these problems, I recommend elimination.  I'll do the first one as an example.

-3x + 16y = 9

-4x + 8y = 12

Multiply the second equation by -2.

8x − 16y = -24

Add to the first equation (notice the y's cancel out).

(-3x + 16y) + (8x − 16y) = 9 − 24

5x = -15

Solve for x.

x = -3

Now you can plug this into either equation to find y.

-3(-3) + 16y = 9

9 + 16y = 9

y = 0

The solution is (-3, 0).

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4 years ago
How to write 79031 expanded form
Jet001 [13]
Seventy-nine-thousand-and-thirty-one


I hope that's right I'm not too sure lol
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Describe how to determine which side of the boundary line should be shaded
iogann1982 [59]
Hello Bri,

If you are talking about a boundary line on a coordinate plane, the best way to determine which side to shade is to start by selecting a coordinate point (x, y) that is not on the line.

Then, substitute those values into your inequality. Simplify the inequality. If the resulting statement is true, then shade the side with your point. If the resulting statement is false, then shade the other side.

Testing a point like this shows the side of the inequality that is the solution and needs to be shaded.

Good luck!
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3 years ago
A circle is centered at the point (-3,2) and passes through the point (1,5). What is the radius of the circle?
Akimi4 [234]

Answer:

5

Step-by-step explanation:

The radius is a distance on a circle from the center to a point on the circle.

We have both of these points describe here in this definition.  Using the distance formula will give us the radius.

\sqrt{(x \text{ difference })^2+(y \text{ difference})^2

The difference between 1 and -3 is 1-(-3)=4.

The difference between 5 and 2 is 5-2=3.

\sqrt{4^2+3^2}

\sqrt{16+9}

\sqrt{25}

5

8 0
3 years ago
Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A
alekssr [168]

Answer:

\theta_{CAB}=128.316

\theta_{ABC}=25.842

\theta_{BCA}=25.842

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

AB= \sqrt{(-3-1)^2+(0-3)^2}=5

BC= \sqrt{(1-1)^2+(3-(-3))^2}=9

CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}

\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}

\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}

\theta_{CAB}=\arccos{-\dfrac{0.62}}

\theta_{CAB}=128.316

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}

\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)

\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)

\theta_{ABC}=\arcsin{0.4359}\right)

\theta_{ABC}=25.842

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

\theta_{BCA}=25.842

this can also be checked using the fact the sum of all angles inside a triangle is 180

\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180

25.842+128.316+25.842

180

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