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DavidQD:

...
By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).
...
~~D~~

Unfortunately, the original question doesn't contain the words "spherical segment", nor any other words that ensure the circle is entirely on the surface of the sphere, so it is possible to interpret it as an incomplete problem with an infinite number of solutions, depending on the assumptions made! However, it seems clear to me that the intention is, indeed, for the assumption to be that the circle lies completely on the surface of the sphere, if only because we get the very nice 5, 12, 13 right-angled triangle that way!

AlanApr 11, 2014

+130561

There are formulas for a lot of different geometric objects under the "Formulary" tab at the top of the page. Look under "Geometry." You will find a "calculator" that will give you the volume when you key in the radius and height.......play around with it for a bit...........

CPhillApr 11, 2014

+130561

15
4 * Σ 3 ^{(n -1)} * (-1) ^(n-1)
n = 4

I can't get this to display correctly...The "15" is supposed to be directly over the summa sign and the "n = 4" is supposed to be directly under it....I think you get the idea......

CPhillApr 11, 2014

Apr 10, 2014

#12

+330

CPhill:

Maybe I'm reading too much into this problem, but I claim that we don't actually know that BC is at a right angle to AB
I guess my point is that we could have a right triangle, but as you see, it might not have to be. Note that ∠BAC is greater than a right angle
In hindsight, the problem only "works" if we assume that AB is perpendicular to BC.......and since we're not given any other information, then DavidQD and Alan would certainly be correct !!!

[center] Definition and Clarification[/center]

By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).

On a sphere, the “Z” axis divides the center of the segment(s) from point (P) to the antipodal point (-P), where the end points for both lines lie on the surface of the sphere.

For clarity, define (P ₀) and (-P ₀) as the implied plane passing through Point (A), the center of the sphere. Define (P ₁) and (-P ₁) as the explicitly described plane passing through point (B), the center of the circle.

To restate: Points (A) and (B) are on the common Z-axis, which is the center of the two parallel planes, segmenting the sphere. Because the planes (lines) are parallel, the AB segment of the Z-axis is perpendicular to both lines. This explicitly defines a 90-degree angle as one of the angles to the triangle.

CPhill’s graphic depicts the intersecting point on the Z-axis, and not the angle at which the plane intersects the Z-axis.

One method to help visualize this: look at the sphere’s presents as “creating” the circle(s) in the plane(s). The concentric circles are always created in planes parallel to the plane that intersects the center of the sphere.
~~D~~

DavidQDApr 10, 2014

+130561

Add up all your test grades so far. Then, multiply the percentage you'd like to maintain times (the number of tests you've taken plus 1).

The difference between the second thing and the first thing tells you what you need to score on the next test

For example, suppose I wanted to maintain an "80" average. So far, on two tests, I've made an 81 and a 76.

Adding the last two things, I get 157. Suppose I have one test left. Then I would multiply 80 times (2 tests taken + 1 left) = 80 (3) = 240.

Then, subtracting the first thing from the second, we get 83. And this is what I need to make on the next test to maintain an "80."

Good luck........

CPhillApr 10, 2014

+130561

12olverabrenda wrote:(-sqrt(65)/9,-4/9) Find cot t

The cotangent is just (-sqrt(65)/9,-4/9)....i.e., (x coordinate value / y coordinate value) = (-sqrt(65)/,-4)

I'm assuming that you want to find the angle with x, y coordinates (-sqrt(65),-4)

OK.....we'll use the cotangent inverse for this

acot((-sqrt(65))/(-4))

Be careful here.....since x and y are negative, then this angle lies in the 3rd quadrant.....we must add this to 180 to get the correct angle...this gives us.....

acot((-sqrt(65))/(-4)) + 180

The answer, of course, is in degrees.....

CPhillApr 10, 2014

Why yes it is! especailly when the soft but hard d**k goes into that soft p***y!

GuestApr 10, 2014

+2354

allan_burnett:
5x + 30y

`
... what do you want to know about this? ...

reinout-gApr 10, 2014

+2354

mariahmorgan143:
-2-5

-2-5 = -7

http://www.mathsisfun.com/positive-negative-integers.html

reinout-gApr 10, 2014

+2354

Guest:
gasoline cost @20 miles per gallon driving from San Francisco, CA to Oklahoma City, OK. estimate fuel cost. 1,630 miles one way

Not sure what gasoline prices in the States are (I'm pretty sure dutch prices are far higher than those in the states), but you can calculate it in the following way;

Since 1630/20 gives the number of gallons you'd use the formula will be

Total fuel cost = (1630/20)*(Price of gasoline per gallon in the States)

edit: I just found that the average price of fuel in the states is 3.596 dollars per gallon.

This would give

Total fuel cost = (1630/20)*3.596 = $293.07

reinout-gApr 10, 2014

+2354

12olverabrenda:
(-sqrt(65)/9,-4/9) Find cot t

I'm sorry but I don't understand your question

Could you rephrase it?

What do you mean by 'find cot t'?

How is the value t related?

Reinout

reinout-gApr 10, 2014

+2354

GSERGY3463TEGRG:
what does pi equal

Look at this page;

http://en.wikipedia.org/wiki/Pi

reinout-gApr 10, 2014

#11

+130561

Not to wear this problem out, but one other thought occurred to me last night........consider the following:

Circle2.JPG

Note that AB = 5 and BC = 12. Thus, Plane "P" (containing B and C, with B being the center of a circle with radius 12 and with C lying on this circle ) would "cut" the sphere in half. What's more, we don't have any triangle at all !!!

Also, note that the radius of this sphere is just 7.

Thus......unless we know the orientation of BC, several interpretations are possible. (As Melody, Alan and DavidQD have pointed out ).

Oh well......I'll shut up now......I know when I've over-stayed my welcome.......

CPhillApr 10, 2014

+2354

An Englishman, an Irishman and a Scotsman walk into a bar.
The bartender turns to them, takes one look, and says, "What is this - some kind of joke?"

reinout-gApr 10, 2014

+2354

untitled:
3 men walk into a bar

... Ouch!

reinout-gApr 10, 2014

#41

+2354

Guest:
I helps to explain it with visuals

1st Pick Reveal Switch?
Nothing Box 1 Yes-Million
Nothing Box 1 No-Nothing
Nothing Box 2 Yes-Million
Nothing Box 2 No-Nothing
Million Box 1-or-2 Yes-Nothing
Million Box 1-or-2 No-Million
As you can see NOT switching ONLY benefits you when you first chose the million (1/3 Chance) and Switching benefits you when ever you choose nothing (2/3). It now comes down to: What is more likely 1/3 or 2/3?

Thank you guest, would you also like to try the new one?
Why don't you sign up so we can identify your posts?

reinout-gApr 10, 2014

#40

+2354

Melody:
I don't know reinout,

If p =0 then I won't confess and we both get 3 years
If p = 1 then I do confess and we both get 10 years.
If p is something else then I think I would have to leave it to the gods because I DO NOT KNOW

Here's a hint: Try to find some formula for the amount of suffering you'll have to deal with.
1 year of jail for your friend amounts to 1 unit of suffering and 1 year of jail for you amounts to 2 units of suffering.

reinout-gApr 10, 2014

#10

+118725

Quote:
Alan: A closer reading of the original shows that the circle is not specified to be wholly on the sphere, in which case Melody and CPhill are correct that the problem is indeterminate without further information.

That is exactly what I was about to say. Thanks Alan.

Quote:
However, I doubt that the problem would have been set without the specification that the circle lies completely on the sphere.

Yes that sounds right too.

MelodyApr 10, 2014

+33666

Melody:
...

. - or am I missing something?

I think so! See my reply immediately above yours.

A closer reading of the original shows that the circle is not specified to be wholly on the sphere, in which case Melody and CPhill are correct that the problem is indeterminate without further information. However, I doubt that the problem would have been set without the specification that the circle lies completely on the sphere.

AlanApr 10, 2014

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