adopim88
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I Need Help With Math

Wed Apr 04, 2007 7:53 am

I hope there are some math geniuses here. I need help...







Please help me and explain how you got your answer. I would appreciate it  Smile
Early bird gets the worm, but the second mouse gets the cheese.
 
graphic
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RE: I Need Help With Math

Wed Apr 04, 2007 8:05 am

41. I think 3/4ths, but I just kinda pulled that out my butt (wanna see it?).
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airfoilsguy
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RE: I Need Help With Math

Wed Apr 04, 2007 8:07 am

If you post this in tech/ops I give it 5 minutes and you will have 3 responders with the correct answer. I used to know but it has been awhile and am too lazy to look it up.
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SpinalTap
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RE: I Need Help With Math

Wed Apr 04, 2007 8:15 am

First question: As the sphere fit perfectly inside the cylinder the length of the cylinder is equal to R
Vcyl = pi*R^2*L = pi*R^2*R = pi*R^3

Vsp = 4/3*pi*R^3
Ratio = Vcyl/Vsphere = pi*R^3/(4/3*pi*R^3) = 1/(4/3) = 3/4


41. The question implies that it is the same for any polygons so I would take the simpliest the square as an example to figure this out. Consider two squares, square 1 and square 2 with lengths (of one side) L1 and L2 and areas of A1 and A2 respectively.
Let L1 = 1
As we know A2 = 2*A1 we can work out L2
A2 = 2*(1*1) = 2
A2 = L2*L2 = L2^2
so L2 = sqrt(A2) = sqrt(2)

Ratio of corresponding sides:
L2/L1 = sqrt(2)

Ratio of perimeters:
(4*L2)/(4*L1) = sqrt(2)

I think this is correct - someone correct me if I am wrong.


42.

Volume of a cone = 1/3(Area of Base)(height) = 1/3*pi*r^2*h
r = 10/2 = 5
V(large cone) = 1/3*pi*5^2*10 = 261.8 cm^3 (4sf)

smaller cone
At the top tip of the cone the diameter = 0
At the bottom of the cone the diameter = 10 cm

The diameter increases linearly from the top tip of the cone to the bottom
D = h (where h is the height from the top) so the height of the smaller cone = D = 4 cm
so we have r = 2, h = 4
V(smaller cone) = 1/3*pi*2^2*4 = 16.76 cm^3 (4sf)

Again someone correct me if I am wrong.

[Edited 2007-04-04 01:19:07]
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GQfluffy
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RE: I Need Help With Math

Wed Apr 04, 2007 8:35 am

Agreeing with Spinaltap on the first one. The radius of the cylinder and the sphere are the same. So just find the volume of both. For the ratio (I'm sittin here with a cold so correct me if I'm wrong  ill  ), just divide the volume of the cylinder by the volume of the sphere. That should give you your ratio...I think...
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kmh1956
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RE: I Need Help With Math

Wed Apr 04, 2007 8:35 am

You won't learn a blessed thing having other people do your work for you. I suggest you go back over your notes and your textbook and try to work it our yourself. If you get it wrong, so be it. Your teacher will at least know you made the effort but may be struggling. this may sound a bit harsh, but it's the truth.
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CastleIsland
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RE: I Need Help With Math

Wed Apr 04, 2007 8:42 am

First of all, this is a test of intuitive thinking, not math, and I hope you will learn how to think like this as we do your homework for you.  

#1: Ratio of volume of cylinder to volume of sphere:

Vol cylinder = (Pi)r^2(h) Note h = 2r (that's the whole purpose of this problem; noting this)
divided by
Vol sphere = 4/3 (Pi)r^3

= 2r divided by (4/3)r.

Answer: 1.5:1 or 3/2.

Quoting SpinalTap (Reply 3):
As the sphere fit perfectly inside the cylinder the length of the cylinder is equal to R

No, it's 2r. And I think you mean height, since length doesn't come into play with a cylinder.

Try the rest on yer own Adopim. Just find the formulae for volume/area calculations, and try to reason it out on your own.

As kmh1956 said: "You won't learn a blessed thing having other people do your work for you. I suggest you go back over your notes and your textbook and try to work it our yourself. If you get it wrong, so be it. Your teacher will at least know you made the effort but may be struggling. this may sound a bit harsh, but it's the truth."

Edited to acknowledge her sage advice.

[Edited 2007-04-04 01:50:19]
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walter747
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RE: I Need Help With Math

Wed Apr 04, 2007 8:50 am

43. Volume = pi * radius 2 * height

3.14 * 22 * 10

Volume = 125.6cm3
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GAIsweetGAI
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RE: I Need Help With Math

Wed Apr 04, 2007 9:24 am

Quoting SpinalTap (Reply 3):
First question: As the sphere fit perfectly inside the cylinder the length of the cylinder is equal to R
Vcyl = pi*R^2*L = pi*R^2*R = pi*R^3

Vsp = 4/3*pi*R^3
Ratio = Vcyl/Vsphere = pi*R^3/(4/3*pi*R^3) = 1/(4/3) = 3/4

We only know that L>2R (slight mistake there I think- I may be wrong- meaning that Vcyl=2*pi*r^3) so we can only say that Ratio>3/2.

Again, I may be wrong in my numbers.
If I'm wrong and SpinalTap is right, this gives Ratio>3/4.


Re 42: what 2 pieces are they talking about? (I know, this may be taking things a bit far, but if it's not specified, you can interpret the text in whatever way you want)




edit: I think my number for the ratio is right: it makes more sense than SpinalTap's, because the ratio calculated thus is bigger than 1 since the cylinder is bigger than the sphere.

[Edited 2007-04-04 02:27:42]

[Edited 2007-04-04 02:27:56]
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Duff44
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RE: I Need Help With Math

Wed Apr 04, 2007 9:45 am

My  twocents  (answers might agree... just seeing if I can still do this shit Big grin)

The first one

Volume of a sphere: (4/3) * PI * r^3
Volume of a cylinder: PI * r^2 * h

Quoting SpinalTap (Reply 3):
First question: As the sphere fit perfectly inside the cylinder the length of the cylinder is equal to R

 no 

In this case, h = 2r, so the cylinder volume is: 2 * PI * r^3

Cylinder/sphere = 2 / (4/3) = 1.5 or 3/2


41

I'll use two right triangles to be different from SpinalTap

A1 = L1 * L1 * (1/2)
A2 = L2 * L2 * (1/2)

A2/A1 = 2 , so (simplified): 2 = (L2^2 / L1^2) ==> L2/L1 = sqrt(2) ~ 1.4142...


Perimeter = L + L + sqrt(2) * L

Since the ratio of the side lengths is the same for all 3 sides, the perimeter also has a ration of sqrt(2) / 1

42

Cone volume = (1/3) * PI * r^2 * h

Upper: r = 2, h = 4
Full: r = 5, h = 10

Vol upper cone = 16.755161
Vol full cone = 261.799388

Vol lower segment = Vol full cone - Vol upper cone = 245.044227


Now go do your own homework....  rotfl 
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SpinalTap
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RE: I Need Help With Math

Wed Apr 04, 2007 9:47 am

Quoting GAIsweetGAI (Reply 8):
Again, I may be wrong in my numbers.
If I'm wrong and SpinalTap is right, this gives Ratio>3/4.

No your right on this one, the cylinder has to be bigger than the sphere!, I should have thought about the physical reality of my result.

I am too use to dealing with diameters. For some reason mathematician prefer radii, engineers prefer diameters.

If I was doing the problem myself I would naturally used diameters rather than radii.
Vcyl = pi()/4*D^2*L= pi()/4*D^3
Vsph = pi()/6*D^3

Vcyl/Vsph = [pi()/4*D^3]/[pi()/6*D^3] = 6/4=3/2
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CastleIsland
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RE: I Need Help With Math

Wed Apr 04, 2007 9:59 am

Quoting CastleIsland (Reply 6):
Answer: 1.5:1 or 3/2.



Quoting GAIsweetGAI (Reply 8):
We only know that L>2R (slight mistake there I think- I may be wrong- meaning that Vcyl=2*pi*r^3) so we can only say that Ratio>3/2.



Quoting Duff44 (Reply 9):
In this case, h = 2r, so the cylinder volume is: 2 * PI * r^3

Cylinder/sphere = 2 / (4/3) = 1.5 or 3/2



Quoting SpinalTap (Reply 10):
No your right on this one, the cylinder has to be bigger than the sphere!, I should have thought about the physical reality of my result.

I am too use to dealing with diameters. For some reason mathematician prefer radii, engineers prefer diameters.

If I was doing the problem myself I would naturally used diameters rather than radii.
Vcyl = pi()/4*D^2*L= pi()/4*D^3
Vsph = pi()/6*D^3

Vcyl/Vsph = [pi()/4*D^3]/[pi()/6*D^3] = 6/4=3/2

Or you could have all just read my post #6.
"People don't do what they believe in, they just do what's most convenient, then they repent." - Dylan
 
vikkyvik
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RE: I Need Help With Math

Wed Apr 04, 2007 11:13 am

Quoting CastleIsland (Reply 11):
Or you could have all just read my post #6.

And miss getting a share of the credit?? Methinks not  Wink

Quoting Airfoilsguy (Reply 2):
If you post this in tech/ops I give it 5 minutes and you will have 3 responders with the correct answer.

Maybe. But a lot of people in Tech/Ops (myself included) would not just type out the answers. Maybe give a formula or two, or a hint, but not the answer.

Of course, it's just an online forum, but I think Kelly said it best:

Quoting Kmh1956 (Reply 5):
You won't learn a blessed thing having other people do your work for you. I suggest you go back over your notes and your textbook and try to work it our yourself. If you get it wrong, so be it. Your teacher will at least know you made the effort but may be struggling. this may sound a bit harsh, but it's the truth.

~Vik
I'm watching Jeopardy. The category is worst Madonna songs. "This one from 1987 is terrible".
 
Duff44
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RE: I Need Help With Math

Wed Apr 04, 2007 11:24 am

Quoting CastleIsland (Reply 11):
Or you could have all just read my post #6.

Never hurts to have a second (or in this case, multiple) set of eyes look at it  eyepopping   eyepopping   eyepopping 
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CastleIsland
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RE: I Need Help With Math

Wed Apr 04, 2007 11:33 am

Quoting Duff44 (Reply 13):
Never hurts to have a second (or in this case, multiple) set of eyes look at it

Not when it comes to my answer, which is truth!  tongue   Wink
"People don't do what they believe in, they just do what's most convenient, then they repent." - Dylan
 
adopim88
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RE: I Need Help With Math

Wed Apr 04, 2007 12:47 pm

Quoting Kmh1956 (Reply 5):
You won't learn a blessed thing having other people do your work for you I suggest you go back over your notes and your textbook and try to work it our yourself. If you get it wrong, so be it. Your teacher will at least know you made the effort but may be struggling

I don't think you understood my request. I wanted someone to help me to understand them (hence why I asked for an explanation). If I had just wanted answers I would have asked for just the answers (or looked them up in the back of the book). I know that giving the answer doesn't help to learn, but giving the answer and a valid explanation can. See, I'm going to school to be a teacher and I understand completely your point, which does make sense. I don't care if I get a few wrong on these assignments, I just can't figure out how to go about them and I want to understand how. I did look through the book and my notes a dozen times at least. And was still stuck. That's when you should ask for help. No one should ever be afraid (or discouraged) to ask for help when they need it.


Quoting Duff44 (Reply 13):
Never hurts to have a second (or in this case, multiple) set of eyes look at it

I thought you hit that right on the nose.


I appreciate your help very much so. Thank you! 

[Edited 2007-04-04 05:50:21]

[Edited 2007-04-04 06:07:35]
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KaiGywer
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RE: I Need Help With Math

Wed Apr 04, 2007 9:56 pm

Quoting Kmh1956 (Reply 5):
You won't learn a blessed thing having other people do your work for you. I suggest you go back over your notes and your textbook and try to work it our yourself. If you get it wrong, so be it. Your teacher will at least know you made the effort but may be struggling. this may sound a bit harsh, but it's the truth.

You have never asked for help in your whole life?

Quoting Adopim88 (Reply 15):
That's when you should ask for help

And when your help is also stuck, ask somebody else...hence A.net...
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vikkyvik
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RE: I Need Help With Math

Wed Apr 04, 2007 10:24 pm

Quoting KaiGywer (Reply 16):
Quoting Kmh1956 (Reply 5):
You won't learn a blessed thing having other people do your work for you. I suggest you go back over your notes and your textbook and try to work it our yourself. If you get it wrong, so be it. Your teacher will at least know you made the effort but may be struggling. this may sound a bit harsh, but it's the truth.

You have never asked for help in your whole life?

You're missing the point.

From the thread starter, it sounded like she wanted us to do the problems for her (which, I may point out, is exactly what ended up happening). Some of us may not think that that is a good way to help.

If we were wrong about that, then so be it, but hey, we're not mind readers  Smile
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KaiGywer
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RE: I Need Help With Math

Wed Apr 04, 2007 10:50 pm

Quoting Vikkyvik (Reply 17):
You're missing the point.

From the thread starter, it sounded like she wanted us to do the problems for her (which, I may point out, is exactly what ended up happening). Some of us may not think that that is a good way to help.

Actually I told her to post. I had helped her with some of the problems, but on these, I was also stuck (haven't had college algebra since spring 2006). Figured someone on here would know how, and then be able to explain it  Smile
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adopim88
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RE: I Need Help With Math

Thu Apr 05, 2007 3:34 am

Quoting Vikkyvik (Reply 17):
From the thread starter, it sounded like she wanted us to do the problems for her (which, I may point out, is exactly what ended up happening). Some of us may not think that that is a good way to help.

If all I cared about was getting the right answer I would have found someone and copied from them (granted if that's all I cared about I wouldn't have taken the extra time because generally speaking I would be lazy). Or for #41 I could have looked up the answer in the back of the book (the odd numbered answers are listed there; the answer listed for #41 was "sqrt2 :1" by the way) I wanted to see how others would go about figuring out the problems, as I had no idea. Please understand, I want to understand the material, which is why I asked for help.

Thank you all for your help! Big grin
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vikkyvik
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RE: I Need Help With Math

Thu Apr 05, 2007 3:42 am

Quoting Adopim88 (Reply 19):
If all I cared about was getting the right answer I would have found someone and copied from them (granted if that's all I cared about I wouldn't have taken the extra time because generally speaking I would be lazy). Or for #41 I could have looked up the answer in the back of the book (the odd numbered answers are listed there; the answer listed for #41 was "sqrt2 :1" by the way) I wanted to see how others would go about figuring out the problems, as I had no idea. Please understand, I want to understand the material, which is why I asked for help.

Oh, I do understand. Wasn't being critical of you - just trying to explain that from the thread-starter, it sounded (to me) like you were just looking for the answers.

No big deal  Smile
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yooyoo
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RE: I Need Help With Math

Thu Apr 05, 2007 3:48 am

Damn, my answer was C
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adopim88
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RE: I Need Help With Math

Thu Apr 05, 2007 4:04 am

Quoting Vikkyvik (Reply 20):
Oh, I do understand. Wasn't being critical of you - just trying to explain that from the thread-starter, it sounded (to me) like you were just looking for the answers.

No big deal

Not a big deal at all. I just wanted to make sure you didn't think I was trying to simply get answers and that you knew I was trying to understand the material in question. Thank you though.  Smile Being a teacher-in-training I do appreciate what you were saying though.  Smile
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YWG
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RE: I Need Help With Math

Thu Apr 05, 2007 4:16 am

If these questions are for a university/college assignment, you may want to think twice about posting next time.

I once posted a question or two up on non-ave. only to received an e-mail from a prof. stating that I was "cheating" and guilty of academic dishonesty. But because I submitted the assignment minus the question and took a 1 mark penalty for a 100 mark assignment and still signed my honesty declaration, he could do nothing so I got away with it. It just so happens when you google a question, anything from a.net forums that matches shows up fon the top of the list.

Just a heads up Big grin

YWG

p.s. I still got an A on the assignment and an A overall in the course
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adopim88
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RE: I Need Help With Math

Thu Apr 05, 2007 4:38 am

Quoting YWG (Reply 23):

I'm not worried. My prof would not consider my questioning to be cheating, he would look at it as asking for help (seeing as he would look it up). For each unit we have a group of assignments. These assignments are worth only 5 points and are handed in the day of the quiz (which was today). No worries. Thank you for your concern though  Smile much appreciated.
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