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If anyone is looking for a tutor in World History or U.S. History ( at any level) I can help you out.


If anyone is looking for a tutor in World History or U.S. History ( at any level) I can help you out.


If anyone is looking for a tutor in World History or U.S. History ( at any level) I can help you out.

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I can do Middle School Algebra tutoring.

 

Currently in 8th, been in Advanced Placement for Mathmatics/Pre-Algebra/Algebra my whole life. (Except 1st Grade, there was no "Math" class.)

 

My timezone is CST, also.

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Hmmm, no Latin? I may give it a try. Just a fair warning, I am not good at it... Yet! So, I can only probably give some general advice or help with sentence structure. But, OH BOY! Does it sound fun! :D 

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@@MetalTao, I've been waiting for somepony knowing Latin for a long time, so you can count me in ;) If you know any links to some good resources for learning this ancient language, then please let me know. I know some basics of Hebrew if you're interested.

 

Anypony speaking Greek, Latin, Hebrew, Arabic or Sanskrit: let me be your student :)

In return, I can teach you one of the subjects I offered somewhere in this thread.

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Okay so basically, I need to learn as much math as I can by December 10th, since I'll be going into college this coming spring term. I understand the 4 most basics aspects (multiplication, subtraction, division, addition), as well as very simple algebraic expressions. I don't understand fractions all too well, and I need to review my decimals, percentages, as well as graphing. I guess you can say that I have an pre-algebra/algebra-1 understanding of math. And I want to strive to reach a basic calculus understanding by the first week of December.

 

Is it possible and can anyone be my little tutor and guide along the way? :3

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Just stating that I can help members who need assistance with College level or lower English. I messaged SCS, but no response as of yet. So, just thought I would put a message here to let those who need it know. ^^

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i can help out possibly with photoshop (or possibly even GIMP for those who cant afford photoshop)

 

on another note...id love to see an art class kinda thing :P

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I'm new here.

 

That's nice but it would be better to do this together in real life

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Omg... best thread ever. I didn't realised this thread, I'll have to check this thread again if I need help. 

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is there anypony familliar with SolidWorks? i use the 2014 version.

i'll explain what i need help with once i find someone interested to help, as it is quite a lof of typing to do so!

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Any one good with Economics? Especially macroeconomics...

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To all who use this thread or have interest in using this thread: while the idea is an amazingly sound one, and I personally loved seeing the assistance offered, we had no way of vetting the quality of assistance you were getting in a Tutoring program and it's pinned status made it easy to assume that this is an official service performed by MLP Forums. This is not the case. MLPF does not validate the quality of assistance that is being offered. There is actually a danger in getting structured tutoring from someone who is not reasonably knowledgeable. Also, everyone learns differently (see Testing 1,2,3). That said, we want to make some changes so that some assistance can be provided.  In order to alleviate these concerns the following changes will be made here. 

 

 

The Topic Title will be changed to focus on homework help

The OP will be adjusted to include a caveat emptor warning

The OP will suggest that you seek IRL assistance as well. 

The OP will remove members subject matching as it comes across as if they are vetted 'experts' in said field. 

 


 

 

From this point forward, if you need assistance in Homework please post specific requests and should someone offer to assist if they can. Again MLP Forums does not guarantee results of the assistance you receive, but we do wish you luck in finding other members who can. 

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is there anypony familliar with SolidWorks? i use the 2014 version.

i'll explain what i need help with once i find someone interested to help, as it is quite a lof of typing to do so!

I am. I've been using it for about 3 years now and got my CSWA last year. I have 2012 at home, but that shouldn't be a problem. Just PM me if you want me to help. I should be able to help.

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I swear I would normally not need to ask, but I'm on wits end here, as I've been cramming in a whole bunch of studying and other homework doings, and this is like... the last problem on my pre-lab that I have to have turned in before lab... and I'm just zonked out.... I've hit the wall.  I probably would know how to do this if I was able to think clearly, but I've been running on 3 hours of sleep per night this week... 

 

Just... please... Can someone tell me how to even this:
 

"Write a balanced half reaction showing how iron (II) oxidizes to iron (III).  Which side of the reaction are electrons needed for charge balance?"

If not, I'll wind up answering it myself in about 4 hours when I get up at 7, and it'll be a bullcrap answer, because I have lab at 8.  Did I say that already, I don't remember...

(Clearly, Miles isn't feeling his best at this point in time... So far gone that he forgot which system his name was...)

~ Kilometers

 

:(

 

---

Edit:

 

I swear I would normally not need to ask, but I'm on wits end here, as I've been cramming in a whole bunch of studying and other homework doings, and this is like... the last problem on my pre-lab that I have to have turned in before lab... and I'm just zonked out.... I've hit the wall.  I probably would know how to do this if I was able to think clearly, but I've been running on 3 hours of sleep per night this week... 

 

Just... please... Can someone tell me how to even this:
 

"Write a balanced half reaction showing how iron (II) oxidizes to iron (III).  Which side of the reaction are electrons needed for charge balance?"

If not, I'll wind up answering it myself in about 4 hours when I get up at 7, and it'll be a bullcrap answer, because I have lab at 8.  Did I say that already, I don't remember...

(Clearly, Miles isn't feeling his best at this point in time... So far gone that he forgot which system his name was...)

~ Kilometers

 

:(

 

Well, I guess my roommate saved me...

4Fe2+(aq) + O2(g) + 4H2O(l) --> 2Fe2O3(s) + 8H+(aq)
The reaction shows oxidation which is the loss of an electron.

 

Yay.

---

The funny thing is,

My roommate was all like:


swag_by_semehammer-d4yspfr.jpg

 

 

"I don't know why it asked this question because it was shown in the lab document above the questions."

And I was all like:


fluttershrug_by_semehammer-d5lg956.jpg

 

 

"I don't know, but there was no way I would have realized that when I was feeling all like this"...

 

 


twilight_crackle_by_semehammer-d4foz7j.j



---

*Sigh*

I lucked out there.

~ Miles

P.S. 

So you have to do the pre-lab first and turn it in before lab --- it can be done individually, or you can do it with your lab group (my group is 4 students including myself)... 2 of them didn't do it, and the 1 that did forgot to print it off... So, I smirked and said "Here, just sign your name on mine, under my signature."  :squee: 

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(edited)

(a) Draw the height of the cone from the apex to the base. The length from the apex to the base of the cup is 20 - 8 = 12 cm. Call this the semi height. By similar triangles, the ratio of the semi height to the base is equal to the ratio of the total height to the r, Or

 

20 / r = 12 / 2.7

Solving for r = 4.5

 

( b ) The volume of the cup is equal to the volume of the full cone minus the volume of the bottom cone. Label b = 2.7 the radius of the base, h = 20 the full height, h_s = 12 the semi height

 

V_cup = V_large - V_small = (pi/3) ( b^2 h_s - r^2 h) = (pi/3) (4.5^2*20 - 2.7^2 * 12)

Edited by BronyNumber42

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(a) Draw the height of the cone from the apex to the base. The length from the apex to the base of the cup is 20 - 8 = 12 cm. Call this the semi height. By similar triangles, the ratio of the semi height to the base is equal to the ratio of the total height to the r, Or

 

20 / r = 12 / 2.7

Solving for r = 4.5

 

( b ) The volume of the cup is equal to the volume of the full cone minus the volume of the bottom cone. Label b = 2.7 the radius of the base, h = 20 the full height, h_s = 12 the semi height

 

V_cup = V_large - V_small = (pi/3) ( b^2 h_s - r^2 h) = (pi/3) (4.5^2*20 - 2.7^2 * 12)

 

Thanks a bunch, really helped.

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Sorry but I don't think I can help with this one, I keep getting 4.05cm

You got further than me, I couldn't remember the method. XD

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I just couldn't get it into 4.5, must have left out some values or something

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Hello homework thread, it's the mathematical idiot again. Just wanted to make sure I was doing this question the right way:

post-29246-0-75607500-1425788987_thumb.jpg

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(edited)

Just wanted to make sure I was doing this question the right way:

Actually I think I can read it. For the last problem, it is asking for a height of 1.8 meters OR LESS, so I think you want 114, not 6

Edited by BronyNumber42

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(edited)

I can't really read it, even after I expand it. Is there a question specifically?

 

Apologies for the bad photo, here's some hopefully better ones: basically I have a list of height ranges and their frequencies, and I have to find the modal class, the mean height, the probability that the height of the first chosen person is more than 1.8 meters and the probability that the height of both the first and second girls are higher than 1.8 meters. I've attempted all the questions but am mostly unsure of question 3aii.

EDIT: I don't know why the photos are rotating themselves: I'm posting directly from my phone.

 

post-29246-0-84455000-1425789518_thumb.jpg

post-29246-0-39877700-1425789612_thumb.jpgpost-29246-0-40824400-1425789689_thumb.jpg

Edited by Sonata Dusk

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(edited)

I don't remember the formulas, so I just use integration for these kinds of problems. :P

 

V = Pi r2 dh from h = 0 to h =20

 

Since r changes linearly with h, you just set up an equation to replace r:

 

r = 4.5 * h / 20

 

so...

 

V  = Pi * h2 * 4.52 / 202 dh

 

V = 4.52 * Pi * h3 / ( 3 * 202 ) from h = 12 to h = 20

 

V = 4.52 * Pi * 203 / ( 3 * 202 ) - 4.52 * Pi * 123 / ( 3 * 202 )

 

V = 1.8746...

 

If you got that, your answer is correct.

Edited by Admiral Regulus

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