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#51 2017-03-26 00:44:33

hummerz5
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Re: Trouble for homeworks? *Post here and use pictures below*

OiRsw3U.png
I'm stealing the credit
via symbolab

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#52 2017-03-26 02:50:46, last edited by SmittyW (2017-03-26 02:51:54)

SmittyW
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Re: Trouble for homeworks? *Post here and use pictures below*

I could be wrong but I think it's like this:

Hidden text

In this image I already expanded the equation. I don't think there is a simple algebraic way of simplifying it, but you have to think differently with infinite limits. The part I circled will always be significantly bigger than the rest of the polynomial when reaching infinity so I believe the rest is negligible. If you focus on that you should get e/pi which is < 1 and therefore converges.

I'd like to know if you have the answer in a solutions manual or something.

Edit: HUMMERZ

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#53 2017-03-26 07:59:03

Onjit
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Re: Trouble for homeworks? *Post here and use pictures below*

SmittyW wrote:

I'd like to know if you have the answer in a solutions manual or something.

It's actually from an upcoming assignment, so I don't have the answers

But thanks for the explanation, it really helped make it make sense

(more than a screenshot from symbolab //forums.everybodyedits.com/img/smilies/wink )


:.|:;

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#54 2017-03-27 00:02:47

Onjit
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Re: Trouble for homeworks? *Post here and use pictures below*

Hidden text

Here's another one that i'm stuck on. any help would be appreciated


:.|:;

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#55 2017-03-27 01:14:54, last edited by SmittyW (2017-03-27 01:17:06)

SmittyW
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Re: Trouble for homeworks? *Post here and use pictures below*

Well how far did you get? The obvious first step would be to set u = x^n and dv = e^-x

You can prove this by setting n=1, n=2, n=3... until you recognize a pattern of solutions that equal n!. That's the way I was taught. I never liked it because I thought this was a single integration problem. Maybe there was another way I dunno. The answers to the first three would be 1, 2, and 6, which makes sense because 3! = (3 * 2 * 1) = 6

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#56 2017-03-27 02:32:59

dcomet
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Re: Trouble for homeworks? *Post here and use pictures below*

why is it awkwardly cropped?
geez, nobody can understand that if they don't have Snow.
brqI36B.png


1xxoWGb.png
X6pVnTA.png
@MAMETCHl on twitter for the pfp artist, @snuffyowo on twitter for the character drawn in the pfp.

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#57 2017-03-27 05:57:48, last edited by Gosha (2017-03-27 05:58:05)

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Re: Trouble for homeworks? *Post here and use pictures below*

1uO0v11.png
saL0gUm.png

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#58 2017-03-28 00:09:11

Onjit
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Re: Trouble for homeworks? *Post here and use pictures below*

SmittyW wrote:

Well how far did you get? The obvious first step would be to set u = x^n and dv = e^-x

You can prove this by setting n=1, n=2, n=3... until you recognize a pattern of solutions that equal n!. That's the way I was taught. I never liked it because I thought this was a single integration problem. Maybe there was another way I dunno. The answers to the first three would be 1, 2, and 6, which makes sense because 3! = (3 * 2 * 1) = 6

Hidden text

here's what i ended up submitting. the integral cancelled itself out so i made up some bs instead


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#59 2017-03-28 17:26:35

912468
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Re: Trouble for homeworks? *Post here and use pictures below*

Onjit wrote:
SmittyW wrote:

Well how far did you get? The obvious first step would be to set u = x^n and dv = e^-x

You can prove this by setting n=1, n=2, n=3... until you recognize a pattern of solutions that equal n!. That's the way I was taught. I never liked it because I thought this was a single integration problem. Maybe there was another way I dunno. The answers to the first three would be 1, 2, and 6, which makes sense because 3! = (3 * 2 * 1) = 6

here's what i ended up submitting. the integral cancelled itself out so i made up some bs instead

I'm late, but here's what i'd do:

Image

vF0MA5o.png

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#60 2017-03-30 01:04:29

Pingohits
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Re: Trouble for homeworks? *Post here and use pictures below*

help i am dying
5RWWE4I.png
KJWVuY2.png
i got the answer for B, which was 1.1x10^7 m/s, but that's double the amount of what the answer key says (5.33x10^6 m/s). Either I did something wrong, or the answer key is wrong (which tends to be many times). c and d requires b to be solved. I only really need guidance for part b, not sure why I included c and d in the picture as well.

thank


791mAP8.png

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#61 2017-03-30 01:56:11, last edited by SmittyW (2017-03-30 02:13:23)

SmittyW
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Re: Trouble for homeworks? *Post here and use pictures below*

Well hoooooooooooooow did you get your answer? If you post your steps then I'll try helping yoooooooou.

EDIT: It was very short so I did it. The answer key is correct (5.33x10^6 m/s). How did you get double?

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#62 2017-04-12 20:55:48

Krosis
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Re: Trouble for homeworks? *Post here and use pictures below*

It was on my test today and nobody could figure it out how to racionalize this fraction but keep the same value

2-(2*sqrt2)/(2-sqrt2)
or this:
2af36c8324f54bf6b78dd0dd648a4d83.png


Krosis.gif

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#63 2017-04-12 21:08:01, last edited by Pingohits (2017-04-12 21:08:58)

Pingohits
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Re: Trouble for homeworks? *Post here and use pictures below*

Krosis wrote:

It was on my test today and nobody could figure it out how to racionalize this fraction but keep the same value

2-(2*sqrt2)/(2-sqrt2)
or this:
http://image.prntscr.com/image/2af36c83 … 8a4d83.png

have you tried multiplying by it's conjugate?

in this case, it's (2+sqrt(2))/(2+sqrt(2))

after that it's just algebra


791mAP8.png

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#64 2017-04-12 21:13:16, last edited by LukeM (2017-04-12 21:24:14)

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Re: Trouble for homeworks? *Post here and use pictures below*

Krosis wrote:

It was on my test today and nobody could figure it out how to racionalize this fraction but keep the same value

2-(2*sqrt2)/(2-sqrt2)
or this:
http://image.prntscr.com/image/2af36c83 … 8a4d83.png

I dont think it can be...
You can rationalise the denominator, but not the whole thing (that might be what you mean, in which case here you go):

      multiply by conjugate   expand     simplify  simplify
2-2√2     (2-2√2)(2+√2)     4-4√2+2√2-4     -2√2
-----  =  -------------  =  -----------  =  ----  =  -√2
2-√2      (2-√2)(2+√2)          4-2          2

Edit: You also forced me to do my homework //forums.everybodyedits.com/img/smilies/big_smile

I was supposed to be revising this too

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#65 2017-04-12 23:33:05, last edited by hummerz5 (2017-04-19 19:00:24)

hummerz5
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Re: Trouble for homeworks? *Post here and use pictures below*

just to join in on this because I have an idea as to what we're talking about

the conjugate is the bottom with the sign reversed.

This is useful because when you go to multiply the original and its conjugate, you'll get two squared terms (First*First, Last*Last) -- this'll get rid of your square roots for you (at least in the denominator)

edit: props to rat for teaching me something. conjugates are conjugates

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#66 2017-04-13 00:56:12

Ratburntro44
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Re: Trouble for homeworks? *Post here and use pictures below*

just to clarify so nobody reading gets confused in a future math class: the conjugate of "x+y" is "x-y", it doesn't inherently mean bottom, but in this case what you want is to take the conjugate of the bottom

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#67 2017-04-19 18:19:58

Gosha
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Re: Trouble for homeworks? *Post here and use pictures below*

1

i do it like this:

1)find 2 bottom Segments:
    12√3 : 2 = 6√3 and 6√3

2

2) use pythagorean theorem (a^2+b^2=c^2)
   find this triangle

3

x^2 + 6√3^2 = 12√3^2
        _________________      ___________       ________      ___
x = √(12√3^2) - (6√3^2) = √144*3 - 36*3 = √432 - 108 = √324 = 18

so Red Line = 18

But , if you think about it, red line can't be longer than 12√3

Hidden text

where is my mistake?

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#68 2017-04-19 18:27:23

Sensei1
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Re: Trouble for homeworks? *Post here and use pictures below*

12√3 is actually longer than 18. 12√3 = about 20,8

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#69 2017-04-19 18:34:49

hummerz5
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Re: Trouble for homeworks? *Post here and use pictures below*

Sensei1 wrote:

12√3 is actually longer than 18. 12√3 = about 20,8

or 20.8 for those of us using regular numbers

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#70 2017-04-19 18:46:31

Sensei1
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Re: Trouble for homeworks? *Post here and use pictures below*

hummerz5 wrote:
Sensei1 wrote:

12√3 is actually longer than 18. 12√3 = about 20,8

or 20.8 for those of us using regular numbers

We use "," in Finland. Does most of the world use "." instead?

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#71 2017-04-19 18:48:10

hummerz5
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Re: Trouble for homeworks? *Post here and use pictures below*

Sensei1 wrote:
hummerz5 wrote:
Sensei1 wrote:

12√3 is actually longer than 18. 12√3 = about 20,8

or 20.8 for those of us using regular numbers

We use "," in Finland. Does most of the world use "." instead?

All I can say for sure is that I use 246,810.1357 and hold ethnocentrism dear, so I naturally shun other ways of doing things.

But iirc spain does a similar numbering method, so you might be closer to standard. Again, we have three pieces of anecdotal evidence tho

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#72 2017-04-19 18:54:05, last edited by LukeM (2017-04-19 18:55:57)

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Re: Trouble for homeworks? *Post here and use pictures below*

English speaking countries use . so thats what is usually used internationally
Im not sure how common , is though, I know a few European countries use it, so now in the UK we arent supposed to use , as a thousands seperator anymore (its now supposed to be 246 810.1357 to avoid confusion)

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#73 2017-04-19 18:57:38

hummerz5
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Re: Trouble for homeworks? *Post here and use pictures below*

destroyer123 wrote:

English speaking countries use . so thats what is usually used internationally
Im not sure how common , is though, I know a few EU countries use it, so now in the UK we arent supposed to use , as a thousands seperator anymore (its now supposed to be 246 810.1357 to avoid confusion)

Perhaps we should employ 246-810*1357 for visual appeal

semi-homework-related:

Is it at all trivial to determine where two spheres intersect, say they have an offset of (x, y, z) -> (1, 2, 3) ... would that be a circle or an oval?

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#74 2017-04-19 19:01:57

Gosha
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Re: Trouble for homeworks? *Post here and use pictures below*

Sensei1 wrote:

12√3 is actually longer than 18. 12√3 = about 20,8

ops, lol

idk why i thought to myself 12√3 is 12 + √3

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#75 2017-04-19 19:27:43

Ratburntro44
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Re: Trouble for homeworks? *Post here and use pictures below*

hummerz5 wrote:

semi-homework-related:
Is it at all trivial to determine where two spheres intersect, say they have an offset of (x, y, z) -> (1, 2, 3) ... would that be a circle or an oval?

http://mathworld.wolfram.com/Sphere-Sph … ction.html
set center of first sphere to (0,0,0), and use then change your coordinates so second is centered at (d,0,0), then this applies

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