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11 minutes ago, CypherHoof said:

Just checked on a copy of libreoffice - if you move the cursor to the line between (say) A and B on the top bar, it turns into a double headed arrow like <--> ... if at that point you click and drag, it resizes the column to the left (A). same thing between the numbers

 

lr.png.1cb364638287eee5a4e85d9c95d91585.png                   ud.png.b75de6f65b4049d499b89b6f8c9b44cc.png

Oh my gosh thank you! That’s exactly what I was looking for! You’re a lifesaver :ButtercupLaugh:

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5 minutes ago, Lucky Bat said:

Oh my gosh thank you! That’s exactly what I was looking for! You’re a lifesaver :ButtercupLaugh:

NP, happy I could help :)

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ᚾᛖᚹ ᛚᚢᚾᚨ ᚱᛖᛈᚢᛒᛚᛁᚴ - ᚦᛖ ᚠᚢᚾ ᚺᚨᚦ ᛒᛖᛖᚾ ᛞᛟᚢᛒᛚᛖᛞ

image.png.1d67db17f637a25cb8070c016012d5cf.png

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  • 1 month later...

Oh look who's back in here again. :sealed: 

I have a research paper due in a couple days. It's on potential planets (other than Earth of course :derp:) that could support life. Does anyone know of any good online sources where I could get some legitimate research info on this topic? I did find a couple things on the National Geographic website but I need more sources. I'm just curious if any of you know of any. Thanks! 

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12 minutes ago, Lucky Bells said:

Oh look who's back in here again. :sealed: 

I have a research paper due in a couple days. It's on potential planets (other than Earth of course :derp:) that could support life. Does anyone know of any good online sources where I could get some legitimate research info on this topic? I did find a couple things on the National Geographic website but I need more sources. I'm just curious if any of you know of any. Thanks! 

Have you tried the NASA official website? Sometimes they post articles for the general public regarding that kind of stuff.

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  • 2 months later...

I'm pretty good in biology, zoology, and environmental science. I could help tutor if needed! Though the latter two are rather unconventional and I know not every school teaches those subjects I figured I could offer anyways. I'm currently at a senior level of high school so all the science I know doesn't reach into college level yet.

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I could use a hand with my Engineering homework:

5. Compute the magnitude of the following complex numbers:

a. 2 + j5

b. e jπ

c. 3ejπ/2

d. j 5/8

6. Simplify to a + jb form:

a. (2+j)/(1+j2) + (1+j)

b. e jπ/2 + j

c. (4+j3)/(1-j)

d. e jπ + (1+j)/(1-j)

While I understand all this in concept, my stumbling block is simplifying to fractional form in Radians. Need some pointers/cheat sheets about that.


 

“Remember that when you leave this earth, you can take with you nothing you have received--only what you have given.”
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It's been a while since I needed this, but let's try.

Disclaimer: I may have made some mistakes, missed a sign or something. If you give my answer to your teacher without verifying I am not responsible for a bad grade.

The magnitude of a complex number of the form (a+bj) can be calculated with Sqrt(a^2+b^2).

4 hours ago, Mirage said:

a. 2 + j5

b. e jπ

c. 3ejπ/2

d. j 5/8

a. Sqrt(2^2 + 5^2) = Sqrt(29)
b. Now this I can interpret two ways, either it is e*j*pi, in which case the magnitude is e*pi or, e^(j*pi), in which case the magnitude is 1. But I'm guessing that this and the other numbers with e in them are polar notation, just that the superscript formatting is lost here.
c. 3e^(j*pi/2) , the magnitude of this is 3. In polar notation, the magnitude is the number that e is multiplied by (in b it was 1) and the angle is what j is multiplied by.
d. OK, I have two ways to read this one as well, it's either j*5/8, in which case the magnitude is 5/8 or j^(5/8). This you can calculate like this: 
image.png.922eec95c4981a7f7ea984df98815f2e.png, in which case the magnitude is 1.

4 hours ago, Mirage said:

While I understand all this in concept, my stumbling block is simplifying to fractional form in Radians. Need some pointers/cheat sheets about that

The idea is this: You need to find the magnitude and the angle. You can find the magnitude by Sqrt(a^2+b^2) and the angle usually by arctan(b/a). However, this fails is a is zero. This is where you have to know three special cases:
0 + x*j, the angle is pi/2
0 - x^j, the angle is -pi/2 or 3pi/4 (they are equivalent)
0 + 0*j, the angle is undefined
all the other cases should work with arctan(b/a).

4 hours ago, Mirage said:

6. Simplify to a + jb form:

a. (2+j)/(1+j2) + (1+j)

1+j2 is either 1+j^2 or 1+j*2. If it's the first one, then 1+j^2 = 1-1 = 0 and you get division by zero, so let's assume it's the second one.
Let's convert 2+j and 1+j*2 to polar notation:
2+j. Magnitude = Sqrt(2^2+1^2)=Sqrt(5). Angle = arctan(1/2) (it does not evaluate to something pretty)
1+j2. Magnitude = Sqert(1^2 + 2^2)=Sqrt(5). Angle = arctan(2)

image.png.46209e715d90015ce18744ef0436c9d2.png

Now, there is a formula with arctan difference. It is: arctan α - arctan β = arctan [(α-β) / (1+αβ)], let's apply it here.image.png.c794c306585d0a9249632ecec897f3e7.png

Still does not evaluate into something nice. Oh well. The number is image.png.a694a7315b158b3ec04cf4d871db006c.png   

Now we need to convert it to a+bi form.

Polar notation m*e^(j*phi) converts into Cartesian notation (?) like this:
a=m*cos(phi)
b=m*sin(phi)
And we get to use a couple of formulas here:
sin-of-arctan.gif
cos-of-arctan.gif

Let's go:

image.png.fc460c775b9d644cbb4bccbf05993419.png

image.png.a2749370678cb1cfb1f5dc3088783894.png

So, 

image.png.022f0f293651a658ab16f6b88d874af6.png

Now we just need to add the remaining part:

0.8 - 0.6j +  (1+j) = 1.8 + 0.4j

4 hours ago, Mirage said:

b. e jπ/2 + j

This one's easier:

a=cos(pi/2)=0
b=sin(pi/2)=1
e^(j*pi/2)=j
j+j=2j
e jπ/2 + j = 2j

4 hours ago, Mirage said:

(4+j3)/(1-j)

Again, this can be j*3 or j^3. Let's say it's j^3 this time:

image.png.bd08b78ef45987d38822904e562084c6.png

Let's try to solve this a bit different, like a normal fraction. Multiply both parts by (1-j). We'll take advantage of the fact that j^2 = -1 and the fact that 1/j = -j.

image.png.7f636bc7186a2adc33147c0afef38b4a.png

4 hours ago, Mirage said:

e jπ + (1+j)/(1-j)

This seems easy, let's try:

image.png.819b1a160bcd7dcb73fd63f7a3a4f27e.pngimage.png.b2c6f3b9ea7845ca7d0849e31a30f49c.png

image.png.cf8720e269174b9e58c447ba8016f01b.png

image.png.3ad74ec6a52141d5ada1f42b4695ca6e.png (this is known as Euler's identity)

image.png.358c8c18af0c840c492e30f3c6905041.png

 

I think I got all of them correctly, but I may have made a mistake with signs or something, You should verify this.

Disregard everything after this line - I made some mistakes and pasted the fixed equations, but I cannot find them among the attachments (because the thumbnails show only a tiny portion) to delete them.

image.png

image.png

image.png

image.png

Edited by Pentium100
Hid the wrong equations
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@Pentium100 Thank you so much! That was much more help than I expected. I shall review and check my work against yours. I'm truly very grateful of your instruction. A true Twilight in our midst!

886353009_1842408__safe_artist-colon-hosikawa_princessluna_alicorn_animated_blackbackground_crown_eyeshadow_female_gif_hoofshoes_horn_jewelry_makeup_mare_open.gif.2c5c3c5e42b5be6312e90f0cf9c1d244.gif


 

“Remember that when you leave this earth, you can take with you nothing you have received--only what you have given.”
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  • 1 month later...
On 2/16/2020 at 12:12 AM, Pentium100 said:

Disclaimer: I may have made some mistakes, missed a sign or something. If you give my answer to your teacher without verifying I am not responsible for a bad grade.

@Pentium100, you are the best!
I keep learning from your math examples. You know that, right? Just want to make sure, your post there never goes to waste!

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9 hours ago, Super Splashee said:

I keep learning from your math examples. You know that, right? Just want to make sure, your post there never goes to waste!

Well, as long as the post was useful at least once, it's good.

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16 hours ago, Bakugou Is My Man ❤ said:

This isn’t necessarily about homework but I am taking an exam tomorrow (boards). It’s a 150 question exam and I have 3 hours to take it at a testing location. I have done my own studying but I am looking for resources that can give tips & tricks on how to pass/get through a multiple choice test. If anyone can give me something, I’d really appreciate it. 

Here is an example of what I am looking for: https://www.businessinsider.com/4-ways-to-outsmart-any-multiple-choice-test-2015-6

I found this: https://intranet.birmingham.ac.uk/as/libraryservices/library/asc/documents/public/Short-Guide-Multiple-Choice.pdf

TBH it... doesn't seem all that much different from a guide for preparing any old exam? Particularly highlighting key vocabulary and concepts, trying to find or make up questions to practice...

It's also crucial to read the questions and understand them - I saw much of that was lacking during my limited time as a replacement teacher. The only significant difference to me, is the ability to proceed by elimination during the test. (I find MCQ to be the easiest kind for that reason, they accommodate my terrible memory.)

The article says to cover up the answers while you read the question, and try guessing before you see the "spoilers" below. But that might not work out for you, if you're like me...

The best piece of advice though, find out whether negative marking will be used. "If your exam penalises wrong answers, only answer questions if you are reasonably confident that you have figured out the correct answer. If negative marking is not used in your exams, answer all questions, even if you have to guess" (it gives you a chance to get the points).


I take writing commissions.

"Nerds build the world, artists decorate it, warriors protect it, leaders talk everyone into doing their jobs." -me, 3 Nov 2017

"That's not a pie, that's a pastry with an identity crisis!" ~Jeric

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  • 3 weeks later...
19 minutes ago, Emerald Heart said:

Like, how did they eliminate those numbers? 

I'll try to explain.

Lets say I want to multiply 5/2 by 4/7. What I could do normally is (5*4)/(2*7) = 20/14, but I then should simplify the fraction - Find the greatest common divisor or 20 and 14 adn divide both of them by it. In this case GCD is 2, so I can divide both numbers by it and get the final result 10/7.

In your example this was done during the calculation, not after.

Let's repeat (5/2)*(4/7) doing that. (5*4)/(2*7), now, what I could do is divide 4 and 2 by 2, getting 2 and 1 so (5*2)/(1*7) = 10/7

This is what they did:

5/12 * 8/7 = (5*8)/(12*7). Since 5 and 7 are prime and 8 and 12 are not (and not co-prime either since they both can clearly be divided by 2), I can simplify the fraction now instead of multiplying and then simplifying, because simplifying 8/12 is easier than 40/84. So GCD(8,12) is 4, 8/12 = 2/3. 

(5*2)/(3*7) = 10/21

If it helps you, you can thing that by multiplying two fractions, you can interchange their numerators and denominators, for example:

5/12 * 8/7 = 5/7 * 8/12. Since 5/7 cannot be simplified, but 8/12 can, then simplify that and get the result.

Or, it is perfectly fine to first calculate 40/84 and then simplify that, but it usually is more difficult to find the GCD of larger numbers.

I hope I answered the right question.

Edited by Pentium100
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31 minutes ago, Emerald Heart said:

Does anyone understand any of this?

That's a weird way to solve a quadratic equation.

Normally I would do it like this:

image.png.321a3034930c8975a9ae3a3996d20c94.png

So, for this equation:

image.png.69e737ff753f98b5dcb0bf63906def64.png

Anyway the idea is that:

When you multiply two sums, you multiply each part of one sum with each part of the other sum.

image.png.b701efc678b2c841e2e4344c4ff37f59.png, squaring a sum looks like this:

image.png.3e7fafa42aba4cb677280520495bfdca.png

So the example has you trying various combinations to find the solution instead of doing the way I normally do.

Edited by Pentium100
Correct a mistake
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3 minutes ago, Pentium100 said:

That's a weird way to solve a quadratic equation.

Normally I would do it like this:

image.png.b36f34ab590bffce7e4e92587f4d742c.png

So, for this equation:

image.png.69e737ff753f98b5dcb0bf63906def64.png

Anyway the idea is that:

When you multiply two sums, you multiply each part of one sum with each part of the other sum.

image.png.b701efc678b2c841e2e4344c4ff37f59.png, squaring a sum looks like this:

image.png.3e7fafa42aba4cb677280520495bfdca.png

So the example has you trying various combinations to find the solution instead of doing the way I normally do.

Awesome! I'll try that out.

Thank you very much! I was getting irritated...


signature.jpg.913c165e4b3ea068a54230ece43f188d.jpg

 

 

Big thanks to @The Wife of Hawks, @Trix or Treatand @Splasheefor these images!

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12 minutes ago, Emerald Heart said:

Awesome! I'll try that out.

Thank you very much! I was getting irritated...

One small correction, it does not matter for this, but it would matter for the general thing:

image.png.7e9dcda5654e9e558ecace47e01aa5a3.png

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On 4/29/2020 at 9:10 PM, Emerald Heart said:

Does anyone understand any of this?

I've tried and tried but I can't get it right.

image.thumb.png.ba6ea433bcc816fe69884765d801b03e.png

It is a method of factoring quadratics (and other polynomials as well) that doesn't require going through that messy quadratic formula. (If you practice it for a while, you might even be able to factor them in your head! :> )
It is based on the fact that if a polynomial has any rational solution (a whole number, like 7, or a fraction, like 2/3, but no radicals, like √2), then it must be a divisor of its constant term. (I can explain you why is this true in my next post if you wish ;) )
So you start by looking at the constant term (49 in this case), and writing down its divisors.
What numbers divide into 49?
Well, it is a perfect square (49=7·7), and it also certainly can be divided by 1 and itself (as every number), so the divisors are (positive as well as negative):
       1,   7,    49
      -1,  -7,  -49
You can check if any of them is a solution by substituting it for x into your equation and checking whether it will produce 0 for the whole thing. (Notice that we add 49 as the last step, so the part z² – 14·z must somehow produce the opposite of it, -14, if we want them to cancel each other out and produce 0 as the final result.) Let's see:
    (1)² – 14·(1)    certainly won't work, because 1² = 1, and we subtract 14, so we get –13, not –49.  Same with
   (-1)² – 14·(-1)   because (-1)² = 1, but (-14)·(-1) = 14, so we get 14+1 = 15, not –49. We need something bigger as a factor. Maybe 7 will work?
    (7)² – 14·(7)    7² is 49, and (-14)·7 = -98, which is twice as big, so it looks promising:  49 – 98 = -49 indeed, so after adding the last 49 it indeed cancels out to 0.
That way we've just found or first factor (and first solution to the quadratic, for that matter) :) That factor will be (x–7), so you can either divide out by it to get the other factor, or use the table as the one you posted above. The table uses the fact that whatever the two factors of 49 are, they must add up to the middle term's coefficient (the one next to z in the first power; or for general polynomials: the coefficient of the almost-greatest power). The middle term has 14 as its coefficient, and we know one part of it: 7, so the other part must also be 7. That's what is shown in the table above. Therefore:
    z² – 14·z + 49  =  (z–7)·(z–7) = (z–7)²

Now why does that all work? (And when it doesn't?) Where doest that table come from?

Let's start with the factored result, in the most general form, in which instead of the 7's, we put some symbols for the solutions, let's say X and Y:
     (z–X)·(z–Y)
Now multiply it out (by the infamous FOIL dictum):
     z·z + (-X)·z + z·(-Y) + (-X)·(-Y)
     z² – X·z – Y·z + X·Y
Collect the z's together:
     – (X+Y)·z + X·Y
and now compare it with the general form of the quadratic (I used the same colors to help you match them):
     + b·z + a
Do you see it?
The product of our solutions, X·Y, matches the constant term a.  This is our 7·7=49.
The negative sum of our solutions, –(X+Y), matches b, the coefficient of z (the almost-greatest power). This is our -(7+7) = -14.
We can summarize it as two formulas, known as Viète's formulas (because he was a famous mathematician using this trick):
    a =     X·Y
    b = –(X+Y)
The table is a short-hand for that.

In case you asked what if there's some coefficient c with the greatest power of z?
Well, just divide everything by it so that you got just a single z² alone, and remember to scale your solutions accordingly:
    a/c =     X·Y
    b/c = –(X+Y)
but the overall process is exactly the same for fractions as for whole numbers.
Although the formulas are still true for radicals, it's harder for us to "factor" them with this method, so this formula is usually applied in schools only when the solutions are whole numbers or fractions. So let me spare you that rAdiCaL mAtHs for now ;)

You can check that this trick works also for higher-degree polynomials, such as cubics, quartics, quintics etc. Just there's more factors to try out.
For cubics, there's always three solutions, so the constant term will be made of three factors:  X·Y·Z, while the coefficient of the almost-greatest power (z² in this case) will be X+Y+Z.
(Notice that it's not negative this time: the signs alternate; for quartics it will be negative again, –(X+Y+Z+W), for quintics positive, and so on...)

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  • 3 months later...

I could use some help with general HTML/CSS/JS ^^'

There's nothing specifically I need help with right this second, but if anyone is proficient with those, I'd love to be able to talk to them and get their insights/views on certain things :3

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  • 1 month later...
4 hours ago, Muffinnz said:

Im afraid to use this thread because people will see my lack of knowledge on a subject

Better to have us see your lack of knowledge, then your teachers/school/grade, etc. Get help when you can! Also, I would never think lesser of you for asking help with homework or anything else

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On 10/28/2020 at 1:18 AM, Muffinnz said:

Im afraid to use this thread because people will see my lack of knowledge on a subject

If it makes you feel better, I had no idea for the longest time that Spain was a part of Europe :ButtercupLaugh:

The point here is that we’re helping you with stuff so it would be weird if you knew everything, you know? We don’t care if you don’t know stuff, we just wanna help.

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3FBC2CD6-82F4-4BE2-9995-20DAD3ED3514.png.4ce87f72cf9bda0d0d5900fa20c489d3.png
 

Boom!

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