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/sci/ - Science & Math

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1285584 No.1285584 [Reply] [Original]

I'm bored.

Post your homework questions and I'll have a go at them.

>> No.1285593

>Homework help
>End of June
>2:30 in the morning

What is wrong with you?

>> No.1285625

>>1285593

Enjoy your shit-tier country

>> No.1285641

>Doing summer courses
>They're all for the teaching certification program for Biology
>Not used to this liberal arts stuff
>Someone either motivate me or kill me

>> No.1285637

>>1285593
>Implying I'm (not as the OP) not taking a course in Operations Research and not doing homework due in 14 hours.

>> No.1285644

If your country is so good why don't you trust your students to do their own homework?

>> No.1285672

isn't h+ just a proton?

>> No.1285679

>>1285672
no it's a proton and a neutron

>> No.1285725

Oh god, that image is a lot funnier than it should be.

>> No.1285732

>>1285679
That's only if it's Deuterium.

>> No.1285737

>>1285732
so h+ is a proton, right?

>> No.1285739

>>1285732
Did you see a neutron being ejected in that picture, because I didn't

>> No.1285766

Let <span class="math">f \in S_2(Γ_0(11))[/spoiler] be the unique normalized eigenform, and define <span class="math">f_1(\tau) = f(2\tau), f_2(\tau) = f(4\tau ), and f_3(\tau) = f(8\tau)[/spoiler]. Show that
a) <span class="math">\{f, f_1, f_2, f_3\}[/spoiler] is linearly independent in <span class="math">S_2(Γ_0(88))[/spoiler].
b) <span class="math">V=span(\{f, f_1, f_2, f_3\})[/spoiler] is stable under the Hecke operators
c) There are exactly three normalized eigenforms in <span class="math">V[/spoiler].

>> No.1285769

>>1285739
ANSWER ME IS H+ A PROTON

>> No.1285784
File: 137 KB, 1440x870, hw.jpg [View same] [iqdb] [saucenao] [google]
1285784

HALP OP!

>> No.1285793

In the context of these particular picture, yes the H+ is almost definitely a sole proton.

But in general if that H is attached to anything and is removed it the resulting H+ would be a neutron and a proton.

>> No.1285802

>>1285769
Yes. If it had a neutron it would be Deuterium.

>> No.1285809

>>1285784
Ask the professor for a hint and then do it yourself.

>> No.1285810

Show the reaction in which serine and tyrosine form a peptide bond.

>> No.1285829

>>1285810
serine+tyrosine=peptide bond.

>> No.1285840

Doesn't H+ hardly ever happen? I always learned that it only happened in special cases and H3O+ was most often what it became.

>> No.1285856

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eddoy jnxxd yslat fytbiflb z oop y yujtw

>> No.1285858

>>1285840
most atoms are full of h+.

>> No.1285877

>>1285840
H+ is a transition form, pretty much anything is willing to take an H+ if they have a primary filled shell element that isn't hydrogen and isn't exceptionally large

>> No.1287202
File: 48 KB, 600x419, 1277646312425.jpg [View same] [iqdb] [saucenao] [google]
1287202

>> No.1287206

never thought of getting trolled that way.

the ass pic is GREAT!!!

*LONG LIVE /sci/*

>> No.1287220

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hdb ssp v onu yqcgptd j vqyb k x lgh gc mpddd fjj

>> No.1287743

Not a troll but...

Would H+ be real?
I mean, it's just a proton, no electrons nor neutrons, how is that even possible?
I call bullshit on that comic.

>> No.1287761

>>1287743
You're an idiot. Water is full of protons all the time. The only way the picture would work would be in solution.

>> No.1287772

derive x^2+2x-3

fuck i hate calculus

>> No.1287773

>>1285810
It's dehydration. Just take that shit out

>> No.1287779

check out my quads

>> No.1287785

>>1287761
>Water is full of protons all the time

No, it is full of <span class="math"> H_3O^+ [/spoiler]

>> No.1287806

>>1287772
(d/dx)(x^2 + 2x - 3) = (d/dx)((x + 3)(x - 1)) = x - 1 + x + 3 = 2x + 2.

>> No.1287814

>>1287806

who the fuck taught you calculus :|

>> No.1287823

>>1287814
Always factorize(?) bro.

>> No.1287827

>>1287814
>implying there's anything wrong with that (even though it's not the "default" method)

>> No.1287828

>>1287785
/facepalm
I didn't say just protons, retard.
Water is never pure.

>> No.1287832

>>1287806

>implying you need to do that much work to get the answer

>> No.1287833

Show that x^2 + y^2 = 3 does not have any rational solutions.

>> No.1287839 [DELETED] 

>>1287833
Just use the quadratic equation.

>> No.1287842

>>1287839


how the fuck u spose to do that when u got to letters?

>> No.1287845

>>1287839
Just do it. Show it to me.

I hope you're not referring to the pq-formula, because we have a polynomial in TWO unknowns here...

>> No.1287853

>>1287828
No sir. You were suggesting that <span class="math">H^+[/spoiler]s were hanging around on their own all the time.

>> No.1287871

>>1287833
Wait, wait, wait. So you are proposing that every time x is rational y is irrational?

>> No.1287879

derive the tan(tan(tan(tan(tan(tan(tan(tan(tan(tan(x))))))))))

>> No.1287884

>>1287871
I'm saying there are no rational numbers p and q such that p^2 + q^2 = 3.

That's all. That would imply that if p were rational and p,q were a solution as above, that indeed q cannot be rational.

>> No.1287887

>>1287833

x=2
y=i

>> No.1287894

>>1287879
cos(x)

>> No.1287899

>>1287887
Since when is the square root of minus 1 a rational number?

>> No.1287906

>>1287894

serious or no?

>> No.1287907

A hard one.
How can you draw a square with the same area as a circle with a ruler and a compass only?

>> No.1287909

>>1287899
imaginary isn't the same as irrational

>> No.1287915

>>1287879

dude that's long.
I'd do it but I'm tired..

sen f(x) / cos f(x) = tan f(x)

there you go.

>> No.1287916 [DELETED] 

>>1287906
Do it yourself, you'll see

>> No.1287928

>>1287916

well reason i posted it is because i cant do it but thanks

>> No.1287929

>>1287899

<div class="math">

i = \frac{i}{1}

</div>

It is rational. :3

>> No.1287942

>>1287915

the fuck am i supposed to do 3with that

>> No.1287939

>>1287909
Ok, you seem to be confused.
Let me clarify:
there's two "kinds" of irrational numbers.
First ones are the algebraic numbers, that's the numbers which are solutions of polynomials over Q. So the square root of 2 is algebraic (x^2 - 2), as is the imaginary unit (x^2 + 1).

Then there's transcendental numbers. Those are irrational but are NOT the solution of a polynomial with coefficients in Q. I think you're confusing irrational numbers with transcendental numbers. This is of course false.
Examples for transcendental numbers are e and pi.

And of course i isn't rational. A rational number is the ratio of two integers. You cannot ever get the square root of minus 1 that way, because the rationals are an ordered field..

>> No.1287954

>>1287942

so you can use the formula of sen and cos,
and don't fuck up with tan.
get it?

>> No.1287970

>>1287954
You mean sin?
>>1287939
i actually didn't really know what I was talking about

>> No.1287978

x is a non zero infinitesimal

explain how <span class="math">x^2[/spoiler] is less than x

>> No.1288030

>>1287978
It's not. x=-.01 => x^2=.001, x^2>x.

>> No.1288035

>>1287884
Yeah, but that is the curve of a circle with radius 3^(1/2) and since there is at least one rational between two irrationals, right?, it would mean that as you follow the curve of the circle x and y would be dancing between the irrationals and rationals, every time x steps into the rationals y steps into the irrationals. Seems crazy to me.

>> No.1288047

>>1287978

x=0.1

x^2=0.01

there you go

>> No.1288066

>>1287978
x^2 < x \iff x < 1, derp?

>> No.1288067

>>1287978
<span class="math">x^2 = 0[/spoiler]

>> No.1288415

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tu gilmk whtplzexhejyegfh d ekr r rtchqhw

>> No.1288411

>>1287833
Let q = a/y, p = x/b with x,y,a,b integers.

Now we have:
<div class="math"> p^2 + q^2 = 3</div>
if and only if
<div class="math"> x^2 y^2 + a^2 b^2 = 3 y^2 b^2</div>
The right side has an uneven number of 3s in its prime factorization, the left side an even number.
You can see this as follows:
Let m and n be the biggest powers of 3 in the prime factorization of x^2 y^2 and a^2 b^2 respectively.
Now if we set k = min(m,n), we can factor out 3^k and have a sum:
<span class="math">3^k (r^2 + s^2) = 3 y^2 b^2[/spoiler] for some r and s.
k is even, since m and n are even.

The sum <span class="math">r^2 + s^2[/spoiler] is not divisible by 3, since at least one summand is not divisible by 3 and both are squares.
(Squares in Z/3Z are: 0^2 = 0, 1^2 = 1 and 2^2 = 1)