help-19

[graceful sand]

The point $a$ is determined by the vector $\vec{oa} = \vec{e_{x}} - 2 \vec{e_{z}}$ The point $b$ is defined by $\vec{ob} = 2 \vec{e_{y}} + \vec{e_{z}}$ The point c lies on the straight $ab$ between $a$ and $b$ with $||\vec{ca}|| = 2 || \vec{bc} ||$ Determine the angle $\wedge{cob}$

[clever fjord]

rainy

[graceful sand]

How do I start this problem?

[odd edge]

@graceful sand Has your question been resolved?

[lavish flame]

Tc v, jaan jna

[graceful sand]

?

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[west surge]

how can i prove this by induction

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[west surge]

<@&286206848099549185>

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[west surge]

<@&286206848099549185>

[west jewel]

Can i see where u got to

@west surge

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[torpid owl]

it's a tedious one

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[fluid kindle]

https://youtu.be/ND9cEdfCFr0?si=FVjJ6340ujSvH--j&t=673

[fluid kindle]

Are there some u subs for def integrals where you don't have replace the bounds?

I'm so confused

I doubt he would post this if this was a mistake.

[ebon harness]

I dont get what you mean by replacing bounds

[fluid kindle]

of the def integral

[ebon harness]

You mean like typing u in place of infinity, and then taking limit as u which goes to infinity?

[fluid kindle]

oh

I don't think I realized that he just did the integral as a indef integral

Can you just do that lol

?

Step 2, he replaced the upper bound of infin with t and we have the limit as t approaches infinity. Step 3, he did a u sub with u = 3x+1 but didn't replace the bounds. And I think it's because he just did the u sub for a indef integral

[ebon harness]

Well it seems like he didnt want to type everything all over again

[fluid kindle]

But he doesn't replace the bounds at all

[ebon harness]

Its like saying integral of 2x is x² + c and taking the boundaries after

[fluid kindle]

with the u sub

[ebon harness]

He has to, for final result

[fluid kindle]

but he doesn't

it remains t and 1 as the upper and lower bound

[ebon harness]

He typed bounds for x

Notice upper bounf is t and lower bound is 1 for x

He doesnt need to change bounds as hes back to using x as variable

Step 3 is irrevelant from step 2

[ebon harness]

Its just to show what integral is equal to (without bounds) like this

[fluid kindle]

https://tutorial.math.lamar.edu/classes/calci/SubstitutionRuleDefinite.aspx

I'm almost positive he needs to

I'm not appreciating why he doesn't need to still

[ebon harness]

He deletes that later

He made that u-sub to help viewer understand why integral is ewual to that

You dont need u sub to type integral like that anyway

[fluid kindle]

In calc 2, we would do u sub.

We have no other way to do it

huh

this example also didn't

[ebon harness]

Not always

If its in the form $\int u.u' du$ you dont need to

[clever fjord]

Cyrenux

[ebon harness]

https://youtu.be/cJpbJKp91Kc?si=8QixPVyHFnuZ8xTg

[fluid kindle]

ty

I blame my school

[ebon harness]

You are welcome

[fluid kindle]

Still stuck on the original problem though :p

[ebon harness]

You can use u-sub (without bounds), revert back to x, then apply bounds again

[fluid kindle]

okay

that looks like what they are doing

I just wasn't sure why that was legal

[ebon harness]

For example if you want to calculate, $$ \int_{0}^{2} x^2 dx$$
Use the the fact that $\int x^2 dx = \frac{x^3}{3} + c$

\ to show that $\int_{0}^{2} x^2 dx = \left [ \frac{x^3}{3} \right ]_{0}^{2}$

[clever fjord]

Cyrenux
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fluid kindle]

gotcha

yeah that's basically just getting the indef integral and then evaluating what you got with the bounds in the def integral.

okay ty

.close

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[narrow raft]

Can someone help me for this ?

[narrow raft]

i think i should replace x 4 by x ^2) ^2

[odd edge]

@narrow raft Has your question been resolved?

[mystic saffron]

t=x^2

$\int^{\infty}_{0}{\frac{t}{1+t^2}dt}$

[clever fjord]

HadarS

[scarlet junco]

so 2t on the numerator

[narrow raft]

ohh

okayy

[narrow raft]

ye

its ln ( 1 + t ² ) ?

[mystic saffron]

Canceled by dt

[mystic saffron]

No

That would be if 1 on numerator

[narrow raft]

why

[mystic saffron]

Definition

[narrow raft]

ln ( 1 + t ² ) ' -+ >> 2 t / 1 + t² ?

u ' / u ?

[mystic saffron]

Multiply by 1/2 and multiply inside

Cuz you need 2t on the numerator

[narrow raft]

we have 2t

[narrow raft]

2x ² on the numerator

[mystic saffron]

No the 2 was taken for dt

[narrow raft]

t = x² ? we have 2t

ohh

okayy

t = x ^ 2 ; dt = 2x

okayy

[mystic saffron]

Yea 2dx

[narrow raft]

dt = 2 * x dx ?

[narrow raft]

okay i understand

1 / 2 ( ln 1 + t ²) ? = 1 / 2 * 2 t / 1 + t²

[mystic saffron]

Yes

You needed the 2 in order to write it as ln so you took it from the outside

[narrow raft]

yep

[narrow raft]

but for be sure

if t = x ²

dt = 2x * dx ?

[mystic saffron]

Right

[narrow raft]

okay thx

[narrow raft]

but i have a problem

2 x ² dx ; we have dt * sqrt ( t ) no ?

2 * x * dx * x

[mystic saffron]

Just noticed it too

I dont seem to know how to solve this

[narrow raft]

i think we should use the arctan (x ) '

[mystic saffron]

tbh i never learned integrals officialy

[narrow raft]

but idk how

ohh okay

[mystic saffron]

Try putting it into photomath or wolframalph

[narrow raft]

i tryed^^

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[sharp flower]

I got 2^24 bytes for this answer to this question, is that correct?

[wheat bramble]

Yup

,w (2^11)(2^13)=2^x

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by lexitorius

[sharp flower]

alr ty

.close

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[compact flume]

Let $\Omega$ be a simply connected set, $f: \Omega \rightarrow \mathbb{C}$ holomorphic such that $f(z) \neq 0$ for all $z\in \Omega$. Let $z_0 \in \Omega$ and $w_0 \in \mathbb{C}$ such that $e^{w_0}=f(z_0)$. Prove that there is some holomorphic function $g: \Omega \rightarrow \mathbb{C}$ such that $f(z)=e^{g(z)}$ for all $z\in \Omega$ and $g(z_0) = w_0$.$\ $
Hint: let $g$ such that $g'=\frac{f'}{f}$ and show that $h=e^{-g}f$ is constant.

[clever fjord]

Casiel368

[compact flume]

I did prove the hint, but the rest isn't obvious at all to me

[manic glade]

How to solve this??

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[compact flume]

This is what I have

I'd like to show that c is 1, but z_0 is the only point I know about. I don't see any more information to use

<@&286206848099549185>

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@compact flume Has your question been resolved?

[compact flume]

<@&286206848099549185>

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[compact flume]

.close

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[sharp zephyr]

idk where to start

[upbeat drum]

find h'(x) first with chain rule

[sharp zephyr]

how

[viscid flint]

,tex .diff rules

[clever fjord]

hayley.jpg

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[sharp oak]

Let's look at the first one.

x - y = 1

• First, we choose an x. The "for all" means we can choose any x we want. I'll choose x = 100 randomly.
• Then we choose a y. The "there exists" means that a y can be chosen, not necessarily any y though. What y would you choose?

The restriction is that x - y = 1

As above, I've chosen x = 100 for you

If I choose x = 100, then you could choose y = 99! Good call.

Now, can you always supply me a y, no matter what x I pick?

Whoops don't know why I said they're natural numbers. You're right, they're integers

And you're right. For all x, there exists a y, such that x - y = 1. So the first one is true.

Let's look at the second:
x - y = 1

• First, we choose a y. The "there exists" means there should be a y we can pick, but not necessarily any y.
• Then we choose an x. The "for all" means that any x can be chosen.

So if we are going with the same strategy as last time, I'll have you pick a y first.

Okay. Now, I should be able to pick any x I want.

I pick x = 3

But 3 - 100 = 1?
I don't think so

It's obviously impossible to allow any x after your choice of y

Note the order is the major factor here

And the "for all x" on the second one makes it false, yeah

"Pick freely then pick specifically" vs "pick specifically then pick freely"

Np, feel free to ask if you have any others!

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[heavy spindle]

How do I make components such that F is minimum

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[sonic sail]

if f(x) = cos(lnx) and g(x) = ln(cosx) how could i prove that f(x) > g(x)?

[civic hound]

in all reals?

i think that not cuz cos(x) could be equals to 0

and ln is no definied on 0

[sonic sail]

Sry i forgot to mention the interval of x

interval of x is e^(-pi/2) , pi/2

[sonic sail]

i think the both functions are defined in the above interval

[sonic sail]

<@&286206848099549185>

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@sonic sail Has your question been resolved?

[sonic sail]

<@&286206848099549185>

sry for double ping

[amber fable]

What was the interval again?

[sonic sail]

this

[amber fable]

Is it e^ (-pi/2) , e^ (pi/2) ?

[sonic sail]

no

[amber fable]

Ok.

[sonic sail]

e^(-pi/2) , pi/2

[amber fable]

Ok 1 min, I will try.

[sonic sail]

Kk lmk if u get some conclusions

[sonic sail]

@amber fable sry for ping but u got anything?

[amber fable]

Nah, I could not. Sorry, but I do not think i would be able to help with this one.

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[placid atlas]

A,B matrices nxn
A is anti symmetric

prove that if ABA is anti symmetric and invertible then B is anti symmetric and n is even

[shy smelt]

Maybe you could take a vector and compare results when left and right multiplying

[placid atlas]

could you please give an example?

[shy smelt]

v * M = M^T * v right?

[placid atlas]

makes sense

vA = -Av

[low locust]

v^T M = M^T v

otherwise dimensions on the left dont fit

[placid atlas]

ye right

what do you think about this?

[low locust]

what is the 2 for

[placid atlas]

there are 2 As

[low locust]

yes. so |A|^2, not 2|A|

[placid atlas]

hahaha sry ye let me fix that

doesnt seem right

still doesnt seems right

[low locust]

well |B|=|B^T| is known already

you havent used anywhere that ABA is antisymmetric

what do you get from that

[placid atlas]

wait how is it already known?

[low locust]

just a general property of determinants

[placid atlas]

and ye Ill try to use it

[placid atlas]

ah right ye it makes sense when I think about it

B is singular I guess

wait no

[low locust]

where is that minus from

[placid atlas]

ABA^t = -ABA

[low locust]

no

[placid atlas]

[low locust]

you mean (ABA)^T = -ABA

can you simplify the left hand side

[placid atlas]

A^2T B^T?

[low locust]

no

you cant just switch the order of matrices

[placid atlas]

A^T B^T A^T

-A B^T -A

[low locust]

yes

one more step

[placid atlas]

A B^T A?

we didnt use the invertblity yet

[low locust]

ok, so AB^T A = -ABA

or equivalently, A(-B)A = A B^T A

wouldnt it now be nice to conclude B^T=-B

[placid atlas]

we can do that?

like I guess we can

[low locust]

well maybe

not in general

we could for example conclude it if A was invertible

[placid atlas]

ye

[low locust]

because then we could multiply by A^-1 from the left and right

[placid atlas]

but we dont know that rn 😭

we can multiplay by (ABA)^-1

but what will we gain from that?

[low locust]

we know that ABA is invertible

now we can use the det

[placid atlas]

|A(-B)A| = |A B^T A| != 0

we would just get |-B| = |B^T|

wait wait

[low locust]

can |A| be zero

[placid atlas]

yee it cant

so its invertable

and -B = B^T

now we need to prove n is even

[low locust]

yes

do you know that |cM| = c^n |M| if c is a scalar

[placid atlas]

ye we need to do that to -1

somehow and them becuase det != 0

-B = B^T
|-B| = |B^T| = |B|
(-1)^n|B| = |B| != 0

n is even\

is that ok?

[low locust]

(-1)^n |B| = |B|

we can divide by |B|

so (-1)^n = 1

[placid atlas]

ye

thanks for the help @low locust

.close

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@mystic saffron Has your question been resolved?

[noble forge]

!status

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2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
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[cursive jackal]

Any short method?

I solved it by putting cos2 theta formula

Answer is B?

[raw gale]

your right

and its the most optimal way aswell

i got min as 1/2 and max as 3/2

1+ (cos (2 theta) / 2 )

just shifted up 1

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[unkempt vale]

why can you just choose to have k=0?

[unkempt vale]

like I get why you would like to not have an extra number floating around but why are you allowed to do that?

[hybrid carbon]

Because all you care about is the derivative of mu(t)

And any K will give you the same mu'(t)

So you might as well simplify things

Or, more broadly, a first order ode will need at least one initial/boundary condition, right?

So one constant, just like a normal integral

At the end, if you have two constants, you can always combine them

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@unkempt vale Has your question been resolved?

[unkempt vale]

@hybrid carbon ahh I remember that now! Many thanks 🙂

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[buoyant jungle]

anyone help with part b ?

[warm knoll]

$ln(a) + ln(b) = ln(ab)$

[clever fjord]

ЖѲTЇЇC

[buoyant jungle]

oh

so ln(x^2-x)=1

[worthy moat]

Yeah

[buoyant jungle]

(x^2-x)=e

how do i find x

[worthy moat]

Quadratic formula

[buoyant jungle]

oh ok'

aight thanks

.close

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[hexed breach]

When shoeing a horse, four nails should be driven into each side of the shoe. To prevent the shoe from slipping, make sure that the difference between the number of nails on the two sides never exceeds one. How many ways can you choose the order in which the nails are hammered?

Why isn't the answer 8! ?

Instead it's 8 * 4 * 6 * 3 * 4 * 2 * 2 * 1 = 9216

So he got 8 choices first. Then he's forced to spike on the other side thus making it 4? But then he got 6 choices? etc.

[green elm]

assuming they're only distinguishing between left and right, and not certain positions on the left and right...

wouldn't a typical sequence look like

LRRLRLLR

in other words, a sequence of 8 digits, each of which is either L or R

(or we could just call them 0 and 1)

[hexed breach]

8!/4!4! ?

[green elm]

so if what i just wrote is true, you know that at most there would be 2^8 ways

since there are 2^8 numbers that have 8 digits where each digit is 0 or 1

but then you have the further constraints

• the number of L's and R's must be equal

• at no point can there be too many L's or R's in a row

(because of the "no slipping" rule)

[hexed breach]

True. My answer only works if it was a straight line without any condition

[hexed breach]

Hmm.. There's 8 holes. So 8 distinct "boxes" but they're numbered [1,2,3,4] and [5,6,7,8]
And the permutation here is also considered, I think

The first set being on the left and other set being on the right

Because you can't put nails on the same element or hole 4 times.

.close

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[frosty gorge]

Can anyone help me to do this/understand what its asking? I'm so confused

[odd edge]

@frosty gorge Has your question been resolved?

[hollow bramble]

Consider some possible examples?

[frosty gorge]

Just any random matrix right

[hollow bramble]

That helps but I'd start considering in R^2

[frosty gorge]

what's that? like a 2x2 matrix?

[hollow bramble]

To satisfy as many equations as variables you need at least 2 equations on 2 variables

And up to infinitely many equations on 2 variables

[frosty gorge]

This is wrong right

As it only has a trivial solution ?

[hollow bramble]

Yes you want a system with non-trivial

Also, this example shows something important by itself you can use for future examples

(I'm just telling you it's not 'wasted work')

[frosty gorge]

ah

do you have any tips to find non trivial or just try random numbers ?

[hollow bramble]

Well, look at this example. What do you notice

[frosty gorge]

i think that homogeneous SLE 2x2 matrix will always give trivial?

[hollow bramble]

No...

[frosty gorge]

if u put a zero in lower right corner?

then it will have non trivial?

[hollow bramble]

You're studying linear algebra

Do you know rank and linear (in)dependence?

[frosty gorge]

i know rank

[frosty gorge]

we have not reached indepdence yet

i am meant to be up to 2.4

[hollow bramble]

Okay sure

[hollow bramble]

In this example you can see that the final solution, the left side is full rank

[frosty gorge]

ye

[hollow bramble]

When you have full rank, and number of equation is equal to the number of variables, the only possible solution is trivial for the homogeneous system

Because you'll always get
1 0 0 0
0 1 ...
0 0 1 ...
...
equal to all 0 on the right eventually

This should be quite intuitive, you can even prove

Therefore...

To get non-trivial solutions

What must you avoid?

[frosty gorge]

avoid less variables than equations?

[hollow bramble]

I don't think that's what I am trying to say.

[frosty gorge]

avoid full rank?

[hollow bramble]

Yes

[frosty gorge]

so I should try a matrix with 3 variables and 2 equations right

i cant think how else you can avoid full rank

if variables = equations then you would always be able to reduce a row down to a leading 1, no?

[hollow bramble]

Why

I don't think so

Not all 2x2 matrices are full rank.

[frosty gorge]

the only scenario I can think is if you had a 0 where a leading 1 should be

but then you would just have 0=0 so it would have infinite solutions

[hollow bramble]

There is an important scenario you're missing

But 0=0 in a way actually does answer the question

[frosty gorge]

i cant think

so if 0=0 then would that be non trivial solutions?

or is it still a trivial solution

[frosty gorge]

Is this horribly wrong lol

[odd edge]

@frosty gorge Has your question been resolved?

[odd edge]

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[dense apex]

i am struggling to express $\frac{3x}{(x-1)(x-2)(x-3)}$ as partial fraction

[clever fjord]

marty

[dense apex]

i have tried to follow the steps in my book but i cannot do it to get their answer

the box in the bottom right is what i got

the first term was correct but the last two aren't

[mystic saffron]

The last fraction should be $\frac{D}{x-3}$

[clever fjord]

- Miles12345

[oblique mango]

yeah i was thinking the same too tbh

[dense apex]

why not (x-2)(x-3) ?

[tepid yacht]

Because the denominators are the roots

[mystic saffron]

That's the case when factors in denominator are repeated

[oblique mango]

in that case the roots are both same

[tepid yacht]

Repeated root is what that is

[dense apex]

i see the difference but the book didn't describe the case when you have three different factors in the denom.

[tepid yacht]

It's the same process as two factors

You just need another fraction

[dense apex]

and then you multiply every variable A, B, D with all the denom. factors besides its own denom. , right?

[oblique mango]

itd be a useful rule of thumb that
if you have something/abc...
separate it as
A/a + B/b + C/c ....

[dense apex]

so here we have D(x-1)(x-2) ?

[tepid yacht]

Yes

[dense apex]

in the full identity 3x = A(x) + B(y) + D(z)

ok cool

[oblique mango]

the base idea of partial fraction is to make integration easier

[oblique mango]

do you think this will make integration easier?

[dense apex]

no

i just thought it was the result based on the rule i followed

thank you, i managed to solve it now

.close

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[sand lichen]

How

Can i get a pointer in what method to use here

[odd edge]

@sand lichen Has your question been resolved?

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@sand lichen Has your question been resolved?

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@sand lichen Has your question been resolved?

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[rapid schooner]

i have 2 euqations, tied together with one variable (T), i need one to be x and one to be y and it to be ploted on a curve, ik u can like interpalate it and make a new equation using coordinats but that dosnt really work for me

[odd edge]

@rapid schooner Has your question been resolved?

[rapid schooner]

<@&286206848099549185>

[foggy tendon]

so you have x and y in parametric form, yes?

x= something t
y= something t
and youre asked to find the relationship b/w x and y?

[rapid schooner]

its more like f(x) = something T

and

G(x) = something T

[foggy tendon]

i got it

what you do is, eliminate T and relate both the equations

[rapid schooner]

hmmm i can try that

heres the 2 eqautions

[rapid schooner]

actually now thinking abt it, it might be more like this

[foggy tendon]

its the same, regardless

be it x or y, be it f(x) or g(x)

[rapid schooner]

heres the 2 equations

[foggy tendon]

think of it like
variable1= something t
variable2= something t

now that variable could be x or f(x) or p or f(p), thats just a notation

[foggy tendon]

i didnt really get the equations clearly

[rapid schooner]

like the pic is blurry?
its x = vt+-4.9t^2
y = ht

[foggy tendon]

ah yes this is much better

so how would you go about eliminating t in both of the equations?

[rapid schooner]

vt+-4.9t^2-x = ht - y ?

[foggy tendon]

here's your aim
t=something x (from eqn 1)
t= something y(from eqn 2)

so from eqn 1 and 2, something x=something y

[foggy tendon]

nope, you cant straight away relate both the equations

you will need to equate it to the same variable

[rapid schooner]

i mean both equal to 0

if u subract y and x

[foggy tendon]

yes yes that isnt wrong

by doing that youre trying to find the solution only

but you will end up with 3 variables in one equation

[rapid schooner]

true

[foggy tendon]

which cannot be graphed on 2 axes

so what you'll need to do is find a way such that you end up with just x and y

your way is not incorrect, but you wouldnt be able to extrapolate meaningful information about it

[rapid schooner]

what if u solve for t then relate the equations?

[foggy tendon]

exactly

[rapid schooner]

do t = y/h and use quadtratic formula for the x one?

[foggy tendon]

that's exactly what eliminating t means

yup ur on the right path

find how t relates with x, and then with y

then you can equate x and y, and that'd be your answer

[rapid schooner]

so i got both equal to t

[foggy tendon]

yes, now you can equate both the expressions

its like saying if B=A, and C=A, then B=C

[rapid schooner]

ye but one is a quadratic formula one is just devison thats not really fun

[foggy tendon]

well they gotta make the process lengthy yea 💀

[rapid schooner]

ima pull out mathway rq see what it says

[rapid schooner]

its not able to solve 💀

[foggy tendon]

xD

[rapid schooner]

if i give it h and v it works then

[foggy tendon]

ah i see

well you're good then, yea?

[rapid schooner]

not realy cuz i need it to be able to change on the go with one formula

i wonder what if i multiply the x side by h which is what y is being devided by

see if that works

that didnt work

wait nvm it did work

but it dosnt work cuz its not the equation that i need

[foggy tendon]

well i cant really help without seeing your work

you've got the basic idea so you'll be good

[rapid schooner]

ur good, i techinically got the proper answer just not the one i was looking for

ima see how interpelation works and see if it can be used as a formulaa

[foggy tendon]

ah alrighty

[rapid schooner]

.close

[odd edge]

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[potent widget]

I CANT FIGURE OUT THE REST

ok i got some more

[rapid schooner]

whats reflexive property again?

[potent widget]

💀

[rapid schooner]

wrong coppy

[potent widget]

i dont think we doing the same thing sir

[rapid schooner]

prob not i did geometry like 3 years ago

[rapid schooner]

but also bisect means to split in half so techinically my math is correct

[potent widget]

but liek

statements

and reasoning

[rapid schooner]

can u show the statements ur givien?

[potent widget]

using these formats you type your own statements

[rapid schooner]

arl

alr

[rapid schooner]

so <ABE = <CBE cuz <ABC was split in half by line BE

then we calssfy ABE and CBE as triangles

[potent widget]

so what would i type in as statement

[rapid schooner]

ye i dont think i did that correct

[odd edge]

@potent widget Has your question been resolved?

[potent widget]

<@&286206848099549185>

[odd edge]

@potent widget Has your question been resolved?

[potent widget]

<@&286206848099549185>

[odd edge]

@potent widget Has your question been resolved?

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[marsh kayak]

trying to solve this ODE through order reduction. but i feel like i messed up somewhere. because the x or v` could be a constant right? to make this zero

[marsh kayak]

after order reduction, i get xe^(-x)v` = 0

but im lost on how to get v

in easier examples, v would just be constant so thats easy to deal with, but im very lost here

<@&286206848099549185>

[odd edge]

@marsh kayak Has your question been resolved?

[marsh kayak]

nope not yet

[odd edge]

@marsh kayak Has your question been resolved?

[marsh kayak]

nope

[odd edge]

@marsh kayak Has your question been resolved?

[odd edge]

@marsh kayak Has your question been resolved?

[odd edge]

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[nocturne barn]

Hi can someone help me

[nocturne barn]

With geometry

[lean willow]

Just post your question

There's no need to ask for help again when you opened a literal help channel rofl

[nocturne barn]

Lol

Simple geometry stuff but I don't rely understand it

Can u help cuz idk why I got it wrong

[odd edge]

@nocturne barn Has your question been resolved?

[nocturne barn]

<@&286206848099549185>

[nocturne barn]

Help plss

Pray

[odd edge]

@nocturne barn Has your question been resolved?

[nocturne barn]

.close

[odd edge]

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[plucky vessel]

i need help with double integral

[swift rock]

solve the inner integral and then solve the outer
consider it as (integral(integral x))

[plucky vessel]

let me send a pic its not that i dont know to explain it i dont have a very good english

I don't know how to convert the ellipse coordinates to polar

<@&286206848099549185>

[tardy lagoon]

I think you should split the integral into two parts

[plucky vessel]

i know that

i just dont know how to solve the -x part

[tardy lagoon]

-x part?

[plucky vessel]

the ellipse part

the second integral its this

i just dont know the first

[tardy lagoon]

I don't think you need to integrate the sin and cos

since you're doing a double integral

you want to find the area right?

[plucky vessel]

yes

but my teacher said when we have curves we should convert to polar coordinates

its easier

but i dont agree

<@&268886789983436800>

[odd edge]

@plucky vessel Has your question been resolved?

[odd edge]

@plucky vessel Has your question been resolved?

[odd edge]

@plucky vessel Has your question been resolved?

[austere gull]

ik im really late for this, but do you still need help for this?

[plucky vessel]

YES pls

i dont know if u understand my question

but im struggling with just that ellipse the rest i know what to do

[exotic lance]

Hello. So you have to solve this with polar coordinates? In this case, that doesn't make it easier 🤔

[exotic lance]

You can isolate r in this equation. Then you can simplify it a bit with sin^2+cos^2 = 1

[plucky vessel]

hm ok

[odd edge]

@plucky vessel Has your question been resolved?

[odd edge]

@plucky vessel Has your question been resolved?

[odd edge]

@plucky vessel Has your question been resolved?

[upbeat drum]

care to elaborate?

[mystic saffron]

What's this "^" sign

[plucky vessel]

r radius

[mellow prism]

Hello

[exotic lance]

Hi. We can help if you elaborate more on where you're stuck right now

[plucky vessel]

Basically I have to calculate the double integral of the region in the photo I sent but I'm not able to convert the Cartesian coordinates to polar coordinates

[quasi sparrow]

https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Algebra_and_Trigonometry_(OpenStax)/10%3A_Further_Applications_of_Trigonometry/10.03%3A_Polar_Coordinates

10.3.1 and 10.3.2

or scroll down for the other direction

[plucky vessel]

ok

.close

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[feral igloo]

I need with this please,

How do i solve this?

[cold viper]

When dividing exponents, if it’s the same exact base it’s just subtracting

[wooden python]

this is not just x^3/x though.

[wooden python]

do you know the difference of cubes identity?

[cold viper]

Sorry

[feral igloo]

Sorry hang on

ontying to reupload a new photo

K.

[feral igloo]

.

[quasi sparrow]

isn't that the same photo

[feral igloo]

-_- bruh

[quasi sparrow]

this is the only relevant part. cropping is your friend

[feral igloo]

Yeah

Does the image come out super blurry to you?

[quasi sparrow]

does this sound familiar?

she's asking if you learned something that would be helpful for your problem

[feral igloo]

Yeah. It was polynomial long division and synthetic division

Seems wrong thro

:/ Whats the steps for solving it?

Ah nvm.

.close

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[feral igloo]

Whats the steps to solve this?

[nimble blaze]

distribution followed with simplification by combining like terms

[feral igloo]

So when i start workong on it, wouldnt it be -3x^3+x-1

[feral igloo]

Oh

Then combine the like terms

How did the 4x not become a 5x?

[nimble blaze]

what 4x

[feral igloo]

Oh mb

[nimble blaze]

there is no 4x here

[feral igloo]

Yeah

Mistaken 4x^2 for it

Ah k. I got it now.

.close

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[native snow]

How would I continue to solve for x?

[odd edge]

Please don't occupy multiple help channels.

[native snow]

The mark scheme goes to tanx=2 but i dont see how

.close

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[rough charm]

let n be a positive integer and define x = (n + 1)! + 2. Prove that none of x, x + 1, x + 2, ..., x + n - 1 are prime

[rough charm]

im honestly just stuck at like figuring out how to think about this

other questions in the lesson are things like "prove that n^3 - n is divisible by 3 for all n e Z" and I think i got that down pretty well (substitute in n = 3k, 3k + 1, 3k + 2 since those are the only possible remainders) so maybe it's like that?

[mortal trench]

Let me define a new variable y

We will write:

y = 1×2×3×4×...×n×(n+1)

We have to show that y+2 though y+n+1 are composite

Which pretty much completes the proof lol

@rough charm

[rough charm]

what do you mean "through y + n + 1"

that was very helpful by the way, thank you so much

[mortal trench]

x = y + 2, do you agree?

[rough charm]

yes

[mortal trench]

So then
x+n-1 = y+2+n-1

Simplify the right side

[rough charm]

x + n - 1 = y + n + 1

[mortal trench]

Do you see?

[rough charm]

imma be honest no not really

[mortal trench]

Do you agree that the numbers x through x+n-1 are the same as the numbers y+2 through y+n+1

[rough charm]

i'm starting to see the picture

[mortal trench]

Now y is divisible by every number k from 2 to n+1

So for each k from 2 to n+1, we have that y+k is composite

[rough charm]

give me a second to logic this in my brain

alright that makes sense

imma be honest im still stuck

i've been trying to reason and it has not been working out all that well

[odd edge]

@rough charm Has your question been resolved?

[rough charm]

imma just ask my math teacher about it

[odd edge]

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[stoic shadow]

Can anyone help me with a math question

[stoic shadow]

Ik im an idiot but can someone help im stuck

I know a and b

but how do u get c

What method do u do ;-;

[cold sage]

for example, 21021 is a third of the answer to (b) so 21*1001
4008004 is 4 times the answer of (a)
and 8008=8(1001)

split stuff up

[stoic shadow]

Ohh alr

[cold sage]

and see what cancels out

[stoic shadow]

so 4008004=1001x404 right?

[cold sage]

4008004 = 4(1001)^2

[stoic shadow]

Also do u know how im supposed to solve a without a calculator, Thats what my teacher wanted me to do but i just did it the long way because my dumb ass couldnt find a method

[cold sage]

like i said, factor stuff and see what cancels

[mystic saffron]

Ywah like this

[stoic shadow]

oh so i just factorise

[cold sage]

on the numerator you have 4(1001)(21)(1001)=4(21)(1001)^2
in the denominator you have 4(1001)^2

[stoic shadow]

Ohh i see thanks i get it

[cold sage]

that result isnt right

[stoic shadow]

I just need to factorise tysm

[mystic saffron]

Not the resultz the thinking

[stoic shadow]

Yall saviours to me

.close

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[eternal grove]

How do I solve this? I know i have to do something with factorising but how do i do that with that high of a degree. Do i do long division?

[livid storm]

use briot ruffini

wait

it's a ugly root

[eternal grove]

Im not allowed to use L hopital

[livid storm]

briot ruffini is not lhopital

$a^3-b^3 = (a-b)(a^2 +ab+b^2)$

[clever fjord]

tales

[livid storm]

What number is acompanying x^2?

[eternal grove]

8

[livid storm]

I'm thinking how to get x^2-2 factors on the denominator xd

[modern sundial]

If that's of the form 0/0 you already know that x-sqrt(2) is a factor, so just use polynomial long division.

[livid storm]

oh

$x^6 -8 -4\cdot x^2 \cdot (x^2 -2)$

[clever fjord]

tales

[livid storm]

use this on x^6-8

and x-sqrt(2) divides x^2-2

[eternal grove]

So i use long division on the denominator witth (x-sqrt(2)) ?

[rotund flare]

wait is that a command for making it do an equation?

[odd edge]

@eternal grove Has your question been resolved?

[eternal grove]

I’ve done long division with the denominator’s polynomial by x-sqrt(2). I don’t see how this will help me

[lament grove]

No need for that. You can use cube of a binomial and difference of cubes like this:

[odd edge]

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[azure musk]

i would like to know why one of them has an index and the other doesn't

[quasi sparrow]

uh

could be anything. ask whoever you got that from

[azure musk]

could be anything? how

[quasi sparrow]

could be typo, could be a sum

depends on context

[quasi sparrow]

bold usually means vector

[azure musk]

we're always talking about the vector

even the one without the component is a vector

it isn't related to the total time derivative?

[quasi sparrow]

are you saying this is a vector?

[azure musk]

yes

[quasi sparrow]

so is bold A a matrix?

[azure musk]

No

it's a quadri vector

[quasi sparrow]

that's just a vector with 4 components?

How does $A_x, A_y, A_z$ relate to $\bf{A}$ ?

[clever fjord]

riemann

[quasi sparrow]

if they're each vectors

[azure musk]

they are its components

[quasi sparrow]

You said $A_z$ is a vector

[clever fjord]

riemann

[quasi sparrow]

what are the components of $A_z$

[clever fjord]

riemann

[azure musk]

i misunderstood

[quasi sparrow]

$\pa_r A_x$ is a vector valued derivative acting on a single function. do you know its definition?

[clever fjord]

riemann

[azure musk]

Yeah i get it now

thank you sm

.close

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[vapid aurora]

what would be the domain of THIS function?

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@vapid aurora Has your question been resolved?

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@vapid aurora Has your question been resolved?

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[gusty blaze]

hi, what is the first step here

[gusty blaze]

i think i can figure it out once i can setup the integral to calculate the mass

[red thorn]

Have you graphed the two curves (or sketched it) yet?

[gusty blaze]

i have not, does that help?

[red thorn]

Yes, it helps a lot

Especially when the curve has a complicated equation

[gusty blaze]

okie so we have this

ok i think these are the formulas i need

[red thorn]

Yes the last one is a bit advanced

[gusty blaze]

so to solve for M we get \
M = $\int_0^2 [(2x-2x^2) + (2x)] dx$\
i think

[clever fjord]

Price

[red thorn]

I think you forgot to square g(x) and f(x)

Wait

Why's there not a square

[gusty blaze]

o do i have a wrong formula

[red thorn]

So are you calculating the moment about the x- or y-axis?

I.e. Mx or My

[gusty blaze]

hmm maybe Mx first, i think i will need to do both to solve the problem

[red thorn]

Since delta(x) is just a constant here we'll ignore that first

So we have 1/2*[f(x)^2-g(x)^2] left (use the difference of squares formula)

[gusty blaze]

so i get \
$2(\frac{x^5}{5}-\frac{x^4}{2}) + C$\
after integrating this?

[clever fjord]

Price

[red thorn]

You don't need the integrating constant for this, but you can include it if you want

I got $\frac{2x^5}{5}-x^4$ though

[clever fjord]

Math Is Fun

[gusty blaze]

hmm we can try wolfram to check

wait no that is the same thing

[red thorn]

Oh yeah my bad

I thought the 2 factor was a 1/2

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[tawdry edge]

Guys

[tawdry edge]

23 and 24

Can someone tell me if this is correct?

Or wrong

Plss

. close

.close

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[fast ivy]

hi so

I already sketched the graph where x>/= 0

But now

They want the inverse ;/

[upbeat drum]

x>=0 is important

[fast ivy]

oh

ohhhh

so

-x^2=-9?

waiit no

[upbeat drum]

no it just means the question only cares about f(x) for x>=0
you can't make an inverse function otherwise

[fast ivy]

o

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[clever ore]

trigonometry question

[clever ore]

Q9 and Q10

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[dense vapor]

-2^2-(-2)^2*(-2)^3

[dense vapor]

Answer is 28, but how?

4-(4)*(-8)

?

[low locust]

-2^2 is -4

[dense vapor]

oh shit

i forgot that its not

()

ty

but i get 16

so i did something wrong

[upbeat drum]

4*8=32

[dense vapor]

my book say

answer is 28

[upbeat drum]

yes

[dense vapor]

but how

-4+4*-8

-4-4X-8

oh i got it

.close

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[eternal cove]

How do i go about this

[upbeat drum]

about doing what to it

[eternal cove]

To get yu^3

One sec ill

[upbeat drum]

you can bring the -1/2 and -2y together, which makes +2/2y=y

and then the middle part is u^3

same thing for the next line

[uncut remnant]

how on earth was anyone supposed to answer this without that context

[eternal cove]

Sorryyy i get it now

Tysm

I didnt see the whole thing

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[marble vale]

hello

[marble vale]

how do i use gauss elimination to get an upper triangular matrix

[low locust]

do you know what gaussian elimination is

[marble vale]

$\begin{bmatrix}
1 & & & \
& 1 & & \
& & 1 & \
& & & 1 \
\end{bmatrix}$

[clever fjord]

lilisworld

[marble vale]

i think so

but i dont know how to use it

[low locust]

what you wrote is the identity matrix

do you know what row reduction is

[marble vale]

yes and now i need to fill the blanks in identity matrix to do the reduction

[low locust]

not sure what you mean

can you apply row reduction on the original matrix

[marble vale]

what's row reduction?

[low locust]

ok so you dont know what it is

do you know what row operations are

[marble vale]

i just know that i have to multiply it by the matrices with 1 on the diagonal

[marble vale]

yes

[low locust]

what are they

[marble vale]

i can interchange, swap, add etc

the rows

[low locust]

yes

what happens if you for example subtract the first row twice from the second row

[marble vale]

something like 0 -1 -2 0

in the second row

[low locust]

yes

what happens if you add the first row twice to the third row

[marble vale]

the third row becomes 0 0 3 1

[low locust]

ok, so now our matrix looks like
$\begin{pmatrix}
2 & 1 & 0 & 4 \
0 & -1 & -2 & 0 \
0 & 0 & 3 & 1 \
0 & -3 & -12 & -1
\end{pmatrix}$

[clever fjord]

Denascite

[low locust]

whats nice about this, is that this is closer to a triangular matrix

because now in the first column you have mostly zeroes

[marble vale]

yes

and now i multiply this matrix with what?

[low locust]

why multiply with anything

technically the row operations we did are as if we multiply from the left by so-called elementary matrices

is that what you want?

[marble vale]

yes but i dont know how to translate the operations to matrices

like this

$\begin{pmatrix}
2 & 1 & 0 & 4 \
0 & -1 & -2 & 0 \
0 & 0 & 3 & 1 \
0 & -3 & -12 & -1 \
\end{pmatrix}
\begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 3 & 0 & 1 \
\end{bmatrix}
$

[clever fjord]

lilisworld

[low locust]

no

we multiply the elementary matrices from the left

[marble vale]

ah yes srry

[low locust]

and it needs to be a minus 3

[marble vale]

yes but i dont know how to get that

[low locust]

well what exactly do you mean with "get"

you just read it off

[marble vale]

how? I dont understand

how do i know i should multiply by -3?

[low locust]

that makes the -1 and the -3 cancel

you want that -3 + c*(-1)=0

c*(-1)=3

c=-3

[marble vale]

ok so if we do the same thing from the beginning

[marble vale]

oh ok thanks now i get it

[low locust]

to answer this now, row reduction and gaussian elimination are two names for this whole process

[marble vale]

ok i didnt know it was the same thing, so just to make sure i understand, if we go back to when you told me to subtract the first row twice from the second row, what does it look like as a matrix?

[low locust]

what do you think?