help-33

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[jolly burrow]

.close

[cold pecan]

its closed

[vast magnet]

real

[marsh citrus]

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[wintry gale]

Why is the determinant for concurrency of lines zero?

[marsh citrus]

@wintry gale Has your question been resolved?

[marsh citrus]

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[wooden lava]

can someone please confirm if this is equivelent to x-2y/2x+y+1

[wooden lava]

I solved it using implicit differenciation

I have 2x-4y-4xy'-2yy' -2y' = 0

[quiet anvil]

your solution differs by a sign from the one written

can you show your work?

[wooden lava]

what I wrote was 2x-4y-4xy'-2yy'-2y'=0

Isolating for y' I got

2x-4y=y'(4x+2y+2)

[quiet anvil]

hmm

let me work this problem really quickly

[wooden lava]

ok thanks

[quiet anvil]

oh

sorry

I missed the negative out in front of the solution

yes. your solution is correct and equivalent

[limpid pond]

-(-x+2y)=x-2y it’s right

[quiet anvil]

and I would argue that it's more simplified.

[wooden lava]

why are we allowed to say -(x+2y)=x-2y?

[quiet anvil]

it's - (-x + 2y)

[limpid pond]

-(-x+2y)=-1(-x+2y)

[wooden lava]

don't we have to multiply top and bottom by -1 then?

[quiet anvil]

-1 = -1/1

so you multiply the top by -1 and the bottom by 1

if you were to multiply by -1/-1 it would be a multiplication by 1.

[wooden lava]

oh ok

I didn't realize we were allowed to do that

[quiet anvil]

(also if you want to target the denominator you can, because -1 = 1/-1 as well)

[wooden lava]

and that it wouldn't change the equatin

[quiet anvil]

well

[supple marsh]

yo

[quiet anvil]

the key is there is already a negative out front

[supple marsh]

i have a question regarding the same matter

if this is closed i can post it here

[wooden lava]

.close

[marsh citrus]

Channel closed

Closed by @wooden lava

Use .reopen if this was a mistake.

[quiet anvil]

No, please open a new channel

[wooden lava]

thanks for the help

[supple marsh]

why is this wrong

[limpid pond]

Put it in a new channel

[quiet anvil]

This thread might lock unexpectedly

please open a new one

bot stuff

[marsh citrus]

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[dusk peak]

What is the priblem ?

[little fog]

who

[marsh peak]

Open a help channel only if you have a question

[fleet fossil]

If you want to help others with problems look in the occupied channels, not in the available channels. If you need help yourself, see #❓how-to-get-help

I assume this can be closed…

.close

[marsh citrus]

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[buoyant folio]

My girlfriend asked me to post these questions here. They are written in turkish but I guess it just asks for the result. Can someone help out with these?

[buoyant folio]

If you guys could write the way you solved them would be amazing, but just the answer is fine too

[limpid pond]

2k+1 is always odd when k is a natural number so should be something like -i+i-i+I so on

[buoyant folio]

Thanks alot. I guess she understood how the first one goes.

[limpid pond]

For this one using log rules
Specifically loga+log+b=log(ab)

[buoyant folio]

Is it possible to get a solving of this one? She said she knows the rule but she just can't solve it

[limpid pond]

lemme check rq

[buoyant folio]

thank you so much

[limpid pond]

Does she know changing in log bases ?

[buoyant folio]

hold on a sec i'll let her talk in my place since it'd be easier 🤣

[limpid pond]

Ok lol

[winter anchor]

Converting every base to 20 might aid

[buoyant folio]

Let me try solving it like that

I cannot solve it, can you please send me the explanation

[marsh citrus]

@buoyant folio Has your question been resolved?

[buoyant folio]

anyone still here?

[limpid pond]

I’m trying it

[buoyant folio]

oh, alright then

thank you

[fleet fossil]

Also trying it. Can’t figure it out. And no free website can give me the steps to the solution either

[buoyant folio]

Yeah... turkish education

and for them these are high school level 🤣

[fleet fossil]

Man, what is this exercise lol

[limpid pond]

fr

[fleet fossil]

This feels like math Olympiad type stuff

[limpid pond]

as least Olympiad feels solvable

[fleet fossil]

Fr

Im making no progress

[limpid pond]

the from is specific I feel like

[fleet fossil]

<@&286206848099549185> cracked logarithm exercise. Answer is ||1||, but how does one prove it?

[limpid pond]

only thing come to mind using a calculator

[fleet fossil]

If I had wolfram alpha plus I could see what the reasoning for the solution is there. When I checked other sites not one of them gave me a reasoning for the solution

[buoyant folio]

Yeah I get it. Thank you anyways for the help ❤️

I'll just leave it open for a bit, maybe someone can try to get the hang of it

[limpid pond]

only idea came to my mind is
250=5*50

[fleet fossil]

I noticed that 20^2/10=40 and 50^2/10=250

And that all prime factors are either 2 or 5

But haven’t managed to find out if that leads anywhere

[slate yarrow]

this is brilliant suggestion i think

[limpid pond]

log(40)=log(2)log(20) too feels like it could help

log(40)=log(4)log(10)

[fleet fossil]

I don’t think that’s true…

log(10) is 1

[limpid pond]

forgot the plus sign

[fleet fossil]

Ah

[slate yarrow]

ok so we can write these like this:
$\log_{20}{50}=\frac{\log_{10}{50}}{\log_{10}{20}}\
\log_{40}{250}=\frac{\log_{10}{250}}{\log_{10}{40}}$

[limpid pond]

damn even the bot couldn’t do it

[elfin berry]

convergence

[slate yarrow]

ah finally

rookie mistake

[limpid pond]

I prefer writing as log but still the same won’t complain

[slate yarrow]

same thing but since we are dealing with different bases i thought it would be better if i specified the base

[limpid pond]

no problem

[lean vortex]

Wallah what's your probelm

[buoyant folio]

this one

[limpid pond]

Tried something like this

[fleet fossil]

Together with a+b=1 this may work

[limpid pond]

That’s clever yo lemme try it

[fleet fossil]

Ngl I would love to see emii‘s teacher try to solve this

[limpid pond]

same

[fleet fossil]

Oh my gosh

I finally got it

[limpid pond]

say it I’m waiting

[fleet fossil]

The idea is to define a=log2 and b=log5

[limpid pond]

that’s what I did ……

[buoyant folio]

i tried doing it like this but i couldnt come up with anything afterwards

[fleet fossil]

Then insert that into the expression and replace b with 1-a

And then you get it all into one fraction

[limpid pond]

hold up there is cooking

[fleet fossil]

And that fraction simplifies to 1

[limpid pond]

Lemme try it

[fleet fossil]

In the end you’ll have (2a^2-7a+6)/(2a^2-7a+6)

Then you have to evaluate the zeroes

And make sure that neither of them is log(2)

[full flume]

hi i've been lurking lmao
i got the answer without any substitutions, should i type that up?

[fleet fossil]

I suppose

[fleet fossil]

Sure

[full flume]

gimme a while x.x

[limpid pond]

…..yk that’s just 1 yea

[fleet fossil]

If the denominator is not 0 for a yes

And in that case the whole expression must be 1

[limpid pond]

let me simplify

[buoyant folio]

thank you for the explanation, i am trying it rn.

[limpid pond]

Yep it is 1

Thank god

Thank god for algebra

[fleet fossil]

This was so unnecessarily complex

[limpid pond]

Welp I’m just simplifying it by force

[fleet fossil]

I dont mean what you wrote

I meant the whole exercise

For Highschool homework

That’s crazy

[buoyant folio]

i solved it too, thank you so muchhh

[fleet fossil]

You’re welcome

[limpid pond]

man around 2 hours to solve it

[fleet fossil]

Keep the channel up a bit longer so citrus can send their answer

Curious to see what they came up with

[limpid pond]

When we just could introduce a simple idea

[fleet fossil]

The crucial part was 1. recognizing that 20,40,50 and 250 factor into 2 and 5 and
2. recognizing that log2+log5=1

Once that was established everything fell into place

[limpid pond]

then naming a variable for log2 and log5

[fleet fossil]

But that was damn hard to get to

[fleet fossil]

Technically you don’t have to do that

But it made writing it out much easier

[limpid pond]

Yep I understand that

[buoyant folio]

i tried solving it by converting everything into log5 but it got so messy

[limpid pond]

Especially the right picture it just clicked

[buoyant folio]

in the end i gave up

[limpid pond]

Specifically when log40=2log2+1

Loved it when everything just got into place

ngl I wish they said some log rules for the question like log5+log2=1

[fleet fossil]

A hint would have been in place, absolutely

At least from the teacher

No high school student can be expected to find that in a reasonable about of time

[limpid pond]

Without hints yea

Fr the best feeling when the expressions were just log2 and log5

[elfin berry]

citrusmunch

[full flume]

hokay that should be reasonably doable. still took me a while to not overthink it lol

[limpid pond]

Love that process but like who would’ve think about factoring
log_40(250)

[full flume]

i tried factoring a lot of things early on, but that ended up being the most straightforward. you just don't see the five or so dead ends i found haha

[limpid pond]

ngl wouldn’t have clicked in brain fr

Tried doing kind of the same process

But met with failure

[full flume]

so much failure... i basically reinvented the base change formula by accident on one of my detours...

🤦

[limpid pond]

Ngl love it when reproving things by accident

[fleet fossil]

That 20^5 / 50 = 40^3 call was crazy

[full flume]

yea at least i knew i didnt make a mistake lmao

[fleet fossil]

Well done

[limpid pond]

Fr

[full flume]

yea that bit was magic.

[limpid pond]

Ngl my process was way too algebraic

[fleet fossil]

How did you find it

[full flume]

i filled up two pages of algebraic substitutions before i quit LOL

[limpid pond]

same

[full flume]

i wanted to get that log - 5 part as a single log_20. division was just prime factorization.

[fleet fossil]

How long did you take in total?

[limpid pond]

ngl the only things I found were
like 250=5x50 and 40=2x20
that’s it didn’t think of simplifying even more

[full flume]

prob about forty minutes? kinda insane for a hs question

[fleet fossil]

Okay glad we weren’t the only ones then

[limpid pond]

Lol

[full flume]

shared trauma 🙏

[limpid pond]

At least we both suffered personality love it

[fleet fossil]

Well I would’ve felt bad if we had taken like an hour and someone else had solved it in 5 minutes 😅

[limpid pond]

fr

[fleet fossil]

But it seems this question really is insane for a hs homework

[limpid pond]

But it least we settled it

[fleet fossil]

Genuinely could be math Olympiad

(Let’s just pretend it was)

[limpid pond]

Not to feel bad about it ok 👍

[marsh citrus]

@buoyant folio Has your question been resolved?

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[marsh citrus]

@potent temple Has your question been resolved?

[marsh citrus]

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[covert thorn]

When doing triple integrals

[covert thorn]

if the function this all a product of the variables

can we split it up into separate integrals

or does it only have to be cartesian

[tepid temple]

What do you mean? can you show us an example?

[covert thorn]

give me a sec

[covert thorn]

that's my other account

oh wait I see

[tepid temple]

no that's not possible

[covert thorn]

it depends on the bounds

only if the bounds are constant right

[fervent rampart]

if the function f(x,y,z) can be factored as f(x,y,z) = a(x) b(y) c(z) and the bounds don't depend on other variables then you can

[covert thorn]

can this be applied to spherical and cylindrical

[fervent rampart]

if the bounds are constant using those coordinates and the function is factorable, then yes

[covert thorn]

okay thanks I'm just making sure

.close

[marsh citrus]

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[marsh citrus]

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[alpine widget]

One message removed from a suspended account.

[alpine widget]

One message removed from a suspended account.

[fleet fossil]

Wdym what is it

[rigid forum]

HI

I NEED HELP

[fleet fossil]

#❓how-to-get-help

[rigid forum]

too hard for me

[fleet fossil]

Is W supposed to represent the W-Lambert function?

[marsh citrus]

@alpine widget Has your question been resolved?

[marsh citrus]

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[daring hinge]

can someone help me break this into a partial fraction

[daring hinge]

<@&286206848099549185>

[quiet anvil]

So consider the degrees of the polynomials in the denominator. You have a degree 4 times a degree 1.

Our solution is going to be:

[
\frac{A(x)}{(x^2 +1)^2} + \frac{B(x)}{x+1}
]

We need to ensure that $A(x) \cdot (x+1)$ and $ B(x) \cdot (x^2+1)^2$ are the same degree and as low a degree as possible

[elfin berry]

OmnipotentEntity

[quiet anvil]

This means A(x) is degree 3 and B(x) is degree 0

[daring hinge]

bro I was trying like A/(x+1) + B/(x^2+1) + C/(x^2+1)^2

will this work?

most probably not

if no why?

[quiet anvil]

You can analyze it like this as well

But B and C need to be both degree 1 polynomials

[daring hinge]

can you please do like this and tell me the values

I got a =-1/2

[quiet anvil]

All in all, this means 5 unknowns and 5 variables

[quiet anvil]

!noans

[marsh citrus]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[daring hinge]

ok lol

no like that I got b = 5/2

but b should be a degree 1 polynomial

[quiet anvil]

Yes, which means you should have:

A/(x+1) + (Bx + C)/(x^2+1) + (Dx + E)/(x^2+1)^2

[daring hinge]

ooo

[stoic slate]

/

[daring hinge]

can you please tell me how to know when we should assume a degree 1 polynomial will be present

I am actually new to this stuff

[quiet anvil]

I already did

#help-33 message

[daring hinge]

ok got it

thank you so much

[marsh citrus]

@daring hinge Has your question been resolved?

[marsh citrus]

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[buoyant jetty]

1.- Let ( S ) and ( T ) be subspaces of ( \mathbb{R}^4 ) where
( S = \langle (1, 3, 2, 0), (4, 1, 0, -1) \rangle ),
( T = { x \in \mathbb{R}^4 \mid x_1 + x_3 + x_4 = 0 } ),
and ( H = { x \in \mathbb{R}^4 \mid 3x_1 + x_2 + x_4 = 0 } ).
Define, if possible, a linear transformation
( f : \mathbb{R}^4 \to \mathbb{R}^4 ) such that
( f(S) \subset T ) and
( \text{Ker}(f) + \text{Im}(f) = H ).

[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[buoyant jetty]

what does it mean that the Im(f) is a subset of H

x1 + x3 + x4 = 0

this is the eq plane T

x1 = -x3 - x4

the condition f(S) subset T means that some of the vectors from S we map them to T

problem is if you also consider Im(f) from the third condition to be a subset of H

then you have that f(s) subset HnT

you see what I mean

3x1 + x2 + x4 = 0 is

x4 = -3x1-x2

(x1,x2,x3,-3x1-x2)

x1(1,0,0,-3) + x2(0,1,0,-1) + x3(0,0,1,0)

ker(f) is also a subset of H

[marsh citrus]

@buoyant jetty Has your question been resolved?

[buoyant jetty]

I found the intersection between S and T, aswell as the one for H and T and S and H, dim(SnT) = 1 , dim(SnH)=1, dim(HnT) = 2

so because f(S) subset T is present

we must map some of the vectors from S to T

but because the last condition says that Im(f) subset H

we must map some of the vectors from S instead to T, to TnH

also because the Ker(f) is a subset of H

that must mean we map some of the vectors from H to the zero vector

since f(S) subset HnT we must map some of the vectors from S to the intersection of HnT, like we can extend a basis of S but we have 2 vectors and we need 4, one vector in H must map to the zero vector, because the last condition implies Ker(f) subset of H, so we have three candidate vectors for a basis of R4, the two vectors from S and one vector from H

[buoyant jetty]

H = <(-1,3,0,0),(-1,0,0,3),(0,0,1,0)>

T = <(-1,0,1,0),(-1,0,0,1),(0,1,0,0)>

also from the rank nullity theoerem we know dim(ker(f)) + dim(Im(f)) = dim(domain) = dim(R^4) = 4 = dim(ker(f)) + dim(Im(f))

okay and we know Im(f) is a subset of H and Ker(f) is a subset of H and we know that summing their dimensions that is 4, like if we only have one vector mapped to the zero vector and have a linearly indepedent basis of R4 between the two vectors from S, one vector from H and one arbitrarily chosen vector that must mean the dimension of the kernel is 1, only containing the Ker(f) subset of H vector, only containing the vector from H that we mapped to the zero vector, since our kernel is of dimension 1 the dimension of the Image of f must be 3, having said that knowing that Im(f) is a subset of H and that HnT is two dimensional , we can choose one vector that is linearly independent to the two vectors from S and linearly independent with the vector from the one vector from H that we map to zero vector (present in the kernel of f) and choose it to map to a vector in H, since HnT is two dimensional and we still need one vector from H since Im(f) is a subset of H

so idk if anyone is following but here is what I propose:
f(1,3,2,0) = (-1,3,1,0)
f(4,1,0,-1) = (-1,2,0,1)
f(0,0,1,0) = (0,0,0,0)
f(v) = (0,0,1,0)
here the first two vectors from S are mapped to the two vectors from HnT, because f(S) subset HnT
the third vector that is a vector from H we map it to zero vector because ker(f) subset H
the third linearly independent to the three past vectors is mapped to a vector of H , because we chose kernel to be one dimensional and Im(f) subset of H

HnT = <(-,1,3,1,0),(-1,2,0,1)>
SnH = <(2,-5,-4,-1)>
SnT = <(3,-2,-2,-1)>

<@&286206848099549185>

[marsh citrus]

@buoyant jetty Has your question been resolved?

[buoyant jetty]

<@&286206848099549185>

[plucky iron]

What u need?

What's the problem

[buoyant jetty]

1.- Let ( S ) and ( T ) be subspaces of ( \mathbb{R}^4 ) where
( S = \langle (1, 3, 2, 0), (4, 1, 0, -1) \rangle ),
( T = { x \in \mathbb{R}^4 \mid x_1 + x_3 + x_4 = 0 } ),
and ( H = { x \in \mathbb{R}^4 \mid 3x_1 + x_2 + x_4 = 0 } ).
Define, if possible, a linear transformation
( f : \mathbb{R}^4 \to \mathbb{R}^4 ) such that
( f(S) \subset T ) and
( \text{Ker}(f) + \text{Im}(f) = H ).

[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[plucky iron]

Hmm

[buoyant jetty]

what does it mean that the Im(f) is a subset of H

the condition f(S) subset T means that some of the vectors from S we map them to T
problem is if you also consider Im(f) from the third condition to be a subset of H
then you have that f(s) subset HnT
you see what I mean

also because the Ker(f) is a subset of H
that must mean we map some of the vectors from H to the zero vector

HnT = <(-,1,3,1,0),(-1,2,0,1)>
SnH = <(2,-5,-4,-1)>
SnT = <(3,-2,-2,-1)>

since f(S) subset HnT we must map some of the vectors from S to the intersection of HnT, like we can extend a basis of S but we have 2 vectors and we need 4, one vector in H must map to the zero vector, because the last condition implies Ker(f) subset of H, so we have three candidate vectors for a basis of R4, the two vectors from S and one vector from H

H = <(-1,3,0,0),(-1,0,0,3),(0,0,1,0)>

T = <(-1,0,1,0),(-1,0,0,1),(0,1,0,0)>

also from the rank nullity theoerem we know dim(ker(f)) + dim(Im(f)) = dim(domain) = dim(R^4) = 4 = dim(ker(f)) + dim(Im(f))

okay and we know Im(f) is a subset of H and Ker(f) is a subset of H and we know that summing their dimensions that is 4, like if we only have one vector mapped to the zero vector and have a linearly indepedent basis of R4 between the two vectors from S, one vector from H and one arbitrarily chosen vector that must mean the dimension of the kernel is 1, only containing the Ker(f) subset of H vector, only containing the vector from H that we mapped to the zero vector, since our kernel is of dimension 1 the dimension of the Image of f must be 3, having said that knowing that Im(f) is a subset of H and that HnT is two dimensional , we can choose one vector that is linearly independent to the two vectors from S and linearly independent with the vector from the one vector from H that we map to zero vector (present in the kernel of f) and choose it to map to a vector in H, since HnT is two dimensional and we still need one vector from H since Im(f) is a subset of H

so idk if anyone is following but here is what I propose:
f(1,3,2,0) = (-1,3,1,0)
f(4,1,0,-1) = (-1,2,0,1)
f(0,0,1,0) = (0,0,0,0)
f(v) = (0,0,1,0)
here the first two vectors from S are mapped to the two vectors from HnT, because f(S) subset HnT
the third vector that is a vector from H we map it to zero vector because ker(f) subset H
the third linearly independent to the three past vectors is mapped to a vector of H , because we chose kernel to be one dimensional and Im(f) subset of H

<@&286206848099549185>

[buoyant jetty]

<@&286206848099549185>

[frozen prairie]

Hi I’m 5 years old

[plucky iron]

Uh..

You're joking right?

[frozen prairie]

Yeah

[buoyant jetty]

Why dont u go and enjoy your youngness

like when I was in highschool

I used to pick my nose all day

and play videogames

[frozen prairie]

How did u know I’m picking my nose

Wait guys

Can you help me with my brothers math question so I can impress him. Just one

[buoyant jetty]

no

[frozen prairie]

:(((

[buoyant jetty]

I cant even help myself with my question

[frozen prairie]

I will tell you anyway.

[buoyant jetty]

like I give a shit

[frozen prairie]

If the hypotanoose is 5 and the ajaycent is 3. What is the opposite

[buoyant jetty]

oppo is your mom

[frozen prairie]

Nvm it’s 4

[buoyant jetty]

good

can u leave now

[frozen prairie]

Ok bye good luck with your stuff

[buoyant jetty]

good luck with urs

aswell

[frozen prairie]

Thanks a lot

[buoyant jetty]

and with ur brother stuff

[frozen prairie]

Yay

He’s graduating uni tmmr

[buoyant jetty]

ask him if he can help me with my algebra

[frozen prairie]

Oh I would but he’s out to buy groceries

That’s why I want to impress him

By doing his homework

He will think I’m a prodigy

[buoyant jetty]

will he?

[frozen prairie]

Uh

I’m 12 and I have learnt the quadratic equation so he will only be impressed if I solve linear equations

[buoyant jetty]

oh

[frozen prairie]

I’m too dumb to learn calculus

[buoyant jetty]

I thought u were 5

[frozen prairie]

No

[buoyant jetty]

how can I know u are not lying again

[frozen prairie]

I can send pic

But I’m insecure

[buoyant jetty]

I do not want

for u to send anything

just leave ok

[frozen prairie]

Why is your about me…

Nvm

Okay bye

[buoyant jetty]

<@&286206848099549185>

[frozen prairie]

Btw I showed my brother the question

Nvm he said he would be wasting his time

[buoyant jetty]

his time is not worth anything

then

[buoyant jetty]

fuck my life and your brother bro

.solved

[marsh citrus]

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[marsh citrus]

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[rich sedge]

can someone help check my work. for f''(x) i got 7sec(x-pi/2)tan^2(x-pi/2) + sec^3(x-pi/2) and for my critical numbers i got pi/2, 3pi/2, 5pi/2, 7pi/2 but idk why my answer is wrong

[wary kite]

$f’(x) = 7\sec(x - \frac{\pi}{2})\tan(x-\frac{\pi}{2})$

[elfin berry]

knief

[wary kite]

then

$f’’(x) = 7\sec^3(x-\frac{\pi}{2}) + 7\sec(x-\frac{\pi}{2})\tan^2(x-\frac{\pi}{2})$

[elfin berry]

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

[wary kite]

so you’re missing a 7

on the sec^3

[rich sedge]

oh

but idk why my concavities are wrong

[wary kite]

$f’’(x) = 7\sec(x-\frac{\pi}{2})(\sec^2(x-\frac{\pi}{2}) + \tan^2(x-\frac{\pi}{2}))$

[elfin berry]

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

[wary kite]

or wait

nevermind

sec(x-pi/2) is never zero

and we can rewrite the inside

using sec^2 = 1 + tan^2

$1 + 2\tan^2(x-\frac{\pi}{2}) = 0$

[elfin berry]

knief

[wary kite]

which is never zero?

how’d you get those critical point

[rich sedge]

tan is sin/cos and sec is 1/cos

cosx = npi + pi/2

[wary kite]

yea ok it’s never zero

[rich sedge]

oh

[wary kite]

we can still find where it’s > 0 and < 0

[rich sedge]

huhh

[wary kite]

or did i graph it wrong

i put it in desmos

[rich sedge]

but then how would we get the concavities/critical points

[wary kite]

hold on we can just turn this into the co functions

the x - pi/2 is annoying

sec(x-pi/2) is csc

and we can use -cot for tan(x-pi/2)

[wary kite]

so this becomes

$f’’(x) = 7\csc(x)(\csc^2(x) + \cot^2(x))$

[elfin berry]

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

[marsh citrus]

@rich sedge Has your question been resolved?

[rich sedge]

.close

[marsh citrus]

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[subtle heron]

How do you find the wronskian when there are repeated roots?

[subtle heron]

Solving the determinant gets me 0 which isnt allowed so is there another root im missing?

[amber birch]

ah, so your two linearly independent solutions are $e^{2x}$ and $xe^{2x}$

[elfin berry]

south's secret twin brother

[amber birch]

you forgot that you needed to multiply by x right

[subtle heron]

why would we need to multiply by x?

[amber birch]

https://math.libretexts.org/Bookshelves/Analysis/Supplemental_Modules_(Analysis)/Ordinary_Differential_Equations/3%3A_Second_Order_Linear_Differential_Equations/3.4%3A_Repeated_Roots_and_Reduction_of_Order

try reading this perhaps

[subtle heron]

ohh okay I get it now thank you very much

.close

[marsh citrus]

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[amber birch]

remember that's what you do whenever you have a repeated root, so if xe^(2x) is taken in the particular solution, you would multiply again by x

x^2 e^(2x)

yeah seems like you missed out on something important while studying DEs

no worries!

[subtle heron]

yea sorry thats my bad i completely forgot

i guess thats what happen when studying for the past week

[amber birch]

it's okay now you know so that other problems don't trip you up

best of luck with your studies

[subtle heron]

Thank you!

.close

[marsh citrus]

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[near elbow]

Guys, did i do any mistake here? I have to solve the equation in 0 to 180 degrees, but My answer is wrong

[vestal sandal]

why is tanx=-0.6248?

[near elbow]

oh wait ure right

[near elbow]

i will fix itt, thankss

[vestal sandal]

nw

other than that its correct

im p sure you get to the 35 degrees ur looking for

[near elbow]

yess i tried itt

i got it

[near elbow]

thankuu smm

btw guys

is my letter c wrong?

my book says the answer is 13/84 but it might be wrong

btw this is the question

[marsh citrus]

@near elbow Has your question been resolved?

[vague grotto]

so far, i can tell you that the book is correct

[vague grotto]

Note that cos A and Sin B are both negative, AND they lie in the same quadrant

Hint: in only one quadrant, they are both negative

[marsh citrus]

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[red nimbus]

Why does the equation 18x+24y=99 have no solutions where x,y are integers

[red nimbus]

I have no clue what the relation is between the gcd and this equation having a solution

[copper raven]

can you factor the left hand side a little bit ?

[red nimbus]

gcd(18,24) = 6 = 18 • (-1) + 24 • (1)

I can factor 6 I guess

[copper raven]

yeah

[red nimbus]

6(3x+4y)=99

[copper raven]

is 99 a multiple of 6 ?

[limpid pond]

Looking good

[red nimbus]

No

But what if I factored 3

[knotty trellis]

a multiple of 6 = not a multiple of 6

[knotty trellis]

then you would get another equation

and you wouldnt be able to conclude whether it has sols or doesnt have sols

[red nimbus]

3(6x+8y)=99

[knotty trellis]

so you would have to try harder

[copper raven]

the gcd tells you the biggest thing you can factor out

[red nimbus]

yup

[copper raven]

it tells you the hardest try

[red nimbus]

is it like

factoring 3 is not enough

[knotty trellis]

yeah

[red nimbus]

as opposed to 6

hmm ok

kinda weird

[copper raven]

if we start here, then you're left with 6x+8y=33

but hey LHS is even

[red nimbus]

waw

ok that clears up

[copper raven]

and then you try harder

[knotty trellis]

it's like trying to solve 4x = 24
and saying "What if I divided it by 2 instead of 4?"

you would get somewhere, but you would still need to try harder

[red nimbus]

Ahhh

ok tyty

I will have to study harder

.solved

[marsh citrus]

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[fierce field]

yoink

[fierce field]

Here i have two functions

one function is $1-4r + \sqrt{1-4r}$ and the other is $1-4r - \sqrt{1-4r}$ and im tryind to see when or if this function is $<0$ so i can do a linear stability test (dynamical systems). My first approach was squaring both sides for both equations but that just created a mess with unwanted solutions and i could decipher which ones were real. As you can see, i squared both sides for the $-\sqrt{1-4r}$ but i used an alternative approach for the 1st image when we have $+\sqrt{1-4r}$

[elfin berry]

SollyPolly

[fierce field]

Does anyone have another other tips i could use? I would prefer to have them both using the same method etc.

[marsh citrus]

@fierce field Has your question been resolved?

[marsh citrus]

@fierce field Has your question been resolved?

[marsh citrus]

@fierce field Has your question been resolved?

[marsh citrus]

@fierce field Has your question been resolved?

[fierce field]

tough life out here

[tight kite]

Let $k = 1 - 4r$, $k+ \sqrt{k}$ is never negative because it's undefined for $k < 0$, and for $k \ge 0$ both terms are positive. For $k - \sqrt{k}$ means $k < \sqrt{k}$ which is only true for values in $(0, 1)$ because taking square root of numbers in this interval makes them bigger.

Finally, if $1 - 4r \in (0, 1)$ then $1 - 4r > 0$ and $ 1 - 4r < 1$, solving you'll have $r < \frac14$ and $r > 0$, which means $r \in (0, \frac14)$

[elfin berry]

kaue

[fierce field]

Thank you kind sir!

I've been struggling to correctly write it out coherently

[marsh citrus]

@fierce field Has your question been resolved?

[marsh citrus]

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[near elbow]

.close

[marsh citrus]

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[marsh citrus]

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[viral wyvern]

i can't seem to get how to apply the system rule on the first image? I'm confused

[marsh citrus]

@viral wyvern Has your question been resolved?

[marsh citrus]

@viral wyvern Has your question been resolved?

[marsh citrus]

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[rich flume]

What do you do to find these?

[marsh citrus]

@rich flume Has your question been resolved?

[rich flume]

<@&286206848099549185>

[amber birch]

ah so the mean is the same as the expected value by the way

so $E(y) = E(0.5w - 255)$

$= E(0.5w) - E(255)$ (linearity of expectation, you can do that with + or - in the expected value)

$= 0.5 E(w) - 255$ (constant multiple rule, mean of constant is just a constant)

[elfin berry]

south's secret twin brother

[amber birch]

now solve for E(w) since you know E(y) from the data

(it reads n = 88, sum of y = 924)
(the spacing in the question is terrible)

for variance you apply the formula $Var(y^2) = E(y^2) - E(y)^2$

you will see this in your textbook as $\frac{\sum y^2}{n} - \left( \frac{\sum y}{n} \right)^2$ btw

[elfin berry]

south's secret twin brother

[rich flume]

Like this?

[amber birch]

,calc 924/88

[elfin berry]

Result:

10.5

[amber birch]

yep, all correct!

hopefully you didn't mistype the variance on your calculator

[rich flume]

Yep, its fine, I checked it

[rich flume]

Is this all?

[amber birch]

oh shit

sorry

[rich flume]

?

[amber birch]

you found the variance of y instead

so you have $Var(y) = Var(0.5w - 255) = Var(0.5w) = 0.5^2 Var(w)$

[elfin berry]

south's secret twin brother

[amber birch]

that's the last step

so the -255 gets ignored cause if you move all the data by the same amount left or right
the spread of the data does not change

and then you square the 0.5^2, cause you are squaring in the variance formula

[amber birch]

as in this

moving the data so you have it ranging from like 100 to 111 for example

[marsh citrus]

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[inner grove]

https://math.stackexchange.com/questions/3308943/distribution-of-arctanx-y

can you explain what we did here in the nominator of the exponent?

[inner grove]

why $v^2 + v^2 (\cos{u})^2$ became $v^2$?

[elfin berry]

redve

[inner grove]

.close

[marsh citrus]

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[inner grove]

https://math.stackexchange.com/questions/3308943/distribution-of-arctanx-y
can someone explain me please, how we got this final result?
It looks like equation $1-\infty$, or something with cosh

[elfin berry]

redve

[spark otter]

should it be e^t between -infinity and 0

[inner grove]

I don't think so. There were absolute value in the equation, but I think I handled it right

[inner grove]

and by substitution in link I got there

[spark otter]

The way they did it here

[inner grove]

with this simplification

[spark otter]

the change of variable t = v^2/2sin^2u is only valid on [0,+infinity)

on (-infinity,0], you can either do t = v^2/2sin^2u, which gives you t in [0,infinity) and e^(-t)

[inner grove]

so we ignore t on interval of negative numbers?

or rather set it to 0?

[spark otter]

no, it's just that the integration bounds are gonna be different

[spark otter]

or you can do t = -v^2/2sin^2u, and t goes from -infinity to 0 and the integrand is e^t

[inner grove]

I will give it a try

[spark otter]

In any case, I think they should have done $\int_{-\infty}^{+\infty}f_{U,V}(u,v)dv = 2\int_{0}^{+\infty}f_{U,V}(u,v)dv$ first

[elfin berry]

rafilou is not not born in 2003

[inner grove]

so we do different substitution in each integral, in the first integral we set t to be positive, and in 2nd integral we substitute it to be negative?

[inner grove]

i mean + and - this

[spark otter]

you are allowed to do t = v^2/2sin^2u on the (-infinity,0] integral too

but the bounds are gonna switch

[inner grove]

but t as a square will never be less than 0

ah

do you mind if I try it now and I close the channel, and then dm you if I get any problems? It might take me some time

[spark otter]

sry I don't take dms, but others can help you just as good as I can

[inner grove]

okay, I will open another channel if I hit the wall. Thank you anyway

.close

[marsh citrus]

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[marsh citrus]

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[river forum]

A swimmer who is moving forward in the water at a constant speed of v = 1.5 m /s
wants to cross a river with a width of b = 100 m that is flowing at a constant speed of v0 = 3 m /s

A) In which direction relative to the water must he swim in order to land on the other bank 300 m downstream? Sketch the possible paths. How long is the swimming time?

B) In which direction relative to the bank do the possible paths run?

[dim zodiac]

Draw a diagram, it’ll always help.

[marsh citrus]

@river forum Has your question been resolved?

[marsh citrus]

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[floral tide]

I don't know how to find the equation

[floral tide]

My Horizontal Asymptote is 2

This is the graph

[smoky spire]

is 4^x + 2 the original function

[floral tide]

yes

[smoky spire]

hang on this is a bit confusing

is verticle shift +2?

[floral tide]

Yes

[smoky spire]

so the original function is 4^x

do u know what the horizontal asymptote is for the original function

[floral tide]

Without the vertical shift, it would be 0

[smoky spire]

write it into an equation

[floral tide]

y=0

[smoky spire]

yes

and it shifted upwards

[floral tide]

y=2

[smoky spire]

yes

[floral tide]

I got it. Thank You

[smoky spire]

np

[floral tide]

.close

[marsh citrus]

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[fast cove]

Could someone help out with this question?

[smoky spire]

circles generally have the equation (x - a)^2 + (y - b)^2 = r^2

[fast cove]

Yup

[smoky spire]

manipulate the equation until u get something like that

[fast cove]

That's the bit I'm a bit confused about

I've got the solutions but I'm not too sure where these numbers came from

[smoky spire]

u mean the 2?

and -1

[fast cove]

Yup

The red underlined

[smoky spire]

its completing the square

(x^2 + 2)^2 would give u x^2 + 4x + 4

(y^2 - 1)^2 would give u y^2 - 2y + 1

so for this question ur trying to find out the constant u would get from completing the square

[fast cove]

Crab claw???

[smoky spire]

idk what crab law is, but just expand

seperate all the x and y

make them into two equations

[fast cove]

Oh aight

[smoky spire]

so like x^2 + 4x = (x-a)^2 + C

and samething for y

[marsh citrus]

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[still temple]

given this find the number for x^5 x y^8 x t

[still temple]

my teacher gave me this formula to apply, and I got from 3/4 from it but gpt said there is none number fit for this

so I want to double check

[marsh citrus]

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[potent cobalt]

Okay, so I have a simple question, but I can't prove it and I am wondering if it is wrong.

[potent cobalt]

Cause all I end up with is that it belongs in to (-1/a, 2/a)

Not (0, 1/a)

[marsh citrus]

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@potent cobalt Has your question been resolved?

[potent cobalt]

.close

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[bleak quartz]

Does my answer properly prove it?
It seems mostly right but the 2nd line from the bottom is congruent to x mod 6, is that an issue?

Theroem 6.2 is

[marsh citrus]

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[low aurora]

How do I identify, visually, the interval of where this is concaving?

[native frigate]

your first interval is right, but theres a second

[low aurora]

(5,7)?

Yeah that's not working either

I looked through the book to find the example. I guess you can't concave downward/upward at the same time, so it's not (1,4) it's (2,4)

[native frigate]

oh i didnt realize 1,2 was supposed to be concave down

could go either way from just looking so i assumed you were right

good work

[marsh citrus]

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[low aurora]

What am I missing with this graph?

[quick moth]

it gives u f prime

not f

lock in

[acoustic lichen]

yeah, big time

[low aurora]

Please try again. You can use the graph of f '(x) to determine where f '(x) is positive or negative. When f '(x) is positive, the function f(x) is increasing, and when f '(x) is negative, the function f(x) is decreasing. A local maximum or minimum occurs where the derivative f '(x) changes its sign.

???

[quick moth]

remember

if f prime is positive

then it makes that f is increase

f prime tells us the slope of the tangent

if the slope of the tangent line is positive, then it must be increasing

conversely

if the slope of the tangent is negative, then it must be decreasing

[low aurora]

So then (0,1) U (3,5)?

[quick moth]

yes

[low aurora]

Originally I put 0,2,4,6 but remembered end-points can't be local max/min

but I think I understand what the feedback means, since you said this is f'

[quick moth]

if

[low aurora]

it's where the sign changes

[quick moth]

f prime is 0

then it means

that the slope of the tangent is 0

if the slope of the tangent is 0, then it must be a local max or min

[quick moth]

ye

if it goes from pos --> neg, then its a max

[low aurora]

I wish I remember when that was covered. Everything just disappears once I leave the classroom

[quick moth]

if it goes from neg --> pos, then its a min

three cases:

if the slope of the tangent line is positive, then it must be increasing

if the slope of the tangent is negative, then it must be decreasing

if the slope of the tangent is 0, then it must be a local min or max

[low aurora]

Was able to answer the next one much more easily with all that info. Now to get on to the new questions:

Thanks

[quick moth]

if its asks anything about concave up or concave down

or points of inflection

if f prime is increasing, then the graph is conave up

is f prime is decreasing, then the graph is concave down

Points of inflection occur where f prime changes from inc <---> dec

[low aurora]

Yes

[quick moth]

oops

no

i wrote that wrong

[low aurora]

Unfortunately they aren't asking me about graphs anymore and gave me a polynomial, so going to read my notes to make sure I do this correctly

[quick moth]

might want to reread what i just wrote if u already read it

oringally was wrong

[low aurora]

I saw

Going to close thread until I run into another block
Thanks again

.close

[marsh citrus]

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[marsh citrus]

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[narrow cobalt]

i need help on all three. i am in grade 9 geometry.

[glass silo]

,rccw

[marsh citrus]

@narrow cobalt Has your question been resolved?

[narrow cobalt]

<@&286206848099549185>

on second thought, i don't need help with 8, just 9.

[wooden shore]

What's the formula for area of a triangle

[narrow cobalt]

A=BH

[narrow cobalt]

but what's the base and height?

[wooden shore]

Any line can be a base

But there's only one height given

[narrow cobalt]

is it 8?

[wooden shore]

Yes

[narrow cobalt]

so i assume the height is 12

[wooden shore]

12 is base

[narrow cobalt]

oh lol my mistake.

i'm kinda tired

[wooden shore]

As u can see here when finding the area of a triangle we are just basically finding the area of a square and dividing it by 2

[narrow cobalt]

so (12x8)/2?

[dull socket]

for 9, its quite easy. Forget the other measurement and just take the height and base. 8x12 is 96 and 1/2 of it is ur area so that is 48

[narrow cobalt]

yeah i got that already.

[dull socket]

Aight u need anything else

[narrow cobalt]

the other one in 9.

[dull socket]

Ok

Lemme see it

[narrow cobalt]

the trapezoid.

[dull socket]

so the formula is a+b/2 h

right so plug in the values

(6+16/2) 7

so thats 22/2 x 7

11 x7

77 units ^2

[narrow cobalt]

can i do something real quick?

[dull socket]

Yeah what

[narrow cobalt]

i had to do something quick for my mom. i'm back.

[dull socket]

Nah ur good

So did u understand it or u need more help understanding

[narrow cobalt]

let me try to figure it out on my own. brb.

[dull socket]

Ok sounds good, just use the formula

[narrow cobalt]

ok i'm done with that one.

[dull socket]

K, hope u understood it

[narrow cobalt]

i do now. thank you. now i just have a few more questions.

[dull socket]

K ping me if u need help, I gotta do my studying to lol

[dull socket]

K leme chek

For the twelfth one, follow the angle length theorems

if its 7 and 8

(7+8) > x

so 15 should always be greater than what ever the x is

so it can be all real positive integers under 15

[narrow cobalt]

alright.

[tired valve]

if the triangle is right then it must follow the pythagorean theorem, and the triplets of pythagorean number are always of the form 3k 4k and 5k, so the triangle isnt right

[narrow cobalt]

thank you all. i'm now done.

[dull socket]

we dont know for sure its a right angle triangle

[tired valve]

a right traignle its one with a right angle

by def

[dull socket]

@narrow cobalt

As your done, this will be closed

.close

Nvm only u can do it

[marsh citrus]

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[marsh citrus]

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[delicate sluice]

Hey guys got a nother fun math problem for yall, I got thi formula, and i want to create one where a,b c and f are all swapped together, so there would be 4 equations essentially. Is there an easy way to type that out into desmos?

[delicate sluice]

so forexable the first equation would look like this

then the second equation it wouldnt be 20f but instead be 20a

and a will be f instead

and third equation

would be 20b

and b would be f

if u get me meaning

should i just manually write 4 equations?

[marsh citrus]

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[split basin]

can someone please help me answer this . im honestly quite lost as how to approach this

[brave marsh]

You want to show that it is a solution. You should compute y'' and y', and then plug that into the equation to see if it is a solution.

[split basin]

ohhh, idky i was so focused on C1 and C2, ill give it a shot now

Is this correct?

[marsh citrus]

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[marsh citrus]

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[split basin]

.reopen

[marsh citrus]

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[crisp sigil]

Can someone help me with this question?

.close

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[still temple]

How do I evaluate the given quantities if it is f(x)-f(3)? I'm stuck and don't know where to start

[rose heart]

can you show the whole problem please

[still temple]

Using the function on the top f(x)=2x-3

I think

[rose heart]

i see

i think they just want you to do 2x - 3 - {whatever f(3) is }

and just leave the answer with the x in it

[still temple]

Does it not equal to f(x) though? sorry

[sick walrus]

Your answer will have x in it

So 2x-3 - ( f(3) ), like wilbur said

Just as you found f(-2) above, find f(3)

[still temple]

Ok thank you

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

Would it end up like this?

[cloud field]

did you subtract f(x) with 3?

[still temple]

Am I supposed to?

[cloud field]

the question after all is f(x) - f(3) no?

[still temple]

😭😭 um

I'm just not exactly sure how to write the equation..

[cloud field]

what is f(x)

[still temple]

3?

[cloud field]

f(x) = 2x-3

f(3) = 3

so f(x) - f(3) is just (2x-3) - (3)

[still temple]

OH

Do I just leave it like that?

.close

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[gloomy plume]

i dont even know where to start

[still light]

You want the probability that x, 1-x, and 0.45 form a triangle if x is uniformly chosen from (0,1)

hmm

so think about what values of x would allow for a triangle to be formed

[gloomy plume]

so the sum of two halves have to be greater than 0.45 ?

(and it would be in each case)

x + 1 - x > 0.45 and x + 0.45 > 1-x and 1 - x + 0.45 > x

like that?

[still light]

right

the first one's always true so we don't need to worry about that

the second and third are actually just opposites basically so you only really need to solve for one of them

[gloomy plume]

i got 2x > 0.55

so x > 0.275

[still light]

actually ignore what I said earlier that's wrong

but yes

[still light]

so this is the lower bound for x

now solve the last inequality to find the upper bound

[gloomy plume]

the upperbound is 0

0.725

so the probability is

0.725-0.275 ?

divided by 1 (which is still same)?

[still light]

yes

[gloomy plume]

ohhhh i see

thank you

[still light]

yw 😄

[gloomy plume]

.close

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[pure dagger]

sorry im stupid

[pure dagger]

eliminate parameter

im rusty

[wary kite]

nice to meet you

[pure dagger]

t = arcsin t so 1-cos(arcsint) but that doesnt work

[wary kite]

nah

don’t do that

square both

[pure dagger]

oh shit

thanks lmao

.close

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[waxen remnant]

if alpha and beta are elements in S_n (n>=3), prove that phi_alpha = phi_beta implies that alpha = beta. (here, phi_alpha is t he innerautomorphism of S_n induced by alpha.)

[waxen remnant]

not sure how to start

if $\alpha$ and $\beta$ are elements in $S_n (n\geq3)$, prove that $\phi_\alpha = \phi_\beta$ implies that $\alpha = \beta$. (here, $\phi_\alpha$ is the inner automorphism of $S_n$ induced by $\alpha$.)

[elfin berry]

lock perpendicularly to the wall

[marsh citrus]

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[normal bough]

how do i solve this?

[proud basin]

remember if its add then you multiply

[normal bough]

oh.

was gonna ask if i should solve it like i did for this.

hm.

huh. now i see what they were talking about lol

-_- so (x+8)(x+4) was the right way...

bruh

that log_4(5) is confusing.

[normal bough]

So it should be, (x)(x+4) = log_4(5) ?

[proud basin]

which question

[normal bough]

x^2 + 4x =

[proud basin]

log_4((x)(x+4))=log_4(5)

[normal bough]

oh ok

ic

hm.

[normal bough]

cancel out log_4 so its just (x)(x+4)=5 ?

[proud basin]

yes

[normal bough]

ight.

x^2 + 4x = 5 ,

Quad form now
x^2 + 4x = 5
-5 -5
x^2 + 4x - 5 = 0

(x+5)(x-1)

x=-5
x=1

[normal bough]

Thank you. Thank you so much for your help. :D

[proud basin]

:D

[normal bough]

.close

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[tranquil ice]

Can some maniac please check if I'm right with 41? Question 50 is the "exercise 10.1.49" in question in question 41.

[tranquil ice]

This is as far as I’ve gotten.

[marsh citrus]

@tranquil ice Has your question been resolved?

[tranquil ice]

Also, if someone happens to get the chance, can they also check this problem for 76?

[tranquil ice]

mm

[marsh citrus]

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[marsh citrus]

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[warped falcon]

$if the amplitude of (z-1+i)=\pi$ and the modulus of z = 2 , then find z$

[elfin berry]

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

[warped falcon]

.close

[marsh citrus]

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[dusky viper]

lot of effort

yikes

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[sudden cove]

is my double integral correct? i cant seem to arrive at a^5 / 20. edit: i forgot to add another r in my equation but erm i think the equation is still wrong

[marsh citrus]

@sudden cove Has your question been resolved?

[harsh falcon]

You need an extra r

dA = r ...

It might be easier to do the integral over the whole quarter circle and subtract that of the triangle

Then you probably have nicer integral bounds

[sudden cove]

yup i added it but eventually i reach a rly complex part

[sudden cove]

thank you! i will try that

like this?

[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.