help-42

[velvet osprey]

anyway

the letter B conveniently hides the place where the small circle juts slightly outside the big one.

[deep violet]

Angle AEB = 90 deg and angle FDB = 90 deg. AB = AF = radius. So DE = DB.

[velvet osprey]

would you like to be taken through a proof that your diagram is shoddy

yes/no

again,

the letter B conveniently hides the place where the small circle juts slightly outside the big one.

[deep violet]

Oh

Answer: No.

I guess I am not getting any help today.

What a waste of time.

||JK, I solved the question.||

[velvet osprey]

mkay

[deep violet]

Thanks.

.clseo

.clsoe

.close

[calm coral]

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[calm coral]

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[warm pewter]

Hey I have a question. Why can a particle move like with a velocity of 5ms^-1 even tho it does not have a force that has the same direction of the direction of motion

[pure kayak]

because theres no forces stopping it from doing so

its velocity only begins to be affected once a force is introduced

otherwise it remains constant, whether that be at rest or not

[warm pewter]

Ohhh I see

Then why in the question there are forces stopping it but they did not say there is a force helping it move up

[pure kayak]

imagine im in a vacuum and i throw a ball, once its left my hand theres no forces acting on it but it still has some velocity v and will keep moving that way if nothing interacts with it

after some arbitrary time i use my magic powers to exert a force on it back in my direction, now it is accelerating towards me, for a while it may continue to move away from me but its speed in that direction will decrease until eventually it starts moving back at me

the force is reducing its speed up the plane but its not like its hit a wall and will just stop instantly

its like a rock sliding on ice, theres some friction, but even if nothing is pushing it anymore it can keep sliding away

[warm pewter]

Ohhhhhhhhhhhhh

Hmmmm

Ok I think I understand a bit

[pure kayak]

if things didnt work like that then if i say, threw a ball, it would just stop the moment it left my hand

[warm pewter]

Ohhhhh I see

But then

What keeps it moving forward

[pure kayak]

momentum, the object contains kinetic energy

[warm pewter]

Ohhhhhh

They are not forces

[pure kayak]

since momentum is conserved it has to keep moving unless something happens

(force is the time derivative of momentum)

no force = no change in momentum = no change in velocity

[warm pewter]

What is conserved?

[pure kayak]

momentum

[warm pewter]

Does it mean momentum is maintained

[pure kayak]

the momentum of a closed system, yeah

that means if no (external) forces are acting

[warm pewter]

Ohhhhhhhhh

Ok

Thanks

I get it now

[calm coral]

@warm pewter Has your question been resolved?

[calm coral]

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[crisp minnow]

Is this correct so far?thanks

[manic vault]

Rather, subsitute $u=\ln(x)$

[potent lotus]

mathisfun

[fringe reef]

idk if this is a typo or not but you can simplify this integral first then substitute

[manic vault]

Since $\frac{x\ln(x)}{x^2}=\frac{\ln(x)}{x}$

[potent lotus]

mathisfun

[fringe reef]

notice that you have x in numerator and x^2 in denominator

[fringe reef]

yes precisely

[crisp minnow]

oh ok

[manic vault]

^

[crisp minnow]

now integrate

[fringe reef]

yes

[manic vault]

Yep

[crisp minnow]

And u =lnx

[manic vault]

Yes

[glad parrot]

You can simplify further dx = ?

[manic vault]

What for

$u=\ln(x)\implies du=\frac{1}{x}dx$

[potent lotus]

mathisfun

[manic vault]

Yes

Keep ln(2)

ln(1)=0

[manic vault]

Wait

du=dx/x

Not dx=du/u

[crisp minnow]

I'm trying to get dx

[manic vault]

Then dx=x du

[crisp minnow]

and then integrate

[manic vault]

No

Remember u=ln(x)

[crisp minnow]

so this

okay

what do I put for u

do I do ln(2)^2/2

-0^2/2

[manic vault]

Yes

[manic vault]

Yep

[manic vault]

GG

[glad parrot]

Are you asked to round ?

[crisp minnow]

Yep 2dp

[glad parrot]

Ok

[crisp minnow]

.close

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[warm pewter]

Don’t know how to do c)

[calm coral]

@warm pewter Has your question been resolved?

[warm pewter]

<@&286206848099549185>

[calm coral]

@warm pewter Has your question been resolved?

[carmine otter]

I can't clearly see the diagram

[carmine otter]

Mhmm alright

As you can see

The pulley experiences

Two types of tension forces acting at an angle

And as you can see the angle is 90-alpha

So the resultant tension acting on it would be the vector sum of those components

R resultant^2 =T^2+T^2 -T T cos(theta)

[warm pewter]

Is it like resolving the forces?

[carmine otter]

Yeah

Btw I'm really confused about one thing

What's the angle between those components of tensions is it 90+alpha Or 90-alpha

[warm pewter]

90-alpha (right?)

[warm pewter]

If I don’t use the formula can I still work it out

[carmine otter]

It's 90+alpha

Lemme explain

[warm pewter]

WHATT

[carmine otter]

The angle isn't just 90-alpha because

The tension forces aren't pulling directly along the incline and horizontally

One is pulling upwards

Along the incline

So it creates an extra angle

So the total angle between the components would be 90+alpha

[warm pewter]

This one is pulling upwards?

[carmine otter]

No following this will make the problem easier

[carmine otter]

Mhm

[warm pewter]

Wait then it would be 180 - alpha

[carmine otter]

No

You're just adding alpha to the 90

°

Lemme make it clear

[warm pewter]

Wait r we doing this angle

[warm pewter]

Okkkk

[carmine otter]

It's 90+alpha because it combines the 90° angle between the components and the angle due to inclination

[carmine otter]

No

[warm pewter]

What is the composition? I am so sorry

[carmine otter]

My apologies it's components

[warm pewter]

Components of both tensions

Is 90 from the horizontal tension and alpha from the inclined tension

[carmine otter]

Yeah yeah

as the both tensions have slight inclantions

You take the components of both

[warm pewter]

But the one on horizontal plane does not have inclination

Oh wait I get it now

I just add them together

Like vertical and horizontal components

[carmine otter]

Wait yeah you're right

We r doing that angle

[warm pewter]

Wait what I am?

[carmine otter]

Yeah

[warm pewter]

Omg ahahahahaha I have no idea

Wait I have a dumb question

[carmine otter]

Aa it's not perfectly straight

[warm pewter]

Why isn’t the force 0 because the tensions cancel each other out

[carmine otter]

It's less than 180°

[warm pewter]

Yeppppp

[carmine otter]

So we take it as 90+alpha

[carmine otter]

Haha if the angle between them is 180° you may be right

[warm pewter]

Ohhhh

[carmine otter]

As the tension produced by the Particle p

[warm pewter]

Ohhhhh

[carmine otter]

Is inclined at an angle

We can't cancel them out

[warm pewter]

this is confusing

I see

[carmine otter]

You got my point?

[warm pewter]

Yesssss

And so the components won’t be equal

[carmine otter]

Oh yeahhh

[warm pewter]

Big brain at 2am

[carmine otter]

Mwhahhaah

[warm pewter]

So hard

[carmine otter]

It's around 1 am here

[warm pewter]

Omg I guess u live near here

[carmine otter]

Mhm

Btw

[warm pewter]

Yess

[carmine otter]

Lets get back to the problem

[warm pewter]

HAHAHAHAHAHA

Getting sidetracked

Yes

[carmine otter]

Alr

So by the formula I have mentioned earlier

[warm pewter]

Okkkk

[carmine otter]

Tres^2 =2T^2 - 2T^2 cos(90+alpha)

[warm pewter]

That is easy then

[carmine otter]

So wait

Whats cos(90+alpha)?

It can also be written as...

[warm pewter]

As

sin(alpha)

Idk

[carmine otter]

• sin alpha

Negative sin(alpha)

[warm pewter]

Ok idk

[carmine otter]

As cos is negative in the second quadrant

[warm pewter]

Yes

[carmine otter]

Mhmm

So

It's

[warm pewter]

Yessssss

[carmine otter]

Tres^2=2t^2 + 2t^2 sin (alpha)

[warm pewter]

Yesssssssssssssssssss

We change sign yes

[carmine otter]

You have already found the value of tension and sin(me)

[warm pewter]

Yes so I just substitute

[carmine otter]

Yeah

Substitute and tell me the answer

[warm pewter]

Okkkkkkk

[carmine otter]

And sin (me) =3/5

[warm pewter]

Wait idk tension

Wait I mean m

[carmine otter]

M?

[warm pewter]

Yes idk the mass

[carmine otter]

Do you see mass here?

[carmine otter]

No need of that

[warm pewter]

No but tension is stuck at 85/25mN

I forgot to write m omg

[carmine otter]

Isn't it 8/25?

[warm pewter]

Idkkkk

[carmine otter]

Bruh

[warm pewter]

Can I dm u tmr or smth

I so sleepy

[carmine otter]

12-4=84?

[carmine otter]

Mhmm okay

Nw

I'm sleepy too

[warm pewter]

WHAT

Omg thanks

Same

[carmine otter]

I forgot that it's 2 am

[calm coral]

@warm pewter Has your question been resolved?

[calm coral]

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[spring gulch]

My calculator is giving me none of these answers what so ever. I have a casio fx-9750giii and I dont know what I have to set it to, to make the plug in work

[spring gulch]

It keeps giving me 2.069160002E-09

current mode is comp

angle is set on degrees and Ive tried it on radian but nothing seems to work on the settings to get any of these answers

I even tried to break down bit by bit in order of operation and still get nothing

I also press the s<>d button and I get no response

[teal drift]

Why are you talking about angles? 🤔

[spring gulch]

Its just my settings

[teal drift]

But here there's no trigonometry 😬

So your calculations are not affected by that

[spring gulch]

I know I have to fiddle around with my settings however due to the test I am taking so I am not sure what setting it making this not work

[teal drift]

What did you type (exactly) on your calculator?

[spring gulch]

I first tried 95e^-0.491(50)

[stuck sequoia]

P is 50

[teal drift]

Ohhh that's why

[stuck sequoia]

not t

[spring gulch]

My teachers guide was awful then

[teal drift]

Yeah she definitely did it the wrong way

[spring gulch]

So 50/95

and use nat log

[teal drift]

Exactly

[spring gulch]

Hold on let me test

Im gonna throttle my teacher

Thanks you guys I waas pulling hairs

[calm coral]

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[swift plinth]

help

[calm coral]

@swift plinth Has your question been resolved?

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[cursive gyro]

Hi this a general question regarding vector spaces

[cursive gyro]

If I have two basis (not necessarily the same) that span the same vectorspace

are the dimensions of both bases equal?

Regardless of whether the vector space is finite or infinite?

I know by the fundamental theorem, if a F vector space V can be spanned by a set S of n vectors, if L is an linearly indepndent set where L is a subset of V, then m=|L| <= n

[glad parrot]

Cardinality of both basis are the same, the dim of vectorspace is the cardinal in question

[cursive gyro]

But the part that trips me is that

[nocturne heron]

oh bah qui voilà-je

[cursive gyro]

m<=n

how would i know m=n?

[drifting seal]

if it spans

[cursive gyro]

I find this really unclear

Can you direct me to a way where I can formally prove such?

[drifting seal]

if a set of vectors is linearly independent and spans the vector space then it’s a basis

this is just by definition

[cursive gyro]

How do you know the cardinality of both bases are equal?

[glad parrot]

Definition

[glass heart]

let L be the first basis and S be the second

and then the other way around

[cursive gyro]

Would I have to find a bijection or smthing to prove this?

[cursive gyro]

This makes sense

thank you

Thank you for everyone that replied

.close

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[warm warren]

Only works on finite ones afaik

https://en.m.wikipedia.org/wiki/Steinitz_exchange_lemma

You have trouble picking a basis in some infinite dimensional vector spaces

[cursive gyro]

ah no wonder

i kept trying to prove it on infinite and it just dont work

[warm warren]

You need a bijection between the sets themselves not just between the basis

[cursive gyro]

oolala

[warm warren]

And I think you run into problems there

[cursive gyro]

that is cool

Would a contradiction proof be good

[warm warren]

And if you were able to find one then you could also construct an isomorphism of vector spaces then in linear algebra they really are the same thing anyway

You also run into the problem of picking a basis in an infinite dimensional vector space

Something about axiom of choice but I’m not sure

[glass heart]

without AoC there are vector spaces without a basis

[cursive gyro]

I will figure that out later but thank you for telling me

[warm warren]

Ah yeah that’s the issue

[glass heart]

AoC is equivalent to the fact that all vector spaces have a basis

[cursive gyro]

Thank you almighty big brain people

[glass heart]

in most infinite dim spaces a basis wouldnt be interesting cause you could never write it down

most notable exception being the space of polynomials as its only countable infinite dim

[calm coral]

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[blazing coyote]

None of these look particularly fun to solve

$

$\int_{0}^{4 \pi} cos^2(t) sin^2(t) dt$

[potent lotus]

What a wonderful world !

[blazing coyote]

This doesn't look like it has a paricularly fun solutiion

[manic vault]

$\sin(t)\cos(t)=\frac12\sin(2t)$

[potent lotus]

mathisfun

[vagrant dune]

bring it to (sin2t)² form

[manic vault]

What I said is implied for that

[blazing coyote]

And then cos^2(t) in terms of cos(2t)

hmm

yeah

that works

thanks

[manic vault]

Wait what

[vagrant dune]

yess

[manic vault]

No just directly use this

[blazing coyote]

cos(2t)+1/2 = cos^2(t)

[manic vault]

$\sin^2(t)\cos^2(t)=\frac14\sin^2(2t)$

[potent lotus]

mathisfun

[vagrant dune]

then use cos2t identity to make the power linear

[manic vault]

Then ezpz integral

[blazing coyote]

yea

thanks

Been a while since I did integrals lol

[manic vault]

Bruh

Do you have a good stash

I wanna try hard ones

[blazing coyote]

yeah, I used a soviet book

[manic vault]

Soviet?

[blazing coyote]

A problem book in mathematical analysis

[manic vault]

Ok

[blazing coyote]

It had hard problems and was cheap

[manic vault]

Pls send?

[blazing coyote]

I don't have a pdf

[manic vault]

😦

[vagrant dune]

what is it called

[blazing coyote]

A problem book in mathematical analysis by GN Berman

[vagrant dune]

thanks

[blazing coyote]

This would be $\int_{0}^{4 \pi} \frac{(1+ cos(2x)}{2} \frac{sin(2x)}{2}dx$

[potent lotus]

What a wonderful world !

[blazing coyote]

Then cos(2x)=u

so -2sin(2x)dx = du

Prett easy from here

[vagrant dune]

no you got sin²x form
use cos2x = 1-2sin²x to make it linear

yeah but substitution also works of course

[blazing coyote]

or $cos(2x)+1 = 2cos^2(x)$ works too

[potent lotus]

What a wonderful world !

[vagrant dune]

yesss

[manic vault]

Or just this 😦

[blazing coyote]

yea

ooh

righ

thanks

$\frac{1}{4} \int_{0}^{4 \pi} sin^2(2t)dt$

[potent lotus]

What a wonderful world !

[vagrant dune]

wait can you explain how

[manic vault]

Because

$\sin(2t)=2\sin(t)\cos(t)$

[potent lotus]

mathisfun

[manic vault]

$\frac12\sin(2t)=\sin(t)\cos(t)$

[potent lotus]

mathisfun

[manic vault]

$\frac14\sin^2(2t)=\sin^2(t)\cos^2(t)$

[potent lotus]

mathisfun

[blazing coyote]

This is $\frac{1}{4} \int_{0}^{4 \pi} \frac{1 -cos(4t)}{4} dt$

[manic vault]

Yeeeeeeeee

[potent lotus]

What a wonderful world !

[manic vault]

Hey

Why do you not use $\cos$

[potent lotus]

mathisfun

[vagrant dune]

yess i was talking about this step

[blazing coyote]

I forget \cos exists

[manic vault]

Lol

[blazing coyote]

so $\frac{ \pi}{4}$

[potent lotus]

What a wonderful world !

[blazing coyote]

as cos(4t)'s integral from 0 to 4π is 0

thanks

I'mma go grab a glass of tea now

thanks

[vagrant dune]

yup

[blazing coyote]

.close

[calm coral]

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[blazing coyote]

For question(c), I suspect it's simply $$\frac{ \int_{-\sqrt{ \pi}}^ { \sqrt{ \pi}} sin(t^2)(2tcos(t^2)dt}{ \int_{- \sqrt{ \pi}}^{ \sqrt{ \pi}} dt$$

[potent lotus]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

[
\frac{ \int_{-\sqrt{\pi}}^{\sqrt{\pi}} \sin(t^2) (2t \cos(t^2)) , dt}{ \int_{-\sqrt{\pi}}^{\sqrt{\pi}}2t dt}
]

so [
\frac{\int_{-\sqrt{\pi}}^{\sqrt{\pi}} t \sin(2t^2) , dt}{2\sqrt{\pi}}
]

I think I messed up the arc length

[potent lotus]

What a wonderful world !

[blazing coyote]

I think this is more like it?
The issue is the bottom integral goes to 0

which makes no sense

[swift laurel]

what did you find for your ds?

[blazing coyote]

ds as in the differential of the equation

rigght

$(-2tcos(t^2), 2tsin(t^2))$

[potent lotus]

What a wonderful world !

[swift laurel]

so then $\odif s = \norm{\alpha'(t)} \odif t$

[potent lotus]

cloud

[blazing coyote]

yea

which is $\int_{- \sqrt{ \pi}}^{ \sqrt{ \pi}} \sqrt{2*4t^2} dt$

[potent lotus]

What a wonderful world !

[blazing coyote]

which is $\int_{-\sqrt{\pi}}^{\sqrt{ \pi}} 2\sqrt{2} t dt$

[potent lotus]

What a wonderful world !

[swift laurel]

where did the 2 come from?

[blazing coyote]

oops

my bad

[swift laurel]

also remember that $\sqrt{t^2} = \abs t$

[potent lotus]

cloud

[blazing coyote]

right, just realised that

that gives me an arc length of $2\pi$

[potent lotus]

What a wonderful world !

[swift laurel]

seems right so far

what about the top integral?

[blazing coyote]

$\int_{-\sqrt{\pi}}^{\sqrt{\pi}} tsin(2t^2)dt$

[potent lotus]

What a wonderful world !

[blazing coyote]

let $2t^2 =u

we then have tdt = du/4

I think I';ll have to break this integral

becaiuse I'm getting identical lower and upper bounds

It can't be zero

can it

[swift laurel]

how did you get that integral?

[blazing coyote]

oops

[blazing coyote]

yea, messed up

or maybe not

the norm of the derivative is 2t

[swift laurel]

remember it's still the absolute value

[blazing coyote]

$\int_{-\sqrt{ \pi}}^{\sqrt{\pi}} 2\abs{t} sin(t^2) dt$

[potent lotus]

What a wonderful world !

[blazing coyote]

This is $2 \int_{0}}^{\sqrt{\pi}} 2tsin(t^2)$

[potent lotus]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

which is 4/2π = 2/π

Which checks out

Thanks

I'd like to do the first part for now

The forward path is given by $\int_{a}^{b} f(\alpha(t)) \norm{ \alpha'(t)}dt$

[potent lotus]

What a wonderful world !

[blazing coyote]

The reverse path would be given by $\int_{a}^{b} f(\alpha( a+b-t)) \norm{ \alpha ' (a+b-t)}dt$

uh, now what

King's rule is the first thing that comes to mind

but as the changes the norm, it won't help here

[potent lotus]

What a wonderful world !

[blazing coyote]

now just using king's rule gives us $\int_{a}^{b} f(\alpha(t)) \norm { \alpha' (t)}$

[potent lotus]

What a wonderful world !

[blazing coyote]

Is this fine

[calm coral]

@blazing coyote Has your question been resolved?

[swift laurel]

seems fine

[calm coral]

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[exotic cosmos]

first of all they used the wrong value for the radius right? lol

secondly, isn’t the answer none of the above?

[coral anchor]

I beleive they've done it right.

Can you explain why you think it is wrong?

[exotic cosmos]

i did tho

[pure kayak]

youre right

[exotic cosmos]

^ what about this part

[pure kayak]

well at least for the radius, havent checked that yet

[exotic cosmos]

yeah well if the outer thickness is 1.5 throughout

then the outer diameter must be + 3 the inner diameter

[pure kayak]

,w 240(6.5-5)

[coral anchor]

yep, +3 diameter, but +1.5 radius

[exotic cosmos]

hold on

[potent lotus]

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Requested by aldrnari_

[pure kayak]

pi

[exotic cosmos]

the diameter of the outer pipe is 13

the diameter of the inner pipe is 10

pi * d = gets you the circumference

[coral anchor]

ok, the diameter is given for the inner pipe

I didn't see that, sorry

[exotic cosmos]

so (13pi * 40) - (10pi * 40)

[pure kayak]

=3pi *40

[coral anchor]

yes, if the diameter for inner pipe is given, then this will be correct

[exotic cosmos]

oh

i can’t add lol

but okay yeah sure

i had that as 240 pi

for some reason in my head

i need to sleep

anyway yeah the difference is constant so his answer is right

he just used wrong values right?

[jolly pilot]

Since the answer doesnt depend on the radius, but only the thickness, he lucked out

[exotic cosmos]

yeah

okay thanks

i gotta learn how to add

[calm coral]

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[rough blade]

in the book: $$z'-1-2i=\frac{1}{\sqrt{2}}[\left\cos_{\frac{\pi}{4} + i\sin(\frac{\pi}{4}}]\right(z-1-2i) = \frac{1+i}{2}z \frac{3+i}{2}$$

[velvet osprey]

tex fail

do you have a picture?

[rough blade]

lemme fix it

$$
z' - 1 - 2i = \frac{1}{\sqrt{2}}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)(z - 1 - 2i) = \frac{1+i}{2}z - \frac{3+i}{2}
$$

[potent lotus]

<rajel />

[rough blade]

so i dont understand how they got the result directly

[velvet osprey]

original problem?

we need to know what z' and z are supposed to be

[rough blade]

well they are just complexe numbers , both are variables

and its just z'(z)

[velvet osprey]

...

!xy

[calm coral]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[rough blade]

do u think that this need an understanding of what is z and z'

[velvet osprey]

yes it does

we need the entire context

[rough blade]

i mean i can just change them with x and y, and assume the exercice is asking to find x in the function of y

[velvet osprey]

just please share the original problem.

[rough blade]

sorry i cant , appreciate your help tho

[velvet osprey]

why not?

[rough blade]

i dont have my phone at the moment

and its in the book

+its french , i guess i understand their solution rn 👍

.close

[calm coral]

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[next hound]

ann speaks French, but glad you got it figured out yourself

[calm coral]

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[wise marten]

what is the formula for this

i dont relaly know what this kind of question

is called so idk what to search for

[calm coral]

@wise marten Has your question been resolved?

[green fox]

need some help

[calm coral]

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[calm coral]

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[cold mason]

Hi, what in my method is wrong for the following problem?

[cold mason]

what I did was finding the integral of g(x)-f(x) with bounds -1 and 1

and adding it to integral f(x)-g(x) with bounds 1 and 2

but I'm not reaching the right answer. Is there a reason for this?

[dull wagon]

show your specific calculations

[cold mason]

ight let me take a picture

then I added 8/3

with 26.5 (26.5 being the result of the second integral)

to get 175/6

oh wait I think I made a mistake in the earlier calculations

when substracting the two functions 🤦‍♂️

lemme try again rq

ok yeah that was indeed the issue, I got the right answer now

.close

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[calm coral]

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[wise marten]

i need to inverse this

[wise marten]

and i cant get the answer

these are the answeres and im getting

(5-4x)/-3x-7

the closest one seems to be A

i dont get why the negatives are reversed

y=(7x+5)/(4-3x)

i begin like that

[pure kayak]

a/(-b)=-a/b is that what you mean

[wise marten]

uhh i think so

is this true?

[pure kayak]

yeah

[wise marten]

ohh

[pure kayak]

multiply by -1/-1=1

[wise marten]

oh then if i got a/(-b)

thats the same as -a/b

ok wait

then i already found the answer yeah

ty

.close

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[calm coral]

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[wise marten]

how do i do this

[wise marten]

is there somethin missing

the statement "his cart can only accomidate fruit" is so weird

[velvet osprey]

it's just to say that there's nothing else he buys or sells

[wise marten]

so how do i do this

[velvet osprey]

but it feels like maybe there's a restriction on like

[wise marten]

this is suppose to be linear thing right

[velvet osprey]

how much weight he can carry in his cart?

and the number for that is missing

[wise marten]

so this is like

unsolveable?

or assume that tehres no limit

could there be an answer

[velvet osprey]

so right now the problem seems to be "find which fruit gives you a greater percentage profit per unit expense and max out on that, stuffing your cart with only that fruit"

but then it seems both types of fruit give 33% profit...

[wise marten]

so ig

hm

[velvet osprey]

so i for one have no idea what they're expecting of you here

there is something missing

[wise marten]

well these are the answers

if i just take any of the 2 fruit

i will gain 600k

450 feels like

the answer if there was a restriction

no way its getting up tot he 800s right?

[velvet osprey]

600k seems to be in agreement with the whole "both fruits give 33% profit" thing

given 1.8M is our capital

we gain 1/3 of that in profit

[wild remnant]

taking m as no of mangoes

g as no of guavas

we have 2 equations

4000m + 3000g is what we need to maximize, as 4000 is profit per mango and 3000 is profit per guava

and we also find one more by the max expenditure

[velvet osprey]

that's one objective function and one constraint

and you will find that profit = 1/3 capital anyway

[wild remnant]

yep, and even if you use the max expenditure equation to find m = 150-0.75g and plug it in you just get 600000 at any rate

so that's the answer ig

[wise marten]

alright

theres no possibility that the answer could be 450k right

or 800k

just to be sure sure

[wild remnant]

there is no possibility to get any other profit amt

you could test this out and its encouraged to do so:

make 2 statements on the problem given m = mango amt and g = guava amt

see for yourself that it's always 60k independent of variables.

try it!

[wise marten]

uhh

[wild remnant]

ok here is a starting point

[wise marten]

what if lets say i wanna make the least ammount of profit possible

[wild remnant]

oh, that's not an option in this question, because its always 60000 no matter what you do

[wise marten]

hm i couldnt get like atleast a 596 even?

[wild remnant]

nope!

i encourage you to make 2 statements and then work it out!

it would be much easier to understand thst way.

okok first of all what's the profit id get from ONE mango?

[wise marten]

4k

[wild remnant]

nice, and what about one guava?

[wise marten]

3k

[wild remnant]

good so if i wanted to check my total cost

i say m is my amt of mangoes

g is my amt of guavas.

what's the formula then?

total profit=?

[wise marten]

(12k)m+(9k)g<=1800k

i think

[wild remnant]

good, that's the equation for cost

and what about profit?

[wise marten]

(4k)m+(3k)g

uhm

idk what it should equal to

[wild remnant]

oh that's what we want to maximize, dw, its not needed for it to be equal to anything

[wild remnant]

but we're interested in this

now find g interms of m

[wise marten]

ok

so

[wild remnant]

use a calculator if u want

it doesnt matter to the logic

i used one 🤷‍♂️

[wise marten]

(9k)g<=1800k - (12k)m?

or do you mean like

make

(4k)m=(3k)g

[wild remnant]

almost there! now just divide 9 from both sides

[wise marten]

alright

[wild remnant]

no no that's without an =, we need g interms of m for that

because a 2 variable expression isnt very nice

[wise marten]

g<=200+(4/3)m

[wild remnant]

nice! ok let's think abt the <=

if i wanted maximum profit

i'd need max expenditure

so for max spending

now evaluate it well

(cause we're thinking abt maximum)

so we have g interms of m.

[wild remnant]

now just plug it in, and see what happens :D

[wise marten]

(4k)m+(3k)(200+(4/3)m)

[wild remnant]

now evaluate it well

[wise marten]

(ok

[wild remnant]

oh, it seems like there's a mistake here

the plus should be a minus

[wise marten]

(4k)m+600k+(4k)m

9oh right

yeah

[wild remnant]

Why? because you shifted things to the other side

[wise marten]

bc it switched sides

[wild remnant]

so that last 4km becomes -4km

[wise marten]

yeah

oh

[wild remnant]

yep

[wise marten]

600k

[wild remnant]

nice :D

[wise marten]

hm

yeah

[wild remnant]

but ofcourse, its the maximum.

[wise marten]

alright ty i get it

[wild remnant]

the minimum is ZERO, if u dont use any money!

[wise marten]

if i spend all my money i cant get less than 600k anyways

[wild remnant]

mhm

so no questions right?

[wise marten]

ok ty so much for taking the time to teach me

yeah i get it

[wild remnant]

good luck!

[wise marten]

like genuenly ty very much you have been very patient

gl to you too

[wild remnant]

k cya

[wise marten]

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[thorn cedar]

hello i need help with a vector problem

[velvet osprey]

image on its way?

[thorn cedar]

find M if AB+MC+MA+CA= null vector (0)

[velvet osprey]

,rccw

[velvet osprey]

what is foicrs

[thorn cedar]

wat

sorry?

[velvet osprey]

you wrote Given foicrs:

[thorn cedar]

thats given facts

[velvet osprey]

handwriting

[thorn cedar]

sorry my handwriting is

[velvet osprey]

anyway what have you tried thus far

[thorn cedar]

i thought about
AB=AB
MC=OC-OM
MA=OA-OM
CA=-BC-AB

still didnt get anything

the question also tells to simplify EA+ED+OA and that gives = OD
and AD+OB=AO

[velvet osprey]

MC+CA+AB = MB

so your vector equation reduces to MB + MA = 0

[thorn cedar]

that may have something to do with it

i think ur wrong

[velvet osprey]

why so

[thorn cedar]

MC+CA+AB=AM

[velvet osprey]

MC+CA+AB equals MB regardless of anything else we know about them

like just tip to tail vector addition

i am telling you to rewrite MC+CA+AB as MB

[thorn cedar]

oh right

so AM=MB?

[velvet osprey]

yes

[thorn cedar]

so the key to this type of questions is write the final 3 vectors as 2 different equations

cz in my series of problems its almost in every single problem

[calm coral]

@thorn cedar Has your question been resolved?

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[outer sedge]

If we have p = 2, phi = pi/3, theta = pi/2, would that be a point in spherical coordinates or a surface?

[velvet osprey]

p? did you maybe mean rho?

anyway you have specified all three coordinates explicitly

that's a point

[outer sedge]

yeah i meant rho

when is it a surface and when is it a point

for example rho = 2 is a sphere

if they specified all three coordinates it's a point?

what about two coordinates

[velvet osprey]

it takes specifying all coordinates to give a point

if only two coordinates are fixed and the third can vary it will be a curve

[outer sedge]

.close

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[outer sedge]

.reopen

[calm coral]

✅

[outer sedge]

what if there's an inequality

like 2 < rho < 3, 0 < theta < 2pi, 0 < phi < pi

[velvet osprey]

this will give you some kind of 3D region

one that's bounded by three sets of coordinate surfaces...

[outer sedge]

ok thank you

.close

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[outer sedge]

doesn't rho = 2 still have a range for theta and phi, but it's not a 3D region?

[velvet osprey]

??

you're asking about several different sets of conditions at once

[outer sedge]

like for a full sphere, 0<phi<pi, 0<theta<2pi

we have inequalities for phi and theta

[velvet osprey]

these are inherent to the coordinate system

to specifically the spherical coordinate system

[outer sedge]

why do we get a 3D region with the other ones

with like 2 < rho < 3, 0 < theta < 2pi, 0 < phi < pi

[velvet osprey]

what other ones

ok like

why do you get a 3d region with

3<x<4, 0<y<11, 7<z<25

it is the same principle here

just that the shape will be different

[outer sedge]

because rho is bounded between two values?

if rho was constant

it would be a surface, right?

[velvet osprey]

yes

[outer sedge]

but theta and phi would still be bounded between two values

[velvet osprey]

you're asking these up in the air and kinda unclear questions one after the other and it is kind of difficult for me to orient myself in what you really want to know

[outer sedge]

thanks
.close

.close

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[outer sedge]

is this the correct way to solve triple integrals?

let's say we have f(x,y,z) = xyz^2, 0<z<3, -1<y<2, 0<x<1

I stick the "x" with dx, the "y" with dy, and the "z^2" with dz

So we have the following:

int(xdx) * int(ydy) * int(z^2 dz)

1/2 x^2 {0x<1} * 1/2 y^2 {-1<y<2} * 1/3 z^3 {0<z<3}

= 1/2 * 3/2 * 9 = 6.75

[eternal shard]

it works here but it doesnt always work out

[outer sedge]

can u give me an example

[eternal shard]

if you had f(x,y,z) = x+y+z then you wouldn't be able to factor

[outer sedge]

I see

wow

ok thanks

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[remote mural]

hello

[remote mural]

isnt it supposed to be F(11;4,6) right?

because we are also taking 12 into account

[velvet osprey]

the gamma distribution is continuous

it is not discrete and not integer-valued

[remote mural]

oh

[velvet osprey]

so no, subtracting F(11; 4, 6) would also give us the entire piece where 11 < X < 12

we don't want that

[remote mural]

yeah

thanks!

.close

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[ornate turret]

hey guys

im new

is this group available for odinary levels?

[velvet osprey]

these help channels in general have no restriction on difficulty or grade level

[ornate turret]

oh

[velvet osprey]

but this particular channel still hasn't cleared, so you'll need to claim your own by posting your question there

read #❓how-to-get-help for instructions

[ornate turret]

Thank you Ann.

[calm coral]

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[wintry echo]

What is a subcube of dimension 1?

[drowsy spruce]

an edge of a higher dimensional cube

just a line

[wintry echo]

And 2 edge form a subcube of dimension 2?

[drowsy spruce]

no

a subcube of dimension 2 must be a square

[wintry echo]

Yes

[drowsy spruce]

you need 4 lines

4 edges

[wintry echo]

Okay thanks

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[opaque yoke]

could someone please help me with this problem

[woven geode]

guys

[opaque yoke]

i made 9^x to 3^2x

and then i did cross multiplied

[woven geode]

ineed math help

[opaque yoke]

after adding them

[calm coral]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[swift dragon]

is this question 12?

[opaque yoke]

yes

sorry i didnt carify

[swift dragon]

!show

[calm coral]

Show your work, and if possible, explain where you are stuck.

[opaque yoke]

3^4x +1 = 34(3^2x)

[swift dragon]

try squaring 3^x+3^-x

[opaque yoke]

ohh

im so dumb i didnt realize it was that type of problem

thanks i got the answer

.close

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[swift dragon]

its actually a surprisingly elegant question

[calm coral]

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[shy nova]

hi

im trying to prove this by induction

and this might me a dumb question but

if i want to prove this verifies for n = 1

i dont exactly know what im proving there

[edgy leaf]

you are proving the base case

[shy nova]

like i know i have to replace n with 1

but just in the rhs?

[swift dragon]

both sides

[shy nova]

in the last term of the left side

right

[swift dragon]

the LHS is a sum: $\sum_{k=1}^n (2k-1)^2$

[potent lotus]

Bonk

[shy nova]

yes

[manic vault]

The left side is $\sum_{k=1}^n (2k-1)^2$ to make it easier

[potent lotus]

mathisfun

[edgy leaf]

in LHS, you use the first term. rewrite it using sigma notation to make it look more clear

[manic vault]

So just plug in n=1

Or you can notice that the last term will be (2(1)-1)^2=1^2, so you know where to end it

[shy nova]

ohh and i should get the same for both sides

[edgy leaf]

yes

[shy nova]

okay i get it now

and for the second step i have to prove it verifies for another natural

so i do n = k

integer

or natural

[manic vault]

n=k+1

[edgy leaf]

natural

your n = k is your assumption

[manic vault]

So say that it is true for n=k

You want to prove for n=k+1

This is induction hypothesis

[shy nova]

yeas

[manic vault]

All you really have to do is mainpulate last term

[shy nova]

okay i think i get it

ill open another ticket if i need more help

thanks

.close

[calm coral]

Channel closed

Closed by @shy nova

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[ionic venture]

Need some help on how to solve this question

[ionic venture]

It's either 0 or 1 or -1 but i forgot how to find the answer

[glad parrot]

Whats the greatest among thoses ?

[ionic venture]

1

[velvet osprey]

well there you have it dont ya

[glad parrot]

There you go

[ionic venture]

thanks

[velvet osprey]

i mean for cos(x) you should know its range

and the same for sin(x)

[ionic venture]

wym range

[glad parrot]

Possible values of y

Its called the range

[ionic venture]

ah ok

I was learning about the unit circle and i remembered all the possible values of cos is 0 , 1 , -1

is that the same for sin or nah

[glad parrot]

No

Or you are wording it wrong

Cos(x) can take any real values

[ionic venture]

ok if it said what is the greatest y value for y=sin(x) what would the answer be

[velvet osprey]

definitely bad wording there

[velvet osprey]

its OUTPUTS all fall in the range [-1,1].

[velvet osprey]

again you should know what the raw sine wave looks like. it's something you learn quite early on doing trig

[glad parrot]

x in R but cos(x) in [-1,1]

,w graph cos

[potent lotus]

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[velvet osprey]

,w graph y=sin(x)

[potent lotus]

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Requested by ann_dec

[ionic venture]

i know what it looks like

[glad parrot]

See how its blocked by -1 and 1 ?

[ionic venture]

this makes more sense thanks

.close

[calm coral]

Channel closed

Closed by @ionic venture

Use .reopen if this was a mistake.

[glad parrot]

Youre welcome

[calm coral]

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[shy nova]

hi

to prove this by induction

will n = k - 1

when i try to prove it for every natural

since its going down by 1

[manic vault]

n=k+1

[shy nova]

so it doesnt change

[manic vault]

Nope

[shy nova]

ok

thanks

.close

[calm coral]

Channel closed

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Send your question here to claim the channel.

Remember:
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[devout meteor]

okay we just learned series and im so confused i have no idea what to do

[upper sparrow]

What type of series do you know about? Do you know how to work the value out of any of them?

[plain bone]

Split the fraction $\frac{a + b}c = \frac ac + \frac bc$

[potent lotus]

King Leo

[plain bone]

$$\sum_{n = k}^\infty \qty(r^n) = \frac{r^k}{1 - r}$$
This only works if $r \in (-1, 1)$. Otherwise, $\sum_{n = k}^\infty \qty(r^n)$ diverges.

[manic vault]

Yeah split fraction then use a/(1-r)

Remember a is the number which is produced when you plug in the lower index

[potent lotus]

King Leo