help-10

[untold raven]

The statement of the theorem is, if an n-dimensional solid is similar to another n-dimensional solid with ratio 1:a, then the n-dimensional volume of the solids are in the ratio 1:a^n.

[supple fog]

Because english is not my first language i am having a tougher time understanding, i understand the triangle example but i do not understand how i can adapt this into my task

from my understanding the top part of the cut pyramid is on the ratio 4:7 with the whole pyramid

so do i need to raise the 7?
( 7^n ) and can i then apply 4/7^n=x/686 ?

so how do i get the n in this situation ?

because i need volume do i do this :

,rotate

[supple fog]

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[obtuse pebble]

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[undone plume]

help

i didnt get any of these answers 😢

[lapis charm]

just calculate the antiderivative of the answers and see which one matches 🙂

[undone plume]

whats an anti derivative

[lapis charm]

integral

but im just kidding

[undone plume]

i got this

it looks wrong 😞

HELP!

[amber dirge]

Personally I'd rewrite it in terms of sin and cos

What are sec and tan in terms of sin and cos

[undone plume]

1/sec

1/sin and 1/cos

[amber dirge]

Well sec is 1/cos

Tan is sin/cos

So you can rewrite secxtanx

As?

[undone plume]

sin/cos

oh

hmmmm

[amber dirge]

$\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}$

[gleaming bronze]

Can anyone solve this problem (70n^3 + 16n^2 - 32) !help

[warm shale]

TayBee

[amber dirge]

!occupied

[obtuse pebble]

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).

[undone plume]

yes

[amber dirge]

So that's not sin/cos

Close

[undone plume]

that is SO SMART

[amber dirge]

But not quite

It becomes $\frac{\sin x}{\cos^2 x}$

Right?

[warm shale]

TayBee

[amber dirge]

So now you can simplify your fraction a bit

[undone plume]

yay

wait how

[amber dirge]

Well what is cosx divided by sinx/cos^2x

How do you divide by a fraction

[undone plume]

multiply….?

😢

[amber dirge]

Yeah flip it and multiply

So now you get

$\cos x \cdot \frac{\cos^2 x}{\sin x}$

[warm shale]

TayBee

[amber dirge]

Which is?

[undone plume]

1/suqnx

1/sinx

[amber dirge]

That's some wild magic you used to cancel the cos's lmao

$\frac{\cos^3 x}{\sin x}$

[warm shale]

TayBee

[undone plume]

WHAT

HOW

[amber dirge]

Well imagine cosx = Q

What is Q times Q^2

[undone plume]

1

ughh im so confused

[amber dirge]

No, you add the indices lol

It's Q^3

x times x is x squared yes?

[undone plume]

oh yeah

[amber dirge]

So x squares times x is x cubed

[undone plume]

yes

[amber dirge]

Excellento

So there you go

[undone plume]

and then x^4

[amber dirge]

Now you have a simple fraction (or should I say... quotient), to differentiate

Do you remember how to differentiate a fraction?

[undone plume]

wait where

yes

ugh i always get confused with log rules

[amber dirge]

$\frac{d}{dx} \left( \frac{\cos^3 x}{\sin x} \right)$

[warm shale]

TayBee

[amber dirge]

Differentiate that

[amber dirge]

If you end up using log rules it has gone wrong lmao

[undone plume]

-sinx(sinx) - cos^3(cosx) / sin^2x

-sin^2x - cos^4x / sin^2x

hooray

[amber dirge]

You're pretty close

[undone plume]

😍🎉

[amber dirge]

I think you differentiated cos^3x wrong

What do you get when you differentiate cos^3x

[undone plume]

wait i know

cosx^3

AND THEN

3cos^2x

[amber dirge]

yes good, 3cos^2x, multiplied by...?

Chain rule(tm)

[undone plume]

no its just x

[amber dirge]

No it's not, it's cosx XD

Let u=cosx

[undone plume]

oh yeah

[amber dirge]

d/dx u^3 = 3u^2 x du/dx

[undone plume]

-sos

-cos

-sin

oops

[amber dirge]

[undone plume]

yay

[amber dirge]

So we get du as $-3\cos^2 x\sin x$

[warm shale]

TayBee

[undone plume]

-sinx(sinx) - (3cos^2x)(-sinx)(cosx) / sin^2x

[amber dirge]

Where's this -sinx(sinx) on its own coming from

What's the quotient rule?

[undone plume]

thats the qutiont rule

[amber dirge]

(vdu - udv)/v^2?

[undone plume]

you said it was right 😢

[amber dirge]

$vdu = \sin x \cdot 3\cos^2 x \cdot (-\sin x)$

[warm shale]

TayBee

[amber dirge]

As sinx = v

And we just worked out du

Does that bit make sense?

[undone plume]

yes

[amber dirge]

Excellent, so that's vdu

[undone plume]

-sinx(sinx) - (3cos^2x)(-sinx)(cosx) / sin^2x

[amber dirge]

udv?

[undone plume]

because

sinx

[undone plume]

wait i meant the trig ratios 😭

[amber dirge]

udv = $\cos^3 x \cdot \cos x = \cos^4 x$

[warm shale]

TayBee

[undone plume]

yes

i agree

[amber dirge]

Okay mb just me misinterpreting your formatting tbf non latex fractions are confusing

but in any case my final fraction before I faff to get it in the form your Q wants is

$\frac{-3\sin^2 x \cos^2 x - \cos^4 x}{\sin^2 x}$

[warm shale]

TayBee

[amber dirge]

Do you agree with that fraction?

[undone plume]

yes

because the chain rule thing

now we can factor it out

[amber dirge]

Ye

[undone plume]

cos^2x(-3sin^2x-cos^2x)

over sin2x

hmmmm

but how do you get rid of the fraction

[amber dirge]

What is $\frac{\cos^2 x}{\sin^2 x}$?

[warm shale]

TayBee

[undone plume]

i dont know

[amber dirge]

Okay, what is sin/cos?

[undone plume]

tan

[amber dirge]

Good, so what is cos/sin?

[undone plume]

tan 2

x

yay

[amber dirge]

That would be sin^2/cos^2

So what's 1/tan

[undone plume]

tan^2x

oh

cot

cot^2x

🎉

[amber dirge]

[undone plume]

YAY

cot^2x(-3sin^2x-cos^2x)

[amber dirge]

Good work

[undone plume]

-cot^2x(3sin^2x+cos^2x)

hooray

[amber dirge]

Flip both signs and you'll be there ll

The -3sin^2x should become + as well

[undone plume]

YAY

oops

yay

[amber dirge]

[undone plume]

thank you

[amber dirge]

Np!

[undone plume]

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[wide flame]

Why we cross out -h with h when multiplying

[worn yoke]

-h = -1 * h

so they crossed out the h and left the -1

[wide flame]

So divison was involed

?

[merry peak]

We’re dividing here

Sk

So

Yeah

[wary badger]

tomato

[wide flame]

So we can cross cancel out for +,- and * and division?

[merry peak]

Addition?

[wide flame]

if the number are the same

[merry peak]

No

[wide flame]

nver

[dire plinth]

the - wasnt canceled

the -h was just split into -1 * h

then canceled the h

[merry peak]

h/h=1

Thats why we can cancel out h

[wide flame]

Thank You everyone

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[flint willow]

Find angle PTR

I don't think I can minus 8 with 4.5 since the TR is like bended

so I'm confused.

[gloomy vector]

it should be fine, since the image doesent always be to scale

[cosmic sonnet]

u can

you have all 3 sides of the triangle

you can use cos rule to find the angle

then find all other angles

pretty straightforward

[flint willow]

but if I minus

I'll get the triangle to a right angle triangle?

[cosmic sonnet]

no

no right angle triangle

do you know the cos rule

[flint willow]

yes

[cosmic sonnet]

ok

you know all 3 sides

one side is 5, one is 5.5 and one is 3.5

see if you can understand where i got those values

now you can use cos rule to find the angle

and from there its just rules of interior angles to find the rest

[gloomy vector]

i think you use the cosine rule for triangle RTQ

[flint willow]

nvm I see it now thx

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[honest narwhal]

Can someone help me please. I have no idea what I’m reading

[obtuse pebble]

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[honest narwhal]

<@&286206848099549185>

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[shrewd trench]

Hello, I understand this question but how would I solve it if it were to be

integral ( 12 sec(3x) tan(4x) dx )

[median dome]

There probably isn’t a clean answer for that

[alpine verge]

,w integral of 12sec(3x)tan(4x)

[warm shale]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by doctormysterio9

[median dome]

Yep

[alpine verge]

well

[median dome]

That one change makes the integral way more complex

[alpine verge]

yep

[modest ingot]

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[shrewd trench]

okay so I don't have to worry about it coming in an exam

[median dome]

Yes you don’t

[shrewd trench]

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[thin cave]

How I'm gon simplify this ?

[haughty coyote]

x verifies (x+2)^2 - 5 = 0

Can you make use of that ?

[thin cave]

Yea i saw similar with x² + 4x -1

[lyric grove]

Can we use binomial here

Yeah.. not a good idea💀

[thin cave]

Should i do a compare with x^2025 + 4x^2024 - x^2023 ?

[rich plume]

no

simply, x^2+4x=1

[haughty coyote]

In most such exercises, 2023 is a random number, so you should remove it / abstract it away
In this case you can factor out x^2023

[rich plume]

make use of this

[thin cave]

then 1 = x^2023 ?

[rich plume]

how would you come to that?

beizer's approach is a similar one

factor out x^2023

[haughty coyote]

x ≈ 0.2 mind you

[thin cave]

ohh yea x^2023 ( x^2 + 4x -1 )

[haughty coyote]

Which actually leads to a much simpler argument that
You're summing 3 minuscule things, so it has to be a minuscule result, very close to 0
The only possible answer is 0

[thin cave]

oh you're right

[thin cave]

I try 2^24 - 2^0

[haughty coyote]

Indeed

[thin cave]

Does it work anything for the rest steps ?

[haughty coyote]

That is a good start

[thin cave]

I just think that can't factor out anything

do i need to do something for the power 24 ?

[haughty coyote]

You nicely rewrote it as a difference of squares

A great way to exhibit some factors

[thin cave]

ignore the 1 as well ?

the 24 i can express as 2^3 x 3

[haughty coyote]

I didn't say that with the intention of it getting ignored

[modest ingot]

i used the fact that 2^6 = 64 and that 2^24 = (2^6)^4

so you consider 2^6 (mod (those answer choices) ) and see which one gives -1 or 1

[thin cave]

why 64 ?

the closer number with divisor ?

[haughty coyote]

If you factor it you'll see why

[thin cave]

(2^12 -1 )^2

factor again ?

[haughty coyote]

Factor correctly first

[thin cave]

after that, (2^12 -1 )(2^12 +1)

<@&286206848099549185>

@haughty coyote you still here ?

[haughty coyote]

There's another difference of squares

[thin cave]

you mean pull power 2 again right ?

[haughty coyote]

You would have been done 30 minutes ago if you just tried that instead of waiting for someone to tell you to try that

[thin cave]

so (2^6 -1 )^4

[haughty coyote]

How did that one happen to be ?

[haughty coyote]

How did you get that from this ?

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[winged gazelle]

i jsut wanna know

that can u assume c1 and c2 tocuhed when c1 is at thee turning point

if so , which bit support that statment

if it isnt turning point, then i got 9/4 x^4 -10x^2 + r^2 =0

can u use discriminant for that

or is doies it have to be in qudratic

[main rose]

I dont think you can. Lest I don't know exactly what you mean by turning point.

You can substitute x^2 = u in the quadratic you mentioned to solve it.

[obtuse pebble]

@winged gazelle Has your question been resolved?

[winged gazelle]

ok

like when dy/dx = 0 i mean

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[low horizon]

I need help with my homework 😭

[low horizon]

I don’t understand it at all😭

It’s homework on distribution, and properties of distribution

[raw torrent]

Just send it

[low horizon]

Noice

[obtuse pebble]

@low horizon Has your question been resolved?

[low horizon]

No

[thick torrent]

send a question people can help with not an entire pdf what

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[timid crane]

I need to verify if these are correct please, thank you and have a good day

[worn yoke]

looks fine

[timid crane]

Thank you

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[bold stag]

The bicycle that you can see in the picture.
Bigger wheel touches point B at the bottom and small wheel touches the point A at the bottom. The small wheel does 1 complete revolution to touch to point B.
The bicycle moves towards point C and the movement will end when both of the wheels touched the point C.
Both of the wheels does this movement with complete revolutions around their center (i.e the number of their revolutions are integers)
What is the distance between point A and B ?

[bold stag]

@vernal wren this is the translated form of the other question.

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[vocal juniper]

Hello guys, I recently did a test on forces, on the multiple choice I got this one wrong

[vocal juniper]

and like i double checked on google rn and they said 9.8 N

so like im just making sure im not crazy stupid

I put 9.8 N, but the test said 9.8 x 10 N

9.8 x 10 would be 98 N and that doesnt seem correct

it even said 1.0 kg so its not 10kg

if i get an answer and it is wrong i can ask my teacher abt this

[cobalt shuttle]

yeah it looks like you should be right

[vocal juniper]

ok ima email my teach

[cobalt shuttle]

good luck!

[vocal juniper]

thanks

[timid silo]

the answer is definetely wrong

you are right

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[obsidian scarab]

can i do this without the hint

[robust sleet]

yea

https://www.youtube.com/watch?v=RUpJUNjGmFI

[pseudo swift]

the coefficients of the matrix aren't constant here tho

maybe it works out idk

[obtuse pebble]

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[obsidian scarab]

Mmm lemme see

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[cedar rock]

really struggling here 😭 (not translating cuz it's rly not important to the actual problem lol)

[cedar rock]

I have to find the area of the light green stuff

so far i found the biggest sector's area but idk how to even begin finding the other 3 since there's no radius

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[cedar rock]

<@&286206848099549185>

[cedar rock]

<@&286206848099549185>

[cedar rock]

.close

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[vapid hedge]

hi

[vapid hedge]

need help with this

I think I am wrong not sure tho

<@&286206848099549185>

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[gloomy vector]

how many permutations are there to arrange 7 red cups, 4 blue cups, 3 yellow cups (all identical)

[gloomy vector]

i got 120120 but idk if its correct or not

i gor that from 14c7×7c4

[wraith wing]

Correct

[iron edge]

✅✅

[gloomy vector]

oh ok

Adam and Bob are playing a game, where they pick balls from a basket where inside of the basket there are 119 balls. They can pick 3-9 balls at once, they each get a turn and the person who takes the last ball will win the game, if Adam is going first and Adam can put balls in the basket. What is the minimum ammount of balls Adam should put to ensure victory?

[merry peak]

Work backwards

[astral silo]

put or pick?

[astral silo]

thats a bit long maybe? idk, i have a trick though

you find what number of balls you can always make from 2 picks when you are the 2nd person picking

[gloomy vector]

im not sure if this is correct, but if i was Adam my strategy would be picking 3 balls at the first round, and whatever Bob picks i try to pick the complimentary ammount so that they both sum up to 12 (so if bob takes 5 i take 7 for example), and i repeat that until it is finished, and since they all sum to 12 you can just say that the "complimentary" cycle thing will always lead to 12, so i got 4, which is 123 balls in the basket

[astral silo]

then you should see where this is going

[gloomy vector]

and if Adam at the start picks 3 that means that whatever Bob does adam will be the one who takes the last

[gloomy vector]

what do you mean?

[astral silo]

like put in or take out?

[gloomy vector]

they play a game where they take the balls out, but the question is saying that adam can put in balls before the game starts

and its asking how many balls to win

[astral silo]

if you put in then u can put like ur answer + 12n where n is a natural number

or put in so much that they can never finish and assume adam can live longer

[gloomy vector]

💀

[astral silo]

cuz there is an infinite amount of answers

[gloomy vector]

the minimum

[astral silo]

oh right

mb

then say -119

*nah ur correct

[gloomy vector]

ok ty

.close

[obtuse pebble]

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[gloomy vector]

.reopen

[obtuse pebble]

✅

[gloomy vector]

does integration rules follow the same as derivation rules (ie chain rule, product rule, ect)

[wary badger]

not quite

it’s backwards

there are techniques to identify product rule and chain rule

to go backwards

[gloomy vector]

uh

[wary badger]

u substitution for chain rule

integration by parts for product rule

power rule has reverse power rule

[gloomy vector]

for $\int \frac{f(x)}{g(x)} dx$ is there a formula?

[warm shale]

Skill_Issue

[wary badger]

well quotient is really just product in disguise but a lot of times there’s some sort of u sub where you let u=g(x)

a common integral is du/u

but other times you could use integration by parts

interpreting it as f(x) * 1/g(x)

[gloomy vector]

my friend wanted help with this but i dont know either, so im asking here

[gloomy vector]

oo ok

[orchid lake]

multiply and divide by cos²x

and you see that xsinx+cosx is the differentiation of xcosx

[oak grotto]

sheesh

[orchid lake]

i hope it helped

[gloomy vector]

so int of 1/cosx^2?

[obtuse pebble]

@gloomy vector Has your question been resolved?

[orchid lake]

yeah sec²x

[gloomy vector]

ok ty

.close

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[gritty portal]

Had no problem doing this by manipulating the inequality, no clue how to prove it using the figure provided though. Any insight anyone could share to help me get started?

[surreal forge]

those four elongated rectangles are meant to be congruent

[gritty portal]

yes they are

[surreal forge]

which means the composite figure is a square

[gritty portal]

that is true, yes

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[vast echo]

Is the answer 44? Every set of 2 adjecent vertices has a plane that passes through 2 other adjecent vertices, plus the planes that pass through each face making 12 total. There are 56 non ordered sets of 3 vertices, 12 of which will be part of one of those already counted 4 point planes, so it should be 44 total by my count

[vast echo]

But I am not confident in my skill

[foggy furnace]

yk i wish i could help you bc youre someone whos like here every day but idk how to do those problems

ping the helpers

[vast echo]

Hasn’t been 15 minutes

[foggy furnace]

ping em

[vast echo]

<@&286206848099549185>

[weary trench]

alright so you had the right idea by subtracting the overcounted planes

[vast echo]

If it’s 50 I’m gonna be so mad

[weary trench]

but you undercounted how much they are overcounted

[vast echo]

Wdym

[weary trench]

you should subtract more than 12

[vast echo]

Why

Wait

I think I see

[weary trench]

alright

[vast echo]

Each would be counted an extra 4 times

Right?

[weary trench]

would be counted 4 times, so an extra of 3 times, yes

[vast echo]

So 8?

[weary trench]

you overcounted each plane 3 times

and there are 12 planes with 4 vertices

[vast echo]

Let me explain my logic

[weary trench]

alright

[vast echo]

Take a plane that is intersecting the 4 vertices that make one of the faces

We include this in the count since it’s intersecting at least 3 vertices

So how when we look at the sets of 3 points, there’s 4 that also result in this plane, one for each vertice missing from the face

So we overcount it by 4

[weary trench]

yep

i think there's just a terminology miscommunication here

[vast echo]

OH I SEE

[weary trench]

you counted 4 times so you overcounted 3 times

hahah

[vast echo]

Yeah lol

So 20

[weary trench]

there we go

[vast echo]

Very hard problem

I think I did pretty damn well there overall

[weary trench]

mhm just gotta be careful where you overcounted

my biggest problem with combinatorics in general lol

[vast echo]

Mhm

Very easy to think you have the right answer when you don’t

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[tawdry pewter]

i dont understand how they got there

so i ended up with 4x^2*e^x^3

okay so now i dont know how to integrate e^x^3 that thing gives u gamma lol

why did they take the two outside and the 2/3?

[red ice]

It's the chain rule

[tawdry pewter]

how does e becomes e^u

[red ice]

So recall that d/dx x^3 = 3x^2

[tawdry pewter]

yes

[red ice]

If we had 3x^2 (e^(x^3)) that would be easy now

Using u = x^3, du = 3x^2 dx

[tawdry pewter]

yes

i am following

so i am thinking it should be (e^x^3)/3x^2?

but that wont work

haha

[red ice]

So now we can multiply this by 4/3 * 3/4

[red ice]

Ah wait you should start with 2x^2 * e^(x^3) according to the solution

Anyways $\int \frac{2}{3} \frac{3}{2} 2x^2 e^{x^3} \ dx$

[warm shale]

south

[red ice]

$= \frac{2}{3} \int 3x^2 e^{x^3} \ dx$

[warm shale]

south

[red ice]

Take the 3/2 and multiply 3/2 * 2

[tawdry pewter]

so the 3x^2 comes from the integration of e^x^3

[red ice]

Yes

[tawdry pewter]

oh that makes sense so for 4x^2/3 we can sub in the limits

and move it outside

[red ice]

Yes, but it's not 4x^2/3 but 2x^2/3

[tawdry pewter]

why?

its y^2

[red ice]

The reason why is that $\frac{y^2}{2} = \frac{(2x)^2}{2} = 2x^2$

[warm shale]

south

[tawdry pewter]

OH crap

lol

[red ice]

ye

[tawdry pewter]

i missed the /2

make sense cheers boss

.close

[obtuse pebble]

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[tawdry pewter]

wait

where did the fraction come from?

3x^2e^x^3

3/2

no idea where that came from?

isnt the integration and diff of e^x^3

the same?

.reopen

[obtuse pebble]

✅

[red ice]

We want to multiply this 2 x^2 e^(x^3) by 3/2 to get a chain rule form right

But then we have to multiply by 2/3 so that 3/2 * 2/3 = 1

And we don't end up changing the integral

[tawdry pewter]

where did the 3/2 come from?

let me show u my wokring out

the 3*2/3 = 2

she got rid of 3x^2

[obtuse pebble]

@tawdry pewter Has your question been resolved?

[obtuse pebble]

@tawdry pewter Has your question been resolved?

[obtuse pebble]

@tawdry pewter Has your question been resolved?

[obtuse pebble]

@tawdry pewter Has your question been resolved?

[tawdry pewter]

The actually answer is 1/4

Why??

[fleet sparrow]

It’s not

[tawdry pewter]

Well, the answers is

1/4(e^-3t - e^-7t)

[obtuse pebble]

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[obtuse pebble]

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[west current]

i was absent for a while and i'm not sure what this question is asking me... could somebody explain it?

[near cairn]

The question is asking you to rewrite the expressions so that they are of the form a* u^n

Where ‘a’ is a constant I assume

@west current

[west current]

sorry! didn't see the messages... I was assuming it has something to do with the chain rule but i'm not sure

[near cairn]

Np👍

[west current]

but then if i use the chain rule on that

what would be the variables

[near cairn]

You don’t need to use the chain rule

You’re just rewriting the expression so it satisfies that format

[west current]

ohh okay

thank you

[obtuse pebble]

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[next crest]

How to do this one?

[next crest]

I found that the triangles are similar but cand find anything past that

q_q

[next crest]

ah, so 12 is to 2x+1 as 5 is to x?

so x = (10x + 5)/12

so x/6 = 5/12 so x = 5/2?

[cunning lava]

yes

[obtuse pebble]

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[tawny kite]

yalll im backkkk

[tawny kite]

after a long break

i dont have questions lol im staying in the channel again tho

@unreal musk im back yayy

[unreal musk]

You're back where did you get to

[unreal musk]

Naughty

[tawny kite]

lol so basically I was gone for i think like a week

i have no idea

i just didnt do math in a longg time cause im also in this software dev course

me and two other people are developing a game

and we're finally done!!!

lol it was like days in the compsci room on our campus

we have other people in the group but they sorta ditched

entire game is self made

i made the sprites and made the soundtracks

another person also made some tracks

but ye

im gonna catch up on math tho.. im really behind😅

[modern onyx]

comp sci major??

[tawny kite]

not sure yet tbh.. i was thinking about comp sci minor

but I really liked the group work and collaboration in compsci

theres like no group work in math lol

[modern onyx]

damn are u undecalred rn?

[tawny kite]

i just compare the answers sometimes with friends

well like rn I have all the credits for both comp sci and math

[modern onyx]

how does switching majors even work

[unreal musk]

Awwww, finally

[unreal musk]

Sounds so cool

[modern onyx]

What game did u make?

[tawny kite]

you own a shop and try to earn money

you lose if you go bankrupt

win if you make a profit after certain amount of days

[modern onyx]

so gamblin

g

[tawny kite]

ye

[modern onyx]

what languae u use

Javascript?

[tawny kite]

its gonna sound really stupid but since its a simple game we didnt use any frameworks at all

just plain old java and javaswing

[modern onyx]

shit javaswing

idk what that is

[tawny kite]

its simple anyways and we dont need databases either

we stored data on excel lmao

[modern onyx]

😟

idk what that is

[tawny kite]

thats ok

i barely know what im doing half the time

@unreal musk you can attest to this right?😂

[unreal musk]

well

[unreal musk]

You get there in the end at least

[tawny kite]

yea its a learning process if im some fcking genuis im dropping out of school lmao

[modern onyx]

its not 😭 Im trying to major in cs

[tawny kite]

trust you're fine

i didnt know what an integral was two years ago

lol you're chillin

im just exposing myself here

[modern onyx]

R u freshmen in college?

[tawny kite]

but yea back to math

[tawny kite]

Im not gonna tell the whole world that info yall can guess

all I can say is that for a fact, I am past the age for general employment in my country lmao

yall take your guesses

nothing else im saying

[modern onyx]

My fault gang

[tawny kite]

i know very informative info right there😎

[obtuse pebble]

@tawny kite Has your question been resolved?

[tawny kite]

with that being said, someone let me know if their company is willing to adopt a person with minimal math and cs knowledge for an internship

perks being, im gonna try my best not to bug yall higher ups with questions and annoying convos

I know yall get enough of that at home if you have kids, perhaps your significant other

jkjk

I can tell you all how I religiously try to "save" money

some will call it spending

but same thinggggg

no but trust i think im good with my finances😊

[obtuse pebble]

@tawny kite Has your question been resolved?

[tawny kite]

currently doing calc

calc is good cause things are usually pretty self explanatory

textbook will have examples to refer to

[obtuse pebble]

@tawny kite Has your question been resolved?

[tawny kite]

no questions yet..

maybe i should just close this channel

[past tendon]

yeah

let other people use it

[tawny kite]

ok

.close

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[gilded needle]

surely there must be more information given

the error could be anything

[lament cairn]

I did the upper bound and lower bound making it 2.565 and 2.575

[lament cairn]

Percent error

[high lily]

what's the difference between those and the measured value

[ruby path]

you don't have a least count?

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[livid onyx]

I need help with this question

[livid onyx]

What should I do after this

[obtuse pebble]

@livid onyx Has your question been resolved?

[obtuse pebble]

@livid onyx Has your question been resolved?

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[placid forge]

question iv)

why is the answer 7/12 and not 5/12 ?

what did i do wrong?

[obtuse pebble]

@placid forge Has your question been resolved?

[placid forge]

<@&286206848099549185>

[obtuse pebble]

@placid forge Has your question been resolved?

[lament cairn]

Thanks

[obtuse pebble]

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[dense lodge]

Bit lost

[pseudo swift]

well how do you check if a matrix is singular ?

[versed stratus]

hint : The Det of a singular matrix is 0

[dense lodge]

Ohhhhhh

Wait so you can just say determinant would be zero and that’s enough?

[pseudo swift]

well you'd have to argue why such a matrix has determinant 0, but it's not too hard

[dense lodge]

Hm interesting

The is my profs answer they’ve done it differently?

[pseudo swift]

I mean that works

they're showing A can have no inverse

[dense lodge]

Ohhh

I kinda get it now

Thanks

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[gleaming karma]

hi! could someone briefly explain what this means?

[gleaming karma]

i mean the second part of the relation

alright nvm i close it, i understood it wrong

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[hushed cairn]

How to approach the rest of this question? I solved a system of equations to get those numbers but idk how to continue

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[wise magnet]

i don't get how he went from the first part to the second

[timid silo]

nvm

[pastel wren]

lnx got factored out

[timid silo]

yes

[wise magnet]

oh

i see

thanks

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[bronze citrus]

how do I do this?

[obtuse pebble]

@bronze citrus Has your question been resolved?

[untold raven]

Are you allowed a calculator

[bronze citrus]

not in exam no, but I just need the method rn, I can do the calculations

[untold raven]

Maybe try partial fraction decomposition

[bronze citrus]

how would you decompose this though

[obtuse pebble]

@bronze citrus Has your question been resolved?

[surreal forge]

i would probably start by noticing something like this

then bounding (||for integers m, n; m = 1 or 2; n = any upper bound of your choosing||)

[bronze citrus]

how did you get this AM <= GM?

[surreal forge]

(the "less than" portion is not actually that important and is only included for clarity
i could have just written ≈ without much loss of meaning)

[bronze citrus]

yes but how did you get 3k^1/3?

[surreal forge]

limit comparison

as k gets fairly large, the constant -1, 0, +1 inside the ^(4/3) powers become negligible

[bronze citrus]

oh yes

[timid silo]

Hey actually Im just entering 12 th but I was still trying this question like thinking its one of sequence and series question so i just wanna know is it? Or are there any 12th topics involved as well

[bronze citrus]

are you indian?

[timid silo]

Yup

[bronze citrus]

guessed it, its a JEE question

[timid silo]

Ik saw it as thumb nail above the img

[bronze citrus]

thats watermark not thumbnail bro lol

so its basically tricky for a normal school 12th student

although I cant solve it too

[bronze citrus]

bro you need to explain how you got this too please

[timid silo]

Dw im a jee aspirant too and sry i just dont use my phone much so got mistaken with watermark and thumbnail

Like for that quesion

I wanna know

Do we need some 12th topics as well

?

To solve it

[bronze citrus]

I dont have the solution, so I dont know

thats why am asking it here

[timid silo]

Which class are u in

[bronze citrus]

13

[timid silo]

Drop?

[surreal forge]

each individual term can be bounded as follows

[bronze citrus]

yes

[surreal forge]

then via integration the above result is derived

[bronze citrus]

hm I think I get it

I will try putting in the limits

[timid silo]

Hey man can ya explain how ur introducing inequalities

[bronze citrus]

its better to move this to dms

[timid silo]

Sure

[bronze citrus]

something is coming off, like I am getting

so ultimately its saying

which is obv not true

[surreal forge]

i believe it is the third term that has the error

[bronze citrus]

these are correct though

I think my calculations might have a error

lemme check them once more

oh wait

I made a mistake

its actually this

wait so the answer is option d for the s999 part

[bronze citrus]

the second inequality for s26 is coming like

3.88 < s26 < 4.27
and the options are s26 <4.25
s26 > 4.25

I think since we cant say for sure so correct option must be d

thank you @surreal forge

[surreal forge]

how did u come to this conclusion?

[bronze citrus]

which conclusion?

that only option d is correct?

[surreal forge]

yes

[bronze citrus]

I made a mistake earlier, I wrote it here, this was the correct form

[surreal forge]

just by contradictory answer choices, we see that
either A or B is correct... And either C or D is correct

[bronze citrus]

so it was like

97. something < s999 < 98. something

[surreal forge]

so one of A or B is also correct

[bronze citrus]

so s999 > 50

[bronze citrus]

but by the inequality we cant say for sure which is correct

[surreal forge]

that doesn't mean neither is correct

[bronze citrus]

well

yeah ur right

[surreal forge]

this is correct

[bronze citrus]

how about a second method
I was thinking of a some kind of progression if we put in values

[surreal forge]

is not necessary

[surreal forge]

u have 3.88 < s26 < 4.27 right

[bronze citrus]

yes

[surreal forge]

then it is extremely reasonable to assume that s26 < 4.25

[bronze citrus]

hmm

[surreal forge]

because of how close the bound is to the upper bound u know for sure is correct

[bronze citrus]

but I am just wondering if the person who developed this question wanted to this question this way

is there any other ways to approch this?

[bronze citrus]

its reasonable but....yk the maker of the question could have given 4.3 if he came the same answer as we did

by this method

[surreal forge]

well maybe better method exists

but i gave u this method bcz i still thought it was sufficiently good enough for a student to confidently arrive at a solution to the problem

for example u can probably prove this rigorously with generalized partials of harmonic series but that seems extremely unnecessary when this relatively simple approach exists

[bronze citrus]

its very clever and easy way to do this for sure

[surreal forge]

(idk why they chose 17/4 it seems weird to me too)

[bronze citrus]

yeah

but thanks for this method though

we reached somewhere 99% to the answer

[surreal forge]

i think i would check ur work here

maybe try bounding with n = 999 instead of n = 998

[bronze citrus]

why thoguh

that seems correct for the options

[surreal forge]

actually n = 1000 makes this more obvious do this instead

[bronze citrus]

oh for approx calculation?

without calculator you mean?

or else it would give the inequality for s1000

we need to solve for s999

[surreal forge]

in here

[surreal forge]

u should try it i am telling u because the answer u have is wrong

choice D is not correct and u will immediately see why if u try n = 1000 into the inequality

[bronze citrus]

oh its wrong?

[bronze citrus]

OH wait my mistake, I forgot to divide it by 2

its coming out to be

49.46 < s999 < 49.96

[surreal forge]

and u still did not try it

[bronze citrus]

bro I tried it

[surreal forge]

i am telling u this bcz 1/2 * 1000^2/3 = 50 exactly

[bronze citrus]

thats why I calculated for 1000 and 998

[surreal forge]

then S1000 is less than 50 which implies S999 is less than 50

[bronze citrus]

saw this shift

[bronze citrus]

#bots message

see this

you can scroll up and down

[surreal forge]

thanks then

[bronze citrus]

ok but the inequality I just sent here

49.46 < s999 < 49.96

shows its less than 50 thogh

[surreal forge]

in actual exam u say u do not have calculator

[bronze citrus]

BRO WHY YOU THANKING ME WTF

I should thank you

[bronze citrus]

well yeah

[surreal forge]

so like 1000^2/3 u can compute immediately without wasting much time

and u can numerically approximate 25^2/3 and see it is close to 17/2

[bronze citrus]

so you were telling me to compare with s1000 to get that s999 is less than that 50 right?

[surreal forge]

that is idea

[bronze citrus]

yep yep I get you

[surreal forge]

reason that is considered is bcz upon reading the question u might notice that 26 and 999 are both 1 less than a perfect cube

trying n + 1 fails for n = 26, but succeeds for n = 999

[surreal forge]

cool

[bronze citrus]

hmmm

thats sus

[bronze citrus]

yea

[surreal forge]

i forget how long students have for JEE maths

[bronze citrus]

for JEE mains (1st stage paper) you get 20 single mcq + do 5 question out of 10 choice of integer answer question
and you have 1 hour for each subject, so you can save some time on chemistry and give maths like 1 hour 15 mins or 1 hour 30 mins

for JEE advanced (2nd stage paper) from which the question is likely, you have to do 6 questions out of 19 in each paper-1 and paper-2, you have 3 hours in each of the papers

so you can give 10 mins for each question basically

to pass the minimum cutoff

if you want a good college you have 5~7 mins/question

[surreal forge]

hm interesting

[bronze citrus]

thanks for the help

I will close the channel now

[surreal forge]

welcome

[bronze citrus]

.close

[obtuse pebble]

Channel closed

Closed by @bronze citrus

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[muted ember]

Hello, does anyone know how to interpret the 6 trigonometric ratios on the trigonometric circle?

[muted ember]

I only understand sine and cosine.

I don't understand the other four.

[obtuse pebble]

@muted ember Has your question been resolved?

[tall smelt]

Tan is y/x, csc is r/y, sec is r/x, cot is x/y

If that’s what you’re asking

[muted ember]

that's not my question, I don't understand the geometrical interpretation

[tall smelt]

Tan=Horizontal/vertical. Like that?

[muted ember]

no, for example the line of tan(x) what does it mean? the result of tan(x) is the length of that line up to the x-axis?

why that line is the cot?

etc..

[obtuse pebble]

@muted ember Has your question been resolved?

[muted ember]

<@&286206848099549185>

[obtuse pebble]

@muted ember Has your question been resolved?

[echo heron]

It means the value of tanθ is the length of the pink segment which is tangent to the unit circle and extended to meet the x-axis

(Of course there is a caveat here as the tanθ may be negative in some cases where θ is not acute; in any case, the length of the pink segment would correspond to the magnitude of tanθ)

@muted ember

Do you understand and agree with what I said

[muted ember]

yes

and the other 3 trigonometric ratios?

[echo heron]

Yes, they would correspond with the lengths of the coloured segments as well

Do you need me to explain why that is true, or are you satisfied with knowing what the illustration is depicting?

[muted ember]

is true

why the secant is that line, the cotangent, etc.

[echo heron]

For tan, if you use the definition tan=opp/adj and apply that to the triangle OBC, then you get tanθ=BC/OB, but OB = 1, so tanθ=BC

The cotangent case is similar but you use the triangle OBA instead

For secant if you use the definition sec=hyp/adj on triangle OBC you get secθ=OC/OB and again because OB=1 you just have secθ=OC

and so on

[muted ember]

Is the OBD triangle the same as the BDC triangle?

[obtuse pebble]

@muted ember Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[plain sparrow]

The area of the base of the cylinder is proportional to the area of its axial cross-section as π:4, find the angle between the diagonals of the axial cross-section

[plain sparrow]

<@&286206848099549185>

[wintry hearth]

hey

[plain sparrow]

hey

[wintry hearth]

i think the angle is 90 degerees

[plain sparrow]

can u show me drawing?

[wintry hearth]

the diagram?

[plain sparrow]

yeah

[wintry hearth]

i can type the explanation , just tell me if u can understand ig?

[plain sparrow]

ok explain

[wintry hearth]

yea so

let

radious of cylinder be r

i hope u know how the axial cross section is like?

[plain sparrow]

yes

[wintry hearth]

yea ok

let the diaonal of the cross section be d

the area of the base of the cylinder is propertiont to pi r^2 (area). and area of axial cross section = 1/2 * pi * r * d

understood till now?

[plain sparrow]

yes

[wintry hearth]

ok then

ratio of areas is pi:4 (given)

so

pi * r^2/ 1/2 * pi * r* d = pi/4

understand?

[plain sparrow]

yes

[wintry hearth]

by cutting common variables

we get

d = 8*r/pi

[plain sparrow]

yes

[wintry hearth]

let us now assume the cylinder is lying flat ok? so the axial cross section forms a rectangle

ok?

[plain sparrow]

ok

[wintry hearth]

so then the axial cross section forms rectangle and diaognals of rectangle intersct at 90 degere. so the angle between the diaonglas of the axial cross section is 90 degrees.

thus it is found