help-39

[tardy reef]

The value of A is 14

But the instant at which you calculate the rate of decrease is when the voltage has dropped to 8V

[livid bolt]

huh

[tardy reef]

This thing that you did

You evaluated V(t) = 8

But in their solution they dont need to calculate that

You solved it to find the time

but since their method directly uses the voltage, they dont need to do any other calculations. They can just plug in that 8

[livid bolt]

but to find the rate of change they need the derivative and the derivative has 14

[tardy reef]

wdym?

[livid bolt]

isnt dis the drrivstive

[tardy reef]

Yes it is

Can you calculate V' / V

[livid bolt]

so wouldnt it be like dis

[tardy reef]

No

[livid bolt]

wait i forgot da k

like dat?

[tardy reef]

Like this

[livid bolt]

OHHHHHHHHHH

OK thanks

damn

wait i have one more question

i cannot for the life of me get this

what al i doing wrong

[tardy reef]

What is dP/dr when the perimeter is minimum

[livid bolt]

wait no part A

[tardy reef]

oh okok

mb

You took r as the diameter 💀

AE = BD = 2r

[livid bolt]

isnt that for the circumference

[tardy reef]

[livid bolt]

oh..

oh..

i see..

...

[tardy reef]

BD is the diameter of semicircle

[livid bolt]

ok thats not fair because

theu shouldve stated r was the radius

you know what i mean

[livid bolt]

...

ok bye

.close

[pearl pond]

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[pearl pond]

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[rich thistle]

did I do something wrong when I integrated it? The textbook answer says it is pi/4

[frozen lantern]

not in the integration

[rich thistle]

this is what the worked sol do

[rich thistle]

whys it that when i dont use u sub i get a diff answer

[frozen lantern]

it is indeed ln 2

either you checked the wrong answer or your book is insane

[rich thistle]

XD

i hope so

okay then thx for the confirmation

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[inland ivy]

what the hell 😭

u = sin^2(x)
1/(1+u^2)

[frozen lantern]

the

okay so what happened here is that they subbed u= sin^2(x) but wrote u^2 in the denom instead of just u

which happens but shouldnt in a book lolol

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[hallow falcon]

Hello, I need help with one maths problem regarding trigonometry in cuboids

[frail temple]

do you have the quesstion?

[hallow falcon]

Hang on

I'm just not sure how to calculate the angle if it is not already in a right angled triangle. The only thing I did was calculate the length of CG

But I just can't see how that would help me - without the right angle

[chrome patio]

!showwork

[pearl pond]

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

[chrome patio]

especially your calculations for CG

[chrome patio]

,rccw

[hallow falcon]

That's what I've got so far

[chrome patio]

the question statement has a typo
you're using the same symbol "H" for two different things. that's not a good practice in math
that should be avoided in any topic and any level of math

cuz you don't wanna confuse your markers, who have to grade hundreds of questions
if you're not patient enough to read into your solution, you'll risk losing marks

just write the name of the segments in the ratio of segment lengths

[hallow falcon]

Okay I'll fix that

[chrome patio]

a general problem solving mindset:
if you're blocked, think about what given condition(s) that you haven't used

[hallow falcon]

You mean like the info in the instructions I haven't used yet?

So, we've got GD - 6cm

[chrome patio]

in a question statement, each given condition has a "reason" to exist, in general

[hallow falcon]

Yeah, I figured

But I just don't see how that can help - in order to calculate the angle we need a right angled triangle somewhere, right?

[dusky willow]

he mean like find Everthing u can find

With the question’s given information

CH and HD would be useful

[chrome patio]

the above two mentioned segments nice exercises, and provide an optional pathway to the desired solution

[hallow falcon]

Sorry, what are segments? English is not my first language when it comes to math terminology

[chrome patio]

But I just don't see how that can help - in order to calculate the angle we need a right angled triangle somewhere, right?
to solve a puzzle, you often have to work from both ends, in two opposite directions

1. "forward" direction: from the "start", use the given conditions
2. "backward" direction: from the "end", also use the given conditions

• to derive equivalent statements
• to derive some necessary conditions for the desired result to hold, so that we can eliminate the cases that don't satisfy those necessary conditions

[hallow falcon]

Okay so I've got the length of HD

What about cH? How do I get that

[dusky willow]

What do u think

[chrome patio]

you can start thinking from the "backward" direction:
the question asks for "angle XYZ"
so that does "angle XYZ" means?
||angle between segments XY and YZ. then you can look into these segments, and ask yourself again, which given condition(s) you haven't used, so that you're blocked?||

[hallow falcon]

Well either way we calculate, we're missing info. If we do a traingle of the base HD we only know 90 degrees. We still need more. And if we did it as AH being the base - same thing, unless the sides are squares, in which case we may get 45°. (Maybe?)

Sorry if I look stupid I'm not a math genious

[hallow falcon]

Well, the 6cm are still quite unused. I think

I'm sure this problem is really simple and I'm just overthinking it

Well 6cm is only useful in two ways, as GD and AH

But GD isn't in anyway connected to CH ?

[chrome patio]

no they're disjoint

[hallow falcon]

What does that mean?

[hallow falcon]

Okay, so GD and CH are disjoint

[chrome patio]

one more condition that you haven't exploited enough

try really diving into my "backward reasoning"

by analysing the literal meaning of the required answer

[hallow falcon]

We've only been given 3 conditions : 21cm, 6cm, and 12°. We've not used 6cm a lot. Could 21 be helpful again?

[chrome patio]

1. try to write in complete sentences (i.e. sentences with complete mathematical meaning) that would make your messages easier to be read (by others and by yourself when you're revising it later)
2. again, look into what's being asked, and try to interpret that, and find some relevant idea(s)

[hallow falcon]

Okay

[pearl pond]

@hallow falcon Has your question been resolved?

[hallow falcon]

Okay so I'll leave it unsolved for now and ask about it to my teacher at school - in my native language which might help a bit. But I want to thank you for your time

[pearl pond]

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[rich thistle]

for d, i get why the bounds can be 1 to 0 but why cant the bounds be 1 to -1

[toxic lichen]

sketch the region?

along with the rotation axis (important)

[dusky willow]

(Sorry just curious how to get -1)

[rich thistle]

yea like when i imagine it rotating along the y axis i imagine both bounds having the same volume

[toxic lichen]

ok well heres the thing

when you rotate around the y-axis

both the left half and the right half sweep out the same region in space

but the trouble is, you're gonna end up sweeping it out twice over

so you get double its volume at the end

that is why you only take one piece!

[rich thistle]

ahhh

ahhhh

i see

thx for the help

.close

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[daring bay]

hi

[daring bay]

so when u have a parallelogram

are the points always labeeld like this

[rough forge]

i dont know of such law

so not necessarily

[daring bay]

oh

okay

because i think i did it wrong in the exam

anyways

so

could u theoretically do

[midnight haven]

Which?

[daring bay]

B-A = C-D

[midnight haven]

AB = CD, AD = BC

[daring bay]

yes but for vectors

[daring bay]

like here

to get to point D

lets say point D is unknown

Couldnt you solve for D then

[midnight haven]

OH

Yes yes it’s true

If you have the rest of the points you can find D if that’s what you’re asking

[daring bay]

yes

i did it wrong in the exam i think

[midnight haven]

Do you have it to show or you don’t need?

[daring bay]

yeah

ok wait

@rough forge dude

😭 😭

ich hav vergessen

so here

i did AB = DC

[rough forge]

seems right

[daring bay]

hello

[daring bay]

hmm

but

also

B-A = C-D

right

[midnight haven]

If you do a -b = c - d I think it works

[daring bay]

wasnt it like this

[rough forge]

Now solve for C

[daring bay]

yeah ok

but wait

woher weiß ich ob

Like how do i know where A B and C is generally

or D

[midnight haven]

Hold on man idk German

[rough forge]

make a rough sketch and from there a good guess

https://www.youtube.com/watch?v=vAKyBrilD78

[midnight haven]

I mean if you need to have it in order ABCD which makes the most sense it’ll be b - a = c - d

[rough forge]

wohl gemerkt, there are many way to construct a parallelogramm

i think das video will helfen

[midnight haven]

You’ll get for C (13,10,12) which makes sense because a and d it’s +1,+2 and +8 difference and so will be c and b

[midnight haven]

Alr

[daring bay]

hm ok

i got an exam firday

so i gotta imagine it in my head

[rough forge]

🔥 day

[midnight haven]

Goodluck

I have exam on Sunday

[daring bay]

?

yeah

?

[midnight haven]

Imagine what?

[daring bay]

like

the points

[rough forge]

bro schau video

du must nichts imaginieren

[daring bay]

ok später danke

ok danke

[pearl pond]

@daring bay Has your question been resolved?

[pearl pond]

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[copper yoke]

Is this right?

[fathom oasis]

I wouldn't say so no

[copper yoke]

Ok! Which parts are incorrect?

[prime bramble]

the second step looks off

[fathom oasis]

Maybe you should solve the integral on pen and paper and see what order you do stuff in

instead of guessing

[copper yoke]

How’s this

[fathom oasis]

that looks better

[copper yoke]

Thank you!

[pearl pond]

@copper yoke Has your question been resolved?

[pearl pond]

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[dawn trout]

hey

[dawn trout]

is the ans 0?

[shy fable]

So its infinity minus infinity which is indeterminate

So you would need to solve it how you would approach an infinity minus infinity problem for limits

[dawn trout]

how can i do that

[shy fable]

Organic Chemistry tutor indeterminate forms guide

I mean lets do some thinking though if you dont wanna watch that

What you can do is multiply by the conjugate

And then take the limit of that

What is the conjugate of (n^2 + 2n)^1/2 - n’s conjugate?

[shy fable]

Hi

Basically like a special multiple of 1

[dawn trout]

wouldnt that become complicated cuz theres a -n

can i divide by n instead?

[shy fable]

Nah dont do that

Do the conjugate

But you can try divide by n and multiply by n

And try l’hopitals

But the conjugate will make it easier

Imo

[dawn trout]

ok

[shy fable]

Sometimes the problem needs to get harder to be solveable

[nimble socket]

do you know what a conjugate is?

[shy fable]

I was about to ask that too

[dawn trout]

lmao ofc its like the opp sign

[nimble socket]

show us

[shy fable]

Ig so send us the conjugate

[dawn trout]

(n^2 +2n)^1/2 + n

is the ans gonna be 1?

[nimble socket]

is it?

calculate it

see what you get

[shy fable]

Yea expand it

Show us your work if you can

[pearl pond]

@dawn trout Has your question been resolved?

[pearl pond]

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[calm wing]

broski it's almost 11pm

i ain't studying at this point

but sure let's take a look

pretty sure this part is wrong

oh

fair enough

i have doctors tomorrow

i'll come after lunch i think

yeah i just come for friends and studying

[wet osprey]

How do you know I-A^t = A - I on line 2

Oh you’re just changing the left side and rewriting the right side from the top

What’s happening on the third line

[pearl pond]

@naive marlin Has your question been resolved?

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[strong current]

can someone explain why the order of this differential equation is 1 instead of 2?

[strong current]

does there need to be parenthesis around the ^(2) in order for it to be order of 2?

$(dy/dt)^(2)$

[jolly parrot]

ScriptedEli

[jovial kiln]

what do you think an order is?

for derivatives?

[barren shell]

The order depends on the derivative order

[strong current]

it's the order of teh derivative

[barren shell]

Not a power of the derivative

[strong current]

for f(x) would be 1st order

[jovial kiln]

the degree and order are different–the degree being the highest "power"

[strong current]

f'(x) would be 2nd

etc.

[jovial kiln]

f(x) is 0th order

[strong current]

oh

whoops yeah

[snow sail]

this is a confusion over notation maybe

[jovial kiln]

d^2y/dx^2 would be second order

[strong current]

but why is it not 2nd order? since it's squared?

[barren shell]

This is first derivative squared

It is of order 1

[snow sail]

$\qty( \dv{y}{x} )^2 \neq \dv[2]{y}{x}$ but $\dv[2]{y}{x} = \dv x \dv{y}{x}$

[jolly parrot]

jan Niku

[strong current]

yeah i think so

like this is 5th order right?

because (5)?

[barren shell]

Indeed

(5) means fifth derivative

[strong current]

but the (dy/dx)^2 is not 2nd order?

[barren shell]

A common notation at least

[snow sail]

no

[strong current]

it would have to be (2)?

[snow sail]

(2) isnt used

[barren shell]

This is first derivative squared

[snow sail]

normally

[barren shell]

(y')²

[jovial kiln]

you start using () after the third order

[strong current]

yeah gotcha

[jovial kiln]

but for leibniz notation, i dont think u use it at all

[strong current]

oh okay

you would just do $d^3y/dx^3$

[jolly parrot]

ScriptedEli

[strong current]

?

[jovial kiln]

yes

[strong current]

oh okay

so just a notation thing

[jovial kiln]

mhm

[strong current]

okay this helps a lot, thank you!

.close

[pearl pond]

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[tight hamlet]

hi

[pearl pond]

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[rare jay]

Hey I was wondering how to set the bounds of intergration for these integrals In the answers they just do it and ther elogic doesnt make sense I get if theres a photo therir sometimes but if a differnt case is there a method to use to identify what to do

[marble sigil]

what other cases are you thinking of? here it's 0 to pi because you only need half a slice to spin around in 3d, and 2 to 0 because that's the value for the u-substition at the previous integral bounds

[rare jay]

right but if I didnt have the drawnig how would I know that

[marble sigil]

for a lot of polar problems you either need to find where the function hits 0 or draw a picture to know what a slice is

[rare jay]

or could I deduce it

right

okay

do u mind if I try find another example real quick

ill be fast!

[marble sigil]

yea that's fine

[rare jay]

Ahh ofc I cant find one

bare wit hme!

[marble sigil]

oh full 3d triple integrals will look different from the slice method yea

you don't need to try and draw a slice you just want the bounds of the radius, z, and angles

[rare jay]

actually sorry I think I phrased my quesitonwrong my understanding isnt the best

I think this is a better question

what is splitting and like why do we do hence when does it need to be done

[marble sigil]

some shapes require two functions to describe

like here the radius has one function for most of it, but get cuts off at the top short

for a) it's because as you go from small to big r, the z is either between the flat top and bottom curve, or two curves, and those are different formulas

[pearl pond]

@rare jay Has your question been resolved?

[rare jay]

right

and r will be a function of theta

[pearl pond]

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[hybrid trail]

,w lim x to infinity (5x^1/2 + 1)/(x^1/2 -7x)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by paulonsteam

[hybrid trail]

how do I solve this?

wait

[jovial kiln]

divide numerator and denominator with highest degree term

[nimble socket]

(denominator overpowers numerator)

[hybrid trail]

which is x right

[tight hamlet]

it's easy

[jovial kiln]

yes indeed

[hybrid trail]

so u get 0 + 0 / -7 + 0?

I was looking at the 1/2 thinking that was the highest term 😐

okay thanks everyone

.close

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[hybrid trail]

f(x) = (sqrt(x^2 + 4x)) / (-x-4)

[hybrid trail]

use lim x to infinity f(x) and lim x to -infinity f(x) to find horizontal asymptotes

I'm so annoyed at myself for not being able to do the algebra properly!

I see square roots and I fold instantly

[flint basalt]

$f(x) = \frac{\sqrt{x^2+4x}}{-x-4}$

[jolly parrot]

Krish

[flint basalt]

just making it easier to read

im not the best at asymptotes but uhm yeah good luck im sure someone else knows

[hybrid trail]

I know the asymptote part I just don't know how to get rid of the damn square root

I've done like 10 problems like this today and I still can't do it myself 😭

[flint basalt]

uhhh L'Hopital's? because wouldnt it be inf/inf when evaluating at x=inf

[jovial kiln]

it is not needed here

perhaps a simpler way is to divide both the numerator and denominator by the leading term

[hybrid trail]

wdym?

my thinking was get rid of the square root

and then u assess the degrees

[signal atlas]

Have you tried squaring it

[hybrid trail]

nah, I'll try rn ig

[jovial kiln]

try dividing numerator and denominator by x too if you want an alternative method

[hybrid trail]

let me ask this follow up question while I work

what is even a conjugate base?

[jovial kiln]

a conjugate base is part of the bronsted lowry acids and base theory

acids and bases come in pairs

[hybrid trail]

lmao not that

for math

good one lol

[jovial kiln]

i dont think there is a conjugate base in math

there is indeed a conjugate though

[versed mica]

-x + 4

is what they mean

[hybrid trail]

okay so just dealing with the top part

sqrt(x^2 + 4x) / x how would I do that?

would I factor out x or smth inside the sqrt

[hybrid trail]

and for squaring it, how would that affect things? (sqrt(x^2+4x)^2

my algebra is cooked

[signal atlas]

numerator becomes x^2+4x denominator because (-x-4)^2

[hybrid trail]

okay so it gets rid of sqrt

[signal atlas]

yeah

[hybrid trail]

why are dividing by x and squaring both viable options?

seems like squaring is the easiest way to get rid of the sqrt

[signal atlas]

We can square it because $\lim (f(x)^2) = (\lim f(x))^2$ assuming the limit is convergent

[jolly parrot]

purururuuriuruin

[hybrid trail]

okay idk what that means but I'll look into it

so I squared and ended up getting x(x+4) / (-x-4)^2 which I turned into

[signal atlas]

In english its "the limit of the squared is equal to the square of the limit"

[hybrid trail]

yeah

x(x+4) / (-*(x+4))(-x-4) is what I ended up getting

so that the x+4 can cancel out and we're left with x/(-(-x-4))

seem about rihgt?

[signal atlas]

yeah

[hybrid trail]

so then it's easy to figure out the asymptote stuff after that

it's just so annoying knowing the work needed to solve but not being able to bc my algebra sucks

I'm really hoping enough practice fixes things

thanks for the help

[signal atlas]

there is another step you need to be careful of. Since we squared, we actually found the limit squared. So if the original limit is 3 that means we will find this squared version to have a limit of 9.

Similarly if the squared version has a limit of 9, then the original limit may be -3 or 3 (because (-3)^2=3^2=9). You can check which one is the correct limit by checking whether the expression you are taking the limit of is positive/negative for large values of x.

[hybrid trail]

O_O

[signal atlas]

Yeah there are a lot of algebra tricks you need to know for limits its a bit annoying

[hybrid trail]

so whenever I perform square or smth I need to apply that to the a in x-->a?

[signal atlas]

yeah

[hybrid trail]

it didn't seem to matter since it was infinity though, I found HA fine. it just applies to when it approaches smth specific?

infinity^2 still infinity ig that's why

[signal atlas]

Hmm I think I might have misunderstood

[hybrid trail]

wait but -infinity

[signal atlas]

You don't do anything to the x -> a I see what you mean now

[hybrid trail]

,w lim x to infinity (sqrt(x^2+4x))/(-x-4)

[jolly parrot]

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[hybrid trail]

,w lim x to -infinity (sqrt(x^2+4x))/(-x-4)

[jolly parrot]

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[hybrid trail]

yeah I think I did something wrong

[signal atlas]

Assume we had $\lim \frac{\sqrt{x^2+4x}}{-x-4} = -4$ (this is not correct but just for the sake of demonstrating what we are doing).

What we are doing is squaring both sides

$\lim \frac{\sqrt{x^2+4x}}{-x-4} = -4$

$\implies (\lim \frac{\sqrt{x^2+4x}}{-x-4})^2 = (-4)^2$

$\implies \lim (\frac{\sqrt{x^2+4x}}{-x-4})^2 = 16$

So we found $(\lim \frac{\sqrt{x^2+4x}}{-x-4})^2 = 16$

So we take the square root of both sides to get

$\lim \frac{\sqrt{x^2+4x}}{-x-4} = \pm(-4)$

So the answer could be -4 or 4, you then just check large values of x (for limit x -> infinity) or small values of x (for limit x -> -infinity) to see which one is correct

[jolly parrot]

purururuuriuruin

[hybrid trail]

I get it, yeah

[signal atlas]

From this you can see the limit of the square is 1, so to find the original limit you find the "square roots" of 1: which can be -1 or 1

[hybrid trail]

yeah, that makes much more sense

[signal atlas]

To identify which one it is you think about what happens when you plug in large values.

For $f(x) = \frac{\sqrt{x^2+4x}}{-x-4}$

numerator is always positive here since its a square root and square roots are never negative

denominator here is negative for large values of x (ex: -1000-4 = -1004)

and is positive for small values of x (ex: -(-1000)-4 = 1000-4 = 996)

positive / negative = negative, so for the x -> positive infinity you know the limit needs to be positive

positive / positive = positive, so for x -> negative infinity you know the limit needs to be negative

[jolly parrot]

purururuuriuruin

[hybrid trail]

positive/negative = negative so x--> pos inifnity needs to be pos?

I don't get why this is negative

,w lim x to infinity (sqrt(x^2+4x))/(-x-4)

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[hybrid trail]

positive numerator makes sense, denominator negative makes sense. so should be negative (I mean I think so at least since the answer is negative), I just don't get why bc we did the algebra and it ended up being x/x+4 which is positive/positive

[signal atlas]

Its because we squared the limit, so the limit is going to change.

So we have $(\lim)^2 = 1$. This is a quadratic though and we know to solve this we would get $\lim = \pm 1$

[jolly parrot]

purururuuriuruin

[hybrid trail]

ah, I see

so we need to test out to see if it's -1 or +1 right

[signal atlas]

Yeah

and we go back to the original limit to figure that out

Maybe you remember when solving square root equations you get "extraneous solutions". Its kinda like that

[hybrid trail]

yeah I don't remmeber that lol

[hybrid trail]

what do we plug in?

large values of x (for limit x -> infinity)?

[signal atlas]

Yeah.

[hybrid trail]

sqrt(x=5) numerator is positive and bottom ends up being -5-4 which is negative

so u get positive/negative meaning the choice has to be the negative option between +- 1?

[signal atlas]

ye

[hybrid trail]

thats wild

I get it though

just a bunch of algebra that I needa hope I remmeber for the examl ol

thanks for the help, really means a lot

[signal atlas]

np

[hybrid trail]

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[orchid heath]

$\sqrt{-1} \times \sqrt{-1}$

[jolly parrot]

SELVATOR

[orchid heath]

We cant say $\sqrt{-1 \times -1}$ right?

[jolly parrot]

SELVATOR

[modest crater]

we can

because -1 * -1 = 1

[vital estuary]

you cant use that exponent law when on complex numbers

[vestal tapir]

we can't say they are equal

[modest crater]

oh yeah

[orchid heath]

So...

$\sqrt{-1} \times \sqrt{-1} = -1$ ?

[jolly parrot]

SELVATOR

[modest crater]

yes

[pearl pond]

@orchid heath Has your question been resolved?

[midnight haven]

we can call this complex number. we introduce complex number to expand real number place.

as u know, this is very strange. how can we redefine about sqrt(-1)?? for completion of the numer space......

[merry stirrup]

√a√b=√(ab) only and only if a,b≥0

[orchid heath]

What if a = 0 or b = 0

[merry stirrup]

Greater then equal to

Than*

My bad

[stoic imp]

❤️

[merry stirrup]

Now it's alright

[merry stirrup]

You can use iota if this condition is not satisfied
Like √-9×√-4=3i×2i=6i²=-6

But don't try to combine the roots

[pearl pond]

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[silver oracle]

How do you find your answer after long division?

Idk where tf I find my answer

It says it =x+12

Is that just based off what's on the top,m

?

[limpid lily]

Yes, it's what's at top, plus the remainder over the original divisor.

So, (x + 12 + \frac{0}{x^2 + 10x - 24}).

[jolly parrot]

Chai T. Rex

[limpid lily]

That 0 is the last 0 on the bottom line.

So, it comes out to (x + 12).

[jolly parrot]

Chai T. Rex

[limpid lily]

That's because if the remainder is 0, you can ignore it.

[pearl pond]

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[pearl pond]

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[trail bough]

Hello.

[trail bough]

can someone please explain to me how to do this

I haven't done this problem yet and I'm a little confused on where to start.

okay so

what I have come up with is sqrt2 / 2

for A

evaluated sin took care of the sin -1

got sin -1 ( - 1/2) = pi/6 substituted it back to get sin (pi/12 + pi/6)
used sum identify for sine and plugged in and substituted and simplified until eventually I got the 3 sqrt/2 + sqrt2/8 and then 4 sqrt2 / 8 = sqrt 2 / 2 and that's for A...

.close

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[cedar scarab]

What does it mean for f:R to R again in this context?

[cedar scarab]

I haven't taken an analysis course in a year now

Is it just saying from x to go to y?

Oh, the question is "Determine if the functions are injective/surjective

[pearl pond]

@cedar scarab Has your question been resolved?

[minor meteor]

f:R to R implies that the provided function is defined for real number inputs only, and the output of that function is a real number too

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[rotund osprey]

pls

i dont know how to apporach

[pearl pond]

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[leaden osprey]

Is this correcr

[pearl pond]

@leaden osprey Has your question been resolved?

[minor meteor]

Yes this is 'correcr'

[leaden osprey]

*correct

Thank you

[minor meteor]

Welcomee

[pearl pond]

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[desert silo]

i need help with this question

[toxic lichen]

``$\ell$im'' is quite the sight...

[jolly parrot]

Ann

[plush moss]

It's a fancy limit

[desert silo]

random things but they werent much useful

[toxic lichen]

ok first off

are you ok with sigma notation or are you allergic to it

[desert silo]

im ok

[toxic lichen]

right

so we are taking the limit of sum[k=1, n] k/(n^2 + k)

[desert silo]

yes

[toxic lichen]

if you try to squint a bit and reason somewhat informally, the general term of the sum looks like k/n^2

question is, how far off does that put us

ie what does $\sum_{k=1}^n \paren{\frac{k}{n^2+k}-\frac{k}{n^2}}$ look like, and can we maybe put some upper bounds on that?

[jolly parrot]

Ann

[desert silo]

the limit of n->inf of this function ?

[compact ridge]

yes

[toxic lichen]

mm yeah kinda

[desert silo]

i think it will be too small in comparison to sum k/n^2

so we can ignore it if x-> inf

[toxic lichen]

n not x

[desert silo]

wrt k/n^2

[desert silo]

yeah mb

[toxic lichen]

ok so

[desert silo]

sum k/n^2

is n(n+1)/2n^2

so lim is 1/2

[toxic lichen]

yes

so you need to place some upper and lower bounds on the "remainder" sum

a lower bound is free: 0

[desert silo]

what is the exact reason for this ?

i just felt like it

[toxic lichen]

mmm its not that it will be small in comparison to the main sum

more that it will be small outright and hopefully maybe approach 0

[desert silo]

ok

thanks

.close

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[fervent grove]

help

[fervent grove]

My question is (sorry the task is german) if the solution is correct because Im pretty sure x3 = 2-x4

and not -x4

I might be wrong but dont I have to solve x3 + x4 = 2
so I put -x4 giving me
x3 = 2-x4 no?

[lilac jackal]

,w rref[[1,0,-2,0,-3],[0,1,2,-3,7],[1,0,1,6,9]]

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[lilac jackal]

from the last row u should get x3+2x4=4 so x3=4-2x4

[fervent grove]

so basically the solution is entirely wrong?

[lilac jackal]

it depends if u made an error in row reduction

[fervent grove]

No thats the solution from last semester exam and im trying to figure out how they got x3 = -x4

so I was assuming their reduction was correct

[lilac jackal]

https://cdn.discordapp.com/attachments/908078029778083872/1351203964200161412/wolf.png?ex=67d98641&is=67d834c1&hm=b17b7e64b0e669838aa182fc3a8676a565f98793de2d2de82c744e4d6273963c&

[fervent grove]

therefore not understandy how they got it to x3 = -x4 since if I use the same reduction they got I get x3 = 2-x4

[lilac jackal]

i used a calculator to get this

[fervent grove]

yeah ik xD so its just wrong what they got?

thats all I need to know

[lilac jackal]

they got x3=-x4?

[fervent grove]

yeah thats their solution

[lilac jackal]

yeah wrong solution or wrong problem

[fervent grove]

yikes

thanks

.close

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[lilac jackal]

np 🙂

[pearl pond]

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[lament pond]

Can someone explain how this equation works? To me it seems to be overestimating the area, because the blue equation would include the area bounded by the cos2angle equation and the x axis, which would also include the area represented by the red equation

[versed mica]

no

it starts at pi/6

not 0

[lament pond]

The lecturer here states that the blue equation specifically represents the "crescent shaped area" but I don't understand

[versed mica]

yep

[lament pond]

Oooh

Wait

[versed mica]

think of standing at the origin and shining a flashlight as your rotate counterclockwise

[lament pond]

Ok i think I see

[versed mica]

the wall for the first pi/6 radians is from the circle

and then after that point the wall becomes the figure 8

[lament pond]

Yeah i see it's the area bounded by the alpha and beta angles

Not the area under the graph

[versed mica]

yep

most people try to do polar integration as they would normally

it is not the same

because our independent variable is theta, not x

changes in theta correspond to rotating a ray/line rather than sliding along the x axis

[lament pond]

Its still a little hard to understand but I get why I'm wrong at least.

Thanks btw I'm gonna keep working at it

[versed mica]

look up ap calc bc polar review videos on youtube

its like

a 50 minute video

towards the 40 min point he gives a nice intuitive explanation

lemme see if i can find it for you

[lament pond]

🙂

[versed mica]

https://youtu.be/TnUsmoWf9eU?si=Jh237S9kYycVsJhs

here you go

@lament pond

[lament pond]

Okay thanks, I'll look through it

And I'll try to reread what you said once I understand more

I don't want to ask u a billion questions

[versed mica]

no worries

[pearl pond]

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[tawny cypress]

i dont understand what even is a semi parabola

[tawny cypress]

what do i do

[rare burrow]

well, they probably touch at a few points

so find points x where (call the first function g and the second f) f(x)=g(x)

(so literally compute f(x)=g(x))

that's the first step

[tawny cypress]

so x-1 = 3sqrt(x-1)

[rare burrow]

indeed

[tawny cypress]

then what

[rare burrow]

solve for x

you're trying to find values of x that satisfy f(x)=g(x)

[tawny cypress]

ohh

[rare burrow]

because then that means the functions share a common point (i.e. "touch" since they are cts functions) at each (x,f(x)) or equivalently, each (x,g(x))

ok i gotta go study for something now :^) if u need help with the integrating part, probably someone else will stop by here.

[tawny cypress]

alr thx

[rare burrow]

it's not too bad as long as you remember that the area you'll get is positive when $h(x)\geq 0$, negative otherwise

[jolly parrot]

00100000

[pearl pond]

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[pearl pond]

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[desert silo]

i tried writing x = pi/2 + h where h->0

so its

what can i do after this ?

[plush bramble]

you can divide top and bottom by h and then recognize this is a ratio of derivatives of log(cos(x * constant)) at x=0 for two different constants

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[inland ivy]

how the hell did anyone think of this

[azure ether]

lol

[inland ivy]

is there some reasonable way of thinking about this which doesn't look like it was fathomed by a crazy person

or just like some motivation as to why anyone would think like this

[hazy pilot]

Probably thinking of matrices as rotations in space

[inland ivy]

i'm not too comfortable thinking about matrices within matrices in the first place

like I understand it but I don't have much intuition about it

[hazy pilot]

I highly recommend this guys video on matrices:
https://www.youtube.com/@visualkernel

[inland ivy]

the first one?

[hazy pilot]

Just all of them about matrices, really helped me think of them as rotations in space. And of course 3blue1brown also has a fantastic video series on them.

[inland ivy]

oh I've seen that one

EoLA

I do know how to think about matrices as transformations

just the matrices within matrices thing throws me off

[hazy pilot]

I haven't had those, but I think they're called tensors

Maybe look up some videos about them to get more of a visual understanding

[inland ivy]

no this is what I'm talking about

block matrices

[pearl pond]

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[pearl pond]

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[full cobalt]

Can somone explain how to find domains with an example

[midnight haven]

you set the bottom equal to 0

[full cobalt]

Equal or the greater than sign

[midnight haven]

you set the bottom equal to 0 to find the asymtotes where x does not equal

[full cobalt]

Is dis a domain qiestion?

[midnight haven]

no

no

[full cobalt]

Fis

Alr

So set x+2 = 0?

[midnight haven]

what does that equal?

[full cobalt]

2+x

[midnight haven]

no it equals -2

x=-2

that is an asymptote

[full cobalt]

How it become negative

[midnight haven]

u subtract 2 on both sides

to isolate x

@full cobalt

[full cobalt]

Oh

Right

Ok so the bottom is now -2

X=-2

@midnight haven

[balmy dove]

finding the domains of a functions is basically finding for what values the function exists, can have values

for yours in the pic up there

as a general rule with fractions , the denominator cannot be 0

[full cobalt]

Greater than 0

<

[balmy dove]

because logically speaking you cant divide something by 0. it would mean it doesnt exist

[full cobalt]

Alr so x+2 >0

[balmy dove]

so knowing that it cannot be 0, you need to find where the bad apple is, for which x the denominator is 0 so you can take it out

[balmy dove]

does it say that in the problem?

[full cobalt]

Na bit u jus said

Cant be 0

[balmy dove]

it can also be less than 0, the function would still exists

just needs to not be 0

[full cobalt]

O jus canr be 0 itself

Ok

[balmy dove]

si to find where its 0

x+2=0, you substract 2 from both sides so x=-2

[full cobalt]

Ok next

I gtg in 1 min

[balmy dove]

now if we go back to the function and check, indeed for -2 the denominator is 0

[full cobalt]

Pls say as fast as u can

[balmy dove]

so it would be R-{-2} the domain

[full cobalt]

R is range

[balmy dove]

R is the uhh

the set of real numbers

[midnight haven]

poppy ur eating fr

[balmy dove]

honestly my professor explained it so bad that i had to take it into my own hands

it was honestly impressive how much sense it made once it was explained well

@full cobalt are you still around?

seems not but if you do come back basically the domain is what values x can take so that the function exists. considering that only for -2 the denominator is 0, then for any other value it will exist therefore the domanin is R{-2} or R-{-2}, in my country both are used idk about the rest. theyre just notations which basically tell you x can take any values from the real numbers set, aside from -2 ( thats what the \ or - means, its just the notation)

[pearl pond]

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[abstract scaffold]

how to do c)

the answer is supposed to be 24

how do you get that

[spare lark]

The least that your venn part can be are 0 right ?

You can't have negative amount of elements in your venn

[abstract scaffold]

yes

[abstract scaffold]

??

[spare lark]

Badly worded i meant my second sentence

Looking at your venn diagram you see that there is three expressions that can fit for oir problem

Thoses in intersection

And so you have that 27-x >= 0 so x <= 27 but for x = 27 the part with 25-x and 24-x are negative and it make no sense

So can't be this one

What about 25-x

Gives x <= 25 but for x = 25 you have that 24-x negative, which is no sense again

Lasting 24-x giving x <= 24 and here no problem with other part of the diagram

[pearl pond]

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[prime bramble]

n in the summation should be fine

you proved it for n = 1, and now you're showing that assuming the statement for case n will imply that it's true for case n + 1

this proof looks fine to me

yes

it makes sense because the summation doesn't use both n as an index and a upper bound

the other side of the = sign doesn't matter

yes, it's fine

[prime bramble]

.

P(1) is true, and P(n) => P(n + 1), so you're good!

you too!

you may close the channel if you're done

[pearl pond]

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[pearl pond]

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[rigid geyser]

would x just be 7?

[rigid geyser]

i proved abd similar to dbc (aaa)

[pulsar lark]

Then how did you conclude x=7?

[rigid geyser]

since i provced that the angles inside are equal and the exact same

idk if im doing stuff right

thanks for the help

.close

[pearl pond]

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[polar prawn]

hey

[polar prawn]

i needs help

but

i dont speak english

[flint basalt]

ask your question instead of saying you need help.

[polar prawn]

its much exercise of maths

but

its in spanish

[unkempt yacht]

send the question anyway and try your best to translate it

[pearl pond]

@polar prawn Has your question been resolved?

[pearl pond]

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[brittle harbor]

Hellow. I'm trying to find the general solution for number 4. The problem suggests we use e^2i instead of cos(2x) and then pick the real part of the result only. I'm unsure about the form a particular solution should take.

[brittle harbor]

Should it take this form?

[steady crescent]

so, I see what you're trying to do, however, you've got a non-homogenous linear ode and so you'll need to solve for two different solutions.

generally in this case we would consider the homogenous solution and the particular solution. They have other names also, but thats just what I tend to call them.

so idk where I was going. But your current for of solution likely wouldn't work

you're welcome to try it

but really your probably going to have to control the exponent factor as well

[steady crescent]

also, you will probs need a coefficient infront of the x^2

[pearl pond]

@brittle harbor Has your question been resolved?

[brittle harbor]

I already found the general homogeneous solution

[brittle harbor]

Right, I forgot it

[brittle harbor]

What do you mean? Why this wouldn't work?

[steady crescent]

Because you don't know how the exponential 'factor' is going to behave. Why is it going to be 2i? Right?

And maybe you have other info which can tell you that it does. But I don't think it does, just from gut feeling anyways

The point of using undetermined coefficients is that you don't know any coefficients of the x terms

[steady crescent]

Yeah, my bad. I don't know what I was going on about. I don't read your question fully. Sorry bout that.

[pearl pond]

@brittle harbor Has your question been resolved?

[pearl pond]

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[glacial sequoia]

is there more context @hot sparrow

do you need to solve it?

[hot sparrow]

obv i do

💀

[glacial sequoia]

did you get it into polynomial form

[oak ivy]

don't we take x^3 as LCM?

[hot sparrow]

nahh, our teacher js dumped this equation and she didnt even teach a single thing cause she was frustrated, she didnt even teach it bru

[oak ivy]

I think you open the brackets and then take x^3 as LCM

[compact bison]

do we need to find the roots, what do we need to do

[oak ivy]

But then we don't get

[hot sparrow]

we need to factorise it:/

[compact bison]

ohk

[hot sparrow]

i dont know a single thing bout this

[compact bison]

u need to do this, and then find a factor by trial and error

[oak ivy]

hmm

then divide?

[compact bison]

is this equal to 0?

[compact bison]

yea long division but it should be equal to 0, if its not idk😭

[oak ivy]

No how

Like if that thing is equal to 0 then ok

[reef marlin]

Factorize this? Hmm...

[compact bison]

yea

[glacial sequoia]

just set it equal to 0; write it in polynomial form and maybe hopefully guess a root

[compact bison]

yup

[hot sparrow]

alrr ill try

[jolly parrot]

neon

[compact bison]

oh wait

sry

[compact bison]

this might not work sry

[glacial sequoia]

,w solve -2x^5 + 10x^4 + x + 11

[jolly parrot]

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Requested by neonperseus

[glacial sequoia]

yeah there's nothing clean here

either we're missing context or your teacher just forgot to give it to you

[hot sparrow]

yea ig

[spiral pivot]

Or the problem isn't to solve it. This isn't an equation, it's only an expression. Is the task to put into a single fraction?

[hot sparrow]

yup

[spiral pivot]

Can we get the original problem with the entire context?

!original

[pearl pond]

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.

[reef marlin]

I'm seeing 2x² + 10x + x⁻² + 11x⁻³.

[glacial sequoia]

bruh I asked you at the start

[hot sparrow]

i already sent the orignal screenshot

[spiral pivot]

Without context

[hot sparrow]

our teacher just sent us a bunch of explressions and nothing more, she told us to factorise it, there was no context, just a word file of which I sent a screenshot

i dont have any context

[spiral pivot]

"she told us to factorise it"

How did she tell you this? Verbally? Or do you have written instructions?

[hot sparrow]

verbally, yea

[spiral pivot]

If written, may we see these instructions?

Ok... Do you remember exactly what she said? Verbatim?

Alternatively, if this is not the first one you have solved, can you provide us with a prompt-solution pair?

[hot sparrow]

its was something like, "you need to factorise these equations", nothing more, just this

[hot sparrow]

its the first one

[grand light]

expand the algebraic expression first

[spiral pivot]

Then I'm afraid I'm at an impasse here. I don't know what is actually being requested in this problem. Perhaps someone else is better able to read the tea leaves than I am.

[hot sparrow]

uh huh, done

i js need someone to teach me from scratch

[oak ivy]

Like if you don't know factorisation why would your teacher give a polynomial of power 5

[hot sparrow]

idk man

[oak ivy]

They should start with 2

[spiral pivot]

The expression you gave simplifies to a quintic. Famously, quintics do not have general solutions, and I trust the other users who have been helping you up until this point that this quintic does not happen to factor nicely.

There is nothing you can do from here without breaking out high level mathematics.

[reef marlin]

Yup. It's a quintic omitting the x³ and x² statements.

[hot sparrow]

oh

...

[reef marlin]

The 11 term is the thing that's the problem here-

[hot sparrow]

so, i just dont have an answer to it?

[spiral pivot]

I double checked and this quintic indeed does not have nice solutions.

It is not factorable.

[hot sparrow]

alright

[spiral pivot]

At least not without introducing a specific special function called a "Bring Radical" and even then your solution is going to be extremely long and tedious

[reef marlin]

You can attempt to factorize it as (-2x⁵ + 10x⁴ + x + 11)/x³. But you'll have to use a lot of little tricks to get anywhere farther than this.

[spiral pivot]

-2x^5 @reef marlin

[reef marlin]

Right.

[hot sparrow]

okey thx everyone, now i gotta close this channel

[reef marlin]

Alright, good luck with the rest of those questions (if there are any left)-

[spiral pivot]

The command is .close

Best of luck!

[hot sparrow]

yes i know:)

.close

[pearl pond]

Channel closed

Closed by @hot sparrow

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[spiral pivot]

Sorry! 🙂

[pearl pond]

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Remember:
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[rigid geyser]

in this diagram, i have to find lenghts BX and XC

[rigid geyser]

i dont know where to start..

[unkempt yacht]

do you know what is sine law?

[rigid geyser]

uh no

ik what sine is but not sine law

the topic is geometric figures

[toxic lichen]

a/sin(A) = b/sin(B) = c/sin(C) ring any bells?

[rigid geyser]

i have already proved abc and cax are similar

[short drum]

U can use similar figures instead

[rigid geyser]

sorry i havent learnt any of that

[toxic lichen]

hm

[short drum]

Ratio of sides is 3 to 2

[rigid geyser]

yeah yeah i know the basics of that

oh u did ab/ca?

i split the triangles into 2 btw

[short drum]

Yeah

[rigid geyser]

triangle BAC and triangle CAX to help me better visualise

[short drum]

U know ax=ac

[rigid geyser]

no sorry

this stuff is apprently advanced for me

cuz im in a special program

[short drum]

Cap is isosceles

Cax

[rigid geyser]

yes yes

oh wait

i thought ac = ax was some formula

yes yes i understand ac = 10 = ax = 10

[short drum]

Oh I was just referring to in the diagram

So we know that bac~cax cause they both have base angle of 80 meaning they’re vertex angle is 20

[rigid geyser]

yep

[short drum]

Ba/ac=ac/cx

We know ba=15 and ac=10

[rigid geyser]

yuh

[short drum]

So we can substitute those back in and get cx=20/3

After thst we also know bc=15

So subtracting cx from bc we get bx

[rigid geyser]

hm im still tryna process

[short drum]

If we substitute the numbers we have into the ratio we have 15/10=10/cx