help-0

[trim wagon]

you should watch this if you don’t understand https://youtu.be/za0QJRZ-yQ4

[vague coral]

Technically (-1,1) isnt wrong but its not enough

[dense moon]

ok

[vague coral]

and the range is [5, +oo)

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[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[alpine sable]

how would I go about finding whats 75% of 1,200? dont need the answer

[slate jolt]

do you know why we call % percent?

(itll help you its not just trivia i promise)

[alpine sable]

nah I dont

[slate jolt]

it comes from french "cent" means hundred

so it means per hundred

75% = 75/100

[chilly pike]

x% means x per hundred

[slate jolt]

so simply

to get x % of something

you do (x/100) * something

here (75/100)*1,200

[fierce prairie]

damn I never actually knew that lmao

[alpine sable]

right and how would you solve it?

[chilly pike]

write it down as fractions

75/100 * 1200/1

and then multiply them

[alpine sable]

🫰

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[alpine sable]

How would I find the exact value?

[wary stream]

Part a: in terms of e
Part b: in terms of ln

[alpine sable]

right

can you explain?

[wary stream]

e^2 = 7.3890
e^2 is more exact
7.3890 is the approximation

What were the steps you took to solve for x, in part a?

[alpine sable]

got it thanks

.close

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[vale lichen]

Hi

[vale lichen]

Honestly idek where to begin on this one

Could someone explain?

Blue is f(x) and red is g(x)

[zinc dagger]

What problems specifically are troubling you? I’ll beat them up

[vale lichen]

All of those in the pic lol

Also this one man, those answers I put are wronf

At least one of them is wrong

[zinc dagger]

So for 3: as x inches to a closer and closer value to 0, f inches closer to 0 and g to 1.5 and 0 times 1.5 is 0

For 4, same thing: f/g inches closer to 0/1.5 which is basically 0

5 same thing: what one place does f seem to go to at x = -1? Plug it into the expression, and u have the limit of what seems to be, as x inches closer to -1

The second photo problem 5, f(0) doesn’t actually have a value at x=0. You’ll notice a hole right there

So r u all good? @vale lichen

[vale lichen]

For 5 in the first one i got the right answer no?

Its 1 because sqrt(3 + f(x)) is sqrt(3 + (-2))

[zinc dagger]

Yup! U got it

Sorry gtg hit the hay

[vale lichen]

All good man thanks for ur help

.close

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[alpine sable]

what does the second one mean?

[warm cipher]

you do the inner function

first

the inverse function\

then the outer one

[ornate condor]

ITS ZAY

[pine kettle]

did you still need help

[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

Please share solution of 2nd one here

[pine kettle]

we aren't going to give you the solution but we can help you solve it

[hollow inlet]

@alpine sable wht did you do for problem 1?

[alpine sable]

I know first one

[alpine sable]

Please share solution of 2nd one

[lone heart]

@alpine sable Has your question been resolved?

[serene junco]

It says all angles have equal measure, do you know what that measure would be?

@alpine sable

Okay I'm not gonna do all the work but I'll give you a couple hints because I'm gonna be going to bed soon

It's a hexagon with six equal angles. This means that each angle is 120 degrees

You can find this by using the formula $S = (n-2)\cdot 180^\circ$, where $n$ is the number of sides and $S$ is the total sum of all the internal angles.

[ocean seal]

tatpoj

[serene junco]

Plugging in n=6 tells you that the total angle sum S is 720 degrees, and then by dividing by 6, you can find that each angle is 120 degrees. (since all six angles have equal measure)

Then, you can draw in the diagonals BF, FD, and DB.

This separates the hexagon into four triangles.

Each triangle is solvable using the Law of Sines and Law of Cosines.

If you work your way through the triangles, eventually you'll find DE and EF.

[lone heart]

@alpine sable Has your question been resolved?

[serene junco]

@alpine sable I gave you an outline of how you could solve the problem. Is there any part in particular you didn't understand?

[alpine sable]

Can u write the steps on a paper

And then explain me

So that I'll get it better

@serene junco

[serene junco]

It's quite a few steps, I'm afraid I can't go through all of them

but

Are you familiar with the Law of Sines and Law of Cosines?

[alpine sable]

Np just share it

[alpine sable]

Yes indeed

[serene junco]

And does it make sense why each internal angle of the hexagon has a measure of 120 degrees?

[alpine sable]

idk

[serene junco]

To find the total angle sum of any polygon, you can use the formula
$$S = (n-2)\cdot 180^\circ$$
where $n$ is the number of sides

[ocean seal]

tatpoj

[serene junco]

So for $n=6$, we get $S=720^\circ$, meaning any hexagon has a total angle sum of $720^\circ$.

[ocean seal]

tatpoj

[serene junco]

Since all angles are equal, each angle has a measure of 720/6 = 120 degrees.

By drawing in these three diagonals, you can separate the hexagon into four triangles

Each triangle can be solved using the Law of Sines and/or Law of Cosines

I labeled the diagonals a, b, and c, but I don't know if it's necessary to find all three of them

But by using a combination of Law of Sines and Law of Cosines, you should be able to solve the triangles and find DE and EF

[alpine sable]

Please share of this

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[vague coral]

,rotate

[desert hare]

How would I answer qustion E b)

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@desert hare Has your question been resolved?

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[alpine sable]

I tried simplifying it:

(a x c/a) + (b x d/b)

and if i do this then i get

ac/a + bd/b

right?

i know the answer to the question is c + d

so i imagine there is some simple way of getting rid of the a in ac/a and the b in bd/b

[hollow inlet]

Yep, you've done all right 👍

[alpine sable]

im just unsure how

[hollow inlet]

wdym

Just cancel it?

[alpine ore]

a•c/a should be ac/a right?

[alpine sable]

yes

i get ac/a + bd/b

[gray isle]

ac and a have a common factor of a

simplify ac/a the same way you'd simplify something like 6/2

[alpine sable]

idk what u mean

but

[hollow inlet]

do you know how to add fraction?

[gray isle]

try not to overthink

[alpine sable]

if i have ac / a

do the a's just cancel eachother?

[gray isle]

yeh

[alpine sable]

oh

ok

i thought i had to do something to get rid of them

[hollow inlet]

[alpine sable]

idk bro this gets a bit confusing cuz i am translating

idk bro

but i dont have to do anything then

idk man

thanks tho

[gray isle]

$\frac{\cancelto{1}{a}\ \ c}{\cancelto{1}{a}} = \frac c1 = c$

[ocean seal]

ℝamonov

[hollow inlet]

demn 🔥

[gray isle]

a bit overkill

[alpine sable]

yea ok great

idk in my mind i just thought that the a and c multiply and they do sommething and i cant just get rid of it

but whatever then

i know how to do it know

thanks

[hollow inlet]

Alr.

🎉

[alpine sable]

.close

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[deep grove]

help please 😭

[keen pasture]

Try to factor out 1/5

[deep grove]

yeah thats what i thought, but idk how with it being under the square root

[keen pasture]

You get 1+0.6x^2 I think?

[deep grove]

but u cant just divide both by 5 while its under a square root what

[keen pasture]

0.6x^2 is not (0.6x)^2

[deep grove]

no i mean

[deep grove]

how do u get this

[keen pasture]

Instead it is [sqrt(0.6)*x]^2

[deep grove]

u cant factor out 5 from the denominator bcs its under a square root right

[keen pasture]

Ups, sorry. You're right. You have to factor out sqrt(5)

[deep grove]

what

oh wait

i get it

yeah nvm

thanks <3

.close

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[deep grove]

.reopen

[lone heart]

✅

[deep grove]

wait

but u cant just

ur saying

$\sqrt{5-3x^2}:=:\sqrt{5}\sqrt{1-\frac{3}{\sqrt{5}}x^2}$

[ocean seal]

imagine

[deep grove]

but if that was true,
then shouldnt
this also be true

$\sqrt{5+3}:=:\sqrt{5}\sqrt{1+\frac{3}{\sqrt{5}}}$

[ocean seal]

imagine

[deep grove]

u cant just factor out the root 5 liek that

@keen pasture

[keen pasture]

That's not what I said

[deep grove]

what did u say then

[deep grove]

how

[keen pasture]

sqrt(5-3x^2) = sqrt(5)*sqrt(1-0.6x^2)

Because sqrt(ab) = sqrt(a)*sqrt(b)

[deep grove]

oh

i see

cool

thnkas

.close

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[lone topaz]

hi i've forgotten how to prove can somebody help me and tell me what to use here

[heady void]

wait

gimme a sec

[trim wagon]

,tex \trigid

[ocean seal]

Deep. (kawaiiCat for emoji)

[trim wagon]

the second one

[heady void]

yes

[gilded vessel]

or if you dont want to use existing formula simply write the tan on the left hand side as sin/cos

then simplify from there

[heady void]

so you would get sec^2(x)/tanx

[trim wagon]

#❓how-to-get-help

[lone topaz]

is this right?

and where do i go from here

[lilac nest]

Try to rewrite sec²(x) in terms of cos first to see whether you can simplify something

It's correct so far btw

[lilac nest]

Remember that sec(x)=1/cos(x)

Ping me if you need more help

[lone heart]

@lone topaz Has your question been resolved?

[lone topaz]

can i still apply that if its sec^2x

[lilac nest]

Yes, you're going to get 1/cos²(x)

[lilac nest]

sec²(x)=sec(x)sec(x)=(1/cos(x))(1/cos(x))=1/cos²(x)

[lone topaz]

nice

[lilac nest]

Do you know how to go on?
Remember that
$$\frac{\frac ab}{\frac cd}=\frac ab \times \frac dc$$

[ocean seal]

Andrea276

[lone topaz]

is this correct

[lone topaz]

i think i got it now

[trim wagon]

[lone topaz]

thank you guys so much

https://tenor.com/view/dog-smile-pet-happy-gif-12068900

.close

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[lean wraith]

Am I on the right track so far. I'm like two pages into trying to find a sub n and I have to think there is an easier way

[pallid scarab]

hello, did you try integration by parts ? integrate cos or sin and differentiate x^n until there is no more x^n term

[lone heart]

@lean wraith Has your question been resolved?

[lean wraith]

I have been using IBP in that exact way but the problem is becoming so complicated now because I am have done IBP about 4 or 5 times

is there something else I can leverage like even and odd function being multiplied together. is there anything down that route that would lead to some terms being cancelled

[pallid scarab]

if you do want to know, you'll notice that x*abs(x) is an odd function, so you don't have to calculate the an terms (because they're all 0)

and thus you also have bn = 2*int(0,1)[f(t)sin(npit)dt]

so you only need 2 IBPs

does that answer your question @lean wraith ?

[lean wraith]

sorry just getting my head around this

this is where I am stuck, do you not have to use IBP four times here

this is where the problem becomes unmanagable for me

@pallid scarab can this integral be done with just two instances of IBP

[pallid scarab]

which integral ? sorry I was dealing with another request

[lean wraith]

no worries

[pallid scarab]

so, remember that x*abs(x) is odd. same is for sin(npix)

[lean wraith]

you mentioned it could be done with two IBPs

[pallid scarab]

yes, so we're looking at 2*int(0,1)[xabs(x)sin(npix)dx]

which is 2*int(0,1)[x^2sin(npix)dx]

so, whenever you do an IBP by differentiating x^n, you end up with x^(n-1)

so if you do an IBP by differentiating x^2 twice, you end up with x^0 = 1

[lean wraith]

man you're a legend

I think I get it now. definitely more work to be done but I am totally seeing what you're getting at

so because both of the functions are odd, they multiply to even. because of this we can say that the original integral can be rewritten as two times the positive side of the integral. is that about right

and that makes the whole thing way way easier to evaluate

[lone heart]

@lean wraith Has your question been resolved?

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[tawny condor]

Could you tell me what vector QR is?

[median agate]

um

not really sure

wait but its a ratio right

[tawny condor]

You know that QR is parallel to PS.

[median agate]

so um 3/4?

[tawny condor]

And QR = 2PS

[median agate]

yeah

[tawny condor]

Where did you get that from?

[median agate]

cuz you add up 3 and 1

[tawny condor]

I was just asking what vector QR is.

We'll get there

[median agate]

so q to x is 3/4

[thorny haven]

we're gambling now

[median agate]

and x to r is 1/4?

[tawny condor]

$\overrightarrow{XR} = \frac{1}{4}\overrightarrow{QR}$

Right?

Oh

Wait

My bad

[median agate]

gimme 1 min to analyze

oh

[ocean seal]

RedstonePlayz09

[median agate]

yeah

[tawny condor]

Just as you said.

[median agate]

yep

[tawny condor]

Now, what's vector QR in terms of b?

Vector PS is b. So what's the relation between QR and PS?

[median agate]

QR is 3/4 + 1/4

i think?

[tawny condor]

No

Notice that the vectors QR and PS are parallel, right?

[median agate]

QR is twice of PS?

[tawny condor]

Yes exactly.

[median agate]

yeah

[tawny condor]

They are parallel so they have the same DIRECTION.

[median agate]

right

[tawny condor]

Now you know that the LENGTH of QR is double the LENGTH of PS.

[median agate]

yep

[tawny condor]

So in conclusion that tells you
$\overrightarrow{QR} = 2b$

[ocean seal]

RedstonePlayz09

[tawny condor]

Now can you tell me what vector XR is?

[median agate]

oh yeah

[tawny condor]

Wait

[median agate]

xr is 1/4 right

[tawny condor]

We don't even need that

$\overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{QR}$

[ocean seal]

RedstonePlayz09

[tawny condor]

Plug in PQ and QR.

[median agate]

so PQ is a

and QR is.. 3/4 + 1/4 is it?

[tawny condor]

No

Vectors are not just numbers

[median agate]

oh

[tawny condor]

We are talking about the vector QR. Not the length.

[median agate]

oh so it's 2 x b

which is 2b?

[tawny condor]

Yes.

[median agate]

so a+2b?

[tawny condor]

$\overrightarrow{PR} = a + 2b$

[ocean seal]

RedstonePlayz09

[median agate]

oh great!

[tawny condor]

Also remember that a and b are vectors aswell. Not lengths.

[median agate]

yep

so the fractions only come in SX i suppose

[tawny condor]

$\overrightarrow{SX} = \overrightarrow{SP} + \overrightarrow{PQ} + \overrightarrow{QX}$

[ocean seal]

RedstonePlayz09

[tawny condor]

That's how we are going to find SX.

[median agate]

can't we go from sr and rx?

[tawny condor]

Yes we can,

[median agate]

oh we dont know sr right

[tawny condor]

but for that we'll need SR.

[median agate]

yep

[tawny condor]

And it's extra work to find SR.

Alright, so first, what's vector SP?

[median agate]

SP is um

-b

[tawny condor]

$\overrightarrow{SP} = -\overrightarrow{PS}$

[ocean seal]

RedstonePlayz09

[tawny condor]

Yes exactly

Just flip SP.

[median agate]

yes!

[tawny condor]

And you get -b

[median agate]

yeah

[tawny condor]

Now, what's vector PQ?

[median agate]

PQ is a

[tawny condor]

Good

And vector QX?

[median agate]

QX is um

3/4 x 2?

because its twice?

[tawny condor]

$\overrightarrow{QX} = \frac{3}{4} \cdot \overrightarrow{QR}$

[ocean seal]

RedstonePlayz09

[tawny condor]

Do you agree with this?

[median agate]

oh

why did you put QR at the end?

[tawny condor]

Each time I write something, make sure you understand it. Go look at the diagram.

Look at the picture again

[median agate]

and the dot is for multiplication right?

okay

[tawny condor]

https://cdn.discordapp.com/attachments/490557019623915520/1023589441681510521/unknown.png

[tawny condor]

Yes

[median agate]

okay

[tawny condor]

So what is 3/4 times vector QR?

[median agate]

we only need QX so why did you put QR?

[tawny condor]

Because we need QR to find QX

[median agate]

oh okay

so 3/4 x 2

[tawny condor]

Do you understand why this is true?

[tawny condor]

What's 2?

Vectors are not numbers

Vector QR is not 2.

[median agate]

uhhh

yep

QR is 2b right

[tawny condor]

Yes.

So what is QX?

[median agate]

um

3/2b?

because 3/4 x 2b

[tawny condor]

Yes

[median agate]

right

[tawny condor]

So in conclusion can you tell me what SX is?

[median agate]

SX is um

gimme a min

is it

1/2b + a?

[tawny condor]

yes

[median agate]

because -b+a+3/2b

a comes from PQ

[tawny condor]

$\overrightarrow{SX} = a + \frac{1}{2}b$

[median agate]

woohoo thanks alot!

[ocean seal]

RedstonePlayz09

[tawny condor]

No problem!

@median agate .close

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[vapid bramble]

what does the | mean in the equation?

[vapid bramble]

let A = element x in real numbers | ??

[tiny vigil]

Left side is the condition for equation on the right. So all x are within real numbers yes.

[vapid bramble]

ohh ok i get it

.close

[lone heart]

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[vapid bramble]

ty

[lone heart]

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[alpine sable]

can somebody explain lowest common denominator and factorization

[wary stream]

https://www.youtube.com/watch?v=O04318jO5gY&ab_channel=BrandonCraft

[runic trench]

lcm - smallest number that is a multiple of 2 or more numbers
lcd - smallest multiple of denominators of 2 or more fractions

[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

thxxxxx

[lone heart]

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[alpine sable]

help

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

.close

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[alpine sable]

Why can't I use u = cos(x)?

then it would make du = -sin(x) so -du = sin(x) and then we get to the negative integral of u^2 etc etc..

[wary stream]

You can, but you would have to deal with a negative sign

[alpine sable]

yeah i mean i did that and my integral was -u^3/3 which is apparently wrong?

[wary stream]

If you did u = sin, no negative signs involved

[alpine sable]

woah

im an idiot

[lilac nest]

Wouldn't you get -u*(du/dx)^2 instead of -u^2 du though? u was equal to cos(x), not sin(x)

[alpine sable]

yeah thats why im an idiot

sorry, my brain is off today

[wary stream]

Plus your u was in terms of cos

[alpine sable]

ye im a goofy goober ty yall

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[lament geode]

im sitting on a math related problem for like two days now,
i have to transform a rcl circut in series to a rcl circut in parallel, i managed to solve it in one direction multiple times, my main problem is that after i transformed the circut i cant transform it again to get the same circut i had in the first place.
here are my condensed notes from today on onedrive: https://1drv.ms/b/s!AmMboXCQRGG_gbAyWtfQiNFBYsJIDg?e=C0XuI9
and also the same pdf enclosed below

[wary stream]

One, circuits is more physics related so you should ask in a physics server
Two, post screenshots, no one here is going to download things because that's sketchy

[lament geode]

alright, here are my screenshots

if its not possible to be solved here, i will try to find another server

<@&286206848099549185> need some help here

[viscid pawn]

Also yeah please go to another channel, this one is Kru's one

#help-7｜zen1thxyz Post it here

[alpine sable]

okay okay, thank you so much

[lone heart]

@lament geode Has your question been resolved?

[lament geode]

<@&286206848099549185> anyone got any idea, why my back transforming doesnt resolve into the anticipated values?

[lament geode]

<@&286206848099549185>

[viscid pawn]

You should ask on the Physics server

[lament geode]

doing that, but there aswell as here nobody seem to help me with my problem. should i close this channel here?

[viscid pawn]

Sorry cant help you

[lone heart]

@lament geode Has your question been resolved?

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[formal wind]

im stuck on this question : If f(x) is a differentiable function and g(x) is a double differentiable function such that |f(x)|≤1 and f'(x)=g(x) . If f2(0)+g2(0)=9 . Prove that there exists some c∈(–3,3) such that g(c).g''(c)<0. please help

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@formal wind Has your question been resolved?

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[daring citrus]

ive divided it and i got a remainder of 6

but ive no clue whether that is c or how to proceed any further

[mellow tusk]

for c put x=-1 cause the combined term would become 0

combined term ie: x+1)(ax+b)

[sullen wedge]

hello

[daring citrus]

ive no clue what that means 😭

so why have you chosen x=-1?

[mellow tusk]

at x=-1 x+1 is 0

[daring citrus]

it doesnt go in evenly?

so no 0?

[wanton tusk]

yesh that’s ur c

ur remainder

[daring citrus]

oh sick

so would i just put it back into the equation they put at the end and thats it?

[wanton tusk]

ya they js wanted the value of c

[daring citrus]

k sick, thanks so much

[wanton tusk]

welcc!

[daring citrus]

.close

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[wooden echo]

I have a two part question

For III and IV, how did we find the values using IVT and MVT without knowing the actual function?

[heady pollen]

f(0)=0 and f(1)=1

[gray isle]

^ and continuous over that interval

for part iii)

similar idea when applying mvt

[weary wyvern]

what values

[wooden echo]

f(c) = 2/3 and f'(c) = 1

[weary wyvern]

2/3 just works because it's in [0,1]

[gray isle]

all you care is that there exists such a value

[weary wyvern]

1 = [f(1) - f(0)]/[1-0]

[gray isle]

you don't know exactly where it/they'll be, just that they exist

[wooden echo]

Ohhh okay. Thank you.

[wooden echo]

It's a T/F type question so I wanted to make sure the values were the "right values"

[gray isle]

wdym by values were the right values

[wooden echo]

Like if they said f(c) = 1/4, would that also be correct or f(c) = 2/3 is the only solution

[gray isle]

is 1/4 also between 0 and 1?

do you know what ivt states?

[wooden echo]

I believe so ? IVT states that there exists a number c in (a,b) such that f(c) = N. Assuming the function is continuous on the closed interval [a,b]

[gray isle]

you're missing a key phrase about N

[wooden echo]

N is any number between f(a) and f(b)

[gray isle]

yeh

[wooden echo]

So then 1/4 is also between 0 and 1?

[gray isle]

yes

[wooden echo]

Thank you

.close

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[lone heart]

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[supple furnace]

16 : (-8) x (-7)

[supple furnace]

how do you solve this?

like what are the steps

[ornate condor]

is x

multiplication

[supple furnace]

yes

[ornate condor]

so

[supple furnace]

it is

[ornate condor]

did u know

-1 * -1 = 1

[supple furnace]

yes i do

[ornate condor]

um ye

so its the same as just

multiplying those numbers ignoring the neg

what do u get

[supple furnace]

no wb th 16 :

what is that

[ornate condor]

??

what did u just say lol

[supple furnace]

hol on

let me take a pic

[ornate condor]

um

[dreamy bramble]

3

[ornate condor]

i think its division

[gray isle]

division

[dreamy bramble]

16/56 i think is answer

[supple furnace]

why don't they put the division mark thinge instead

[ornate condor]

why tho doesnt it go from left to right

• and / have the same prior no?

[supple furnace]

that shit confused the shit out of me

[supple furnace]

@dreamy bramble

[dreamy bramble]

it does

16/(-8*-7) = 16/56

: == /

[gray isle]

that's not going from left to right

[ornate condor]

oh

i get it now

,w 16/(-8)*(-7)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Circle#0025

[gray isle]

that's inserting your own parentheses on a whim

[dreamy bramble]

bruh

r u bad at math

[ornate condor]

..

[supple furnace]

math is complicated, you gotta think outside the box

i thought it was 16/56 at first

[gray isle]

$\div$ is nearly as dodgy.

[ornate condor]

u are

goated

WHAT THAT

[ocean seal]

ℝamonov

[ornate condor]

oh

hahaa

[supple furnace]

🤣

[ornate condor]

yea so we use

$\frac{1}{0}$

[ornate condor]

[supple furnace]

$1+1$

[ocean seal]

daddy goat

[gray isle]

seeing as they're capable of displaying horizontal fraction lines, the choice is deliberate to test your understanding of the order of operations

[ornate condor]

ohh

[weary wyvern]

just notation

[supple furnace]

corrent

[ornate condor]

in the question options themselves hahaha

[supple furnace]

i will think harder next time

.close

[lone heart]

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[tawdry whale]

What does this equate to?

[tawdry whale]

Is it sqrt of 2?

[alpine sable]

set it equal to x

then substitute it

[tawdry whale]

Substitute what?

[dreamy bramble]

oh yeah mybad

i failed

[tawdry whale]

Xd

I know it's sqrt of 2

Butthats by brute forcing it

[alpine sable]

$x = \sqrt[3]{2\sqrt[3]{2\sqrt[3]{\dots}}}$

then

[ocean seal]

Jester

[alpine sable]

$x = \sqrt[3]{2x}$

[ocean seal]

Jester

[alpine sable]

and solve for x

[tawdry whale]

Got it ty

!close

.close

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[alpine sable]

whart

how does this work

im getting 3.27*10^44

[earnest garden]

What have you tried?

[alpine sable]

I did 1.97x10^22/6.022x10^23

10^23 is much larger than 10^22

so my answer makes sense to me

I dont understand how it evaluates to 3.27x10^-2

[earnest garden]

How did you get 10^44?

[alpine sable]

a calculator

[keen plinth]

let's try this with a calculator

,w scientific notation 1.97 * 10^22/(6.022 * 10^23)

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 夢雪#2250

[alpine sable]

hmm

ok then

thanks

[keen plinth]

please put brackets around your expressions properly

when using a calculator

[alpine sable]

yeah that was the problem

.close

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[vestal talon]

did I write this set notation right?

[serene junco]

,rotate

[vestal talon]

my answer key doesnt use union in its answer

[serene junco]

Your set notation is not quite right

But it should still have a union

What does your book say?

[vestal talon]

(-inf,-2) , (1,inf)

that left side on mine should be a negative 2 i messed that bit up

[serene junco]

What exactly do the directions say?

[vestal talon]

just to solve the inequality and give answer in interval notation

[serene junco]

Yeah, I guess they just used a comma instead of a union sign

It's a little less formal

But you're not going to get marked wrong for using union

But

You will get marked wrong for including the interval (-2,1) in the middle

[vestal talon]

is that only for when its equal to on either side?

or both?

like (-inf,-2] and [1, inf)

[serene junco]

The square bracket means you're including -2 and 1

which we don't want to do here

You wrote

(-inf,-2) U (-2,1) U (1, inf)

But the inequality is only true for x < -2 and x > 1

Not between -2 and 1

[vestal talon]

so is the line going away from eachother?

[serene junco]

Yes

[vestal talon]

whats a circumstance where you would use a middle set between two unions?

[serene junco]

Practically never honestly

If you do (-inf,-2) U (-2,1) U (1, inf)

That includes all real numbers except for -2 and 1 specifically

[vestal talon]

oh so it includes everything between -2 and 1 okay

just not -2 and 1 themselves

[serene junco]

Yes because (-2, 1) is the interval between -2 and 1

[vestal talon]

that makes a ton of sense okay

well thats better thank you tat

[serene junco]

sure thing 👍

[vestal talon]

.close

[lone heart]

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[ornate condor]

so for any non-zero mero f on D, f'/f is its log deriv?

[viscid pawn]

Wouldnt be helpful sorry

[keen plinth]

yes

just chain rule

take derivative of log f(z)

[ornate condor]

hm hb the

simple poles at zeros n poles of f

oh ye

i get it now

conceptually

thanks

all u

❤️

[ornate condor]

all u

thanks

.close

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[mild bluff]

Hello

[mild bluff]

How do I do this

<@&286206848099549185>

[prisma zephyr]

what is a good equation to describe when to vectors are parallel?

[mild bluff]

Ima be honest I have no clue what the hell im supposed to do

[prisma zephyr]

hmm

u know the cross product?

[mild bluff]

no?

[prisma zephyr]

oh

[mild bluff]

bro I started a levels 2 weeks ago and I literally missed the lesson I was supposed to be taught this

so I’m completely lost

[prisma zephyr]

okay thats shit, ik how u feel

well first, your equation should have something like a+b

because this is the resultant your looking for

the crossproduct is a product between vectors which spits out another vector perpendicular to the others

and it returns to 0 , if both vectors lie on each other

[mild bluff]

And c must be a multiple of a+b right?

Coz they’re parallel

[prisma zephyr]

yes

so youve got to ways to describe this:

the first: $(\vec a+\vec b)\times\vec c)=0$

[mild bluff]

i have no idea what any of that means 😭

[prisma zephyr]

okay then maybe the second equation helps more:

$\vec a+\vec b=k\vec c$

[ocean seal]

Willst-du-Stress-Tensor

[prisma zephyr]

where k is any number but not zero

then solve for p

like u said c is then just a multipile of a+b

[mild bluff]

Is p -3/2

that’s what I got idk if I’m even close

[prisma zephyr]

idk I need to calc it myself

but -3/2 seems like a nice answer

yeah but thats true

[mild bluff]

Oh wait

I was doing a dif question

[prisma zephyr]

oh xD

[mild bluff]

This one

Same type of question just dif numbers

I got p= -3/2 for this

HOLY SHIT

i think I got it right

tysm :>

[prisma zephyr]

nice np I havent rlly done anything^^

[mild bluff]

moral support if nothing else ♥️

[prisma zephyr]

<3

I recommend to u 3b1b its a youtube channel

[mild bluff]

I’ll check it out :>

[prisma zephyr]

he nicely explains how to get a little intuitive when doing linAlgebra with cool animations

then have a nice day bb^^

[mild bluff]

3Blue1Brown?

Is he the one with the square bouncing video that equates to pi?

if so then I’ve seen some of his stuff, I’ll definitely watch more lol

you have a nice day too (:

[lone heart]

@mild bluff Has your question been resolved?

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[junior warren]

how would i go about evaluating this

[plain flame]

divide top and bottom by x^3

[junior warren]

i see

[open urchin]

alternatively spam lhospital until you get the answer

[junior warren]

then i end up with all the fractions with x in denominator

[open urchin]

but those go to 0

as x goes to inf

so they disappear

[junior warren]

and then i think i can just eval from there right?
yes

exactly

[sour dove]

https://tenor.com/view/anger-management-jack-nicholson-yes-duh-nods-gif-5222702

[junior warren]

so smart

[open urchin]

im gonna send l'hospital to the hospital

[junior warren]

divide by big number=small number

[open urchin]

fr

[sour dove]

lol he didn't even make up the rule himself. He bought it off of someone

[open urchin]

that's like every mathematician ever

[junior warren]

newton moment

[sour dove]

yes exactly. One thing you can take away from this is that the highest degree polynomial will be dominant

[junior warren]

right

[sour dove]

if your highest term is in the denominator, it diverges

[junior warren]

so jus try dividing by the highest degree

[open urchin]

yes

[sour dove]

if it's in the denominator, the limit is 0

and if they're equal, then the limit is the coefficient of the numerator divided by the coefficient of the denominator

[junior warren]

oh true

because something like

3x/2x

is just 3/2

[sour dove]

bingo 😉

[junior warren]

wow

[sour dove]

3x^2/2x^2 is also 3/2

[junior warren]

yeah

x cancels

[sour dove]

so if you have something like (4x^5 + 2x + 1) / (2x^5 - 4x^4 + 2x^3 + 6x^2 + 1), then the limit is 4/2 = 2

just a shortcut 🙂

[junior warren]

now what if i had the same question but the limit was -infinity

[sour dove]

the leading polynomial dominates as x approaches infinity

[open urchin]

also you can think of all the terms of the polynomial with a degree smaller than the degree of the polynomial are useless compared to the biggest term

[open urchin]

same answer

[junior warren]

oh right

just the negative version of it?

bc still really big number

[open urchin]

no

just same answer

[junior warren]

oh true

[open urchin]

wdym by negative version?

[junior warren]

because it goes to 0

no i was not thinking straight

[open urchin]

relatable

[junior warren]

3/x whether its negative or positive

goes to 0

lol

ok ty everyone

.close

[lone heart]

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[junior warren]

stuck on how to evaluate this. do i divide by x^2 or sqrt(x^2)

[dawn hamlet]

sqrt(x^2) which is x

[junior warren]

is that only when the limit is positive

[dawn hamlet]

Yeah

[junior warren]

ok

so when i expand it

how do i deal with sqrt in the numerator

[dawn hamlet]

Thats why i wrote x

[junior warren]

yes

in my numerator i now have

(11/x+8x)^1/2

did i expand wrong

[dawn hamlet]

Wait i am dumb, thought numerator is the bottom 😂

[junior warren]

L

i cant break it into sqrt(11) and sqrt(8x^2)

right?

[dawn hamlet]

No

[junior warren]

hm

[dawn hamlet]

U should divide the inside of the sqrt by x^2 not x

[junior warren]

oh rigth

because when i move the x under the root

i have to square it to be equal

okso

now my numerator is

sqrt(11/x^2+8)

can i evaluate that from therre?

i should be able to?

[dawn hamlet]

Yes u can

[junior warren]

now what if the limit was negative

as that is the next question

i cant evaluate at that point right

bc sqrt of negative

[dawn hamlet]

Then x = -sqrt(x^2)
so u will have to "add" a minus sign in front of the sqrt

[junior warren]

actually wait in this instance

it wouldnt change anything right

because those terms are still all going to 0?

[dawn hamlet]

Only some of them

[junior warren]

the 11/x^2 is unaffected

bc squared is a positive

so that goes to 0 anyways

but on denom

6/x

goes to 0?

as well

[dawn hamlet]

Remember that u have the 3 so it is not zero on the denominator

[junior warren]

yes

but that term

goes to 0?

[dawn hamlet]

Yes

[junior warren]

so it ends up being the same limit?

[dawn hamlet]

What?

[junior warren]

when it goes to +infinity

my limit was sqrt(8)/3

but when it goes to -infinity

shouldnt it also be

sqrt(8)/3

[dawn hamlet]

No, it should be -sqrt(8)/4

[junior warren]

oh

but in the term sqrt(11/x^2+8)

wouldnt the term with x just end up being 0?

[dawn hamlet]

U divided the bottom and top by x right?

[junior warren]

yes

[dawn hamlet]

But x is negative so x =/= sqrt(x^2)

[junior warren]

ohhhhh

yes

1/x when x is negative will be

negative

even if x is really big

its just a really small

negative number

[dawn hamlet]

So u get x=-sqrt(x^2)

And thats where u get the minus sign from

[junior warren]

right

ok that makes sense

and that negative has to get carried through all the expansion and simplification

and thats why its there at the ends

cool

ty

.close

[lone heart]

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[mint estuary]

Yo

[mint estuary]

Help

[rustic coral]

With what? It looks complete

[mint estuary]

I didn’t draw that line

It says know how to….

Idk how to

So help

[rustic coral]

Oh

You can just Google that tbh

[mint estuary]

I did

Don’t understand

[rustic coral]

Anywhere specific?

[mint estuary]

Huh

?

Yo

U there

.close

[lone heart]

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[shut bison]

Anyone help for question 12? Just need to knw how to set up the formula

[mint estuary]

Ok

[harsh swallow]

You set up a set of equations

You give the amount of each gear teeth a name

For example gear 1 has A amount of teeth

[mint estuary]

That

Then

[shut bison]

This is what I did

Urs seems simpler

Can you finish going thru yoira

[mint estuary]

Me?

[harsh swallow]

In terms of g1, g2, and g3 what is gt?

[shut bison]

Yes plz

Gt is total

[harsh swallow]

Total is 107

But total can als be written in terms of the teeth on gear 1 and gear 2 and gear 3

[mint estuary]

Idk

Idk how to setup equation bruh after that

I would just guess

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@shut bison Has your question been resolved?

[lone heart]

@shut bison Has your question been resolved?

[lone heart]

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[pearl lantern]

How would i take the laplace transform of this function

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[deep grove]

the area between the curves, im not sure what it would be

[deep grove]

would it be f(x)-g(x)

or g(x)-f(x)

and why

[waxen flame]

Which equation is above the other?

[deep grove]

fx i think

[waxen flame]

To get a positive result, you want to subtract the smaller value from the larger value.

[deep grove]

so f(x)-g(x) here?

[waxen flame]

Yes.

[slender gull]

If confused you can do either of the two and consider the absolute value after that.

Should work also.

[deep grove]

oh

[waxen flame]

In the bounded region, f(x) > g(x) so you want to subtract g(x) from f(x).

[deep grove]

im a bit confused on how subtracting even gives the right area can someone explain that

[waxen flame]

You are learning integrals right now?

[deep grove]

yeah