help-39

[midnight haven]

[rustic gate]

.close

[pearl pond]

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[midnight haven]

wait ayo this can be a new party channel

[rustic gate]

.reopen

[pearl pond]

✅

[rustic gate]

[midnight haven]

lets do the question game agian

[rustic gate]

[vital estuary]

https://tenor.com/view/high-intelligence-specimen-gif-25869526

[midnight haven]

#help-5 message @rustic gate

a bunch of shit is like

breaking down

[rustic gate]

[midnight haven]

<@&268886789983436800> the help channels are breaking

[rustic gate]

time to call in mnipple

[midnight haven]

[cinder flower]

hey maybe help 21 will break

[midnight haven]

if only

[midnight haven]

@rustic gate it got the name now

.close

[pearl pond]

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[midnight haven]

wtf is happening

[rustic gate]

[lethal thorn]

Hi everyone

I need help with math

[sharp smelt]

[pearl pond]

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[kindred perch]

Hi, how do i factorise (5-x-2y)(x+2y)-6?

[kindred perch]

i was planning to move the 5 out somehow

but idk what to do

wait nvm

.close

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[shell brook]

Why did we not let a-2d and a+2d?

[cursive wraith]

you need four numbers

so what would your other 2 numbers be

[shell brook]

I don't get it

[cursive wraith]

"Why did we not let a-2d and a+2d?" instead of what?

[modest sage]

if you let a-2d and a+2d

[shell brook]

they let a-3d and then istead of going to a-2d they went a-d

which i didnt understand

hope that clears it up

[modest sage]

then the four numbers don't form AP

[cursive wraith]

the progression between all numbers is 2d

(a-d) - (a-3d) = 2d

(a+d) - (a-d) = 2d

[modest sage]

becasue a-2d and a-d's difference is not 2d but d

[cursive wraith]

(a+3d) - (a+d) = 2d

[shell brook]

oh

ig that makes sense

tyvm

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[slender atlas]

someone help

[daring aspen]

(AP-BP) = 2a is a form of hyperbola

where A and B are focii coordinates

[slender atlas]

how do you know its hyperbola

[slender atlas]

help

[pearl pond]

@slender atlas Has your question been resolved?

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[wicked viper]

Prove Binomial coefficient
C(n,0)+C(n,4)+C(n,8)+... = 2^(n-2) + (√2)^(n-2) cos(nπ/4)

I am stuck at the beginning

[plush bramble]

what class is this for

[pearl pond]

@wicked viper Has your question been resolved?

[pearl pond]

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[midnight haven]

When considering the line integral for a vector or scalar field over some parameterised curve defined by $\6{\vj r}t$, what exactly are the compositions [
\6{\vj F}{\6{\vj r}t} \tss{and} \6F{\6{\vj r}t}]
performing/entailing?

[midnight haven]

where F (arrowed) and F are the vector and scalar fields respectively
like i dont understand what exactly this composition is doing

edited my question for better clarity hopefully

[timid spindle]

It describes the field as you're going along your curve

[limber oasis]

it's qylo day today init

[plush bramble]

need a max count of how many channels a helper opens in a rolling 24 hour period

[midnight haven]

so if F(x,y,z) is some surface (lets consider the scalar field) the composition is restricting our function's range to follow that of the curve's path?

[timid spindle]

Yeah, in that case it's a path on the surface

[midnight haven]

i thought the path by definition has to be on the surface?

wait no let me reread that slowly

[timid spindle]

Here's a helpful but obnoxiously slow gif

[plush bramble]

oh i like that gif

[vital estuary]

thats the wikipedia one right

[timid spindle]

Possibly, I just remember it and search "line integral gif"

That and the radians one are my favs

[midnight haven]

wait what is the red line

[limber oasis]

r(t)

[midnight haven]

i thought that is the blue one C

[timid spindle]

The red is c, the blue is "f(c)"

Your curve is in the xy plane

[midnight haven]

oh i was confused because they showed C right next to summoning the blue line

[timid spindle]

Im only thinking about z =f(x,y)

[midnight haven]

but yeah i understand that

but like for vector fields what is this analgously

[timid spindle]

Eh harder to visualise

You have a curve in 3 space and f(r(t)) is just containing the information of the value of the vector field at all point on your curve

Oh acc here's another gif for 2d vec field

[midnight haven]

so in a sense f(r(t)) (vector) is restricting our vector field's range to like along the path?

oh

[midnight haven]

oh boy

wtf is happening now

lemme eyeball this

[timid spindle]

The dot product is measuring "how aligned is your velocity with the vector field as you travel along the curve"

[midnight haven]

right

but whats the point in the composition F(r)

like cant we totally just find the dot product F with r'

[timid spindle]

You need to know both the value of the vector field at a point on your curve and your velocity vec at that point

At all points along your curve

[midnight haven]

ah ok because we dont want like some random value of the vector field somewhere else

in bumfuck nowhere

[timid spindle]

The former is F(r(t)) and the latter is r'(t)

[midnight haven]

ok this is starting to come together

[timid spindle]

Works the same for a 3d vec field but the visualisation becomes messier

[midnight haven]

yeah i can definitely imagine

but okay wait

more question

so like

imagine a one dimensional vector field

,, \vj F = \6M{x,y,z}\vc\I

[timid spindle]

https://mathinsight.org/line_integral_vector_field_introduction

[midnight haven]

for example

[timid spindle]

Has a good interactive applet

[midnight haven]

oh okay thanks will check that out later

[timid spindle]

I wouldn't call that one dimensional, just all the vectors are pointing up or down

The dimension just refers to the ambient space

[midnight haven]

dont u mean right or left?

i mean

[timid spindle]

Right yeah

[midnight haven]

ok those are like pretty vague directions

[timid spindle]

However you wanna imagine 3 space

[midnight haven]

yeah

[timid spindle]

The direction i points in for you

[midnight haven]

but like all the vectors lie on the x axis no? its kinda fucked because the scalar is a multivariable function of 3 variable

actually how does that make sense

[timid spindle]

It is assigning a vector to every point in 3 space

So you can imagine going to some point in 3 space and F tells you which way to face

[midnight haven]

OH wait so the vectors lie on R^3 but like its just the direction itself is like one dimensional

in a sense

[timid spindle]

E.g. a 3d vector field might model what directions particles in water are going to flow depending on where they are

[timid spindle]

Yeah sure the possible directions only have one dimension, even though its a 3d vector field

[timid spindle]

This could be like a water stream in a pipe

[midnight haven]

as in this is kind of whats happening

[timid spindle]

I highly recommend googling 3d vector field and finding some interactive geogebra thing or something

Then you can spin that mfer around and see what's up

[midnight haven]

i really would love that

lmfao

[timid spindle]

https://www.geogebra.org/m/u3xregNW

Since you asked so nicely

[midnight haven]

LOL

geogebra makes my pc go brr but lets see

yeah ok i see

hmmmmm

[timid spindle]

The most simple ones will just be putting constants in the PQR slots, everything will point in one direction

They get cooler once they depend on xyz

[midnight haven]

wait so zac

so when plotting like a vector field

like uh

how tf do i put this

[timid spindle]

Youre just plotting some sample points if that's what you're gonna ask

[midnight haven]

for Pi + Qj + Rk, (P, Q, R) determine the actual (x,y,z) coordinates of the vector and the unit vectors determine the direction?

[timid spindle]

Yeah

It's just describing a vector the usual way

A linear combination of ijk

But your linear combination can depend on the xyz of where you are in space

[midnight haven]

ok i see

i feel like this all should be very simple for me but because im rushing thru so much rn its somehow harder for no reason

[timid spindle]

Honestly just mess around on geogebra for a while and I think it'll click

[midnight haven]

i gotchu king

[plush bramble]

i gotchu too

[midnight haven]

LOL

[timid spindle]

Put in different vec fields and try and see why is has that shape because of what you put in

[midnight haven]

nice one riemann

[timid spindle]

E.g. if you want less magnitude as you get further away from center you could divide everything by 1/(x²+y²)

[midnight haven]

yeah good idea

also like vector fields are like obviously defined over all the reals right

[midnight haven]

this is just showing like

discrete bullshit

ok im assuming continuity and a smooth path and whatever

[timid spindle]

Then you've just invented gravity

[midnight haven]

[plush bramble]

and electrostatics

[midnight haven]

is this in xy?

[timid spindle]

2d yeah

[midnight haven]

ok yeah makes sense

because like

[timid spindle]

Hopefully it's clear what the 3d analogue is

[midnight haven]

coloumbs law is like uh

1/r^2 (insert bullshit here)

so i guess that makes sense

same for gravitational

[timid spindle]

Yeah so electric/gravitational fields really are vector fields

[midnight haven]

riiight

actually

yk im speedrunning this rn for my applied electromagnetism course

[timid spindle]

Okay then definitely get to grips with this lmao

[midnight haven]

yeah im trying

it heavily sucks because uh

we didnt actually learn it in calc 3

its super weird because uh

our calculus 3 is a course that actually includes both calculus 2 and calculus 3 together

so like

from infinite series to the multivariate stuff

but ofc they cant fit that much content in so they had to stop after like triple integrals

[timid spindle]

You'll do it in calc 12

[midnight haven]

every passing day i regret not knowing linear algebra

ITS ALL LINEAR ALGEBRA GODDAMN IT

okay whateve

[timid spindle]

That is also unfortunate

Come to my uni I'm tutoring it

[midnight haven]

ok hook me up with the visa bestie

alright enough CRANKERY

until another time my friends

thanks a bunch yall especially zac

.close

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[dusk otter]

Can I receive help with this question? I will send my solution so far

[dusk otter]

I know that I need to show how it holds for n + 1

I am just unsure how to show this because it does not seem to be working out, like I can't get the left side to equal the right side

[pearl pond]

@dusk otter Has your question been resolved?

[pearl pond]

@dusk otter Has your question been resolved?

[modest sage]

hi again

I'll give a tip for mathematical induction problems

It would be more helpful if you write statement or equation for n and n+1

In this case,
t_n = 2t_{n-1} + 1
t_{n+1} = 2t_n + 1

keep looking it then you will find the way

[dusk otter]

Hello : )

[dusk otter]

Alright I try and figure something out : )

[dusk otter]

I wrote my equation for n + 1 which was t_n+1 - 2t_n = 1

I rearranged a bit

Dk if that is necessary

[modest sage]

Can you find the relation between the two equation?

n and n+1

I think they'll be related

[dusk otter]

l

I tried this out but not for sure

Seems that they both equal to one

Dk if that helps

well wait yeah that's obvious

[modest sage]

and don't forget the given relation

p(n) is given and there is another given relation

p(n+1) is relation that you want to claim

[dusk otter]

Do I need to add something to p(n) to show that it equals p(n +1)?

[modest sage]

yes probably

p(n) only will not enough to solve that

[dusk otter]

would this work at all?

I was thinking to simplify it and prove that it is equal to 1

[modest sage]

3t_n - 2t_{n+1}?

[dusk otter]

sorry no -1

[modest sage]

ok

[dusk otter]

Like in the subscript of the 2t

it should be n - 1

[modest sage]

keep going

[dusk otter]

I dk how : (. I may be overthinkng this but how do you simplify subscripts?

like can you only simplify subscripts that are common

like this?

[modest sage]

yes

you're almost done

[modest sage]

yessss

[dusk otter]

yayy!!

Thank you for the help!

[modest sage]

😄

[dusk otter]

I have another similar one I am working on . I am going to apply the same techniques and I just want to ask if it is correct since there are nosolutions in my homework unfortunately

Which is why I have to ask so many questions all the time T_T

[modest sage]

That’s ok

But I have to sleep now…

[dusk otter]

oh no problem!

You can reply when you are available

or i can ask another server

[modest sage]

Thanks

Its 6 am in my country lol

[dusk otter]

No problem!

[pearl pond]

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[dusk otter]

.close

[pearl pond]

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[pearl pond]

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[neon pier]

how would i solve these?

[neon pier]

for the second one, does it matter that one of them has a multiplicity of 2

does the multiplicity even matter for any of them

and can i just factor (x^3 - x)

[west sapphire]

x^2 is nonnegative, so if the inquality is gonna hold, then exactly one of the other factors has to be <= 0 and the other has to be >= 0

[neon pier]

i dont understand what youre saying

oh wait

one of them has to be positive and one has to be negative right

so that it's always less than or equal to 0

but how does that help like

hmm wait let me work it out a bit

i got
x^2(x+1)(x-1)(x-x)(x-1)
for the factors

for the x-x can it just be 0

or is just the number that we put in

it's jujst the number right

that doesnt even make sense

x=x

wait do i just not include it

so the crucial point are 0, -1, 1

@west sapphire bitte help me

i got that it's always less than or equal to 0

-infinity < x < infinity?

guys what did i do wrong

<@&286206848099549185>

oh isee

it's x(x+1)(x-1)

is it -1 < x < 0

do i need the equal signs

guys for the second one i think it's
-infinity <= x <= -7

[subtle shard]

yes as i can see

but

-1 ≤ x ≤ 0

[neon pier]

icic

[subtle shard]

because it is 0 when x = -1 and 0

[neon pier]

what about the second one?

[subtle shard]

yes -∞ < x ≤ -7

For minus infinity, it is not written as greater than or equal to

and for infinity too

[neon pier]

icic i wasnt sure abt the infinity either

thanks so much

[subtle shard]

you are welcome

[pearl pond]

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[cosmic mist]

Hey! I'm currently making exercises for a group project for my linear algebra class. I made this question on ideals. Can someone tell me if it looks good and the solution makes sense?

[cosmic mist]

And also if the question itself makes sense or not

[pearl pond]

@cosmic mist Has your question been resolved?

[cosmic mist]

<@&286206848099549185>

[spiral granite]

the question makes sense, however "-f = -0 = 0 = f" in 2.1.2 is suspicious

[cosmic mist]

how so?

[spiral granite]

this is supposed to hold for any f in I, so it cannot be correct. I suppose you meant that for constant terms, not the polynomials themselves

[cosmic mist]

ah i see, you're right that could be ambiguous

[pearl pond]

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[kindred geyser]

Hello I don’t understand well the question

[kindred geyser]

Argumenter is argue

And a(k)/b(k) is k!/2^k

[plush bramble]

Yea there's not enough context

[vivid vigil]

send the full question

[kindred geyser]

It is the full question

[vivid vigil]

that's part e

we need to know what $a_k$ and $b_k$ represent

[jolly parrot]

aurelianus

[kindred geyser]

I just said it

[plush bramble]

You follow the pattern from a-d

[kindred geyser]

Yea ?

Which criteria should I use ?

Do I need to use the Alembert one ?

Hmmm

[plush bramble]

Did you try using induction

[pearl pond]

@kindred geyser Has your question been resolved?

[kindred geyser]

Idk how to use it

[pearl pond]

@kindred geyser Has your question been resolved?

[pearl pond]

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[wise sparrow]

how were they able to simplify it ot such? for example how did the 5 turn into 25

[brittle mantle]

pick things in common from both adding parts

$3x^2(5x^2+1)^6 + 60x^4(5x^2+1)^5$

[jolly parrot]

Aza

[brittle mantle]

what do they have in common?

$3x^2(5x^2+1)^5$, no?

[jolly parrot]

Aza

[brittle mantle]

so factor that out

$3x^2(5x^2+1)^5 [ (5x^2+1) + 20x^2 ]$

[jolly parrot]

Aza

[brittle mantle]

just sum things

$5x^2 + 20x^2 = 25x^2$

[jolly parrot]

Aza

[wise sparrow]

so basiclly do we just

$(5x+1)(5x+1)$

[jolly parrot]

Daddy Goat

[pearl pond]

@wise sparrow Has your question been resolved?

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@wise sparrow Has your question been resolved?

[wise sparrow]

.close

[pearl pond]

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[timber elbow]

Does anyone know how to solve this. I'm so lost

[pearl pond]

@timber elbow Has your question been resolved?

[pearl pond]

@timber elbow Has your question been resolved?

[pearl pond]

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[inner peak]

can anyone help me with p2?

[tired violet]

!noclopen

[pearl pond]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[tired violet]

the length of the arc is just t isnt it?

since the radius of the circle is 1

[inner peak]

ahh

trying to find BD

[tired violet]

yeah isnt it t

[inner peak]

yea its right burt can u explain how??

[tired violet]

the arc length of a sgement of a circle is given by rtheta where theta is the angle of the arc in radians and r is the radius

how to derive it? i mean its kinda how radians are defined i guess

[elfin summit]

the circumference is 2pi radians times the radius

[tired violet]

you could use an integral mayb e if you wanted to

[elfin summit]

by changing the angle, you can get the length of an arc

[inner peak]

could u explaon mathemtically how pls i dont understand all these words sorry lol

so r theta r = 1 theta = t

[tired violet]

yes

[inner peak]

oh ok

thanks

[tired violet]

again its just how radian measures are defined

the circle is 2pi radians

[inner peak]

yea i never used the radians formula

i always just converted it to degrees

[tired violet]

stop doing that

youre in calculus

[inner peak]

ya bad habit

[tired violet]

stop

lmao

learn radians

theyre so much better

and more intuitive

[inner peak]

true

[elfin summit]

radians>degrees except for application problems

[tired violet]

think about it like this, degrees abritrarily divide the circle into 360 pieces, why?? idk ease of computation in acient times or smth

[inner peak]

hm yea ill do it all in rads from now

[tired violet]

radians divide the circle into units of pi which is a natural unit since it relates the diameter and circumference of the circle

pi is literally the ratio of the circumference and the diameter of the circle

a unit circle has circumference 2pi

there are 2pi radians in a circle as well

thats another way of looking at it

[inner peak]

true

hmm

[tired violet]

radians are essentially the circumference

of a unit circle

in your problem you can see that clearly since t is the length of the arc

literally just the angle is also the length of the arc

like i said earlier it is because that is essentially how radians are defined

[inner peak]

yea got it thank you

[pearl pond]

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[drowsy leaf]

square root of 11 is 3.317

Is my calculator wrong

[lone shale]

I tgink ur calc rounded

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[fervent fulcrum]

this is the answer key... to use the nth term test u have to include the alternating part though right

[fervent fulcrum]

how do u know the limit as n approaches infinity of the entire thing though

[drowsy leaf]

This is NOT math 😭

[fervent fulcrum]

wdym

[drowsy leaf]

is this calculus

[fervent fulcrum]

yes

[drowsy leaf]

Why is there an infinity sign in math

[worldly glacier]

well (-1)^n just gives you the sign

it's alternating signs

anyway

you can ignore the suns

and notice that it doesn't go to 0 as n -> inf

[fervent fulcrum]

to use the nth term test

if u don't knwo the exact limit

can u just state it doesn't equal 0 if its obvioujs like this

[worldly glacier]

yeah ofc

[fervent fulcrum]

wait so like

[worldly glacier]

the limit is (-1)^n 1

for some k \in N

which is like not 0

[fervent fulcrum]

it kinda keeps alternating but will never equal 0

[worldly glacier]

mhm

[fervent fulcrum]

thank you!!

that makes sense!!

[worldly glacier]

yeah like you would expect like

something close to +1, -1, +1, -1, 1,..

in the end or sm

close to infinity

[fervent fulcrum]

ok cool

[worldly glacier]

it won't be exactly 1 but close to 1

[fervent fulcrum]

like it will get kinda close to 1

[worldly glacier]

mhm

[fervent fulcrum]

but like alternating

ty!!

.close

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[prime bramble]

do you know the formula for the volume of a cylinder?

[charred badger]

Took the exact words out my mouth. To a T lmao.

[copper grotto]

yeah isnt it pi r ^ 2 h

[prime bramble]

right

[prime bramble]

my bad

[charred badger]

We're all here to learn and help others!

[prime bramble]

so V = pi*r^2*h

[copper grotto]

so would i do 3.147^2

[prime bramble]

we can rearrange that formula for h, and then plug in what we know to get a numerical answer

[copper grotto]

wait idk the h

[copper grotto]

huh 😭

[prime bramble]

that's the point

[copper grotto]

wait i dont get it

[prime bramble]

1846 = 3.14 * 7^2 * h

[copper grotto]

how do i figure out the height

?

[charred badger]

Move everything to one side of the equation, leaving only h on one side of the equation.

[copper grotto]

im so sorry but what do u mean by that

[charred badger]

So when solving for a variable, you generally want to make things easier on yourself by taking the variable you are solving for, and separating it from the rest of the equation.

for example, an equation such as h*pi+30=20, you would want to simplify by moving pi+30 to the other side of the equation, leaving h alone. Remember to pay attention to the signs as they swap sides of the equation.

[copper grotto]

oh okay i think i get it

is the answer 154 cm?

[charred badger]

That doesn't sound correct. Let's just start by figuring out how to move things so that h is on it's own on one side of the equation

[copper grotto]

Okay

[charred badger]

so when you are trying to move expressions from one side of the equation to the other, you want to cancel out everything on the side of the equation with h, except for h itself.

[copper grotto]

wait so

if u do

1846/pi x 7^2

would u get the answer?

[charred badger]

That would set the side of the equation with h to only have h, yes. From there, you can simplify and divide to solve for h.

And remember to round your answer according to the question.

[copper grotto]

oh alright so that would be 1846/153.86

and then

12

rounded up it would be 12?

is that correct 😭

[charred badger]

I believe so. Your answer should be 11.99 or so, which is then rounded to 12.

[copper grotto]

yeah 12 is one of the options

thank youuu

according to the rules ur supposed to limit to one question right

[charred badger]

And if you'd like to check your answer, you can fill in h as 12 in the original equation, then multiply it out and if your answer is near 1846 (to account for your rounding), then your answer is most likely correct.

[copper grotto]

oh alrightt

[charred badger]

Sorry?

[copper grotto]

I had one more problem to solve

Its a different topic in math tho

should i send it here or just make a new

thing

[charred badger]

Oh you mean the server rules. Yeah you can go ahead and close this channel and open a new one.

[copper grotto]

Okay

[lone shale]

Nooo

[copper grotto]

what 😭

[lone shale]

Sorry wrong chat

[prime bramble]

!done

[pearl pond]

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

[prime bramble]

@copper grotto

[copper grotto]

oh mbmb

.close

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[latent tendon]

Sorry, poste this in another channel but it closed and I can't find it anymore so I'm reposting it here. But anyway, I'm really stuck on number 20. We are supposed to use factoring but I'm not sure how to since this one is different to what we have done

[grizzled dust]

,rotate

[grizzled dust]

FYI: you can always go to the search box in discord and type from: (your username) to find a channel like that, click your username when prompted

[latent tendon]

Thanks!!

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[sage bane]

How do I continue solving this legrange multiplier equations? The question is the above, we want to determine the min and max volume according to the surface area and length constraints. I think I’ve done everything correctly and I’ve got two cases but I’m not sure how to continue with case 1

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[sage bane]

not solved

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[swift lodge]

Question 10

[swift lodge]

What I have tried

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[gleaming plank]

cota + cotb = cotacotb-1

use this fact

simply the lhs and then use this fact

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[twilit harbor]

hi

[twilit harbor]

Can I get some verification for this

its an openstax exercise #126 but they don't include the answer https://openstax.org/books/calculus-volume-1/pages/3-3-differentiation-rules#fs-id1169739304919

anyone ?

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[undone pivot]

I need help with this function question please which I dont know how to get with only multiple choice questions

[undone pivot]

Would be great if you explain the steps for it

[pearl pond]

@undone pivot Has your question been resolved?

[vagrant dirge]

Well we can split it up and find out the domain for both

Then find a common domain and thats our domain

Think about it, can we have -inf terms in the first one? What about infinity? What happens if we add a large positive number to the first square root

[pearl pond]

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[undone pivot]

so what would be the asnwer?

I need steps for it

if you dont mind

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[paper sorrel]

for part 4 i forgot what I did, when it says neither and not or do you multiply?

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[paper sorrel]

Just wondering how to do part 4

.close

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[upper delta]

How can we proof that $\frac{-2x^{3}-3x^{2}+1}{\left ( x^{3}+1 \right )^{2}}$ is equal to $-\frac{2x-1}{\left ( x^{2}-x+1 \right )^{2}}$?

[jolly parrot]

Bennxy

[upper delta]

I couldn't find a step where we should get the other denominator

Long division won't work here either

[midnight haven]

yo

have you tried setting them as equals? and seeing where that leads?

[plush bramble]

-1 is a root of both numerator and denominator

[midnight haven]

oh ;-;

[plush bramble]

so divide top and bottom by x+1

if -1 is still a root, repeat

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[rocky flume]

I’ve simplified it to (5^x) times 5 ^log5(x)
But I don’t understand have 5 ^log5(x) = x. Can somebody help me fix this flaw in my knowledge

[plush bramble]

https://www.mathsisfun.com/algebra/logarithms.html#:~:text=In its simplest form%2C a,So the logarithm is 3

[rocky flume]

Thank you

[plush bramble]

if a=5, then raising 5 to both sides of the bottom equation gets the top

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[pearl pond]

Please don't occupy multiple help channels.

[smoky saffron]

How do you know which part is discontinuous?

nvm

.close

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[plain pulsar]

how do i go about solving this

[versed mica]

is this a test

[plain pulsar]

no

its homework

[calm wing]

hint: cos(x) is bounded and e^3x goes to 0

[plain pulsar]

how would i express that as an answer

[pearl pond]

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[rough forge]

what artemetra wrote is basically it

e^(3x) going to 0 and cos(x) is oscillating betwewn -1 and 1

so you have basically 0 • [-1,1] = 0

[plain pulsar]

so the answer is zero

[rough forge]

yea

due to artemetras explaination

[plain pulsar]

i see

.close

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[white cape]

Hello

[white cape]

Differential equations

once again

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@white cape Has your question been resolved?

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[distant marlin]

Can someone do this

[distant marlin]

Maybe write it down on a Jamboard

[west sapphire]

try multiplying/dividing by the conjugate?

[distant marlin]

Specifically the numerator?

[strange quiver]

Are you allowed to use l'Hopitals rule

[distant marlin]

Haven’t learned that yet

Introductory to calculus

I’d assume hospital rule is calc 1?

[strange quiver]

Yeah

Can you see that the numerator must be 0?

When x = 0

[distant marlin]

Indeed

Can’t we just rationalize

The numerator and denominator

[strange quiver]

You can try

Oh i got it

[strange quiver]

First use this to find out b

Then "rationalize" to find a

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[frank monolith]

Hello, I need help with a group theory question:

[west sapphire]

wait what?

i assume they mean "is a subgroup of (Z13\{0}, x)"

not +

[frank monolith]

let me see

thats incorrect

[strange quiver]

It might mean addition modulo 13

[frank monolith]

I translated it from spanish

[west sapphire]

but Z13\{0} is not a group under addition

[strange quiver]

Oh wait it's integers nvm

[frank monolith]

it should be a . instead of +

[west sapphire]

yea ok good

you can prove it using the usual test for whether something is a subgroup

[frank monolith]

So here's the table

U13 would have 13 elements here?

since there's a 1 in every row

[west sapphire]

12 elements in fact

0 is not in U13

[frank monolith]

yes 12,

[frank monolith]

0 should have a line above it, btw

[west sapphire]

yea, understood

any thoughts on how to prove that U13 is a subgroup?

[frank monolith]

So, what would this contain?

I mean you could just look at the tables

[west sapphire]

by definition that's all the nonzero numbers in Z13

so everything except 0

[frank monolith]

so its the same group?

[west sapphire]

in fact it will turn out that U12 equals Z13\{0} (you already saw that from the multiplication table)

but you only need to show that it's a subgroup of Z13\{0}

which is easier in general

[frank monolith]

You can easly see from the table (ignoring the 0 row and column) that all elements got an inverse, and from the way we define it the identity element is contained in U13. So that's enough for those two parts I imagine

And for proving closure you could also refer to the table?

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[pearl pond]

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[delicate solstice]

idk how to solve this integral anymore

[delicate solstice]

it’s supposed to be trig substitution but it’s killing me

if someone could help me run it down entirely, it’d be awesome

[pearl pond]

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[delicate solstice]

<@&286206848099549185> sorry for the ping, i’m just wondering if anyone has any tips

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@delicate solstice Has your question been resolved?

[pearl pond]

@delicate solstice Has your question been resolved?

[inner granite]

Search this on google "integration of (-10)/(x^2 + 2x + 2) dx".

[earnest yoke]

https://www.integral-calculator.com/

it turns into a standard integral for arctan

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[hazy arrow]

For 8c I got the same as 8b

.close

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[dense crater]

ive done this type of question before but now i just can't remember how to do it, can someone help?

[waxen talon]

Is that a right angle

[dense crater]

yes

[midnight haven]

try to get relations using those two right angled triangles having same base

[waxen talon]

Can you first name the sides and find the angles.

[midnight haven]

or just use the triangle with 40deg inclination

if you just want h

[dense crater]

oh right thank u

[waxen talon]

Doing this should be the first step in my personal opinion, it helps in thinking of possible ways to solve it.

Do you know of the sine law?

[dense crater]

yes

[waxen talon]

Did you name the triangle?

It also helps avoid confusion what am I supposed to say "The big triangle"?

[dense crater]

lmao sory

oh im redrawing the triangle i drew the wrong one mb

the wrong problem

[midnight haven]

lol yeh i think too

and name them too

[waxen talon]

Good

Now consider triangle AEB

We can solve it in 1 step with the sine law

[dense crater]

but 200 is not the length of eb tho

[waxen talon]

Oh

Now you changed it

Amazing

[dense crater]

yeah lmao

i drew the first triangle wrong sorry

[waxen talon]

It's fine

[midnight haven]

You should also draw units next to the numbers if there are any given

use tan for three angles and get relations and solve

are you given tan50, tan40 ?
or calculator is allowed ?

[dense crater]

yes its given and calculator is allowed

[midnight haven]

good then

[waxen talon]

You actually only need to use 2 angles

Still haven't named one side

[midnight haven]

yeah two are good cause have two variab.

wisely pick the angle

[dense crater]

like this?

[waxen talon]

This is a lot harder this ways

[dense crater]

lol i think my tutor taught me to do it somehow similar to this. is there a simpler way?

[dense crater]

if it was tan shouldnt it be h/200+y tho?

[waxen talon]

Let me see

[waxen talon]

Why?

[dense crater]

because tan = opp/adj?

.close

[pearl pond]

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[waxen talon]

Tan is opp over adj

[dense crater]

nvm lol i got it

its tan60=AE/x
=> AE= x.tan60
tan40= AE/(x+200)
tan (40) = x.tan(60)/x+200

[pearl pond]

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[sullen leaf]

could someone help

[sullen leaf]

with this?

[pearl pond]

@sullen leaf Has your question been resolved?

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[livid bolt]

hi

[livid bolt]

why is contrapositive of a inequality like that

what happened to not?

[sharp smelt]

congrats on getting the active role!

[livid bolt]

that is not good...

it means i struggle so much i have to come here

so much

that i get this role

[livid bolt]

do u know the answer

[sharp smelt]

No, never taken a class on this, sorry

[livid bolt]

u didnt do proofs!?!?!

LUCKY

[sharp smelt]

I did proofs, but never learnt about contrapositives, converses of statements etc

[livid bolt]

o

but those are included in proofs no?

[zinc pewter]

Do you know what the contrapositive of a statement, P => Q is?

[livid bolt]

yes

not q => not p

[zinc pewter]

Ok so if we have (P>5) -> (B <= 10) what would we do to find the contrapositive

[livid bolt]

uh...

to be honest idk unfamiliar with inequalities

[zinc pewter]

Its just as before, we negate each side

[livid bolt]

o

so

B>10 => P<=5

[zinc pewter]

Not quite, say it in words, if P is not bigger than 5, then what can we conclude about P?

[livid bolt]

P<5

[zinc pewter]

Is 5 bigger than 5?

[livid bolt]

no

[zinc pewter]

So what do we have?

[livid bolt]

B>10=>P<5

but what about the = to

[zinc pewter]

this is right

[livid bolt]

huhh

[livid bolt]

how come in their answer one of it has an equal to

[zinc pewter]

Do the same thing for B ( i ment to call this Q oops lol)

OH sorry sorry

[zinc pewter]

I misread this

[livid bolt]

o lole

[zinc pewter]

Ok so you know that 5 is not bigger than 5, we agree on that?

[livid bolt]

yea

[zinc pewter]

So if P is not bigger than 5, could P be 5?

[livid bolt]

no

P<5

[zinc pewter]

is 5<5

[livid bolt]

and then B is not smaller than or equal to 10

[livid bolt]

no

[livid bolt]

oh wait

yes

[zinc pewter]

So when we negate P>5 what do we get

[livid bolt]

P<=5

[zinc pewter]

exactly!

So could you do the same thing for the statement about B?

[livid bolt]

B>10

[zinc pewter]

Perfect

[livid bolt]

ok thx this reallt helped me understand!

i have a test tomorrow!!

do u have any tips

[zinc pewter]

So does the solution from before make sense now? When we negate an inequality we flip the direction, and either include or exclude equality (depending on if it was allowed before)

[zinc pewter]

Symbols can be confusing so try to say the statement out loud if you get stuck, apart from that just try to stay calm and remember what youve learned

[livid bolt]

yes!

[livid bolt]

ah ok my test is on

circles and proofs

😔😔

i have trouble figuring out counterexamples

its supposed to be like common knowledge but i am not smart

[zinc pewter]

Im sure youll be fine, just break problems down into small parts, look at each part individually, youll nail it i have faith

[livid bolt]

aw ok thx u?

!

i an sleeping now

WISH ME LUCK

FIRST TEST OF THE YEAR

.close

[pearl pond]

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[sinful timber]

Need help

[sinful timber]

Would this be correct

[finite jolt]

Yeah

[pearl pond]

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[sinful timber]

Need help with vector valued function

[sinful timber]

This is what I ended up with but I’m not sure if its correct

<@&286206848099549185>

<@&286206848099549185>

[modest valley]

omg what did u do

[sinful timber]

?

[pearl pond]

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[pearl pond]

@sinful timber Has your question been resolved?

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[lyric oyster]

is this correct? 2cos(pi/2)x

[gentle hatch]

one small detail

$2*\cos(\frac{\pi}{2}*x)$

[jolly parrot]

nico

[lyric oyster]

isnt that the same thing tho

[gentle hatch]

no be careful you wrote that the x was outside the parenthesis

[lyric oyster]

oh lmao how do i tell if it goes in or out

[gentle hatch]

to get a cos function, the x need to be influenced by the cos so the x need to be inside the parenthesis

[lyric oyster]

ohhh

i see

thank you

[pearl pond]

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[hearty flame]

is there a way to figure this out without just guessing

[near echo]

you could generate it in a graph editor and drag the nodes around
https://csacademy.com/app/graph_editor/

on paper (e.g. an exam) you would just draw the steps one at a time

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[ancient gull]

what does real valued arithmetic mean?

[prime bramble]

i suppose arithmetic with real numbers?

[near echo]

do you have more context

[ancient gull]

I have this question to do, I'm confused on what the real valued arithmetic part means

[near echo]

i wouldn't worry too much about it. i would assume they mean you shouldn't be using complex number arithmetic

it seems that you could compute this in various number systems, but they want you to use the real numbers

[ancient gull]

does it have anything to do with bidmas? I just kind of pasted it in with no brackets etc

[near echo]

i think it refers to what i said above, but i cant be certain

[ancient gull]

Which other number systems are there?

[near echo]

complex numbers, real numbers, and many others

[humble lintel]

They just don't want you to use something like a MATLAB symbolic toolbox. Frankly, if you are asking this question, you won't know enough about MATLAB to have this be a problem that would arise.

[near echo]

ok phew someone here knows matlab

forget i said anything

[ancient gull]

Yeah, I just started today

[ancient gull]

lmaoo

.close

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[pearl pond]

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[foggy venture]

I feel like I know how to do it but I don't

[sterile coyote]

Well, we can start by noting the points.
the y-intercept is another point of the parabola, namely (0,-2).
That is between the points (-8,-1) and (2,-1), thus the parabola is convex (the vertex will be the global minimum). Let then f(x)=ax^2+bx+c. For x=0 we have f(0)=-2=0+0+c.
For x=-8 we have f(-8)=-1=64a-8b-2 and for x=2 we have f(2)=-1=4a+2b-2.

that is a system of 2 linear equations:\

[foggy venture]

im a bit confused

[sterile coyote]

a general parabola will be in the form of $f(x)=ax^2+bx+c$

[jolly parrot]

🇵🇸Mína🔆

[sterile coyote]

you're just trying to find the parameters

[foggy venture]

would it be possible to use vertex form or factored form to solve this?

or can you only solve it using standard

also where do you go from here

c=-2?

[sterile coyote]

of course

[foggy venture]

so we have ax^2+bx-2?

[pearl pond]

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[pearl pond]

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[snow sonnet]

-1

can someone explain how to find the x or y intercepts without drawing a graph or using a calculator if possible?

[sharp smelt]

what's the x intercept defined to be

[snow sonnet]

Where the equation crosses the x axis?

[versed mica]

which means what

in terms of math

[snow sonnet]

The x axis must have a y of 0

[versed mica]

yes the x intercept is the x value that satisfies y=0

[snow sonnet]

Ohh ok, now I’m just mostly confused with how to handle the log

[haughty kraken]

is there another value next to the log? i see a - sign

[snow sonnet]

-1

[haughty kraken]

o oke

[versed mica]

so first add 1

then what should u do

[haughty kraken]

yeah, then divide by -4 to make the log by itself

so its -1/4 = log(x+2)