help-39

[restive wedge]

and just go on as normal

or wait is subtituing just it

[cloud zephyr]

yea, you can do that
like you can substitute
y=x+2
then
with some basic algebra you can get
x→-2 becomes y→0

find an unoccupied channel

[restive wedge]

you should go to one of the open help channels

help 4 8 and 38

(ones without names at the end)

its ok

[cloud zephyr]

it's alrighty

[restive wedge]

answer is just 2 ig

cool

[cloud zephyr]

i think not

[cloud zephyr]

lemme check

[restive wedge]

how so?

i got

$2\cdot\frac{2sin(\frac{y}{2})}{y}\cdot\frac{sin(\frac{y}{2})}{y}$

[cloud zephyr]

2(1-cos(x+2))/(x+2)²
=2(2sin²((x+2)/2))/(x+2)²

[restive wedge]

where did the /2 come from?

[cloud zephyr]

(x+2) is the doubled angle

so we have to half it

$2\cdot\frac{2\sin^2\left(\frac{x+2}{2}\right)}{(x+2)^2}$

[restive wedge]

oh agian i forgot to put 2 infront of the sin

[jolly parrot]

Biscuity

[cloud zephyr]

hehe

[restive wedge]

tell me about this doubled angle thing i dont get it

[cloud zephyr]

oh sure

like before, we had
cos(2x)=1-2sin²x
so
this time we have
cos(x+2)=1-2sin²((x+2)/2)

[restive wedge]

OHHHH

IC IC IC

[jolly parrot]

veryhuman

[cloud zephyr]

yep

[restive wedge]

is something like this right?

[cloud zephyr]

yep

[restive wedge]

y is x=-2+2 meaning y is 0

[cloud zephyr]

yea

[restive wedge]

oh wait no

the bottom ys

[cloud zephyr]

?

[restive wedge]

those will be 0s to

[cloud zephyr]

yea

[restive wedge]

is the answer just 0?

[cloud zephyr]

nah

[restive wedge]

wait

[cloud zephyr]

recall the sin(x)/x when x tends to 0?

[restive wedge]

yes it becomes 1

but

this time its

sin(x/2)/x

[cloud zephyr]

yea

[restive wedge]

it has no effect even if theres a /2?

alright

[cloud zephyr]

it has

just like you did something like
sin(3x)/x when x tends to 0 =3

[cloud zephyr]

what will this be

[restive wedge]

1/2?

OK

1

shouldbe answer now

[cloud zephyr]

yep

[cloud zephyr]

ooo

[cloud zephyr]

what will you do this time?

[restive wedge]

ok

rn this is on my mind

[cloud zephyr]

nah, I don't think we need that

[restive wedge]

oh alr

tan is

sin/cos

does that do anything

[cloud zephyr]

yep

1-tan(x)
=1-sin(x)/cos(x)
=cos(x)/cos(x)-sin(x)/cos(x)

[restive wedge]

1/cos-sin/cos?

oh

OH WAIT OOPS

ok glad i was in the right direction just messed up the top part

$\frac{\frac{cosx-sinx}{cosx}}{sinx-cosx}$

[jolly parrot]

veryhuman

[cloud zephyr]

yep

[restive wedge]

ok i forgot how fractions on fractions work tbh

[cloud zephyr]

oh okay

first hint: fraction line work as division

[restive wedge]

i forgot if i should multiply bottom to top

[cloud zephyr]

so, it will become

[restive wedge]

oh wait

(cosx-sinx)(sinx-cosx)/cosx

[cloud zephyr]

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

[jolly parrot]

Biscuity

[restive wedge]

oh

alr

[autumn stag]

https://tenor.com/view/jerry-eating-popcorn-watching-you-watching-drama-the-chat-watching-the-chat-gif-2227940436337602914

[restive wedge]

cosxsinx-cos^2

[cloud zephyr]

you dont have to expand it 😛

[restive wedge]

wait

[cloud zephyr]

note that:
cos(x)-sin(x)= -(sin(x)-cos(x))

[restive wedge]

ok yeah that ik

just

should i

my brai is very confused

[cloud zephyr]

why?

[autumn stag]

yea why?

[cloud zephyr]

is it the fraction part?

[restive wedge]

could i just

[autumn stag]

oh

[restive wedge]

dp

[autumn stag]

i got you

[restive wedge]

-1/cosx

[autumn stag]

so what you need to do

is

[cloud zephyr]

yea, you can just sub π/4 in

[restive wedge]

OH

i was wonderin if i was allowed to do that

[autumn stag]

yea no

[cloud zephyr]

if it's not undefined, you can

[restive wedge]

-2/rt2

[cloud zephyr]

so, the final answer is?

[restive wedge]

this

[cloud zephyr]

can be further simplified

[restive wedge]

since cos π/4

just rt2?

[cloud zephyr]

-√2 yea

[restive wedge]

i keep forgetting x/rt(x) = rt(x)

[cloud zephyr]

okay, i gtg~

cheers! and have a nice day!

[restive wedge]

alright

[restive wedge]

have a nice day too

pretty sure i can do the alst one on my own so

/end

[autumn stag]

no problem

if you ever need my help again just ask

[restive wedge]

how do i close this channel xD

oh wait if ur here sure why not

[autumn stag]

i dont know

look

go to uber

[restive wedge]

ok idk how to help with logs lol

hm what do i do witht aht cos^3

nvm got it 3/2

.close

[pearl pond]

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[pearl pond]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[proud vapor]

what’s the equation for this function?

[torn hawk]

This is a cubic polynomial

Whose roots are: -3, 2, 5

So our equation is (x+3)(x-2)(x-5)=0

I hope you can expand it

[proud vapor]

what

[regal herald]

thats not the whole equation

[proud vapor]

yea im confused how do i write it

[torn hawk]

Bro, what

[regal herald]

theres a constant out front

what you have has the same roots but its a different graph

[torn hawk]

Is it possible?

[regal herald]

plug in the 1 for example

[upper drum]

x³ – (α + β + ɣ)x² + (αβ+ βɣ + ɣα)x – (αβɣ) = 0

[regal herald]

its a(x+3)(x-2)(x-5), you just need to find what a is

[proud vapor]

????

[torn hawk]

Let's see if both equations are same or not

I expand mine, then will do according to azo's merhod

[upper drum]

if you don't want to simply, keep roots in the equation
assume alpha, beta, gamma as -3, 2, 5

[torn hawk]

wdym assume

They are the roots

See graph

[upper drum]

yea

it is ik

[regal herald]

this method also misses the scalar out front

same roots different graph

its why the (1,8) is there, so you can find what it is

[upper drum]

aha yes that's a good point

[proud vapor]

so..what’s the equation

[upper drum]

you know how to expand right?

[torn hawk]

Hey azo, you correct man

[upper drum]

yea azo is correct

[torn hawk]

When I put 1 in (x+3)(x-2)(x-5), we get 16

While we should get 8

[upper drum]

yup

[regal herald]

just wack a half on it

[proud vapor]

0.5(x+3)(x-2)(x-5)

[upper drum]

which gives a=1/2

[torn hawk]

Oh lol

[proud vapor]

right?

[upper drum]

correct

[torn hawk]

Yes

I was so close and still so far

Thanks azo for opening eyes

[proud vapor]

lmao

[upper drum]

same

aha it's confusing as she also got same pfp, i thought I send that lmao msg 💀

[torn hawk]

Hehe

[upper drum]

i literally tried to delete it

you can close the channel, if you don't have any more queries, by doing: .close

[proud vapor]

.close

[pearl pond]

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[pearl pond]

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[visual canyon]

is this correct?

[upper drum]

yes

acpid creating multiple channel

just keep this one and ask here

[visual canyon]

oh? I thought it would be good to make and then release since my initial question was answered

I will keep this one then

[upper drum]

yup

[visual canyon]

I'm asking as I solve, if I get any wrong I will go into my final friday with bad info and i'm riiiiight on the cusp of a and b so im trying to give myself the best shot at an A possible

sorry for the incoming 50000 questions in advance

[upper drum]

I am going to sleep

[visual canyon]

do your thing! There are thousands of members

thanks for your help today

repasting the question for simplicity

is this correct?

[sterile plover]

yes it is correct

cos(180+x)=-cos(x)
cos(180-x)=-cos(x)
cos(30 degree)=sqrt(3)/2
just let x=30 to see

to prove these you can use cos(a+b) and cos(a-b) identity

[visual canyon]

awesome

thanks

hows this one?

.close

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[jade kite]

$Given\ x = \frac{5}{2\sqrt{k}}, y=e^{-\frac{x\pi}{\sqrt{1-x^2}}} For\ what\ range\ of\ k\ is\ x \leq 0.7\ and\ y \leq 0.07$

[jolly parrot]

senoune

[jade kite]

I started by solving for x finding that $K \geq 12.176$

[jolly parrot]

senoune

[jade kite]

I'm stuck on step after that

.close

[pearl pond]

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[visual canyon]

how is this?

[spare lark]

typo to first line, sin not sine

Otherwise its good

[visual canyon]

great ty!

.close

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[urban scarab]

how do I calculate the gradient of l2 norm given a vector R^N

[urban scarab]

so far I know how to get the square root out using chain rule

I end up with this thing inside of ((sum here)/2)^-1/2

idk how to get the gradient of a sum 😭 , very basic but I can't find it anywhere

[errant cedar]

what’s the gradient in that case?

[urban scarab]

gradient with respect of x^2 + other (x^2)

x1^2 + x2^2 + x3^2 n times

the point is to get the gradient of the entire thing

[errant cedar]

k the gradient is a vector in R^n, more simply an ordered list of n quantities : the partial derivatives with respect with each x_i

[urban scarab]

yeah

[errant cedar]

well you just gotta fiend each number individually

[urban scarab]

hm?

[urban scarab]

can you elaborate, I really don't know what to do past the ((sum)/2)^-1/2

[urban scarab]

OH

u saying that I should just take the partial derivative of a single x_i then thats the gradient

mb

I was so confused when I saw a sum in there

but that sum is just x^2 right?

[errant cedar]

it’s a component of the gradient one number in the list imo try to do it for n = 2 then generalize if you feel confused

[urban scarab]

I see, tysm

that makes sense

[urban scarab]

ty.!

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[loud chasm]

i need help, i am undsure how to slove this

[sonic iris]

shift the sine or cosine function to make it equal to the graph

[loud chasm]

okay

is there anything elese to do about it?

am not sure

.close

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[sonic iris]

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:transformations

[pearl pond]

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[sterile turtle]

hi i need help with part b

[sterile turtle]

nvm

.close

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[torn hawk]

You got it?

I was just about to type answer lol

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[median lintel]

ive got some data that i had to figure the Q1 and Q3 for

[median lintel]

i used excel to solve it

but then did it by hand

there is a massive difference in the answers

which one do i trust?

and my answers are:
Q1: 727.65
Q3: 1102.6

[pearl pond]

@median lintel Has your question been resolved?

[median lintel]

<@&286206848099549185>

[summer gorge]

excel

[median lintel]

Why though?

[summer gorge]

its better at these stuff

like even big companies use it for these

[median lintel]

Uh okay

[pearl pond]

@median lintel Has your question been resolved?

[pearl pond]

@median lintel Has your question been resolved?

[pearl pond]

@median lintel Has your question been resolved?

[pearl pond]

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[real rose]

can any1 plz asap help
for this and this, i understand why to do 4^2 - f(y)^2 for the 1st pic

[real rose]

why dont we do it with this one

where its 6^2 - f(y)^2

vs just f(y)^2

pls @ me ifany replies

[pearl pond]

@real rose Has your question been resolved?

[midnight haven]

Because the guy in the video’s one goes through (0,0)

[midnight haven]

You have to subtract the line from x=4 bc ur antidiffing with respect to the y axis

And you don’t want all the volume

[real rose]

ive seen like inner and outer radius

but if so would that be like x = 2 as inner and x = 4 as outer

since thats not the case since we do 4^4 (inner) afaik

[real rose]

so if theres anything that has a space between the axis and the actual area /function, you use the outer bound (in this case 4) and the function
i just thought to use it for the video guy's one because as you move up the y values, a space does appear between the axis and graph?

vs just a one around the y

[midnight haven]

If the graph is enclosed by the acid you don’t need to do the extra work

[midnight haven]

Wdym by this?

[real rose]

^ like pi * integral (R^2 - r^2) to find the volume - called the washer method or whatever

[real rose]

but besides the point im more confused about ^

the 2nd line i typed

[real rose]

yep so like

but for this one, i just see this and think... space

[midnight haven]

Ohhh are you saying it’s not fully rotated around the axis?

[real rose]

yea

so i see that its not, and i think to use the other method

but it seems that as long as a point that ur finding the volume around is touching the axis, its fine? is what im understanding

i.e. goes through 0,0

[midnight haven]

Wait do you have a question where it’s not fully rotated

[midnight haven]

Doesn’t have to be touching

[real rose]

isnt y = sqrt(x) one?

i.e. this around y

[midnight haven]

Yeah that’s fully rotated

[real rose]

wait

isnt this fully rotated because its this

now

i thought that became

[real rose]

if its these 2 right?
is it the 1st or 2nd if its rotated about the y?

as in the volume your tyring to find

or does it depedn on how it words it

[midnight haven]

1st i think

[real rose]

becuase thats like this right? but were doing the 2nd pic

so is it because it SPECIFIES its around the x axis and x = 4

[midnight haven]

You subtract the shaded part in pink

OHHH WAIST

[real rose]

waitttt

lmao

[real rose]

yea for this one your finding the shaded are you not? the pink shading you did

but in a circle rotated

[midnight haven]

Yeah you’re finding shaded

Wait why is 4^2 inside the integral @next dove

😬😁

[real rose]

lol

then i/we still dont know which one it is if its these rotated around y?
unless it depends on how the question is worded

[pearl pond]

@real rose Has your question been resolved?

[pearl pond]

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[tender siren]

for more rational and easy solving of questions or problems, can I somehow relate the products and the sums of logs of numbers with the same bases? for example

log x + log y and (log x) . (log y)

it would help for conversion of these terms in equations as the product doesnt have a direct formula but the sum does, i.e, log x + log y = log xy.

if you need more context, please inform as I myself think this wouldnt be enough, its rather a discussion over theory rather a direct question help, but its help indeed

[tender siren]

I am sure this will cause confusion lol

[tulip ore]

ab is too inherently different from a + b to have any direct connection
so the same thing applies with (log x)(log y) and log x + log y

for identities that involve ab and a + b, you should consider googling vieta's formulas or simon's favorite factoring trick

[tender siren]

any other factors or searches should I consider?

[tulip ore]

theres also that (a + b)/(ab) = 1/a + 1/b

[tender siren]

it sure is different but intuition sure does say its as equal as same too

[tulip ore]

yea if youre not reading them

other than that I dont see anything that can relate a + b and ab together

[tender siren]

thats right though, yeah

[tender siren]

hm, want to know the origin question that made me ask this?

[tulip ore]

go for it

[tender siren]

its rather quite simple, prove:

ab + bc + ca = abc

with a = log x xyz; b = log y xyz; c = log z xyz;

its rather a 12th grade question, and surely if I will solve it in the prescribed way I would cancel out a lot of terms, but I used a product wise analogy and I just thought how could I get that product form into a sum

that question in itself, isnt a problem, but this relation, as for me, is.

[tulip ore]

yea thats too different to count

maybe you should divide both sides by abc

then rewrite a, b, and c as fractions using change of base formula

[tender siren]

yup

but still, it would be much simple if I could implement a direct connection to products and sums, if I get that, a same analogy could be applied to quotients and differences

let me go through the google's you talked about though, lets see

hm, now that can make something, heh

[tender siren]

dont you think inversing the terms and reversing back to original can also work if I relate each term directly to parent equations of a, b, c?

or am I clumsily breaking some huge laws of solving equations?

[tulip ore]

you were looking for a rule that connected adding to multiplying

(a + b)/(ab) = 1/a + 1/b not only fits that in a capacity

but directly answers your question

(ab + bc + ca)/(abc) = 1/a + 1/b + 1/c

which simplified

[tender siren]

made me smile instantly, heh

[tender siren]

it needs a little for shaping and yeah, that is my answer, lol

[tender siren]

but will this work too?

[tulip ore]

no

you dont seem to be seeing how these sorts of problems work

beyond the textbook problems, there are no longer consistent methods to solve, only reliable and predictable ones

[tender siren]

thats seems to be the case

[tulip ore]

you saw that (a + b)/(ab) = 1/a + 1/b worked, not because its a catch-all method for any a + b and ab, but because it fits the structure of the problem so nicely that it can be recognized

[tulip ore]

what youre saying here is asking for a consistent method

its not a huge law but its a common misconception

you know any 2D geometry problem can be written in coordinates

[tender siren]

a method that is resilient with the structure of logs themselves

[tulip ore]

but that doesnt mean itll be rewarding or interesting to solve them that way

[tender siren]

yup

[tulip ore]

there isnt one

you find out that theres something, or you dont

you have to find the method that works for each specific problem, they will all have a separate method for each, the problem is really to find that method out

[tender siren]

hm, what if I say I am asking for a 'relation' that is resilient with the structure of logs themselves rather than a specific 'method' that works for all questions?

[tulip ore]

I told you the three that I knew

[tender siren]

because yes, what you say is true and all problems need different methods depending on their structure, but I can form them if that relation is there

[tulip ore]

my god do you type anything original

[tender siren]

thats why I am satisfied!

[tender siren]

I am a memer, what can you really expect from me

[tulip ore]

anyways I told you the three that I knew but other than that, there likely isnt anything

if it could apply to logs, but take the general form of a+b and ab (instead of something specific to logs like a log law), then it would not be a "log rule", it would just be a general rule relating a+b and ab

for something related to logs, the log laws already do enough to carry you through most problems

so looking for a specific log law wont really do you as much as looking for a law that can help you factor

those rules involve things like (a+b+c)(ab+bc+ca) and you can google those

[tender siren]

I can relate quadratics with logs with that form and I guess, make a way for those problems

[tender siren]

i mean yeah, if I keep simplifing already factored and simplified stuff, its gonna be a super factor

but thanks, I guess I have some for confidence now to do those more

[tulip ore]

alr and gl

one more thing

you noticed the law that really helped here essentially had nothing to do with logs but still allowed the logs to simplify

so just because its unrelated doesnt mean it wont help

so really you have a lot of laws that deal with logs that can help you here rather than just the log rules

[tender siren]

yup

[tender siren]

anything that is there as material, always does help in someway or the other, imo

[tender siren]

yup, I just need to make that structure, rather, method to do it, right?

[tulip ore]

you need to recognize what structure can work to see which identities or properties can apply

then again

theres problems where you have to take a long time to get much of anywhere, those dont have an immediate recognizable property, but they do have ways to make progress
one recognize leads to another until you stumble upon the final answer
often for big or at first unapproachable problems, you have to recognize small steps that also apply instead of big ones

[tender siren]

small steps, big insights, thanks again

do I need to write any commands?

or will this close on itself?

[tulip ore]

you can type .close or .solved to close it yourself
if not, it will close by itself if you purposefully dont respond if the bot pings you whether the problem is solved or not

[tender siren]

hm

then as its solved, thanks to you, I should close it

.solved

[pearl pond]

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[pearl pond]

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[midnight haven]

hi all, I’m attempting to learn how to write proofs as I’m taking a theory of computation course next semester that’s very proof heavy. in high school, we skipped over proofs as we started learning the topic when Covid hit, and thus there is a gap in my mathematical knowledge. Could someone tell me if my solution to this rather simple proof, is correct, and if I have the right idea?

[midnight haven]

After reading it again, I believe I made a mistake on line 3..

[pearl pond]

@midnight haven Has your question been resolved?

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[young plover]

Hello! I have a question about set theory from a real analysis textbook and I would like to check my work

[young plover]

The question reads: Let $$A_n$$ be the set of positive integers divisible by $$n$$. Find $$\bigcup_\gamma A_n$$ and $$\bigcap_\gamma A_n$$

[jolly parrot]

Eater of Bees

[young plover]

whoops, I meant to use n instead of gamma there

Anyway, point is, I'm pretty sure the union is just $$\mathbb{N}$$ and the bottom is the empty set, but I'd like to check my work

[jolly parrot]

Eater of Bees

[young plover]

Oh shoot, forgot a detail again, the original problem statement starts the indexing at 2, so its actually $$\bigcup_{n=2} A_n\hspace{0.2 in}\text{and}\hspace{0.2 in} \bigcap_{n=2} A_n$$

[jolly parrot]

Eater of Bees

[young plover]

right, so my new answer is $\mathbb{N}-{1}$ and ${}$

[jolly parrot]

Eater of Bees

[young plover]

since the union is just every number divisible by any prime (which is all of them except 1), and the intersection is every number divisible by every prime (which is none)

[sharp quest]

is n any natural number?

[young plover]

yes,

[sharp quest]

oh, starting at 2, then yes you are correct

[young plover]

awesome, thank you

[pearl pond]

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[pearl pond]

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[midnight haven]

Is this good

.close

[pearl pond]

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[trail linden]

i dont get the concept of an altitude…

like i only get triangle ABC with it being intersecting an opposite side

from the triangles vertex

[steel canopy]

Basically altitude is kind of height of the traingle

U start at one vertex and draw a line perpendicular to the opposite side of the vertex

@trail linden

[trail linden]

any vertex, not just the top vertex

Also, are all triangle medians, altitudes and angle bisectors at the same time?

[steel canopy]

Yes

[steel canopy]

Nop

Altitude is the one I said

Median is u start from one vertex and draw a line which touches midpoint of opposite side

Angle bisector basically cuts the angle into two equal parts

For equilateral triangles, they are the same

[trail linden]

yeah

but the altitude has a leeway such that it only has to touch the opposite side; not necessarily the midpoint right?

[steel canopy]

Nop

Not exactly opposite side also

U can extend the side

It must be perpendicular to the opposite side

[trail linden]

oh i gotchu

but its possible that an altitude can also be a median?

[steel canopy]

Yes

Equilateral triangles

Isosceles triangles also

[trail linden]

But are both altitudes and medians angle bisectors?

[steel canopy]

Not everytime

But there are possibilities

[trail linden]

so its sometimes true

[steel canopy]

Yes

[trail linden]

gotcha; thanks

.close

[pearl pond]

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[steel canopy]

Anytime

[pearl pond]

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[midnight coral]

can someone help me solve the second dervorive of

[midnight coral]

just need to find the second

here is the first

[next dove]

where is it?

[midnight coral]

the x()^-1/2

[next dove]

okok found it

[naive hemlock]

x/(x^2+4)

Can u click and send a pic of fresh page

Imma write on it

[midnight coral]

idk where to start i used the product rule

okay

[naive hemlock]

Hint:use quotient rule

[midnight coral]

would it work with product?

[next dove]

yes

[naive hemlock]

Yes but u need to pay extra attention

Imma do the product ok

[midnight coral]

ok

i’ve gotten this so far

[naive hemlock]

This is by product rule

[midnight coral]

like that’s the awnser?

[naive hemlock]

Yes

[midnight coral]

i’ll be honest

i used quotient rule

[naive hemlock]

Okok

[midnight coral]

cause it just got so confusing

n i’m here

[naive hemlock]

U want to check your answer

Okok wait

[naive hemlock]

,w d/dx (x/√(x^2+4))

[jolly parrot]

Results provided by WolframAlpha

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

Requested by oppailol.

[naive hemlock]

Check it

[midnight coral]

the awnser is right

[naive hemlock]

Okok

Good

[midnight coral]

but the steps is locked for me

[naive hemlock]

Enjoy

Ya correct steps

I checked

[midnight coral]

was what i doing wrong

it’s locked

i can’t access

[naive hemlock]

What

Idk

[midnight coral]

okay wait

i got it

finally lol

tysm

.close

[pearl pond]

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[pearl pond]

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[forest sun]

i am lost with this problem

[naive hemlock]

Yo

Nah I'd win

[forest sun]

sup

[naive hemlock]

Sup

Oh

We need graph of slopes

[forest sun]

i think its the second line but i don't know how to check

[naive hemlock]

Find a point

[light helm]

location of the stationary point

[naive hemlock]

Where the derivative slope is zero

Lol

We both said the same things

[forest sun]

?do we need the y values

i assume not as there are none

[elder mica]

the derivative of the parabola intersects the x axis when the parabola has 0 slope

where in the graph do you think the parabola have 0 slope?

[forest sun]

around -1

[elder mica]

alright and which line intersects -1?

[forest sun]

like the x axis?

or the acutal point on the parab

[elder mica]

nope, like the given lines

either black, blue, orange, or green

[forest sun]

third one?

i am color blind

[elder mica]

no, just say the colorr

which line intersects -1 at the x axis??

[forest sun]

wdym

i think its the third line?

[light helm]

the x-axis, yes

[forest sun]

as so then the second

[elder mica]

uhm, just say the color?

[forest sun]

i am color blind

[elder mica]

oh alright

from left to right then

[forest sun]

top to bottom

second

[elder mica]

okay so pick blue in the options

[magic wharf]

OIUYTGFueyiwod

[forest sun]

nice its correct

[magic wharf]

hi

[forest sun]

then this one?

[light helm]

similar idea

[elder mica]

alright

when do you think does the middle curve or the odd curve have 0 slope?

[forest sun]

at 4, 1.5

so 2?

[elder mica]

nope

actually nvm

yup

[forest sun]

could you tell me the color please?

[elder mica]

the orange one

[forest sun]

let me checl

check

[elder mica]

those p sets aren't really colorblind friendly aren't they??

[light helm]

both 2,3 had the same intercepts
did you consider something else when choosing 2

[forest sun]

yes

😭

[forest sun]

i did the increase/decrease thing

so 2 looked right

[forest sun]

never thought i'd have that problem in calc

[light helm]

ok. that's fine

[forest sun]

thanks for the help

.solved

[pearl pond]

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[pearl pond]

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[sinful frigate]

,w solution of {lambda x+y+z=1,x+lambda y+z=0,x+y+lambda z=0}

[jolly parrot]

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Requested by bill220220_58022

[pearl pond]

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[proven matrix]

given a coordinate and a velocity vector (2-dimensional, but should be the same as 3 dimensional with a 3x2 matrix), and the coordinate and mass of a central body, is it possible to determine the ellipse of the objects path, and possibly its coordinates as a parametric function of time, according to keplers laws of motion

[pearl pond]

@proven matrix Has your question been resolved?

[pearl pond]

@proven matrix Has your question been resolved?

[pearl pond]

@proven matrix Has your question been resolved?

[pearl pond]

@proven matrix Has your question been resolved?

[pearl pond]

@proven matrix Has your question been resolved?

[proven matrix]

i think it isn’t an ellipse

i might just resort to numerical methods

thanks guys o7

.close

[pearl pond]

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[regal copper]

Very hard algebra problem. Find E

[gray girder]

3

[pearl pond]

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[prime bramble]

is this a test?

[normal dragon]

No

Homework

@prime bramble

[prime bramble]

I see.

regardless, I suggest you be much more clear with which question you need help with, and what exactly you're having trouble with

[normal dragon]

18

Question 18

Please

@prime bramble

[pearl pond]

@normal dragon Has your question been resolved?

[normal dragon]

Someone help

[ornate lagoon]

there are 2 dots over x
that means acceleration right?

@normal dragon

the first part is easy, draw the free body diagram and write net force:

$F_{net} = F-kv^2$

$ma = F-kv^2$ ---(i)

terminal velocity will be achieved when the body attains equilibrium

$F = kv_{terminal}^2\implies k = \dfrac{F}{v_{terminal}^2}$

substitute this in (i)

$ma = F\left(1-\left(\dfrac{v}{v_{terminal}}\right)^2\right)$

[jolly parrot]

[ᴛʜᴇ ᴇᴍᴘᴇʀᴏʀ]

[ornate lagoon]

v_terminal = 270km/h = 75m/s

the other parts require integration

[pearl pond]

@normal dragon Has your question been resolved?

[pearl pond]

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[dusty mango]

uh any hints?

[pearl pond]

@dusty mango Has your question been resolved?

[dusty mango]

@signal atlas did you delete something

<@&286206848099549185>

[signal atlas]

Yeah thought I had something but I lost track of the constraints. Could prob smartly brute force though since there aren’t many possibilities

[pearl pond]

@dusty mango Has your question been resolved?

[pearl pond]

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[midnight coral]

idk how i’m calculating this wrong but the answer is 1.75 and I keep getting the answer that Shawn on the screen

[tender gate]

idk who shawn is but you can start by decomposing the intergral by : int(a-b) = int(a) - int(b)

[next dove]

that 0.078 is supposed to be approx 0.0078

[midnight coral]

but it shows this on my calculator

[next dove]

what did you typed in it?

[midnight coral]

like does this mean i have to multiply it by 10

[next dove]

to the -3 power

so 0.0078..

[midnight coral]

i have no idea what that means

it means 3 numbers?

[tender gate]

you can multiply by 10^3

[next dove]

shift to 3 decimal places

[midnight coral]

oh

so when i get that

what do i write in the calculator again to get the number i said?

u

[light helm]

you messed up your signs

[midnight coral]

where

[next dove]

after second step

in second last line

[midnight coral]

oh yeah

but i noticed that

still got it wrong

[midnight coral]

how that the addition become into subtraction

because it’s two negative numbers dividing each other making it into a posative

[light helm]

but you have a - sign on the outside of the ()

[midnight coral]

yeah cause it’s the formula

[light helm]

so you effectively have 3 negs

[midnight coral]

like u minus it

no two negs cause of the subtraction and the one after

like i could also write it like this:

[light helm]

and there's another - in the denom

[midnight coral]

2 and 3 makes it a posayive

it’s two negative numbers dividing each other

[light helm]

the - on the outside distributes to all terms inside

yes, but you still have the - on the outside

$-(a+b) \redneq -a + b$

[jolly parrot]

ℝαμΩℕωⅤ

[midnight coral]

oh

so like this

[light helm]

no

that's your old image

[midnight coral]

i just changed it

[light helm]

actually that's worse

[midnight coral]

dayum

[light helm]

you changed the wrong part
the part that was originally valid is now also wrong

[midnight coral]

it’s just that

here is what my school did

this makes sense

[midnight coral]

but they didn’t do that in the example

like can we follow that for now cause ik the signs switch

but we can also do that later

[light helm]

you can do stuff as long as its valid

you don't have to follow what they did exactly

you were doing fine up until the end

its just that you didn't distribute that - sign on the outside properly

they seemed to first simplify everything inside the [] which is fine

if you want to go that route,
the sign on the outside stays as a -

it was actually a bit had to see whether some of your + signs were + or -
as they were semi-erased

focussing on just this part

depending on personal preference you can go:
$$-(-2.5 + 0.125)$$
or
$$+(2.5 - 0.125)$$

[jolly parrot]

ℝαμΩℕωⅤ

[pearl pond]

@midnight coral Has your question been resolved?

[midnight coral]

okay

i finally understood that

tysm

😭😭😭

.close

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[glacial sequoia]

How would I go about doing (c)

[glacial sequoia]

For (b) I interlaced terms of (1/n) and (1 + 1/n) and I think I should somehow extend that pattern

[rustic gate]

how

[glacial sequoia]

have a sequence (1/2, 1 + 1/2, 1/3, 1 + 1/3, 1/4, 1 + 1/4, ...)

the terms of odd indices converge to 0 and even indices converge to 1

[rustic gate]

i thought you were doing c

[glacial sequoia]

yeah

I was wondering if I could somehow extend this

[rustic gate]

well in b you couldnt actually use 0 or 1 in the sequence

[long gorge]

you can do it that way

[rustic gate]

so you're forced to approach it

but in c you can just use 1, 1/2, 1/3, and so on

[glacial sequoia]

Oh

So like

(1, 1, 1/2, 1, 1/2, 1/3, 1, 1/2, 1/3, 1/4 ...)

[rustic gate]

sure

[long gorge]

sounds about right

[glacial sequoia]

well that was simple

thanks

:)

.close

[pearl pond]

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[long gorge]

btw say we have countably many sequences converging to a, b, c, ...

a1 a2 a3 ...  converges to a
b1 b2 b3 ...  converges to b
c1 c2 c3 ...  converges to c
.
.
.

then you can put them all in one big sequence by going along the diagonals: a1, b1, a2, c1, b2, a3, ...

so you can do it even with the "does not contain" condition

[rustic gate]

just add like 1/πn to the sequence

[long gorge]

true

but the infinite interlacing idea works

[pearl pond]

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[glacial sequoia]

.yeah this was the interlacing thing I was talking about

[pearl pond]

Please don't occupy multiple help channels.

[eager yoke]

sorry

i’ll close the other

[eager yoke]

am I correct?

[regal herald]

missing a bracket, but sure, it should work

[pearl pond]

@eager yoke Has your question been resolved?

[pearl pond]

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[shell socket]

hi what is abcd here?

[latent quail]

Constant

[shell socket]

oh yeah okay thank you

.close

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[brisk steeple]

In external division of a line segment, if the ratio is negative, does placing the '-' sign on the numerator and not the denominator and vice versa, have an effect on the answer?
In Cartesian plane.

[upper drum]

it shouldn't affect the final answer

wait you edited?

[brisk steeple]

No, before you replied.

[brisk steeple]

Yes

[upper drum]

it shouldn't

the ratio i think will turn out to be same

[brisk steeple]

Yes, I guess my peers have a confined imagination.

I can imagine both ways.

-m/n and m/(-n)

Thanks

.close

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[brisk steeple]

hi

I am back

[pearl pond]

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[lusty patio]

0∫6 |x-4|dx

Im supposed to solve the problem above, but im a bit confused on the interaction between finding the derivative of an ablosute value and the integration. Am i supposed to intigrate first? if so, how does that even work? if not, how do i intigrate the absolute value of the denominator after i solve for absolute value?

[light helm]

do you mean
$$\int_0^6 |x-4| \dd{x}$$

[jolly parrot]

ℝαμΩℕωⅤ

[lusty patio]

yes

i dont know how you did that 😅

[light helm]

its recommended that you split this into two integrals, applying the piecewise definition of the abs val

[latent quail]

Draw

[light helm]

(or do it geometrically)

you "could" integrate it directly but that would involve the sign(um) function

[lusty patio]

peicewise definition?

[light helm]

which will be a little tedious

$|\text{this}| = \begin{cases} \text{this}\ &\text{if } \text{this} > 0 \
-\text{this} \ &\text{if\ } \text{this}\leq 0 \end{cases}$

[jolly parrot]

ℝαμΩℕωⅤ

[lusty patio]

ohhhhhhhhhhhhhhhhhhhhhhhhhh

i see.

... no i dont.

i split the functuon into $int_0^4$ and $int_5^6$ ?

[jolly parrot]

Craw

[light helm]

not quite

why 5 to 6 for the second integral

[lusty patio]

if x < 4, the output is negative, if x = 4 the output is 0, and if x > 4 the output is positive, right?

[light helm]

yeh

[lusty patio]

so three integrals?

[light helm]

no, two is sufficient

[lusty patio]

im used to doing the absolute values in the u/|u| method

and i think my brain is stuck on that

[lethal palm]

another way u could do it is by simply graphing it, then add of the area of the 2 triangles

[lusty patio]

so would i do:
$\int_0^3 x-4\ & \int_4^6 x+4$

[jolly parrot]

Craw
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[light helm]

no

why 0 to 3

[lusty patio]

its negative then flips positive

[light helm]

you're not flipping the sign properly either

first by using the bounds
0→3
4→6
you're missing the region for 3→4

[lusty patio]

0 to 4, 4 to 6?

[light helm]

and based where you're integrating, you should have either
x+ 4 or -(x+4)
(not x-4, you don't just change the sign in between)

no need for 4→4

that will trivially be 0

why 1→4

now you're missing the part from 0→1

why are you skipping integers

[light helm]

upon the edit, yes

[lusty patio]

0 to 4 -(x+4), 4 to 6 (x+4)?

then integrate them both seperately?

[light helm]

yes

[lusty patio]

but theres still the dx, which means i have to derive afterwards? or am i mistaken?

[light helm]

no

dx indicates what you're integrating with respect to

[lusty patio]

ohhhhhhhhh

so i need to find the difference between the lowest and highest bounds of the two seperate integrals, then find the differnece between the difference of the two?

[light helm]

no

you add the results from each integral

[lusty patio]

so find the difference individually then add the results or add then add?

[light helm]

after splitting
first interal + second integral
just follow what the sign says, evaluate normally

[lusty patio]

but theyre finite intigrals, so shouldnt i solve them by differentiating the bounds?

[light helm]

no

you should not be differentiating anytyhing

for each integral
after finding the antiderivative
eval at upper bound - eval at lower bound

[lusty patio]

so, -4 and 2?

[light helm]

where are those values coming from

[lusty patio]

OHHH

it would be 4 not -4

right?

then it would be 4 + 2 = 6?

[light helm]

how are you getting 4

[lusty patio]

((4-4)-(0-4))

[light helm]

what were the steps before that

[lusty patio]

because the bouds are 0 to 4 and the equation is -(x +4)

[light helm]

what were the steps before that

what was your antiderivative

[lusty patio]

OH RIGHT

I FORGOT abOUT THAT

would it cancel out the negative in front of -(x+4) or no?

[light helm]

wdym

would what cancel

[lusty patio]

if i intigrate, the function.

or no

[light helm]

no

[lusty patio]

okay, just making sure

so -32 +54?

[light helm]

can you show all your work

[lusty patio]

(-(4^2 + 4(4)) + (0^2 +4(0))) and ((6^2 +4(6) - (4^2 +4(4)))

[light helm]

1 sec

[lusty patio]

so (-32 + 0) and (60-32)

[light helm]

sry, i mistyped a sign earlier

[lusty patio]

so 28 for the second one my bad

[light helm]

said (x+4), -(x+4) earlier
but that was supposed to be
(x-4) or -(x-4)
did a poor copy paste job

[lusty patio]

OHHHHHHHHHH

so 0-0 and -12 - 0?

[light helm]

can you please show your work

[lusty patio]

((4^2 - 4(4)) - (0^2 - 4(0))) and (-(6^2 - 4(6) + (4^2 - 4(4)))

[light helm]

starting with the antiderivativves

[lusty patio]

int0^4 (x-4) int4^6 -(x-4)

which then becomes

[light helm]

you mixed the (x-4) and -(x-4)

[lusty patio]

?

[light helm]

is (x-4) positive or negative from 0 to 4?

[lusty patio]

so i need to do both ints twice, one for each (x-4) and -(x-4)?

[lusty patio]

negative?

so int0^4 -(x-4) int4^6 (x-4) ?

[light helm]

yes

now integrate those

so i need to do both ints twice, one for each (x-4) and -(x-4)?
they'll be simliar, the antiderivatives themselves will just be negatives of each other

[lusty patio]

(-(4^2 - 4(4)) + (0^2 - 4(0))) and ((6^2 - 4(6) - (4^2 - 4(4))) ?

[light helm]

what your antidervatives

can you please show ALL your work

[lusty patio]

int0^4 -(x-4) + int4^6 (x-4)

[light helm]

• sign in between your integrals

what was the next step

[lusty patio]

-(x^2 - 4x)|0^4 + (x^2 - 4x)|4^6

[light helm]

you didn't integrate correctly

integral of x isn't x^2

[lusty patio]

x^2/2

[light helm]

yes

[lusty patio]

god-

im rusty

now i know why nothing worked

10?

[light helm]

can you please show all your woprk

[light helm]

looks ok

[lusty patio]

its 1 of the 4 answers

i think ill go with it

am i allowed to curse?

[light helm]

try to refrain from doing it

[lusty patio]

alrighy

only 7 more questions to go...

thanks for the help

.close

[pearl pond]

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[pearl pond]

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[lilac sluice]

hi

[lilac sluice]

how does the cos term just disappear?

[light helm]

got converted to (-1)^j

[prime bramble]

cos(k*pi) = (-1)^k

[pearl pond]

@lilac sluice Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

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Send your question here to claim the channel.

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

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[silver sequoia]

,w lim _{x \to \infty} {x^4 sin (1/x) + x^2} ÷ {1+x^3}

[jolly parrot]

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Requested by crazeie

[silver sequoia]

How do you do this..

[faint scaffold]

Have you learned l’hospital’s rule?

[silver sequoia]

No

[silver sequoia]

I was taught using trigs

[acoustic path]

would not recommend l'hopital on this

[silver sequoia]

But i dont know exactly what to do

[silver sequoia]

What should i do..

[acoustic path]

have you heard that sin(x) is proportional to x for very small values of x

[faint scaffold]

Or Taylor expansion?