help-19

[lavish jackal]

do u need to add it by something?

[vivid wyvern]

We should name the vertices to make it easier to explain

The whole length (green) is 8

Blue+red=green

So u need to find the red part

[lavish jackal]

5?

[vivid wyvern]

5 is what

Wait lemme draw it and explain

[lavish jackal]

sure

[vivid wyvern]

Okay so
Tell me if u agree that AB=CE

[lavish jackal]

yes I can agree

both are 8

right?

[vivid wyvern]

Yes

[lavish jackal]

ok so how would u get the red line

its 8 and something correct?

[vivid wyvern]

The red line is GD

[vivid wyvern]

Yeah we need to subtract from 8

[lavish jackal]

I got 4.5 and it says correct

3x3 9 and divide by 2

[vivid wyvern]

U understood how to get that tho right

[lavish jackal]

yea i do

[vivid wyvern]

Ok good

Move on to next one

[lavish jackal]

so i believe gd is 4.5 and im finding for the blue shape

[vivid wyvern]

Find this now

[vivid wyvern]

Blue shape is wrong btw

I mean the shading is wrong

Area will be same

[lavish jackal]

green shape?

[vivid wyvern]

Yes fine the one i marked

Here this traingle should be blue

Not the other one

[lavish jackal]

like this?

i think both are 3

[pulsar oak]

[lavish jackal]

it let me have two 3s

wont*

[vivid wyvern]

Remove the other one

[lavish jackal]

the down one?

[vivid wyvern]

Yea

[lavish jackal]

done

[vivid wyvern]

It's not included in shaded region

Alr now work on the triangle

Can u find it's height and base

[lavish jackal]

height is 3 and base is 5?

[vivid wyvern]

Nope

Height is fine

Base is wrong

How did u get 5

[lavish jackal]

8-3

[vivid wyvern]

Y would u do 8-3

Alr check this out again

[vivid wyvern]

Here

[lavish jackal]

but i see 3 tho

[vale vortex]

i didnt read any of thje convo

i just filled in what i saw

just in case u guys alr did this

[vivid wyvern]

AC=BE,right?

[lavish jackal]

yes

so its 2?

[vivid wyvern]

We didnt need all lengths

[vivid wyvern]

Yeah

[lavish jackal]

bruh what i was looking at lmao

[vale vortex]

i lokey just wanted to do it anyway

[vivid wyvern]

Fun ik

[lavish jackal]

that means its 3

[vivid wyvern]

Alr AH=CD

Ryt

[vivid wyvern]

Yea

Now u plug in the value in formula

[lavish jackal]

its 3

is there something left

[vivid wyvern]

Yeaha the rectangle

[lavish jackal]

oki

[vivid wyvern]

Uk the formula

For finding area of rectangle right?

[lavish jackal]

yes

i got 10

[vivid wyvern]

Good

Summing all these up should give u the answer

[lavish jackal]

that worked thanks a lot 🙂

[lavish jackal]

.close

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[frosty coral]

how do i use the graphing calculator to determine the values of b and c

[quasi sparrow]

What's b and c

[frosty coral]

the questions

a.

b.

c.

use the graphing calculator to determine at which values of t does ther abbit population fall below 6000

[quasi sparrow]

Oh I thought they were variables

Set R=6000 and find the minimum t>0 that satisfies the equation

[frosty coral]

how do i do that on desmos graphing calculator

could u show me

[quasi sparrow]

Nah

[frosty coral]

oh

well

i did this

but i dont know what im supposed to do with that

[quasi sparrow]

Do this

[quasi sparrow]

Find the point on the graph

[frosty coral]

alr now what

[quasi sparrow]

Satisfying the R(t) = 6000 = ... equation

Did you write that equation down

[frosty coral]

uhhh

I dont understand what you mean by satifying the R(t)=6000

[quasi sparrow]

You only have one equation for R(t)

[frosty coral]

like this

?

[quasi sparrow]

Yes

[frosty coral]

so 2.5 and 7.5 are the two values?

[quasi sparrow]

Is the rabbit population below 6000 between those two values?

[frosty coral]

yes

[quasi sparrow]

Then yes

[frosty coral]

how do i determine the values of the wolf population at a minimum

what is that question trying to ask

[quasi sparrow]

Minimum means smaller than or equal to all other values of the function

For example the minimum of f(x) = x^2 + 1 is 1 at x=0

[frosty coral]

ohh

so like maximum and minimum

in a sin or cos graph?

but we are fidning the minimum only?

would the minimum be 600

because the y level is 1200

and the amplitude

is

600

so without the y level the minimum is 600

-600

and with the y level you do

1200-600

so its 600 right?

yeah i think so

thank you

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[pulsar tiger]

a.) Given these conditions, we know that $\epsilon > 0, \exists \delta > 0$ s.t if $|x - c| < \delta$, then $|\frac{f(x) - f(c)}{x - c} - f'(c)| < \epsilon$. We know that we are going to have the delta-neighborhood of $ c - \delta < x < c + \delta$, so we can have some interval: $f'(c) - \epsilon < \frac{f(x) - f(c)}{x - c} < f'(c) + \epsilon$. Now, we can set some epsilon to obtain a delta such that $\epsilon = \frac{f'c}{2}$. Now, we know from the problem that $f'(c) > 0$, so we have some interval $0 < \frac{f'c}{2} < \frac{f(x) - f(c)}{x - c} < \frac{3f'c}{2}$. We know that $x - c > 0 \implies x > c$ and $f(x) - f(c) > 0 \implies f(x) > f(c)$. So, this would would prove what we are looking for.

For part b.) We can employ the FTC by taking the chain rule of this to obtain: $f'(x) = F'(x^2)2x - F'(x^3)3x^2 =\cos(x^4)2x - 3x^2 \cost(x^6)$

[pulsar tiger]

Is this right?

[clever fjord]

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

[summer wave]

yeah

[amber schooner]

cos(x^6)

get rid of the t after the \cos

[odd edge]

@pulsar tiger Has your question been resolved?

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[odd edge]

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[errant stream]

lmfao

[crystal charm]

uh what happened

[odd edge]

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[latent meteor]

Given a line g and three points I,J and any point Z which is noch on the line g in projective field. How do i construct a line l through Z which is prepandicular to the line g?

[odd edge]

Please don't occupy multiple help channels.

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[sage helm]

I think the author made a mistake in the description, and I want to verify that I'm right. He says that matrix O changes from the standard to to the column basis, but it's the other way around: from the column basis to the standard basis.

[pastel orbit]

you are correct

[sage helm]

Thank you

🙂

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[mystic saffron]

hi

[mystic saffron]

is there any standar methodes for integrating e^f(x) and its not elemnetry

[quasi sparrow]

no

[mystic saffron]

then how ?

[quasi sparrow]

how what?

[mystic saffron]

how to integrate

[quasi sparrow]

why do you think there should be a solution?

[mystic saffron]

idk i have to find solution

[quasi sparrow]

!original

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[mystic saffron]

i need to find general solution

to solve such integration

[quasi sparrow]

picture or screenshot

[mystic saffron]

its just e^f(x)

[mystic saffron]

find general solution than can solve integrate e^f(x) dx whatever f(x) can be , and the integration is not elemnetry , from real number to real number

that is everythink

[safe vapor]

where did you get this question from?

[quasi sparrow]

so why do you "need" to find a general solution

[mystic saffron]

college project

[quasi sparrow]

welp find a better project

[safe vapor]

Not every integral has a general solution. That includes a subset of e^f(x) integrals. Consider f(x) = -x^2.

[mystic saffron]

think of it , its amazing idea

[mystic saffron]

this is a nightmare

[crystal charm]

I don't really see a use of it though

[safe vapor]

Yeah so amazing that you might even win a fields medal if you figure it out

[mystic saffron]

what is that medal ?

[quasi sparrow]

absolute waste of time

[crystal charm]

it's a prize medal every 4 years

think of it as the nobel prize but for maths

[safe vapor]

I was making a joke. The problem you're proposing in its current form is unsolvable, a solution does not exist. The joke was that if you are able to solve it (you will not), you would have significantly advanced the field. The reality is that you'll just waste your time if you try, and so you should find a project that is actually possible to complete.

[mystic saffron]

true not every integral is solvable , any thank you

.close

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[night bone]

is this a biconditional theorem

[night bone]

or can there be a fundamental set of solutions with duplicate eigenvalues

[forest sky]

the eigenvalues don't necessarily have to be distinct as long as there are n linearly independent eigenvectors (i.e. A is diagonalizable)

but if A is defective then the solutions will look different

[odd edge]

@night bone Has your question been resolved?

[night bone]

ty

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[brisk compass]

hi tried making my own polar artwork image, can someone help me double check if i did good for in regards to the rubric, honestly wasnt aware of what i was doing..

[mystic saffron]

well, im not too smart but based on the rubric i think you did good

[brisk compass]

all i needed to hear lol, thank you

[mystic saffron]

i would verify with like atleas 3 other people

[brisk compass]

okie ill keep it open!

[mystic saffron]

not hard ngl

well the color part is ez

u have 3+ equations

nice work

[brisk compass]

i mean like according to the rubric it's an A but tbh it was so easy its kindve scaring me

[mystic saffron]

lol

i hate when tests are ez they make me overthink them

[brisk compass]

litreally the last assignemnt was having to make a drawing out of like parabolas and stuff and im still working on that its stressing me out

im tryna make a bear

[mystic saffron]

waht

what grade are u in?

[brisk compass]

moving out to college in 12 days but had to do a math class over the summer because i didnt do it in highschool

i go to a CSU now but its just IM3

[mystic saffron]

dang gl

[brisk compass]

thankss

[mystic saffron]

CSU is what colorado or chicago?

[brisk compass]

California State University

[mystic saffron]

oh

mb

[brisk compass]

its like a group of ones

[mystic saffron]

i was thinking

[mystic saffron]

cool

meanwhile im tryna get into MIT or Stanford

im not even that old to go to college

[brisk compass]

lol im not smart enough for either

my gpa was like 2.4

[mystic saffron]

oh

depends on school

aight ima sleep

[odd aspen]

@brisk compass still need help?

[mystic saffron]

she needs someone to check her work based off of rubric

[odd aspen]

ooh

[brisk compass]

yes what ronin said

[mystic saffron]

that

[brisk compass]

it was just weirdly easy

just wanna make sure i'm not missing something

[mystic saffron]

why are u the only one here with a help channel

[odd aspen]

I think is all good you just need to fulfill the link and statement

[mystic saffron]

ye

i dont get how puttting the link is very important

[odd aspen]

fr

[mystic saffron]

here they give u a 0 if u dont do it

and are like WOMP WOMP

at your face

[odd aspen]

In my place we dont do the digital

we use the manual

[mystic saffron]

oh god

ur living the dream

[odd aspen]

lol

[mystic saffron]

my eyes hurt from working all day

[mental lotus]

if this is the image you want graded according to the rubrik, sad news, this is not a Polar plot (what they wanted). You have used a cartesian plot

[mystic saffron]

no

thats not it

[mystic saffron]

that is

[mental lotus]

oh, ok

I just saw this image after the discussion and it still looked unresolved

Yea, first image is all good

~~ I'd give full marks for all the sections except the first and last part (since first part is dependent on how she submits it, and the content relevant to the last part is not included here) ~~

[mental lotus]

#help-19 message I read this again. She wants to make a bear. So it gotta be the one I replied to... The first one must be the one the teacher provided as a sample template or smthn.

[mental lotus]

So yea, to reiterate, you got a cartesian plot, so its no good for submission

[mystic saffron]

:o

[odd edge]

@brisk compass Has your question been resolved?

[odd edge]

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[pliant nexus]

i recommend graphing out the region being integrated first

have you tried doing this?

[raven finch]

yea , as best as i can by hand

the z-y part kinda sketchy tho

[pliant nexus]

ok so given y

can you tell me the sets of equations bounding the region being integrated in terms of x,y,z?

so just looking at a cross section at a certain $y=c$

[clever fjord]

Arnavutköy

[limpid grail]

ok hi

solve

[raven finch]

wdym by that, x=4-2z

[limpid grail]

3x(squared)+y=0

[pliant nexus]

create your own channel

go to #❓how-to-get-help

[raven finch]

thank you

back to this

[pliant nexus]

uh okay lets do some process of elimination

[limpid grail]

oh sorry

ok

[pliant nexus]

no problem

why does options A and C not work?

[raven finch]

i mean thru proccess of elim, i know its A or C

[pliant nexus]

sorry freudian slip

i mean only

[raven finch]

?

[pliant nexus]

okay yes you know its A or C

[raven finch]

well the x part of having y^2 lower bound isnt going anywhere

[pliant nexus]

ok sorry let me just graph this out

[raven finch]

and x(z) = 4-2z

so those are the musts

[raven finch]

ok

is there a way to do this jus by inequalities?

and identifying the min and max of the bounds

cuz i feel like that way helps as a solid double check when i feel sketchy abt my graphing

[pliant nexus]

yes there is

uhh

so the max for z is going to be 2-x/2 where x is minimized right

[raven finch]

yes i think

zero to 2-x/2 so yea

[pliant nexus]

yes, so what is the minimum possible value of x

[raven finch]

y^2

[pliant nexus]

in terms of y

good

so the max of z is 2-y^2/2

[raven finch]

OH SH , i jus replace

ohhhhh

wait is this method full proofed

cuz i jus figured it out recently and been abusing it ngl

[pliant nexus]

yes, this will always work for "nice" functions (a.k.a. everything you are dealing with unless you are taking a graduate math class)

[limpid grail]

oh

[raven finch]

ok ok , yea nah im jus 1st yr eng

[raven finch]

does this work in spherical as well

cuz its kinda sketchy with like phi and stuff

[pliant nexus]

uh yes, this works whenever you are just switching the order of integration around

[raven finch]

why dont they teach this , and always jus give us the stuggle of graphing it and imagining

[pliant nexus]

to be technical, the jacobian of the matrix is going to be just some version of the identity matrix, with rows and columns switched round

[pliant nexus]

uh dealing with when something is minimized and maximized can be easy to make mistakes

[modern sundial]

I think it helps you know what to look for if you understand what it means to integrate out a variable, e.g., if you integrate out the z's there will be no more z's remaining in the integral.

[raven finch]

i see, imma use both for safekeep

[pliant nexus]

yes, this

[raven finch]

uhm , integrate out?

like cancel it off early on?

[modern sundial]

So if you have say, dzdxdy, you know that z is an inequality (or possibly independent of) x and y, x is a function of y alone, and y is always going to be a range between two numbers

[raven finch]

yes

oh wait i kinda got it

[modern sundial]

Yeah so y is easy, you know it is -2 < y < 2

and you know whatever x is, you know it must be some f(z,y) < x < g(z,y) (in your problem)

[raven finch]

this is prly the first thing i should keep in mind before even trying any inequality

ok thank you both

[odd edge]

@raven finch Has your question been resolved?

[odd edge]

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[rare roost]

i cant seem to get the answer correct

i’ve used the formula l = rxtheta

and rearranged it

[wild glacier]

so

[novel hinge]

try

[wild glacier]

For theta did you take it in radians or degrees

[novel hinge]

nevermind im blind

conversion matters too

did you convert from degrees to minutes properly?

[rare roost]

i rearranged formula to angle= arc length/radius

[wild glacier]

Yea sure so

That angle

[rare roost]

angle = 25/10

2.5

[wild glacier]

Did you take it to be degrees

so 2.5 degrees is what you did?

[novel hinge]

2.5 is in radians

[wild glacier]

Yes

[rare roost]

2.5 is radians,

[wild glacier]

okay

[rare roost]

how owuld i convert that into degrees

[novel hinge]

now you convert

[wild glacier]

multiply with 180/pi

[rare roost]

180/pie x 2.5??

[novel hinge]

yes

[wild glacier]

pi btw

[rare roost]

just a question on my calculator how do i keep 180/pi x 2.5 in exact value form??

[rare roost]

lol

ty

[novel hinge]

you shouldnt

they asked you to round your answers

[rare roost]

i know

but out of this context

how do I??

because when im finding eg the arc length i need it in exact form

but my calculate just gives me simplified

[novel hinge]

i see

[wild glacier]

wdym by exact form

[novel hinge]

as in your calculator just gives you the decimal places?

[wild glacier]

?

[rare roost]

yes

[rare roost]

fraction form

[wild glacier]

Why do you even need fractions

[rare roost]

i know like how to do it mentally, but i dont want to keep doing that each time

can calculators not do that?

keep it like that

[wild glacier]

I think you can change it in the settings

[rare roost]

any idea how to??

[wild glacier]

Okay let me

[novel hinge]

to be fair its very basic simplification.

[wild glacier]

Okay so that

[rare roost]

ok sure, but when numbers get bigger

[wild glacier]

Do you have a S<=>D button?

[rare roost]

i do, but nothing happens when i click it

[novel hinge]

well

it's not going to be bigger than what the calculator can fit would it

[honest elk]

Justmultiply the numbers then divide by pi by yourself

[rare roost]

ok ty guys

🙂

.close

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[hallow lion]

hello i need help with this

[hallow lion]

im guessing you do keep change flip?

[wild glacier]

Why is there an x=10

[hallow lion]

idk

thats what has me stumped

[unkempt lichen]

so judging from answer choices

[tribal ingot]

thats probably a typo

[unkempt lichen]

i suppose that was meant to be a +?

[tribal ingot]

if i had to guess its probably a +

[wild glacier]

I think x=10 is not there

[tribal ingot]

but yeah keep change flip works

[wild glacier]

So it would be only x^2-25/x+10 /x+5

[tribal ingot]

actually

i think that might've supposed to be a -

[hallow lion]

why?

[wild glacier]

yes that might be it

[novel hinge]

noo

its a +

[wild glacier]

because x+10 does not give an answer matching the options

Either it has to be x-10 or that should be ignored

[tribal ingot]

yea

[novel hinge]

nevermind im stupid

[wild glacier]

yes so x-10

[wild glacier]

It is not x=10 , it is x-10

[tribal ingot]

probably a typo

[odd edge]

@hallow lion Has your question been resolved?

[odd edge]

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[narrow creek]

are these identities wrong? ramanujan wrote them in his letter to hardy but i cant seem to see why they are correct

[narrow creek]

these are just a geometric series and the answer is completely different

[viscid flint]

if you get a different result then they're not right ig

[narrow creek]

i refuse to accept that the great ramanujan was wrong

[viscid flint]

then maybe it's a printing error? what do you get for the results / what do those radical thingos evaluate to?

[narrow creek]

@real trellis thank you

its so strange how ramanujan got the wrong answer

or i have to be reading it wrong

but its the correct letter to hardy, and the expression gives the wrong decimal values

[viscid flint]

confused by why everything gets put over 1

[narrow creek]

yea i guess we can ignore that part

he could have removed that

[viscid flint]

,w x = e^-2pi; 1 / (1 - x)

[clever fjord]

Results provided by WolframAlpha

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Requested by kitten.in.a.teacup

[viscid flint]

ok i deserved that

,w 1 / (1 - e^(-2pi))

[narrow creek]

well i get what you're trying to do and yes that gives the correct answer

[clever fjord]

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Requested by kitten.in.a.teacup

[narrow creek]

BUT LIKE

why is ramanujan wrong

its just a simple geometric series

he cannot be wrong, he is perfect

[viscid flint]

,w ( sqrt( (5 + sqrt5)/2) - (sqrt5 + 1)/2 ) * e^(2pi/5)

[clever fjord]

Results provided by WolframAlpha

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Requested by kitten.in.a.teacup

[narrow creek]

WHYYYY I CANT BELIEVE RAMANUJAN CAN BE WRONG

his answer isnt even the same decimal

i thought he rewrote it somehow

[crystal charm]

I mean he is human, so he still makes mistakes like the rest of us (not to discredit him though, he was definitely very insightful)

[viscid flint]

i'm seeing a fairly similar identity elsewhere

[narrow creek]

NO BUT LIKE this is a simple geometric series, even i know the answer

how would ramanujan get this wrong but he gets far harder series correct

i cant believe that truth

i just wont

[viscid flint]

this is the one i see

which does look similar in some ways but not the same

ohhhh

okay

so it's a typographical error

this explains the /1 business

where did you find this original image?

[narrow creek]

WHATTTT

THANK GOD RAMANUJAN REMAINS CORRECT

wait so its literally a printing error?

ohh

[foggy wagon]

I guess the + being “in” the denominator was them trying to show that

[narrow creek]

oh wow

@viscid flint THE 1s at the bottom of the fractions represent the continued fraction too

i was wondering why they were there

[viscid flint]

still a printing error imo

[narrow creek]

thankfully destiny remains the same, and ramanujan remains perfect

i thought for a second that ramanujan.. was wrong

[viscid flint]

https://mathoverflow.net/questions/288410/what-did-ramanujan-get-wrong

[narrow creek]

yeah.. apparently its the notration but its just so confusing they should have printed better

[narrow creek]

no, it cant be

no

[foggy wagon]

Just tore down his whole worldview

[odd edge]

@narrow creek Has your question been resolved?

[narrow creek]

what if

what if he got it wrong on purpose

he surely knew if it was wrong

[viscid flint]

probably not

every mathematician makes errors

that's why we talk to other mathematicians

[odd edge]

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[narrow creek]

maybe you're right

its just such an unebelievable idea if ramanujan was wrong, like i was nervous in real life

[odd edge]

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[slow cradle]

question

[slow cradle]

for this

is it possible

that

different values for x for tax

leads to a different value for sin

[wooden python]

did you mean to ask:

Is it possible that different values of x for tan(x) lead to a different value of sin(x)?

or did you mean sin(2x) at the end

[slow cradle]

yes

[wooden python]

mmmkay so which is it.

is it sin(x) or sin(2x)

[slow cradle]

sin2x

[wooden python]

for sin(2x), no, you won't get any different values.

[slow cradle]

why

[wooden python]

$\sin(2x) = 2\sin(x)\cos(x) = 2 \cdot \frac{\sin(x)}{\cos(x)} \cdot \cos^2(x) = \frac{2\tan(x)}{\sec^2(x)} = \frac{2\tan(x)}{\tan^2(x) + 1}$

[clever fjord]

Ann

[slow cradle]

i dont understand

is there not a simpler way

[wooden python]

im showing you that sin(2x) can be expressed entirely in terms of tan(x).

[slow cradle]

ok

[wooden python]

which means that you will not have any ambiguity or sign bullshit when finding sin(2x) when tan(x) is known

[slow cradle]

wait so

since

all these tan values

ok

so

all these tanx values are the same

plugging into formula

equals the same sin2x

wait so how did this guy do it graphically

[wooden python]

that looks like a difficult graph to use

but your video solved for x

even though as i showed there's a way to avoid doing that

[warped glacier]

well the key step is $2x = \frac{2 \pi}{3} + 2 \pi n$

[clever fjord]

south

[warped glacier]

since you know that sin(x) has a period of 2pi

[slow cradle]

but tbh

i dont have ur method memorised

as in the proof

[slow cradle]

yh

sin2x should have a period of pi right?

[warped glacier]

then you know there must only be one value of sin

[warped glacier]

indeed

[wooden python]

you weren't supposed to memorize it.

[slow cradle]

as in sin2x = 2sinxcosx

[slow cradle]

why so?

[warped glacier]

as in, $\sin 2x = \sin \left(\frac{2 \pi}{3} + 2 \pi n \right) = \sin \left(\frac{2 \pi}{3} \right)$

[clever fjord]

south

[warped glacier]

for all integer n

[slow cradle]

wait

i didnt get that part

oh wait

[warped glacier]

take sin of both sides here

[slow cradle]

hold on

i get it

[warped glacier]

cool

[slow cradle]

cos

wait a sec

i dont

mb man

[warped glacier]

okay, what exactly don't you understand?

[slow cradle]

this

this statement i dont get

liek

i get the periods

and everything

i get how tanx= root 3

has multiple solutions

but i dont get

how sin2x has one solution

wait

ok

so the tan x solutions

are er

pi away

and sin x

has a period of pi

so that means

sin2x can only equal one thing

ok

i think i got it

is my thinking

correct?

[warped glacier]

sin(2 x) has a period of pi

but yes that reasoning is correct

[slow cradle]

yh mb

forgot to write that

ok

[warped glacier]

so then if you have $\tan(x + \pi)$ = \sqrt{3}$

you also get $\sin(2(x + \pi)) = \sin(2x + 2 \pi) = \sin(2x)$

[clever fjord]

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

[warped glacier]

in other words, it doesn't matter which solution to tan x = sqrt(3) you choose

[slow cradle]

yh exactly

[odd edge]

@slow cradle Has your question been resolved?

[odd edge]

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[shrewd trellis]

Did I do this integration wrong

[nimble blaze]

yes

[narrow crypt]

[ f(x) = \sqrt{2}]

[clever fjord]

k

[nimble blaze]

overthinking / improperly applying power law

[narrow crypt]

is this what u're integrating, js to be sure?

[shrewd trellis]

Yep

[narrow crypt]

sqrt(2) is a constant

[nimble blaze]

you're attempting to apply (reverse) power law as though you have $\red{x}^{1/2}$, which isn't what you have here

[clever fjord]

ℝαμOmeganato5

[unkempt lichen]

isn't this a constant

[shrewd trellis]

Is the reverse power rule the reverse chain rule

[narrow crypt]

U can think of this as ${f(x) = \sqrt{2} \cdot 1 = \sqrt{2} \cdot x^0}$

[tepid pelican]

It's the reverse power rule as its name suggests

[clever fjord]

k

[nimble blaze]

$\int \red{x}^n \dd{\red{x}}= \frac{\red{x}^{n+1}}{n+1} + c,\ \ (n\neq -1) \ \
\int \blue{k}^n \dd{\red{x}} \redneq \frac{\blue{k}^{n+1}}{n+1} + c$

[narrow crypt]

Wonderful

[shrewd trellis]

So the reverse power rule the anti differentiation method

[narrow crypt]

What

😭

[shrewd trellis]

Is*

[narrow crypt]

It’s one of the formulas for integrating (“anti-differentiating”), yes

[weary pelican]

would it help putting the x in dx red too?

[clever fjord]

ℝαμOmeganato5

[weary pelican]

x is the variable of integration

k is constant with respect to x

[nimble blaze]

the first line is (reverse) power rule
the second line is the misapplication of that (trying to apply it when the base isn't x; the variable of integration which is wrong)

taking the square root out of this, e.g.
$$\int 100 \dd{x} \redneq \frac{100^{1+1}}{1+1} + c$$

[clever fjord]

ℝαμOmeganato5

[shrewd trellis]

Wouldn’t it be respect to n in omegas example

[nimble blaze]

wdym

n is also a constant

[shrewd trellis]

Oh ok nvm then

[nimble blaze]

the variable of integration is indicated after that d

[shrewd trellis]

So I need to solve it in this form

[nimble blaze]

that rule just tells you when its applicable

if you want to apply the reverse power rule in full,
#help-19 message

but there is the simpler constant integration rule

$\int a \dd{x} = \what$

[clever fjord]

ℝαμOmeganato5

[shrewd trellis]

This is a bit advanced for me imo, I might be dumb

[nimble blaze]

lets bring this back up

U can think of this as ${f(x) = \sqrt{2} \cdot 1 = \sqrt{2} \cdot x^0}$

[clever fjord]

ℝαμOmeganato5

[nimble blaze]

apply (reverse) power rule to the x^0

[shrewd trellis]

Oh yes I understand that

[shrewd trellis]

So x ^2/2?

[nimble blaze]

do you mean $x^{(2/2)}$ or $\frac{x^2}{2}$

[clever fjord]

ℝαμOmeganato5

[shrewd trellis]

first

I mean second

[tacit wasp]

...

[nimble blaze]

how are you getting that?

[shrewd trellis]

Wdym?

[tacit wasp]

Literally, how did you get that result?

[shrewd trellis]

I got that by anti differentiation

[nimble blaze]

you reached a result, i'm asking for how you got that

how are you anti differentiating

what rules are you applying,
and what exact calculations are you performing

[shrewd trellis]

Added one

[nimble blaze]

to what

[shrewd trellis]

Divided by that result

[tacit wasp]

Using reverse power rule? Which values did you use for n?

See, have you done it on a piece of paper?

[shrewd trellis]

Ill show my working out

[tacit wasp]

Better, yes

[shrewd trellis]

Actually, it’s just x

[nimble blaze]

yes, the antiderivative of 1 or x^0 is just x

so applying that, what would be the antiderivative of sqrt(2) or sqrt(2) * x^0

[shrewd trellis]

Which is 2^1/2 * x + c?

I’ll just write it down

[nimble blaze]

no need to put the sqrt(2) in exponent form

$\int \sqrt{2} \dd{x} = \sqrt{2} \gray{\cdot} x + c$

[clever fjord]

ℝαμOmeganato5

[nimble blaze]

or more generally
$$\int a \dd{x} = ax + c$$

[clever fjord]

ℝαμOmeganato5

[nimble blaze]

no

this isn't what you typed just now
and is wrong

why are you still trying to apply power rule to 2^(1/2)

[shrewd trellis]

hmm

So we don’t find the anti derivative of sqrt 2

[nimble blaze]

that's the ultimate goal

you can find its antiderivative of that by viewing it as
sqrt(2) x^0
and applying the power rule to the x^0,
and not touching the sqrt(2) because of constant multiple rule

if you're integrating a product, you shouldn't be integrating the multiplicands indivually anyway

$\int f(x)g(x) \dd{x} \redneq F(x)G(x) + c$

[clever fjord]

ℝαμOmeganato5

[shrewd trellis]

Ohhhh

fair enough

[nimble blaze]

the simplest way to do this is the constant integration rule i've mentioned multiple times
$$\int a \dd{x} = ax + c$$

[clever fjord]

ℝαμOmeganato5

[nimble blaze]

if you want to explicitly use power rule, then after getting
$$\int \sqrt{2} x^0 \dd{x}$$
the relevant rules from here are:
$$\int a\cdot f(x) \dd{x} = a \int f(x) \dd{x}$$
$$\int x^n = \frac{x^{n+1}}{n+1} + c, \ \ (n \neq -1)$$

[clever fjord]

ℝαμOmeganato5

[shrewd trellis]

I seeeee, I’m pretty sure I have some how to solve this now

thanks 🙏

.close

[odd edge]

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[analog basin]

Number of positive integral solutions to
x1 + x2 + x3 +3x4 + 3x5 + 5x6 = 21 is?
is case-work the only method here or is there a shorter one?

[odd edge]

@analog basin Has your question been resolved?

[sturdy cape]

Case-work's probably the fastest way; start by choosing x6, then seeing what options are available for x5, and then seeing for each x5 what options there are for x4, and so on

[weary pelican]

casework is not as bad as you think, as was said, because there's only two choices for x6

and then you can probably split in terms of the value of x4+x5

[odd edge]

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[mystic nova]

What have you tried?

[pliant nexus]

have you drawn a diagram with the lines and the currrent points?

please also draw the circumcircle in such a diagram if you haven't already

[mystic nova]

We don't actually need the circumcircle here, using the Euler line we can find the exact centroid coordinate

[pliant nexus]

finding the line given the centroid is still decently complicated

oh wait

you know the slope of the third line given the orthocenter

[odd edge]

@frail mauve Has your question been resolved?

[mystic nova]

Yeah, dot product equal 0 should get it done

If we do this we may have to check if the equation we found is actually true i think

Nah it's better to find 2 unknow point and do dot product equal 0 twice i guess

[odd edge]

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[soft sorrel]

hi, i have a question about limits as the variable approaches infinity, if i have the product of 2 functions, one that approaches 0 and one that approaches infinity, can i say that the end result will be 0? or will it be unbounded? this is the question that made me think about it:
$\lim_{n\to\infty} n sin(\frac{\pi}{n})$

[clever fjord]

dis_da_mør

[tiny pendant]

wait sorry

[faint knot]

$0\times\infty$ is considered an indeterminate form

[clever fjord]

mtt

[faint knot]

you can have $\lim_{x\to\infty}\frac{c}x\cdot x=c$ and $\lim_{x\to\infty}\frac1x\cdot x^2=\infty$

[clever fjord]

mtt

[faint knot]

in both cases, this is a 0 times an infinity as x approaches infinity

but on the left, it converges to the number c for any choice of c

and on the the right, it diverges to infinity

[soft sorrel]

i see

[loud bluff]

You can do a proper sin notation by using \sin

[soft sorrel]

the question is in one that says to use l'hopital, and i think this means i should rewrite it as $\frac{\sin(\frac{\pi}{n})}{\frac{1}{n}}$, but i'm not sure where to go from there

[clever fjord]

dis_da_mør

[soft sorrel]

deriving the denominator doesn't seem to result in something easier to solve

as $n^{-1} \to -n^{-2}$ doesn't seem better, is this unsolvable?

[clever fjord]

dis_da_mør

[soft sorrel]

wait but if i do l'hopital again i get
$\frac{-\pi^2 \sin{\frac{\pi}{n}}}{2 n^{-3}}$

[clever fjord]

dis_da_mør

[soft sorrel]

will this converge to $\frac{-\pi^2}{2}$

[clever fjord]

dis_da_mør

[mystic nova]

This has the form $\lim_{n \to 0} \frac{sin(n)}{n}$

[clever fjord]

Alexis_Fx

[mystic nova]

Better not to use L'Hopital here

[soft sorrel]

the question is in one that says i have to

it also says the limit can not exist

[mystic nova]

We prove (sin(x))'=cos(x) using this so using L'Hopital it would be a loop

[soft sorrel]

ok

so it's unsolvable (at least by l'hopital)?

[mystic nova]

rewrite it as $\lim_{n \to \infty}\frac{\pi\sin(\frac{\pi}{n})}{\frac{\pi}{n}}$

[clever fjord]

Alexis_Fx

[mystic nova]

$n \to \infty \Rightarrow \frac{\pi}{n} \to 0$

[clever fjord]

Alexis_Fx

[mystic nova]

You know $\lim_{n \to 0} \frac{sin(n)}{n}=1$ right?

[clever fjord]

Alexis_Fx

[soft sorrel]

that's not in my formulae sheet so i don't think i can use it

[mystic nova]

You could, but in the end you will end up using lim sin(x)/x

[soft sorrel]

alright

thanks

[mystic nova]

you can prove this using L'Hopital

[soft sorrel]

right

[mystic nova]

But it's a loop, maybe your teacher doesn't care

[soft sorrel]

yeah thanks

.close

[odd edge]

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[odd edge]

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[tall owl]

Lim=1

[lavish jackal]

!original

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[mystic nova]

[unkempt lichen]

wat = problem

[odd edge]

@tall owl Has your question been resolved?

[odd edge]

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[shrewd trellis]

can soemone help me with 3

i thought i understood but no

[faint knot]

whats the derivative of 4x

[shrewd trellis]

4

[faint knot]

what about sqrt(2) x

[shrewd trellis]

1/sqrt2

[faint knot]

sqrt(2) is a constant

it doesnt change

d/dx sqrt(2) x is just sqrt(2)

[shrewd trellis]

oh ok

[faint knot]

does that make sense

[shrewd trellis]

this doesnt

[faint knot]

so whats d/dx 1.4 x

[shrewd trellis]

u said sqrt (2) is a constant so it becomes 0

[faint knot]

...

4 is also a constant

doesnt mean d/dx 4x is 0, is it

[shrewd trellis]

it does

[faint knot]

thats not how product rule works

d/dx 4x is not just 0 * 1

[narrow crypt]

[ \sqrt{2} = \sqrt{2}\cdot x^0]

[clever fjord]

k

[faint knot]

what is making you think d/dx 4x is 0

you just said it was 4 earlier, and d/dx 4x is 4

[mystic nova]

isn't this antiderivative? why everyone talk about derivative here

[shrewd trellis]

u stated 4 earlier but d/dx 4x is 4

[faint knot]

yes???

[shrewd trellis]

yea

[narrow crypt]

ftc, ig

[shrewd trellis]

its anti derivative

[faint knot]

youre the only one saying d/dx 4x is 0

you cant calculate antiderivatives until you learn how to calculate derivatives

you at least need to know what d/dx sqrt(2) x is

so that we can find the antiderivative of sqrt(2)

d/dx 4 is 0
d/dx 4x is 4
these are different

what is making you think that d/dx 4x is 0, specifically

[shrewd trellis]

1/sqrt(1)?

[shrewd trellis]

No its 4

[faint knot]

please stop guessing 1/sqrt(1) and answer the question

why did you say 0

this might be for a big reason that we can fix

[shrewd trellis]

I said 4

[faint knot]

???

[shrewd trellis]

I thought you meant whats d/dx of 4

[faint knot]

alr thats ok

so lets try a few more things before d/dx sqrt(2) x

to fully convince you of how to d/dx sqrt(2) x correctly

first, whats d/dx 1.4 x

[shrewd trellis]

1.4

[faint knot]

what about d/dx 1.41 x?

[shrewd trellis]

1.41

[faint knot]

d/dx 1.4142135623 x?

[shrewd trellis]

1.4142135623

[faint knot]

d/dx (1.4142135623...) x?

[shrewd trellis]

I think I get the point now

1.4142135623...

[faint knot]

alr, so d/dx sqrt(2) x is?

[shrewd trellis]

sqrt(2)

[faint knot]

very good

so the antiderivative of sqrt(2) is?

[shrewd trellis]

0?

actually

sqrt2x

[faint knot]

d/dx sqrt(2) is 0, yes

[shrewd trellis]

sqrt(2)x

[faint knot]

plus?

[shrewd trellis]

+c

[faint knot]

yep

sqrt(2) x + C

try not to get tricked by the sqrt, its over a number

[shrewd trellis]

so

[faint knot]

if it doesnt change, it can be ignored for the derivative

[shrewd trellis]

we dont have to manipulate it

[faint knot]

yes

the explicit purpose of that question was likely to make you overthink

[shrewd trellis]

into a power form

[faint knot]

its a good check

[shrewd trellis]

alr thanks man, appreciate it

.close

[odd edge]

Channel closed

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Use .reopen if this was a mistake.

[faint knot]

np

[odd edge]

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[craggy meteor]

Hiiii, i need help abouttt this questionn relating to surds

[craggy meteor]

i know it might be more simpler with a calculator but since its asked not to, how do i answer this easily?

[wooden python]

well first off can you give the unsimplified form

[craggy meteor]

It got really complicated when i tried multiplying with denominator🥹

[wooden python]

for the height, that is

[craggy meteor]

Unsimplified??

[wooden python]

yes unsimplified

like before you do ANY simplification

what should the height be

[craggy meteor]

Is it something like dat?

[wooden python]

$A = \frac12 bh$ therefore $h = \frac{2A}{b}$ yes.

[clever fjord]

Ann

[wooden python]

so yes that's what i was asking you for

now here is a suggestion:

before you try to deal with the root, factorize 30 + 12sqrt(5) as 6(5 + 2sqrt(5)).

this gets you $h = \frac{12 \sqrt{5}}{6(5 + 2\sqrt{5})} = \frac{2 \sqrt{5}}{5 + 2 \sqrt{5}}$

[clever fjord]

Ann

[wooden python]

do you understand what i said just now @craggy meteor Y/N

[craggy meteor]

Wait im abt to read everythingg

[craggy meteor]

OhHhh so i should have made it simplerrr so it can be easier to do?

[wooden python]

well if you ask me then yes you should've

but not because this particular step in this particular circumstance had any air of obligation to it

but rather because you should always try to make your own life easier

smaller numbers are easier to work with than larger numbers

ya feel?

[craggy meteor]

i get whattt u meannn nowwww 🤣😭

[wooden python]

you can even go a bit further and cancel out $\sqrt{5}$ from num and denom so you get $h = \frac{2}{\sqrt{5}+2}$.

[clever fjord]

Ann

[craggy meteor]

OhhhHh i didnt know you could dividee sqr tooo but im afraid that ill do it wronglyy🤣😭

[wooden python]

you can cancel out anything nonzero

[craggy meteor]

I seee i seee 😂 okeeyy, ill try to doo itt now with this wayy

.close

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[mystic saffron]

Hi, I’ve found the domain of x(because of the square root and demonator) : [-1/2,0)U(0,1/2]

I’ve also found the zeroes by squaring (0, 6/13) but apparently 6/13 is a pseudo-zero.

How can I reason its elimination?

[mystic saffron]

Please tag me

[nimble blaze]

how did you get 6/13

[mystic saffron]

1-3x-sqrt(1-4x^2) = 0
-sqrt(1-4x^2) = 3x-1
Squaring both sides,
1-4x^2 = 9x^2 -6x + 1
0 = 13x^2 -6x
0 = x(13x-6)
x= 0 or x = 6/13

[nocturne brook]

x=6/13 is the intersection between -sqrt(1-4x^2) and -(3x-1), the negation of the right-hand expression

[nimble blaze]

eliminate by considering range of the square root

[nocturne brook]

I didn't verify this actually lol

well anyway, it could be

[nimble blaze]

3x-1 is positive when x=6/13
but -(1-4x^2) is non positive

personally, I'd use conjugates when approaching this:
multiply both sides by 1+sqrt(1-4x^2)

[mystic saffron]

How’d you get that?

[mystic saffron]

How?

[nimble blaze]

I mentioned it immediately afterwards

[nocturne brook]

in your procedure, you squared both sides. squaring does preserve equality, but it doesn't reflect it: 3^2 = (-3)^2 does not imply 3 = -3

so if you have two expression a(x) and b(x), and you found x=r satisfies a(x)^2 = b(x)^2 but not a(x) = b(x), it ought to be the case that a(r) = -b(r)

it should satisfy the other version

[mystic saffron]

-sqrt(1-4x^2) = 3x-1
Okay plugging in 6/13 doesn’t work.

But am I supposed to keep an eye out at every step in fear that something like this will happen?

[nimble blaze]

be aware of processes that may lead to extraneous solutions

as described above

squaring square roots is quite common

[mystic saffron]

Okay that makes a lot of sense

Can you explain this in reference to the equation we have here, I’m having a little trouble applying it here

[mystic saffron]

That would be whenever we have a function that comes with its own additional domain. Log, sqrt, anything else?

[nocturne brook]

in our context, a(x) = -sqrt(1-4x^2) and b(x) = 3x-1. the solutions to the equation a(x)^2 = b(x)^2 are precisely the solutions for at least one of a(x) = b(x) or a(x) = -b(x) --- it could be both

[mystic saffron]

Like 0 was for both

[nocturne brook]

x=6/13 was a solution for a(x)^2 = b(x)^2, but not for a(x) = b(x). therefore, it must be the case that x=6/13 satisfies a(x) = -b(x)

well, no actually

-(3x-1) at 0 is 1, unlike -1, being -sqrt(1-4x^2) at 0

anyway, that's just why you find 6/13 popping up, but not why it's not a solution

[mystic saffron]

So whenever I’m squaring both sides, I have to circle back and check?

[nocturne brook]

yup

[mystic saffron]

I’ve done a ton of practice problems before this, how come this issue never come up before?

[nocturne brook]

maybe the examples were particularly nice

[mystic saffron]

Probably Yeah XD

Okay

Thank you so much

I was very confused, but now I have an easier time digesting how just squaring could give a pseudo solution

Thank you so much again:)

.close

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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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[jade ice]

hi.. i was wondering if anyone could provide me solution to this..

[jade ice]

it seemed easy to me

but i couldnt solve it

[fluid tundra]

find an explicit way to describe the set A \ B

[jade ice]

uh i couldnt understand the meaning of the word here "explict"

[fluid tundra]

then simply describe to yourself the set A \ B

[jade ice]

ohh

so u mean we break it into two sets first multiples of 3 and then multiples of 5

uhh

i couldnt understand what u meant

@narrow crypt

[fluid tundra]

to clarify, is your answer supposed to be a simplified fraction?

[jade ice]

yes

[fluid tundra]

is this question ai generated?

[jade ice]

yes

[fluid tundra]

...

[jade ice]

i asked ai for some questions to practice

[fluid tundra]

never do this ever again

get your questions from a legitimate source

[jade ice]

..??

[fluid tundra]

you can use ai to ask for help, but do NOT ask it to generate questions

[jade ice]

oh