help-36

[onyx peak]

oh maybe i could use IVT

[opal pelican]

What motivated this question?

[onyx peak]

what exactly

[steep hare]

nvm

[opal pelican]

What made you think of it may be of help

[onyx peak]

idk, it just came to my mind somehow while doing unrelated stuff

[steep hare]

its just algebra

or pre-calculus

[opal pelican]

That's real analysis though

Or well, calculus some of y'all may call it

[iron dawn]

can anyone give me science sv link

[patent compass]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tropic crest]

You mean #network

[woven ledge]

like this?

[onyx peak]

oh, i can just let r = -1/2, then the sum is 2/3 and the prod is over 1.
r = -0.999 the sum is 1/2 and the prod is 0

[onyx peak]

Yeah exactly

[woven ledge]

its like around -0.72

[onyx peak]

oh cool

i wonder if that constant has a nmae

[grizzled tusk]

<@&268886789983436800>

[woven ledge]

well it would have to not be algebraic first!

or some combination of e and pi...

[onyx peak]

its probably too random to have a name

thanks though

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[azure terrace]

for part a)
i showed H is in the set, using that H is non-empty and ab^{-1} is in H.

the mark scheme shows that I should've used H is closed under multiplication, the identity is in the set and the inverse.

is it okay for me to use the first method or is it better to use the mark scheme method? or does it even matter

[ornate spade]

the way you did is fine

[azure terrace]

perfect thank u

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[tropic crest]

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[robust palm]

bro

where the fuck does the mu come from

[vale mist]

ap:pb

[rugged merlin]

it is provided in your problem statement

[robust palm]

lambda = l
mu = u

AP = l/(l+u) * AB
AP + OA = OP

this only has lambdas in the numerator

[vale mist]

$\mu (\vec{OP} - \vec{OA}) = \lambda (\vec{OB} - \vec{OP})$

[soft zealot]

notasnugglebugz

[vale mist]

rearrange to

$\vec{OP} = \frac{\mu \vec{OA} + \lambda \vec{OB}}{\lambda + \mu}$

[soft zealot]

notasnugglebugz

[vale mist]

and substitution

[robust palm]

okay makes sense

[vale mist]

ok done?

[robust palm]

yeah ty

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[grizzled socket]

how do i find the asymptotes of this hyperbola

[grizzled socket]

y = b/a(x) and y = -b/a(x)

[severe hawk]

well, you're given a vertex and one of the focii of the hyperbola

with that you can compute both focii and thus their distance, which is 2a

[grizzled socket]

yeah the c is 13

c^2 = b^2 + a^2

13^2 = 12^2 + b^2

169 = 144 + b^2

25 = b^2 b = 5

equations are 5/12(x) and -5/12(x) ?

[severe hawk]

what are you calling c, b and a there?

[grizzled socket]

c is the distance between the middle and the focii

a and b is x^2/a^2 - y^2/b^2 = -1

-1 here because its horizontale

[severe hawk]

this only applies when the major axis is the x axis

how do you need to change that if your major axis is y?

[grizzled socket]

wait how do u know if the major axis is y

[severe hawk]

.

[grizzled socket]

oh cause of the -1 ?

[severe hawk]

and because of the graph

both focii and vertex are in the y-axis

[grizzled socket]

ohh

why did they put b/a(x) here and -b/a(x)

the focii and vertex are on the y axis

and its -1

[severe hawk]

because it's the solutions for a DIFFERENT exercise

see that in your case you have the vertex being (0,12), not (0,5)

[grizzled socket]

ye

so they used b/a(x)

and we're doing a/b(x)

so the difference is that the y value is different?

[severe hawk]

and what they are calling b and c, probably

[grizzled socket]

i call it the same way

[severe hawk]

they are calling a to the horizontal, b to the vertical, and c to the distance from center to focii

not a the distance on major axis and b on minor axis

[grizzled socket]

a is 12

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@grizzled socket Has your question been resolved?

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[left trail]

What would the indeterminate form be here? I just see undefined / undefined

[drowsy epoch]

inf/inf

[compact laurel]

But the limit doesn't exists for either the numberator, or denominator

[left trail]

how so? it is (1/0)/(1/0)

[drowsy epoch]

from the left you get +oo/+oo from the right you get -oo/-oo=+oo/+oo

that is one of the indeterminate forms

[left trail]

isn't it other way around ? from the right it would be +infinity / + infniity

[compact laurel]

Oh that's interesting. So, even though neither of the limits exists for the numerator, or denominator since (LHR does not equal RHR). Since approaching from the left, or from the right both yield an indeterminate form, L'Hopitals still applies. That's cool.

[left trail]

+infinity / + infinity = -infinity / - infinity? we can factor out the negative and cancel?

[drowsy epoch]

lhopital applies for oo/oo or 0/0

even if it was -oo/oo it doesnt matter since you could factor out -1

[left trail]

infnity has -1 in it?

[left trail]

also what about this?

[left trail]

nvm infinity has everything

[compact laurel]

You might call me stupid #star, but that is actually cool

[drowsy epoch]

https://www.desmos.com/calculator/9aarhkoxua

[final saddle]

@left trail Has your question been resolved?

[hazy bay]

the limit exists

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[raven sage]

how to do this

[raven sage]

there are these rules which i dont understand

[willow cliff]

is the problem saying show this?

[raven sage]

what

[willow cliff]

like is your problem/question you need to solve just to show that inequality

[void valley]

Find the domain first, then square both sides

[raven sage]

whats a domain

[void valley]

note that this equation cannot be satisfied for any arbitrary x, e.g. you cannot have a negative number under the square root

[raven sage]

i know that much

[void valley]

So first of all x^2 - 3x + 2 >= 0

you can also use what you've sent

i.e. consider when g(x) < 0 and g(x) >= 0

[raven sage]

ah i know what a domain is its just called differently in my language

[raven sage]

i dont get it

[void valley]

Okay, have you already determine it then?

[raven sage]

no...

[void valley]

Can you solve x^2 - 3x + 2 >= 0?

[raven sage]

using the quadratic formula?

also i have to turn it into something like (x-1)(x-2) >=0

x=/=1 and x=/=2

wait no

[void valley]

yes

Now plot the schematic graph and read the solution

(or use the sign chart)

[raven sage]

it can be equal to 1 and 2

since its >=

[void valley]

?

this is an inequality, not an equation

the solution is supposed to be an interval

[raven sage]

shouldnt it be > instead of >=

[void valley]

No

[raven sage]

then how is it not 1 and 2

included i mean

how are they not included

in solutions

[void valley]

It must be >= since it's allowed to have zero under the square root

[void valley]

They are

[raven sage]

you said yes to this

[void valley]

To this*

This is true and it doesn't imply x can't be 1 or 2

[raven sage]

what to do from here

[void valley]

e.g. for x = 1 you have (1-1)(1-2) = 0 * (1-2) = 0 >= 0 ---> TRUE

[raven sage]

it cant be 3 for example right

i get what you mean

no wait it can be

it cant be anything less than 1

[raven sage]

oh it cant be anything between 1 and 2

[void valley]

ye

[raven sage]

but it cant be 0.5 for example right

[void valley]

it can (1/2 is less than 1)

[raven sage]

ohhhhh

i see

[void valley]

so the solution is

[soft zealot]

Modus

[raven sage]

no

from 0 to 1

i think at least

no youre right sorry

since two negatives give a positive

[void valley]

It's our domain

Not the answer yet

Whatever you come up with now has to be in this range

[raven sage]

whats the next step

[raven sage]

so g now

x-2 < 0

[void valley]

Notice that if x - 2 < 0, the inequality is true because LHS is always nonnegative

and hence:
smth nonnegative > something negative ---> ALWAYS TRUE

this is why we do it

[raven sage]

what is lhs

[spark palm]

ls

[void valley]

Left Hand Side

(of the equation)

[raven sage]

elaborate a bit more please

[void valley]

Can't you solve this

[raven sage]

x<2

do you mean from minus infinity to 1

is always true

[void valley]

it says f(x) >=0 and g(x) < 0

so you have to take the intersection now

[raven sage]

so this is the answer right

discluding the right part

[void valley]

To this part, yeah

[raven sage]

thank you

[void valley]

But it's not the final answer yet

[raven sage]

whats the step after it

[void valley]

now you have to consider what happens if g(x) >= 0

[raven sage]

what do i need to consider

x>=2

[void valley]

yes and then sqrt(x^2 - 3x + 2) > x - 2 becomes x^2 - 3x + 2 > (x-2)^2

[raven sage]

elaborate a bit more

[vital crag]

a > b implies a^2 > b^2 when both a and b are nonnegative which is true when x - 2 >= 0

[raven sage]

so what do i need to do next

[final saddle]

@raven sage Has your question been resolved?

[raven sage]

<@&286206848099549185>

[vital crag]

solve for x on the right

[amber vector]

@raven sage

[raven sage]

yes im here

let me solve it

[amber vector]

ill give you good idea

[raven sage]

im listening

[amber vector]

u have f(x)>g(x)

study the sign of f(x)-g(x)

using derivatives way

ill write it in hand so that ull understand

@raven sage

[raven sage]

im here

[amber vector]

you will derivate f(x)-g(x)

yull find the derivative is always negatives so that will make f(x)-g(x) strictly decreasing

[leaden moon]

Idt this is a calculus exercise 😭

[amber vector]

but it'll lead to the point

[raven sage]

thank you

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[leaden moon]

@tribal ocean

[tribal ocean]

oh

oops

[raven sage]

its not homework i promise

this is for practice

[tribal ocean]

@raven sage as long as it makes sense though

They're basically asking you to split the logic by cases whether x - 2 is positive or negative

then solve for each case

and the reason we want to split is so we can square in one of the cases

you can't square for all x

[raven sage]

thank you

[timber leaf]

Delete the message

[raven sage]

what do i do now

i wanted to analyse that

man why are you doing this to me

i asked this question previously as well

like a month ago

its not homework i promise damn

[main mirage]

Well

Let's look at what you have so far:

As Modus said, your inequality is now
x^2 - 3x + 2 > (x - 2)^2

[main mirage]

This is from here.

[raven sage]

ok i see

[main mirage]

Uhuh

So

Any ideas?

[main mirage]

How should we go from here?

[raven sage]

let me do it for a min

you get that x is larger than two

x>2

[main mirage]

Alright nice!

So,

What's the 'final answer' to our question?

Or whats the conclusion?

[raven sage]

is it this

is it blank what

[main mirage]

Yeah I cant read it

[raven sage]

oh wow

[main mirage]

I mean you can just type it :)

[raven sage]

discluding 2

so (2, plus infinity)

[main mirage]

Yep 👍

[raven sage]

is that the final answer

[main mirage]

Yeah. The answer you put down excluding 2, that's right.

[raven sage]

is g(x) always >= when we square both sides

[main mirage]

Wdym

Oh

[raven sage]

ignore the red part

[main mirage]

I'd say yes, if you include domain restrictions.
If you have sqrt(f(x)) >= g(x), then it is true that
f(x) >= g(x)^2.

There is an assumption that f(x) is non-negative 👍

[raven sage]

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[steep canyon]

I am currently stuck at this problem, ive thought about using a gaussian sum but then out of convenience switched to doing it in code. i dont really understand why my question is regarded as wrong.

[steep canyon]

maybe if i use 2n-1 instead of n, for the gaussian sum that would atleast guarantee its an odd integer. but to calculate it i would have to apply it on the non-sum aquivalent part as well and i would need to proof by induction ig?

(2n-1)^2*

[bold turtle]

Sorry, your code did generate an answer, so what's your concern here?

[steep canyon]

that it is wrong, which it supposedly is

[bold turtle]

"supposedly", because...?

[steep canyon]

when i type in my answer it is regarded as wrong. its a website

im typing it without exponential form ofc

[rare girder]

why is your upper bound 359500?

[steep canyon]

does 2 steps each time, i would exceed 719k otherwise, wouldnt i?

[shadow marlin]

There's a problem with your "induction" step here

[scarlet sequoia]

and by doing exactly 359500 steps, what is the last number you end up summing

[steep canyon]

718999 the last odd number i think

[steep canyon]

what exactly?

[bold turtle]

Test at a smaller case

What happens when you run from i:= 1 to 5?

[steep canyon]

it works ive tested it

[bold turtle]

No but what do you get

[steep canyon]

for i:=3 it gives 35 for i:=5 it gives 165

…

[bold turtle]

!occupied - read what the bot that just replied to you is saying:

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[pallid meadow]

Okay, sorry!!! 🙏🙏🙏

[steep canyon]

not i:= lol but n

i:=1 to 3 / 5

[shadow marlin]

Ok I misread, it seems that the induction is correct

[scarlet sequoia]

Btw I computed it for myself, there is indeed something wrong with your end result

[sand barn]

1 to 3/5 . I thought induction was for natural numbers

[steep canyon]

ik thats why i was asking

[steep canyon]

not a rational, 3 and 5 seperately tested, mb for poor articulation

[sand barn]

n(n-1)/2 and other formulae are for integer n only

[sand barn]

oh sorry

i was trying to help

[scarlet sequoia]

What happens when f,g are int64 instead of real

maybe computation errors arise from there, idk

[steep canyon]

you would be absolutely correct, tyvm

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[wind adder]

Someone teach me calc form scratch

[tropic crest]

uhh i dont think we can

however you can learn it yourself

[honest carbon]

That's what school is for

[tropic crest]

and this

[wind adder]

thanks for the advice

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[honest carbon]

Or you can just take a calculus textbook. I recommend Baby Rudin.

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[radiant igloo]

???

[sonic crystal]

whats your doubt

would you like to be interrogated

[radiant igloo]

what do i do here?? i tried like all combinations i could think of 1,2,3,4 but none seem to fit

same for squares and cubes of each and their patterns

[late gazelle]

is this apart of the unit your doing? Or is it recreational / competition stuff

[radiant igloo]

competition stuff

[late gazelle]

mhm kk

[shadow hornet]

is that the area or?

[rugged merlin]

definitely something about multiplication i reckon cuz these numbers become bigger very quickly

[radiant igloo]

its a Logical Reasoning puzzle ;-;

[rugged merlin]

oh i got it

i think i do

[late gazelle]

if it had something tangible to work with besides "find the ?" it would be at least approachable, but i think its just trial and error until something clicks

[rugged merlin]

@radiant igloo hint: ||you mutiply something by 3 and then add that by something||

[radiant igloo]

omg

that is so stupid

bruh

HOW IS THAT A PATTERN HOW WAS I SUPPOSED TO FIND THAT

1+4x3=13, 2+1x3=5, 3+2x3=9, 4+3x3=13?

[rugged merlin]

in the country i am in they have an intelligence quotient section in the standardised university entrance exam

[shadow hornet]

uhhh i think it is not 13

[rugged merlin]

it is 13

[rugged merlin]

thats correct

[shadow hornet]

i thought it is 11

[radiant igloo]

well yeah it is 13 the answer key also says 13

[rugged merlin]

so this stuff is all too common

[radiant igloo]

i mean apart from that crappy pattern here ig

anyways thanks!

.close

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[solar lichen]

Yo I have no idea why I’m getting this wrong

Someone go through steps pls

[severe verge]

what's your reasoning for your answer

and if you want to know the correct way it should be doing u-sub on sin(3x)

[solar lichen]

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[thorn meadow]

Just need a little helppp...

Okay so I don't quite understand how the bisector has anything to do with proving that the the two triangles are congruent..

I also don't particularly understand why SSS, AAS, SAS, etc. help to prove a triangles congruency.

so basically.. I don't quite understand any of it 🙁

[surreal vault]

do you know what it means when two triangles are congruent

[thorn meadow]

uhh the angles and sides are equal?

[surreal vault]

alright

we can say two things are congruent if they are (roughly speaking, im sure theres a more exact term) exactly the same

[thorn meadow]

okay

[surreal vault]

so for example if you see a triangle with 3, 4, and 5 sides then see another triangle with 3, 4, and 5 sides

[thorn meadow]

they are cogruent?

[surreal vault]

yes

now theres four main congruency rule theorem things that show two triangles are congruent

the first one is SSS or side-side-side which youve used above, it basically just says that if two triangles have the same three side lengths they are congruent

[thorn meadow]

uh huh

[severe hawk]

congruent includes similar triangles, not only equal ones, no?

[surreal vault]

oh okay i didnt know that

huh?

[severe hawk]

i meant to question it, not state it

sorry, phone writing is slow

[surreal vault]

oh okay i was just a little confused since im not really good at geom as well

[severe hawk]

im not sure

[thorn meadow]

uhh

[severe hawk]

anyways, if triangle 1 and triangle 2 are congruent, you can get from one to the other by reflection, rotation or moving it (not sure if scaling too. those would be similar, not sure if congruent)

[shadow hornet]

this is true, but similar doesnt imply equal

[thorn meadow]

so basically the triangles are just duplicates of each other?

[severe hawk]

yes

just looked it up, congruency doesnt include similarity; its the other way around

[thorn meadow]

similarity includes congruency?

okay

okay... but you know how in the question it says, 'AQ is a common side in both ▵AQB and ▵AQC' what does knowing the common side prove?

[severe hawk]

that they have one equal side

since it's literally the same one

[thorn meadow]

but why do they have to go through with all the other stuff if its as easy as just saying that one of the sides are the same??

[severe hawk]

since Q is bisector, you also know an equal angle

[severe hawk]

cuz tou need 3 things being same, not 1

[thorn meadow]

ohhhh

[severe hawk]

and since isosceles triangle, you also know a second angle is equal and/or side

[thorn meadow]

so like Q is bisector for...

[severe hawk]

angle on A

so both half angles on A are equal

[thorn meadow]

so its like there is an angle at the top with a line in the middle

[severe hawk]

yes

angle BAC

divided in BAQ=QAC

[thorn meadow]

OHHH

Wait...

so

becuase we have proof of the shared line in the middle, and we also have the angles at the top.. and we also have the equal sides on the outside (BA and CA) thats three things to prove that the triangles are congruent???

[severe hawk]

correct

in this case, criteria SAS

side-angle-side

with the angle being the one between sides

[thorn meadow]

YAY

THANKS

Uhh sorry...

One more question!!!

Same triangle as before btw

[severe hawk]

since congruent, all angles and sides are equal

[thorn meadow]

okay 👌

actually.. waitt

Umm

How would you prove that AQ is perpendicular to BC??

And how prove Q is the midpoint of BC??

[final tangle]

apply cpctc

[thorn meadow]

sorry what is that??

[lapis sail]

Corresponding Parts of Congruent Triangles are Congruent

[thorn meadow]

woah um

[lapis sail]

its reasoning

its the reasoning for when u proof two sides/angles are equal of triangles that u have already proven to be congruent

[final tangle]

can you identify which sides of triangle AQB correspond to which sides of triangle AQC

and same thing for angles

[thorn meadow]

uhh

Which sides of AQC correspond to AQC?

[final tangle]

whoops

AQB

[thorn meadow]

Ohh

[final tangle]

made a typo

[thorn meadow]

Well... AB = AC (they are equal)
⦣QBA = ⦣QCA (Iscocelles triangle so angles are equal)

Ugh I can't finish this now... is it fine if I leave this thing open?

[final tangle]

channel will auto close after a while if you're not here

how long will you be gone for

[final saddle]

@thorn meadow Has your question been resolved?

[final saddle]

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[final saddle]

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[robust sequoia]

can someone help me go through this exercise 🥀

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[robust sequoia]

1

the first question is easy but it's the first time I've seen questions like the following

[tropic crest]

Sorry to interrupt but <@&268886789983436800> sean combs pfp

[marble cedar]

Tf is sean combs pfp

[bold turtle]

Yh - while that's not an explicit rule, it is in contravention of #changelog message

[honest gust]

diddy

[tropic crest]

This

[bold turtle]

[see the changelog message I linked]

[marble cedar]

So diddy = sean combs pfp

[tropic crest]

Diddy is sean combs yes

But can we please stop and let the mods deal with this

[marble cedar]

Oh i see

[marble cedar]

But with what?

[tropic crest]

The guy

[marble cedar]

For the diddy pfp?

[tropic crest]

Ye

[marble cedar]

U cant be serious mate

[proper dagger]

tell that to the mods.

[tropic crest]

I mean it violates #changelog message

[bold turtle]

@marble cedar Read the changelog message I linked

[proper dagger]

the #changelog post was made for a reason.

[bold turtle]

/ that larp larp linked

[bold turtle]

regarding the question, and I'm afraid I'm gonna commute so I won't be around in a bit - do you recall how to test whether a function's injective (for question 2)?

[proper dagger]

you're welcome to modmail if you think it's absurd but I don't think it's productive to debate about this in a help channel.

[marble cedar]

But arent those for pedophilia accusations specifically?

[tropic crest]

@marble cedar

[lyric obsidian]

take this elsewhere please and thank you

[spare summit]

just let the mods decide and let the helpee (and future helpers) have his stuff.

[marble cedar]

No Im not even per se saying its absurd Im saying ur applying the rule incorrectly

[proper dagger]

then the mods will advise.

[marble cedar]

I guess

[robust sequoia]

I opened a channel and now there's a whole debate for a pfp seriously. Even in the changelog there's only mentions of calling someone else or making jokes using Sean comb or the other guy. I'm going to open a channel later this one made the question go up

.close

[final saddle]

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[grave trout]

You know that the question gets pinned right?

[robust sequoia]

most people would come and see others debating and probably wouldn't stay any longer, at least that's what I think. Please let this debate stop

[grave trout]

Its related to the diddy pfp

You should probably change it when your on this server

[grave trout]

Dont think the debate will stop

But speaking of which, no moderators actually said anything

Seems safe ig

[final saddle]

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[hard blaze]

Is it considered mathematically invalid to rewrite the second expression like I did in red?

[polar spruce]

in³?

[hard blaze]

Basically what I'm asking is if cubing the denominator only is invalid, or am i required to cube the whole expression?

[polar spruce]

you cube the inches

or else you'll get a length as the result

and the imperial system isn't that weird

[proper dagger]

cube everything.

[hard blaze]

ohhh im illeriate i didn't read the question properly

the question wants cubic inches

[proper dagger]

1 cb. in. is equivalent to (2.54cm)^3. 1 in. is not.

[hard blaze]

Okay i figured out my mistake

[hard blaze]

yeah i realise that now

thanks guys

.close

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[drifting geyser]

My professor gave me that problem to solve but I just couldn't figure it out, I know that 10^(log(x)) = x so I used this so that 10^(log(2+sqrt(r))) is 2+sqrt(3) but I just couldn't understand where to go to eliminate the sqrt(3)
Obs: I simplified the whole log expression to just log just so it helps (the question is under the draft)

[plucky rover]

Girl my eyes hurt trying to read that can you increase the contrast a bit

Or maybe type out the question

[drifting geyser]

sorryy

idk if in white helps

but the question is:

[plucky rover]

Very slightly

The question is the last line?

[drifting geyser]

yes

[plucky rover]

Suppose $f(x) = 10^{1+x} + 10^{1-x}$. Show that $f(\log(2+\sqrt{3})) \in \mathbb{Z}$

[soft zealot]

Zavier 🌺

[drifting geyser]

it's "if f(x) = 10^(1+x) + 10^(1-x) then show that f(log(2+sqrt(3))) is integer

[plucky rover]

Rational?

You've written Z

That's integers

[drifting geyser]

oh yeah

I mixed up the words

sorry

it's integers

[plucky rover]

Let's write $f(x) = 10 \cdot 10^x + \frac{10}{10^x}$

[soft zealot]

Zavier 🌺

[plucky rover]

Then we factorize out the 10 to get

$f(x) = 10\left(10^x+\frac{1}{10^x}\right)$

[drifting geyser]

uhm, okay

[soft zealot]

Zavier 🌺

[plucky rover]

Now let's sub in log(2+√3)

[drifting geyser]

yeah but then we'd get to the same place I was

I have no idea what to do with this

[plucky rover]

Well, rationalize

[plucky rover]

See this was because I couldn't read what you had sent

Are you familiar with rationalization

[drifting geyser]

sorryyyy

[plucky rover]

Don't be

[drifting geyser]

yeah I just multiply by 2-sqrt(3)

[plucky rover]

Do that then

[drifting geyser]

and then I'd get here

[supple mantle]

Wait

[plucky rover]

Uh no, you should get 4-3 on the denominator

[plucky rover]

Also rationalize here

[drifting geyser]

ohhhhh

that's the reason!!

[plucky rover]

Where did the 10's come from

[drifting geyser]

I was with the 4+3 not 4-3

[drifting geyser]

It's not 10, I mixed up again... wow

it was 2

okay

then the sqrt of 3 would be gone and it'd be 10(4) and it's40

okay

I got it!!

thanks!!!!

.close

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[plucky rover]

. @hasty knoll post your question here

[hasty knoll]

aiit

[hasty knoll]

sorry for slowness

how is one suppose to differtiate a absolute function

[tiny gorge]

you could try simplifying it first

sqrt(y^2) = |y|

[hasty knoll]

thne

then

[deep condor]

think of $|y|$ as a piecewise function,
$$|y| = \begin{cases} y & y \geq 0 \ -y & y < 0 \end{cases}$$
then differentiate each case

[soft zealot]

haseeb ♥

[hasty knoll]

i can visual that function in my head

yes

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[hasty knoll]

.close

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[gray niche]

HELPPP

[final saddle]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[gray niche]

Is this right

[drowsy epoch]

,rccw

[drowsy epoch]

No

[gray niche]

Is this one right

[obtuse grail]

Yes this is right

[drowsy epoch]

Yes

[drowsy epoch]

This is right

|-3| is multiplied not added

[gray niche]

Ty

❤️

[final saddle]

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[final saddle]

Channel closed

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[stable viper]

This is the full q

[stable viper]

I’m confused on how they did this

[final tangle]

Conjugates, rationalise denominator

[stable viper]

Where did the 3 in the denominator go?

[final tangle]

you didn't multiply the denominator properly

result should be
2 - 1
not +

[stable viper]

Ohh. Thanks sm

.close

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[lime crest]

i cant seem to get any info out of the construction ;-;

are (ABC), (AMN), gamma tangent?

[final saddle]

@lime crest Has your question been resolved?

[lime crest]

<@&286206848099549185>

[severe verge]

Menelanus theorem

Let F be the intersection of MP and AD; F' be the intersection of NQ and AD

then prove FD/FA = F'D/F'A

[severe verge]

Do you see angle BDP = CDQ?

<@&268886789983436800>

[lime crest]

yes

[lime crest]

whys this sufficient?

[severe verge]

it's easy to see that F and F' both lies on segment AD

and this ratio means that F and F' lies on the same position on AD

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@lime crest Has your question been resolved?

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[rain sentinel]

is this how we classify SPs

or do we take the 2nd derivative

[still wyvern]

stationary points include point of inflection right??
for that u need 2nd derivative and equate it to 0

[rain sentinel]

but is my method correct

[still wyvern]

ah yea
for finding maxima and minima its correct

[rain sentinel]

why

[still wyvern]

cuz derivative means slove of the function at some point basically

at max and min for most polynomials

the slope is 0

hence derivative is zero

[rain sentinel]

but I didn't take the secon derivative

[still wyvern]

2nd derivartive accounts for the slove of the function that u get after 1st derivaton

basically like imagine if u have x square then dy/dx will be 2x
means at any point x you will have slove of x sq as 2x
on 2nd derivative it'll go to 2
means the slope of first derivative is always 2

for maxima and minima u never almost need 2nd derivative

[final saddle]

@rain sentinel Has your question been resolved?

[rain sentinel]

I mean do we classify using the hessian matrix here

or can we use a different method as shown?

[still wyvern]

i am not really familiar with hessian matrix

using differentiaiton is pretty effective

[silver maple]

You would need to use the hessian matrix

[still wyvern]

why?
the question is just asking for maxima minima and inflextion point no?
thats just highschool diffferentiation

[silver maple]

The function consist of two variables, so you can't apply single variable calculus

[rain sentinel]

could we just evaluate at the points

[still wyvern]

ooohh wait i didn;t see it mb

[silver maple]

How are you going to determine if a point is max, min or inflection

[rain sentinel]

because the highest points are +/-1/2

[silver maple]

How do you know what is the highest or lowest values of F

I guess you could take an inspection based approach

I see what you are doing now

[final saddle]

@rain sentinel Has your question been resolved?

[rain sentinel]

.close

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[patent compass]

23

[patent compass]

23

. close

.close

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[tropic crest]

Holy self solve

[patent compass]

Uhh

I am onto it

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[timber leaf]

I know that for $f(x)=\arcsin{(\frac{1}{\sqrt{1+x^2}})}$, the derivative can be found using chain rule, is there other way to solve this though?

[soft zealot]

Nefer

[timber leaf]

@ when reply, thanks!

[scarlet sequoia]

There is most likely is a "trick involved" way to find it too

[supple mantle]

Triangle

[scarlet sequoia]

Like x = tan(theta) is what I'm thinking

[supple mantle]

Yeah

[timber leaf]

;-; So how can this be done?

[supple mantle]

Triangle like before

[timber leaf]

but triangle for which one?

the previous one is for tan y = x

[supple mantle]

sin(y) = 1/sqrt(x^2 + 1)

[timber leaf]

mhm

[supple mantle]

Hyp = sqrt(x^2 + 1)

Opp = 1

[timber leaf]

then f(x)=arcsin(sinx)

[supple mantle]

Hmm

[scarlet sequoia]

Arcsin(sin(y)) i think

[supple mantle]

Lemme scratch it out on a note rq

[timber leaf]

\section{Question 3}
a) First, recall the derivative of trigonometric functions:
\begin{gather}
\frac{d}{dx}(\arccos(x))=\frac{-1}{\sqrt{1-x^2}}\
\frac{d}{dx}(\arcsin(x))=\frac{1}{\sqrt{1-x^2}}\
\frac{d}{dx}(\arctan(x))=\frac{1}{1+x^2}
\end{gather}
Suppose we have:
\begin{gather*}
f(x)=\arcsin{(\arctan{(\arccos{x})})}\
\frac{d}{dx}[f(x)]=\frac{d}{dx}(\arcsin{(\arctan{(\arccos{x})})})\
f'(x)=\frac{1}{\sqrt{1-(\arctan{(\arccos{x})})^2}}(\frac{d}{dx}(\arctan{(\arccos{x})}))\
f'(x)=(\frac{1}{\sqrt{1-(\arctan{(\arccos{x})})^2}}(\frac{1}{1+(\arccos{x})^2}(\frac{d}{dx}(\arccos{x}))))\
f'(x)=(\frac{1}{\sqrt{1-(\arctan{(\arccos{x})})^2}}(\frac{1}{1+(\arccos{x})^2}(\frac{-1}{\sqrt{1-x^2}})))\
f'(x)=\frac{-1}{\sqrt{1-(\arctan{(\arccos{x})})^2}(1+(\arccos{x})^2)\sqrt{1-x^2}}
\end{gather*}

[soft zealot]

Nefer

[timber leaf]

This is for my part a, doesn't related but I didnt solve using triangle

pretty tedious, so I am looking for a faster method

[supple mantle]

Triangle should shorten the calculus route

[obtuse grail]

why want another way

[timber leaf]

Dude is asking why

https://tenor.com/view/whatldo-whatldoo-gif-21433439

[obtuse grail]

?

whats wrong with that im curious

[timber leaf]

I think 1 +x^2 there can be simplified in a right triangle

Such that a side is x, a side is 1

b) Now, suppose that in a right-angle triangle, we have an angle $\theta$ such that:
\begin{gather*}
\tan{\theta}=\frac{\text{opposite}}{\text{adjacent}}\
\text{(Let opposide side be $x$, while the adjacent side be 1)}\
\text{Hypotenuse =} \sqrt{\text{adjacent}^2+\text{opposite}^2}=\sqrt{x^2+1}
\end{gather*}

Ok i got sth like this

[soft zealot]

Nefer

[timber leaf]

What can I do next from here?

[supple mantle]

$\sin y = \frac{1}{\sqrt{x^2 + 1}} \ \
\dv x \sin y = \dv x \left( \frac{1}{\sqrt{x^2 + 1}} \right) \ \
y' \cos y = -\frac{2x}{\sqrt{(x^2 + 1)^3}} \ \
\cos y = \frac{x}{\sqrt{x^2 + 1}}$

[timber leaf]

Now i need to find sin theta?

[soft zealot]

VulcanOne

[supple mantle]

Then you solve for y' by substituting cos y

[timber leaf]

oge let me do it

[supple mantle]

@vital crag thanks btw

[novel kettle]

i think doing that gives the cleanest solution. you get f(x) = arccot(x) for positive x and then it's just a standard derivative

[timber leaf]

b) Now, suppose that in a right-angle triangle, we have an angle $\theta$ such that:
\begin{gather*}
\tan{\theta}=\frac{\text{opposite}}{\text{adjacent}}\
\text{(Let opposite side be $x$, while the adjacent side be 1)}\
\text{Hypotenuse =} \sqrt{\text{adjacent}^2+\text{opposite}^2}=\sqrt{x^2+1}
\end{gather*}
Now, to find $\sin{\theta}$:
\begin{gather*}
\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\
\sin \theta = \frac{x}{\sqrt{x^2+1}}
\end{gather*}
Now, we substitute back into the given equation, such that:
\begin{gather*}
g(x)=\arcsin(\frac{1}{\sqrt{1+x^2}})\
g(x)=\arcsin(\sin x)\
\frac{d}{dx}[g(x)]=\frac{d}{dx}(\arcsin(\sin x))\
g'(x)=\frac{1}{\sqrt{1+(\sin x)^2}}(\frac{d}{dx}\sin x)\
g'(x)=\frac{\cos x}{\sqrt{1+(\sin x)^2}}\
\end{gather*}

[soft zealot]

Nefer

[timber leaf]

Isn't it the same, i said 1 + x^2 there

ahh

b) Now, suppose that in a right-angle triangle, we have an angle $\theta$ such that:
\begin{gather*}
\tan{\theta}=\frac{\text{opposite}}{\text{adjacent}}\
\text{(Let opposite side be $x$, while the adjacent side be 1)}\
\text{Hypotenuse =} \sqrt{\text{adjacent}^2+\text{opposite}^2}=\sqrt{x^2+1}
\end{gather*}
Now, to find $\sin{\theta}$:
\begin{gather*}
\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\
\sin \theta = \frac{1}{\sqrt{x^2+1}}
\end{gather*}
Now, we substitute back into the given equation, such that:
\begin{gather*}
g(x)=\arcsin(\frac{1}{\sqrt{1+x^2}})\
g(x)=\arcsin(\sin x)\
\frac{d}{dx}[g(x)]=\frac{d}{dx}(\arcsin(\sin x))\
g'(x)=\frac{1}{\sqrt{1+(\sin x)^2}}(\frac{d}{dx}\sin x)\
g'(x)=\frac{\cos x}{\sqrt{1+(\sin x)^2}}\
\end{gather*}

[soft zealot]

Nefer

[supple mantle]

And "let opposite side be 1, while the adjacent side be x"

[timber leaf]

I think i made big mistake there, hold up

[timber leaf]

b) Now, suppose that in a right-angle triangle, we have an angle $\theta$ such that:
\begin{gather*}
\tan{\theta}=\frac{\text{opposite}}{\text{adjacent}}\
\text{(Let opposite side be $1$, while the adjacent side be $x$)}\
\text{So that, we have:} \tan \theta = \frac{1}{x}\implies\theta=\arctan{\frac{1}{x}}\
\text{Hypotenuse =} \sqrt{\text{adjacent}^2+\text{opposite}^2}=\sqrt{x^2+1}
\end{gather*}
Now, to find $\sin{\theta}$:
\begin{gather*}
\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\
\sin \theta = \frac{1}{\sqrt{x^2+1}}
\end{gather*}
Now, we substitute back into the given equation, such that:
\begin{gather*}
g(x)=\arcsin(\frac{1}{\sqrt{1+x^2}})\
g(x)=\arcsin(\sin\theta)\
g(x)=\theta\
g(x)=\arctan\frac{1}{x}\
\frac{d}{dx}[g(x)]=\frac{d}{dx}(\arctan\frac{1}{x})\
g'(x)=\frac{1}{1+\frac{1}{x^2}}(\frac{d}{dx}\frac{1}{x})\
g'(x)=\frac{1}{1+\frac{1}{x^2}}(-\frac{1}{x^2})\
g'(x)=\frac{-1}{x^2+1}
\end{gather*}

[soft zealot]

Nefer

[timber leaf]

Finally locked in

[rocky tusk]

i mean this just isn't complete

the method he suggested assumes x > 0

[supple mantle]

[timber leaf]

wdym

🫩

[rocky tusk]

the derivative is not -1/(1 + x^2) for x < 0

[timber leaf]

why not?

[rocky tusk]

if sin y = 1/sqrt(x^2 + 1) then you can't just say "oh i drew a triangle so therefore the other side is x and so cos y = x/sqrt(1 + x^2)." we actually have cos y = |x|/sqrt(1 + x^2) in this case

[timber leaf]

but x cant be negative here? for a right triangle

hmm

[rocky tusk]

😭

brother

the problem doesn't assume we are working in a triangle

[timber leaf]

homie its 1am

[rocky tusk]

it assumes we are working with real valued functions

[supple mantle]

The triangle does require that x is positive

[rocky tusk]

no

take the side length to be |x|

problem solved

[timber leaf]

Ahhh true

This homie got a point

what if x = 0

[supple mantle]

Oh ye the side length needs to be positive

Thanks for correcting my mistake

[rocky tusk]

you should try the tan substitution as well

you'll have to do some domain/range work though

[timber leaf]

derivative of 1/|x| is what guys

[timber leaf]

i did do that in my work

@rocky tusk @supple mantle Ok final solution

I go to sleep now

Thanks guys

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[rocky tusk]

the last line should have x/|x|

x^2/x = x

[timber leaf]

Wut

.reopen

[final saddle]

✅ Original question: #help-36 message

[rocky tusk]

when you transitioned to the last line you dropped an x

[timber leaf]

Ah yes

Naisu

Thanks again i sleep now -.-

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[unborn meteor]

where did k+1 come from I keep getting k+12

[proper dagger]

looks like it got cut off there

also, the augmented matrix's closing bracket is missing, which also hints at the text being cut off

[unborn meteor]

oml

good call lmao

thx i thought i was missing smthng

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[hard blaze]

Why is u taken as 0 in the answersheet to part b? didn't we calculate it to be -3 in part a?

[reef pike]

u know?

all cases of factorization?

[crimson zenith]

u changes with the time

for a, it's u at time t=0

for b, it's something else

[hard blaze]

why is it 0 tho?

[crimson zenith]

'initial'

[hard blaze]

im sorry i dont understand

[crimson zenith]

because we are finding the initial velocity

and initial means t=0

[hard blaze]

no in question b it asks to calculate the maximum distance upstream

here we are calculating displacement

so idk why u is 0 here

[crimson zenith]

oh

im not sure sorry

[hard blaze]

No worries, thank you for trying to help.

[final saddle]

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[viscid topaz]

i've done 3a (XY = (-2, 3, 6) and |XY| = 7) but i'm not sure what is being asked for 3b. is it asking for a parallel unit vector?

[proud igloo]

well how would you get a unit vector?

[charred agate]

Its asking for a unit vector, unit vector is a vector with its mod equating to 1 & the required unit vector is supposed to be in same direction as XY vector ( yes , you are right )

[proud igloo]

what country uses mod? normally we say norm

[viscid topaz]

ah, that would make sense. let me check my notes, because i'm pretty sure it ends up being something like XY divided by |XY|

[charred agate]

|| India, by Norm we mean like, 'Norm of vector XY is 7?' ||

[charred agate]

Yes, because when you take the norm of that vector that you get by driving XY with its norm, it's new norm should be 1.

[viscid topaz]

i'm not sure what norm means, but it does in fact check out with my notes lol

thanx for the help

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[viscid topaz]

woah, there were messages completely missing on my end. that's weird

[charred agate]

What do you mean?

[viscid topaz]

i didn;t see that other users messages at all until i rebooted my app because i knew something was off

[charred agate]

Interesting.

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[halcyon glade]

hey guyss, i need help with understanding smth. what on earth are eigen values of a matrix?????

[radiant igloo]

so basically

when you take a matrix A of order n

[halcyon glade]

Of order A? dont matrices usually have a number as their order?

[radiant igloo]

sry mb

typo

we basically equate |A-xI|=0

the roots of x here are eigen values

[halcyon glade]

dw abt lmaoo

okay i dont get it... what is the purpose of an eigen value? what to do they do?

[sonic crystal]

https://youtu.be/PFDu9oVAE-g?si=mTru6n9X8AxKXoj0

[radiant igloo]

these values are basically there so that they can be transformed

[royal gust]

Eigenvectors and eigenvalues come in pairs.

Eigenvectors are the vectors that, when A acts on them, they stay in the same direction.

Eigenvalues are the stretch factors that A applies to their corresponding eigenvector

[radiant igloo]

okay yeah that

[sonic crystal]

youll probably learn more from this video than anyone here can tell you over text

[radiant igloo]

(its a bad idea to take up too many channels at once)

[halcyon glade]

aaaa alr alr thanks so much!

[timber leaf]

isn't eigenvalue is just kinda like the dilation factor of the eigenvector

[halcyon glade]

ummm alr so i realised my foundation is super weak. i havent yet learnt them as representation of vectors in class lol

okay so theres a problem that i have thats been bugging me and idk what to do about it lol. lemme see if i can send it

Q 18

i dont think i need to learn what eigen values are to answer this...

[radiant igloo]

okay so

can we break P into 2 matrices such that one of them is an identity matrix? if yes then what type of matrix would the other one be?

[halcyon glade]

... i dont know??? i dont even understand the question

[radiant igloo]

we basically need to find P^5 here. I3 is a 3x3 identity matrix and we are given Q-P^5=I

[halcyon glade]

if You mean that im supposed to break P into the from I + B then im assuming be would have the lower part below the diagonal as 3 9 and 3

[radiant igloo]

exactly

[halcyon glade]

Okay great!

[radiant igloo]

can you verify if the given matrix(B) is nilpotent?

[halcyon glade]

ummmm yes if i sqaure it then im pretty sure we would get 0

[radiant igloo]

(N^2 is not 0 here)

[halcyon glade]

no wait if we sqaure it wed get 9 and if we cube it we would get 0

[radiant igloo]

exactly

[halcyon glade]

yeah yeah mb

[radiant igloo]

do you remember binomial expansions?

[halcyon glade]

yes i do

[radiant igloo]

can we expand (I+B)^5 here

(as P=I+B)

[halcyon glade]

That would be I + 5B + 10 B^2 +10B^3+5B^4+B^5

[radiant igloo]

exactly

[halcyon glade]

So eveything after B cube is gonna be 0

[radiant igloo]

exactly

can we compute P^5 from here now?

[halcyon glade]

Okay yayy! ummmm i got that

[halcyon glade]

wait alr alr yeah i think so

do i try it out on my own and come back if i have doubts?

[radiant igloo]

we already found B and B^2 earlier

[radiant igloo]

yea

[halcyon glade]

Yes yes we did

Okay okay brb

[radiant igloo]

(well i gtg now, if you do have any other doubts you can ping the helpers by doing <@&286206848099549185> )

[halcyon glade]

okayyy Thanks so much for your help omg

I GOT IT!!! WOOO

Also quick q, how do i know that im supposed to split matrix P as (I+B)?? like ig im trynna find out why we did that

<@&286206848099549185>

can someone help me out

[deep condor]

no need to ask for help, just ask your question :)

[leaden moon]

13k people got pinged because of this single message 💀

[halcyon glade]

lmao alrr

[halcyon glade]

oh shit

[halcyon glade]

could somone help me out with this?

[leaden moon]

well not yours, yours is necessary

the one im replying with

[halcyon glade]

nah he was jus trynna help me out, i just started using dc and this is the my first time on this server so it was pretty helpful

[leaden moon]

ic

[halcyon glade]

yeah

could u help me out with my prob?

[leaden moon]

yea i can't do some matrices stuffs

im sorry

[halcyon glade]

lmfaooo yeah thast valid. dw abt it, thanks anyway

[deep condor]

the idea here is that we want to split P into a nilpotent matrix and something else, then when we compute powers of (I + B)^n there will be fewer terms (why?).
we can see ones on the diagonal and we know that I is never nilpotent, so we can subtract P - I and hope that the matrix we get, B, is nilpotent

[halcyon glade]

okay so this isint a fool proof method? we are basically just guessing ??

[deep condor]

if you can find a nilpotent matrix in the decomposition, then this is a faster method to solve the question

faster than computing P^5 by hand, anyway

[halcyon glade]

hmmmm alr i get the idea

tysm for your help!

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[late gazelle]

I did pt.a but idk how to do pt. B

[late gazelle]

.close

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[late gazelle]

.reopen

[final saddle]

✅ Original question: #help-36 message

[late gazelle]

@tidal birch did you have something ? Bc I’m still stuck

I got this but idk, it’s not correct

[final saddle]

@late gazelle Has your question been resolved?

[tribal ocean]

i think you might be conflating circle and sphere?

do you mean circle or sphere?

[late gazelle]

Circle

[tribal ocean]

then its asking for the circle on the plane

i guess?

[late gazelle]

Yea

But idk how to make a circle tilted, so my classmate said to do the above working but idk why it “works”

[tribal ocean]

the intersection of the sphere and plane is the circle

Right

[late gazelle]

Yea

[tribal ocean]

because the points on the circle are still the same radius away

so they are part of the sphere

[late gazelle]

Mhm

[tribal ocean]

then what you did is valid

[late gazelle]

Uh huh, but idk why making x in terms of y and z make it true

[tribal ocean]

Each equation corresponds to one of the two surfaces

if they are both true

then it corresponds to their intersection

so it "selects" all points that are on both surfaces

if you solve their system

[wooden ferry]

Anybody know how to apply the calabi yau manifold?

[late gazelle]

Open your own help channel and someone will help you

[wooden ferry]

ok, sure

[tribal ocean]

or what did you mean here exactly

[late gazelle]

I’m my final line of working

I take the equation of the plane and solve for x

Then sub that into my sphere equations

[tribal ocean]

yes

then you can rewrite the sphere equation to look nicer

but itll still be a circle

[late gazelle]

I got both on desmos, but both are 3D

Their intersection is 2D

[tribal ocean]

yes

x depends on y and z

y and z are your parameters

if x depends on y and z it doesnt add a dimension

[late gazelle]

Ahh ok

[tribal ocean]

a dimension has to be independent

its basically the number of independent variables

[late gazelle]

Mhm kk

That makes sense

[tribal ocean]

When i first learned about that, i visualized that as once you fix y and z, the value of x is fixed

which means the surface has no space to move on that third axis

[late gazelle]

It only has “space to move” on the domain of the intersection circle in this case right

[tribal ocean]

yeah

well actually wait

[late gazelle]

Open your own help channel

[tribal ocean]

i guess the circle is a 1d curve tho technically?

its just that y and z have two possibilites for each other

so sorry theres only one parameter

the other one can be different by sign

only

[late gazelle]

Well the answer is in terms of y and z

So I think what we already have is fine

[tribal ocean]

i meant that the intersection is a one-dimensional space

a curve

[sly forum]

What plane is it talking about, one of the co ordinate plane?