help-33

[halcyon surge]

yes i know , my problem is in how to make table. i said i forgot how

i just wanna ask is that enough? or i should like do more

[lone heart]

You need more

[halcyon surge]

yes how

[lone heart]

Derivative is the slope. You want to find the slope at 1

So you need to make a table with values around 1

[halcyon surge]

can you give me an example to make it clear to me

[lone heart]

You made a (small) table with the value of 1

[halcyon surge]

yeah

[lone heart]

Now, you use other values

[halcyon surge]

what is the other values

[lone heart]

Whatever x values you want

Like 0

2

Etc

Just a table

[halcyon surge]

uhh i got it

thank you very much

[full leaf]

Yeah isn't this just mean value theorem?

[halcyon surge]

yes but in this question i should estimate it numerically

there is another i should use this theorem for

@lone heartthank u man u r the best ❤️

[full leaf]

Yeah but the mean value theorem states that f'(c)=(f(b)-f(a))/b-a (for a given c between a and b)

And if you pick a,b such that they're close enough to 1 (one greater and one lower) it should be a good numerical estimation

[halcyon surge]

i understand u but i prefer to solve it with a table

[full leaf]

Yeah, dw, but I'd recommend having mean value theorem in mind usually

[halcyon surge]

i will try it all. thank u buddy ❤️

[marsh citrus]

@halcyon surge Has your question been resolved?

[halcyon surge]

.close

[marsh citrus]

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[halcyon scaffold]

a) What topic of math is this
b) How do you solve it

if it helps you can use a scientific calc

[red nimbus]

a) trigonemtric equations
b) use sin(2x) = 2sin(x)cos(x) and factor cosine, use zero product theorem

[halcyon scaffold]

is there an easy way to do that on a calculator

[halcyon scaffold]

or do you have to memorise

[red nimbus]

yea you can use a graphical calculator

[red nimbus]

keeping some trig identities in mind is helpful

[halcyon scaffold]

okok, how do you do this one?

its just these two questions ive never learnt b4

[red nimbus]

make use of the identity

sin²θ + cos²θ = 1

Solve for sine

[halcyon scaffold]

alright ill try them, thanks

.close

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[hoary marsh]

im pretty sure you're supposed to simplify the first equation to 3^2x+3^-2x=34

not sure where im supposed to go from there

someone please help me with this

[tight furnace]

What's (3^x+3^-x)^2

[hoary marsh]

sorry i had to do somehting

thanks though

i didnt realize it was this type of problem

.close

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[marsh citrus]

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[daring wadi]

I am working on software for rendering polyhedral graphs. this category of graphs happens to be undirected, unweighted, and dense.
i am trying to efficiently enumerate the faces / peripheral cycles of these graphs.
I've found and implemented a satisfying solution for enumerating chordless simple cycles (https://arxiv.org/pdf/1309.1051), but notably, this solution does not exclude cycles which partition the graph into disconnected subgraphs, such as the three cycles of length 4 which hug the meridians of an octahedron.
previously, i had been using eulers polyhedron formula to stop solving once 2 + E - V cycles had been found, but this is insufficient for edge cases in which these non-face cycles might be found before face cycles. i can provide more detail on that if desired.
am i missing something obvious? trying to research planar embedding makes my head spin,,,, id deeply appreciate any help 🙏

[marsh citrus]

@daring wadi Has your question been resolved?

[daring wadi]

if this question is outside of the scope of channels like these just lmk. im new here so i'm not sure what kinds of questions are appropriate to ask

<@&286206848099549185>

[marsh citrus]

@daring wadi Has your question been resolved?

[livid cipher]

What is the angle shown by a red line?

I believe it is theta but quite not sure.

<@&286206848099549185>

[vagrant frost]

I also think it is theta, if the auxiliary line you drawed is perpendicular to the horizontal line.

[marsh citrus]

@daring wadi Has your question been resolved?

[marsh citrus]

@daring wadi Has your question been resolved?

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[still temple]

Bolzano-Weierstrass in my book is "A bounded sequence of real or complex numbers has a convergent subsequence"
.
so i started with there exists a subsequence (p_n_r) with p_n_r \in E s.t lim n->inf p_n_r -> L. If i could show p_n_r ≠ L i'd be done and i'd say L is a limit point but the only thing i know abt (p_n_r) is that its monotonic so my problem with saying p_n_r ≠ L is what if its something like (5,5,5,5,5,5,5,.....)

pls help 😭

[glass silo]

A subsequence of what sequence?

[still temple]

oh right an interval is not a sequence idk why i said that

[tired oxide]

you are right in needing to choose a good sequence

you didn't use all the hypotheses given to you

about the set E

(like, there is a sequence in E that is not like (5,5,5,....), or even has that kind of behavior on its tail end)

[still temple]

does sth like this work

let p_n be an infinite squence: for all n\in N, p_n \in E

by bolzano weirstrass theorem there exists a monotonic subsequence p_n_r -> L

[tired oxide]

you need to guarantee that p_n is not constant

or like

that you can choose a p_n that isn't constant

and even more than that, you need to guarantee that none of the p_i is actually equal to L (this is the problem that you mentioned earlier, because we want the sequence to demonstrate that L is actually a limit point of our set E)

(actually you only need that infinitely many p_i are not equal to L, and then you can do some work to find a subsequence such that none of them are equal to L but have L as a limit)

so the question is now: how are we going to choose our "good" sequence p_n?

[still temple]

can we say sth like let p_n be an infinite sequence st no tail is constant

[tired oxide]

how can we say that there is such a sequence in E?

[still temple]

because its infinite?

[tired oxide]

yes, and if you want to make that part of it formal, you might use induction (you can just take each p_i to be different from all the previous ones at each step).

so now you have the p_n that is not eventually constant, and that's good.

[still temple]

yep

now we pick our convergent monotone subsequence with bwt

[tired oxide]

yep okay, say that's p_n_r -> L as you wrote previously.

there is now only one potential problem with this sequence p_n_r.

[still temple]

it could be constant

[tired oxide]

we took care of that

[still temple]

what if p_n = 1,4,1,5,1,6,1,7,1,8,....
and p_n_r = 1,1,1,1,1,1....😭

[tired oxide]

if you choose your sequence right, then you won't run into this probem. you can actually choose p_n such that p_i \neq p_j if i \neq j

$p_i \neq p_j$ if $i \neq j$

[elfin berry]

differential smayded lie algebra

[still temple]

okay

[tired oxide]

this will come from the justification using the fact that E is infinite

[still temple]

okay

[tired oxide]

so that part is not a problem with the subsequence p_n_r. what we do care about though, is that some of the p_n_r might be equal to L, and we do not want that

[still temple]

so can we say p_n_r ≠ L as if it were then for some r>=N p_n_r =L so p_n_r+1 = L and so on and it gives a constant tail which is a contradiction?

[tired oxide]

well, there is no actual contradiction that we can derive (p_n_r+1 doesn't also have to be L, so your reason would not hold). we might actually have that one of the terms in our sequence is equal to L. what we want is to find a subsequence of p_n_r such that we skip over any that are equal to L. (the limit will still be L because subsequences of convergent sequences still have the same limit). how do we know that we can do this?

we can also recap or take a step back to look at the big picture at any point if you want or feel like you need me to go back

[still temple]

so sth like p_n_r_k = x \in R : x = p_n_r for all r\in N, x≠L

[still temple]

yes please a recap would be helpful😭

[tired oxide]

okay let me go wash my hands with warm water because I'm freezing oops and then I'll come back and say what we've done so far and what we want to do

[still temple]

oh lord take your time

[tired oxide]

okay so here is what we've done:

the first thing to do is to recognize that it suffices to show that there is some $L$ in $\mathbb R$ and a sequence $(p_n){n=1}^\infty$ such that $\lim{n\to \infty}p_n = L$

we used the fact that $E$ was infinite to get a sequence, say $(p_n)_{n=1}^\infty$ such that no two $p_i$ are the same.

we then used Bolzano-Weierstrauss to get a subsequence $q_r = p_{n_r}$ that had a limit $L$.

we are about to trim the sequence down further, which means we are about to take another subsequence $s_k = q_{r_k}$ such that no $s_k$ is equal to $L$. and then we will verify that $s_k$ demonstrates that $L$ is a limit point of $E$, and then we will be done.

[elfin berry]

differential smayded lie algebra

[still temple]

yesss sounds good

[tired oxide]

the key to the last bit is, only one of the $q_r$ could possibly be equal to $L$

[elfin berry]

differential smayded lie algebra

[tired oxide]

so let's say $q_{r^}$ is equal to $L$, which is the worst case, then we can define $r_k = k$ if $k < r^$, and $r_k = k+1$ if $k \geq r^*$.

[elfin berry]

differential smayded lie algebra

[tired oxide]

then $q_{r_k}$ is the sequence that skips over the problem term.

[elfin berry]

differential smayded lie algebra

[still temple]

yesss

[tired oxide]

the thing to fill in is why this part is true, and then explain why the new sequence is what exactly you want (or why it implies that L is a limit point)

and then you'll be done.

[still temple]

because no two terms are equal

the new sequence converges to L and no term is equal to L

[tired oxide]

yes this is the relevant part

and also that all the terms in the sequence are in E

[still temple]

yess

thank you so much

[tired oxide]

np. real analysis can be a struggle sometimes lol

[still temple]

ive been liking it

but this was kinda depressing

[tired oxide]

yeah it's really tough when you're in the weeds to actually be able to step back and put everything together, and that's something that you gain over time (and problems that you find miserable are pretty much the only way to get better at that unfortunately)

[still temple]

theres more 🥰🥰🥰🥰🥰

i will give them more thought and try my best to get them done

thank you for being so patient

❤️

.solved

[marsh citrus]

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[unborn osprey]

I'm back, I never quite figured this one out

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be differentiable, and $f(x,y) = 0$ whenever $x^2 + y^2 = 1$ Show that there is atleast one point $(x_0, y_0)$ such that

$\frac{\partial f}{\partial x}(x_0, y_0) = \frac{\partial f}{\partial y}(x_0, y_0) = 0$

[elfin berry]

Bean Man

[unborn osprey]

There are certainly points where df/dx is 0 and points where df/dy is zero by Rolle's theorem after restricting f to the closed unit disk, but I'm not sure I can find a point where both are zero

[marsh citrus]

@unborn osprey Has your question been resolved?

[unborn osprey]

I’ll just ask in the analysis chat

.close

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[dawn heart]

What's 16274929 divided by 234

[smoky spire]

use a calculator

[dawn heart]

Because it bugs me as a first grader

No I'm not cheater

But ok

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[ocean flame]

This Is A Qustion Of India's Enginering Entrance Exam JEE Main from 10 April 2023 Shift 1, I searched the whole internet for this question but the same answer and method is used. First Of All They have can straight away canceled x and y in tangents eqns as if they are equal, Then They Have Directly Substituted the coordinates in the eqn of circle which gives distance btw the circumference and the point which wrong cause asked is Distance btw the two point of contacts on the curves. Please Help Me

[dawn heart]

The answer is 34

I think

https://images.app.goo.gl/YUMN2unLoGY2mEGbA

This image clarifies your linear based question with y value, z value, and quetiont value. You should know by now under math regulations the multiplication based tables are useful for situations like linear, exponents, and rules. This page also explains how physics and science works as the years go by as space and earth have different times due to black holes, and how gravity will pull to make everything around you one singularity

Hope that helps

[marsh citrus]

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[dawn heart]

Good luck

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[dawn heart]

What?

[marsh citrus]

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[chilly flower]

I am unsure how 3^x-2 gets me -10

[chilly flower]

Photo math showed me this but I am still unsure how x-2 turns into x+4

[true parcel]

do you know this rule?

[chilly flower]

I do not

What would be equal to B and what would N and M be equal too

And what is this rule for?

[true parcel]

if $a^x = a^y$ then x = y

[elfin berry]

Astar777

[true parcel]

lets say a is 3

so $3^x = 3^y \implies x=y$

[elfin berry]

Astar777

[true parcel]

here you first make the base same (3), 9 is 3^2

so it becomes $3^{2(x+4)}$

[elfin berry]

Astar777

[true parcel]

then using the rule above, you do x-2 = 2(x+4)

[chilly flower]

Okay I am very dumb when it comes to math I really am just trying to pass my college algebra class so bear with me

How would I make 3^x-2 into -10

[true parcel]

.

[chilly flower]

I am doing logarithmic equations and do not understand

[hallow finch]

you’re solving for x, not trying to transform 3^(x-2) into -10

[chilly flower]

The answer is -10 I just want to know how I get there

[bitter terrace]

If you have 3^x=3^y, x=y

Look at what they said above for this

[hallow finch]

@chilly flower if you have 3^a = 3^7, what’s the only value of a that could make this true

[chilly flower]

7

[hallow finch]

yes

now what if it’s 3^(a+2) = 3^7

[chilly flower]

A=5

[hallow finch]

yes

this is the idea behind the rule $b^a = b^c$ implies $a = c$

[elfin berry]

Nacho Boi

[hallow finch]

whenever the bases are the same, the only possible way for the whole expression to be the same is if the exponents are the same

so when we have $3^{x-2} = 9^{x+4}$, we want to write this in a way such that they have the same base

[elfin berry]

Nacho Boi

[hallow finch]

because then we know that the exponents are equal

@chilly flower does everything I’ve said so far make sense?

[chilly flower]

It should but it doesn’t, I get what you are saying that they need to equal each other but I don’t get how 3^x-2 =9^x+4

[hallow finch]

so we’re being told that 3^x-2 = 9^x+4, it’s like how sometimes you’re told y=3x+4, it’s just your given information

[chilly flower]

In the equation I am only given 3^x-2 and somehow the answer I need is -10

I know the answer is -10

[hallow finch]

we’re almost there

do you understand the rule $(a^m)^n = a^{mn}$?

[elfin berry]

Nacho Boi

[chilly flower]

I don’t understand a thing about this but at this point if it comes up on the test I will just skip it

Thank you for your help

[hallow finch]

this is likely the majority of what the test will be

exponent rules

[chilly flower]

That is what my exam is over

[hallow finch]

11. is just testing your understanding on a few basic exponent rules

[chilly flower]

Okay I am a moron I get it now

Thank you

[hallow finch]

you’re not a moron, math can be difficult

[chilly flower]

Nope moron

.close

[marsh citrus]

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[marsh citrus]

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[flint sleet]

how to find side length x

[flint sleet]

$\text{Find side length:} x$

[elfin berry]

Amir

[cobalt widget]

And angle z?

[flint sleet]

I have angle Y but not Z

[cobalt widget]

Hmm

Is angle x 90?

[flint sleet]

Ni

No

[cobalt widget]

Idk then sry lol

[idle lagoon]

is y 28.8

[flint sleet]

No XYZ is 74 degrees

[idle lagoon]

youll need someone else’s help on this i suck at this stuff 😭

[still temple]

are you allowed to use sine rule ?

/ cosine rule

[flint sleet]

yes

[still temple]

use it ?

[marsh citrus]

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[cunning wedge]

hello

[cunning wedge]

i have a doubt

but idk if its math----
its physics actually-

[random cosmos]

just send the question TOT

[cunning wedge]

why do i get different answers on using Q = delU + delW and using Q = nCpdelT?

[random cosmos]

must be an error somewhere

[cunning wedge]

ok gimme 5 mims, ill send my working

@random cosmos

[random cosmos]

i have been summoned

[cunning wedge]

fr

[random cosmos]

i must say, ur pen is at it's last stage of living

[cunning wedge]

my hand's broken, i cant press the pen any harder 💀

shouldn't have played football during exam days

[random cosmos]

oh damn, get well soon

anyways back to the question

[cunning wedge]

yea

[random cosmos]

are u sure ur delU is correct?

[cunning wedge]

i mean

ig-

lemme check

yea looks fine

[random cosmos]

okay then, im unable to see the sign convention properly, recheck that ig

[cunning wedge]

the formula has 1-gamma in denominator

thats why u end up with negatives

@random cosmos r u indian

[random cosmos]

....

[cunning wedge]

no

not like that

cuz this is sm

indian formulae

idk if its there in ur country

[random cosmos]

i am indian...

[cunning wedge]

i mean, jee aspirants cud relate only ig

ahca

[random cosmos]

HOW DID U-

[cunning wedge]

acha

😭

BRO THEN

[random cosmos]

[cunning wedge]

U SHUD KNOW THIS AAAH 😠

[random cosmos]

man tbh, i suck in physics

[cunning wedge]

😭

jee 2025 or ug?

[random cosmos]

im in college...

[cunning wedge]

oh ok

[proud arch]

This is like family reunion but worse

Lol

[random cosmos]

exactly

[cunning wedge]

😹

answer the doubt

please

aaah

i cant spot any erorr

error

sorry

@random cosmos

i used the formula wrong 💀

its gamma-1 in denominator 💀

[random cosmos]

😭

i knew it

[cunning wedge]

😔

ok nice meeting u didi

how do i close this

ticket

[wary kite]

!done

[marsh citrus]

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

[random cosmos]

.close

[wary kite]

nice try

.close

[marsh citrus]

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[random cosmos]

DUH

[wary kite]

😹😹😹

[marsh citrus]

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[wary kite]

set the limit equal to L then take ln(L)

[elfin berry]

Allophane

[small stream]

Yea, i have already done that, but then???

[wary kite]

$\ln(L) = \frac{1}{\ln(1+x)} \cdot \ln(\lim_{x\to 0} \frac{\ln(x+\sqrt{1+x^2})}{\ln(1+x)})$

[elfin berry]

knief

[wary kite]

does that say ln^2?

[small stream]

Ye, you mean in the rhs denominator?

[wary kite]

yea the image is blurry

[small stream]

Yea lol

[wary kite]

ok and does it say - ln(1+x)

[small stream]

Yep

[wary kite]

$\ln(L) = \lim_{x \to 0} \frac{(\ln(\ln(x+\sqrt{1+x^2}) - \ln(\ln(1+x)))}{\ln(1+x)}$

is that what you had

no yours was 1-x

it says 1+x

[elfin berry]

knief

[wary kite]

then

so they have $\frac{\ln(x+\sqrt{1+x^2})-\ln(1+x)}{\ln^2(1+x)}$

[elfin berry]

knief

[small stream]

Yep

[small stream]

Typo

[wary kite]

this isn’t even true though

so we must’ve made the same mistake somehow or theirs is wrong

[small stream]

Maybe my mistake

I don't know where tho

Just want to know how did they got the rhs

nvm i got it

[marsh citrus]

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[wary kite]

i’m curious how?

[small stream]

.reopen

[wary kite]

.reopen

[small stream]

They use the important limit

[wary kite]

for e?

oh

[small stream]

Ye

[marsh citrus]

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[small stream]

This is strange because I thought the results obtained by using that important limit and directly taking the log should be the same

[wary kite]

yea and it looked so similar too

i briefly thought of using that limit but it wasn’t in the usual form

so i kept banging my head against the wall doing it the way you did

[small stream]

But it's obvious that sometimes we have to use the logarithmic method, sometimes we have to use this important limit method, and sometimes I really can't distinguish when to use this and when to use that

These two methods should get the same result, but in this problem, they get different results. We can only get it through the method of the important limit, but the logarithmic method cannot get us thru it. This makes me feel confused

[wary kite]

hmm

strange

i’m not sure

[small stream]

😿

Ok thanks

[wary kite]

yea if i helped at all lol

[small stream]

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[severe wasp]

im learning about limits and i dont get the idea of LHL=RHL to check if its continous and where u do +h and -h and equate, i dont get it

[severe wasp]

no

i dont get it

2?

yes

2 itself?

oh wait so like 0??

[glossy flint]

No, why would it be 0??

[severe wasp]

i dont get it

[glossy flint]

Well, you can't do limits if you don't know how to read a graph

[severe wasp]

def

any resources i can refer online??

oh

alrr

[marsh citrus]

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[whole scroll]

can anyone explain why the roots of x-4 is diffrent in this board? cause when i tried chatgpt it came up with diffrent answers

[blissful scarab]

wdym

whats the qn

[whole scroll]

qn?

[blissful scarab]

question that u are doing on the board

[whole scroll]

x<-3 to x=-4

why is the factor diffrent

[whole scroll]

look at this

[whole scroll]

and the factor of x-4

factors

they are completely difftent

diffrent

nvm i found the answer

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[topaz adder]

Hi there! How do I find the antiderivative of this rational function?

[vernal forge]

try taking u=sqrt(t)

and then simplifying

[topaz adder]

WDYM? Can you show me?

[vernal forge]

do you know u-substitution?

[topaz adder]

No.

Why?

[vernal forge]

like let u=sqrt(t), du=1/(2sqrt(t)) dx

this kind of thing

[vernal forge]

it's a nice method to evaluate this integral but you can do it without it

[topaz adder]

Hmm...

How about no?

[vernal forge]

you can split the fraction

[topaz adder]

Okay.

Lets do that.

I know how to split the fraction amongst the other parts.

[vernal forge]

good

what do you get?

you should get ||2sqrt(t) - 4/sqrt(t) + 3||

[topaz adder]

2t/sqrt(t)-4/sqrt(t)+3sqrt(t)/sqrt(t).

Lol close enough.

[vernal forge]

good, you can simplify it to get to what i wrote

[topaz adder]

Where's the 2t?

[vernal forge]

$\frac{2t}{\sqrt{t}} =\frac{2\sqrt{t}\cancel{\sqrt{t}}}{\cancel{\sqrt{t}}} = 2\sqrt{t}$

[elfin berry]

artemetra

[topaz adder]

How dod we get two square roots of t?

[vernal forge]

t = (sqrt(t))^2

[topaz adder]

Oh okay.

I get it now.

[vernal forge]

awesome

[vernal forge]

now rewrite everything in terms of power of t

recall that sqrt(t)=||t^(1/2)||

[topaz adder]

Uh huh.

And the way to bring it up to the top.

Is by making it t^-1/2?

[vernal forge]

yep

precisely

[vernal forge]

do you know how to take it from here?

[topaz adder]

No...

What next if I might ask?

[vernal forge]

recall that antiderivative of the sum is the sum of the antiderivatives

(basically you can evaluate each separately and add them up after)

[vernal forge]

how would you find the antiderivative of $x^a$?

[elfin berry]

artemetra

[vernal forge]

for some a

[topaz adder]

x^a+1/a+1?

[vernal forge]

yep

[topaz adder]

Oh okay.

[vernal forge]

do that for each term and you are done

[topaz adder]

What about 4?

Or 4/sqrt(t)?

How would that be resolved?

[vernal forge]

4/sqrt(t) = 4t^(-1/2)

exact same rule

[topaz adder]

I see.

Would it not be like 4*-*t^(-1/2) though?

Like not multiplied?

To 4?

[vernal forge]

(4-t)^(-1/2) or 4-(t^(-1/2))?

in any case it is neither

[vernal forge]

or like what are you asking

[topaz adder]

Yeah I thought bringing it up would make it a +/- to a constant.

Not make it as a coefficient.

[vernal forge]

the coefficient is not affected whatsoever

[topaz adder]

So it would be multiplied TO the number then?

If that makes any sense?

4 x sqrt(t)^-1/2?

[vernal forge]

yes, that's what being a coefficient means

[topaz adder]

Okay.

Thank you!

[topaz adder]

Yeah I know how to do the process of finding the derivative through powers.

I just didn't know how to handle it as like a more complex rational function.

Ty.

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[flint jolt]

108

I used expansions

And i got k=2 but it is incorrect

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[still temple]

Is this correct?
Cause
x^2 - 7x + 6
x^2-7x+6=(x-=6)(x-3)
For the first one

x^3+3x-18
x^3 + 3x - 18 = (x+6)(x-3)

(x² - 7x + 6)(x² + 3x - 18) = (x - 6)(x - 1)(x + 6)(x - 3)

[still temple]

So C

[smoky spire]

yes

[still temple]

nice

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[still temple]

would it be -4 0 1

[still temple]

from zeros in order

least from greatest

[amber birch]

yes

[marsh citrus]

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[still temple]

(2x +1) is a factor of w(x)
, whose graph is shown.

Is the multiplicity of (2x+1)
Odd

[still temple]

Cause odd is the only on e that would be odd

[amber birch]

yes, you can see that cause the sign changes at x = -1/2

so that matches the shape of x^3, x^5, x^7 etc. locally at x = -1/2

[still temple]

Thx

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[abstract hull]

Can someone check and tell me if this is correct

[abstract hull]

Rhs was too perfect I'm second guessing myself

[marsh citrus]

@abstract hull Has your question been resolved?

[proud arch]

Hi @abstract hull

The solution you have presented is correct

Great job

I just wanted to remind you that

Tan^-1( A) + cot^-1(A) = pi /2

Where A Is some real number

You got the same format here

(x^2+y^2)^1/2

Is some real number

Differentiation of constant is 0

Just another idea of solve your question

Though what you have done is absolutely fine

Hope this helps @abstract hull

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[rugged marten]

Can someone help me make practice questions based of numbers 1 and 7 on this test about Sets, Counting, and Probability?

[marsh citrus]

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[hallow pier]

Need help.

[hallow pier]

This can be expressed as

-2<x<2

right?

[proud arch]

Hello @hallow pier

Yeah you are right

[hallow pier]

@proud arch

Can you help me with this one

[calm harbor]

He said it’s correct.

[hallow pier]

how would you express this

as a restricted enviroment

my guess

is

(x-delta, x+delta)

but is this correct?

[marsh citrus]

@hallow pier Has your question been resolved?

[marsh citrus]

@hallow pier Has your question been resolved?

[rugged marten]

No

[rugged marten]

.reopen

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@hallow pier Has your question been resolved?

[full flume]

what do you mean by "restricted environment" ?

[hallow pier]

@full flume

[full flume]

hmm so are you deleting zero? would you not use a union as they have it in their example?

[hallow pier]

well, i'm supossed to have a neightbourhood of p = x?

[full flume]

i don't know the full question, but yeah a deleted neighbourhood about p on the interval (a, b) would look like (a, p) U (p, b)

p in (a, b) ofc

[hallow pier]

well

i need to express this as a deleted neighbourhood if possible

is it possible tho?

[full flume]

a deleted neighbourhood needs something to be missing/deleted

as you have it, it's a neighbourhood about zero

it's not clear to me whether you're expected to delete zero.

maybe that's a reasonable expectation? idk we can ask someone with more topology education haha <@&286206848099549185>

[hallow pier]

lemme translate you the exericse

[hallow pier]

Determine the neighbourhoods characterized by the indicated inequalities. Express them using interval notation and, where possible, neighborhoods. Draw a graph for each case.

[full flume]

ok it seems you don't need to delete any values for this case then

you have an interval (-11, 11)

[hallow pier]

yeah

[full flume]

i mean that's it

i guess you can say it's a neighbourhood about 0

[hallow pier]

@full flume when can I express them using neighborhods?

[full flume]

maybe something might be lost in translation here

Express them using interval notation and, where possible, neighborhoods

i would say in words:
(-11, 11) is a neighbourhood about 0

idk if there's a better answer sorry

[marsh citrus]

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[molten ferry]

is there an easier way to find the direction of a cross product without having to do any physical work with right hand rule

[marsh citrus]

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[jolly sigil]

Did I draw the area correctly?

[marsh citrus]

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@jolly sigil Has your question been resolved?

[marsh citrus]

@jolly sigil Has your question been resolved?

[jolly sigil]

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[sharp aurora]

Could someone please explain to me why there needs to be a Y(S)?

[brave marsh]

That's just what the laplace transform of a derivative is.
$$\mathcal{L}(f''(t)) = s^2 F(s) - sf(0) - f'(0)$$

[elfin berry]

Azyrashacorki

[sharp aurora]

Where did s^2 come from?

Where did -0 come from?

Where did -s*1 come from?

[brave marsh]

That's what the formula is. You can find it by using the Laplace transform definition and doing integration by parts

[silk python]

im also confused

[sharp aurora]

oh ok

[brave marsh]

In this case your initial conditions give you that y(0) = 1, whence the -s * 1 term

And y'(0) = 0, whence the - 0

[sharp aurora]

Oh ok I will check the formula sheet

[marsh citrus]

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[vast seal]

,rotate

[vast seal]

Could somebody please help me answer part c

[still temple]

let x=(a+b-c)/2, y=(b+c-a)/2, z=(c+a-b)/2

substitute in b

[vast seal]

Why does it make reference to a triangle

Is that not needed for the question?

[quiet anvil]

Triangle inequality.

[vast seal]

Ohhhhhhhhh

I kinda see it

But how do u get that into final answer

Actually ill give it a go and see if i can do it myself using triangle

Thanks for help

[quiet anvil]

Yw!

[vast seal]

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[half stump]

really confused on how to do this question

[pearl haven]

do you know about the identity sin²x + cos²x = 1?

[half stump]

yeah

[pearl haven]

just use that

[smoky spire]

no need for that

u know how sin is opp/hyp

this one is a speical pythag number

[half stump]

3 4 5 triangle

[smoky spire]

yes

[half stump]

oh

ok thanks

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[still temple]

when finding the basis for KerT and RngT in a linear transformation, why do they have the same matrix? i was told to think about the nullspace and column space of a matrix but i am still lost. thank you!

[static quarry]

what do you mean by same matrix?

[still temple]

in this example, when finding the basis the KerT and RngT the matrix you do RREF is the same

im not quite sure why that is and how it relates to nullspace and column space of a matrix

i understand that KerT is nullspace and RngT is column space but why do they yield in the same matrix?

[fervent rampart]

the matrix is associated with the linear transformation by T(x) = Ax

[still temple]

could to explain that a bit more

why does column space and nullspace have the same matrix?

[fervent rampart]

all linear transformations on R^n have an associated matrix, where applying the linear transformation is the same as multiplying by the matrix

so they come from the same matrix because they relate to the same linear transformation

[still temple]

what if it was P2?

would it be the same?

[fervent rampart]

in the case of vectors not in R^n, you could apply the same concepts by expressing them as coordinate vectors with respect to a basis

[still temple]

what does this have to do with nullspace and column space having the same matrix when finding a basis in a linear transformation?

[fervent rampart]

because then the linear transformation has an associated matrix

[still temple]

could you explain that a bit more?

[fervent rampart]

if you can express the vectors in a given space as coordinate vectors wrt a given basis, then we can treat them as vectors in R^n, and all linear transformations in R^n have an associated matrix, where applying the transformation T is the same as multiplying by the matrix A, i.e. T(x) = Ax

the image of the linear transformation is the same as the column space of the matrix (multiplying a matrix by a vector is the same as taking a linear combination of its columns), and by definition the null space of the matrix is the same as the kernel of the linear transformation

[still temple]

just to clarify, when finding the basis for KerT and RngT (nullspace and column space) they have the same matrix because we are essentially multiplying the matrix by A to apply the transformation?

[fervent rampart]

if we have a vector x, then applying the transformation T to x is the same as multiplying by the matrix A, i.e. T(x) = Ax

it's a general theorem that all linear transformations on R^n have an associated matrix for which this is true

[still temple]

how would that apply to this example?

[fervent rampart]

if we can find the matrix associated with the linear transformation with respect to given bases, then any relevant information about the linear transformation may be found from the matrix (because they are equivalent), including a basis for the image (column space) and kernel (null space)

[still temple]

so the matrix is only equivalent when the transformation is in R^n?

[marsh citrus]

@still temple Has your question been resolved?

[fervent rampart]

any other vector space can be represented in R^n as coordinate vectors wrt a basis, and multiplying by the matrix is equivalent to performing the linear transformation on the coordinate vectors

so although it is more roundabout the matrix is equivalent to the transformation even for other vector spaces

[marsh citrus]

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[desert gyro]

Hello, I am following a solution for a "difficult" inequality and I believe I spot an error. Here is the problem + solution:

[desert gyro]

Isn't the derivative of f(t) computed wrong ?

It should be $\frac{1}{16t}-\frac{1}{2(1+3t)\sqrt{1+3t}}$

[elfin berry]

nomemory

[desert gyro]

Do I need to get back to calculus class ?

[fringe iron]

It should be $\frac{1}{16t}-\frac{3}{2(1+3t)\sqrt{(1+3t)}}$

[elfin berry]

Katrro

[desert gyro]

So the solution is wrong ?

[fringe iron]

it seems like it since their derivative is wrong

[desert gyro]

Well, the derivative

Anyway, creative solution if not wrong...

Thanks @fringe iron , usually they don't mistakes like this, so it's suprising to see that, that's why I wanted to double check

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[abstract hull]

Thanks

[marsh citrus]

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[wary yew]

Prove $c>1$ then $c^n > c$
\newline
Proof: $c >1 \implies c^n > \ldots > c> 1 \implies c^n >c$

[wary yew]

Can you check if this is correct?

I’ve applied transitivity property repeatedly

Is there a more striaghtforward way, or is it this?

[elfin berry]

Dubs

[knotty charm]

This proof relies on the fact that $x^n > x$ for $n \in \mathbb{N} \backslash {0,1}$ and $x > 1$

Which is quite easy to prove on it's own, but should be noted regardless.

[lucid zenith]

you just restated the problem with x instead of c

[knotty charm]

oh i'm dumb

but either way, there's something about c that means certain algebraic manipulations are allowed in the statement $c^n > c$

[elfin berry]

aeiou

[wary yew]

this is what I did, $c^3 > \implies c^2$

then repeated proceeded the same

[lucid zenith]

you're essentially doing induction

which is fine

but you should write it out explicitly

[wary yew]

assume c^k > c , then prove c^{k+1} >c ?

[lucid zenith]

yes, and you also need a base case

[knotty charm]

yeah

[wary yew]

does induction always require base case?.

[lucid zenith]

yes

without a base case you haven't proven anything

[wary yew]

I don’t completely see how

In the inductive step, we are using base case right

Unlike in recursion, I see the need of base case

Is it proved so to show it is atleast true for something to start with

[lucid zenith]

Let your claim be f(n). So for example in this case your claim could be c^n > c

[knotty charm]

the inductive step (P(k) implies P(k+1)) is just an implication on arbitrary points. Without a first true statement P(start) you can't obtain any other true statements.

[lucid zenith]

With this you're proving $f(k)\implies f(k+1)$

[elfin berry]

kheerii

[lucid zenith]

but you still don't know if f(k) is true for any k

[wary yew]

We are assuming f(k) is true right?

[lucid zenith]

it could be the case that f(n) is not true for any n

[lucid zenith]

in which case this doesn't tell us anything

[lucid zenith]

you are proving that if f(k) were to be true then f(k+1) must be true

[wary yew]

Without base case we are doing ex falso quodlibet?

ie from a flasehood anything follows

[lucid zenith]

i'm not sure what that means

[knotty charm]

without a base case we're just proving that if something is true, then something else is true

[wary yew]

Why is induction only allowed in N?

I’m sorry if my questions are stupid

[lucid zenith]

it isn't!

[wary yew]

Trying to see why induction works in general

[lucid zenith]

you can use induction on any countably indexed set

in fact there's also a way you can induct on the reals (though I don't quite know how it works)

[knotty charm]

Induction is defined on any set with a defined successor function (and on R but that's above my level)

[lucid zenith]

you can think of induction as a set of dominos

by proving f(k) implies f(k+1) you're setting up all the dominos

but you still need the base case to push over the first domino

[wary yew]

That’s the analogy im trying to connect, please go ahead

[wary yew]

I see, if the base case is true so is the base case + 1, so is base case + 2,…?

like that?

[lucid zenith]

indeed

[wary yew]

Wow

[lucid zenith]

if I prove that f(1) is true, then I can use my induction step to say that f(2) is true, which I can use to say that f(3) is true, and so on

if I don't know f(1) is true I don't know anything at all

[wary yew]

I see it now

Thanks a lot

[marsh citrus]

@wary yew Has your question been resolved?

[marsh citrus]

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[obsidian mountain]

Can someone just tell me which technique or methodology do I need to know to solve this type of questions?

[sweet pawn]

the first thing i'd say to do is to solve for y in each of these constraints

then graph

[obsidian mountain]

Hmm makes sense

Thanks:)

And use options right 🤔

Like graph everything and just get a rough idea

[sweet pawn]

i forget the name of the theorem

but the optimal solution to linear programming problems comes at the vertices of the 2D region formed by the constraints

[obsidian mountain]

I see.. I'll look into it..

Thanks..

.close

[marsh citrus]

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[sweet pawn]

it's called the fundamental theorem of linear programming

[obsidian mountain]

Wow that's really helpful now .

I'll study it

[marsh citrus]

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[pallid elbow]

can someone help on 14a? i have no idea on how to get to the answer i've been trying for 30 minutes. i know the height is 2 root 3

[jade edge]

,rotate

[jade edge]

Find the length and the height of rectangle

the length = the side of triangle - 2x

the height just use the ratio of 30-60-90 triangle

[pallid elbow]

would it be tan(60)*x = root 3 x?

(root 3*x)(4-2x)

ohhhh thank you so much!

idk how i missed out

that

it’s so obvious

.close

[marsh citrus]

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[pallid elbow]

thank you so much

[marsh citrus]

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[dull beacon]

From Khan academy:

y = Asin(Bx-C) + D
y = Acos(Bx-C) + D

From the web:

y = Asin[B(x-C)] + D
y = Acos[B(x-C)] + D

Which of these are correct?

[chilly spoke]

are A, B, C, and D arbitrary constants?

if not, show the original problem

[dull beacon]

not really a problem, I saw khanacademy is saying the formula should be this way and from web I find it's something else

so was confused which one to use

[dull beacon]

sinusoidal formulas

[knotty charm]

Both work, but I'd use the top pair because otherwise C changes with B

[dull beacon]

A is for amplitude
B to determine period
D for the midline

[dull beacon]

you mean, khanacademy's?

[dull beacon]

i've learnt these stuff like a year ago

[knotty charm]

yeah

khanacademy's i mean

[dull beacon]

now I'm tryna recall things from my notes, so a bit confused to two different equations

[chilly spoke]

ah, I thought it was a differential equation problem

[dull beacon]

it's okay 😉

.close

[marsh citrus]

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[marsh citrus]

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[pulsar crescent]

Prove that in a self-complementary graph (G) with n vertexes ,
n=4k or n=4k+1

[pulsar crescent]

And it n=4k+1 prove that there exists a vertex with the degree of (n-1)/2

[marsh citrus]

@pulsar crescent Has your question been resolved?

[pulsar crescent]

<@&286206848099549185>

[marsh citrus]

@pulsar crescent Has your question been resolved?

[marsh citrus]

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[whole zealot]

hi guyss !! quick question, 570 degrees' negative coterminal angle is -150 right?

[whole zealot]

and when u convert 325 degrees to radian measure, it's 65pi/36? that's all thank you

[unreal oxide]

,calc 570 - 360

[elfin berry]

Result:

210

[whole zealot]

hii that one is my positive coterminal angle ^^

im unsure about my negative

[proud arch]

This is negative

Co terminal angles

[whole zealot]

ohhh

[proud arch]

570 - 360

210

[unreal oxide]

,calc 360 - 210

[elfin berry]

Result:

150

[unreal oxide]

So 150 is correct

[proud arch]

For positive you add 360

[unreal oxide]

570deg = 210 deg = -150deg

[whole zealot]

thank you guyss ❤️

[proud arch]

This one is correct too

[unreal oxide]

,calc (325/360)*2

[elfin berry]

Result:

1.8055555555556

[whole zealot]

TYY!!

[unreal oxide]

,calc 65/36

[elfin berry]

Result:

1.8055555555556

[unreal oxide]

Yes

[proud arch]

Please .close the channel if you are done using it @whole zealot

[whole zealot]

okay okay thank youu

.close

[marsh citrus]

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[marsh citrus]

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[pulsar crescent]

Prove that in a self-complementary graph (G) with n vertexes ,
n=4k or n=4k+1

[pulsar crescent]

And it n=4k+1 prove that there exists a vertex with the degree of (n-1)/2

[marsh citrus]

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

@pulsar crescent Has your question been resolved?

[marsh citrus]

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[still temple]

can someone help me with this its probably simple but im confused asf

[forest kayak]

So what do you not understand?

[still temple]

everything well like i do 3.14 * 8 squared * 9 and idk if thats all i need to do or if im doing it right

[forest kayak]

r is the radius not the diameter. And the radius is half of the diameter.

A composite figure is the figure of both the green and the blue object.
So they want the volume, what you do is calculate the volume of the green object and calculate the volume of the blue object. After calculating both you just add the volumes together

[still temple]

oh.

alright thank you

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[smoky valve]

hi, i can't understand how to find m, can someone please help me? Thanks a lot!!

[knotty trellis]

Can you show the whole question?

[tawdry rampart]

I don't think this is enough to solve for m

[proud arch]

Using this you can just find a range of m

Which, from what I can see is m>2

[smoky valve]

ah yes i meant it, but i can’t clearly understand how to get that

[knotty trellis]

1 <= x <= m/2

so 1 <= m/2

[proud arch]

If a is greater than b and b is greater than c

Than c is also greater than a

[proud arch]

Does it make sense?

[smoky valve]

Yes, and why did we only choose m > 2? Like where did the equal go?

[proud arch]

I couldn't figure out how to write greater then equal to

Sorry for that

In the images I have written it tho

[smoky valve]

I see

But somehow in the key answer I got it is only “m >2”

[proud arch]

Then it's an error with the answer key

m=2 is definitely possible

And should be included

[smoky valve]

yes I think that so

[proud arch]

Please discuss this with your teacher

Anything else?

@smoky valve

If not please .close the channel

[smoky valve]

Actually, the whole question is “The domain of the function y = is a segment on the number line if and only if…”

Does this help with the final result i got?

[proud arch]

What does "is a segment on the number line mean"

@smoky valve

[smoky valve]

I'm sorry, I’m using a translator so it may be not clear

[proud arch]

It's ok

Tell me your interpretation of it

What do you think It means

[smoky valve]

Determine the range of values for m so that the function y =… is defined over its domain

Is this clearer?

[proud arch]

Yeah

I still think m=2 is viable

The thing inside the root can be zero

So I believe I will stick with the answer we got

What do you think @smoky valve

[smoky valve]

Ah maybe it’s because of the translation

I've just discussed with my teacher

But anyways thanks a lot for your help

[proud arch]

What did they say

[smoky valve]

You know the ( ) and [ ] ?

Idk what’s it in english

[proud arch]

Yeah

[smoky valve]

In the question it requires the ( )

So it’s only > not equal

[proud arch]

I wonder why

Y is defined at m=2

So it should be included in the domain

I don't get it

Give me your teachers contact
I will talk

Lol

[smoky valve]

unfortunately he doesn’t know english TT)

We can add and talk about this later

[proud arch]

Sure

[smoky valve]

I'm having an exam tomorrow 😦

[proud arch]

My Brain will bother me

Till I get an answer

[proud arch]

Best of luck

[smoky valve]

lol same

[proud arch]

You will do well

[smoky valve]

i've just sent a request

See ya tomorrow:))

[proud arch]

See ya too

[smoky valve]

.close

[marsh citrus]

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[hasty briar]

help

[tawdry rampart]

Ping helpers

Lol

[hasty briar]

i can?

<@&286206848099549185>

oh

[tawdry rampart]

Lol

Tho wait 15mins before doing so

!15m

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[hasty briar]

okayy

[proud arch]

Is the answer 10

???

[hasty briar]

yes

explain

pls

[proud arch]

Just assume z= x+iy