help-13

[hushed folio]

Of course, visualization will never be useless

But visual proofs aren't really a thing

[green crow]

It seems like the supremum of B is some number between b and a

[hushed folio]

Uhh

I mean b and a arent exact values

They represent all elements of B and A

[green crow]

I don't know what I'm doing but did I do something right here atleast

I don't know if I can solve this problem, been on this thread for 2 hours already

[cedar kiln]

@green crow Has your question been resolved?

[hushed folio]

i gave u a proof

since $B = {b : b , \text{is a lower bound of} , A }$

[wraith dagger]

hiidostuff

[hushed folio]

that means that $\forall a \in A$ and $\forall b \in B$, $b \leq a$

[wraith dagger]

hiidostuff

[hushed folio]

now we show that sup(B) is in B

[green crow]

If supB is in B then would that mean supB≤a

[hushed folio]

let $\Phi$ represent sup B and and $\Psi$ represent inf A

[wraith dagger]

hiidostuff

[hushed folio]

now we choose an arbitrary $\epsilon > 0$. Since $\Phi$ is the LUB of B, it follows that $\Phi - \epsilon$ is not an upper bound of B, and so, there exists an x in B such that $\Phi - \epsilon < x$

[wraith dagger]

hiidostuff

[hushed folio]

that make sense?

just using the idea of the least upper bound

[green crow]

I haven't seen that definition yet

[hushed folio]

i mean if Phi is the least upper bound

then any number less than phi is not an upper bound

which means theres an element of the set that is greater than that number less than phi

ok continuing the proof

Since $x \in B$, we know x is a lower bound of A (since thats how B is defined). Since $\Psi$ is the greatest lower bound of A, it follows that $x \leq \Psi$.

[wraith dagger]

hiidostuff

[hushed folio]

By transitivity we then know that $\Phi - \epsilon \leq \Psi$

[wraith dagger]

hiidostuff

[hushed folio]

and since epsilon can get arbitrarily close to 0, it follows that $\Phi \leq \Psi$

[wraith dagger]

hiidostuff

[green crow]

I don't understand phi- epsilon is not an upper bound of B

[hushed folio]

yes but its a lower bound of A

which is what that inequality means

[hushed folio]

from this inequality tho

we know that the sup B is less than the greatest lower bound of A

meaning that sup B is a lower bound of A so sup B is indeed in B

[green crow]

If epsilon greater than zero then phi-epsilon > phi?

[hushed folio]

no

its less than phi

btw, which college course are you taking that this is one of the first times ur exposed to proofs?

this seems pretty complicated for someone being introduced to proofs

[green crow]

I'm not taking a course, I'm trying to learn on my own

[hushed folio]

i see

i definitely wouldnt start with real analysis

if ur new to proofs then this is above ur paygrade at this moment in my opinion

i myself didnt start with this stuff

i did proofs like "Prove that if n is odd then n^2 is odd"

[green crow]

People from math discussion said real analysis is a good place to start

[hushed folio]

nah id start with discrete

then move to real

[green crow]

What proofs are there for discrete

The only proof I understood was proving square root of 2 is irrational from analysis textbook

[hushed folio]

yeah thats discrete id say

ill give some examples

could u prove directly that if n is odd then n^2 is odd?

[green crow]

i think ive seen that proof before

[hushed folio]

yep

then theres contradiction proofs

like the sqrt(2) one

contraposition as well

[green crow]

What other proof do I need to know or would be a good exercise

[hushed folio]

i dont know any proofs textbooks by memory

but i would imagine you could look some up

thought to be honest, discrete math proofs are pretty different from analytic ones

you have to be a lot more comfortable with arbitrary differences like less than vs less or equal

[cedar kiln]

@green crow Has your question been resolved?

[cedar kiln]

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[small barn]

in the 2, 3 and 4th examples when integrating

why is the number factored out of the integral -2/3, 2/9 and -1/3

shouldn’t it be 2/3 because of the 1/(a+1) rule

[hushed folio]

The inside of the square root has a negative x

So by what is basically the inverse chain rule

We gotta remember to multiply by a negative

[small barn]

what’s that?

[hushed folio]

I mean it's not explicitly a thing

I'll show u tho

[small barn]

okok

[hushed folio]

For the second integral

Do a u-sub where u = 2-x

[small barn]

can we just do for the last one for example

[hushed folio]

Sure

[small barn]

(1-2u)^1/2

[hushed folio]

Do a t-sub since the variable is already u

t = 1-2u

[small barn]

integrating that we get 2/3(1-2u)^3/2

[hushed folio]

We dont get that

[small barn]

why not tho?

[hushed folio]

Do the substitution and you'll see why

Let t = 1-2u

[small barn]

alr

so sqrt(t)

[hushed folio]

What do we do from here

Yep

[small barn]

t^1/2

[hushed folio]

But we need to change du to dt

[small barn]

oh yeah

[hushed folio]

So for t = 1-2u

[small barn]

so dt = -2 du

[hushed folio]

Differentiate both sides with respect to their own variables

Yep

Which means -1/2 dt = du

[small barn]

yep

[hushed folio]

And we can plug in now

We also have to change the bounds with respect to t

Initially we had $\int_{-1}^{0} \sqrt{1-2u} \dd u$

[wraith dagger]

hiidostuff

[small barn]

how Do we do that

yes

[hushed folio]

Well the -1 on the bottom means u = -1 there

[small barn]

An then t+1

[hushed folio]

So since t = 1-2u

We plug in -1 for u

So 1 - 2 * -1

Which is 3

[small barn]

whatttt

yes

[hushed folio]

And then for the top we get 1 when we plug in 0 for u

[small barn]

how do we know that

[hushed folio]

So changing it to a t integral

[hushed folio]

Because the integral is with respect to u

[small barn]

oh yeah

so it’s equal to -1 and 0 ig?

ohhhh

okok I get it now

[hushed folio]

Mhm

[small barn]

but the solution skipped allat

[hushed folio]

And then changing to t we get the integral from 3 to 1

[hushed folio]

We will come back to it

[small barn]

yep

[small barn]

so there’s a quicker way or no?

[hushed folio]

But now we have $\int_3^1 -\frac12 \sqrt{t} , \dd t$

[wraith dagger]

hiidostuff

[small barn]

yep

[hushed folio]

So let's integrate this

[small barn]

let’s take -1/2 out

[hushed folio]

Yep

[small barn]

we get 2/3(t)^3/2

[hushed folio]

Mhm

But the 2/3 is multiplied by -1/2

[small barn]

yep

so -1/3

[hushed folio]

So we get -1/3

[small barn]

okhhh

[hushed folio]

And now we can change everything from t back to u

[small barn]

yep

[hushed folio]

So $-\frac13 \sqrt{t} = -\frac13 \sqrt{1-2u}$

[wraith dagger]

hiidostuff

[hushed folio]

And even tho we changed the bounds earlier

We can keep em the same now that we switched back to u

[small barn]

yes

so is there a quicker way to this,

[hushed folio]

Does it make sense why we can just directly integrate this

[small barn]

idk

[hushed folio]

Yes, if the inside of the square root is a linear function

[small barn]

I get the substitution way

I can’t see any other quicker way

[hushed folio]

Divide the integral by the derivative of the linear function

[small barn]

hmm ok

so

[hushed folio]

So lets do a general example

$\int \sqrt{4-3x} , \dd x$

[wraith dagger]

hiidostuff

[small barn]

yep

so

wait I forgot the derivative rule for this

[hushed folio]

Take the derivative of 4-3x

[small barn]

oh

-3

[hushed folio]

Yep

[small barn]

so it’s -1/3

[hushed folio]

Now integrate as if you were ignoring the inside

The inside of the square root i mean

[small barn]

-1/3(2/3(4-3x)^3/2)

[hushed folio]

Mhm

[small barn]

so -2/9

[hushed folio]

And so we get -2/9 * the stuff to the 3/2 power

That make sense?

[small barn]

yep

can u help with another example

[hushed folio]

Sure

[hushed folio]

I see

We basically want to multiply the weird square root stuff by some additive conjugate

In order to get the difference of squares

[small barn]

how do we get sqrt(x-1)-sqrt(x) in the second step

[hushed folio]

Well

Do u agree that $(\sqrt{x} + \sqrt{x+1})(\sqrt{x} - \sqrt{x+1}) = x - (x+1) = -1$

[wraith dagger]

hiidostuff

[small barn]

yes

[hushed folio]

We basically want to get rid of the square roots in the denominator

And instead have them in the numerator

[small barn]

ohhhhh I got it

thx

and one more time pls

[hushed folio]

Sure

[small barn]

in the second one

[hushed folio]

It's the same thing happening in the second one

[small barn]

how they take out the 1/5

[hushed folio]

Well think about what they are multiplying in the denominator

[small barn]

ohh yeah

[hushed folio]

They're canceling the square roots to get a constant

[small barn]

yes

[hushed folio]

Hope i was able to help

Intro calculus can be a pretty boring and rule-filled part of math so sorry if this stuff is hurting your brain

[small barn]

it’s fine

can I add you

ur explanations are extremely clear 👍

[hushed folio]

Id just ask here again as I won't be online at this time very often

I live in the USA

[small barn]

ohh I see

im in Australia

[hushed folio]

It's 4:30 am

Makes sense lol

[small barn]

lol yep

[hushed folio]

If i end up helping u again tho, I'm curious what u want to study in uni

[small barn]

prolly med

[hushed folio]

If ur doing math then I should probably be a lot more rigorous with my explanation

[small barn]

oh wait one more thing

[hushed folio]

Ah nvm then

[small barn]

yeah

im only 14 as well I got plenty of time to master this shi

[hushed folio]

Truth

[small barn]

in the first example here

[hushed folio]

Tho calculus doesn't come up that often in med

[small barn]

when they subsitute the bounds

why do they do -2/3*-1^3/2

[hushed folio]

0+1 = 1

[small barn]

I mean like after the crackers

brackets

[hushed folio]

Yes

It's bc when u substitute 0 in

[small barn]

I don’t see any reason to take that out

oh

[hushed folio]

They are subtracting that part after the parentheses

[small barn]

it’s not taken out

yeah

that whole part is the second bound

ahhhh

[hushed folio]

Yep

[small barn]

okok cool

[hushed folio]

They just associated everything when u plug 1 in

[small barn]

.close

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[zinc depot]

with taylor

[swift olive]

Try getting the taylor series of each term first

[cedar kiln]

@zinc depot Has your question been resolved?

[cedar kiln]

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[dry mason]

how to integrate this?

[blazing dune]

by parts

[dry mason]

how do you integrate e^x^2

[blazing dune]

that is the neat part, you don't just integrate that

what do you think is needed to be with that expression so you can integrate it

[twilit escarp]

Hmm

[dry mason]

yeah idk 😭

[steel crest]

substitute a = x^2

ig

[blazing dune]

that would work too

[blazing dune]

don't you see that having an x multiplied by it would allow you to integrate it

[dry mason]

im not sure how

[hushed folio]

Do a u-sub

Let u = x^2

[swift olive]

Yea you can do that then proceed to do IBP

or just IBP completely

[dry mason]

yeah how do you use by parts completely

[hushed folio]

Well you need to define two functions

One to differentiate and one to integrate

[dry mason]

yes

and then

[swift olive]

its the form int of u dv = uv - int of vdu

[hushed folio]

Let u = x^2 and dv = xe^(x^2) dx

[dry mason]

yeah how to integrate the dv

that's what i don't understand

[hushed folio]

U sub

u = x^2

[dry mason]

hmm

[swift olive]

Ok start from the beginning, use u sub where u = x^2 so du = 2x

[dry mason]

wait i think i got it

thanks a lot everyone, appreciate it

.close

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[normal dock]

Hello, I am struggling with this question
As somebody whos currently trying to learn mathematics as a beginner I am trying to learn how to solve it from chatgpt but the AI's responses become more stupid as i ask harder questions

Actually, for this question i dont get what kind of 'union' or 'intersection' we are looking for. Can somebody at least give some ideas to how to approach? This is not my homework and im not trying to make you people solve any kind of assignments, this is just a sample test for my upcoming exam.

[steel crest]

ok

so

you want the set of numbers that belongs to A_n for some natural n, in the first one
you want the set of numbers that belongs to A_n for all natural n, in the second one
you want the set of numbers that belongs to the conjugate of A_n for some n, in the third one (where by conjugate i mane belong to reals, but dont belong to An)
you want the set of numbers that belongs to B_n for all n, in the third one
you want the set of numbers that belongs to B_n for some n, in the fourth one

[normal dock]

thanks for explaining, but for example how do we solve the first one?

[steel crest]

ok

[normal dock]

gpt made me realize that the A_n getting closer to 1, 5

and it says it is [1, 5] but when i asked why isnt it (1, 5) it said 'youre right' which was a random question..

yes listening 🙂

[steel crest]

you are right

in this case

all number between 1 and 5 will be in some A_n, yes?

[normal dock]

yes

[steel crest]

and 1 and 5 will not be in A_n for any n, right?

[normal dock]

like for any given numbers?

[steel crest]

for any n, yes

[normal dock]

yes

[steel crest]

also all numbers smaller than 1 and greater than 5 will never be in A_n for any n, right?

[normal dock]

yes

so 1 and 5 will never be in A_n in any case

that means there is no way to make this set have 1 and 5

[steel crest]

yes

so

the union will be

(1,5) only

[normal dock]

sorry but what the hell are we looking for here

thats where im lost

ok (1, 5) makes sense but to some point

[steel crest]

we are looking for the value which are in A_n for some n

[normal dock]

generally i would be looking for some values that determines whether some equation is true or not

[steel crest]

here it is

[normal dock]

but here it is a bit different

for me at least

[steel crest]

there exists n such that
1+1/n <= x <= 5 - 2/n

[normal dock]

so n is the input

what is x then

[steel crest]

x is the input

we are looking for x such that there is an n

which satisfies it

[normal dock]

lets say the input is 6 then why is it wrong

[steel crest]

because

[normal dock]

sorry i know im asking stupid questions

[steel crest]

not stupid

[steel crest]

there is no natural number n, such that 1+1/n <= 6 <= 5-2/n

[normal dock]

lets say
1+1/n <= 4 <= 5-2/n
and 4 satisfies it
where does the n come from?

if it isn't the input how do we assign any values to t

[steel crest]

we do not assign values to it

but check if there exists a value of it which worls

in this case

yes

setting n = 2, allows this to work

as

(1+1/2<=4<=5-2/2) is a correct statement

[normal dock]

hmm

but n = 1 wouldnt work

[steel crest]

right

we only look for all n work when doing intersection

when doing union we check if any n work

[normal dock]

so is it ok if i just dont care about the value of n and ignore it?

[steel crest]

yes

you just need to be able to show if there is an n, not what the n is (tho in certain cases the simplest way to show there is an n, is to find the n)

[normal dock]

(1+1/2<=4<=5-2/2)
this perfectly makes sense but where did the <= come from

because the initial thing doesnt have anything like that

[steel crest]

the definition of the [1+1/n,5-2/n]

is that

[a,b] is the set of all values, x, such that a<=x<=b

[normal dock]

so for the second one does it apply?

[steel crest]

as in?

[normal dock]

for Bn

[steel crest]

no

Bn is of the form [a,b)

this means all values

x such that a<=x<b

[normal dock]

oh alright

so [ means included, and here it means equal

when we have not included which is the paranthesis, it is not equal but only greater/lesser

when you said no i thought there is a whole different thing but at least this makes sense

[steel crest]

hm

[normal dock]

thanks for your help, i started to grasp the overall thing a little bit

then it is time to

.close

[cedar kiln]

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[cedar kayak]

On a German motorway, black cars are four times more common than
cars of other colors. Most black cars are driven by men: 60% of men drive black cars, while 20% of women drive
cars of a different color. We observe the cars on the motorway. State the probabilities in percent, rounded to whole numbers.
What is the probability of observing a man driving a
black car?
What is the probability that the car is not black?
We see a black car. What is the probability of a woman driving it?

1. P(B|M)=P(B∩M)/P(B)
   0,8*0,6/0,8=0,6

2. 1/1-4/5=1/5

3. P(W|B)=P(W∩B)/P(B)
   0,8*0,4/0,8=0,4

Can someone tell me if this is correct?

[cedar kayak]

The is an x between 0.8 and 0.6 and between 0.8 and 0.4

[cedar kiln]

@cedar kayak Has your question been resolved?

[cedar kayak]

<@&286206848099549185>

[cedar kiln]

@cedar kayak Has your question been resolved?

[cedar kiln]

@cedar kayak Has your question been resolved?

[cedar kiln]

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[obsidian dawn]

Hi so I have a question about mathematical induction regarding sequences.

[uneven quartz]

Okay, just ask

!justask

!da2a

[cedar kiln]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[obsidian dawn]

finding the screenshot

one second

I got to assume n = k + 1

But I don’t know how to manipulate it the same way as say a series

[dire geode]

show all your work

[obsidian dawn]

I know you’re supposed to separate k from Uk+1, and then make it equal to the RHS but I don’t know how to do it

oh and d = 600

<@&286206848099549185>

@dire geode do you have any ideas?

[dire geode]

factor out an r to get it to look like u_n

you'll add and subtract something to keep the fraction on the right the same

r^k -1 = r^k - r + r - 1

[cedar kiln]

@obsidian dawn Has your question been resolved?

[obsidian dawn]

wait can u write this down?

[dire geode]

wut

[obsidian dawn]

do you mean r^k-1 or r^k - 1

[obsidian dawn]

like on paper

[dire geode]

no

$r^k - 1 = r^k - r + r - 1$

[wraith dagger]

riemann

[obsidian dawn]

are you talking about teh fraction

but then what about after this step?

[dire geode]

this only applies to the numerator in the fraction

[dire geode]

you were supposed to factor out an r first from both terms

[obsidian dawn]

wait how?

[dire geode]

this is just u_1 * r^k

[obsidian dawn]

but how do I do the induction bit

which r am I supposed to factor out?

[dire geode]

you factor and do algebra above until you get u_k

you can write u_k and work backwards to get u_k+1

[cedar kiln]

@obsidian dawn Has your question been resolved?

[cedar kiln]

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[fluid pawn]

exo 114 pls i couldnt solve/understand the last 3 qs

[lusty pawn]

a direct ss might be better

[lusty pawn]

oh god

i'm sorry i don't understand that language

[crimson sedge]

english

if you translate it

maybe we can help

[fluid pawn]

oops its in arabic ill see if i can translate it

[crimson sedge]

ok

[fallen heath]

You know that area of a ∆ is 1/2 base • height right?

[dull oxide]

Arya I didn't know you read arabic

[fluid pawn]

yes

[fallen heath]

The question asks you to write area of ∆s AA'C and AA'B in two ways (by using the perpendiculars h once, and next using "d" and the respective bases)

[fluid pawn]

oh thx but those r not the questions that im working on i already solved it

[fallen heath]

For example, [AA'C] = 1/2 db = 1/2 A'C • h

[fluid pawn]

.

[fallen heath]

Oo

Didn't see +_+

[fluid pawn]

Exercises 114 The center of the circle drawn inside a triangle is a point of convergence)) Prove that: The heights in a weighted triangle (AC BC) in terms of ARC KCB b Conclude that is the reference (Cumy) (tan) The center of the circle drawn inside this triangle and point C The height passing through point 8 intersects (AC) at 1 and the height passing through the point) intersects (AD) at M give similar results for naphthalene/f. The intersection of the heights in it and) the center of the circle surrounding the triangle. The goal of this problem is to find three real numbers such that or or (0) is a reference (C) (8.3) (α) is weighted to: d) Show that the point meets the heights in the weighted (C). ,tany) (B, tan) (A, taner) [AB] [AC] |BC| And the middle ones, respectively. (1) The center of the circle drawn inside the triangle bisector of the angle BAC intersects ACI at A son of A equal distance from the two parallels (AN) and (AC). It is symbolized by the fact that the axes of the triangle ABC are at altitudes to this distance B. The length of the height passing through the point... in triangle MN b through n from 10 as a weight to 0 and. A Use the result (1) d). The product that 0 is the center of the circle surrounding the triangle ABC is a reference A and C are attached to coefficients that are requested to be set. (this is what i got when i translated it but im sure its not 100/100 accurate)

[fallen heath]

So why did you not do part 1?

@fluid pawn

[fluid pawn]

i did it

[fallen heath]

=_= which part do you need help with, specifically

[fluid pawn]

part 2 questins a b and g

question2 of part 2 to be clear

[fallen heath]

Alr

[fallen heath]

Me neither (@_@;)

Anyways, @fluid pawn for the first part you need to show that perpendicular bisectors of the midpoints of ∆ABC are altitudes for the medial triangle

Which is trivial because, by midpoint theorem PN || BC and so perpendicular bisector of BC is basically a perpendicular drawn from midpoint M of BC onto PN.

Similarly for others

[fluid pawn]

ahh yes thank u

[fallen heath]

If you can explain what you did for part 1d) I can help you with 2b). But 2c) is basically asking for you prove that the intersection of these altitudes of the medial triangle MNP or quite simply the intersection of the perpendicular bisectors of ABC is the circumcenter

That is, distance from this point to any other vertex would be equal

[fluid pawn]

i still havent done part 1d but i solved a similar question in this same exo and this is what i did: i wrote them as vectors then used علاقة شال and kept putting in the points that i needed till i got it

[cedar kiln]

@fluid pawn Has your question been resolved?

[fluid pawn]

u know what imma ask a teacher. thank you for your work really appreciate it @fallen heath

.close

[fallen heath]

Okie.

[fluid pawn]

.close

[cedar kiln]

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[cedar kiln]

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[latent jewel]

i found x as 20

<@&286206848099549185>

[dire geode]

Show your work

[latent jewel]

ignore the other stuff, the work is inside the box

@dire geode

[dire geode]

I don't see why you set 5x-4 equal to 2x+56

[latent jewel]

why would it be equal to 180?

[dire geode]

What's equal to 180

[latent jewel]

a supplementary line

[dire geode]

I don't see a supplementary line

[latent jewel]

okay, so that's why it's equal

now, go on?

[dire geode]

You should be using isosceles triangle properties

And the sum of angles in a triangle adds up to 180 deg

[latent jewel]

so 5x-4+2x+56 = 180?

then solve that way?

@dire geode

[dire geode]

These are only two of the three angles in the triangle

Do you see the isosceles triangle?

[latent jewel]

yeah

[dire geode]

You're given two of the three angles

[latent jewel]

so there's 2 of the 5x-4's?

[dire geode]

The third one can be inferred

Yes correct

[latent jewel]

okay okay

thank you

keep this channel open though as of now

[cedar kiln]

@latent jewel Has your question been resolved?

[cedar kiln]

@latent jewel Has your question been resolved?

[cedar kiln]

@latent jewel Has your question been resolved?

[cedar kiln]

@latent jewel Has your question been resolved?

[cedar kiln]

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[untold egret]

can some1 help

[slate lintel]

what are "these"?

[fallen heath]

Can you factorise 6 - 1 for me? Assume a³ = 6, b³ = 1

[untold egret]

I dont even know how to

[cedar kiln]

@untold egret Has your question been resolved?

[raven hemlock]

(6(1/3) -1)*(36(1/3) +6(1/3) +1)

the (1/3) is power of 1/3

[fallen heath]

Yeah, so 7/(³√36 + ³√6 + 1) = 7/5 * (³√6 - 1)

[cedar kiln]

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[narrow scaffold]

i need help with finding the x and y intercepts and the hole for this function

[cedar kiln]

Please don't occupy multiple help channels.

[sand cradle]

What's unclear?

[narrow scaffold]

i plugged zero in for the x's this equation to find the x and y intercepts, this is what i got

[sand cradle]

This is the y-intercept, yes

[narrow scaffold]

so the y is -1?

okay

im confused on how i should find the x intercept

if my notes say i should plug in zeros..?

or do i set it to zero

[sand cradle]

What properties does the graph have at a x intercept/how does it looks like graphically?

[narrow scaffold]

it passes through the x-axis(?)

[sand cradle]

Yes

(or touches it)

[narrow scaffold]

ohhh

yes

[sand cradle]

The x-axis has the equation y = 0 (or if you'd like, g(x) = 0)

[narrow scaffold]

so i would set it like this?

because i thought that was my vertical

asymptote

[sand cradle]

x = 4 is where it passes through the x-axis, as you stated

That's the x-intercept

[narrow scaffold]

ohhh vertical is set the denominator to zero(?)

[sand cradle]

Yes

[narrow scaffold]

oh i see

one second

[sand cradle]

Because there, you will have something like

,w plot 1/x

[wraith dagger]

Results provided by WolframAlpha

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Requested by kepe_5235

[sand cradle]

In this example, the vertical line at the origin (x = 0) is a vertical asymptote; the function has a hole there

[narrow scaffold]

ohhh

[sand cradle]

Yes

.. so?

[narrow scaffold]

im unsure about this because my graph doesnt pass 16

[sand cradle]

x^2 is not your graph

[narrow scaffold]

ohh

its 4..?

[sand cradle]

Your graph is (-4x + 16)/(x^2 - 16)

[narrow scaffold]

square root 16?

[sand cradle]

This is just an equation you got somewhere

You can solve it

[narrow scaffold]

okay one second

[sand cradle]

Well, you want to find all $x$ that satisfy $x^2 = 16$

[wraith dagger]

Kepe

[narrow scaffold]

do i set that equation to zero?

[sand cradle]

Yes

Which x squared give 16

[narrow scaffold]

4

[sand cradle]

and?

[narrow scaffold]

im not sure

[sand cradle]

-4

[narrow scaffold]

oh

[sand cradle]

Ok, now you got x = 4 and x = -4 as your vertical asymptotes

But remember that at x = 4, you said there is also a zero?

[narrow scaffold]

so thats a hole?

[sand cradle]

If this happens and the multiplicities in the numerator and denominator for the zero match, it will be a removable discontinuity

That's what you call hole, I assume

[narrow scaffold]

i believe so

so there is no x intercept

?

because setting the numerator to zero gave me 4

and my vertical is also at 4?

[sand cradle]

In a case like this, you will be able to simplify your function f(x) because you can factor out (x - zero) from both numerator and denominator

[narrow scaffold]

so one of them cant be graphed

[sand cradle]

$f(x) = \frac{-4x + 16}{x^2 - 16} = \frac{-4(x - 4)}{(x+4)(x-4)} = \frac{-4\cancel{(x - 4)}}{(x+4)\cancel{(x-4)}} = \frac{-4}{x + 4}$ for $x \neq 4$!!

[wraith dagger]

Kepe

[narrow scaffold]

ohhh factoring it first

[sand cradle]

Yes, if something is a zero of both numerator and denominator, you can always do this

[narrow scaffold]

ohh so there is so x

[sand cradle]

Because you can factor out zeros

[narrow scaffold]

int

[sand cradle]

So your graph will look like the graph of $\frac{-4}{x + 4}$, just that at $x = 4$, you will have a hole (you draw it with a little circle)

[wraith dagger]

Kepe

[narrow scaffold]

does this look right?

the horizontal at 0, hole at 4 and y intercept at -1?

no x int or vertical

[sand cradle]

Didn't we have a x int?

[narrow scaffold]

oh

-4?

[sand cradle]

Ah, nope

We had only one value for the x intercept, 4

The +4 and -4 was for the vertical asymptotes (out of which 4 falls away and becomes a hole)

[narrow scaffold]

ohh so the vertical can still sit on -4?

[digital cliff]

(remember to check the y coordinate of the hole)

[sand cradle]

There is still one vertical asymptote at x = -4, yes

[narrow scaffold]

for this i would have to plug f(4)?

[digital cliff]

in the simplified form

[narrow scaffold]

oh okay one second

-4/4+4?

so -0.5?

thank you so much!!!

im done ✅

[sand cradle]

It needs to look like this

[sand cradle]

Looks good

[cedar kiln]

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[versed fulcrum]

Hello

[versed fulcrum]

Why is the first equation non-linear?

[dusk goblet]

cos y

[rigid schooner]

We say an ODE is linear if there's a nonconstant function multiplied to y

[rigid schooner]

That just means it's inhomogeneous

[versed fulcrum]

What's a nonconstant function?

[rigid schooner]

In general, a linear ODE is of the form ay'+by=f(x) where a and b are strictly constants

There exists functions a(x)y'+b(x)y=f(x) that are first order

[versed fulcrum]

Are constants the same as parameters?

[hidden mural]

isn't f(x) y + y' = 0 also linear

[dusk finch]

cos(y) doesnt seem like a legitimate part of f(x) to me

[versed fulcrum]

I think this is right too

[dusk goblet]

yes

[hidden mural]

yes in a way

[versed fulcrum]

"A DE that contains no products of terms involving the dependent variable(s) is called linear"

What does products of term mean

Dependent variables can't be multiplied with independent variables?

[sly swan]

how tf is the last one linear

e^x is not linear

[dusk goblet]

only dependent variables matter

[hidden mural]

maybe a more rigorous definition is there exists a function L such that the DE can be written as
L(x, y, y', ...) = 0
where L is linear in every argument except the first

[sly swan]

can someone tell me how to solve it to

[hidden mural]

you can have x*y and still be linear

[versed fulcrum]

So in conclusion, in the first equation, x is the variable that makes the equation non-linear?

What

[hidden mural]

no i think the x doesn't matter in linearity

only y and its derivatives

[versed fulcrum]

So why is the first equation non-linear?

[hidden mural]

y = dependant variable

[hidden mural]

cos(y)

[versed fulcrum]

What's wrong about that

[hidden mural]

cos(y1 + y2) != cos(y1) + cos(y2)

[dusk goblet]

^

[hidden mural]

cos is a non-linear function

[versed fulcrum]

Ohhhh

I see

Thank you so much guys!

.close

[cedar kiln]

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[spark palm]

This is a dumb question, but i have a 150ft rope and need to figure out the distance between 2 green lines, I'm not sure how id do that. the red line is the rope. (this is in a game, im debating someone on something.)

[digital cliff]

not enough information here to say really

[spark palm]

Hm ok.. what info would you need?

[sick gazelle]

can we assume angle thru shadows?

between the two ahh trees thingy

[spark palm]

I mean, its possible i think.

[sick gazelle]

still more info would be appreciated

[spark palm]

What kind of info do you need?

[sick gazelle]

the angle

the distance btw them is obv more than 150ft

[digital cliff]

you could probably get a reasonable-ish guess, if you measure the screen and convert the scale using the rope, i doubt you need anything that specific

itll be off though

[spark palm]

Wouldnt it be less then 150ft? the bottom line (the red one) is 150ft, the green lines are inside of that.

[sick gazelle]

wut about the angle between them?

hypotenues is always greater na

[spark palm]

Alright, ill try this. i dont need to be exact tbf.

I should be abel to get it from here, thank you guys.

.close

[cedar kiln]

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[autumn moss]

How to do 3(g-f)(3)

[autumn moss]

Is it 3g(3) - f(3)

Or would you do

3(g(3) - f(3))

[oblique flare]

this

[autumn moss]

Oh thank you

I was just confused

[oblique flare]

nah it's ok

[autumn moss]

Cuz (g -f) is the same as g() - f()

[autumn moss]

K thanks

Insanely quick response!

.close

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[loud jacinth]

u can do it

measure rope through your screen and like create a scale .. then measure the distance between and then convert to rope...\

[cedar kiln]

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[turbid pond]

I'm struggling to find the r of the geometric sequence

[dreamy void]

,rccw

[celest flume]

write everything in the form you have below

[turbid pond]

This form?

[celest flume]

the a1*r^(n-1)

[turbid pond]

Well that's the part I'm stuck at

[oblique flare]

he meant like a2 term write it as
a2=a1 r^2-1

and for the other terms

[turbid pond]

How do I find a1?

It's not given

[loud jacinth]

is the answer 36 power 1/3

[turbid pond]

I don't know the answer

[loud jacinth]

oh well

[turbid pond]

It equals to 36

Not power 1/3

[loud jacinth]

no i mean

the answer..

but what i did was

[turbid pond]

No I don't have the answer

I couldn't find the answer

[loud jacinth]

i assumed a2 to be ar^2 a3 to be ar^-1 a4 to be ar and a5 to be ar^2

then put in the equation and a gets cancelled

[loud jacinth]

hi

[turbid pond]

Hi

[loud jacinth]

do u get it

[turbid pond]

No

[loud jacinth]

just assume

a get cancelled

cuz both numerator and denominator will have a in their terms

[turbid pond]

Ohhh

Yes I get it noe

Now

[loud jacinth]

try it , it will get cubic if u use standard terms

,w (x^3 + x^2) / (x+1) = 36

[wraith dagger]

Results provided by WolframAlpha

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Requested by paci2007

[loud jacinth]

it will be 36 to power 1/2

[turbid pond]

I think I kinda get it but don't understand this part

a canceled out

I might be missing something

[celest flume]

factor an r^2 out of the denominator

[turbid pond]

ohhh

okay thanks i get it now

.close

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[exotic laurel]

can someone explain me how the 3_
x'2 goes out

[void sand]

what do you mean by "goes out"?

[exotic laurel]

i know the answer but i dont know how 3_
x'2 becomes 6x-3

like how does it transform

[void sand]

you may use 3/x^2, by the way

[exotic laurel]

3/x'2 is 6x'-3

how

[void sand]

I don't know what the question is

can you show the context?

[hollow kiln]

türk müsün

[exotic laurel]

türküm

[void sand]

right, okay

[hollow kiln]

türkçe sor

[dull oxide]

That's a new way to write fractions

[exotic laurel]

ya bu

[dull oxide]

$\frac{3}{x^2}$

[wraith dagger]

SWR

[exotic laurel]

bu nasıl 6x'-3 diye çıkıyo

[void sand]

you need to use the power rule to differentiate this

[exotic laurel]

yani

[void sand]

,, \frac{3}{x^2} = 3x^{-2}

[wraith dagger]

ourfallenstars

[hollow kiln]

bunda üssü bir azaltıp başa düşürcen

[void sand]

now, apply the power rule $\frac{d}{dx} x^n = nx^{n - 1}$

[wraith dagger]

ourfallenstars

[exotic laurel]

evet şeyi biliyorum türevi

bi dk

3/2x oldu

dimi

başa attık

[hollow kiln]

$-2 \cdot 3 \cdot x^{-3}$ oldu

[wraith dagger]

haygiya

[exotic laurel]

bi dk beynim bi alsın

almıyo valla

[hollow kiln]

:D

[exotic laurel]

malım biraz

mat hiç çalışmadım da

yapıyorum normalde türev sorusuda

bunu a

[hollow kiln]

$x^5$ türevini alsan $5x^4$ olmuyo mu

[wraith dagger]

haygiya

[exotic laurel]

evet

oluyo

[hollow kiln]

$4x^7$ türevini alsan $28x^3$ olmuyo mu

[wraith dagger]

haygiya

[exotic laurel]

evet

oluyo

[hollow kiln]

e o zaman

[hollow kiln]

bunun türevini alsan

-2 başa düşcek 3le çarpılcak

[exotic laurel]

3/2x

[hollow kiln]

üssü de bi azalcak gene

[hollow kiln]

bu olcak

[exotic laurel]

bende bölmede sıkıntı olmuş

o türev şeysini anladım da

3/x de

problem olmuş

[hollow kiln]

3 çarpı 1/x

1/x olunca $x^-1$ oluyo

[wraith dagger]

haygiya

[exotic laurel]

eevet

[hollow kiln]

$1/x^2$ olunca $x^{-2}$ vs.

[wraith dagger]

haygiya

[exotic laurel]

3/x2 olunca 3x'-2 oluyo

[hollow kiln]

aynen

[exotic laurel]

tamamdırrr

çok sağ ol

[hollow kiln]

👍 rica ederim

.close yazarsan kapanıyo kanal

[exotic laurel]

.close

[cedar kiln]

Channel closed

Closed by @exotic laurel

Use .reopen if this was a mistake.

[exotic laurel]

@hollow kiln

burda ne diyo

e üzeri y türev

eşittir

y türev çarpı e üzeri y mi

[cedar kiln]

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[exotic laurel]

can someone explain me

[exotic laurel]

this part

[unkempt breach]

yes

[exotic laurel]

how does it become lnx + x.1/x

i understand it until

[dire geode]

product rule

[exotic laurel]

lny= x.lnx

[unkempt breach]

So they differentiate both sides

[exotic laurel]

like

[unkempt breach]

On the left, you get the chain rule

[exotic laurel]

im really bad at math

[unkempt breach]

On the right, you get the product rule

[exotic laurel]

i dont know the rules names but i know the things but

[eternal gale]

No you got it

[exotic laurel]

can u just explain to me how (x.lnx)' becomes 1.lnx + x.1/x

$can u just explain to me how (x.lnx)' becomes 1.lnx + x.1/x$

[wraith dagger]

SirKuzey

[dire geode]

,tex .diff rules

x = f(x)

[exotic laurel]

ik this

[dire geode]

then follow the right side

[exotic laurel]

OHHHHHHH

[exotic laurel]

what happend here

y'/y then y is gone

,tex .diff rules

[wraith dagger]

SirKuzey

[exotic laurel]

.diff rules

,help

[wraith dagger]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[dire geode]

you see how the left side went from y' / y to just y' ?

how did they do that ?

[exotic laurel]

yeah but explain incase

[dire geode]

no you

[exotic laurel]

. it with y

[dire geode]

yea so do it to both sides and susbtitute y

[exotic laurel]

subtitute means

what

like

.

?

[dire geode]

use this to replace y

[exotic laurel]

y'/x'x = 1.lnx + x.1/x

[dire geode]

should do this first

but it's fine either way

[exotic laurel]

y'/x'x = 1.lnx + x.1/x is this corect

[dire geode]

yes

[exotic laurel]

Okay

then

. x'x with both sides

correct?

[dire geode]

yes

[exotic laurel]

okay how does

1.lnx + x.1/x becomes x'x (lnx+1)

with .x'x

like

i cant do the . problem

[dire geode]

what do you think it should be

1 * anything = anything

and a * 1/a = a/a = 1

[cedar kiln]

@exotic laurel Has your question been resolved?

[livid elm]

,help

[wraith dagger]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[cedar kiln]

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[crimson sedge]

In the graph shown, MP is the mediatrix of BC.
If AP=CD=3 and m ABM=82°, calculate BC

[dull oxide]

pretty

[crimson sedge]

mm

yeah, why

[dull oxide]

why what? I was just saying that the line work and handwriting is pretty

[crimson sedge]

ohh okay, that's great, thx
I thought it was pretty bc you found something haha

[dull oxide]

que significa "mediatriz" en ingles?

[crimson sedge]

oh it's a segment that divides another in half

and it's perpendicular

[dull oxide]

ah

[crimson sedge]

I have to put that

[dull oxide]

"perpendicular bisector"

[crimson sedge]

yeahh

[dull oxide]

You can use law of sines for this triangle

[crimson sedge]

right
even tho it's not included on the topics of this exercise, I'll use it

[dull oxide]

wild. There might be a smarter way but I do not see one yet

[crimson sedge]

i'm supposed to use congruent triangles, this is the topic, it's normally tricky, at least for me

idk if I should do that, the segment above is 24/7 and the right one is 15√2 / 7

[cedar kiln]

@crimson sedge Has your question been resolved?

[dull oxide]

@crimson sedge you mixed up your k and 7k lengths

[crimson sedge]

yeah, I fixed it right after

I noticed

I'll send it

again

[dull oxide]

@crimson sedge, find HC in terms of k

I wrote x by mistake

there

[crimson sedge]

ok

7k - 24/7 ?

[dull oxide]

why -24/7?

[crimson sedge]

mm 3 + 3/7 = 24/7

all is 7k

[dull oxide]

oh you did it that way. ok

then yes

Now, find HC in terms of MC

[crimson sedge]

sm like 5V2/2 k csc 82 ??

like wdym

that equal

5V2/2 k csc 82 = 7k - 24/7

?

[dull oxide]

yes

[crimson sedge]

ohh okay, i'll do it

is k 1

idk

the problem is that I can't directly put it in a calculator to verify

if k = 1 then the answer is 5√2

but idk

[dull oxide]

yes

[crimson sedge]

thank you very much, it was interesting

[dull oxide]

you need to find an exact answer to sin(82) then

[crimson sedge]

mm as a fraction

?

[dull oxide]

yes

oh

one moment

[dull oxide]

It's not exactly 1

[crimson sedge]

mm

[dull oxide]

I didn't verify myself, I just assumed the 7to1 ratio was accurate

but it's not

[crimson sedge]

it could be 24/49

[dull oxide]

you'll have to redo most of your calculations

[dull oxide]

It's not

Just leave it in terms of sin(82) or cos(82)

there's no exact answer

[crimson sedge]

soo

it's not 1

?

[dull oxide]

no

It's slightly less than 1

[crimson sedge]

mm but then maybe it wouldn't be in the options?

[dull oxide]

Oh right, it's multiple choice. That's unfortunate because that is not an exact answer

[crimson sedge]

yeah, maybe I just need to approximate

[dull oxide]

$5\sqrt{2}$ is the closest, but not exact

[wraith dagger]

SWR

[dull oxide]

Maybe. But the problem never said that. It's extremely misleading

[dull oxide]

How did you even get 7k here btw?

[fading grail]

f of x =3 to the power of x\

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[robust adder]

I dont understand the question

[digital cliff]

what about it confuses you?

[cedar kiln]

@robust adder Has your question been resolved?

[robust adder]

Parallel to the y-axis

[digital cliff]

thats a vertical line

they have the form x=a where a is some constant

a horizontal line has the form y=a where a is again some constant

[cedar kiln]

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[crimson sedge]

shouldn't this be (x-10)(x+2) or am i wrong?

[dusk goblet]

yea idk why they wrote that on the right when they wrote the correct thing at the bottom