help-0

[marsh rapids]

.reopen

[lone heart]

✅

[marsh rapids]

that will be simpler

[fluid ingot]

Is this correct?

@marsh rapids

[marsh rapids]

it's n times that

for the sum

[fluid ingot]

Oops right, for the first term right?

[tardy tapir]

how did you get 1/n

[hidden widget]

wait wut?

also the n²/n²+n

[tardy tapir]

its correct

[marsh rapids]

termwise ues

[tardy tapir]

1/n is not

[marsh rapids]

but not the whole sum

[fluid ingot]

Okay what do I change here

[tardy tapir]

1/n should be that

[hidden widget]

ok i get it

[marsh rapids]

find a bound for each term, then sum and get the bound for the sum

[fluid ingot]

Ohh okie I’ll try

[fluid ingot]

Wait is it just 1 then?

[tardy tapir]

no

how did you get 1/n

[fluid ingot]

I removed the 1, 2, 3… from the denominator of the terms

We have n/n^2 left

[fluid ingot]

But I forgot to multiply this by n later

So that gave 1

[tardy tapir]

yes thats correct

[hidden widget]

i have a doubt

[tardy tapir]

yes?

[fluid ingot]

💀💀😭 for me this entire question is a doubt

[hidden widget]

shdnt it be n²/n²+1 and n²/n²+n? or im i missing smthn?

[tardy tapir]

its not should

1/n * n is also correct

yours is also correct

[hidden widget]

but theres no 1/n, it starts from n/n²+1

[fluid ingot]

Like this?

[tardy tapir]

you have to find something thats greater than that sum

[marsh rapids]

yes

[tardy tapir]

and its limit is same as the first one you found

[fluid ingot]

Which one would be better for this question? And how do I figure that out

[marsh rapids]

then summing lets you apply the squeeze theorem

[tardy tapir]

i have done this question in the past, and i used 3909's one

but imo your works too

[fluid ingot]

Is the limit just 1? 😭

[tardy tapir]

yes

[fluid ingot]

Ohh

[fluid ingot]

Okay I’ll try both

I’m not closing this channel yet in case I get stuck

Thank you very much yall for ur help

[lone heart]

@hidden widget Has your question been resolved?

[lone heart]

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[outer palm]

hi i have 2 quesitons 🗣️🗣️

[outer palm]

1. what the hell is a solution, i don’t get this at all

2. confused ab how to find x

help please 😜😜

[tepid plaza]

integration?

[outer palm]

i have no idea

[minor needle]

I guess

[alpine sable]

Help I am stuck 😭

[minor needle]

they want you to check if the coords of the point satisfy equation of the line

[outer palm]

bro this is my channel thing 😭😭

[remote heron]

a solution here means the point is on the line

[minor needle]

if so, then it's called "a solution"

[strange zenith]

X solution is when you plug it in all the equations are true

[outer palm]

ohh?

[outer palm]

okay

[remote heron]

!help

[lone heart]

Please read #❓how-to-get-help

[outer palm]

do i grab any random number for x?

[remote heron]

how do you interpret the pair of numbers?

[strange zenith]

so u plug in for (x, y)

[remote heron]

what does (0, 2) mean

[outer palm]

do you mean 2,0?

2 is x, 0 is y

[remote heron]

yea

so when they give you an equation like y=3x+2

on this line is every point where the equality is true

[outer palm]

ohh

so

[tepid plaza]

1 answer is 10

[craggy spire]

Boy a bumbo

[outer palm]

BITCH

THATS THE REASON IM GETTING HELP😭😭

[remote heron]

sorry u got the honor of being the first help channel

so u get the riff raff

[minor needle]

#❓how-to-get-help , I'll help you

[outer palm]

oh my god

[remote heron]

okay if i tell you that (0,0) is a solution you can check

you plug in x=0 and y=0 into the equation

and youll get 0 = 3*0 + 2

or 0 = 2

[outer palm]

Yeah okay

what do i do with those numbers

[remote heron]

with 0 and 2?

well does 0=2

[outer palm]

yeah

no

[remote heron]

right

so (0,0) isnt on the line

[outer palm]

OHH

THAT MAKES SO MUCH SENSE NOW

WAIT TY OMG😭😭

[lone heart]

@outer palm Has your question been resolved?

[lone heart]

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[echo scroll]

Please answer me, Im learning sin cos tan and im really confused

[echo scroll]

So the sin for A also known as Sin A = Opp/Hyp which in this case would be 16.28/17=0.95 (b=17). Say we had to find b however, we would use Sin 65.56 = 16.28/b and find b correct but what confuses me is that Sin A is not equal to Sin 65.56. Please help asap

better pic

<@&286206848099549185>

[lone heart]

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.

[near apex]

You can apply those rules only in a right angled triangle. You can't apply them on a random triangle. At least not trivially, there's more nuance to that.

[echo scroll]

oh.

[near apex]

Those rules, i mean sin(A) = Opposite/Hypotenuse

And other ratio ones.

[echo scroll]

ok so can u help with this question

[near apex]

What do you need to find?

[echo scroll]

I know I have to do Sin B * (a/Sin A), but I would appreciate a much more detailed explanation

[near apex]

Ah.

So, You want to understand where a/Sin(A) = b/Sin(B) comes from ?

[echo scroll]

not exactly

I understand its a rule or theorem or smth

a = 12

Sin A = a/c

How would u find c

[near apex]

This is not true.

[echo scroll]

what

isnt is opp/hyp

confusion

[near apex]

As i said, those ratio formulas are only applicable in a right-angled triangle.

You can't use them for any triangle.

[echo scroll]

OH THE WHOLE DAM SIN THING ONLY WORKS IN RIGHT TRIANGLES

I SEE

how would I calculate Sin A then

could you explain that

[near apex]

It works in others as well. Just not that simply. As i said, there's a bit more nuiance to that.

[near apex]

You are given the angle A.

Use a calculator to find sin(69°).

[echo scroll]

ok

1 sec

[echo scroll]

oh so the Opp/Hyp is invalid for non right triangles

[near apex]

Yes.

[echo scroll]

so

so in that case I would get (12/ sin 69)sin 36=b

i got 7.58

[near apex]

Yes

[echo scroll]

answer is 7.56

how

maybe cuz i round decimals

thanks

i was really confused

btw whats ur elo

in chess

[lone heart]

@echo scroll Has your question been resolved?

[echo scroll]

I have a different question, for this question, I would have a/ Sin A = b/ Sin B. Since we are trying to solve for the angle of B, the way I believe we solve since is solve for Sin B and find the angle measurement from there. By doing this we get (Sin B) = (Sin A)b/a

nvm I forgot to do the inverse of sin

[lone heart]

@echo scroll Has your question been resolved?

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[fierce pebble]

Hello

[fierce pebble]

How do I put "9 is added to one-fifth of d" in math expression"

Would it be 9+ 1/5d

[somber eagle]

yeah

[fierce pebble]

I'm not so sure that's why. Thanks

[somber eagle]

as long as the d isnt in the denominator there

[fierce pebble]

So it's not d/5

[somber eagle]

it is d/5

im saying it isnt 1/(5d)

[fierce pebble]

Oh it's the same thing 💀forgot mate

[somber eagle]

just in case you were meaning that

[fierce pebble]

Thank you

.close

[somber eagle]

np

[lone heart]

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[craggy crystal]

help

[craggy crystal]

help its linear algebra

<@&286206848099549185>

sorry

[umbral star]

remember that $[\mathbf{x}]_B=\mathbf{B}^{-1}\mathbf{x}$, where $\mathbf{B}$ is the matrix $\begin{bmatrix}\mathbf{b}_1&\mathbf{b}_2\end{bmatrix}$

[ocean seal]

qianqian07

[lone heart]

@craggy crystal Has your question been resolved?

[lone heart]

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[minor needle]

well

10^(-2) cancels out

and it becomes ln(x/1.5)

= ln(x) - ln(1.5)

[pallid scarab]

maybe not since ln(10) is more like 2.3

[minor needle]

typo

maybe

[pallid scarab]

yes it shouldn't matter

if you're dividing cm by cm, same as dividing mm by mm

so you should have 12/(ln(10) - ln(1.5)) * (ln(x) - ln(1.5))

OH I KNOW

since you're doing in CM instead of MM

your answer x will be in cm

whereas your correction will give x in mm

yes

Your formula for V(x) is correct if you input x in cm

Because their formula for V(x) is correct if you input x in mm

np

[lone heart]

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[clear minnow]

hello

[clear minnow]

If we assign a value of X as cardinal, and its values within the set, saying that X= infinity. What exactly happens to the set if we put it in such notation;
If |X| = Inf, what happens to the set? Are we just measuring size, can I give specific values within the set X specificity to solve for some value within it?

<@&286206848099549185>

[barren bolt]

I’m pretty sure that infinity isn’t a cardinal, cardinal would be like the aleph numbers, countable, uncountable infinites etc

And I guess all finite cardinals

Also I’m not so sure what you are even asking

[dawn barn]

if so then Cardinal must also be infinity

[barren bolt]

If a set is infinite then there are elements so you could find elements in the set

[clear minnow]

Yes I can set it as an infinte, though I'm trying to solve within the set X, that there exists F which satisfies Y, in which Y =F+1 and F=Y-1 and both are within the set X, but F must have an infinite set before it for it satisfy Y.

[clear minnow]

Wait would F and Y be cardinal numbers?

[barren bolt]

Do you have an actual problem statement

Like from a textbook or a problem set?

Cause I’m very confused by what you’re writing tbh

[clear minnow]

I've written out the problem,
I also rewrote it;
There is an infinite Setof opening doors from a first door (We assign this series as a Set of |X| = N ~ or X ={n+1}::
There exists a door Y within the set of X; And for Y to open the previous door must be opened.::
We then label the door before the set Y, as the set F. |F| = {Y-1}::
For door F and Y there exists an infinite set of doors opening before them (I disagree with this point as you have to validate that this is the case but I digress::
If you start from the beginning of X; it then exists that F and Y cannot be satisfied or exist within this set as it is supposed that an infinite number of doors is needed to satisfy this condition.
If Y is an "original" or "current" number (so that it is different or from all numbers within X and isn't F) within the infinite set of X; And that F ⊃ Y; and that there is an assumed previous event before Y;;then;; Y cannot be satisfied without F and the infinite set isn't possible*?
Conclusion::
Y cannot satisfy the previous event of F; As F is an infinite series before Y; and therefore Y cannot exist, so that the infinite set of F before Y is not possible as F needs to satify an infinity number of open doors?

[dawn barn]

well there are an infinite number of numbers, is there any Arithmetic Progression? cause an infinite number of relations can be drawn out

[barren bolt]

Why did you rewrite it? Is it not in English?

[clear minnow]

Just such the set X is being constantly opened, and I put that as X=n+1; can I just use bijection to solve?

[clear minnow]

It's solving for a paradox of an assumed infinite regression for F,

P1: There is an infinite hallway with an infinite number of doors, each labeled sequentially.

P2: For 'Door Y' to be opened, all the doors before it, in the sequence must be opened.

P3: 'Door F' precedes 'Door Y' within this infinite sequence.

P4: To reach 'Door F', one would need to open an infinite number of doors.

P5: If one were to attempt to open the doors sequentially starting from the first door, 'Door F' and consequently 'Door Y', would never be reached, as there are an infinite number of doors to open before them.

P6: If 'Door Y' represents a current event, and all preceding doors represent prior causes or events, 'Door Y' can never be reached if there's an infinite regress of causes or events.

C: The existence or occurrence of 'Door Y' contradicts the idea of an infinite regress in a causal sequence, making the infinite regress infeasible.

We need to solve that Y can exist without F being satisfied.

[barren bolt]

So you’re trying to prove that there are only finitely many doors before Y?

[clear minnow]

either that or, you don't need F to satisfy Y.

As it's within the set X, besides P4 causing this problem

[dawn barn]

I just woke up from sleep, and it's 4 am here, I have to be a little fresh, then I will see the problem again

[clear minnow]

I've been on this problem for a while 💀

Don't worry

[barren bolt]

Well from what I can see, isn’t it enough to simply map the doors from the first door to Y onto a finite subset of the natural numbers

And can’t you do that by defining a linear order on X with the first door as the Least element and then a door is less than another if it is opened before that door

It seems to me that in this way you already have shown that any door has only a finite sequence of doors before it

[clear minnow]

It would seem that way, though I'm having trouble with actaully writing it out.

I've set X=n+1, n+2, n+3 . . .
and F = Y(X) - 1 ~ F = Y-1
and Y = F(X) + 1 ~ Y = F+1
We can biject Y and F onto the set of X so that they can satisfy themselves?

@barren bolt

Or that Y exists within a finite set of X, and can be satisfied as stated for having a finite number of X before Y

[barren bolt]

If you can define a well order on the set X then there is a theorem that relies of the axiom to dependant choice which states that any strictly decreasing sequence of elements must terminate after a finite number of steps

So if you can define that well order then that would be enough as far as I’m concerned

[clear minnow]

alrty, just gotta find the proper notation for it xD @barren bolt thanks

[barren bolt]

I think it can be defined by x<y if y is open implies x is open

Then I guess prove that this relation is a well order on the set X

[clear minnow]

we wouldn't need to satisfy F -> Y because F is = to X in the sequence. So that F and Y can be solved as that instead of the original problems notation

[barren bolt]

Ok but isn’t F before Y?

[clear minnow]

yeah, you need F to satisfy Y/

And so F can = X in that X = n+1

and F = Y-1

[barren bolt]

If F is equal to X then how can F be a door contained in X?

Also how could there be an element behind it Y which is also open

I’m very confused by your notation tbh

I agree that F and Y are sets as all numbers are sets, but as far as I understand F and Y are both in X

So why would F equal X?

I don’t know what you mean

Also isn’t X infinite? So why would X=n+1?

[clear minnow]

I may be forgetting another variable, as X is the set of opening doors. But maybe I could've put X = Inf and the sequence of opening doors is Z = x+1?

[barren bolt]

I thought X is the set of all doors in this sequence

And Y is just any door and F the door preceding Y

So what is Z?

Also you’re trying to prove that the set of all doors preceding Y is finite, no?

[clear minnow]

Wouldn't we need to differentiate from open and closed doors?
We can put the value of X = Inf
And Z = X + 1
Then Z can be differentiated from X?

yes I am

[barren bolt]

So X is all the opened doors and Z is X union one door that is closed?

[clear minnow]

Z is the opening door(s)

X is the set of doors (closed or not changed)

I think I almost got it;

[barren bolt]

Ok but then Z should be a subset of X

so why is it X+1 as you put it

I assume you’re in a university class?

[clear minnow]

Because we need to put an event of opening doors as such, so that there is an event that be satisfied for F.

I mean we could just put this all over X and call it a day

[clear minnow]

ye

[barren bolt]

Is this a foundations/set theory course?

[clear minnow]

ye

[barren bolt]

Did your prof give this problem in an assignment? If so can I read the problem verbatim?

When you say stuff like X=inf then Z=X+1 it confuses me

I’m assuming that you are talking about the cardinalities and not the sets

[barren bolt]

This was copied directly from your prof?

[clear minnow]

no I rewrote it

[barren bolt]

Ok

[clear minnow]

from this

also no sorry I misread what you stated, I'm not in a course atm

[barren bolt]

Oh ok just using a textbook or smth?

[clear minnow]

but It's a problem I'm trying to solve 😭

[barren bolt]

Or did you come up with this situation

[clear minnow]

I've been using wiki 💀

[clear minnow]

It was a paradox presented by someone, and I'm trying to solve it.

I almost, accidently put X as a series lolol

[barren bolt]

The original is with the P1, P2, etc?

[clear minnow]

mhm

[barren bolt]

So in fact the paradox assumes that there are infinitely many doors before Y

So all we can prove or disprove here is that this paradox is logically consistent

Also another thing I notice is that the paradox talks about physically opening the doors

And that being a reason we cannot reach Y

[barren bolt]

You are trying to show that F can be open without Y being open? But that is a contradiction to the premises 2 and 3

So I’d say you cannot prove that

So what is the purpose of this paradox? Are they trying to argue that infinite regress isn’t possible

Or that it is?

[clear minnow]

infinite regress isn't possible;

[barren bolt]

Ok well I agree with that

In fact infinite regress is used to prove things by contradiction

Like if you can demonstrate that an infinite regress exists due to an assumption then you can usually reject that premise

[clear minnow]

hm

I really want to solve for Y though :c

[barren bolt]

You keep saying this but I’m confused what you mean

It’s not like Y is a number unless you consider the position of it in the sequence to be it’s number

And if so, you know from P4 that Y is infinite

P4 says that you need to open infinitely many doors to reach F and then to reach Y you need to open F

[clear minnow]

mfw infinites in infinites

F verifies Y, and F is needed an infinite of X to verify itself. therefore F and Y doesn't exist.

[barren bolt]

This argument seems very strange all around, like are P5 and P6 major premises that are derived from P1-P4 or are they just to be taken for granted

P5 and P6 and even P4 to an extent seem strange to me

Because P4 assumes that F is infinitely far along

And then P5 assumes that someone is physically going to open each door???

And P6 sounds like this is all supposed to be some strange analogy

And then the C is just that infinite regresses cannot occur???

I’m sorry for wasting your time but I don’t think I can help, I barely can understand what you are trying to prove

[lone heart]

@clear minnow Has your question been resolved?

[lone heart]

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[clear minnow]

It's alr, I'm not sure either

[lone heart]

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[fallen verge]

Do you remember what condition is required for a series to converge?

[bronze knoll]

no

[fallen verge]

|r|<1

Do you know what r is?

[bronze knoll]

no

but im assuming its the # outside the ()

hi

[lone heart]

@bronze knoll Has your question been resolved?

[lone heart]

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[rich basin]

Could someone please clarify to me about taylor series
Why does |x| < 1?
wouldn't it still be valid for x to be any value?

[alpine sable]

The statement "|x| < 1" specifies that the absolute value of a variable x is less than 1. This means that x can take on any value between -1 and 1, excluding -1 and 1 themselves

[rich basin]

why is this true?

[alpine sable]

so the reason why this is true is based on the definition of absolute value. The absolute value of a number is its distance from zero on the number line. When we say "|x| < 1," we are saying that the distance of x from zero is less than 1 unit. Therefore, x can be any value within the interval (-1, 1), but it cannot be equal to -1 or 1.

idk if this make sense

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[barren portal]

Let S be set of positive rationals, with operation, a * b= 2ab. Does this form a group?

[barren portal]

I verified closure, associativity, identity

i think the identity element is 1/2

is the inverse b=1/4a?

[sour dove]

a * 1/2 = 2(a)(1/2) = a. Identity looks good!
a * 1/(4a) = 2(a)(1/(4a)) = 1/2 = e. Inverse looks good!

I can't remember if you need to specify that 0 can't be an element, but if you were to plug in 0 you'd get an undefined inverse. It's been a minute since I've done any of thise so verify that but your identity and inverse look good to me

[lone heart]

@barren portal Has your question been resolved?

[barren portal]

0/1 isn’t a positive rational right

so we are safe ig

[sour dove]

right on

everything looks good then

[sour dove]

oh my b I didn't see the underlying set

yeah you're good to go then, nice work!

[lone heart]

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[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[ebon torrent]

3

[lone heart]

Show your work, and if possible, explain where you are stuck.

[young finch]

why didnt you just add √(2x+3) to both sides and square

[ebon torrent]

so I can just write 6x - 5 = 2x + 3?

[young finch]

yes

you made a mistake 2nd line

its -8x

the -2 should have been squared roo

but this is a very inefficienr way to do it

[ebon torrent]

oh

yea its the way I was taught now but thanks for bringing it up the mistake

[young finch]

who the hell taught you to do it like that

they need to be fired

[ebon torrent]

the -2 that should have been squared would turn to positive 4 right?

[young finch]

why not do this though

[young finch]

yes

and rhs is (2-8x)^2 right

[ebon torrent]

yea so just the -2 should have been +4 I did write the 8x in my note book and I got the answer right this way as well

thanks a lot

[young finch]

but you could have done this so much easier lol

[young finch]

doing this gives you 2 as well

[ebon torrent]

its just what I need to do now , thanks for showing me the easier way

.close

[lone heart]

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[trail python]

Need help in learning/steps how it could be rearrange into this, thanks

[tender mason]

help

[tardy stag]

that's kind of a weird thing they did, they solved for this particular x

[trail python]

oh

ok i think i got it. thanks

.close

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[drowsy mountain]

2x+3y=-12

I have tried putting it in slope-intercept form and am not sure if I am write or wrong.

[mental flame]

looks correct 👍

[drowsy mountain]

alright

thx

[mental flame]

np

[drowsy mountain]

.close

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[drowsy mountain]

.close

.close

it isnt working

[young finch]

it already closex

wait for the bot to process

[lone heart]

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[somber coral]

Find the values of m and b that make the following function differentiable.

Could anyone guide me through the question please?

[echo socket]

Clearly, the function is already differentiable for any x other than 2, so we have to make sure that it is differentiable at x = 2

First of all, we would like the sided limits of f(x) to be equal there, right?

So, we set 4 = 2m + b

[somber coral]

Oh does that mean function isnt differentiable if the x is 2?

[echo socket]

No, depends on the value of m and b here

[somber coral]

hmm okay

[echo socket]

I just meant for x > 2 or x < 2, didn't want to exclude x = 2

[somber coral]

oh i see

[somber coral]

yeah this is probably the part i didnt get ;-;

cause I learned what limit is and what derivative is seperately

[echo socket]

Can you imagine what the graph of the function would look like if the one-sided limits do not agree somewhere?

[somber coral]

uhh.. im not sure about that

[echo socket]

Something like this

So, it would be discontinuous there

[somber coral]

ohh i get it

[echo socket]

And discontinuity implies that it's not differentiable there

[somber coral]

yeah

[echo socket]

That's why we want f(x) to be continuous at x = 2

So we set the one-sided limits equal to each other

(We would also need to make sure that those one-sided limits are equal to f(2), but here it's already taken care of)

Okay, so continuity at x = 2 happens when 4 = 2m + b

[somber coral]

wait how did you get that 4 = 2m + b?

oh do I get it after setting the limits equal to each other?

[echo socket]

Yeah

And just say $\lim_{x\to2^-}f(x) = \lim_{x\to2^-}x^2 = 4$ and $\lim_{x\to2^+}f(x) = \lim_{x\to2^+}mx+b=2m+b$

[ocean seal]

A Lonely Bean

[echo socket]

Now, we need to make sure that the one-sided limits of f' also agree at x = 2, as it would imply differentiability at that point

The derivative of x^2 and mx + b are 2x and m respectively, right?

[somber coral]

yeah derivative of x^2 is 2x but isn't the derivative of mx + b m?

[echo socket]

Oh yeah I made a typo there my bad

[somber coral]

oh okay

[echo socket]

So we also set 2 * 2 = m

Meaning m = 4 and now we can solve for b as well

Using the first equation that we derived, we get b = -4

Here is what the graph of f looks like with these values of m and b

[somber coral]

hmmm sorry... but I don't really get what is going on since this one

so what do i get from setting the limits equal?

[echo socket]

Continuity

[somber coral]

4, and 2m+b are the points that limits equal?

[echo socket]

4 and 2m + b are the one-sided limits and we set them equal

[somber coral]

oh so 4=2m+b

[echo socket]

Yes

[somber coral]

what do I do with this equation though?

[echo socket]

That's just one linear equation in m and b, obviously it would be insufficient to solve for those unknowns, so we look for another condition that makes f(x) differentiable at x = 2

We made sure that the values of x^2 and mx + b are the same, now we need to make sure that their slopes also agree

[somber coral]

ah i see

[somber coral]

so since x^2 was 2x, derivate mx+b has to be the same?

[echo socket]

mx + b differentiated* has to be the same, yeah

So 2x and m should be equal at x = 2

[somber coral]

so it should be 2(2)=4 m=4?

[echo socket]

Yeah

And we can now solve for b as well using 4 = 2m + b

[somber coral]

4=2(4)+b b=-4

is it right?

[echo socket]

Yes

[somber coral]

Ohh so we got everything required m=4 and b=-4

[echo socket]

Yeah

[somber coral]

thank you very much for the help

.close

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[steel olive]

radius of the circle is 8

distance between the blue and green parallel lines is also 8

centre of circle lies on the blue line

why is the graph like this

[analog falcon]

like what? can you elaborate

[steel olive]

shouldnt the circle be just touching the red and green lines

because its radius = distance between parallel lines

[analog falcon]

they're not vertical lines

[serene junco]

the distance horizontally like you have in yellow there is 8, but usually when we talk about the distance between two lines we mean the shortest distance between them

and you could draw a shorter distance by moving perpendicular to the lines

if the green and red lines were just touching the circle, the shortest distance between the three lines would be 8

[steel olive]

wait wasnt the formula of distance between parallel lines |c1-c2|

wait holy shit it is |c1-c2|/root a^2+b^2

bruh

the last 2 month

s

[serene junco]

that'll work lol

[steel olive]

i have been using formula as |c1-c2|

my god

my entire life is a lie

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[crimson carbon]

Find the primitive function of \(\abs{x^2\cos(x^3)}\), i.e., \(\int\abs{x^2\cos(x^3)}\dd x\).

[crimson carbon]

So would i use the hybrid function definition of absolute value or write abs as sqrt(blah^2)

[tardy stag]

i'd use the defn of abs

[crimson carbon]

so then i'd have to integrate x^2 cos x^3 and -x^2 cos x^3 ye?

[tardy stag]

yea

well

no

when is x^2 cos(x^3) > 0?

i mean yes you'll have to integrate both of those but how you sew them together is important

[crimson carbon]

So i gotta find x^2 cos x^3 >= 0 and x^2 cos x^3 < 0 as per the definition

so x^2 cos x^3 >= 0 implies x^2 >= 0 or cos x^3 >= 0 right? or am i using the nonexistence of nontrivial zero divisors incorrectly, or partially incorrectly

[tardy stag]

yall gotta ping me

think about how you can multiply two things to get a positive number (but also think about what x^2 can be)

[crimson carbon]

1/3 *3x^2 is the derivative of the composed function

[lone heart]

@crimson carbon Has your question been resolved?

[crimson carbon]

Mb back had to do some things

[crimson carbon]

As in (-1)^3x^2cos(x^3)

[lone heart]

@crimson carbon Has your question been resolved?

[lone heart]

@crimson carbon Has your question been resolved?

[pale kestrel]

To do the same for |2x|, you note -x^2 + A for x<0 and x^2 + B for x>0 work.

Now the key is to ensure the antiderivative works for the whole domain R is it needs to be differentiable at the join, with the correct derivative.

This one is relatively simple for |2x| as theres only 1 join

[pale kestrel]

That's why identifying this is important

[pale kestrel]

Youve got to identify the joins and then make them all work by using the correct constants of integration for each 'bit'

[crimson carbon]

I see

[lone heart]

@crimson carbon Has your question been resolved?

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[wispy trellis]

Anyone know how to find the x and y intercept on a table in math isn’t it when x =0 whatever y is, is the intercept and vice versa

[wispy trellis]

Like when x is 0 here does that mean y intercept = 4

[echo socket]

No, -2 * 0 + 2 is not equal to 4

[wispy trellis]

So how to find the x intercept with a table

[echo socket]

Look at the y value which is below the number 0 among the x values

[wispy trellis]

So y = 2

And x = 1?

[echo socket]

Yes, that's the y intercept

Yeah and x = 1 is the x intercept

[wispy trellis]

Ok Ty

.close

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[wispy trellis]

.reopen

[lone heart]

✅

[wispy trellis]

Oh wait also if on the table there is no 0 say for x what calculation do I do to figure it out

[echo socket]

For y intercept you simply take the equation, replace x with 0 and solve for y

And vice versa

For x intercept you replace y with 0 and solve for x

[wispy trellis]

Ohhh yeah thanks again

.close

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[hasty adder]

hey there, can anyone explain how the answer to 5 is -12 VDC?

i thought it would be zero?

apologies for the scribble on the diagram, you can ignore that

also apologies if this is the wrong discord for this, cant find anywhere else

[lone heart]

@hasty adder Has your question been resolved?

[alpine sable]

.gg/physics

[hasty adder]

oh thank you!

hello, i am very bad at prefixes

i do know the formula

f = 1/2pi/sqrt LC

$f = /frac{1}{2pi/sqrt{LC}}$

[ocean seal]

Pirate

[hasty adder]

$f = /frac{1}{2pi/sqrt(LC)}$

[ocean seal]

Pirate

[hasty adder]

turns out im bad at latex too lol

[buoyant storm]

$f = \frac{1}{2\pi\sqrt{LC}}$

[ocean seal]

orthogonal

[hasty adder]

ah yup

so i need to go back to basics

how would 20mH and 50uF convert into just hertz after the use of this formula

i do know the product of LC is in the power of -9

and by converting 1000 to 100 within the square root

i can reduce that power to -8 then take the square root of both 100 and 10^-8

actually ill send a photo of my work one second

i know it ends up becoming 500/pi

but i am confused how it becomes just hertz instead of kilo hertz

[tardy tapir]

kilo would be if you divided by 1000 at the end

but you didnt

[hasty adder]

oh i see

it was after the fact i divded 1000 by 20

[tardy tapir]

btw that bear is cute lol

[hasty adder]

that it became just hertz?

thanks it's line friends lol

[tardy tapir]

you can confirm the hertz by the dimensions of L and C

[hasty adder]

can you elaborate on what that means

[tardy tapir]

are you familiar with dimensions?

like

dimensional formula of velocity is $[LT^-1]$

[ocean seal]

Dissrupt

[hasty adder]

oh this is entirely new to me

[tardy tapir]

oh

[hasty adder]

it sounds useful tho

how does it work

[tardy tapir]

it is a different topic

[hasty adder]

oh i see

[tardy tapir]

basically we break the quantities into basic or fundamental quantities

like mass, length, time, current

like velocity is displacement/time, and displacement is length basically

[hasty adder]

got it

[tardy tapir]

so velocity is L/T

[alpine sable]

It is just a way of expressing and checking formulas

[tardy tapir]

yup

[hasty adder]

gotcha gotcha

[tardy tapir]

and forming formulas too

[hasty adder]

okay i see where i made the mistake also on my calculation

[tardy tapir]

(without constants)

[hasty adder]

by converting 10^4 to 10000 it becomes a power of just 1

so basically hertz

not sure if that process of thinking is wrong tho

[tardy tapir]

basically the product of units of self inductance and capacitance is time

[hasty adder]

s^-2 right

[tardy tapir]

what?

[hasty adder]

idk that's what someone told me yesterday here in this discord

[tardy tapir]

oh got it

yes yes

[hasty adder]

seconds to the power of -2?

1/t he says i think

[tardy tapir]

not -2

just 2

[hasty adder]

oh i see

why is it square though

and not s^-1?

if it's 1/t

and t = s

[tardy tapir]

well the unit comes out of the root

and becomes t

so it must be t^2 inside the root

[hasty adder]

oh i see so it just becomes t

[tardy tapir]

yeah

[hasty adder]

ooooh okay gotcha gotcha

[tardy tapir]

in your formula, its in denominator

so in the end t^-1

[hasty adder]

right right

okay thank you so much!!

this helped me a lot

[tardy tapir]

this topic is alternating current right?

[hasty adder]

direct current

oh wait

sorry

yes alternating current

AC electronics

[tardy tapir]

i hate that topic

[hasty adder]

yeah it's a lot to digest for sure

the book doesnt even explain how it got the answer

just tells you the answer and that's it

[tardy tapir]

lol

[hasty adder]

anyway thanks for everything

really appreciate the help

[tardy tapir]

no prob mate

[arctic juniper]

.close

@hasty adder Close this by typing '.close'.

[lone heart]

@hasty adder Has your question been resolved?

[lone heart]

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[modern sedge]

<@&268886789983436800>

[near apex]

In which base?

[modern sedge]

Message history

[lone heart]

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[lone heart]

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[azure remnant]

i know that there is a way to prove it with a diagram, but is there any way to prove it without a diagram?

[alpine sable]

u need to construct the diagram , even if there is no diagram given in the question , i guess so

[azure remnant]

okok

[mellow grail]

AFAIK, no

[azure remnant]

what does afaik mean

oh okat

nvm

[mellow grail]

As far as I know

[alpine sable]

which grade are u?

[azure remnant]

how do i approach this?

[azure remnant]

Is the law of our side, if I say ay?

[alpine sable]

nvm

[azure remnant]

sorry im gr9

funny quote lmao

[alpine sable]

that's fine

[azure remnant]

i think so at least

[alpine sable]

i am 10

[azure remnant]

coolio

i'm kinda dumb

so can you help with this

[azure remnant]

^

how would i approach this

[alpine sable]

yes give me a sec

isn't it , angle in a semi circle is 90 degrees

i am not sure about it even

[azure remnant]

wut

[alpine sable]

one sec

[azure remnant]

<@&286206848099549185>

sorry for ping

but help w this problem please

[azure remnant]

^

[buoyant storm]

do you agree that angle ADC=angle ADB+angle BDC

[alpine sable]

sum of opposite angles in a circumscribed quadrilateral equals 180

[azure remnant]

yes

[buoyant storm]

yes what is angle ADB? note that AB is a diameter

[azure remnant]

oh fuck got it

thanks

kinda sthupid

sorry

[buoyant storm]

this also works of course

multiple ways to go about this

[alpine sable]

sorry i couldn't

[azure remnant]

wait but how find BDC

[alpine sable]

bdc = bac

because circumscribed quadrilateral properties

[azure remnant]

how

oh

we weren't taught this lmao

idk why teacher put it on hw

[alpine sable]

you can prove it easily tho

bdc = bac = 1/2 boc

[azure remnant]

oh yeah similar triangles

because you split BAC

and BOC when bisected is equal to BAC

nice

[alpine sable]

oh shoot my bad it's inscribed quadrilateral not circumscribed

sorry

[lone heart]

@azure remnant Has your question been resolved?

[lone heart]

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[alpine sable]

We also know that the surface area of the water A, is given by A = pi*(20h-h^2), so what is the rate of change of water surface area when h = 12 in terms of pi?

[alpine sable]

I've managed to calculate that dA/dh = pi*(20-2h)

so by right, substituting h as 12 into dA/dh will find me the correct answer

but the answer indicates that rather, dA/dh * 0.5 is correct, why?

[tardy tapir]

dont forget that you need to differentiate with respect to time

you need dA/dt

[alpine sable]

ohhhhhh

[tardy tapir]

you are given dH/dt

[alpine sable]

so 0.5 is the dh/dt

makes sense

thank you

.close

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[naive ruin]

I'm not sure if my process is right

[tardy tapir]

there was 2/3 with t^3/2

when u integrated it, it just disappeared

[naive ruin]

Don't I have to use the power rule for integrals

add 1 to the exponent and bring the exponent to the denominator?

I'm more concern with getting the inicial values correct

Where to plug in inicial velocity and incial position to solve for C

I see my mistake on the second part

just fixed it

Still concern if I'm plugging in the initial values for velocity and position correctly though

[winter night]

Have somebody here tried Kumon?

[naive ruin]

huh?

[lone heart]

@naive ruin Has your question been resolved?

[naive ruin]

No

<@&286206848099549185>

[lone heart]

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[cerulean thistle]

I have data with a relationship pattern between the dependent and independent variables like the following graph. what is the estimation method suitable for modeling it?

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[native cloud]

Can someone help me this?

[native cloud]

Why does $\lim :_{x\to :\infty :}\left(sin\left(x\right)\right)$ diverges
while $\lim _{x\to \infty }\left(\frac{1}{sin\left(x\right)}\right)$ the limit of this does not exist?

[ocean seal]

LE SSERAFIM

[tacit arch]

diverge and limit does not exist are sometimes the same thing in some textbooks

[marsh rapids]

some say "goes to infinity" to be more explicit than diverges, but this one isn't even the case

[native cloud]

Ohhh so does that mean all sin, cos, tan functions diverges?

[marsh rapids]

it doesn't mean it, but it's true

at infinity at least

[native cloud]

Ah okay, thank you!

[tacit arch]

all periodic functions diverge as x -> inf

no wait that's wrong

[marsh rapids]

nonconstant

[tacit arch]

constant functions are periodic

[worn fox]

constant functions lol

[tacit arch]

alright thank you for the correction

[native cloud]

Ohh thanks!

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[wise dirge]

If Thomas used $\frac{2}{3}$ on a smartphone, and $\frac{1}{5}$ on shoes, in terms of x, how much has Thomas used?

[ocean seal]

紅卫兵

[wise dirge]

that would be $\frac{2x}{3}+\frac{x}{5}$?

[ocean seal]

紅卫兵

[wise dirge]

or $$\frac{13x}{15}$$

[ocean seal]

紅卫兵

[native cloud]

I don't understand the question

[wise dirge]

what would the expression of how much Thomas used if all of Thomas money = x

sorry if i phrased it a little weird

[lone heart]

@wise dirge Has your question been resolved?

[near apex]

Yes, that's correct.

[wise dirge]

thank you

because my textbook said something else

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[formal rampart]

evaluate where a is constant. only use limits

[lone heart]

Please don't occupy multiple help channels.

[formal rampart]

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[robust current]

I’m not sure how to tackle this problem

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@robust current Has your question been resolved?

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@robust current Has your question been resolved?

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[pearl ice]

Greetings

[pearl ice]

I'm not sure if this count as any type of mathematics although it is chemistry lmao. (Feel free to dismiss me if you must) but any help step-to-step would be appreciated here.

[vapid shuttle]

you're dismissed

[pearl ice]

Yes sir

.close

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[vapid shuttle]

no wait

.reopen

[worn fox]

try the chem server in #old-network

[lone heart]

✅

[pearl ice]

o

[vapid shuttle]

yes try the chem

but I was also being sarcastic

[pearl ice]

oh lmao

[vapid shuttle]

I'm sure someone can help with this here probably

[pearl ice]

I didn't wanna be annoying but thanks guys

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[slow lance]

I have no idea how to solve

[mental flame]

you can start by setting up a system of equations, do you know how to do so?

[slow lance]

Well, I tried to do 2J + 3S =225

And then 1J = Sx3

But I'm not sure how I would solve

[mental flame]

thats a good start.

do you know how to solve systems of equations in general?

[slow lance]

I'm learning currently but I'm confused

[mental flame]

ok np, have you heard of the substitution method?

[slow lance]

2(Sx3) +3s = 225? And yes i have just not sure how I would apply

[mental flame]

well that is one way, but alternatively, notice you can replace 3s with j

[slow lance]

So 2J + J = 225? So 3J = 225? Meaning 225 / 3 is J?

[mental flame]

yep, and what is 225/3?

[slow lance]

75 so J = 75

[mental flame]

yep, now what do you think you can do?

[slow lance]

Divide 75 by 3 to get 25 which = S

[mental flame]

perfect, great job 👍

[slow lance]

I'm a little confused how we figured out S = 3J

Oh wait I see because it says in the equation. So how would I write the full algebraic equation?

[mental flame]

you mean J=3S?

[thorny patio]

This equation yields that

From your setup

1J = J

S*3 = 3S

Therefore

J = 3S

[slow lance]

So would I write 2J + 3S = 225 and then a separate equation that states J = 3S? I'm just wondering if there's a way to incorporate it into one equation

[thorny patio]

It would be called a system of equations

Equations being plural

You have multiple equations that are simultaneously true in a system

If i just said J = 3S alone there are infinite combinations to satisfy that expression

But then you say

In addition to that, I also require this other constriction:

2J + 3S = 225

In terms of how you write it a typical way is to use a big curly bracket

{

And then write each equation on top of each other

[slow lance]

Okay I think it's starting to make sense to me, since we were given the value of 3S = J we then can derive that from total by dividing by 3

[thorny patio]

{ x + y = 2
{ x - y = 0

[thorny patio]

Looks vaguely like this on paper but 1 big curly bracket

[thorny patio]

Yea you were given additional information in the form of another true statement which you could bring into the first equation

You can have systems of 3 or more equations

[slow lance]

Okay, that makes sense

So to clear it up,
2j+3s=225
3s=j
So 3j = 225
225 ÷ 3 = j
j÷3 = s

J=75
S=25

[thorny patio]

Yea that's it

[slow lance]

Thank you for the help

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[mystic vortex]

have you made any progress?

[unkempt parrot]

no

please help me

[mystic vortex]

I actually have to go now sorry

I'd start with some rules of exponents

and try to simplifiy the left side

[alpine sable]

hello

[unkempt parrot]

can u help me?

[alpine sable]

Yes i can

Do you know what

[unkempt parrot]

ok the question above

[alpine sable]

$$ \frac{x^4}{x^2}$$ simplifies to

[ocean seal]

Brandon H

[alpine sable]

do you know what $$x^4*x^2$$ simplifies to?

[ocean seal]

Brandon H

[alpine sable]

what is another way to write $$x^{-2}$$ ?

if you can answer these three "easier" problem you should be able to apply teh same concept to your orginial one

[unkempt parrot]

ik this

[ocean seal]

Brandon H

[alpine sable]

What is the answer

[unkempt parrot]

ik this

[unkempt parrot]

x 6

i think

[alpine sable]

$$x^6$$ that is correct

[ocean seal]

Brandon H

[unkempt parrot]

i was correct

[alpine sable]

what about this one

[unkempt parrot]

x2

?

[alpine sable]

Correct

[alpine sable]

last one?

[unkempt parrot]

i dont even know this

bro im dumb

[alpine sable]

$$x^{-a} = \frac{1}{x^a}$$

[ocean seal]

Brandon H

[alpine sable]

you should had known this because you were able to solve the other two

because

$$x^a * x^b = x^{a+b}$$

[ocean seal]

Brandon H

[alpine sable]

and if

[unkempt parrot]

yeah

[alpine sable]

$$\frac{x^4}{x^2} = x^4*x^{-2}$$ then $$ \frac{x^4}{x^2} = x^{4 + (-2)}$$

[ocean seal]

Brandon H

[alpine sable]

they are in #discussion

id:customize

[alpine sable]

back to the orginial problem

[unkempt parrot]

aight

[alpine sable]

is it possible to simplify the denominator and numerator in this problem?
if it is how would you do it?

[unkempt parrot]

ohh

cancel it out

[alpine sable]

That is one way

[unkempt parrot]

so i cancel it all out to get the values of a b c

[alpine sable]

yeah you can do that

but do you understand why that works?

[unkempt parrot]

no not really

[alpine sable]

$$\frac{9^{12} * 7^4 * 5^8 * 7}{7^2 * 9^5 *5 * 9^3} $$

[ocean seal]

Brandon H

[alpine sable]

You can simplify the numerator by grouping all of the like terms

$$\frac{9^{12} * (7^4 * 7)* 5^8 }{z} $$
which simplifies to
$$\frac{9^{12} * (7^5)* 5^8 }{z} $$

[ocean seal]

Brandon H

[alpine sable]

you can simplify the denominator by doing the same thing

$$\frac{z}{7^2 * (9^5 * 9^3 * 5 } $$
which simplifies to
$$\frac{z}{7^2 * (9^8) * 5 } $$

[ocean seal]

Brandon H

[alpine sable]

if you rearange the both parts so like terms are above and below each other

$$\frac{5^87^59^{12}}{5 *7^2 * 9^8) } $$

and you should know that $$\frac{a}{b} * \frac{c}{d} = \frac{ac}{bd}$$

[ocean seal]

Brandon H

[alpine sable]

which means that $$\frac{5^8 *7^5 *9^{12}} {5 *7^2 *9^8 } = \frac{5^8}{5} *\frac{7^5}{7^2} *\frac{9^{12}}{9^8}$$

[unkempt parrot]

thank you

I got it

u are the best

[ocean seal]

Brandon H

Brandon H

[alpine sable]

so many latex mistakes. hope you understand the process tho

[lone heart]

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[rain birch]

hi yall

[rain birch]

i have a question

from a khanacademy video

https://www.khanacademy.org/math/multivariable-calculus/greens-theorem-and-stokes-theorem/greens-theorem/v/green-s-theorem-proof-part-1

timestamp 1:33

how come he writes $dr = dx \hat{i} + dy \hat{j}$?

[ocean seal]

jriver

[rain birch]

shouldn't it be $dr = \sqrt{dx^2 \hat{i} + dy^2 \hat{j}}$

[ocean seal]

jriver

[rain birch]

thanks in advance!

[gleaming ridge]

dr, or more commonly dl (differential length)

There are two

$$\vec{dl} , , and , , dl$$

[ocean seal]

LUNA

[gleaming ridge]

$$in , 2D: ,, \vec{dl}=dx , \hat{i} + dy , \hat{j}$$

[ocean seal]

LUNA

[gleaming ridge]

but dl without arrow is its magnitude

using pythagorean theorem as usual for any vector

$$you , get: dl=||\vec{dl}||=\sqrt{dx²+dy²}$$

[ocean seal]

LUNA

[rain birch]

ohhhh that make sense

one's a vector and one represents the magnitude

thank you!