help-13

[hearty lily]

<@&286206848099549185> help me out pls

[hearty lily]

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[nimble oxide]

Help

[nimble oxide]

How this give me a positive number it must -18?

[crimson delta]

(-3)^2 = (-3)*(-3)=9

-3^2 = -(3^2)=-(3*3)=-9

[gloomy nacelle]

A number squared always gives you a positive number

[nimble oxide]

Ok thankyou

@gloomy nacelle @crimson delta

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[oblique lynx]

Hey, I have this set of vectors called B1

[dire geode]

ouch that race condition

[oblique lynx]

this channel is claimed

[oblique lynx]

can you delete it?

oh god

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[unborn needle]

What are the supremum and the infimum of Q in R?

[unborn needle]

Are they even defined?

[crimson sedge]

usually they are defined to be +/- infinity

https://mathworld.wolfram.com/Supremum.html

[unborn needle]

Oh thanks again! 👍

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[viscid scaffold]

how can i find the domain and range of this

[hollow minnow]

Let’s do the domain first

What values of x can’t we use here

[viscid scaffold]

oh anything that makes x = 4

restriction

yeah

2, -2

[hollow minnow]

Yeah

So the domain is $\mathbb{R} - {-2,2}$

[viscid scaffold]

oh i see

[wraith dagger]

Pure

[viscid scaffold]

also also what about range

would it just be y >= 0 since its the |x| reciprocal

[hollow minnow]

Yeah the denom is always positive

So y>0

[viscid scaffold]

oh i see

okay one more thing

sqrt(36-4x^2)/9)

can the denominator inside the square root be moved outside it

[hollow minnow]

Yes

sqrt(36-4x^2)/3

[viscid scaffold]

oh so you distribute the sqrt into the /9

is it?

[velvet mortar]

$√(a/b) = √a/√b$

[wraith dagger]

Die

[viscid scaffold]

ooh

thanks

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[frozen raft]

what does subscript n C subscript x mean in the binomial probability distribution function?

[dire geode]

binomial coefficient

https://en.wikipedia.org/wiki/Binomial_distribution#Definitions

[frozen raft]

im sorry but im unable to parse out what wikipedia is stating

[dire geode]

you just need the definition of n C x

the formula is there

[frozen raft]

can you you show me step by step what to to with nCx?

n=6 x=0.93

[dire geode]

this is the definition. what don't you get ?

[dire geode]

show the whole problem. that's probably not your x in n Choose x

[frozen raft]

(nCx×(0.93^3))×(1−0.93)^(6−3)

[dire geode]

show the whole problem

the original, not your typed

also that's not x=0.93. that's p=0.93

[dire geode]

read the definition very carefully

[frozen raft]

whoops x=3

thank you i beleeive i have gotten the correct answer

0.005517889

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[dense creek]

.reopen

Find the sum sum of the following arithmetic sequences

[dense creek]

1, 0.5, 0, -0.5, -1,... n=21

[hollow minnow]

do you know the formula for the sum of an arithmetic sequence?

[dense creek]

Sn equals n divided by 2 [2a1 plus (n-1)d]

[hollow minnow]

Yes

You are trying to find the number of terms?

we know a=1

d=-0.5

Don’t know n

[dense creek]

I know the n

@hollow minnow

@hollow minnow excuse me are you solving my problem

<@&286206848099549185>

[hollow minnow]

What exactly are you trying to solve then?

Oh

n=21 got it

Then just plug things into the formula

Do some arithmetic

[dense creek]

Its actually 22

[hollow minnow]

Okay

[dense creek]

@twin cape

[hollow minnow]

Find the common different

Then just use the formula

[dense creek]

Its 4

I mean

[dense creek]

Its actually 0.5

[livid hound]

how are you getting 0.5 for the difference

[dense creek]

I minus them

[livid hound]

what exactly where you minussing that lead to 0.5

which question are you actually doing atm

[dense creek]

Find the sum of the following arithmetic sequences

[dense creek]

For ex. You put 1 and minus the 0.5 at 4 times

[livid hound]

oh, you're doing the original question you posted and not the one in the image (purple )?

[dense creek]

At least my question are many like 6

[livid hound]

you forgot your -sign

[dense creek]

Oh i forgot

[livid hound]

0.5-1 = -0.5

[dense creek]

Yes

[dense creek]

Yep

@livid hound

[livid hound]

did you fix your mistake?

also don't reply ping, toggle it off

and if you happen to reply ping don't send a ping on top

[cedar kiln]

@dense creek Has your question been resolved?

[dense creek]

I think its actually correct

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[untold glen]

Hey trying to solve this by trig substitution

[untold glen]

The bottom solution is not by trig but normal u substitution, u = x^2 + 9

[hollow minnow]

+C

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[prime fjord]

can someone explain how to do this to me?

[leaden snow]

if x is a function of y, you will have to use the chain rule

[shut locust]

Chain rule mate

[prime fjord]

so like 2y(x^2+y^2)?

[leaden snow]

no

[shut locust]

I would recommend

Writing x as x(y)

x as a function of y

[prime fjord]

so just 2x+2y?

im strugglin

[shut locust]

No, write x as a function of y first

Do you know your chain rule?

[prime fjord]

i guess not LMAO

so x^2+y^2)2x?

[shut locust]

No

Hmmm....do you know this:

d/dx f(g(x)) = f'(g(x))g'(x)

[prime fjord]

honestly I dont remember doing that, let me try it

2y+2x(dx/dy?

[shut locust]

Yeap

You got it

[prime fjord]

hahahah thanks man your fomula pushed me there

i appreciate you! ❤️ 😘

[shut locust]

You should remember it teah

It's the chain rule

[prime fjord]

yeah i was completly blanking out ugh thanks again

[shut locust]

Welcome. Have fun mate

[prime fjord]

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[winged pewter]

b+3 basically inf + 3 which is just inf

then arctan inf is pi/2

but the 2 that came before it

is comepltly ignored?

is this an error or

[hollow minnow]

Should be just pi

[winged pewter]

nice

thanks

[hollow minnow]

[winged pewter]

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[urban quartz]

how u do this

[valid yacht]

how far have you gotten?

[urban quartz]

nowehere bro im stuck

[valid yacht]

assume the smaller integer is $x$, then what will be the next consecutive integer?

[wraith dagger]

n11

[valid yacht]

Do you know about successors and predecessors (?)?

[urban quartz]

no

can u help me answer them because Its due in 15 mins

[valid yacht]

When you add one ($1$) to a number, you get its successor. And when you subtract one ($1$) from a number, you get its predecessor.

[wraith dagger]

n11

[valid yacht]

Thats quite a lot of time for this, you needn't worry

[urban quartz]

its 10 queestions like this

its not the only one

[valid yacht]

Did you read this?

[urban quartz]

yes

[valid yacht]

Follow that logic, and think what will be the successor of $x+1$?

[wraith dagger]

n11

[valid yacht]

wow

I need help

I meant x, not x+1 lol

[urban quartz]

like i said, can u help me answer, because Im focused on finishing it on time then ill learn it

[valid yacht]

You aren't even thinking. You could type the answer quicker than you typed this.

What'll be the successor of x?

Add one to it!!

[urban quartz]

2

43?

[valid yacht]

When did anyone say x is 1

[urban quartz]

1x

[wraith dagger]

n11

[valid yacht]

thats your first answer

[wraith dagger]

n11

[valid yacht]

Do you get that?

[urban quartz]

ok got it

[urban quartz]

thats how I rewrite it for #help-2

#2?

[valid yacht]

Now for the second part, we need to write the problem algebraically

[urban quartz]

ok

[valid yacht]

Problem is: Product of $x$ and $x+1$ is $42$.

[wraith dagger]

n11

n11

[urban quartz]

ok

[valid yacht]

That is it

For the third question, you just need to solve the quadratic equation

[urban quartz]

x^2+x=42?

[valid yacht]

$x^2+x-42=0$ is the equation

[wraith dagger]

n11

[valid yacht]

Yes yes

[urban quartz]

ohh need to subtract 42

ok

[valid yacht]

So you can factorise it or apply the quadratic formula

[urban quartz]

how u do that

it says i need two numbers

[valid yacht]

You'll get some values of $x$, and for each of the eligible value, you'll get a corresponding $x+1$
where $x$ is the smaller number, and $x+1$ is the larger number

[wraith dagger]

n11

[cedar kiln]

@urban quartz Has your question been resolved?

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[sand patio]

I cant seem to get past this part

[cedar kiln]

@sand patio Has your question been resolved?

[sand patio]

<@&286206848099549185>

[sand patio]

<@&286206848099549185>

Pls

[dire geode]

Multiply the top and bottom by the conjugate of the denominator

[sand patio]

That's the problem, I don't know how to conjugate, I looked up a vid but it doesn't seem to work here

[dire geode]

The conjugate of a+sqrt(b) is a-sqrt(b)

[dire geode]

Read
https://www.cuemath.com/numbers/conjugates-and-rationalization/

[sand patio]

I did this but the solution doesn't matchup with the solution of the problem

[dire geode]

where did the sqrt(3) terms go?

[sand patio]

Wdym?

[dire geode]

i mean where did they go

from the left to right

[sand patio]

Oh,
Like this?

[dire geode]

2*sqrt(3) is not the same as sqrt(6)

[sand patio]

-sqr(3)-sqr(3)

[dire geode]

-a-a = -2*a for all numbers a

[sand patio]

So would -sqr(3)-sqr(3) = -3?

[dire geode]

try it in a calc

[sand patio]

-1.126

[dire geode]

,calc -sqrt(3)-sqrt(3)

[wraith dagger]

Result:

-3.4641016151378

[dire geode]

stop trying to simplify it

2 * sqrt(3) is already most simplified

or use 2 = sqrt(4)

[dire geode]

how did you get 2 in the numerator

[dire geode]

and

[sand patio]

My b

[dire geode]

sure that looks right. you can also use this to simplify

[sand patio]

I dont see how tho?

Are you sure this can be simplified?

[dire geode]

yes

[dire geode]

use a=sqrt(3) and plug into your calculator both sides

[sand patio]

1.73?

[dire geode]

are they the same?

[sand patio]

I dont thinkni understand

[dire geode]

you should be getting this. you're not using you calculator correctly

,calc sqrt(3)

[wraith dagger]

Result:

1.7320508075689

[sand patio]

So do I use that? I can't use calculators on the exam so I don't know how I would ever do this on there

[dire geode]

you're just convincing yourself they're the same right now because you don't believe me

[dire geode]

this is the simplification you don't believe

[sand patio]

So it equals 2?

Sry it's really hard to understand typed stuff

[dire geode]

all i'm saying is type these two expressions in your calculator

-sqrt(3) - sqrt(3)

and

-2 * sqrt(3)

no one is saying you need to know how to calculate these in your brain on a test

if you don't want to do it then i don't know how to explain how adding two of the same term works to you

[dire geode]

you need to close your parentheses in the first image

you typed
-sqrt(3 - sqrt(3))

,calc -sqrt(3 - sqrt(3))

[wraith dagger]

Result:

-1.1260325006105

[dire geode]

sqrt(a + b) is not the same as sqrt(a) + sqrt(b)

review PEMDAS https://www.cuemath.com/numbers/pemdas/

[sand patio]

I still don't see what this has to do with anything

[dire geode]

which one is simpler

[dire geode]

this should make sense now

[sand patio]

Oh, so they become -2sqr

[dire geode]

do you know what sqrt(2) means?

like the definition of the symbol $\sqrt{}$

[wraith dagger]

riemann

[sand patio]

Not really no

Oh

[dire geode]

you probably should read this then
https://www.cuemath.com/square-root-formula/

[sand patio]

Nvm

[sand patio]

Does that mean it simplifies down to this?

[dire geode]

.

[dire geode]

no. if your numerator changes in your calculator, then you've done something wrong

,calc 1 - sqrt(3) - sqrt(3) + 3

[wraith dagger]

Result:

0.53589838486225

[dire geode]

which is clearly not 7

[sand patio]

Where did you get 7?

[dire geode]

this is really all there is. i don't know why you're trying to do more

[dire geode]

.

,calc 4+3

[wraith dagger]

Result:

7

[sand patio]

Do you have some paper? I really can't understand it typed

[dire geode]

$-a - a = -2a$

[wraith dagger]

riemann

[sand patio]

So its -2*a?

[sand patio]

Oh, so do I round it?

Cuz the answer for the two -sqr3s is a decimal

-sqr3-sqr3 = -2 ?

Or -sqr3-sqr3= sqr-2?

Or -sqr3-sqr3= -2(sqr3)?

[cedar kiln]

@sand patio Has your question been resolved?

[cedar kiln]

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[cloud jay]

I am practicing Implicit Differentiation. My answer is the boxed one and the correct answer (according to the book) is the one on its right. I’m wondering where the excess dy/dx came from.

[cloud jay]

so is it $\frac{dy^3}{dx} = 3y^2\frac{dy}{dx}$ ?

[wraith dagger]

aab

[valid yacht]

Yes, by chain rule

[cloud jay]

ahhh i see!

$\frac{dy^3}{dy} \times \frac{dy}{dx} = 3y^2\frac{dy}{dx}$

[wraith dagger]

aab

[cloud jay]

right ?

[valid yacht]

Indeed

[cloud jay]

cool

thanks

.close

[cedar kiln]

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[valid yacht]

npp!

[cedar kiln]

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[errant meadow]

again with the rational zero theorem

[errant meadow]

x⁴-3x³+x²+4=0

what's the answer for this????

imma get a Q in my class fr

[crimson delta]

well what does the rational zero theorem say

[errant meadow]

what does that mean?

[crimson delta]

you mentioned the rational zero theorem. what is that. what does it say

[errant meadow]

ohhhh

The Rational Zeros Theorem says: If P(x) is a polynomial with integer coefficients and if is a zero of P(x) (P( ) = 0), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x).

[spice tundra]

what's the dif between rational zero theorem and the factor theorem

[crimson delta]

well they say different stuff?

[spice tundra]

i'm sure they do

[crimson delta]

ok so p is a factor of 4 and q is a factor of 1

what are the options for p and q then

[spice tundra]

(x-2) is a factor

[crimson delta]

dont tell answers

[spice tundra]

i'd just divide the polynomial by x - 2 to factorise the rest of it

[crimson delta]

this question is not about you

[spice tundra]

who's got your tits in a twist

[crimson delta]

the whole point of the rational zero theorem is to figure out that x-2 is a factor

[spice tundra]

i'm certain he can figure out that if you sub in P(2) it gives 0, therefore (x - 2) is a factor

[spice tundra]

what's the rational zero theorem

sounds like it does the same as the factor theorem

[crimson delta]

no

[crimson delta]

here they even said it for you

dont help in a question about the rational zero theorem if you dont know what it is

[spice tundra]

rational theorem takes into account the leading coefficient?

and in this case, it's just one

why is it needed

what does it accomplish that factor theorem doesn't

[crimson delta]

well because it might not always be 1?

[spice tundra]

but it is 1

[crimson delta]

it tells you which options to check for the factor theorem

[spice tundra]

ohh

okay

[crimson delta]

how do you know to check 2 here

[spice tundra]

logical

[crimson delta]

you don't want to guess a shitlot of options

you only want to guess a small list if possible

like if you had 3x^3-4x^2-4x-7 or something, you'd never guess naturally that 7/3 might be a root. but the rational root theorem tells you to test 7/3

if you have further questions, get your own channel

now back to the OP

@errant meadowyou still there?

[errant meadow]

ye

[crimson delta]

.

[crimson delta]

.

[errant meadow]

for p i got 1,2,4 and q i got just 1

is that what u were asking?

[crimson delta]

yes

but don't forget the negatives

so the rational zero theorem now says that if you have a rational zero, it has to be one of {1, 2, 4, -1, -2, -4}

and now you can test those

[errant meadow]

OHHHH yeahhh ok forgot about the negatives

ok ill test those

idk if did it wrong but i still got no zero

[crimson delta]

then you did a mistake

2 is a zero

[errant meadow]

what did i do wrong?

i feel so dumb rn

[crimson delta]

the double minus shouldnt be there

3(2)^3=24

not -24

[errant meadow]

OMGGGGG

THAT MAKES SO MUCH MORE SENSEEEE

ok my problem is solved

thank you person

[crimson delta]

ok do you know how to continue from here? now that you know 2 is a zero?

[errant meadow]

yes sir

the synthetic division?

right

[crimson delta]

yes

[errant meadow]

okokokok

[errant meadow]

nvm ok wait idk what going on its not equalling to zero

[crimson delta]

dunno I don't know that particular notation of polynomial division. but you certainly made a mistake somewhere

the second coefficient is -3

and the coefficient of x is 0

[errant meadow]

ohhh ok

yeah now its equal to 0

was just a little confused cuz for the first part to make 2 a 0 i took out the -3

and made -3 a 3

or maybe my calculator just did that putting a - sign for no reason thing

[cedar kiln]

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[odd lily]

My question wasn't answered yesterday

[hollow minnow]

A bit hard to read but

I think you made a mistake when you said sin(3pi/2+x) = -sin(x)

[cedar kiln]

@odd lily Has your question been resolved?

[odd lily]

I tried to use anti-supplementary angles there

What should I do instead then?

[cedar kiln]

@odd lily Has your question been resolved?

[hollow minnow]

$\sin{(\frac{3\pi}{2}+x)}= \sin{(\pi +(\frac{\pi}{2} +x))} = -\sin{(\frac{\pi}{2}+x)} = -\cos{x}$

[wraith dagger]

Pure

[cedar kiln]

@odd lily Has your question been resolved?

[odd lily]

what characteristic did you use to make the pie dissappear from step 2 to 3

anti-supplementary angles?

[crimson sedge]

I'm not Pure, but here is a visualization if that helps. And yes I think they used anti-supplementary angles.

[odd lily]

yeah it seems like it

didn't know the method of seperating the pie though

[crimson sedge]

it's just fractions; 3π/2 = π + π/2

[odd lily]

Yeah I know, just didn't know you could use that to you advantage here

[crimson sedge]

ah I see.

[odd lily]

I'll try the exercise again and let you know if I can do it nw

@crimson sedge could you explain to me why pie/2 + x equals the complementary angle characteristic, isn't it supposed to be pie/2 - x

[crimson sedge]

yes, π/2-x is the complementary angle characteristic. I wrote -sin(π/2+x) is the same sin as sin(3π/2+x) and visualized π/2+x in the second sketch, which would be in the second quadrant if x ≤ π/2.

as to the antisupplementary part: (3π/2+x)-(π/2+x)= π or 180º

[cedar kiln]

@odd lily Has your question been resolved?

[cedar kiln]

@odd lily Has your question been resolved?

[cedar kiln]

@odd lily Has your question been resolved?

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[crimson sedge]

I’m trying to figure out how I graph this on the horizontal axis

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[cobalt flicker]

Anyone

[cobalt flicker]

AB and CD are two chords of a circle meeting at O. If O Is midp of each of chords, prove that AB and CD mist Br Diameters

.close

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[steel sorrel]

couldnt any function be made surjective by restricting the codomain to the range?
Or vice-versa made not surjective by increasing the codomain beyond the range?

[cedar kiln]

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[cobalt flicker]

Hello

[cobalt flicker]

Anyone active?

[rugged palm]

[cobalt flicker]

Thank god

You r a life saver

AB and CD are two chords of a circle meeting at O. If O Is midp of each of chords, prove that AB and CD mist Br Diameters

@rugged palm

[rugged palm]

What have you tried?

[cobalt flicker]

Uhh

Im sending pic

So this is my plan.....

[rugged palm]

I don't download files, can you send an image?

[cobalt flicker]

I use mobile

So it automatically converts

[rugged palm]

Ok now I see it

So what do you think?

[cedar kiln]

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[trail phoenix]

The refractive index of water is 4/3 and that of glass 3/2.Calculate the refractive index of glass with respect to water?

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[desert cosmos]

inequality

[desert cosmos]

inequality

[eternal comet]

,w inequality

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Remavas#3737

[eternal comet]

There you go, happy to help.

[desert cosmos]

help me

@eternal comet

<@&286206848099549185>

[stuck wave]

hello can somebody help me

[desert cosmos]

what

<@&286206848099549185>

[stuck wave]

this

but its in french

[turbid anvil]

This is ur ques ?

[stuck wave]

these are my homeworks

[turbid anvil]

Ayo i don't understand french

[stuck wave]

well i can translate it

[dusty hazel]

Get a channel of your own.

#❓how-to-get-help

[stuck wave]

done

ok

now can i get some help

[dusty hazel]

I don't know, wait in your channel to figure out.

[stuck wave]

ok

[cedar kiln]

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[prime fjord]

hey guys, can someone help me solve for Y? I got X pretty smoothly but now im blanking on what to do

[prime fjord]

.close

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[marble ermine]

hello, how to prove prove that $\lim_{x \to 0} \frac{1-cos(x)}{x^2}=1/2$

[south tundra]

$ from both sides

[marble ermine]

if the trigonomic form of 1-cos(x) is 2sin²(x/2)

[wraith dagger]

lilisworld

[south tundra]

Rewrite (1 - cosx)/x^2 as 2sin^2(x/2)/4(x/2)^2

Which is the same as 1/2 * (sin(x/2)/(x/2))^2

[south juniper]

Are you required to use cos(x) = 1 - sin^2 (x/2)?

[marble ermine]

no

[marble ermine]

wait

[south juniper]

Then try multiply/dividing by 1 + cos(x)

[marble ermine]

@south tundra so you mean $\frac{1-cos(x)}{x^2}=2\frac{sin^2(x/2)}{4(x/2)^2}$ ?

[south tundra]

Yes

[marble ermine]

uhm ok

[south tundra]

\sin looks nicer btw

and \cos

[wraith dagger]

lilisworld

lilisworld

[marble ermine]

where does the 4 come from?

[south tundra]

x^2 = 4x^2/4 = 4(x/2)^2

[marble ermine]

@south tundra ok so we get 1/2 because the limit of sin²(x/2)/(x/2)² is 1?

[south tundra]

Yes

[marble ermine]

ok thank you very much 👍

.close

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[long stag]

How do you go about finding the sum of this power series?

[long stag]

The answer is

But I am unsure of how to get there

[cedar kiln]

@long stag Has your question been resolved?

[south juniper]

@long stag Consider splitting it into two sums. One will be geometric and for the other one, use $\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ for $|x| < 1$

[wraith dagger]

Sup?

[long stag]

Thank you i will try this!

So I tried it non-geometrically and it gives the correct answer

So

[south juniper]

actually i just realized

for both you have to use sum x^n = 1/(1-x)

lol

[long stag]

$\sum\limits_{n=0}^{\infty} x^n = \frac{1}{1-x}$

[wraith dagger]

yxceed

[long stag]

Yea aaa

So we have
$\frac{12}{1-x}=12x^n$ and $\frac{x}{(1-x)^2}=x^nn$

[wraith dagger]

yxceed

[long stag]

Added together equals
$\frac{12-11x}{(x-1)^2}$

[wraith dagger]

yxceed

[long stag]

@south juniper OMG thank you so much. Even tho you said smth wrong at the beginning it helped me realize i could just split it up. Thanks a lot

[south juniper]

np

[long stag]

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[still tinsel]

hello

[still tinsel]

A hand of seven cards is dealt from a standard deck. How many of these would contain three spades and four hearts?

[fair geyser]

it's made of 3 spades and 4 hearts, so you multiply 13c3 adn 13c4

[mental trail]

Choose 3 spades among the possible spades. Choose 4 hearts among the possible hearts

[still tinsel]

thanks

How many ways can you choose up to three items from a set of seven different items

is anyone available

.close

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[robust drift]

the question is solve simultaneously x^2 + 6 > 5x and x^2 >= 1, how do i know hcih whay the <> signs point when i solve?

[long arrow]

generally, you can interpret in geometrically, x^2 + 6 > 5x -> x^2 - 5x + 6 > 0 means where the graph of y = x^2 - 5x + 6 is above the x-axis

also after finding solution you can imagine this as set on number line

"and" word means take intersection of sets

[robust drift]

thanks

.close

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[marble ermine]

Hello
Can someone re explain me why is is true ?

[marble ermine]

The "-->0" means the limit if x approach+infinity

[rain drift]

hi there. Yes. So when you have something like (1 + x) raised to some power $\lambda$, the leading exponent will be $\lambda$

[wraith dagger]

MellowDramaLlama

[mighty drift]

it's less than x^(alpha - lambda), which goes to 0 by the inequality given

[rain drift]

so for example $\lambda = 2$ is $(1 + x)^2 = (1 + 2x + x^2)$. $\lambda = 3$ is $(1 + x)^3 = x^3 + 3x^2 + 3x + 1$. Notice how whatever $\lambda$ is, it's the leading power?

[wraith dagger]

MellowDramaLlama

[mighty drift]

what about lambda = pi ?

[marble ermine]

Yes

[rain drift]

So since $\lambda > \alpha$, then the leading power on top will always be smaler than the leading power on the bottom

[sharp chasm]

isn't it the other way around

[rain drift]

whoop

[mighty drift]

yes

[rain drift]

yes thank you

[wraith dagger]

MellowDramaLlama

[rain drift]

lmao thanks

[mighty drift]

this argument doesn't quite work when lambda isn't an integer though

[marble ermine]

And now if it's the leading power, what does that imply?

[sharp chasm]

the bottom will always be a larger number than the top

[rain drift]

nah it would still work. If $\lambda = \pi$, then we can say that $\alpha = \pi - 1$ and the argument still holds

[wraith dagger]

MellowDramaLlama

[rain drift]

it doesn't have to be integers

[mighty drift]

the leading term thing doesn't make much sense when there's no binomial theorem

[marble ermine]

So if the bottom is larger its limit is Always 0?

[sharp chasm]

it approaches 0

it isnt 0

[mighty drift]

also you don't get to choose alpha

[mighty drift]

this argument is just a lot simpler

[sharp chasm]

alpha < lambda

[rain drift]

I know you don't get to choose alpha but I was just using it as an example

[mighty drift]

you still don't choose it

[mighty drift]

an example that's so convenient it's almost the only one that works is a terrible example

[marble ermine]

What's the proof of that im dumb

[rain drift]

as an example. I know you don't choose it in the argument. I'm using it for a visualization

[eternal comet]

Well, $(1+x)^\lambda$ asymptotically behaves as $x^\lambda$. Just wondering how to show that rigorously.

[wraith dagger]

Remavas

[mighty drift]

just squeeze theorem

[sharp chasm]

-> means it approaches zero

it'll not be zero but an extremely low number approaching zero

[marble ermine]

Yes but why is that so

[sharp chasm]

lambda > alpha so the bottom will always be a larger number than the top

if you divide a high number with an even higher number then the answer comes close to zero

[marble ermine]

Mmmhh it makes sense

So for example if f(x)<g(x) then f(x)/g(x) approach 0

?

[sharp chasm]

that depends on the formula i guess

since f(x) = 1 and g(x) = 2 holds but does not approach zero

[dark silo]

Or f(x) = x and g(x) = 2x

[marble ermine]

So why does it work above?

[dark silo]

Do you understand why, say, $\lim_{x\rightarrow \infty} \frac{x}{x^2} = 0$?

[wraith dagger]

RSB

[marble ermine]

Yeah because it's another form of 1/x

[dark silo]

Hmm, how about $\lim \frac{x+1}{x^2+2}$?

[wraith dagger]

RSB

[sharp chasm]

exactly, 1/ infinte approaches 0

[marble ermine]

Well now you can find the limit by rewriting it factorization but

[dark silo]

Wait, can you? I tried to avoid that haha

[marble ermine]

By factorizing it but it is not equal to 1/x

[dark silo]

$(x^2+2)$ isn't factorable I think

[wraith dagger]

RSB

[sharp chasm]

you cant factorize it

[marble ermine]

It's an interminate form so it's not comparable to x/x^2

[sharp chasm]

with limits you divide by the highest denominator power in this case x^2

[marble ermine]

I meant above you factorize by x and also on the botom

So you get rid of the x above yeah but anyway

[sharp chasm]

you can't do that

you get (x+1)/x

that is 1+1/x

[dark silo]

I think the real answer to your question goes back to the definition of limit

In this case $\lim_{x\rightarrow \infty} f(x) = 0$ means that $\forall \varepsilon, \exists x_0$ so that $\forall x > x_0, |f(x)| < \varepsilon$.

[wraith dagger]

RSB

[dark silo]

That's a very technical answer. The intuition is like this: a function like $\frac {x+1}{x^2+2}$ gets smaller and smaller as $x$ gets bigger; there's no number that it always stays bigger than.

[wraith dagger]

RSB

[marble ermine]

Isn this true?

[sharp chasm]

divide by highest denominator so divide everything by x^2

[eternal comet]

still 0/1 = 0

[marble ermine]

Yup... But the main argument in the first question was that the function on the bottom was Always bigger than the one on top, so i asked if it was Always true in general and no:(

[dark silo]

Right, it's not enough for the function on the bottom to be always bigger

[marble ermine]

Mm ok

[marble ermine]

So i missed something

[dark silo]

We can certainly cook up bizarre functions where factoring tricks won't work, like idk $\lim_{x \rightarrow \infty} \frac {x+(\sin \ln x)^2}{x^2}$

[wraith dagger]

RSB

[marble ermine]

But now, what did i miss

[dark silo]

What are you confused about?

Ah, do you still want to know why $\lim \frac{x^\alpha}{(1+x)^\lambda} = 0$ when $\alpha < \lambda$?

[wraith dagger]

RSB

[marble ermine]

yes exactly, you told me the reason was because (1+x)^âlpha was bigger than x^alpha, but it confused me

the denominator being bigger is not anough

or im missing something

[dark silo]

OK, I might write this proof like this: $x^\alpha \leq (1+x)^\alpha$. Therefore $\frac {x^\alpha}{(1+x)^\lambda} \leq \frac {(1+x)^\alpha}{(1+x)^\lambda} = (1+x)^{\alpha-\lambda}$. We know $\alpha-\lambda$ is negative so this goes to 0

[wraith dagger]

RSB

[dark silo]

That means the sequence $\frac{x^a}{(1+x)^\lambda}$ is bounded below by 0 and above by something that converges to 0.

[wraith dagger]

RSB

[dark silo]

The rest is squeeze theorem

[marble ermine]

yeah it's better like this, but it means that it was not that obvious?

[dark silo]

I think it is not totally obvious

But there's a general intuition you build up, like, the top 'looks like" $x^\alpha$ and the bottom 'looks like' $x^\lambda$.

[wraith dagger]

RSB

[dark silo]

So if you've been doing this for awhile, you start to just recognize things like that and it feels more obvious

This kind of problem is exactly what helps build that intuition!

[marble ermine]

actually yes now it seems obvious

[dark silo]

🙂

I guess we did a good job explaining it then haha

[marble ermine]

thanks 🙏

[marble ermine]

yes

.close

[cedar kiln]

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[feral peak]

Hi , does anyone has idea on this problem ? I don't really know where to start with

[red pumice]

A B and C being subsets of that fancy letter

[eternal comet]

I guess they want you (in a malformed way) find the lowest natural number which is a multiple of 2, 3 and 5.

Assuming {multiples of y} means the set of natural numbers up to whatever which are a multiple of y.

[feral peak]

ohh I see .
let me try it out

[cedar kiln]

@feral peak Has your question been resolved?

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[crimson sedge]

someone please help

[cedar kiln]

Please don't occupy multiple help channels.

[crimson sedge]

please tag me

[gaunt hamlet]

What have you tried @crimson sedge

[crimson sedge]

so i did L(x)

and got 5+5(x-0)

but the answer is supposedly 5-5x

i dont understand why its - instead of +

[gaunt hamlet]

Explain your answer

[crimson sedge]

so first i did f(0) + f'(0)(x-0)

f(0) = 5

f'(0) = 5

[gaunt hamlet]

Show your work for this

[crimson sedge]

5(e^-0) = 5(1/(e^0)) = 5/(e^0) = 5/1 = 5

[gaunt hamlet]

The derivative of 5e^(-x) is not 5e^(-x)

[crimson sedge]

use product rule irght

right

[gaunt hamlet]

Chain rule

[crimson sedge]

why not product

how do u know when to use which

[gaunt hamlet]

Because there is no product

[crimson sedge]

its 5e^-x

[gaunt hamlet]

5 is just a constant so you can take it out of the derivative

[crimson sedge]

ohh

so it will be 5(e^x)(-e^-x)?

[gaunt hamlet]

No

Whats the chain rule

[crimson sedge]

@gaunt hamlet

<@&286206848099549185>

nvm i got it ty

[cedar kiln]

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[heady salmon]

so i added the pi fractions and i got 3pi over 6 but how do i add the 5 after?

[gaunt hamlet]

What's f(3pi/6)?

[heady salmon]

arnt i suppose to add pi over 3 to pi over 6?

then add 5 after

[gaunt hamlet]

It's f(x + π/6) + 5

So you find x + π/6

Plug that into f

Then add 5

[heady salmon]

how would i do that?

would it just be x pi over 6?

[gaunt hamlet]

You found x + π/6 was 3π/6

What's f(3π/6)

[heady salmon]

Do its decimal?
but what happens to the 5?

[gaunt hamlet]

What's f(x)?

[heady salmon]

sin

[gaunt hamlet]

What's sin(3π/6)?

[heady salmon]

1

then add 5?

[gaunt hamlet]

What's 1 + 5?

Yes

[heady salmon]

so 6

k i see thanks

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[umbral current]

How do I show that a function is continuous on an interval?

[thick cipher]

Which function,

[umbral current]

I think showing the function straight up would be dishonest

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[desert lichen]

lim(n^2*sin(2pien!))(n->inf)
i have not any ideas

[toxic moat]

oh my

that is so small

$\lim_{n \to \infty} n^2 \sin (2 \pi e n!)$

[desert lichen]

yea

[toxic moat]

feels like dne to me

:c

wait actually

[desert lichen]

Okay, can you help me with something else?

[toxic moat]

no

soul stealer

no way

[desert lichen]

hah

okey

some one can explain me how to solve it?

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[cedar kiln]

@magic laurel Has your question been resolved?

[magic laurel]

<@&286206848099549185>

[magic laurel]

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[crimson sedge]

can i get some help, i did a b and c

[red pumice]

,rotate

[red pumice]

,rotate

[keen elbow]

bruh

[red pumice]

,rotate

[crimson sedge]

for a i got a = 2 and b = 60
b i got (x-5)
c i got (x - 5) (2x - 1) (x+6) (x - 2)

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

um

<@&286206848099549185>

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[merry tinsel]

could someone explain how they go from step 1 to step 2

[sharp sinew]

sorry

[merry tinsel]

no worries

[sharp sinew]

hmmm well u have to dy/dx on both sides right

[merry tinsel]

the dy/dx is y prime

[sharp sinew]

yea im just making sure i get this right

[merry tinsel]

the thing is

how do we get rid of the y prime on the right side

i dont understand what theyt did

they sorta got this from nowhere

[merry tinsel]

would you know?

[sharp sinew]

im trying to write it out

[merry tinsel]

ok this is the original question

[sharp sinew]

hmmm

[merry tinsel]

just if u need it

[sharp sinew]

okay if u differentiate both sides

what do u get

[merry tinsel]

this

i just need help simplifying it

[digital cliff]

they expanded the RHS, moved all the dy/dx terms to the left, factored out dy/dx, then divided by the LHS to leave dy/dx on its own

[merry tinsel]

i dont get how they simplified that last part

[sharp sinew]

yea

[merry tinsel]

wym

wait

[sharp sinew]

si factor out with the 2x * dy/dx

[merry tinsel]

like this?

[merry tinsel]

what

[digital cliff]

from your first line, move all the terms with y' to the left and everything else to the right

then factor out y' on the left and divide both sides by whats left to get y' alone as the subject

[merry tinsel]

so like this?

[digital cliff]

the sin term moves to the left by addition not division

[merry tinsel]

how so?

oh

lol

im dumbass

should've just moved that first

then done the rest

[merry tinsel]

the dervative of 2xe^y

would that be 2xe^(y)*y'

[digital cliff]

it would be 2e^y+(2xe^y)*y'

[merry tinsel]

?

wait

how

?

[digital cliff]

product rule

[merry tinsel]

omg brah ah hall naw wtf man

[merry tinsel]

https://www.youtube.com/shorts/Y9531sxAf2A

[digital cliff]

[merry tinsel]

i got y'=-x-1/x

that right?

[digital cliff]

you only asked me this, idk what the full Q is

[merry tinsel]

this is what they got

thats the same thing right?

[digital cliff]

yes

[cedar kiln]

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[haughty vapor]

Noah makes a $900 payment for his new car. He will continue to pay $240 each money. If the car's original price was $5,460. How many months will it take for noath to pay off his near car?

[haughty vapor]

Dont know how to do this

[dusty hazel]

$240 each money? Month you mean?

[haughty vapor]

yes

[dusty hazel]

Use a variable for the number of months.

Anyways, there's no interest whatsoever right?

[molten vine]

probably not

interest in problems like these is like the air resistance in high school physics problems

It's always ignored

[haughty vapor]

what

nO

I just have to make a algeebraic solution

[dusty hazel]

yeah.

let's say it takes x months for noah to pay the money completely

[haughty vapor]

yes

[dusty hazel]

And now the money Noah needs to pay is not 5460, rather 5460-900

Cause he's paid 900 already

[haughty vapor]

4560

[dusty hazel]

Now, if he pays 240 in 1 month. How much will he pay in x months?

[haughty vapor]

4560

what

[dusty hazel]

uh

[haughty vapor]

oh

19 months

[dusty hazel]

,calc 4560/240

[haughty vapor]

yeah 19

[dusty hazel]

Lol alright.

[haughty vapor]

,help

[dusty hazel]

Bet.

[haughty vapor]

bruh

[dusty hazel]

The bot's dead, I almost forgot.

[haughty vapor]

oh ok well what next

[dusty hazel]

wdym what next, didn't you figure how to do it already

[haughty vapor]

how do I turn it into a equation

[dusty hazel]

First, you didn't answer how much will Noah pay in x months?

Think of it like this, if he pays 240 in one month. How much will he pay in 2 months?

[haughty vapor]

480

[dusty hazel]

How much in 3?

[haughty vapor]

700 prob

720

[dusty hazel]

lol no

Yeah 720

So how much in x months?

[haughty vapor]

so just 4560

[dusty hazel]

Yes 4560, but let's say we don't know 4560 for now

[haughty vapor]

ok

[dusty hazel]

Like, you did 240 * 2 for two months

240 * 3 for three months

What should you do for x months?

[haughty vapor]

240x?

[dusty hazel]

Exactly.

[haughty vapor]

oh

ok so

[dusty hazel]

So...?

[haughty vapor]

like 240x - 900 = 5460 or something like that

wait

[dusty hazel]

Did we not subtract 900 from 5460?

[haughty vapor]

yeah