help-13

[primal lodge]

just a confirmatory question

but i would just make this a rational inequality right?

[wraith dagger]

hiii

hiii

hiii

[primal lodge]

then i have to consider the case where numerator < 0 and denominator < 0?

but denominator < 0 gives 2^{4-k} < 0 which is never true

[tropic oxide]

made your own life complicated there.

[primal lodge]

you mean i could just cross multiply?

[tropic oxide]

do you want to know how to do this without headache or do you insist on doing it specifically this way

[primal lodge]

i would like to know how to do it without headache, but if possible i would like to clarify if i'm approaching this rational inequality right

just for future references

maybe not so much for this question

[tropic oxide]

this isn't a rational inequality

this is an exponential inequality

[primal lodge]

oh yeah whoops

[primal lodge]

if i'm dealing with this

would i have to consider the case where denominator = 0

as well as numerator = 0 and then work on that basis?

i'm guessing the non-headache way was just to do 2^{1-k} < 2^3

which implies {1-k} < 3 -> -2 < k

[tropic oxide]

you are, once again, going down a painful route.

you could write your ineq as $2^{k-1} > 2^{-3}$ right off the bat and solve it in seconds!

[wraith dagger]

Ann

[primal lodge]

oh yeah lol

whoops

i didn't think of that

thanks

[primal lodge]

btw could you confirm the "general approach" i used

like is the process as said?

The answer comes to be right ofc

just want to clarify

[tropic oxide]

your 'general approach' is probably okayish but poorly suited to this scenario...

[primal lodge]

okay sure

thanks for the alternative way

and clarifying my query

.close

[cedar kiln]

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[jolly sable]

#help-18 message

[jolly sable]

@kindred storm

Rank is possible only for square matrix

So maximum rank is 4

That's why I cut the option first

[crimson sedge]

wdym

are you sure that notation stands for rank?

it think itb about the size of the matrix

[cedar kiln]

@jolly sable Has your question been resolved?

[jolly sable]

100%

P sign is rank

Rank of AB will be equal or less than rank of A rank of B whatever minimum is

[crimson sedge]

no, i meant the first possible option

i meant, thank you for telling me what p stands for

[jolly sable]

Which notation

[crimson sedge]

but why did you cut the first option

[jolly sable]

I is identity matrix

Yes. I cut the identity matrix

[crimson sedge]

can you send me again the exc pls

[crimson sedge]

oh okay

sorry, i misunderstood the question before

yup, the first one doesn't apply here, they aren't square matrices, give me a moment pls

[jolly sable]

Moment is over?

[cedar kiln]

Helpers are just people volunteering their time to help you. Be polite.

[jolly sable]

I am joking

Who is this behind bot?

[crimson sedge]

lmao 😂 i forgot how to do this, maybe i need more time

lets see if someone else can help

<@&286206848099549185>

[clear berry]

yes

Like iv is also correct but the last one is that most apt

update, a bit confused between iv and last option

[jolly sable]

This is the proper question

[clear berry]

oh well then it's iv

cause rank(AB) <= min(rank(A), rank(B))

[crimson sedge]

I think I have it

nvm numbily already answered

[jolly sable]

Answer is given 2

[clear berry]

BRUH, I didn't read half the question

It's already given that BA = I4

[jolly sable]

So?

Rank of BA <=rank of B.rank of A

[crimson sedge]

after thinking for a while, maybe i think i have the reasoning

[cedar kiln]

@jolly sable Has your question been resolved?

[jolly sable]

Elaborate

[crimson sedge]

GKAOGKAWKOTKAO rwaofpgk3p0l6-3o6-3op6lghlh- Im either BLInd cuz i dont find a way to join these things and have no idea about the statmentes provided above.cAN You wait 30 minut pls

or <@&286206848099549185> can someone help pls :3

[clear berry]

*sighs*

[clear berry]

alright so we have that BA = I4

I4 means identity which is invertible so it's full rank ie rank(BA) = 4

which means min(rank(A), rank(B)) = 4

also notice that their maximum rank can be 4

since both maximum and minimum is 4. Both their ranks must be 4

which implies rank(AB) = 4 as well

[crimson sedge]

I think I know how to do it if rank(BA)=rank(rref(B)rref(A)) which I'm unsure for a reason

in an easier way

I'm gonna assume rank(BA) = rank( rref(B)rref(A) ) = 4

[cedar kiln]

@jolly sable Has your question been resolved?

[crimson sedge]

I think I have it, but... I also reached a conclusion im not sure if its true

this conclusion: ~ If A size 7x4 and B size 4 x 7 and BA=I, then rank(A)=rank(B) ~

Let B be a 4x7 matrix and A be a 7x4 matrix, It's given that BA = I_4. First, we know that rank(BA) = rank( rref(B)rref(A) ) = 4. From BA=I_4 we know that there are either or 4 independent columns in B or 4 independent rows in A, Because rank(M)=rank(M^T) for any matrix M, if that's the first case then there would also be 4 independent rows in B, or otherwise 4 independent columns in A. Let's analyze the first case. Because there are 4 independent columns in B, rank(B)=rank( rref(B) )=4. At this point B could have 4 of their colums be independently out of the possible 7. Because swapping columns does not affect the rank of the matrix, we will now set B' to be an adjoint matrix [ I | 0], and we could think of swapping rows does not affect the rank of the matrix too, but its not necessary since entries above, below, right or left of a pivot are always 0, so doing this step already organizes our rows. Reorganizing A in the same way would lead us to A' = [ I/0 ]. Therefore, because A' has an I matrix adjointed, it would imply A has also 4 independent columns, either case rows. Now we can do the same argument for the case rank(A)=4 and rank(B) is unknown. Because both arguments lead to same conclusion in both directions, we can conclude strictly that A and B have 4 independent columns. Now we can only use the fact that A has 4 independent columns to conclude that the columns of AB are just linear combinations of the columns of A with the coefficients along columns of B. Therefore, rank(AB) = 4

this argument shows the case 4-independent 4-independent or viceversa, so Rank(A)=Rank(B) additionally as additional conclusion?

[cedar kiln]

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[icy anchor]

Hey I jus loss the channel for help

[icy anchor]

I was talking to someone on

[dire halo]

?

Send w

Q

[icy anchor]

It was how to

Shade the area of the triangle formed between the y axis and two straight lines. Calculate the area of this triangle

[dire halo]

Send

Photo

[icy anchor]

Hey guys

Can anyone help me

Sketch a graph

I don't get

How to shade the area of a triangle formed between the y axis and two straight lines

And find the area for it

[clear berry]

[icy anchor]

Yes

So I dk which triangle to use

On the question

Which triangle between the y axis and two straight lines?

[dire halo]

R those the 2 lines?

[cedar kiln]

@icy anchor Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Do u know Chinese remainder theorem

[south tundra]

Have you tried applying Euler's theorem?

[crimson sedge]

Hint: || consider doing it mod 7 and mod 19 first||

[jolly sable]

yes

[crimson sedge]

Is that saying too much idk

[jolly sable]

2^132=1mod133

[south tundra]

That is not what Euler's theorem says

[jolly sable]

2 should be remainder

[south tundra]

No

Can you state Euler's theorem?

[jolly sable]

see

[south tundra]

.

[jolly sable]

a^p-1=1mod p
a and p gcd should be 1

2 and 133 gcd is 1

[south tundra]

That is not Euler's theorem

[jolly sable]

i am saying fermat theorem

[south tundra]

It is $a^{\varphi(n)} = 1 \mod n$

[jolly sable]

i solved it with fermat

[wraith dagger]

A Lonely Bean

[south tundra]

Fermat's little theorem is useless in this case until you consider different modulos

Because 133 is not prime

[south tundra]

So it's better to go with this

Where phi is the Euler's totient function

[jolly sable]

okay let me use it

2^108=1mod133

[south tundra]

,w 133 * (1 - 1/7) * (1 - 1/19)

[wraith dagger]

Results provided by WolframAlpha

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Requested by alonelybean

[south tundra]

Yeah

[jolly sable]

next?

2^25?

,w 2^25 mod 133

[wraith dagger]

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Upgrade to WolframAlpha Pro!

Requested by arjunn5589

[jolly sable]

@south tundra

What next?

[south tundra]

There is your answer

[jolly sable]

I have a idea

what if we seperate like prime numbers

[south tundra]

May work as well yeah

[jolly sable]

but how will we do 2^25 in the exam?

I got 128 by seperate them

and use of fermat

[south tundra]

Just subtract 128 from 133

[jolly sable]

then

[median holly]

fermt

fermats

[jolly sable]

read the full convo

[median holly]

you can directly apply fermats

[jolly sable]

yes i did

i am asiking how will we get answer by lonely bean method of euler

i meant we are stucked at 2^25

[median holly]

lonely bean?

you mean totient?

[jolly sable]

yeah

we have only two choice

[median holly]

you would get

[jolly sable]

so use of fermat

or just get 2^25

[median holly]

2^25= 1 (mod 133)

[jolly sable]

lmao

[median holly]

now u raise both side to power 5

2^128 = 1 (mod 133)

[jolly sable]

,w 2^25= 1 (mod 133)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by arjunn5589

[median holly]

awk

lemme find thje totient

whats the totient

[jolly sable]

108

[median holly]

wait cant use fermats

thats not prime

[jolly sable]

2^108=1 mod133

[jolly sable]

we can if we seperate

[median holly]

ok got it

[jolly sable]

7*19

[median holly]

now what you do

[jolly sable]

but i want to solve it by euler fully

[median holly]

2^7

is

-5 mod 133

okay?

[jolly sable]

what

[median holly]

2^7 is 128

128 is -5 mod 133

[jolly sable]

okay

[median holly]

multiply 128 both side thrice

so

2^(108 +21) = -125 mod 133

fine?

[jolly sable]

you are doing a long method

[median holly]

you cant

simplify

further

[jolly sable]

we are here 2^25=mod133

[median holly]

this isnt long

yes and in 2 steps

you have answer

im literally now using properties of modulo

[jolly sable]

2^7* 218=mod(719)

[median holly]

yeah u use crt and fermats

but if u want to do with totient

[jolly sable]

i did

and got the answer

[median holly]

after that point

[jolly sable]

i am trying to use of euler

[median holly]

u cant simplify

totient will only get you to that step

[jolly sable]

.close

[cedar kiln]

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[tawdry mesa]

,w 4x^2 -(3x+2)^2 factorise with explanation why it's equal to -(x+2) (5x+2)

[wraith dagger]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Jasblox#7891

[tawdry mesa]

4x^2 -(3x+2)^2 factorise with explanation why it's equal to -(x+2) (5x+2)

[median holly]

fundamental identity of

a^2-b^2

[thin roost]

Difference of square?

4x^2 = (2x)^2

(2x)^2-(3x+2)^2

[wraith dagger]

_basudev

[tawdry mesa]

but how did we find 5 in (3x+2)^2

[median holly]

huh?

if you add them what do you get

[median holly]

in this form

[tawdry mesa]

its 3x+2

not 3+2

[median holly]

its a+b

a is 2x

b is 3x+2

[tawdry mesa]

oooohh

[thin tusk]

4x² - (3x+2)²
4x² - 9x² - 12x - 4
-5x² - 12x - 4
-1(5x² + 12x + 4) 
I just dont like negative numbers so took -1 common

[tawdry mesa]

yes i get it now thank you!

.close

[cedar kiln]

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[cedar kiln]

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[grand quarry]

I found on a textbook: for every x != 0 and y two real number proof that x^2+y^2 > 0.

[grand quarry]

it seems so obvious that i don't know what to proof

[mental trail]

so only x is != 0

(not necessarily y)

[grand quarry]

not necessarily

[mental trail]

yeah sorry I formulated it wrong

[grand quarry]

we know that x^2 >0

and y^2 >= 0

[mental trail]

so y^2 >= 0

so x^2 + y^2 >= x^2

[grand quarry]

yes

so x^2 + y^2 >0

[mental trail]

so there's your proof

[grand quarry]

but why putting that on a textbook -_-

anyway ty

I was just not sure if I missed something xD

.close

[cedar kiln]

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[tame lance]

hi guys i was wondering how can P(A U B) = 1 help us in this question

[versed fulcrum]

What's P(A|B)?

[tame lance]

P(A|B) = P(A and B) P(B)

[lunar lynx]

It's P(A|B) = P(A and B)/ P(B)

Not multiplication

[tame lance]

oh yeh mb i forgor the divide

[lunar lynx]

Okay.

[wraith dagger]

Enemagneto

[tame lance]

yes but how can i get using this formula P(A), P(B) and P(A and B)?

i need P(B) for P(A|B) to get P(A and B)

[lunar lynx]

Well, make use of the information given in the question.

[lunar lynx]

Firstly, try to find a relationship between p(A|B) and p(B|A).

[tame lance]

P (A|B) P(B) = P(A and B)
P (B|A) P(A) = P(A and B)

P(A|B)P(B) = P(B|A)P(A)

[lunar lynx]

Nice

Now, put values of p(A|B) and p(B|A) in that equation.

You will find a relationship between P(A) and P(B).

[tame lance]

i got P(B) / P(A) = 9/16

so P(B) = 9 and P(A) = 16?

[cedar kiln]

@tame lance Has your question been resolved?

[tame lance]

.close

[cedar kiln]

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[prisma gull]

Is option b correct?

I listed all the bijections from {1,2,3} -> {1,2,3}, and saw the function where 1->3, 2->1, 3-> 2 has the property 2

[versed fulcrum]

Uhh

I think we might need more context

[prisma gull]

sure, I'm trying to answer the above question

I think symmetry group of three letters is same as set of all bijections from {1,2,3} to {1,2,3}

so I listed all the 3! elements in the set, ie 6 different functions,

where the operation is composition

[tropic oxide]

yup

[prisma gull]

was my approach correct?

ie to to check composition manually?

or is there something simple?

suppose if I ask is x^12 = e?

is it easy to answer?

[tropic oxide]

did you mean "which elements x ∈ S_3 satisfy x^12 = e"?

[prisma gull]

yes

[crimson sedge]

ya he meant that

[tropic oxide]

all of them

because |S_3| = 6

[crimson sedge]

ya

[tropic oxide]

i asked op not you tho

[prisma gull]

how does that help us here?

i'm pretty much a starter

[tropic oxide]

i think i saw you ask about the proof of lagrange's theorem

it's either the theorem itself, or a corollary thereof, that for any finite group G and any element x ∈ G, x^|G| = e

[prisma gull]

so if |G| divides k, a^k =e also?

[tropic oxide]

of course

[prisma gull]

makes sense

[tropic oxide]

exponent laws still work just fine in a group

as long as all your exponents are integers obv

[prisma gull]

makes sense

[prisma gull]

i don't think it's me

started group theory less than a week ago

thanks again.

suppose I collect those elements a, such that a^2 =e

and put in a new set

will it form a group under same operation ?

@tropic oxide

[tropic oxide]

will it?

if ord(a) = 2 and ord(b) = 2, is it also true that ord(ab) = 2?

[prisma gull]

inverse does exist

no

[tropic oxide]

well, that should answer your own question.

[prisma gull]

it's given abelian

[tropic oxide]

...

[prisma gull]

how does that change?

[tropic oxide]

it changes everything lmao

why did you choose to hide this info

[prisma gull]

inexperience

mb

[tropic oxide]

if a^2 = e and b^2 = e does that imply (ab)^2 = e?

now that you know G is abelian

[prisma gull]

yes

[prisma gull]

aabb=e

or abab= (aa)(bb)= e*e= e

i guess so

[tropic oxide]

yes

[prisma gull]

let me verify the group axioms here

since e*e= e, e is in H, so identity is there

associativity hopefully yes,

inverse is g*g, given

so group

is that okay?

[cedar kiln]

@prisma gull Has your question been resolved?

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[cedar kiln]

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[oblique prawn]

how to find total distance traveled using position vs time graph

[dire geode]

find all points where direction changed and sum all individual paths

[oblique prawn]

do i have to use distance formula to find the arc length

[violet night]

post the original question

[oblique prawn]

assuming the graph is composed of just straight lines

[dusk finch]

You are not trying to find arc length, distance traveled is just change in position

Post the graph, it will be better to show it on a particular example

[oblique prawn]

ok i just found this off google

let’s say this graph represented the position of an object

[dusk finch]

okay, so first try finding what distance did it travel going from A to B

dont overthink

[oblique prawn]

is it just 15

[dusk finch]

now B to C

and then C to D

and sum it all up

[oblique prawn]

do i not have to find the length of the lines

[dusk finch]

No

[oblique prawn]

Why

[dusk finch]

Why yes? When it was at point A, it had position 0. When it came to B, it's position changed to 15. So the distance travelled was 15

no arclength

just change in position

[oblique prawn]

so would the total displacement be like that as well

you would do final position-initial position

[dusk finch]

displacement is final position - initial position

total distance traveled may differ from displacement

e.g. when you go 10 steps forward, 4 steps back and another 5 steps forward your total displacement is 10 - 4 + 5 while total distance travelled is 10 + 4 + 5

[oblique prawn]

ok ty

.close

[cedar kiln]

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[cedar kiln]

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[vale plaza]

32x-12x divided by 143x

[unkempt saddle]

20/143x

[vale plaza]

!help

[cedar kiln]

Please read #❓how-to-get-help

[mighty shuttle]

!status

[cedar kiln]

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

[vale plaza]

1

[mighty shuttle]

First find 32x-12x

What is it?

[vale plaza]

20x

[lusty dome]

like terms combine

[vale plaza]

It is minus.

[lusty dome]

note that you could verify this by factoring out x to get x(32 - 12) = x*20 = 20x

[vale plaza]

That's good.

But I know it is 20x.

What is the next step?

[mighty shuttle]

So you have to now find 20x/143x

Does that give you any ideas?

[vale plaza]

Well Can you cancel the x?

[mighty shuttle]

Yes!

[vale plaza]

Well I've Canceled The x Now.

What Now?

[mighty shuttle]

That's your answer

20/143

[vale plaza]

Well thank you for helping me.

[mighty shuttle]

Np

[vale plaza]

.close

[cedar kiln]

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[obsidian coral]

What's the goal?

[dire geode]

,tex .exp rules

[wraith dagger]

riemann

[dire geode]

use uh product rule

[cedar kiln]

@eager wyvern Has your question been resolved?

[eager wyvern]

how does that help me tho

[dire geode]

answers this question

[eager wyvern]

how

[dire geode]

did you read the product rule

[eager wyvern]

oh

so it is 5^x time 5^1

[cedar kiln]

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[tawdry mesa]

factorize 16x^2 - 36/16

[mental trail]

!status

[cedar kiln]

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

[tawdry mesa]

I don't know where to begin

[mental trail]

Do you know the formula for a²-b² ?

[tawdry mesa]

yes

?

yes i do know the formula

[mental trail]

Ok

Do you see how it could apply here ?

What would your "a" and "b" be?

[tawdry mesa]

a = 16x^2 and b: 36/16 ?

[mental trail]

Be careful as we're looking for a²-b², not a-b

[tawdry mesa]

or a = 16x and b: 16/8 ?

hmm

[mental trail]

We're getting a bit closer but still not there

So yes we want a² = 16x² in a first part

So what is a ?

[tawdry mesa]

oohh so a= 8x

[mental trail]

Close

But 8² =...

[tawdry mesa]

64

4²?

a= 4x

so a²= 8²

hello?

.close

[cedar kiln]

Channel closed

Closed by @tawdry mesa

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[tawdry mesa]

.reopen

[cedar kiln]

✅

[tawdry mesa]

factorize 16x^2 - 36/16 I have begun but got stuck midway

i know now that a= 4x
so a²= 8² is that right?

anyonee?

[mental trail]

Sorry

[tawdry mesa]

np

[mental trail]

a = 4x is correct

[tawdry mesa]

yay

[mental trail]

Then a² = (4x)² = 4² x²

Which is indeed 16x²

As for b, we want b² = ?

[tawdry mesa]

36/16

or -36/16

or something else

[mental trail]

Yes +36/16

Because -b² = -36/16

And so b = ... ?

[tawdry mesa]

b= 16/4

no

6/4

[mental trail]

Correct !

[tawdry mesa]

yyaay

[mental trail]

And so now 16x² - 36/16 = a²-b²

What now?

[tawdry mesa]

so a² = 4²x² and b² = 36/16

[mental trail]

Yes

And a²-b² = ?

[tawdry mesa]

is (a-b) (a+b)

[mental trail]

Yes

And now we can replace a and b by their actual values

[tawdry mesa]

a = 4x b= 6/4

(4x-6/4)(4x+6/4)

Thank you so much!

[mental trail]

One last thing

Maybe simplify 6/4 into a smaller fraction

[tawdry mesa]

but its 1.5

[mental trail]

Yes

We can find a smaller denominator to express 6/4

Do you know how it can be written?

[tawdry mesa]

root?

no

idk

ooppsssss

lol

ofc its 3/2

haha

Ty bye

[mental trail]

Np

[tawdry mesa]

.close

[cedar kiln]

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[tawdry mesa]

.solved

[jade creek]

Hello! I need help. On function and non-function

[jade creek]

Just giving me examples

And I identify

It and if you could use a verbal descriptions

For example

If a person is watching a football broadcast streamed to a million people, is that function or not a function?

I don't have a specific problem because my teacher has a specifically given me any practice. So I'm asking you before my test. This is going to be easy. It's just I want to have more practice from some people that are

Better in the topic than me

[buoyant latch]

Well, what is a function

[jade creek]

Alexis

The relationship

Or a relation?

Between the x and y

Each set having one input and output

You could use the VLS or the vertical line test to see

If it's a function or not a function

The rule is if it passes more than once

Then it's not a function if it only goes through once it is a function

[buoyant latch]

A function is simply a mapping between 2 sets, the input and output set, where each element of of the input set is mapped to at most 1 element in the output set

[jade creek]

Yes, I already know

What I'm saying is just to give me examples

Like the one I gave you earlier

Or even just a whole worksheet

Or something like that

[buoyant latch]

That’s not what you said when I asked what a function is

[jade creek]

Since I never got any practice problems

[jade creek]

It's along the lines what you said

I'm just going based off with my teacher saying

[buoyant latch]

What you said was not precise and not always true

[jade creek]

.close

[cedar kiln]

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[blazing aurora]

The question is in this image. Due to the phrasing of the question, it appears that there would only be one answer, but I am unable to limit it to only one. Is there any other method?

What I did:

r = b2 - 1 / b + 1, r = 1 / b2 - 1
therefore (b2 - 1 / b + 1) = (1 / b2 - 1)

This can have 3 solutions. What am I doing wrong?

(I know that this is a silly question but I am confused :( )

[mental trail]

maybe the idea that there are 3 solutions is an illusion. Do you know what b^2 - 1 can be factored to?

[blazing aurora]

I think I see what you mean

Forgive my algebra skills but would it be (b+1)(b-1)

and as the first term is (b+1) r would be (b-1)?

[mental trail]

your algebra skills need not be forgiven for this is correct :)

[blazing aurora]

thank you!

[mental trail]

r = b-1 yes

[blazing aurora]

I will take another crack at it with this information

I have a result of 1.205

I believe this to be correct, thank you for the help

[mental trail]

👍

[blazing aurora]

.close

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[fiery yoke]

i cant seem to ever understand why if we root both sides in the final step the plus minus goes to the root 10

why is it plus minus root 10 instead of just root 10?

if possible could someone explain conceptually as possible ;-;

[mental trail]

the same way that when you have for example 4

both 2^2 = 4

and (-2)^2 = 4

if you want the rigorous justification :

[fiery yoke]

i know but why dosent the x turn into plus or minus

[mental trail]

oh the last line is the same as $\pm x = \sqrt{10}$

[wraith dagger]

rafilou2003

[mental trail]

but here's the rigorous way to show that :

[fiery yoke]

😃 ?

whaaaaaaaaattttt the heeeeck

[obsidian coral]

Because for $\sqrt{x}$, the domain is $x \geq 0$ but with $x^2$ the domain is all real numbers hence $\pm$, idk if that answers your question

[fiery yoke]

what does the upsie down question mark mean?

[obsidian coral]

It was supposed to be x >= 0

[fiery yoke]

ohh

[obsidian coral]

But latex messed up the formatting

[fiery yoke]

gotchu

[mental trail]

x >= 0 yes

[wicked brook]

The function $f(x) = \sqrt{x^2}$ is equivalent to saying $f(x) = |x|$. As a result you should consider both plus or minus.

[wraith dagger]

Zander

dldh06

[mental trail]

yeah, $x^2 = 10 \Longleftrightarrow |x| = \sqrt{10}$

[wraith dagger]

rafilou2003

[mental trail]

(over the real numbers)

[fiery yoke]

but I feel like I did many questions like that where I took the square root of both sides and didnt end up having the plus minus sign

[mental trail]

maybe because you knew x was positive

[fiery yoke]

is that because my teacher didnt ask for that or beccasue there are some problems where it is not needed?

[fiery yoke]

o like if there was a limit next to the equation going x>0?

[mental trail]

uh not limits

[wicked brook]

you can ignore a negative from context of a problem: for example if it's asking you how many apples Abe has, it cannot be negative.

[obsidian coral]

How I remember is, if you manually have to apply the sqrt, ie you wrote it, it's plus/minus, but if it's already there before you do the problem, then it's just plus

[fiery yoke]

tysm for all 3 of yalls 😭 im so much less confused now lmao

bye!!!

.close

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[crimson sedge]

$\lim x-->x \frac{1}{x}- \frac{-1}{senx}

[crimson sedge]

$\lim x-->x \frac{1}{x}- \frac{-1}{senx}$

$\lim{0} \frac{1}{x}- \frac{-1}{senx}$

[wraith dagger]

Kangaroo

$\lim{0}  \frac{1}{x}-  \frac{-1}{senx}$

[crimson sedge]

$\lim_{x\to 0} \frac{1}{x}- \frac{-1}{senx}$

[wraith dagger]

Kangaroo

[crimson sedge]

I need help with this problem, I need to apply to L'Hôpital's rule

[mental trail]

so you know l'hôpital applies to a quotient

[crimson sedge]

Yes so I write it as

first

[mental trail]

also did you put one too many "minuses"?

[crimson sedge]

$\frac{-x+senx}{(senx)x}$

[wraith dagger]

Kangaroo

[mental trail]

this is good ok

[crimson sedge]

maybe is better on the initial from

[mental trail]

but i would use the fact that sen(x)/x has limit 1

[mental trail]

no this form is better

[crimson sedge]

okay

[crimson sedge]

Mhh, why?

[mental trail]

write this as

$\frac{-x + sin(x)}{x^2}\cdot \frac{x}{sin(x)}$

[wraith dagger]

rafilou2003

[mental trail]

the thing on the right has limit 1, so we can discard it if we're looking at the limit in 0

as for the term on the left, what does l'hôpital give us?

[crimson sedge]

Sorry why limit 1?

[umbral dew]

whats sen btw

[crimson sedge]

sen is sin in spanish

[umbral dew]

oh, okay

[crimson sedge]

Yeah it is confuse

[umbral dew]

yeah

[crimson sedge]

The right thing would be sin(1) no?

[mental trail]

? no

[crimson sedge]

So why limit 1?

[mental trail]

$\lim_{x\to 0}\frac{sin(x)-0}{x-0} = ?$

[wraith dagger]

rafilou2003

[crimson sedge]

Oh okay

[mental trail]

it's cos(0) which is 1

because sin'(x) = cos(x)

[crimson sedge]

Because x->0

[mental trail]

so 1/(all of this) is still 1

[mental trail]

so the term on the right is 1

(of limit 1 at least)

[crimson sedge]

And when that happens sen f(x) = f(x)

So x/x =1

[mental trail]

i didn't understand this but ok

[crimson sedge]

Haha

it is on my book in that way

But my book is more like a student guide than a math book

but the answer is 0 so it is correct

Oh no

I reach an indetermination not a solution : V

. close

.close

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[runic prawn]

first 2 are definitely true, but how do you check the third one

[cedar kiln]

@runic prawn Has your question been resolved?

[flint plinth]

see if you can find a counterexample

[cedar kiln]

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[sterile grove]

What is the probability that the car is repaired under budget given that it is ready on time?

[sterile grove]

I don't know why I'm struggling with this

I keep getting stuck with odds that are way lower than they should be

last time I got like 13%

waaait

I feel like it's coming to me

nope

0.6554 this time

I know I'm closer to getting it but I'm still doing something wrong

YESSSSSS

.close

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[cedar kiln]

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[sharp kestrel]

how come these are the same

and how would I know that

[royal loom]

what is the period of cotangent

[flint plinth]

and what's the difference between -17pi/3 and 4pi/3

[sharp kestrel]

21?

[sharp kestrel]

idk

[flint plinth]

what happened to the pi

and the /3

[sharp kestrel]

21pi/3?

[royal loom]

well you should probably find out

[sharp kestrel]

so what do I do with this

[royal loom]

21pi/3 simplifies to what

[sharp kestrel]

oh

7

ok ty

.close

[cedar kiln]

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[royal loom]

what

[sharp kestrel]

what

[royal loom]

7?

[sharp kestrel]

.open

[royal loom]

.reopen

[cedar kiln]

✅

[sharp kestrel]

21/3?

[royal loom]

it doesn't simplify to 7

why are you ditching the pi

[sharp kestrel]

idk

21pi/3?

[royal loom]

simplifies to what

[sharp kestrel]

pi/7

[royal loom]

no

[sharp kestrel]

7/pi?

idk

[royal loom]

well that's because you are randomly guessing

no

$\frac{21}{3}\cdot \pi=?$

[wraith dagger]

Austin

[flint plinth]

try writing it out and calculating it

don't be afraid to use a pen and paper

[sharp kestrel]

7pi

[royal loom]

yes

7pi

[sharp kestrel]

ok now what

[royal loom]

What is the period

of cotangent

[sharp kestrel]

sorry idk what that is

[royal loom]

Trig functions are periodic

meaning that they repeat themselves over and over again

how often they repeat themselves, that is the period

,w plot y=sin(x)

[wraith dagger]

Results provided by WolframAlpha

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Requested by austinu

[royal loom]

sin(x) is periodic

it repeats itself every 2pi

[sharp kestrel]

ok

[royal loom]

look at the graph

verify this for yourself

the period of sin(x) is 2pi

so if you had

[sharp kestrel]

ok

[royal loom]

sin(x) and sin(2pi+x)

these are the same

[sharp kestrel]

ok

[royal loom]

so find cotangents period

and use the difference in between the two points you are evaluating

to make a judgement

[sharp kestrel]

,w plot y=cot(x)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by iwnlrino

[sharp kestrel]

pi?

[royal loom]

yup

[sharp kestrel]

ok

[royal loom]

so cot(-pi/2) and cot(pi/2) are the same

why?

because -pi/2 is pi away from pi/2

[sharp kestrel]

ok

[royal loom]

you can do the same thing forever

cot(-pi/2) and cot(pi/2) and cot(3pi/2) and cot(5pi/2) are all equal

why?

[sharp kestrel]

they are all the same distance away

[royal loom]

pi distance away

specifically

[sharp kestrel]

yea

[royal loom]

yes

so if you are pi apart

or 2pi apart

or 7pi apart (this should remind you of your original question)

they will be the same

[sharp kestrel]

so how do I get to 4pi/3

[royal loom]

as you said earlier, the distance between -17pi/3 and 4pi/3 is 7pi

so this tells you that the cotangent of both are equal

[sharp kestrel]

ok how would I get to 4pi/3 if I wasnt given it

[royal loom]

What do you mean by get to it?

[sharp kestrel]

like how would I know they are equal

[royal loom]

because they differ by pi in distance

anything that differs by an integer multiple of pi

will be equal

in cotangent

[sharp kestrel]

but do i have to memorize that

[royal loom]

memorizing the period is useful

yes

[sharp kestrel]

ok

is that it?

[royal loom]

unless you have more questions yes

[sharp kestrel]

ok

.close

[cedar kiln]

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[royal loom]

you're welcome

[obsidian coral]

Thank you

[cedar kiln]

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[abstract wing]

I have no idea how to even start this problem 😭

[cedar kiln]

@abstract wing Has your question been resolved?

[foggy merlin]

f and g are linear functions so you can write them like this : f(x) = ax+b, g(x) = cx+d, with a and c the slopes
a=2 so f(x) = 2x+b on (-1,+oo)
g(0) = -2, so g(x) = cx-2 on (-1,+oo)

f(x) > g(x), therefore 2x+b > cx-2
🤔

[flint plinth]

you can already determine some information about the slope of g given the first two sentences

if the slope of g were less than 2, then eventually f and g would cross somewhere in (-infinity, 1)

[cedar kiln]

@abstract wing Has your question been resolved?

[cedar kiln]

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[next spruce]

I don't know how to solve this😭 The blue is the given and is being found. The red is my attempt at least

[next spruce]

I'm not sure how to get the domain

[nimble mulch]

(x+2)² is not x²+2²

it's x²+4x+2²

so you end up with x²+4x+9

[next spruce]

what's the solution to this? learning~

[nimble mulch]

do you mean the domain?

or what solution

[next spruce]

no like how it went to x^2 + 4x

[surreal cave]

wdym?

[next spruce]

the bot earlier answered the question i think

[surreal cave]

yea that was me, you're having trouble expanding? :)

[next spruce]

yeah, I'm a bit unfamilliar with this

[nimble mulch]

(x+2)² = (x+2) * (x+2) = x² + 2x + 2x + 2² = x² + 4x + 2²

[next spruce]

They didn't teach it at my school before

[surreal cave]

$(x+2)^2=(x+2)(x+2)=x(x+2)+2(x+2)$

[wraith dagger]

MrFancy

[next spruce]

ahhh foil method?

is that right-

[surreal cave]

there are often many methods to expanding/factoring, find which way works best for you :)

[nimble mulch]

onto the actual question, how could you get the domain of x²+4x+9?

[next spruce]

how..

[surreal cave]

is there any number that you could plug into x^2+4x+9, that would make it undefined?

[next spruce]

0?

[surreal cave]

why would 0 make it undefined?

[next spruce]

it's not a real number-

[surreal cave]

0 is very real mein freund

[next spruce]

testing my stock knowledge IEJSFOIJFES

[surreal cave]

can you tell me what $0^2+4\cdot0+9$ is equal to?

[wraith dagger]

MrFancy

[next spruce]

9

[surreal cave]

notice all I did was put zero into the x's in the quadratic before ^

[surreal cave]

and is that undefined? :)

[next spruce]

nope

[surreal cave]

ok so that means 0 is in the domain

is there any number you think that would make the function undefined?

[next spruce]

ahhh, that's how you check

-9 then

[surreal cave]

why would -9 make the function undefined?

(-9)^2+4*(-9)+9=81-36+9=54 and 54 is very much defined :)

[next spruce]

joke wait

[surreal cave]

Hint: there is no number that makes x^2+4x+9 undefined

[next spruce]

ack

[surreal cave]

that means the domain of this function is every real number

do you know how to express that in interval notation? :)

[next spruce]

ah infinite

[nimble mulch]

Consider it this way, the domain are all values for x for which the function is defined, therefore instead search for all cases for which the function is not defined. This includes division by 0, taking the root of a negative number, 0⁰ etc

[next spruce]

this one

[surreal cave]

not quite

just putting infinity doesn't tell us a whole lot, do you know interval notation?

[next spruce]

no...

I'm in the learning process of functions haha

[surreal cave]

ok that's fine, if I had the interval: [-3,5] do you know what that means?

[next spruce]

uhh

sorry no

[surreal cave]

it means the range between -3 and 5

so -2 is in that range

1 is in that range

but 69 isn't

nor is 420

[next spruce]

ohhh

[surreal cave]

that make sense? :)

[next spruce]

yes yes

[surreal cave]

so would 1/2 be in the interval [-1,1]?

[next spruce]

yes

[surreal cave]

yep! ok so do you feel comfortable with interval notation so far?

[next spruce]

yes!!

I understand the concept now

[surreal cave]

ok very good, so I have one more thing to add, there is a differnce between [] and ()

and the difference is important

[] means the end points are included

so if the interval is [-3,5] then -3 and 5 are included in the range

but if we used parentheses

[next spruce]

oooooh

[surreal cave]

they are not

make sense?

[next spruce]

yupp

[surreal cave]

ok, so what would the interval notation be for all real numbers?

[next spruce]

hmm

these 2

[surreal cave]

yes very good! but one final question do we use square brackets or parentheses?

[next spruce]

dying, and the test is tomorrow

square brackets

[surreal cave]

why?

[surreal cave]

hey you're learning pretty quick

5 mins ago you didn't even know interval notation

[next spruce]

they are included within the range, and it's infinite?

[surreal cave]

but infinity is not a number, so it relaly cant be included itself

[next spruce]

ohh

[surreal cave]

so whenver there's an infinity in interval notation

we use parentheses, make sense? :)

[next spruce]

yupp

[surreal cave]

ok, so finally, what would be the domain of this function?

[next spruce]

this with ()

it can be interchanged right?

[surreal cave]

yep (-inf,inf)

[surreal cave]

wdym by interchanged?

[next spruce]

like (inf,-inf) and vice versa or is there just supposed to be 1?

[surreal cave]

we always go from left to right on the number line, good question :) negative to positive

[next spruce]

icic

[surreal cave]

there we go, hope I helped you not only answer the question but understand math a little more in an intuitive manner

[next spruce]

OMG THANKS SO MUCH ISOEJFOSEFJ

I've been stuck on this question since yesterdau

[nimble mulch]

neat introduction 👍

[next spruce]

thanks for being patient

[surreal cave]

if you have any other question I'd be happy to help otherwise you can type .close to close your channel and I hope you have a very good evening mein fruend

[next spruce]

.close

[cedar kiln]

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[cedar kiln]

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[old monolith]

Hey all

[old monolith]

I'm a bit confused about something

Green's theorem is supposed to let me treat this as a double integral over a region

Simple enough

But every example solution I'm looking at is saying something about "having to break it into two curves, and find the line integral of each"

Is the whole point of Green's Theorem not to treat the problem as a double integral?

...

Am I supposed to be working backwards? Starting with the double integral, converting to a curve, and then solving for the line integral?

[tropic oxide]

yes, that's exactly what you're supposed to do here.

[old monolith]

Ahhhhhhhhh

[tropic oxide]

the double-integral interpretation comes directly from the fact that you're computing area

[old monolith]

And because, once its framed in the context of a line integral, it isn't smooth

[tropic oxide]

namely that the area of a region is the integral of the constant 1 function over it

[old monolith]

Excellent

Thanks!

[cedar kiln]

@old monolith Has your question been resolved?

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[polar flame]

could someone walk me through ∫ arctanx dx

[polar flame]

or help me get started with it. it's not homework, I'm just sitting I'm bed and I'm realizing I have no idea how to do it

[prisma gull]

do parts

[polar flame]

mm how would you split it in parts?

[hollow trail]

1 * arctan(x)

[polar flame]

🫢

that's right huh

[prisma gull]

Yep

[polar flame]

thank you

.close

[cedar kiln]

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[lunar lynx]

Hello,
I have $f(x) = 4x^{3} - cx^{2} + 2cx -4c$.
I am trying to show that for any real and positive value of c, f(x) doesn't have three positive, distinct, and real zeros.

I have narrowed it down to showing that $f(m)\cdot f(n) > 0$, where m and n are real zeroes of the f'(x). I have already ascertained that m and n are both positive for f'(x) under restrictions on c. For real zeroes of f'(x), we have constricted c values to be greater than 24.

Now, doing this last part seems highly computational(since m and n are in terms of c, and are in form of roots that we get from quadratic formula) so i have, obviously, not proceeded. I was hoping if someone could enlighten me with a way of making it simpler.
Any other way of approaching the problem is welcome as well. Overall, i appreciate any useful input whatsoever.
Thank you.

[wraith dagger]

Enemagneto

[cedar kiln]

@lunar lynx Has your question been resolved?

[violet flume]

i am bad at these kind of problems but

could you make some kind of argument with the inflection point of the cubic

since its so much easier to find exactly where it is

i know thats not really helpful

maybe something like this? https://math.stackexchange.com/questions/3440575/what-is-the-main-relationship-between-the-inflection-points-and-the-roots

just spitballing, hope you have luck

[muted bear]

The pattern of coefficients seems sus

maybetry multiplying by x+2

[lunar lynx]

original polynomial?

[muted bear]

Yes

[lunar lynx]

I brought a geometry problem into algebraic form so "sus" is inevitable. 😶

[lunar lynx]

Let me try.

[muted bear]

Dont xy problem please

Whats the original q

[lunar lynx]

Yeah. I can show that but i still wanted to learn this way as well. It's a proper translation of the problem with no missing info(i think).

@muted bear

[muted bear]

Interesting

[lunar lynx]

Honestly, i'm bad at geometry so i couldn't think of anything else but to do what i did.

Woah. Multiplying with (x+2) seems good. Terms of x^2 and x vanished.

But i don't really see what to do with that.

[muted bear]

What polynomial do you get?

[lunar lynx]

$4x^{4} + (8-c)x^{3} -8c$

[wraith dagger]

Enemagneto

[muted bear]

Hmmm

Why do the roots need to be distinct?

[lunar lynx]

This was nice to learn but i can't see how that helps me here. Thanks, jan Niku, nevertheless.