help-13

[merry heath]

carry on

[crimson sedge]

you could memorize but i different question of same type would confuse you

[crimson sedge]

yup

.

uk what P(a intersection B ) means

[merry heath]

is this correct

[crimson sedge]

....

give me a sec

Nope

That P(A|B) = P(A intersection B)

thats not P(A|B)

thats P(A intersection B)

[merry heath]

can you visualise it

the intersection one

[crimson sedge]

...

me?

[merry heath]

anyone

[crimson sedge]

that depends on question

and most of the times, yes

[merry heath]

short answer I do not

[crimson sedge]

..

[merry heath]

oh

[crimson sedge]

i do not understand what youre trying to say

nvm

[merry heath]

is the intersection just P(a) X P(b)?

[crimson sedge]

have you learnt about sets yet?

[merry heath]

yes but I don't necessarily remember everything

[crimson sedge]

that is only when the given sets are independant

not always

[merry heath]

what does this symbol mean?

[crimson sedge]

a subset

B is a subset of A

for the second one

it is true only when two events are independant

[merry heath]

and how do we know that it is independant?

[crimson sedge]

.

[crimson sedge]

it will be given in the question

[crimson sedge]

this

[merry heath]

nevermind I see it now

[crimson sedge]

P(a int b) = P(a)xP(b)

[merry heath]

I fully unstand ii)

[crimson sedge]

..

nice

[merry heath]

okay but honestly

[crimson sedge]

for third one

[merry heath]

still blur on i)

hold on

let me send something

[crimson sedge]

Kay

[merry heath]

I assume this is correct yes?

[crimson sedge]

.

thats the very first thing i toold you

[crimson sedge]

Yes it is correct

.

this happens when

[merry heath]

I was going off this formula

[crimson sedge]

..

[merry heath]

so does that mean this is true?

[crimson sedge]

yes

[merry heath]

and if yes, this is what I had trouble understanding

[crimson sedge]

damn

think ab it

[merry heath]

I have

still no idea

[crimson sedge]

dm if you want to

imma go and eat for now

or

<@&286206848099549185>

[merry heath]

alright ty for helping so far

[crimson sedge]

btw for the third one

they have to be mutually exclusive events

so

P(a int b) = 0

and thus the answer is zero

[merry heath]

got it

[crimson sedge]

nice

[merry heath]

alright I think I got it ty

.close

[cedar kiln]

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[crimson sedge]

my pleasure

[cedar kiln]

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[whole adder]

,rotate

[whole adder]

!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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[whole adder]

1

ty

forgot how to do this

[clear umbra]

gaussian elimination has something to do with matrices

so you want to make those equations into a matrix

[whole adder]

yea ik

but I think u have to sub t or sth

bc three unknowns but 2 eqns

[clear umbra]

no?

[lusty grotto]

do u have the system written down as a matrix?

[whole adder]

uhhh

how wrong is this

[clear umbra]

why did your first row change

[whole adder]

idk help

[lusty grotto]

what did u do to get the second matrix

[clear umbra]

the elementary operations used^

[whole adder]

row echolean (can't spell)

[lusty grotto]

yeah can u tell exactly what u did?

[whole adder]

i think R2 - 3R1

then R1 - R2

does this make sense

[lusty grotto]

yes it makes sense

[whole adder]

so then what do i do next....?

🥲

[lusty grotto]

R1 - R2 should be
1 - 0 -2 - (-1) 1 - 1 3 - 1
1 -1 0 2

[whole adder]

huhhh

I don't understand R2

[lusty grotto]

this is okay

u subtracted 3R1 from R2 and u get R2 as u have written in the second matrix

[lusty grotto]

for this u must subtract the R2 u just got, from R1

[whole adder]

ok, what's the next step then?? 😶‍🌫️

[whole adder]

done

[lusty grotto]

but uve done the subtraction wrong

[whole adder]

really

[lusty grotto]

this is what it must be

[whole adder]

nah i just did -3 wrong in R1

idk cuz this is from my tutor and i haven't learnt from school yet

so im just kinda fonfused

confused

[lusty grotto]

yes -3 is wrong, that must be -1 there

[whole adder]

yes but i don't see anything wrong with my R2

[lusty grotto]

R2 is correct

[whole adder]

anyways let's say i got to ur step, what to do next?

[lusty grotto]

so u have -y + z = 1, and then x - y = 2

[whole adder]

it says hence tho

thanks anyway

[whole adder]

oh so I don't get numbers

[lusty grotto]

u dont because there arent enough equations

[lusty grotto]

so u can let z = t or something, like u said here

and then in terms of that z, u can solve for x and y

[whole adder]

how do i use it to do (b)

it says "Hence"

meaning i must use ans from (a)

wait OK I got it

thanks @lusty grotto

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[barren parcel]

I’m going to take a test today and I’m going to ask here because it’s a lot easier to get explained things but what are functions they teach you originally like exponential linear and all that and how do you change the y-intercept and all that on the equation

[brazen dagger]

You should not ask things here during the test. I think that's against the rules

You can clarify after test

[barren parcel]

Im going to

[brazen dagger]

[barren parcel]

It’s 6:33 am school starts at 7:30

😭

[barren parcel]

?

[brazen dagger]

^^

[barren parcel]

It says I’m going to take a test

And that test is in like

An hour and a half

[brazen dagger]

You can ask doubts before your test starts. But during the test you can't ask questions here as that can be considered cheating

[barren parcel]

Yeah the test hasn’t started dude 😭

[brazen dagger]

Then ask your doubts instead of wasting time

[barren parcel]

??

What are the main functions and how do you change them

Or more so

Translate them

[barren parcel]

Referring to the first algebra ones such as linear, quadratic, and expontential

[barren parcel]

⁉️

[clear umbra]

wdym translate them

[brazen dagger]

What you are asking is broad , you gotta be more specific

[barren parcel]

Up 2 right 3

Like if I had an exponential equation

Or I’m not sure of another example

And I were to want to make the y-intercept down 2 and translate it right 3

How would I change the equation to do that

[brazen dagger]

Ohhhh stretching graph

[clear umbra]

to change the y intercept, you add the function with some constant

so f(x) + 2 means up 2

horizontal movement you change its input

f(x-2) means right 2

f(x+2) left 2

[barren parcel]

So if I were to want to change it it would be

f(x)=2^(x-2)+2

[clear umbra]

so i guess thats for an exponential function 2^x

[barren parcel]

Yes

[clear umbra]

thats moving it up 2 and right 2

[barren parcel]

Alright thanks that helps is it the same for the other equations?

[clear umbra]

yes

[barren parcel]

Alright thanks all I wanted a refresher on 🙏

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[worn moss]

Hey! Sorry To Ask some simple questions, but I cannot understand questions (2), (3), and (4) Can somone help me?

[hot crag]

construct one line parallel to l and m for each of the intermediate angles i guess

[worn moss]

yeah, i got that, but i dont know what to do after that.

[hot crag]

use the properties of alternate interior angles

and cointerior angles

[worn moss]

ah i remember

[cedar kiln]

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[sonic thicket]

I dont understand the different between these two functions

[sonic thicket]

Please ping when reply

[void sand]

well, one of them is not a function

[sonic thicket]

Should I think of it like a piecewise function?

Arent both like that though

Im confused on why the graph changes depending on where the modulus is

<@&286206848099549185>

[cunning sparrow]

well your absolute function is on different variables

[sonic thicket]

How would I go about plotting it on my own

If its |y|

|y| = -3

[cunning sparrow]

you can think of it as a piecewise function

[cunning sparrow]

this is not

possible

by definition of the absolute function you wont get anything

[sonic thicket]

What would the ranges be?

[sonic thicket]

You are right, the function isnt negative there

But, how are we getting the negative value

[cunning sparrow]

sorry, I dont really understand what you are saying

like the negative value of y?

[sonic thicket]

Should i rewrite it as 1 - |y| = x?

And when y is < 0

it becomes 1 + y = x

?

[cunning sparrow]

yea in essence that works as a piecewise function

so for y<0, y=x-1, and for y=>0, y=1-x

[sonic thicket]

Would I be able to make it piecewise on the basis of x? like when x > c, y = 1 - x or something?

[cunning sparrow]

you can see that the x-1 and 1-x are conjugates

[cunning sparrow]

yes you can

just put a constant inside the absolute function

so abs(x-c)=something

if x=>c, x-c is non negative

[sonic thicket]

I meant when $|y| = x - 1$, Could I make this a piece wise on the basis of x, or is that not possible?

[wraith dagger]

Adarsh

[cunning sparrow]

hmm I dont think so

your piecewise is on the basis of y here

[sonic thicket]

I see

Thanks a lot!

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[sonic flicker]

a, b, c are linearly independent vectors.
r = xî + yĵ + zk̂. Express r as a linear combination of a, b, c.

well I did
r = αa + βb + γc

now let n1 = b x c n2 = a x c n3 = a x b
so
α = r⋅n1/a⋅n1
β = r⋅n2/b⋅n2
γ = r⋅n3/c⋅n3
and finally
r = (r⋅n1/a⋅n1)a + (r⋅n2/b⋅n2)b + (r⋅n3/c⋅n3)c

now my teacher said that from this result we can derive the cramers rule or this is basically the cramers rule, which is the thing i dont understand
like, am i supposed to put n1 n2 n3 and then expand the whole thing
but even so how is this related to cramers rule

[cedar kiln]

@sonic flicker Has your question been resolved?

[sonic flicker]

<@&286206848099549185>

[cedar kiln]

@sonic flicker Has your question been resolved?

[native spade]

yes

https://cdn.discordapp.com/banners/552979169667973139/a_77acf61c999d9158d903fdbccf6b85eb.gif?size=512

https://cdn.discordapp.com/banners/774987205721980928/a_fc64e99ff20d4c55e0108cf417ed25ae.gif?size=512

[sonic flicker]

what

[cedar kiln]

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[crimson sedge]

what is the probability that the 50th card drawn from a deck of 52 cards is the 4th ace?

[crimson sedge]

i did probability of getting 3 aces in 49 cards = (4c3*48c46)/(52c49)

and then out of the rest 3 theres 1/3 chance of it being ace

so multiplying it comes close to 1/5

but my question is.. the probability of 50th card being an ace itself is 1/13?

so shouldnt the odds of that being the last ace be lesser

[drifting marlin]

,w (4C3)*(48C46)/(52C49)

[wraith dagger]

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[drifting marlin]

useless

,w binom(4,3)*binom(48,46)/binom(52,49)

[wraith dagger]

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[drifting marlin]

this times 1/3 is about .07, less than 1/13

[crimson sedge]

OH thank you! also last question can i just multiply the 1/3 with it? are the events independent?

[drifting marlin]

it's less about independence and more that there's some hidden conditioning being done

what you're calculating is P(3 aces in first 49)*P(50th is an ace | 3 aces in first 49)

you can read the original question as P(3 aces in first 49 and 4th ace on the 50th) and then you're just applying the definition of conditional probability

[crimson sedge]

got it thanks ☺️

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[dapper quarry]

Hello, i need help

[dapper quarry]

why are my answers incorrect?

[drifting marlin]

your first box has 2 pis

[dapper quarry]

I think the first one is a mistype

how about the second and 3rd one?

[drifting marlin]

those i have no idea but i imagine whoever wrote the questions messed up

oh wait, lol i can't read

you're supposed to answer in cm, not m

[dapper quarry]

ohh

so im supposed to do conversion

r = 200 cm
h = 200 cm
?

That was correct, Nice. Thanks:)

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[floral tundra]

I'm not sure how to do this

[last mango]

Do you know the condition for a quadratic equation to have distinct real roots?

[floral tundra]

yeah

the

discriminant

must be 0 or greater

wait

(k+1) is supposed to be the coefficient of x^2?

[last mango]

Yes

[floral tundra]

but

oh

I see

[last mango]

Notice the roots have to be distinct

[floral tundra]

meaning?

[last mango]

What happens when D = 0

[floral tundra]

1 root

oh distinct so

so like

it cant be the same number? ig

idk

[last mango]

Yea

[floral tundra]

oh I see

let me use the formula brb

wait

how am I supposed to use the formula

when its k+1

[crimson sedge]

a = k+1

[last mango]

What's the formula for D

[floral tundra]

b^2 - 4ac

[last mango]

Write down a, b and c first

And then plug it in

[floral tundra]

16 - 36k - 36

-36k - 20

[last mango]

What's b?

[floral tundra]

-4

-36 i meant

for k

[last mango]

You sure?

[floral tundra]

ohh

-4k

oh damn

so its

-36k^2-36k

?

[last mango]

Uhh I havent tried

[floral tundra]

but yh

thats what i got for the D

so what should I do next

[last mango]

Why first term negative

[floral tundra]

because I multiplied by -4k

[last mango]

Can you mention your a b and c

[floral tundra]

k+1

-4k

9

[last mango]

Ok

b^2 - 4ac yeah?

[floral tundra]

nah wait

[last mango]

Try computing it again

[crimson sedge]

dude

[floral tundra]

i got

[floral tundra]

16k^2-36k-36

[crimson sedge]

(-x)^even no., is positive
(-x)^odd no., is negative

Yes

correct

[crimson sedge]

find roots of k

[floral tundra]

Okay

[crimson sedge]

This would be equal to 0 or what?

They gave distinct roots

[floral tundra]

idk

thats the discriminant

[crimson sedge]

Yes

discriminant = 0

whats the thing?

1 root exists

[floral tundra]

how do we know it equals 0

[crimson sedge]

discriminant > 0

[crimson sedge]

if it is equal to zero

[last mango]

It's not

[crimson sedge]

yes

[crimson sedge]

.

[oak bane]

why is discriminant zero

[crimson sedge]

for distinct roots

[floral tundra]

omds

1 at a time..

[oak bane]

is the qustion to find the roots of the polynomial

[oak bane]

this

[floral tundra]

@last mango what should I do next

[crimson sedge]

It is not zero now, but if it is zero, then the roots of the equation are equal and real

[crimson sedge]

this is the discriminant

it must be greater than 0

[oak bane]

tyler is correct equate the discriminant to be >0

[last mango]

You found out the discriminant and you know you have to make D > 0 for distinct real roots

[floral tundra]

okay

[last mango]

Can you solve this: 16k^2-36k-36 > 0

[crimson sedge]

divide it by 4 on both sides to make it easier

[floral tundra]

4k^2 - 18k - 18

[crimson sedge]

then, x = (-b +- discriminant)/2a

[crimson sedge]

no

[floral tundra]

oh mb i divided by 2

[crimson sedge]

36/4 ≠ 18

[crimson sedge]

Yup

[floral tundra]

4k^2 -9k -9

[crimson sedge]

Yes and that is greater than 0

[floral tundra]

okay

[crimson sedge]

Yup

[floral tundra]

should I factorise?

[crimson sedge]

Yess

genius

[floral tundra]

🤣

ok ok

wait

[crimson sedge]

okay, do it on a paper and send a pic if you want to

[floral tundra]

ok so

[crimson sedge]

yes?

did you factorise it?

[floral tundra]

2(2k^2-9k-9)

[crimson sedge]

Bruh

HOWW?

how is that even possible???

[floral tundra]

wdym

[crimson sedge]

you want help?

[floral tundra]

I edited it

[crimson sedge]

still wrong

[floral tundra]

it isnt tho

[crimson sedge]

how is it correct?

if you take 2 common, you take it from all three of them

and why would you ever want to take 2 common

whats the use

ok ok lemme help you

[floral tundra]

whats common between

4 and 9?

[crimson sedge]

nothing ig

and you don't need to take common

did you not learn factorisation yet?

[floral tundra]

(4k+3)(k-3)

[crimson sedge]

=>4k²-9k-9
=>4k²-12k+3k-9
=>4k(k-3)+3(k-3)
=>(4k+3)(k-3)

[crimson sedge]

close

but how

[crimson sedge]

these are the steps

[floral tundra]

ik

[crimson sedge]

bruh

[floral tundra]

so k

is

[crimson sedge]

is?

[floral tundra]

-0.75

and 3

[crimson sedge]

no

range

[floral tundra]

-0.75 < k < 3

[crimson sedge]

ig

k>-0.75

k>3

[floral tundra]

k is larger than 3

ah

[crimson sedge]

ig

[floral tundra]

this is wrong

[crimson sedge]

yes

how

[floral tundra]

idk lol

[crimson sedge]

bruh

k<-0.75

[last mango]

Now that you factorized it

[crimson sedge]

why

idk

bruh

@last mango

it's correct tho

[last mango]

You have to find k for which (4k+3)(k-3) > 0

[last mango]

Yup

[crimson sedge]

divide by k-3

[last mango]

So the way you find the range of values for k is

You have 3 options
k < -0.75
-0.75 < k < 3
k > 3

[floral tundra]

ok

i was thinking the middle one at first

[last mango]

The nature of the sign of the expression would alternate between these ranges

[crimson sedge]

nvm

[last mango]

So you can pick any and find its behavior

[floral tundra]

how come

[last mango]

Hmmm

[crimson sedge]

magic

[last mango]

Do you know inequalities?

Like if ab > 0

then either both a and b are positive

Or both negative

[floral tundra]

yes

[last mango]

Ok so if you try to make both of them positive

It would mean 4k + 3 > 0 as well as k-3 > 0 yeah?

[last mango]

Remember we're trying to solve this

[crimson sedge]

dude continue

[floral tundra]

oh I see

yeah

[last mango]

Yeah so

[floral tundra]

oh I understna

stand

[last mango]

that can happen when k > -3/4 as well as k > 3 which means k > 3

[floral tundra]

since 4k+3 > 0

k > -3/4

[last mango]

3*

Yes!

Oh

k*

Ok XD

[crimson sedge]

yup

[floral tundra]

but thats stil wrong

[last mango]

No no

[crimson sedge]

nuh uh

[floral tundra]

i meant

[crimson sedge]

.

[floral tundra]

k greater than -3/4

[last mango]

It's not that straightforward

[floral tundra]

oh ok

carry on

[crimson sedge]

ye ye

[last mango]

Ok lets take an easy example

[floral tundra]

sure

[last mango]

Say (k-2)(k-4) > 0

Ok?

[crimson sedge]

ok

[last mango]

Now we have k< 2
2 < k < 4
k > 4

Three options

[floral tundra]

yh

[crimson sedge]

yes

[last mango]

If you want to just solve it boom then

[floral tundra]

what? 😭

[last mango]

You just need to know that the first range is positive, the second is negative and it keeps alternating

[floral tundra]

boom then

[floral tundra]

uh is this always the case?

[last mango]

yeah unless the root repeats

[crimson sedge]

wavy curve method

[last mango]

Yea ^^

But lets forget that ig

[crimson sedge]

Yup

[last mango]

We go to our example and try to understand in simpler ways hopefully

[crimson sedge]

Yup yup

[last mango]

So uh (k-2)(k-4) > 0

[crimson sedge]

@last mango topper.. genius.. prodigy

[last mango]

for ab > 0 we know it only happens either if you have:
a>0 as well as b>0
OR
a<0 as well as b<0
oki?

[crimson sedge]

oki

[floral tundra]

yes

[last mango]

@floral tundra Lmk if you struggle somewhere, I'd stop

[crimson sedge]

nah its alr

[floral tundra]

clear so far

[crimson sedge]

continue

[last mango]

Lmao dude

[floral tundra]

he didnt ask u 😭 🙏

[crimson sedge]

ik

[floral tundra]

carry on

[crimson sedge]

yup

[last mango]

So lets try both the cases
First we go with a>0 as well as b>0

[crimson sedge]

mhm

[last mango]

So k-2>0 AS WELL AS k-4>0

[crimson sedge]

mhm

[floral tundra]

k > 2

k > 4

[crimson sedge]

aisha genius

hi

doubtt

[last mango]

Yeah but we have to satisfy both the conditions

[crimson sedge]

yes

[last mango]

If you need help unrelated to this, refer to #❓how-to-get-help

[crimson sedge]

aww

[floral tundra]

carry on

[crimson sedge]

yup

[last mango]

If k needs to be greater than 2 as well as k need to be greater than 4

[crimson sedge]

yup

[last mango]

Then it means it has to be greater than 4

[crimson sedge]

yup

[last mango]

Cos let's say if we pick 3

Then 3 > 2 Sure

[crimson sedge]

it won't satisfy both

[last mango]

But 3 > 4 nope

Yea

[crimson sedge]

yup

[last mango]

It wont satisfy both right?

[crimson sedge]

k>4

[last mango]

So thats why it has to be greater than the bigger number of the 2 and 4

[crimson sedge]

the other way around, k<2

[last mango]

So k > 2 AND k > 4 => k > 4

Do you get this?

[crimson sedge]

yes

[floral tundra]

I get it

[crimson sedge]

carry on

[floral tundra]

now lets do the actual question

[last mango]

Oki

[crimson sedge]

Oki

[floral tundra]

Oki

[crimson sedge]

aww

[last mango]

So (k+0.75)(k-3) > 0

[floral tundra]

brb 3 min

sorry

[crimson sedge]

its alright

[crimson sedge]

.

k>3

[last mango]

So for the first case: ab>0 => a>0 as well as b>0

k > -0.75 as well as k > 3 and since we need to satisy both, it has to be k > 3

[crimson sedge]

.

yup

second case a<0 and b<0

[last mango]

for the second case: ab>0 => a<0 as well as b<0

k < -0.75 as well as k < 3 and since we need to satisy both, it has to be k < -0.75

[crimson sedge]

k<-0.75 and k<3

k<-.75

[last mango]

So yeah together we have the entire range of values for k
k < -0.75 or k > 3

[crimson sedge]

(-inf to -.75) U (3 to inf)

???

[floral tundra]

oh so

[last mango]

Yeah

[floral tundra]

i understand

[crimson sedge]

Yup

[crimson sedge]

good

[floral tundra]

wait

no i dont

[crimson sedge]

ok

lol

[floral tundra]

so it has to satisfy

both

[crimson sedge]

yup

[floral tundra]

but

it does

cuz

[last mango]

Which one are you talking about?

[floral tundra]

if k > -0.75 and k > 3

k can be bigger than both

[crimson sedge]

ab>0

a>0,b>0

a<0,b<0

[floral tundra]

so it cant be the first one

[last mango]

Only if it's bigger than 3

[floral tundra]

cuz that would be ab < 0

[floral tundra]

explain it again please

why is k smaller than -0.75

im confused bcs

we got the answer to be

[last mango]

Because remember

[floral tundra]

k is greater

[floral tundra]

yeah?

[last mango]

Greater than ?

[floral tundra]

-0.75

[last mango]

No

Ok so

[crimson sedge]

No

[last mango]

Can we do the (k-2)(k-4) for simplicity? then u just switch 2 with -0.75 and 4 with 3

[floral tundra]

ok

[crimson sedge]

ok

yo yo yo

lets do (k-2)(k+4)

[last mango]

So if we want ab to be > 0 then we either need
a>0 AND b>0
OR
a<0 AND b<0

[crimson sedge]

.

imma just stfu, let her understand

[last mango]

So now first case is a>0 AND b>0

Where k-2 >0 as well as k-4> 0

k-2>0 is true for all k > 2

k-4>0 is true for all k>4

[floral tundra]

that works

[last mango]

But since we want both of them to be true simultaneously

[floral tundra]

k just has to be greater than 4

[last mango]

yea

NOW second case

a<0 as well as b<0

Because that is also part of the solution right

[floral tundra]

yh

[last mango]

So we need k-2<0 as well as k-4<0

First is true for all k<2

And second is true for all k< 4

Since we want both to be true together

It's true for all k< 2

[floral tundra]

ok

[last mango]

So we found out we can either put any k which is less than 2

Or any k that is greater than 4

Right?

But not in between

Because if you put any k which lies in between 2 and 4 (like 3) then it would make the expression negative

But we want it positive (>0)

[crimson sedge]

🌊, for aisha

[last mango]

1st half

[floral tundra]

yeah

anything below 2

would work

[last mango]

2nd half

Together they make the entire solution

[floral tundra]

ok so

k has to be below 2 whilst also greater than 4?

[last mango]

(-inf,2) U (4,inf)

No no

They are separate conditions

[floral tundra]

okay

[last mango]

Like yk

[floral tundra]

but both part of solution?

[crimson sedge]

either this or that

[floral tundra]

ah okay

[last mango]

ab>0 is true if:

[floral tundra]

yhyh i get it

[last mango]

a>0 AND b>0
OR
a<0 AND b<0

Notice the OR

It's not AND

Yeah

[crimson sedge]

AND = intersection

OR = U (union)

[last mango]

yup

[crimson sedge]

.

[floral tundra]

ok

[crimson sedge]

U

union

if you're doubts are cleared

please do close the session

[floral tundra]

so

[last mango]

If you get this, i think you can also get the answer to your original question

[floral tundra]

now with our question

[crimson sedge]

bruh

[last mango]

It's fine. let them be haha

[floral tundra]

k>3 and b>-0.75

[crimson sedge]

.

[crimson sedge]

them?

me?

[last mango]

Aisha

[floral tundra]

I think he means me

but anyways

[crimson sedge]

ohh

[floral tundra]

look

[floral tundra]

so this

[last mango]

No

[crimson sedge]

b?

whats b

[floral tundra]

thats the thing u litteraly just did

this or that thing

[last mango]

Oh ok so are you solving first half? like the a>0 and b>0? if so, then yess

[floral tundra]

yes

[crimson sedge]

yes

[last mango]

k>3 AND k>-0.75 => k>3
OR
k<3 AND k<-0.75

[crimson sedge]

k>3

k<-0.75

[last mango]

Now what about the second one

[floral tundra]

what was the 2nd half again

sorry

[last mango]

here

[crimson sedge]

yup

[floral tundra]

thats the first half no?

[crimson sedge]

both

[floral tundra]

okay

[last mango]

There are 2 lines

One above AND is solved

One below is left

k<3 AND k<-0.75 this one

[floral tundra]

so

k < 3

[crimson sedge]

no

k<-0.75<3

k<-0.75

[floral tundra]

ok

[crimson sedge]

yup

[floral tundra]

k has to be bigger than 3

tho

[last mango]

Thats another part

So look

[floral tundra]

OH

[crimson sedge]

k is bigger than three(first part)

[floral tundra]

so there is 2 parts

ok listne

listen

[last mango]

We have two terms

ab

[floral tundra]

2 parts

[crimson sedge]

yup

[last mango]

either make both of them positive

or both negative

[floral tundra]

1 part the solution is k>3 and the other is k<-0.75

[last mango]

YES!

[floral tundra]

too easy

[last mango]

But the final answer

[floral tundra]

😭

[last mango]

They have to be mentioned with a union

[crimson sedge]

Yup

[last mango]

(-inf,-0.75) U (3,inf)

[crimson sedge]

true

[last mango]

Union means it can belong to either of the ranges

[crimson sedge]

yup

[floral tundra]

okay so

[last mango]

inf = infinity

[crimson sedge]

open brackets only

[floral tundra]

did u really just

omds

ok so

whats next

[crimson sedge]

thats it

thats the range

of k

for the eqn to have distinct roots

[last mango]

Uh thats the answer basically lol

[floral tundra]

this?

[last mango]

yup

[crimson sedge]

yup

[floral tundra]

this is what

is in my book

[crimson sedge]

same

literally the same thing

[crimson sedge]

.

[floral tundra]

I just said that

and u said

"But the final answer"

...

.close

[cedar kiln]

Channel closed

Closed by @floral tundra

Use .reopen if this was a mistake.

[floral tundra]

thanks btw tho

[crimson sedge]

his pleasure

[crimson sedge]

well the common way of writing it is what he told

[cedar kiln]

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[wheat herald]

could anyone tell me what g'(x) is equal to ?

[sonic thicket]

Chain rule

Do you want to confirm your ans?

[wheat herald]

im not sure about what exactly am i supposed to do here

the f is annoying

[sonic thicket]

Can you differentiate f(x)?

[woven crescent]

im guessing since youre not given the equation for f (also i cant read... french i think?) youre just supposed to give the derivative of g(x) in terms of f(x)

or does that say at point 3?

[wheat herald]

Let me translate it

Deteremine the equation of the curve Cg at the abscissa point 3

[wheat herald]

yes

[woven crescent]

does it give you anything else?

[wheat herald]

only this

[cedar kiln]

@wheat herald Has your question been resolved?

[smoky cliff]

So

Use chain rule

And just leave f'(x) as it is

[cedar kiln]

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[azure violet]

can i get help?

[azure violet]

determine the limit

i found that the limit is 8
however
can i verify it by doing this=

this is h->0

but im wondering if i can add 1 and -1 to verify

the teacher wants us to verify

and not with derivative

can i please get help with this?

I dont have that much time before the due date

[true zinc]

no you can't

[cedar kiln]

@azure violet Has your question been resolved?

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[fossil berry]

Someone check whether they are correct for me.

[crimson sedge]

yeah

[fossil berry]

thanks!

.solved

[cedar kiln]

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[honest saddle]

Does $\mathbb{R}(i) = \mathbb{C}?$

[wraith dagger]

bboy

[crimson delta]

yes

[honest saddle]

cheers

.close

[cedar kiln]

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[cedar kiln]

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[static knot]

heelp with this pls

[cedar kiln]

@static knot Has your question been resolved?

[cedar kiln]

@static knot Has your question been resolved?

[cedar kiln]

@static knot Has your question been resolved?

[static knot]

.close

[cedar kiln]

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[fringe night]

why did my teacher not get 1/1+sin(x)

[fringe night]

in the video he was like "take the derivative of the top, then take the derivative of the bottom"

but that isnt how derivatives work

you have to use the quotient rule right?

[patent ivy]

He isn't taking the derivative of the entire fraction

He's using l'hopital's rule

[fringe night]

oh

ok thanks

.close

[cedar kiln]

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[cedar kiln]

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[wary sonnet]

I need a walkthrough on how to solve this please, if it can be solved. I have two points: (282.02 , 2.6096x10^-4) and (286.15 , 1.887122x10^-3).
Is there something I can do to create a logarithmic function that passes through both?

[wary sonnet]

Sup 👋

[cedar kiln]

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[gleaming silo]

if an entire quadratic equation is >= 0 means it either has no REAL roots or exactly 1 root

[gleaming silo]

am I right?

[warped coyote]

‘or 1 real root of multiplicity two,’ but yes

[gleaming silo]

.close

[cedar kiln]

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[austere hull]

this like ai which is dumb

[gleaming silo]

someone promoting it to me

[tepid sail]

not really AI has gotten pretty good

[warped coyote]

It’s still not completely reliable

[tepid sail]

ofc not, but neither are ppl

[austere hull]

still sounds like it is correct but has stupid mistakes lol

[warped coyote]

I find it is more confidently wrong than people are, especially in a semi public place like here where if you give a wrong answer someone will likely chime in

[warped coyote]

For example your ai response is actually wrong, but it looks convincing

In the third case if the discriminant of a parabola is >0 there are two distinct roots. There cannot be a repeated root

[cedar kiln]

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[hoary charm]

I have a group theory question

[hoary charm]

I'm trying to prove if n = 2, 4 , p^r, 2p^r then U(n) is cyclic

Ive looked at some proofs and the part I keep getting stuck on is

The justification for U(4) being cyclic

Particularly where this part came from

[cedar kiln]

@hoary charm Has your question been resolved?

[cedar kiln]

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[eager hemlock]

Let ( a_n = \int_{0}^{n\pi} x|\sin x| , dx ), where ( n = 1, 2, \cdots ), how to find the expression for ( a_n )?

[wraith dagger]

Scoria

[cedar kiln]

@eager hemlock Has your question been resolved?

[oak bane]

can u use integration by part, let u=x v'=sinx, im not sure if the absolute value is important

[patent ivy]

i think you can just find $\int_{0}^{n\pi} x\sin x , dx$ by IBP and then flip the signs every $\pi$

[wraith dagger]

Luke

[patent ivy]

Try finding $\int_{(n-1)\pi}^{n\pi} x\sin x , dx$ for odd $n$

[wraith dagger]

Luke

[patent ivy]

and flip the sign for even $n$

[wraith dagger]

Luke

[eager hemlock]

I don't quite understand [flip the signs for every pi] ..could you maybe elaborate a bit?

[cedar kiln]

@eager hemlock Has your question been resolved?

[patent ivy]

Sorry just saw this

When is x sin x positive?

It should be positive from 0 to pi, 2pi to 3pi, 4pi to 5pi

And so on

It should be negative from pi to 2pi, 3pi to 4pi, and so on

$\int_{0}^{\pi} x\sin x , dx-\int_{\pi}^{2\pi} x\sin x , dx+ \int_{2\pi}^{3\pi} x\sin x , dx-\cdots$

[wraith dagger]

Luke

[patent ivy]

Try evaluating each term

[eager hemlock]

$\int_0^{\pi} x \sin x , dx = \left. -x \cos x \right|_0^{\pi} + \int_0^{\pi} \cos x , dx = \pi$

[wraith dagger]

Scoria

[patent ivy]

Do you see why we keep alternating between adding and subtracting

[eager hemlock]

ye since |sinx|

$\int_{k\pi}^{(k+1)\pi} x (-\sin x) , dx = -\int_{k\pi}^{(k+1)\pi} x \sin x , dx.$ for k odd

[patent ivy]

Yep

[wraith dagger]

Scoria

[patent ivy]

Now find $\int_{(k-1)\pi}^{k\pi} x\sin x , dx$

[wraith dagger]

Luke

[patent ivy]

For k odd and k even

[eager hemlock]

$\int_{(k-1)\pi}^{k\pi} x \sin x , dx = -x \cos x|{(k-1)\pi}^{k\pi} + \int{(k-1)\pi}^{k\pi} \cos x , dx = -[k\pi \cos(k\pi) - (k-1)\pi \cos((k-1)\pi)] + [\sin(k\pi) - \sin((k-1)\pi)]$

[wraith dagger]

Scoria

[eager hemlock]

@patent ivy I'm not sure if this is the right track..

[patent ivy]

Notice how the sins go away since they are multiples of pi

Let’s say k is odd

Then the expression becomes

[eager hemlock]

for k= odd, this integral is -k\pi+(k-1)\pi-0

[patent ivy]

I think it should be $(2k-1)\pi$

[wraith dagger]

Luke

[eager hemlock]

hmmm

why

[patent ivy]

cos(kpi)=-1

And cos((k-1)pi=1

[eager hemlock]

k\pi+(k-1)\pi+0

brain moment yes

[patent ivy]

How about for even k?

[eager hemlock]

for k= even, this integral is -[k\pi+(k-1)\pi]+[0-0]

[patent ivy]

So we get -(2k-1)\pi

[patent ivy]

So it turns out this becomes $\pi+3\pi+5\pi+\cdots$

[wraith dagger]

Luke

[eager hemlock]

yes ..

thank you

btw any other easier method?

Ok thank you so much luke ! @patent ivy

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[patent ivy]

To be honest I’m not sure 😅

[patent ivy]

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[vernal sentinel]

hi guys im watching a math video about differential equations, but i don't understand where they get y^2/2 on the fifth row

[vernal sentinel]

and the x^2/2 as well

[wraith dagger]

ren

[hot crag]

same goes for x and y

[vernal sentinel]

ohh that's the rule?

[hot crag]

it's just an integral-

don't start with differential equations if you haven't done integrals yet

[vernal sentinel]

ooohhh okay okayy

thank you so much!!

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[blazing ridge]

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[halcyon junco]

If i had to determine the domain of this graph, I would say (-3, -1,2) U (0,9)

[halcyon junco]

what do you guys think

[normal cipher]

!occupied