help-19

[frozen sail]

and p is given here

[viral bone]

one sec

alright ye it was in my notes, i thought it was

i had to go like i said in another chat on here

theres a lot of videos on it

https://www.youtube.com/watch?v=EblrLncv7a0

i think this is a good one

i gotta go tho so you might wanna open another ticket if you dont find the answer this way ^^

hope it helps

have a nice day ! <3

[frozen sail]

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[kindred umbra]

So I’m trying to understand transformations and I can’t find a good video explaining basic vertical stretch and compressions all the videos I see have curves like a quadratic but I’m working with y=x a v shape. Can someone help me understand how the a(vertical stretch) and compression affect a graph with pictures?

[thin kelp]

y=x is a straight line?

[kindred umbra]

No it’s a v

[odd magnet]

perhaps you mean y=|x|

[kindred umbra]

Y=|x|

Yes

[odd magnet]

my recommendation would be to fire up desmos or wolframalpha or whatever and just fiddle with plotting y=3|x| and y=|3x| and stuff

[thin kelp]

red line represents the original

well i should have left the 2 and 0.5 outside the | | but whatever

you see that the 2 essentially doubles the y value at each point

[kindred umbra]

So it adjusts the y intercept?

[thin kelp]

not the intercept

all Y values

look at x=1

[kindred umbra]

OH

So it would raise all by the stretch value so I can place a dot on a raised y and then bring it back to the vertex. What about horizontal

Is there a way to display that in that program

I can’t seem to have the bars around the x on my calculator

[thin kelp]

might be called 'abs'

short for absolute value

[kindred umbra]

Where do I find the bars to place?

[thin kelp]

impossible to tell with that photo

[thin kelp]

yeah looks like u dont have it

[kindred umbra]

Would you mind showing the compression on your program?

[thin kelp]

vertical compression?

[kindred umbra]

Horizontal

So is there a vertical stretch and a vertical compression or is it just one of those?

[thin kelp]

ah so i actually messed up a horizontal dilation looks like |ax| and a vertical dilation looks like a|x| but in this situation they both do the same thing

if you look at the green line thats a vertical compression, all the Y values are being halved

a horizontal compression can be determined by 1/a

[kindred umbra]

So when reading a graph how do I tell the difference? If they are the same

[thin kelp]

they wont always be the same

if its outside the brackets its a vertical dilation but if its within the brackets its a horizontal dilation

[kindred umbra]

Can you show one with an equal How do I tell if it’s a stretch or compression. Is a stretch just only used for horizontal?

Also I’m talking about when reading a graph

You can’t see the equation how would I determine that

[thin kelp]

so look at this photo a vertical stretch, stretches the values, if you look at the blue line, its just the red line but stretched, a compression flattens the graph, if you look at the green line it is just the red line but flatter

both stretch and compression are used for horizontal and vertical dilations

if you are giving a graph without an equation, it will show the original equation and what it looks like after the strech or compression,

or it will show you a graph and from memory you have to remember what the original looks like and then determine if its been stretched or compressed

[kindred umbra]

So when the v gets lower that’s a compression but when it gets taller it’s a stretch?

Blue is a stretch and green is compression?

[thin kelp]

so onto horizontal compressions and stretches, like i said you can determine it by 1/a where 'a' is the coefficient of x. If you look at the blue line the 'a' value is 2 so it have a dilation of 1/2 which means that every Y value is being pulled closer to the Y axis.

[thin kelp]

correct

[kindred umbra]

Awesome got it. Thank you so much

[thin kelp]

all good

[kindred umbra]

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[lucid quiver]

help a homie out 🙏

thanks

[odd magnet]

I'd probably just jam this into the quadratic formula

[lucid quiver]

didnt work 😭

[odd magnet]

Didn't work how?

It'll be messy but I imagine a bunch of it will cancel down.

[wooden python]

there is a sneaky shortcut here actually

[lucid quiver]

pls send\

[wooden python]

notice the sum of the coefficients is 0

[lucid quiver]

kinda yea

[wooden python]

this lets you say that a particular value of x is a root

wdym "kinda"

there's no "kinda" here

[lucid quiver]

like as in x2 = 0?

[wooden python]

??

no that isn't what i said

[lucid quiver]

Ill do my research

pls keep going

[lucid quiver]

wait no I get in now

[lucid quiver]

it doesnt, cuz square root ruins it

[violet tulip]

Open the brackets see what you get

Ann is right, the sum of coefficents (which are (b-c) (c-a) and (a-b)) is 0

b-c+c-a+b-c = 0

Thats kinda hinting that its better to open the brackets and take in common some stuff which should make life easier

[wooden python]

because the sum of the coefficients is 0, you ||immediately get that x=1 is a root||

[lucid quiver]

is that just a fact?

[lucid quiver]

right ok

[wooden python]

you should verify it for yourself

[lucid quiver]

answer is right

[wooden python]

DO NOT take my word for it

[violet tulip]

Also do this first. (b-c)+(c-a)+(a-b) = 0 => c-a=-(b-c)-(a-b)

Should help

[wooden python]

unnecessary

[violet tulip]

Necessary if he is going to do it my way, otherwise he wont get anything useful

[lucid quiver]

just got it, u had to expand, factorse then factorise again to factor x-1

thanks @violet tulip @wooden python

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[violet tulip]

Well I guess we all solved it somehow xd

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[mystic saffron]

is this fine to do? cross multiplying the denominator to the other side then getting the derivative?

[mystic saffron]

if so why does my answer differ a lot from the answers i get in, say derivative-calculator.net?

[violet tulip]

What would be the derivative of the right hand side if you didnt multiply and divide by the denominator?

[mystic saffron]

this is from the site mentioned

[hybrid ginkgo]

what happened to your y^2 on the lefthand side in your first line?

[mystic saffron]

0?

wait do you mean the right hand side after the equal sign?

[hybrid ginkgo]

yes the first step

[mystic saffron]

here?

[hybrid ginkgo]

yes

[mystic saffron]

starting from the second line i took their derivatives

x² to x

2y to 2yprime

is that what you meant? sorry if i misunderstood https://cdn.discordapp.com/emojis/575016270135558153.png?size=48

[hybrid ginkgo]

its where the y^2 from the very top goes. also isnt the derivative of x^2, 2x? or did you simplify after that?

or is that not part of the problem?

[mystic saffron]

OH

sorry

yes please ignore that

its not part of the problem

[hybrid ginkgo]

alright.

[mystic saffron]

so uh, can i do this?

[hybrid ginkgo]

so you swithc it around and the derivative of both sides. how did x^2 become x?

im pretty sure you can do this strat sure.

[mystic saffron]

i took its derivative with the power rule?

OH SHIT

FUCK IM DUMB

[hybrid ginkgo]

should be 2x yeah

[mystic saffron]

bro my head just

thank you

i'll try to reanswer this

my mind farted omg

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[thin kelp]

not ur only mistake

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[graceful sand]

Can someone tell me what im doing wrong in this long division problem? Im trying to find the gcd of those 2 polynomials

[graceful sand]

well, calculator says im right, however the gcd should be 1

how

[lucid quiver]

what happened to the n here?

[graceful sand]

wdym the n?

[lucid quiver]

is this x or n?

[lucid quiver]

?

[graceful sand]

1

its the remainder of the division above that

2x+1

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[scarlet grove]

i need to find the resultant force and i have no idea where to begin

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[scarlet grove]

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[sacred jacinth]

can anyone help me find a?

[sacred jacinth]

i know this has to do with the definition of the derivative but im not entire sure what f(x) is

[old hollow]

f(x) is the function you're deriving

[sacred jacinth]

would f(x) be sqrt(x)

[old hollow]

Hmmm I don't think you're supposed to use integer as the x when you're doing definition of derivative

If it's sqrt(x) then it'll stay √x instead of 2

[sacred jacinth]

im just tryin to relate it to the formula below it

[prime loom]

Remember the limit will only give you the derivative of a function at a point with respect to a

[sacred jacinth]

so i have to try and figure out what x-value that point has?

[prime loom]

Yes or in other words

f’(a)=sqrt(2)

Or maybe I’m confusing things but I think that’s right

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[sacred jacinth]

i had to multiply the numerator and the denominator by the conjugate of the numerator and solve it like a normal limit

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[olive knot]

when do i use polar form to get limit of a function ? is it only when (x,y) -> (0,0) ?
can i use polar coordinates if (x,y) -> (a,b) where a,b != 0 ?
if yes can u give an example and show how ?

also how can i be sure if there is path that i am not taking whose limit is different from the previous paths?
like is path 1, path 2, path 3 ... so on give me same limit, but there could exist infinite paths right ? what is one in those infinte paths give a different limit ?
how can i be sure ?

[olive knot]

is there a intermediate value theorem for multivariable calculus ?

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[past temple]

Can i get help with this

[quasi sparrow]

Use u substitution

[past temple]

Ok

what would I substitute

the e^x?

[quasi sparrow]

Try it

[past temple]

ok

it didn't do anythinjg

[fervent hound]

what's the derivative of f'(e^x) ?

[past temple]

e^x is the same even if we get the derivitive

ok

oh

the whole thing

9?

[fervent hound]

so u don't even need to do u-sub

[past temple]

?

so du is 9?

No U-sub?

[teal pagoda]

Guys, i need help. What is integral of x sin x cos^2 x dx. Please

[sullen ferry]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[past temple]

isn't it a definitive integral?

[quasi sparrow]

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[fervent hound]

oh srry

[past temple]

the e^x just dissapears?

[fervent hound]

u need to find antiderivative (indefnite integral) of that function first

then use fundamental theorem of calc

[past temple]

The antiderivitive of e^x is e^x

[fervent hound]

substitute u=e^x

it's easier that way

[past temple]

ohh ok

let me try it

ok

I solved for dx = 1/e^x

Which cancels out the e^x

[quasi sparrow]

You should have a du somewhere

[past temple]

yeah

I moved it to the side

(1/e^x)e^x(f''(u))du

e^x cancels out

leaving me with f''(u) du

[fervent hound]

can u solve the integral now ?

[past temple]

FTC

F'(1) - F'(0) ?

Oh

Am i supposed to change the numbers on the integral?

the integral is f'(u)

oh

[fervent hound]

yes

[past temple]

Im lost on that part

[fervent hound]

$u=e^x \x=1 \implies u=e^1=e$

[clever fjord]

Adam Chebil

[past temple]

ok

so for 0 it would be 1?

it's 0 to 1

1 = e

[fervent hound]

do the same for 0

[past temple]

0 = 1?

e^0 = 1

f'(e^x) FTC: 1 to e

The x is what gets substituted right?

[fervent hound]

$\int_{1}^{e}f''(u) \dd u=f'(e)-f'(1)$

[past temple]

f'(e^e) - f'(e^1)

ok

Yeah got that

and then I put u in there

and then do ftc

[fervent hound]

why e^e ??

[past temple]

once we do ftc we plug in the values e and 1 into f'(e^x)

u = e^x

[clever fjord]

Adam Chebil

[past temple]

we don't plug in u back?

[fervent hound]

u do that in indefinite integrals

[past temple]

ohhh ok

thank you very much

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[plain vigil]

Let X be a random variable uniformly distributed on ] 1, 1[. Find density
of the random variable |X| and calculate P(|X| > 1/2)

[grizzled tide]

am i tripping or is ]1,1[ just only 1

[plain vigil]

]-1;1[ sorry about this mistake

[grizzled tide]

do you know how to do density of random variable

specifically for a uniform distribution

[plain vigil]

The integral of
f(x) from -1 to 1 should equal 1?

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@plain vigil Has your question been resolved?

[grizzled tide]

yes, and it should be uniform in this case

aka its a flat line

you dont need any calc to do this probem, you shoul dbe able to just find the line

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[clear tree]

Hello please I need help in section "a"

more preciselly in "ii"

how did we sketch it...

<@&286206848099549185>

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[clear tree]

<@&286206848099549185>

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[river arrow]

Quick question, for the provided problem I have a question that says the doctor can choose both x and y, what is the doctors optimal choice of y given this assumption? I solved the partial derivatives wrt x and wrt y for Ud(y). Would I then solve for x ad plug into the partial derivative wrt to y to get an answer?

[odd edge]

@river arrow Has your question been resolved?

[river arrow]

<@&286206848099549185>

[quasi sparrow]

Try it and show your work

[river arrow]

This is what i have so far, I just want to make sure Im doing the correct process

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[civic folio]

if my root is -3/5 how do i rewrite it into a factor

[civic folio]

would it be

5x+4?

[fathom crystal]

If x=-3/5 causes y to be 0

Then (x+(3/5)) would be a factor

[civic folio]

ik

[fathom crystal]

Since -3/5 plus 3/5 would cause result in zero and multiplying by that would cause everything to be 0

Right the fraction

Sorry I got u wrong for a minute

[civic folio]

well

i know that its a factor

[fathom crystal]

5x+3 would be right

[civic folio]

i was just rewriting it

to make it look simpler

[fathom crystal]

Yes yes I'm sorry I misunderstood

Simplifying 5x+3 would go like
5(-3/5)+3
5(-3/5) = -3
-3 + 3
0

So we still get there in the end
The factor will indeed be 5x+3

[civic folio]

,rotate

[civic folio]

why is my answer wrong?

[distant nova]

it isn't

[summer cradle]

that looks right

[civic folio]

okay

symbloab

symbolab just said something else so i wasnt sure

ty

.close

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[civic folio]

,rotate

[civic folio]

is this right so far?

,rotate

[civic folio]

how can we tell when and where to involve “k” into the general equation?

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@civic folio Has your question been resolved?

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[steel notch]

how can i convert this imaginary number to cos representation −
2.389533i + 0.390128

[steel notch]

if you tell me a website or something that does this its fine

the original function is cos(3t+ 45)

i am basically working with this circuit but thats another story

[toxic monolith]

there is such a formula:

$z=ai=a\cdot e^{i\frac{\pi}{2}}=a\left( \cos\frac{\pi}{2}+i\sin\frac{\pi}{2} \right)$

[clever fjord]

Joanna Angel

[toxic monolith]

where z = ai, is a purely imaginary complex number

I suspect, that you should formulate your question more precise, too.

[steel notch]

im sorry

i am a bit confused too

so i need to convert that number into a real one

this is a phasor

what i have

idk if that makes sense

[toxic monolith]

there is also a formula, to convert complex number z = a+bi, into its trigonometrical form

$z=a+bi=\left| z \right|\left( \cos\varphi+i\sin\varphi \right)$

[clever fjord]

Joanna Angel

[toxic monolith]

where

:

[steel notch]

let me try it

thank you

[toxic monolith]

$\left| z \right|=\sqrt{a^{2}+b^{2}}\\\cos\varphi=\frac{a}{\left| z \right|}\\\sin\varphi=\frac{b}{\left| z \right|}$

[clever fjord]

Joanna Angel

[toxic monolith]

but i do not know whether it is useful for, you

probably yes

[steel notch]

so basically

well i got the right angle

but the wrong magnitude

i mean i got double the magnitude

maby i made a mistake during calcualtion

but seems strange

maby i forgot to divide by 2 somehwere

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[forest marlin]

im confused a little bit

<@&286206848099549185>

[stone yoke]

I have a doubt tho can I dm you ?

[forest marlin]

sure

<@&286206848099549185> last guy has gone

[pearl flume]

bayes theorem

P(A|B)=P(A n B)/P(A)

i think

[forest marlin]

right

[pearl flume]

no

no its P(A|B)=P(A n B)/P(B)

sorrye

[forest marlin]

what would be the aswer then, ive never done this before but got set it for homework

[pearl flume]

so P(A|B) means probability of A given B

[forest marlin]

k

[pearl flume]

P(A|B) = (7/16)/(12/16)

=7/12

[forest marlin]

7/12 is the answer

[pearl flume]

no

wait lemme think about it

[forest marlin]

np

[pearl flume]

oh wait sorry misread the problem

it asks for what the probability theyre not in A

so we want probability A' given that theyre in B

so P(A'|B)

P(A'|B)=P(A' n B)/P(B)

P(A' n B)=5/16

P(B)=12/16

P(A'|B)=(5/16)/(12/16)

P(A'|B)=5/12

[long patio]

the person is picked at group B

we have 2 conditions

inside group B and NOT inside A

[long patio]

this is the correct answer

[pearl flume]

i had the correct working...?

sorrye im not confident with probabilitye

[forest marlin]

so 5/12

[pearl flume]

90% sure

[forest marlin]

u know more than me, its like speaking chineese

[long patio]

(not related to the question just a formula)

[pearl flume]

ye okie im not on crack

[forest marlin]

ok here we go

OMG you sexy mf its rigth

wow thanks i even asked chat gpt before and he got it wrong

[pearl flume]

lmao

[long patio]

chatGPT sucks at math

[pearl flume]

dont ask chatgpt maths

theyre horrid

i feel theyre sometimes good at abstract problems

but generally dont rely on them

for maths

[long patio]

if you want to know the explanation and why the answer is like that feel free to ask

though you dont really need a formula for that as it's just venn diagram you can solve it by just looking at it

[forest marlin]

yep if your 200iq like youself xd

but thanks guys !

[pearl flume]

np

[long patio]

np

[forest marlin]

.close

[odd edge]

Channel closed

Closed by @forest marlin

Use .reopen if this was a mistake.

[forest marlin]

.reopen

[odd edge]

✅

[forest marlin]

crap is all i can say

i have another question the same ish

<@&286206848099549185>

[open knot]

just to clarify, shall it be read as 9 people went to Andy?

different countries have different ways of describing venn diagrams

or did 9+5 people go to Andy

[forest marlin]

yep 9 people went to just andys 5 both and 11 just donnas

[open knot]

kk

so under the restriction that a person went to Andy

how many people do you have that fulfill that restriction?

[forest marlin]

9?

[open knot]

well 9+5, since the people who went to both must have also gone to Andy

[forest marlin]

oh ye

[open knot]

so you have a total of 14 people

[forest marlin]

ye

[open knot]

now, how many of those 14 people did not visit Donna?

[forest marlin]

9

[open knot]

--> 9/14

9 out of the 14 people who went to Andy did not visit Donna

[pearl flume]

you can use the equation as well

we'll be using P(D'|A) meaning what is the probability of not visiting donna given that they picked andy's buffet

[long patio]

iam personally against using any equations for venn diagram

[forest marlin]

i think this one is different a lil easyer

[pearl flume]

i see

idk i prefer using equations

i cbb thinking rationally

[open knot]

but yeah generally this would be formally better

[long patio]

generally yes

not all questions will give you a venn diagram

[forest marlin]

realistically i dont see why i need to learn this just for 1 exam then never in my life am i going to do this

[long patio]

wait wrong picture

ok yeah if you don't have a venn diagram you will use different approaches

nvm

[forest marlin]

right thanks again bro

[long patio]

probability is straight forward, the best thing about it is that there are no trick questions
I wish you good luck in your exam

[forest marlin]

can we go over one more please

[long patio]

one more question?

sure

[forest marlin]

so here, in total 8 people had a starter correct

[long patio]

ok so

what is the sample space?

what do you think it is?

[forest marlin]

what does that mean a sample space?

[long patio]

all possible chances

whether right or wrong

[forest marlin]

that they ordered a starter

[long patio]

we have 4 blocks there

3 of them in the intersection between ordered a starter and order a dessert

meaning that those 3 ordered both
we have 5 people that orders only a starter
9 people that ordered a dessert
4 people that ordered neither

the sample space in this situation is literally all of those people inside the restaurant
whether they ordered starters or not

so basically the sum of all people

4+9+5+3

[forest marlin]

right ok i get that

21

[long patio]

we want people that didn't order a starter

ignore probability for now

think logic

[forest marlin]

k

[long patio]

how many people didnt have a starter at all inside the sample space (restaurant)

[forest marlin]

9+4?

[long patio]

exactly

[forest marlin]

ok

[long patio]

in this example we have no conditions

meaning that we don't pick from a specific group of people

we just pick from the entire restaurant

so what is the probability of the person not getting a starter?

[forest marlin]

ok

[long patio]

no problem if you don't know, I want you to be able to solve everything your own after that

[forest marlin]

13/21?

[long patio]

yep that is the correct answer

[forest marlin]

Acctauly?

[long patio]

yep thats literally it

[forest marlin]

oh right not to hard then

[long patio]

for the other examples above we had conditions

like a condition that we only pick from B for example

that means that the sample space for that condition is B itself, and not everything like the example above

[forest marlin]

okayyy

[long patio]

I hope that made things easier for you

[forest marlin]

it did thanks daddy teracura !

[long patio]

Please dont say that.

lmao

[forest marlin]

hehe xd

thanks 🥲

.close

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[gloomy cypress]

hello

[gloomy cypress]

does anyone know the correct ans to this

i ticked those two boxes and i got it wrong

[mystic saffron]

you have to understand what an identity is

"an identity is an equality relating one mathematical expression A to another mathematical expression B, such that A and B (which might contain some variables) produce the same value for all values of the variables within a certain range of validity"

do you think this holds true in your case

[gloomy cypress]

yes

so this is an identitiy

is that right

[mystic saffron]

yes

[gloomy cypress]

oh i see

and the reason is that it is an algebraically valid statement right

bc that seems to be the only one that makes sense

in this case

[odd edge]

@gloomy cypress Has your question been resolved?

[odd edge]

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[warped phoenix]

how can I identify from the graph

[odd edge]

@warped phoenix Has your question been resolved?

[odd edge]

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[floral orchid]

im not very good at doing this for a long time

my calculator gets flooded

[lime turret]

Whats the problem is coming in doing this question

[floral orchid]

i know the formula but it fills up my calculator

[lime turret]

?

[floral orchid]

500+(500/2)+(125/2)+(62.5/2) etc

i do that continuously

[lime turret]

Just apply the formula bro

Ain't it will be 500+500/2+500/4+...

On applying the formula it will be 500/(1-1/2)

Means maximum depth is 1000feet

[floral orchid]

my bad i wasnt really sure what r was in the equation

[lime turret]

R is common ratio

Which is 1/2

[floral orchid]

.close

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[odd edge]

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[mystic saffron]

If 7 cosecA-3cotA=7 then prove that 7cotA- 3cosecA= 3

[cobalt raptor]

first convert cosec to 1/sina

sinA*

[mystic saffron]

i did

[cobalt raptor]

1/sinA-3cosA/SinA

[mystic saffron]

yes

[cobalt raptor]

are you in class 10?

[mystic saffron]

yep

[cobalt raptor]

same bro

[mystic saffron]

indians ?

i applied the convert to sin cos trick

isnt working sadly

[cobalt raptor]

yeah im indian

[mystic saffron]

yes

noice

m in 12th xd

[cobalt raptor]

boards are from feb 23

[mystic saffron]

ah

good luck

[cobalt raptor]

10th class

[mystic saffron]

can u help me

[cobalt raptor]

pb 1 from 27 dec

how to get 90+ in 10th boards?

[mystic saffron]

ez

idk but can u help me solve that

just use ncert

dont go for other bullshit

yt lectures etc

and solve previous year papers

alr but pls solve the question :"")

[cobalt raptor]

i can do it

[mystic saffron]

pls

[cobalt raptor]

7cosecA-3cotA whole square = 49

[mystic saffron]

tried that 🥲

[cobalt raptor]

7
cosec
θ
−
3
cot
θ

7
(
7
cosec
θ
−
3
cot
θ
)
2

49
On simplifying, we get
49
cosec
2
θ
−
42
cot
θ
.
cosec
θ
+
9
cot
2
θ

49
⇒
49
(
1
+
cot
2
θ
)
−
42
cosec
θ
.
cot
θ
+
9
cot
2
θ

49
⇒
42
cosec
θ
.
cot
θ

58
cot
2
θ
....(1)
Now, let
7
cot
θ
−
3
cosec
θ

x

Thus
x
2

49
cot
2
θ
+
9
cosec
2
θ
−
42
cosec
θ
.
cot
θ
⇒
x
2

58
cot
2
θ
+
9
−
58
cot
2
θ
(using (1))
⇒
x
2

9
⇒
x

3

i did the answer

[mystic saffron]

goddam the lines

[cobalt raptor]

sorry

[wide jackal]

,, 7
cosec
θ
−
3
cot
θ

7
(
7
cosec
θ
−
3
cot
θ
)
2

49
On simplifying, we get
49
cosec
2
θ
−
42
cot
θ
.
cosec
θ
+
9
cot
2
θ

49
⇒
49
(
1
+
cot
2
θ
)
−
42
cosec
θ
.
cot
θ
+
9
cot
2
θ

49
⇒
42
cosec
θ
.
cot
θ

58
cot
2
θ
....(1)
Now, let
7
cot
θ
−
3
cosec
θ

x

Thus
x
2

49
cot
2
θ
+
9
cosec
2
θ
−
42
cosec
θ
.
cot
θ
⇒
x
2

58
cot
2
θ
+
9
−
58
cot
2
θ
(using (1))
⇒
x
2

9
⇒
x

3

[clever fjord]

naxtisy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cold urchin]

lol what is this

[cobalt raptor]

bot

[mystic saffron]

where did 49(1+cot2theta) come from ?

ohh

cosec

steps so hard to read

,, 7
cosec
θ
−
3
cot
θ

7
(
7
cosec
θ
−
3
cot
θ
)
2

49
On simplifying, we get
49
cosec
2
θ
−
42
cot
θ
.
cosec
θ
+
9
cot
2
θ

49
⇒
49
(
1
+
cot
2
θ
)
−
42
cosec
θ
.
cot
θ
+
9
cot
2
θ

49
⇒
42
cosec
θ
.
cot
θ

58
cot
2
θ
....(1)
Now, let
7
cot
θ
−
3
cosec
θ

x

Thus
x
2

49
cot
2
θ
+
9
cosec
2
θ
−
42
cosec
θ
.
cot
θ
⇒
x
2

58
cot
2
θ
+
9
−
58
cot
2
θ
(using (1))
⇒
x
2

9
⇒
x

3
TeXit
BOT,,

[clever fjord]

TOBI .')
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[mystic saffron]

.close

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[mystic saffron]

Do you guys have any idea on what theyre trying to say?

[open knot]

dym the theorem?

[mystic saffron]

mmhm

I dont get what theyre trying to say by "if 1 2 3 ... not satisfied"

[open knot]

if you consider any point of a graph, let's take the given graph sample above and look at x=6

if you try to get the left-wise limit, it yields 4

since f(x) approaches 4 as x->6 from the left

likewise f(x) approaches 8 as x->6 from the right

however these two limits are different

[mystic saffron]

Ahhh

[open knot]

from that follows that the limit doesn't exist

that'd be (1) and (2) are fulfilled

however (3) is not fulfilled

what they mean to express by "if 1, 2 or 3 is dissatisfied":

you'd first check whether the limit exists from the left. If not ==> there can't be a general limit at that point

[mystic saffron]

its only fulfilled when both are getting infinitely close to the same L no?

[open knot]

ys

[mystic saffron]

Ahh I see I see

[open knot]

it explains the procedure of showing that there is a definitive limit

[mystic saffron]

thank you so much

[open knot]

if it's undefined from the left => no limit

undefined from the right => no limit

defined on both sides but not the same limit => f is discontinuous at that point

meaning there is no limit

[mystic saffron]

where can I look more into that^^

like a yt video

what do I look up?

I think I get it somewhat but id like to have a very clear idea on it

[open knot]

hm I don't have specific channels in mind, but you can look for "left-hand / right-hand limits (Calc 1)"

[mystic saffron]

Thank you!

thats actually really nice because we can use calculators in the quiz

take care!

I appreciate it

[open knot]

thee too 🦇

[mystic saffron]

.close

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[gentle lantern]

I don't understand what I did wrong. I did (-6) - 6 = -12 to get the integral within the [-2, 2] interval, but it's incorrect (?) Why?

[mystic saffron]

Pls help me convert this.

[mystic saffron]

please use different channel

this one is occupied

K

[gentle lantern]

I did the exact same method with another problem, and it turned out to be correct. Is the WebWork broken or am I doing something wrong?

[mystic saffron]

its correct

can you check the solution ?

for this question

[gentle lantern]

I don't have the solutions with me, they're only published once the homework deadline is met. 😭 But thank you for the confirmation.

I think I'll email my teacher then.

.close

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[violet tulip]

Discrete math

[violet tulip]

Ignore what the problem wants from me

How do I calculate R^2?

I know that its R o R

but I dont know how to continue from here

[hasty dome]

(a, c) is a member of R^2 if and only if you have (a, b) and (b, c) in R.

[violet tulip]

[hasty dome]

So, if you have R = {(a, b), (b, c)}, R^2 = {(a, c)}.

Basically if R is a set of "jumps", R^2 is all the ways you can go by jumping twice.

[violet tulip]

wait

if a,b is form R then a+b is even

[hasty dome]

So, you can jump from a to b, then from b to c. That's two jumps, so a to c is in R^2.

[violet tulip]

same for b,c

but how does it turn into a,c

[hasty dome]

That's a generic definition.

[violet tulip]

so we consider from a to b one step

then same for b to c

and then we make a big jump from a to c

thats R^2

does it mean R^3 would be one more such jump?

[hasty dome]

Right, that would be any path of exactly three jumps.

[violet tulip]

right

okay how do I apply it to get R^2?

[hasty dome]

For your specific R, you have (any even number, any even number) and (any odd number, any odd number). So, when you start at an even number, you can jump to any even number, but no odd numbers.

I mean any in A.

And if you start at an odd number, you can only get to another odd number in one jump.

[violet tulip]

no why

why do we get any even number in A

[hasty dome]

Well, it says a + b is even.

If a and b are integers, their evenness must be the same.

[violet tulip]

so both should be even or both odd

gotcha

[hasty dome]

So, after two jumps, it's the same thing.

You can't start at an even and jump twice to an odd number.

[violet tulip]

yeah

so

[hasty dome]

So, let's pay attention to only even numbers.

[violet tulip]

no why

[hasty dome]

Because it lets us deal with one part at a time.

We'll deal with the odds later.

[violet tulip]

ah okay

[hasty dome]

So, you can do one jump to get from any even number in A to any other even in A.

[violet tulip]

yep

[hasty dome]

So, doing another jump won't improve that or make it worse.

[violet tulip]

its still even

[hasty dome]

Right, you start with any even in A, you jump to any even in A, then you jump to any even in A.

So, R^2 = R for the evens.

[violet tulip]

no

[hasty dome]

Why not?

[violet tulip]

Hold on

so when we do the jumps

we get from a,b to a,c

[hasty dome]

Well, we jump from a to b and from b to c.

[violet tulip]

yeah

thats two jumps

[hasty dome]

Right.

[violet tulip]

so technically b becomes c

if b is 2

what will c be

[hasty dome]

It can be any even number in A.

[violet tulip]

there is no next even in A

[violet tulip]

shouldnt it be the one thats a jump away from b?

[hasty dome]

All evens are a jump away from b.

[violet tulip]

so if it was 2, 4, 6

we jump from 2 to 4

thats one jump

but from 2 to 6 its two jumps

[hasty dome]

No, you can do one jump from 2 to 6.

That's because 2 + 6 = 8, which is even.

[violet tulip]

ah I see

how will it work in A then if 2 is the only even

its going to be always 2

[hasty dome]

Then, you can jump from 2 to 2.

Because 2 + 2 is even.

[violet tulip]

yeah

Okay go on

🎙️

[hasty dome]

OK, so if you can jump from any even to any even, doing two jumps doesn't add anything.

It doesn't matter what the middle even number is.

[violet tulip]

yeah

[hasty dome]

You can do 2 to 2 to 6 or 2 to 6 to 6 or whatever.

[violet tulip]

mhm

[hasty dome]

So, R^2 = R for the even numbers in A.

[violet tulip]

ehh

[hasty dome]

Does it make sense why?

[violet tulip]

sort of

Lets do it for odds too, should get better

[hasty dome]

OK.

So, one jump can go from any odd number in A to any other.

Because odd + odd = even.

So, it doesn't matter which odd you pick for the starting or ending.

[violet tulip]

7 is odd

9 is odd

9+7 is even

yeah

[hasty dome]

Right, so you can jump from 7 to 9.

Two jumps is like any odd to any odd and then from the second odd to any odd.

So, you start at any odd and you end up, after two jumps, at any odd.

But that's the same thing that one jump let you do.

[violet tulip]

so its basically the same as for evens

[hasty dome]

Right.

So, R^2 = R for everything.

[violet tulip]

so this is wrong

[hasty dome]

Yes.

[violet tulip]

after | it should say a+b is even

[hasty dome]

Right.

[violet tulip]

Gotta practice this more to remember

Thanks a lot man

all the best

[hasty dome]

You're welcome.

You too.

[violet tulip]

.close

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[olive burrow]

Can any of u correct for me a problem ?

[mystic saffron]

[mystic saffron]

Which one?

[olive burrow]

Wait I have to write the prob in English
Cs I have it in Arabic

[mystic saffron]

Yeah no problem

[olive burrow]

The numbers from one to 99 are written on the board.How many ways are there to choose 50 numbers so that at least two numbers whose sum is 100 and two numbers whose sum is 99? @mystic saffron

We can't choose the same number twice but we can choose three numbers two of them sum to 100 and two of them sum to 99

[mystic saffron]

Hmm

[olive burrow]

For example 50.49.51

[mystic saffron]

Let me think

We have to choose 50 numbers from 1-100

At once

And in those 50

There should atleast be two numbers whose addition equal 100 and , another pair of two numbers whose addition = 99 ??

[olive burrow]

I will write u the result I found if it os not similar to urs I will send u the explanation of my answer

[mystic saffron]

Yep

[olive burrow]

@mystic saffron

[mystic saffron]

Explain it, i dont think the result can be that big

[quasi sparrow]

,rotate

[olive burrow]

Ok I will try to write it in a paper

[mystic saffron]

Cant it directly be
(99C50) - 1
As (All possible combinations of selecting 50 numbers out of 99) - (the case where you take 1.2.3....50) as any pair addition cant be 100

[olive burrow]

But how can u be sure that all cases have two numbers sum to 100 and others to 99

[mystic saffron]

See

Two ways other selections can go -

Case 1) 1.2.3.4....49.51 skipping 50, then skipping 51 and so on

Case 2) 2.3.4....49.50.51. and so on

Its obviouss in case 2 and you always will have additions to 99 and 100

In case 1 , i dont know how to explain it over text

Yeah see

You'll always have a number greater than 50 in both the cases

@olive burrow yes or no?

[mystic saffron]

And its conjugate will always be present too

See

Now for a case

Where we have 75

There will be 49 cases where it will have a conjugate, ie, 25 {for the addition of 100 rn were discussing, for simplicity}

There will be one case tho, where 25 isnt present

There you just use 76+24 or 74+26

Brb

[olive burrow]

I think I got it thank u

[mystic saffron]

Uhh

By my method or there was some other result?

Just for the sake of curiosity..

[olive burrow]

Idk
But I still wonder if my method can be correct

[mystic saffron]

If you can explain it to me,i may be able to answer that

[olive burrow]

Ok I will do so just a minute pls

[mystic saffron]

Yeah

[olive burrow]

I just realized it is false

Thank u for the help

[mystic saffron]

Umhmm

Guessed so

[mystic saffron]

Youre welcome

.close if youre done

[olive burrow]

.close

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Channel closed

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Use .reopen if this was a mistake.

[olive burrow]

.reopen

[odd edge]

✅

[olive burrow]

@mystic saffron I just realized u have to substruct 2

Bcs

[mystic saffron]

Why..?

OH RIGHT

50.51.52.53....99

Thanks sorry

I missed that

We needed two numbers to make a pair

[olive burrow]

Yeh ig I didn't explained it like that

[mystic saffron]

What were you saying then?

[mystic saffron]

Umhmm

[olive burrow]

Same but just u see here
We have 49 pair in every sum
So if we are choosing 50 numbers we are absolutely choosing two of the same pair only two pairs that have different numbers I mean 99 is not included in the sums of 99 and 51

[mystic saffron]

Valid

Thats what i did

Think of pairs rather than numbers

[olive burrow]

But 50 too in not in the sums of 100 so we have to substruct 3

[mystic saffron]

Hmm?

Example of the 50 numbers you will take?

Oh you meant to say we cant take 50 twice to make 100 ?

So we need to subtract once more?

[olive burrow]

50 . 49 . 1 .2....... 48

[olive burrow]

Exactly

[mystic saffron]

Read this

1st case would be 1-50
2nd would be 49-99

Now if you take example,
48,50,51,52,....99

You can make 100 using 48+52

[mystic saffron]

Thats same as 1.2.3....49.50

We already subtracted that once

[olive burrow]

I got this yeh right

[mystic saffron]

As we are asked for combinations, not permutations

Permutation would have been not at all hard

Just multiply with 50!

Yeah

[mystic saffron]

Feel free to ask more

Any more doubts?

[olive burrow]

Just a moment I'm trying to find out why I have made the mistake by the method I used

[mystic saffron]

Sure

[olive burrow]

How 49-99
This case is true not false
Cs 49.50.51...99 satisfies the condition

[mystic saffron]

Right

Mb

Earlier i was thinking about 50-100

And just like the dumb person i am , reciprocated it to 49-99

Yeah so
99C50 - 1

[olive burrow]

Sorry I disturbed u

[mystic saffron]

Aye why

Dont be sorry you didnt

I was more than happy to help

[olive burrow]

That's kind from ur part
Thnx
Have a good day

[mystic saffron]

Night tbh

Its already midnight , i have a lot to do, it will be 3 am before i sleep

[olive burrow]

Yoo so good night
God help you inshallah

[mystic saffron]

You too!!

[olive burrow]

.close

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[cinder pilot]

im not sure how to start this

[tame kraken]

Probably de moivre's theorem

[cinder pilot]

any other suggestion

[tame kraken]

why not de moivre's?

It's a pretty standard approach to compute large powers of complex numbers

[green elm]

equivalently, write z in polar form so it's trivial to take powers

[cinder pilot]

ur right

[green elm]

in this case, you could first compute z^2 manually, and then take the result to the 10th power

(because z^2 is pure imaginary, so it's obvious what happens when you take it to the 10th power)

[tame kraken]

Oh nice!

[cinder pilot]

regarding exponents: say if i have (-1)^100 how would i know the sign of the solution will be positive or negative?

[green elm]

(-1)^2 is 1, right?

[cinder pilot]

yes

[green elm]

so if you have an even power, say 100, you can write the power as 2 x 50

then (-1)^100 is (-1)^2 to the 50th power

and (-1)^2 is 1

so it's 1 to the 50th power

which is just 1

in general, (-1) to an even power is 1, and (-1) to an odd power is -1

[cinder pilot]

got it so any number negative n to an even number as a the exponent will result in a positive n value and odd exponent will result in negativ n value

so something like (-2)^51 would result in a negative value

(-2)^52 would result in positive value

[green elm]

yes

simply because -2 = -1 times 2, and taking -2 to the 51 power is the same as taking -1 and 2 to the 51st power and multiplying the result:
$$(-2)^{51} = (-1)^{51}(2)^{51} = -(2)^{51}$$

[clever fjord]

Bungo

[cinder pilot]

ty

.close

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[marble vale]

Hello, if $f(x) = x - \cos(x)$, how do I transform the problem into a fixed-point problem using a function g to find the zero of f ?

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[clever fjord]

lilisworld

[marble vale]

can i just write g(x) = x + f(x)?

or which g might work?

[stark coyote]

fixed point would be f(x) = x wouldn't it?

x = x - cosx

[summer cradle]

she wants a zero of f, so a fixed point of h(x) = cos(x)

[marble vale]

yes thanks

but how do you find the h pls?

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[weak kiln]

can someone help with this question

[odd edge]

@weak kiln Has your question been resolved?

[fair prism]

!status

[odd edge]

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

[weak kiln]

1

i think for b both equations have to have the same gradient?

[fair prism]

That is true for both of them

If you want to make them have the same gradient, it would be a good idea to put them in the form y=mx+b

infinite solutions means theyre the same line. no solutions means parallel but NOT the same line

[weak kiln]

y= -3x/m + 5/m, y=(m+2)x/5 + m/5

is that right?

[fair prism]

first one is missing a negative sign, 2nd one needs to be completely rechecked. Treat (m+2) as a single thing

wont ever need to be split

[weak kiln]

y= -3x/m + 5/m, y=(m+2)x/5 + m/5

is that better

[toxic monolith]

I don't want to disturb , but for such systems of equations you can also use Cramer's theorem:

$\left{ \begin{array}{cl}
ax+by=e &\
cx+dy=f &
\end{array} \right.\\D=\begin{vmatrix}
a &b \
c&d
\end{vmatrix},\text{ }\text{ }\text{ }D_{x}=\begin{vmatrix}
e &b \
f&d
\end{vmatrix},\text{ }\text{ }\text{ }D_{y}=\begin{vmatrix}
a&e \
c& f
\end{vmatrix}\\1^{\circ }\text{ one solution: }\Leftrightarrow D\neq 0\2^{\circ }\text{ infinitely many solutions}\Leftrightarrow D=D_{x}=D_{y}=0\3^{\circ }\text{ no solutions}\Leftrightarrow D=0\wedge \left( D_{x}\neq 0\vee D_{y}\neq 0 \right)$

[clever fjord]

Joanna Angel

[weak kiln]

im so confused

[toxic monolith]

I gave you a ready-made recipe = procedure, perfect for linear systems with a parameter, like yours, they work very quickly in each case

[weak kiln]

.close

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[fringe jungle]

my teacher told me there is a problem with this, but can’t find it, she made it purposefully this way so i know there is a problem

[ocean bramble]

show the full instructions

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[cinder pilot]

would this be c?

[weary pelican]

Yes, computing the characteristic polynomial gets you (X-1)² + 6 = (X-1-isqrt6)(X-1+isqrt6)

[cinder pilot]

this one would be e?

I have a set of questions that ive completed but want to verify ive gotten the correct answer for

[weary pelican]

Post them all together for faster results

Yes it's e

[cinder pilot]

ok

36: c
37: a
38: a
39: d

40: e
41: b
42: e
43: c
44: e