help-27

[sand dove]

all real numbers?

an uncountable sum hardly makes sense

[ember patrol]

Pain

My end goal is to arrive at $f(x)\eval_{-\infty}^{\infty}= 0$

[woven radish]

KySquared

[sand dove]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[sand dove]

might require to know what f is

[ember patrol]

It’s not a problem I was given persay, I’m looking at Fourier transforms and was asked to verify why $\mathcal{F}{\frac{dx}{dt}} = -i\omega F(\omega)}$\
f is just an arbitrary function

[woven radish]

KySquared

[ember patrol]

So the last part I need to show is $xe^{iwt} \eval_{\infty}^{\infty} = 0$

[woven radish]

KySquared

[ember patrol]

I figured since e^iwt has magnitude 1, I only needed to consider x evaluated at those limits

And all I know about f is that $\abs{f(x)} \eval_{-\infty}^{\infty} =0$

[woven radish]

KySquared

[ember patrol]

@sand dove

[sand dove]

uh small problem

|f(x)| has no limits between +inf and -inf

unless you meant cauchy principal value

[ember patrol]

?
But f(x) is real. It should eventually die off

[sand dove]

?

well |f(x)| = |x|

[ember patrol]

^

We are assuming f(x) has this pro- OHHHH

Even though f(x) has that property, that doesn’t tell us anything about f(x)

.close

[devout snow]

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[devout snow]

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[humble sapphire]

Can someone double check my work

[glossy star]

looks all good!

[humble sapphire]

Everything makes sense?

[devout snow]

@humble sapphire Has your question been resolved?

[restive river]

@humble sapphire your hand writing is insanely neat

[humble sapphire]

No it’s not

[restive river]

its actually neat LOL

[humble sapphire]

The pencil is half broken

Can someone double check

I’m literally rushing lol

[fleet belfry]

yup looks all good to me

[humble sapphire]

Okay perfect ima do the last page in like an hour

Thanks guys

.close

[devout snow]

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[wild imp]

hi

[wild imp]

i tried to solve this

by solving the integral from 0 to 3

and i got c = 1/21

however this is the solution

i dont understand why their method works (or why my method is wrong for that matter)

[sacred hound]

I mean, your problem is discrete

You use the axiom of probability to show that the sum is 1.

Your method is more on the line of determining the c if your input is from [0,3]

which has infinitely many elements

so your x (given your method) can be 1.3214 or 2.1124

[wild imp]

i see, but are they not asking for a function here?

or what is the function they ask for?

[sacred hound]

I mean... they are asking for c. they already give you a function

[wild imp]

yes, i mean what function is this

what does

[sacred hound]

that function is probability mass function

[wild imp]

how is that connected to a random variable

[sacred hound]

it assign probability to the number defined by random variable

like you assign x = 0 with 4/30

[wild imp]

but the values are from x not X

[sacred hound]

your random variable can be 0 with 4/30 probability

your x is realization value of X

[wild imp]

realization?

[sacred hound]

like, you already know what it is (in terms of value)

like you can roll a dice. You can define your random variable from the face of the dice. I don't know what it is but once I roll it, I know the value

and I know that it will be 1 with probability of 1/6 and so on

but I don't know what it is yet in the form of random variable

[wild imp]

ohh i see now. and it has to add up to 1 because 0 to 3 is all the possiblr values?

[sacred hound]

yes

[wild imp]

okay, thanks a lot

.clos

.close

[devout snow]

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[devout snow]

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[storm current]

I know I have to use modulus

but I'm not sure how to execute this problem when subbing d into the second equation

Can anyone help me with this please

[devout snow]

@storm current Has your question been resolved?

[patent birch]

the property that if a|b, then a|b*k where k is an integer helps here

[storm current]

not sure i follow

[patent birch]

if some integer is divisible by 3, then 2 times that integer is also divisible by 3

so if 2a + 4b + 3c + d is divisible by 5, any multiple of it is also divisible by 5

anyways if you get stuck someone just asked this in another channel

[storm current]

oh

which channel

[patent birch]

#help-29

[elfin hill]

You can also multiply the first equation by three mod 5

[storm current]

.close

[devout snow]

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[devout snow]

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[tiny oyster]

If the tangent for the function f(x) = e^x at ( 0 , 1 ) intercepts both the x and y axis at point A, B
Then the area of triangle ABO where as O is the origin equals :

[tiny oyster]

I'm having trouble visualising this

I want to send my try on this and please correct me lighty

Lightly*

Ik bad photo be patient with me 👏

[tiny oyster]

What did I do wrong here

[tender cobalt]

did you find the equation for the tangent line at (0, 1)?

[tiny oyster]

Equals 1?

[brittle inlet]

I'd first find the equation of the straight line containing the tangence point

[tiny oyster]

Cuz e^x?

[brittle inlet]

Then find the coordinates of A and B, draw the triangle and calculate the area using vectors

[tiny oyster]

I know you're helping but please chill I'm trying to understand this not to just figure out the answer 😭🙏

[brittle inlet]

Alright, sorry if I went a bit too fast there

[tiny oyster]

You're fine 🙏

So first I need to find the tangent line using the point slope form? Or something else?

[brittle inlet]

Well, the points A,B and (0,1) must be the in the same straight line, do you get why?

[brittle inlet]

Yeah, that's what I'd do first, be it using the point slope form or any other

[tiny oyster]

Because it intercepts both x and y?

OK wait wait I need to know something first

Can you visualise it for me on a graph? I know it won't be accurate because it has e but you can change that

[brittle inlet]

Alright, brb

[tiny oyster]

I'm just not getting how the function and tangent are drawn

[tiny oyster]

Tysm 🙏

[brittle inlet]

These are the graphs of the function and the tangent

[tiny oyster]

Okay so the blue is the tangent of e^x at ( 0 , 1 )

This is what I got

[brittle inlet]

It's more purpleish but yeah

[tiny oyster]

y = x + 1

That's the tangent equation

[brittle inlet]

Yup

[tiny oyster]

I would say indigo

[brittle inlet]

Fair enough

[tiny oyster]

OK so got the equation

Now what that I got it

[brittle inlet]

You were asked about the area of the triangle OAB right?

[tiny oyster]

Wait nvm sorry

Let me fix that

[tiny oyster]

Wait when it says it intercepts both x and y at A and B

But it also intercepts ( 0 , 1 ) no?

[brittle inlet]

Yup, was about to tell you that

(0,1) is one of the vertices of the triangle, a straight line can only intersect each axis once throughout the entirety of its domain

[tiny oyster]

Does that mean I have 3 points? First is 0 , 1, second is B, third is A

Is that correct?

[brittle inlet]

Nope, that means (0,1) is either B or A

B and A are the points of intersection, each one for an axis

[tiny oyster]

OH

Ahuh

[brittle inlet]

Then the third vertice of the triangle is the origin(as specifically stated by the problem)

[tiny oyster]

So what does it specifically mean by origin, (0 , 1) or 0 , 0?

0 , 0

[brittle inlet]

The origin is always (0,0)

[tiny oyster]

Probably

[brittle inlet]

Generally denoted as O

[tiny oyster]

Yeah okay got that when I remembered that 0 , 1 is either A or B

[brittle inlet]

Then you need to calculate the last point and then find the area

[tiny oyster]

I will come back in a bit I gotta do something sorry

BRB

[brittle inlet]

Ok

[tiny oyster]

So this is what the triangle looks like

[brittle inlet]

Something like that Ig

But you still need B, and since you know it's the intersection with the x axis

You already know y=0

[tiny oyster]

Bruh I changed location like 5 times doing this is why its taking so long mb 😭

OK so I plug 0 into y = x + 1

To get the point

Now that I have all points I can just do 1/2 * h * b ?

Okay got it

Wait do I

How is an area fucking negative

I'm lost

@brittle inlet sorry can you help or?

<@&286206848099549185> 😭

Let write my issue again cuz no one wants to read ik

.Close

Oh?

It's not closing

Divine sign

[olive snow]

Its .close

No maj

[tiny oyster]

Shit I thought it was help 😭

[olive snow]

Oh you need help ?

I thought you was done

[tiny oyster]

Yes

[granite bough]

feller was about to give up on us

[olive snow]

Aight let me see

[tiny oyster]

Nah just the guy who was helping stopped sending

I moved locations too much so I took long responding 😭, I don't blame him for not responding

[granite bough]

the value of the area under the line is negative, but an area cannot be negative, if you get a negative area you'd subtract that part from the area

or something im sorry i really should have payed more attention in school

it is something like that though

you can have a negative area in the sense that it means you'd subtract that area from something

but like you say you can't have a negative area

[olive snow]

The are of the triangle is negative ?

[tiny oyster]

It came out negative for me

[granite bough]

yeah well in this case you can use an absolute value

[tiny oyster]

I😭😭😭

[olive snow]

What did you do ?

[tiny oyster]

Aight give me a second

[granite bough]

if they ask for an area you need to take the absolute value

just turn any negative positive

and you will have your area

[olive snow]

b*h/2

[tiny oyster]

Yes that

[olive snow]

h=1 and b =1

[tiny oyster]

b is negative 1 no?

[tiny oyster]

Oh

[granite bough]

you could say that too yep but you'd have to turn it positive

[tiny oyster]

If I can just do that then really I got it right

[olive snow]

b is the length of the base

So its |-1|

[tiny oyster]

No matter the sign?

[olive snow]

= 1

[granite bough]

yeah when we talk about areas

[tiny oyster]

Oh OK makes sense

[granite bough]

they will always be positive except for very very few exceptions that come way way later on

if it would've integrated you could have gotten a "negative" area, but again you cant have a negative area so you just take the absolute value of that. if they wanted the value of the integral the answer would be negative

i think that's how it goes

sorry for not being able to express myself in solid math terminology

[tiny oyster]

Thank you so much both of you

I get it now

🙏

.close

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[summer sage]

yo i need some clarity on two equation formulas

[summer sage]

first is for solving

two intersecting chords within a circle

is it ab = cd

[umbral shuttle]

depends what you're saying ab and cd represent

[summer sage]

yeah the first one

[devout snow]

@summer sage Has your question been resolved?

[brittle inlet]

Sorry, was busy

[devout snow]

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[stiff hill]

hi I’m not sure if I did the limits right.

[stiff hill]

<@&286206848099549185>

[devout snow]

@stiff hill Has your question been resolved?

[devout snow]

@stiff hill Has your question been resolved?

[devout snow]

@stiff hill Has your question been resolved?

[devout snow]

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[scenic surge]

for step 4 shouldnt it have been 0 <= theta <= pi?

[scenic surge]

it says above xy plane

[boreal helm]

0 to π is half circle

Hint

[scenic surge]

oh above xy plane meant positive z?

[boreal helm]

,w yes

[scenic surge]

smh alr ty

[boreal helm]

Welcome

[scenic surge]

.close

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Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[sterile smelt]

(2y - 5)^2

[devout snow]

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[devout snow]

@mighty acorn Has your question been resolved?

[astral smelt]

what do you need help for

do you still need help @mighty acorn

[dense lynx]

for a curve f(x) rotated about the x-axis, yes this formula is correct

[astral smelt]

yeah, its 2pi times the integral of the function times the arclength between the endpoints

[devout snow]

@mighty acorn Has your question been resolved?

[dense lynx]

@mighty acorn what part of your question has not been resolved?

[devout snow]

@mighty acorn Has your question been resolved?

[devout snow]

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[astral smelt]

help please

[astral smelt]

<@&286206848099549185>

[visual hazel]

maybe find what v cross x is

[astral smelt]

thats what i cant do

idk how

[lusty sapphire]

!15m

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@astral smelt Has your question been resolved?

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[ocean knot]

Could somebody explain to me why the answer to c is npi/2??? Ive solved a and be but I have no idea how that got to c?

[ocean knot]

the answers to a and b are

[rustic laurel]

bouta waffle but i think its becasue cos(pi/2) = 0, that means that you would only count for the value of x ?

[ocean knot]

but if cos(pi/2)=0, that means it would be f(x)=cos (x)

and if you sub -x

so f(-x)= cos(-x)=cos (x)

doesnt that mean its even

[runic prawn]

c is just anything that's not a or b

[winter patrol]

can you show exactly what they've given as the answer for c)

[runic prawn]

it's here

[winter patrol]

ah, discord preview cut it off

[winter patrol]

they said NOT equal to

[ocean knot]

thats what i was thinking

[runic prawn]

f is neither odd or even iff (f is not odd) AND (f is not even)

not odd means not b and not even means not a

[runic prawn]

yeah and that agrees with their answer

[ocean knot]

i just dunnno why its npi/2😭

[topaz beacon]

you have npi makes the function even and pi/2+npi makes the function odd

[runic prawn]

a number of either form here can be written as npi/2

[topaz beacon]

all of those numbers are of the form npi/2

[runic prawn]

m*pi = n pi/2 where n = 2m

pi/2 + m pi = n pi/2 where n = 2m+1

[ocean knot]

ohh okay

i kind of get it (no really) but i get the idea😭so the answer is just saying it cant be the answer for even or odd, and wrote it as npi/2 bc the answers to both are in the same form? (idk why tho)

[winter patrol]

if you list out the values from each set
0, pi, 2pi, 3pi ...
pi/2, 3pi/2, 5pi/2 ...
combining them
0, pi/2, pi, 3pi/2, 2pi, 5pi/2, 3pi...
same idea for the negatives
you'd get all the integer multiples of pi/2

[ocean knot]

OHHHHH

I GET IT NOW

THANK YOU SO MUCH😭

yall are saving my life 🙏😞

[runic prawn]

yeah basically both sets intertwine

one between the other

[ocean knot]

👍

if i was asked that in an exam id be screwed

but thanks again!

.close

[devout snow]

Channel closed

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[runic prawn]

np

you'd probably be fine in an exam if you said not a or b tho

[winter patrol]

you could still write excluding the sets from a,b

and get some marks without the final simplification

[ocean knot]

ohh okay phew

[devout snow]

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[hexed shell]

!help

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[polar chasm]

Hi, this channel is now yours. If you have a question, you can ask it now

@hexed shell

[hexed shell]

its a puzzle challanged by my friend does it count?

involves god damn binary

i need lat 50 bucks 🤑

[polar chasm]

Sure, if it's related to math

[polar chasm]

did your friend allow you to ask other people for help? Lol

[devout snow]

@hexed shell Has your question been resolved?

[hexed shell]

is binary math or no

[polar chasm]

Yeah

just send the puzzle, if it doesn't fit in here, we could redirect you to the right server

but i'm pretty sure you are at the right place

[hexed shell]

Its a puzzle in game tho

check this out

I need to make "Binary" and then convert it into decimals after, im pretty sure you have to start on the top right and read it like opposite your reading a book

[polar chasm]

Do you just need to convert it to decimal system? (Normal numbers)

[hexed shell]

no i need to convert it into binary and decimal

both

[polar chasm]

alright

[hexed shell]

it swithces every 3 minutes tho

cause 50 bucks is fifty bucks

[polar chasm]

can you write the sequence of lights as 1s and 0s

[hexed shell]

Alive should be 1 and dead is 0

[polar chasm]

yeah, but im not sure how to read it

is it like 11100?

the one on right

or 101 111 101 011 010

[hexed shell]

Actually, thats the thing. I dont know 💀

i think

you have to start

from top right

[polar chasm]

that would make it start with 0

that's not usual

but it would then be
010 011 101 111 101

if you just need to pass the game, you can convert it like this

,w 010 011 101 111 101 binary to decimal

[woven radish]

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

Requested by methisalwaysright

[polar chasm]

okay nvm lol

,w 010011101111101 binary to decimal

[woven radish]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[polar chasm]

yep

010011101111101
Otherwise you would read it from right to left and multiply each digit by corresponding power of 2.

1 * 2^0 + 0 * 2^1 + 1 * 2^2 + 1 * 2^3...

[devout snow]

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[lilac gazelle]

hey does anyone know why this equation works?

[lilac gazelle]

why is bn equals to that whole lotta thing

cuz when I try to write bn in terms of cn I get bn+cnx{bn-1}

[hollow ice]

Its just multiplying and dividing by a number done multiple times

[cold bone]

check the denominator of first term and numerator of next term

[lilac gazelle]

so like it cancels out?

[cold bone]

yes

[lilac gazelle]

ah I see

[cold bone]

you can say

[lilac gazelle]

thx man

/close

.close

[devout snow]

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[devout snow]

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[lavish umbra]

@hollow iceI read what u typed but I don't get based upon what does he make up this formula
FG = Tr on G ... etc. etc.

[devout snow]

@lavish umbra Has your question been resolved?

[hollow ice]

T R on G is just the force exerted by the rope on the gymnast

so in the frame of reference of the gymnast, since she is stationary, the forces must balance

so that force is equal an opposite to her own weight

because, any more force and she accelerates up, less and she accelerated down

weight is just mg, so we have the value 490N from that

In short, the forces have been balance on all the bodies in the solution. Each body has been considered separately and all the applied forces and their reactions are tracked. Since everything is in equilibrium, the net force is 0. So unknown reactions are found using the free body diagrams

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[manic meadow]

How do i find the angles for A,B and c

[brisk panther]

this looks like one of those notes from the slenderman game

[restive river]

can you please send a clean & well labelled diagram?

[cold bone]

2,8 = 28/10 right?

[brisk panther]

no its 2.8

[brisk panther]

my spanish isnt good

[cold bone]

then what's that

[brisk panther]

does 2.8 cm refer to the radius

[manic meadow]

This is portuguese

[brisk panther]

mb

[cold bone]

?

[manic meadow]

Not spanish

[manic meadow]

Oki

[brisk panther]

?

@manic meadow

[manic meadow]

I Saw that

[manic meadow]

Yes

[brisk panther]

well from this you can find the side lengths right

just add the radii

then use the law of sines

[manic meadow]

Here in europeu we do in centímeters

[manic meadow]

Thanks

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[foggy terrace]

need help with part ii)

[devout snow]

Please don't occupy multiple help channels.

[foggy terrace]

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[floral lodge]

I am a bit confused if my understanding of conditional probability is correct. When talking about probability of an event, $P(A) = 0.3$, that means it has a 30 procent chance of happening in regards to the sample space $\Omega$. Does that mean thatr $P(B | A)$ is the probability relative to $A$ instead of $\Omega$? In other words, we are only interested in $B$ at the same time as $A$ occurs?

[woven radish]

Pen

[drifting pewter]

exactly

[floral lodge]

Phew, thought I was going mad for a second as my classmate is disagreeing

Thank you, Odina ❤️

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[lyric gyro]

2+2?

[restive river]

75

[lyric gyro]

Ohh okay

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[lavish umbra]

is it normal that I do not understand the solution

[drifting hearth]

which part specifically do you not understand?

[devout snow]

@lavish umbra Has your question been resolved?

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[feral sun]

The answer is incorrect but i believe i integrated it all correctly

this is the right answer

but with fubini's theorem the answers should be the same

[tiny panther]

you changed the bounds incorrectly

in your region you have (I believe) 0 <= y <= x <= 2

so you should be integrating w.r.t x with bounds y and 2, not 0 and y

[feral sun]

forgot to send this too

[feral sun]

This is how i found the bounds

is the y incorrect?

or x incorrect

[tiny panther]

I believe you're actually right, and the 'right' answer is the one with the mistake, now that I've actually seen what E is supposed to be

[feral sun]

ok thank you very much

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[radiant rampart]

Hi guys, I need some help understanding the highlighted line, why do we need to generate 0?

[stone stump]

it will become clearer later

keep reading a bit

[devout snow]

@radiant rampart Has your question been resolved?

[radiant rampart]

I'm not sure that it explicitly states why but I think i can make a guess now. Is it because when you generate the 0-vector (where all entries are 0), you effectively have a system of linear equations that all equal 0, which makes it easier to find a general solution?

<@&286206848099549185>

[radiant rampart]

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[uncut token]

Can someone explain to me as if I was 10 how to turn a polynomial function of second degree from a general form to a factorise form?

[devout snow]

@uncut token Has your question been resolved?

[uncut token]

<@&286206848099549185>

[timber pebble]

what do you mean general form and factorize form

like $ax^2+bx+c$ into $a(x-x_1)(x-x_2)$?

[woven radish]

jan Niku

[uncut token]

yes

[timber pebble]

okay, so you know the quadratic formula I'm guessing

[uncut token]

ye

[timber pebble]

okay, so you can just apply that

write $a \qty( x^2 + \frac ba x + \frac ca)$

[woven radish]

jan Niku

[timber pebble]

now do you know the roots?

[uncut token]

wait u got me lost

[timber pebble]

okay, lets start with $ax^2+bx+c$

[woven radish]

jan Niku

[uncut token]

ok

[timber pebble]

I want to make the coefficient in front 1

for no particular reason

i anticipate it might make our life easier later

[uncut token]

I could give u my equation if that helps

[timber pebble]

okay

[uncut token]

one of them

[timber pebble]

i mean, we can solve it in general

but sure

[uncut token]

2x^2 + 4x - 6

[timber pebble]

so first, factor off the 2

because its easier to work with stuff that has a leading coefficient of 1

[uncut token]

is the product-sum method valid?

not sure if that what u call it in english

[timber pebble]

yea

3b1b made a video on this, right

https://tenor.com/view/3blue1brown-quadratic-equation-math-mathematics-gif-25677768

[uncut token]

it not what I have as a formula

[timber pebble]

its rewritten

whats your form

[uncut token]

ac is the product and b is the sum

and then I find what matches

[timber pebble]

oh yea

i mean, this will work when it works

its up to your ability to figure it out

i still think its easier to factor off the 2, first

[uncut token]

ok so, if I factor 2,
it would be
2(x^2 + 2x - 3)

[timber pebble]

yea

can you solve it?

[uncut token]

I did this earlier and erased it because I was stuck at a part

I’ll give u the full details in a few mins

ok so I did (b/2)^2 which gave me (2/2)^2 which result was 1

and then I did 2(x^2 + 2x + 1 - 1 - 3)

but this is the part I start getting confused

[timber pebble]

oh, maybe i dont understand this method

i thought you meant

the sum is +2

and the product is -3

[uncut token]

the sum would be 4 and the product would be -12

because ac = -12
and b = 4

[timber pebble]

ac = -12?

but a is 1 and c is -3

and b is 2

[uncut token]

it this one

a is 2

c is -6

[timber pebble]

okay but then why did you factor out 2

[uncut token]

that was after we did any change

[timber pebble]

now you're working on two different polynomials

[uncut token]

oh no no, I was just following ur advice

cause u told me u would factor 2 here

[timber pebble]

i would, and then work on x^2 + 2x -3

[uncut token]

ye, and when I worked on it, I got 2(x^2 + 2x + 1 - 1 - 3)

it make sense so far

[timber pebble]

i dont think i understand this method

sorry

[uncut token]

oh it all good

in french it called complétion de carré

I don’t know if it might help

[timber pebble]

its just not clear if you mean by inspection or specifically using the coefficient

which i guess you mean the second

[umbral shuttle]

completing the square?

[timber pebble]

but as far as im aware if the roots are idk w and z

[uncut token]

yes

[timber pebble]

oh

OH

completing the square isnt the form you said though-

[uncut token]

it wasn’t?

[timber pebble]

do you want form $a(x-x_1)(x-x_2)$ or $a(x-h)^2 + k$

[uncut token]

I want the first one

[woven radish]

jan Niku

[timber pebble]

right so completing the square is only useful for getting the first form indirectly

completing the square is usually nice because it tells you the bottom or top of the quadratic

the vertex

but it doesnt directly provide you the roots

and the roots are what you need to factor

[uncut token]

oh so would u say it pointless in this situation to use it?

[timber pebble]

its a good technique to know and its important

its just not useful here as much as other methods, i think

im sorry i should have gathered thats what you were using earlier

[uncut token]

I should be the one apologizing for not giving clear instructions

[timber pebble]

its all good it dont matter

so usually like $2(x^2+2x-3)$

[woven radish]

jan Niku

[timber pebble]

i think normally, someone would try to guess the roots here

but you could use other techniques too

[uncut token]

how do u “guess” it?

[timber pebble]

do you know how to solve by inspection? i mean, just trying to guess

you know the product of the roots is -3

and you know the sum is +2

so, you try to think of two numbers that give you these results

can you think of two numbers that multiply to -3 and add to +2?

[uncut token]

3 times - 1
3 - 1

[timber pebble]

yea

so then we know the roots, they should be x=3, and x=-1

then we can write it in the factored form, given the roots

its $2(x^2+2x-3) = 2(x-3)\qty(x- (-1))$

[woven radish]

jan Niku

[uncut token]

Ohh

only one thing

how is it x - 3

when it was positive

[timber pebble]

it works like this

[uncut token]

oh wait nvm I understood it

[timber pebble]

the roots are where $2(x-3)(x+1)$ is 0

[uncut token]

it just the formula

[timber pebble]

yea

[woven radish]

jan Niku

[timber pebble]

well, we want the polynomial to be zero where x=3

so if you put in x=3, and then subtract 3

[uncut token]

wait actually no,

[timber pebble]

that creates a 0 factor

it makes the whole thing 0

[uncut token]

in the textbook the answer is 2(x + 3) (x - 1)

[timber pebble]

yea

wait what 👀

sorry, what was the original polynomial again

[uncut token]

f(x) = 2x^2 + 4x - 6

[timber pebble]

,w psolve 2x^2+4x-6=0

[woven radish]

Results provided by WolframAlpha

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

Requested by janniku

[uncut token]

wtf😭

ai is crazy

[timber pebble]

right, sorry, im being dumb

im over complicating it really

we could just write out $(x+x_1)(x+x_2)$

[woven radish]

jan Niku

[timber pebble]

we know $x_1+x_2 = 2$ and $x_1 x_2 = -3$

[woven radish]

jan Niku

[uncut token]

ye

[timber pebble]

and thats it

x_1 = 3 and x_2 = -1

$(x+3)(x-1)$

[woven radish]

jan Niku

[uncut token]

ohh

okok, I think I got it from here

thank you so much🙏

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[dense hemlock]

i don't really know how to start this, i've tried some ways but they're wrong

[winter patrol]

can you show what you tried

[boreal helm]

Ok

[dense hemlock]

1. get the lengths of AB which is 8

but i have to determine the ratio between the areas of ABC and ABQ

which i dont really know how to do

[winter patrol]

did you make a diagram/sketch of the figure

[dense hemlock]

nah i can't really visualize what the question is asking

which is why i'm struggling mostly

usually for things like nuumber theory it's ok but i'm weak in geometry

[winter patrol]

try making one

starting with the triangle,
then placing points P,Q,R

[dense hemlock]

ok

this is what i see

but idk if it’s accurate

i thought i could go from AB=8 to finding the height, which is 10

but idk what to do from here

[winter patrol]

sry, I can see a way to approach this ATM, you may need to wait for someone else

[dense hemlock]

that's fine

thank you for your time nonetheless

[patent birch]

im also doing it right now but i think theres something to do with the height from p to c

anyways this week seems way harder than the other weeks

[dense hemlock]

Are you in the CEMC thing too??

[patent birch]

yes

[dense hemlock]

oh i see

[devout snow]

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[stable cedar]

any hints as to how to approach this problem?

[dense lynx]

notice how this number is really close to 40000

[stable cedar]

yeah but we want integers

[patent birch]

W guess and check

[solid osprey]

my first thinking was making it (a+b)^2-2ab

[patent birch]

anyways you can square root 40501 and notice that its close to 201, and when you square 201, its 40401

which just happens to be the original minus 100

[solid osprey]

oh it only wants 2 pairs bruh

[dense lynx]

yes... so the idea is you guess an integer close to 200

[solar goblet]

the engineer way

[dense lynx]

well it's not even "guessing" since you should know what 201^2 evaluates to without doing any math

[stable cedar]

oh

[solar goblet]

i wonder if the fact that 40501 = 101 * 401 helps

[stable cedar]

diff square maybe

na

wait

[solid osprey]

thats a^2-b^2 unfortunately

[stable cedar]

oh

201^2 + 10^2

works

[dense lynx]

no

[stable cedar]

sorry edited it

yea not sure what im supposed to do

i tried doing smth with the fact that 40501 = (10^2 + 1)(20^2 + 1) and after some sophie germain thing it led to

like 181 * 221 + 500 idt thats

helpful tho

<@&286206848099549185>

[visual cradle]

hi

[quasi star]

Hi guys

[empty inlet]

Hi

[devout snow]

@stable cedar Has your question been resolved?

[devout snow]

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[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[edgy frost]

Why hello! I'm trying to understand the ramanujan summation. Can someone explain how we got from step 4 to 5?

[polar chasm]

Remember that 1 + 2 + 3 + 4 + .... = C

[edgy frost]

Yeah but 1/4-c doesn't equal -4c

[polar chasm]

It's a derivation

[jagged harbor]

supposedly they want B to be 1/4, maybe in an earlier step

sorry, earlier derivation

[edgy frost]

How can it be 1/4

[polar chasm]

each step has to follow from the ones above it

[polar chasm]

What even is B?

[jagged harbor]

from the image, B is 1-2+3-4+...

[polar chasm]

Oh, I see

i'll quickly find some derivation of that

wait a moment..

[edgy frost]

Okay tysm

[jagged harbor]

I think there was some nonsense assumption like 1-1+1-1+... should be 1/2 that they use to derive B

showing my true colors here, this is all nonsense

[polar chasm]

Yeah, that's an important point

this is all non-sense

because by saying "C = 1 + 2 + 3 + 4 + 5...", you already make the assumption that it converges to some number, but it actually doesn't

B = 1 - 2 + 3 - 4 + 5 ...
2B = 1 - 2 + 3 - 4 + 5 ...
+ 1 - 2 + 3 - 4 ...
2B = 1 - 1 + 1 - 1 + 1 - 1 ...

[edgy frost]

Yeah haha

[grand siren]

A = 1 - 1 + 1 - 1 + 1 - 1 + ...
A = 1 - A
2A = 1
A = 1/2

[woven oak]

.

[polar chasm]

@lusty sapphire is that at least somewhat right explanation? Or is my explanation also nonsense

[polar chasm]

Hence 2B = 1/2

and B = 1/4

[jagged harbor]

well we can say C isn't even a real number

[grand siren]

"1 + 2 + 3 + 4 + 5 + ..." just isn't a meaningful object, at least with the standard definition of summation

[jagged harbor]

we're suggesting that C is the limit of the sequence of partial sums of natural numbers

this limit doesn't exist in the reals, hence C isn't real

[grand siren]

same as how expressions like 1/0, or sqrt(-1) if you're working within the real numbers, just don't really evaluate to anything

[lusty sapphire]

This is the key point

This is outside of the idea of summations to infinity as we normally do them

[polar chasm]

Yeah, that's why we can't use standard rules of arithmetic to manipulate it

and algebra

[lusty sapphire]

Numberphile, of course, describes it

[jagged harbor]

in fact the rules don't apply whenever a series doesn't converge absolutely

we can't permute the terms of a series freely because the resulting series post-permutation may not converge to the same point

[polar chasm]

In fact i think there is a theorem that says that by rearranging it, we can make it converge to anything

(or diverges)

[jagged harbor]

it's true

[signal crag]

look at the cesaro sum for one type of sum

that allows you transform a divergent sequence to converge

for instance, the divergent grandi series can be manipulated by taking its cesaro sum to give us a value of 1/2

[jagged harbor]

digression but grandi's series is a really good example of something that does not need to be named after anyone lol

[devout snow]

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[hybrid maple]

Hey i need help with a proof for linear algebra

[hollow ice]

swap Djk and Dkj and show that it evaluates to (DtD)t

[hybrid maple]

We did a similar proof to this but i’m lowkey a bit lost

[devout snow]

@hybrid maple Has your question been resolved?

[devout snow]

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[steep cipher]

Need help with sequences

[steep cipher]

4,8,20,56,_

[dapper fable]

take the difference between those terms

[chrome isle]

It's a wierd pattern
When you take 4 common out you get 1,2,5,14
1+ 3^0 = 2
2+3^1 = 5
5 + 3^2 = 14

I'd say take 4 common out then subtract the sequence with itself shifted one place

To obtain a gp

[soft umbra]

Mb, keep going

[dapper fable]

are you trying to to find the general form or the sum?

[steep cipher]

I found it thx

Yall

[chrome isle]

Wlc

[steep cipher]

.close

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[mild sorrel]

Please explain how Q is the incenter of triangle ABC?

.

[hollow ice]

Its the intersection of angle bisectors

thats the definition

you can see the angles B/2 and C/2

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[humble wren]

Assignment: Write the rational exponents in radical form. Simplify if possible.

1. 2^1/5
2. 10^-1/2
3. 18^1/4
4. (-3)^1/3
5. 9^1/8
6. 2^2/5
7. 27^5/3
8. 10^4/5

[humble wren]

my answers in 1 is 5√2, 3 is 4√18, 5 is 8√9, 6 is 5√2^2, 7 is 243 and 8 is 5√10^4

are these correct?

I don't understand how to do 2 and 4 though

[chrome isle]

x^-1 = 1/x so 10^-1 = 1/10 now try 2 by yourself

[frozen aurora]

i assume you mean $\sqrt[5]{2}$ and not $5\sqrt{2}$

[woven radish]

artemetra

[humble wren]

yeah

[humble wren]

do you mutiply it 1/10 by 10/10?

[chrome isle]

Huh?

What do you mean?

[humble wren]

oh wait

the final ans is 1/10?

[chrome isle]

For the second que?

[humble wren]

yeah

[chrome isle]

No

[feral agate]

false

[frozen aurora]

wut

[feral agate]

$x^{-n}=\frac1{x^n}$

[woven radish]

kheerii

[frozen aurora]

i think you meant (1/n)^x

[chrome isle]

Fck I fcked up sorry
n^ -x = 1/n^x

[chrome isle]

Yea I fcked up

[humble wren]

so like, (1/10)^2/1 then 1^2/10^2?

then 1/100?

[chrome isle]

(1^ 1/2 )/( 10 ^ 1/2)

So 1/√10

[humble wren]

huh?

[devout snow]

@humble wren Has your question been resolved?

[scenic surge]

why 2/1

[scenic surge]

only the sign changes for the power

so it would be 1/2 instead of -1/2 when you bring the whole thing down

thats how they got 1/sqrt(10)

[devout snow]

@humble wren Has your question been resolved?

[humble wren]

OHH

i thought you would need to like find the reciprocal of -1/2

[scenic surge]

nah

[mild sorrel]

I mean how they form BQ forms the angle bisector

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[lavish umbra]

is this right

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[lavish umbra]

No

[slow thunder]

Q: ABCD is a rectangle, <DBC is 55 degree.The diagonals AC and BD intersect at point O.Find <AOB.

[mild sorrel]

repost yo question

[slow thunder]

ABCD is a rectangle

[mild sorrel]

ik

draw a diagram

n sow us

[slow thunder]

<DBC is 55 degree

we dont know AOB

AC and BD intersect at point O

at the middle

So we have to find AOB

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[rose charm]

Do the graphs around double zeros ever form in quadrants 3 or 4? As, in will double zeros ever yield negative y values? If so, how?

I'm confused because I know you can't get a real negative value from squaring something, but if it's possible to get a negative graph from double zeros, then I'm confused how the function makes that happen.

[rose charm]

would it just be a negative "a" value attached to it? like f(x) = -(x-h)^2

oh wait, i can just use desmos to find out. my bad yall

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[mild sorrel]

i

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[ashen vault]

let
$
A

\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & 0 & 0 & \dots & 0 & 0 \ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 & \dots & 0 & 0 \ 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \dots & 0 & 0 \ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \ 0 & 0 & 0 & 0 & \dots & \frac{1}{2} & \frac{1}{2} \end{bmatrix}
$ and let $x = \begin{bmatrix} x_1 & x_2 & \dots & x_n\end{bmatrix}^T$.

How to prove, that
$
\lim_{k \rightarrow \infty}
\prod_{i=0}^{k}
s_i A x
$
where $s_i \in (0.8, 1.2)$
will have all vector elements close to the same?

[woven radish]

redve

[ashen vault]

I use similar algorithm to "smooth up" a vector, and I wanna formally prove it smooths up the vector

and finally got this limit

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[ashen vault]

<@&286206848099549185>

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[west mulch]

hello , if i know the equation of a line in space , can i find the equation of the planes that their section is that line ?
and how ?

[dense lynx]

there are infinitely many pairs of planes such that their intersection is the same line

[west mulch]

@dense lynx yeah i know but how we can find 2 of them that aren't parallel with each other ?

[dense lynx]

you can plug the parameterized version of the line into the equation of a plane Ax + By + Cz = D and find values of A, B, C, D that fit

[west mulch]

@dense lynx thanks mate, have a good one

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[old yacht]

Hi I need help and understanding what operations to do and in what order

I know how to solve for a already

a) P(x) =-0.4x^2+55-12

Although when it comes to section B I kind of start getting confused can you help me understand what order and how to do it please

<@&286206848099549185>

[dusky coral]

You just use 100 as x

You put 100 into formulae R, C and P instead of x

[old yacht]

Can you walk me through it? I'm sorry I tried but I must have messed it up

<@&286206848099549185> can someone help me walk through this so I have notes? Sorry I want to know step by step so I don't mess up please 🥺

Thank you for the help Seb just need little more info cause I mess up easy

[soft umbra]

Please don't ping helpers unless your post hasn't been visited for 15 minutes

[old yacht]

Oh I waited 15 I thought sorry!

[warm breach]

R(x) = 59x - 0.4x²
you would get R(100) by keeping x=100 i.e.

59(100) - 0.4(100²)
for simplification you can solve it as
100[59-(0.4)(100)]

same thing goes for C(100)

and once you get R(100) and C(100), subtract them and get P(100)

[old yacht]

Okay ❤️ thank you very much I will try now 😄

[warm breach]

if
f(x) = 10x
then what is
f(10)?

i just want to make sure you understand the main part behind it

[old yacht]

1000?

100

Oops

XD

[warm breach]

100 is correct

alr i think you got it

[old yacht]

Haha ty Vansh 💜❤️

[warm breach]

np

[old yacht]

.closep

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[dusky coral]

Sorry I didn’t see 😭 you can ping me personally next time.

[old yacht]

@dusky coral no worries ty for the help!

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[faint gorge]

,, 2\pi \cdot \frac{11}{\sqrt{7}} \int_0^{28} x^{\frac{1}{4}} \cdot \sqrt{1+ax^{-\frac{1}{2}}}

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

Maybe subbing the term inside the root might lead to something

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

Let $u = 1+ax^{-\frac{1}{2}}$

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

brb

ok so then

,,\frac{du}{dx} = -\frac{1}{2} \cdot a \cdot x^{-\frac{3}{2}} \Leftrightarrow dx = - \cdot \frac{2}{a} \cdot x^{\frac{3}{2}}: du

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

,,- 2\pi \cdot \frac{11}{\sqrt{7}} \cdot \frac{2}{a} \int_1^{1+\frac{a}{\sqrt{28}}} x^{\frac{1}{4}} \cdot x^{\frac{3}{2}} \cdot \sqrt{u} : du

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

,, x^{\frac{1}{4}} \cdot x^{\frac{3}{2}} = x^{\frac{7}{4}}

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

Let's see if the substitution helps

no

we have to write x^7/4 in terms of u

$u = 1+ax^{-\frac{1}{2}} \Leftrightarrow x^{-\frac{1}{2}} = \frac{u-1}{a} \Leftrightarrow x = \frac{a^2}{(u-1)^2}$

oh I forgot a

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

,,- 2\pi \cdot \frac{11}{\sqrt{7}} \cdot \frac{2}{a} \cdot a^\frac{7}{2} \int_1^{1+\frac{a}{\sqrt{28}}} \frac{1}{(u-1)^\frac{7}{2}} \cdot \sqrt{u} : du

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

ok i hope so far so good

I am thinking if that's the good way

that's what I am thinking

Maybe do another substitution

but first let me see if the previous stuff is correct

ok should be fine

Maybe let's try $z = \sqrt{u} \Rightarrow \frac{dz}{du} = \frac{1}{2z} \Leftrightarrow du = 2z : dz$

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

,,- 2\pi \cdot \frac{11}{\sqrt{7}} \cdot \frac{2}{a} \cdot a^\frac{7}{2} \cdot 2 \cdot \int_1^{\sqrt{1+\frac{a}{\sqrt{28}}}} \frac{z^2}{(z^2-1)^\frac{7}{2}} : dz

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

z²-1 = (z-1)(z+1)

hold on

Maybe a trig sub

😂

getting worse and worse

z² - 1 we could make use of sec²(x) - 1 = tan²(x)

but is there another way other than trying

Let $\sec\theta = z \Rightarrow dz= \tan\theta\sec\theta : d\theta$

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

,,- 2\pi \cdot \frac{11}{\sqrt{7}} \cdot \frac{2}{a} \cdot a^\frac{7}{2} \cdot 2 \cdot \int_0^{\arcsec(\sqrt{1+\frac{a}{\sqrt{28}}})} \frac{\sec^2\theta}{\tan^7\theta} \cdot \tan\theta \sec\theta : d\theta

,,- 2\pi \cdot \frac{11}{\sqrt{7}} \cdot \frac{2}{a} \cdot a^\frac{7}{2} \cdot 2 \cdot \int_0^{\arcsec(\sqrt{1+\frac{a}{\sqrt{28}}})} \sec^3\theta \cdot \cot^6\theta : d\theta

didnt help shit

ah no i did a silly

[woven radish]

𝔸dωn𝓲²s

𝔸dωn𝓲²s

[faint gorge]

,w sec(x)cot(x) = csc(x)

[woven radish]

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[faint gorge]

,,- 2\pi \cdot \frac{11}{\sqrt{7}} \cdot \frac{2}{a} \cdot a^\frac{7}{2} \cdot 2 \cdot \int_0^{\arcsec(\sqrt{1+\frac{a}{\sqrt{28}}})} \csc^3\theta \cdot \cot^3\theta : d\theta

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

,w cot^2(x) = csc^2(x)-1

[woven radish]

Results provided by WolframAlpha

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Requested by adoniszgks

[faint gorge]

,,- 2\pi \cdot \frac{11}{\sqrt{7}} \cdot \frac{2}{a} \cdot a^\frac{7}{2} \cdot 2 \cdot \int_0^{\arcsec(\sqrt{1+\frac{a}{\sqrt{28}}})} \csc^3\theta (\csc^2\theta - 1) \cdot \cot \theta : d\theta

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

Let $w = \csc\theta \Rightarrow d\theta = -\frac{dw}{\cot\theta \cdot \csc\theta}$

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

,,- 2\pi \cdot \frac{11}{\sqrt{7}} \cdot \frac{2}{a} \cdot a^\frac{7}{2} \cdot 2 \cdot \int_{\arccsc(\arcsec(\sqrt{1+\frac{a}{\sqrt{28}}}))}^{\infty} w^2(w^2-1) : dw

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

This never exists lmao

not the only way

maybe there is a much smarter way

did you create that integral?

what surface area formula

,w Integrate[x^1/4Sqrt[1+a/Sqrt[x]],x]

[woven radish]

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[faint gorge]

I see

wolfram suggested to first simplify the square root

which ahhh