help-43

[dusty moat]

why would it be 4D?

[timber cipher]

The way I like to think about it:
Every point on that circle "owns a complex number" given by f(z)

[dusty moat]

sure

[timber cipher]

Where the circle is 2D, and the complex number that each point has is another 2D

[dusty moat]

so if the function was like x+y+z=2+i on the circle at point 1,1 the imaginary part would be just i?

[timber cipher]

Let's say the function was f(z) = z²

And we want to know what's happening at the point z = 1 + i

f(1 + i)
= (1 + i)²
= 1 + 2i - 1
= 2i

So the point 1 + i "owns" the complex number 2i

The circle might exist on 1 + i, so the 2i gets summed into the integral there

[dusty moat]

i dont understand what f(z)=z^2 means

like i get its a function

but

shouldnt it also be defined by an x and y

[timber cipher]

There's nothing more. It's a function. It takes complex numbers, and returns complex numbers.

You might also think of it as f(x + iy)

[dusty moat]

that makes less sense

[timber cipher]

We often use z = x + iy as a notation in complex

[dusty moat]

so y is the imaginary part normaly?

[timber cipher]

Where z is a complex number, x and y are reals

x usually the real part of z, y usually the imaginary part

But this can change always of course

[dusty moat]

wait so if my function f(z)=z^2 how would i find z at x=0 y=1 would it just be z=0+1i ?

[timber cipher]

Yeah

[dusty moat]

ok but i still dont understand why the intergral is 0.

[timber cipher]

Anyway you're asking why this integral always happens to be 0. Note this is a non-trivial result called the Cauchy-Goursat theorem

[prisma vapor]

writing the function as a f(x, y) = a + bi is actually an important step in one of useful theorems for holomorphic functions

[timber cipher]

Integrating f(z) over any simple closed curve, where f(z) is complex differentiable everywhere in that closed curve, gives 0

[dusty moat]

ye why thats my question

[prisma vapor]

note that being complex differentiable everywhere is a very strong condition

[timber cipher]

I admit, I don't have a nice way to think about why! I wish I did haha.

[prisma vapor]

and it only works for strictly complex functions

[dusty moat]

i figured

[prisma vapor]

real-valued functions cannot be holomorphic

unless constant

[dusty moat]

what does holomorphic mean

[prisma vapor]

everywhere complex differentiable

[dusty moat]

is complex differentiable different from regular

[prisma vapor]

yes

for real functions you usually differentiate with respect to variables

so in directions parallel to the axes

in complex case it has to be differentiable this way for ANY direction

[dusty moat]

i mean if i had sin(x) and i changed the direction of the axis wouldent it still be differentiable?

so would that be complex

differeientable

[prisma vapor]

i wrote that badly

by direction i mean on any curve

and the limit in definition of derivative has to give the same result every time

this can be broken if you differentiate it as R2->R2 function

[dusty moat]

so like instead of just a change in x its both a change in x and y?

nvm

im done for today my brain hurts. thanks for the help i understand it a little better

.close

[compact pewter]

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[compact pewter]

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[mossy shadow]

How is this wrong

[kind crane]

,w partial fractions (4x^2-5x+3)/((x+2)(2x-1))

[boreal girder]

Results provided by WolframAlpha

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

Requested by rie.mann

[kind crane]

Why do you think it's wrong

[mossy shadow]

Not the same as answers

[kind crane]

Show

[kind crane]

Can you show the problem too

[mossy shadow]

12

[kind crane]

Yea looks wrong

You can verify it's wrong by graphing both the original function and the book's solution in desmos

[mossy shadow]

These answers always wrong

🙏 thanks

.close

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[compact pewter]

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[kind surge]

I just need some clues

PQ=QT=32 and TS=SR=12.5

[dense mason]

First you might need to find QS

then you find distance between P and R

[compact pewter]

@kind surge Has your question been resolved?

[compact pewter]

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[kind surge]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[kind surge]

QS will be 32+12.5=44.5

PR is 2r

[compact pewter]

@kind surge Has your question been resolved?

[dense mason]

Then. From S you draw a line that perpendicular to PQ, name this intersection W

[kind surge]

Whattt??

@dense mason

And??

[dense mason]

Bro

At S

That is not S

[kind surge]

It is from S

[tall pike]

nah, its from T

[tall pike]

yay

[kind surge]

And??

Please give me clue quickly i have very less time

[tall pike]

can you find QW, and hence find WS

[tall pike]

ok

[kind surge]

QW is not given

[tall pike]

can you FIND

[kind surge]

It is perpendicular

I guess no

it doesn't divide equally

QP

[tall pike]

Note that WP = SR

[kind surge]

How can you say?

[tall pike]

(property of rectangle)

[kind surge]

Ohh great

SR is 12.5

QW is 35-12.5

22.5

[tall pike]

yes, method correct

[tall pike]

no, answer wrong

32

[kind surge]

I can find now by pythagoras

[tall pike]

32-12.5

[kind surge]

Ohh 32

19.5

[tall pike]

yes

got it?

[kind surge]

QS^2=QW^2+WS^2

But but

How will it help me to find radius after this

[tall pike]

what is PR w.r.t. radius

[kind surge]

2r

[tall pike]

there

[kind surge]

Ohhhhhhhhh i got it now

Thanks

🤗🤗

Thanks

Thanksss

.close

[compact pewter]

Channel closed

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[kind surge]

Ohh wait

[tall pike]

yes?

[kind surge]

which book did you read for geometry? And for practice

[tall pike]

textbook

[kind surge]

i meant ... name?

[tall pike]

i have not other idea

[tall pike]

um... sorry can't help with that

[kind surge]

My school board books are not advanced

[tall pike]

try Khan Academy

[kind surge]

So just randomly asking...so i can study on my own

.close

[tall pike]

and mathisfun.com for clarification

[dense mason]

correct

Not correct ratio tho but thats fine

[kind surge]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[kind surge]

By curiosity i wanted to know....QOS angle

can I find it?

[dense mason]

not enough info given

what circumference you got btw

correct one is 40 pi

[kind surge]

yeah i got it same

[dense mason]

nice

[kind surge]

Actually Somewhere it is written if we have two parallel lines

and tangent is drawn

It makes 90° at center

[dense mason]

wow

[kind surge]

So just connecting this thing to this problem

and if it is 90° then i can use this

so tell me please how it makes 90°

loooking fir that proof🤣

[dense mason]

Well not enough info is given

so cant prove yet

[kind surge]

@dense mason

[dense mason]

Correct

[kind surge]

Hmm

Can you explain what they did?

I didn't get what angles they took

[dense mason]

Your teacher did this?

So they prove TQO and TOS are similar triangles to get side ratios

Then they just substitute the side lengths in to find radius

[kind surge]

But how they are similar?

90° one side common

[dense mason]

They share same line OT, angles T = 90 degree and TOS=TQO

Why would you solve this way lol this is not robust at all

[kind surge]

Tos=Tqo how?

[dense mason]

Use pythagorean is easier

[kind surge]

Yeah but i am curious in diversity

[dense mason]

Suppose theta is angle TQP, then show triangle PQT = QTO,

[dense mason]

Diversity is good but you need clean way and efficiency, not long way

[kind surge]

PQT=QTO whatttt

[dense mason]

1+1=2, not use something like 1/10*10-5+5=1

[kind surge]

Hang on

[kind surge]

?

[dense mason]

My bad QOT and OPQ

Show them equal

For angle TSR i dont see how it can be 180-theta

[compact pewter]

@kind surge Has your question been resolved?

[compact pewter]

Channel closed

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[cinder lynx]

The least common multiple (LCM) and greatest common divisor (GCD) of p q are 84,700, and 2, respectively. How many pairs of p q have LCM and GCD of 84, 700, and 2, respectively?

[cinder lynx]

plz solve this question for me

[dense mason]

what have you done

[rancid crypt]

We can't, but we can help you.

What have you tried?

[cinder lynx]

84700x2=2^3x7x5x11^2 and member of 84700x2=48

48/2=24 there is no choice

[rancid crypt]

Let's see, we're told that the greatest common divisor (GCD) is two, so both p and q are divided by 2. We're told to look for even numbers. We need to find how many even numbers p and q we have, so we can express them this way: p = 2a, q = 2b

[small mason]

the sensible thing to do first is to prime factorize those 3 numbers you're given, right
which I don't see u've done

[cinder lynx]

ok

[peak smelt]

are there 3 numbers or 2 ?

[small mason]

84, 700 and 2

[cinder lynx]

it's 84,700, and 2 i'm sorry

[peak smelt]

loos like a typo ? only mentions LCD and GCD

[small mason]

i'm not paying attention

no im not paying attention to whats wrong with what i said

[small mason]

oh i missed this

ok

you can have lcm and gcd of 3 numbers though, but fair enough

[peak smelt]

oh, hm because here they did 84700x2

[small mason]

ok ignore what ive said then, you already prime factorized

lcm(p, q) = 84700
gcd(p, q) = 2

How many pairs p, q fit this

that's the question right, to clarify?

[cinder lynx]

yeah

[small mason]

this is a bit messy, restate what you're trying to say perhaps

[cinder lynx]

I use LCMxHCF=AxB

[small mason]

whats the prime factorization of 84700

[small mason]

yes thats right

2^2 . 5 . 7 . 11^2 ?

[cinder lynx]

84700=2^2x7x5x11^2

[small mason]

lcm(p, q) = 84700 = 2^2 . 5 . 7 . 11^2
gcd(p, q) = 2

pq = lcm(p, q) . gcd(p, q) = 2^3 . 5 . 7 . 11^2

so that's what you've done so far right

[cinder lynx]

yes

[small mason]

So you know their product is 2^3 . 5 . 7 . 11^2

what's the next thing you're trying

[small mason]

what's this

[cinder lynx]

pair of p q have LCM and GCD of 84,700, and 2

[small mason]

i dont get where 48 comes from

[cinder lynx]

member of 84,700x2

[small mason]

factors?

but its not 48

[cinder lynx]

yes

[small mason]

4 . 2 . 2 . 3 isn't it?

oh thats 48

i don't get why you divided 48 by 2

[cinder lynx]

pair

[small mason]

24 pairs of factors. got you now finally

but you know not all 24 of these may work?

some may fail, right?

[cinder lynx]

yes i know

[small mason]

so what is your step now

to make sure they have the correct lcm and gcd

you need to think sensibly how 2^3 . 5 . 7 . 11^2 must be split

[cinder lynx]

um.

[small mason]

for example 1 and 2^3 . 5 . 7 . 11^2 definitely does not work.

[cinder lynx]

it much either p or q divisible by 2

[rancid crypt]

Think about MCD.

[rancid crypt]

Yes, you're right.

[small mason]

(i apologize for taking so long to understand your working earlier )

[cinder lynx]

I love math so much

[rancid crypt]

❤️

Do you have more questions?

[cinder lynx]

wait

this is Entrance exam for grade 7

[rancid crypt]

Ok, do you understand how to do?

[cinder lynx]

yessssssssssssssssssss

[rancid crypt]

Do you have more questions?

[cinder lynx]

no

[rancid crypt]

Have a nice day, you can type .close.

🙂

[compact pewter]

@cinder lynx Has your question been resolved?

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[rough bronze]

i am at a standstill on this question

[rancid crypt]

Do you know the theory and what ideas you have?

[rough bronze]

the theory and ideas i had are already stated. thats why i am at a standstill

from the 3 expressions its clear somehow tension is W but i dont get how

[rancid crypt]

Block P moves up the slope thanks to the tension of the rope. Right?

[rough bronze]

correct yes

oh and the tension of the rope is pulled down by W

[rancid crypt]

Right!

[rancid crypt]

Do you get it now?

[rough bronze]

so the 3rd expression would be wrong right

[spark wigeon]

this whole thing looks unphysical

[rancid crypt]

Yeah, haha.

[rancid crypt]

Right.

[rough bronze]

okayy thankss

[rancid crypt]

Do you have more questions?

[rough bronze]

nopee got it

.close

[compact pewter]

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[rancid crypt]

Have a nice day!

🙂

[spark wigeon]

I can't solve it

[tired bear]

how did he get it

[compact pewter]

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[stone agate]

Hello, Ans given is wrong?

[short ferry]

it looks right to me

why do you think its wrong

want a hint on how to solve it?

[rancid crypt]

Yeah for me it looks right.

[small mason]

i agree it is right if you count the polynomial 0 to "touch" the x-axis "once"

[rancid crypt]

Yeah, it touches.

[stone agate]

a = b = c = 0

[rancid crypt]

Why did you think it’s wrong? Do you have any doubts?

[short ferry]

okay yeah, this is the 0 polynomial exception

[stone agate]

Yep, isn’t it touching infinitely many times because zero polynomial has infinitely many roots

[small mason]

that depends on your perspective

[rancid crypt]

Yeah, it’s an exception.

[stone agate]

Should question have mentioned this?

[small mason]

yes

[rancid crypt]

Yeah.

[short ferry]

You can bring it up to your teacher, if your answer was marked wrong and your teacher is reasonable, you should get your marks back

[small mason]

are you happy the question is correct otherwise?

[stone agate]

I have asked my teacher about the case where a + b + c = 0 and n = 0 and they have given incorrect explanation

[rancid crypt]

It can be used as an argument.

[stone agate]

Hence I wanted to ask

[stone agate]

When I have said this

[rancid crypt]

What did he tell you?

[stone agate]

They replied with “If you take n = 0, you get 0 = 0. The problem is that it’s a meaningless statement as nothing is being plotted in that case.”

But 0 = 0 is not meaningless

Isn’t it saying that it is true for all x

[short ferry]

yeah, it is

[rancid crypt]

It’s correct.

[small mason]

yes something is being plotted, wrong answer

[short ferry]

you should be plotting y = 0

[small mason]

the line y = 0 is a plot

[stone agate]

What is correct? His explanation?

[short ferry]

your explanation

[rancid crypt]

[stone agate]

Yep and I’m not understanding what he’s saying

It just seems wrong

What logic did he use

Even if he’s wrong

I’m still trying to understand

Isn’t the plot gonna be fully shaded

Since 0 = 0 is always true

[short ferry]

My guess would be that he isnt used to seeing stuff like 0 = 0 in equations and so he wasnt sure how to interpret it and hence gave it no interpretation at all

[stone agate]

[small mason]

there is no 0=0 in question

what

[stone agate]

I think

[small mason]

oh nvm

when u find the roots

[stone agate]

He means f(x) = 0

Where f(x) is the polynomial

[small mason]

when you find the roots of f, you are doing 0 = 0

aka all x work

[stone agate]

Yep so 0 = 0 is meaningful

[small mason]

0 = 0 <=> x in R

[stone agate]

Yep yep okay so how to fix this question

What to add

[small mason]

just make c not 0

[short ferry]

require a,b,c to not be all 0

[stone agate]

But a + b + c = 0

Can also happen

If n = 0

[small mason]

???

just make c not 0 and the question is fine

[short ferry]

oh, n should prolly be a positive integer too, otherwise its not a polynomial

[stone agate]

If you pick n = 0 then you get the polynomial to be a + b + c

If you set that to 0

[small mason]

well the question does say 'the polynomial'

[stone agate]

That also gives you 0

[small mason]

ugh, i suppose

[stone agate]

Yeah but also n is an integer then seems useless

[small mason]

then make the question 'the non-zero polynomial'

that's the simplest fix

[stone agate]

Yep thought about this

What about the first line

Delete?

N is an integr

[small mason]

no need

[short ferry]

I'd at least make it positive integer

[stone agate]

If u leave it without changing it

Will there be an edge case still

[short ferry]

not really, the fact that its a non-zero polynomial covers it

[small mason]

if n is not a positive integer that expression is not a polynomial (apart from 0, which is fine)

[short ferry]

but like if the line is there, it better be something meaningful. "n is an integer" is almost as useless as "n is a number"

[stone agate]

There is no path that one can take while solving the problem which will lead to the evaluation of 0^0, right?

Just thinking of that as well

[stone agate]

Yep

[small mason]

0^0 is 1 who cares

[stone agate]

Reason I’m checking is because of the last answer

Seems like at any ambiguity

You can almost select

Option D

[small mason]

the intent of the question is clear enough to myself

[stone agate]

It is board exam

So they care about objectivity

If it was a class exam

Then I have no problems

[small mason]

well i care about doing math tbh

[stone agate]

Yeah same but I don’t want to get marked wrong

[rancid crypt]

If you still have question you can discuss them with your teacher.

[small mason]

a teacher rarely marks their students' exams

[stone agate]

No he is confused himself

[small mason]

a problem you come across will be one everyone comes across

in a national exam or something

[stone agate]

Yep

But there’s still ambiguity

[rancid crypt]

Yeah. Normally are specific teacher.

[stone agate]

In this national exam

And the existence of option D

Makes it

Sad…

[small mason]

and that can be appealed

[stone agate]

You know there was a pemdas question too

Apparently

You had to use implied multiplication

?????

[small mason]

i dont think they're gonna screw everyone over when it comes to a national exam

[stone agate]

Board doesn’t even care

But okay for the math of it

[rancid crypt]

Personally I will use your opinion and if they mark incorrect you can appeal. But it will be strange.

[stone agate]

There will be no case where we have to evaluate 0^0 right?

Do you guys see a case on that ..?

[small mason]

0^0 = 1

[stone agate]

[small mason]

if you argue 0^0 is not 1

then i argue x^0 is not a polynomial

so case closed

[stone agate]

Yeah but everyone calls that a zero order polynomial

Isn’t there contention

On what 0^0 is

[small mason]

if x^0 is a polynomial, then 0^0 = 1

[stone agate]

I don’t think there’s any contention on this

[small mason]

there is no other recourse

[stone agate]

I guess

Okay so just

Add Nonzero polynomial

Can leave the first line about n there

And question is fixed

So answer is now B?

Just clarifying

@small mason @short ferry ?

Sorry for ping but since u guys were already helping

Just wanted to finally conclude

[rancid crypt]

Yea. You can.

[daring axle]

yeah if u want to make the least amount of changes

[stone agate]

Okay thanks 👍

[compact pewter]

@stone agate Has your question been resolved?

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[regal pewter]

im not sure how to finish off this question, i know that i would need to use the log form of arcosh which is ln (x + root(x^2 - a^2)) but im not sure what 'a' would be in this case

[winged lion]

where did you get this formula

[regal pewter]

the one with ln?

[winged lion]

yes

[regal pewter]

from my exam board's formula booklet

[winged lion]

oh well

[small mason]

cant you directly apply identity

use defn of cosh

[regal pewter]

like with the e's?

[small mason]

yes

[winged lion]

you could write technically (5/3)/1 and 1/1

[small mason]

but i guess thats long winded maybe

[regal pewter]

okk i'll try that

[small mason]

but do note that:

2cosh(x) = e^x + 1/(e^x)
and this is a quadratic in e^x which you can solve to figure out the formula for arcosh in terms of ln

[regal pewter]

got it thanks!

.close

[compact pewter]

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[compact pewter]

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[small mason]

.reopen

its over

[copper sierra]

ya

[peak smelt]

your power is useless here

[small mason]

.shutup

[peak smelt]

https://tenor.com/view/pepe-gif-18261463

[compact pewter]

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[spark dagger]

x^2 + y^2 = 25
xy + x - y = 7 ​

[spark dagger]

Simulatious equations 😄

[rancid crypt]

What-s your question?

🙂

[spark dagger]

i have no clue how to do it im not used to questions like that

how do i solve it

?

btw the ^ is squared

[rancid crypt]

Do you want to solve for x and y, right?

[spark dagger]

yes

[rancid crypt]

First we must clear one of the two variables, right?

[spark dagger]

yes

[rancid crypt]

Ok, then replace, right?

[spark dagger]

ye

dont u rearange both equations?

1st one srry

not both

i mean 2nd lol

[rancid crypt]

Yeah, you-re right.

What do you have now?

[spark dagger]

x(y+1)-y=7

[rancid crypt]

Can you show me your steps?

[spark dagger]

sure

wait u just group the terms

then divide each side by y +1 to make x the subject

so x=y+1/y+7​

right?

[rancid crypt]

The clearance of $x = \frac{y+1}{y+7}$ is not correct because the $y$ you are subtracting is not divided. For the grouping trick to work, you need the left side to be factorizable.

[boreal girder]

Ga³¹Br³⁵I⁵³9000✞

[spark dagger]

oh

...

[rancid crypt]

Np.

[spark dagger]

i understand it now thanks

.close

[compact pewter]

Channel closed

Closed by @spark dagger

Use .reopen if this was a mistake.

[rancid crypt]

Have a nice day!

🙂

[compact pewter]

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[lofty mountain]

does anyone knows about moments in statistics

[covert crater]

what about them?

[lofty mountain]

i don't get it how the formulas are formed

in it

i know that it has mean, median, mode, s.d. and kurtosis

let me show you something

what mu represents

?

[covert crater]

Pretty sure mu here is just the central moment for rth order

Basically it's showing the derivation of the relationship between central moments (μ_r) and the arbitary moment (μ_r)'

So how to transform moments taken about an arbitrary origin A (where d_i = x_i - A) into moments taken about the mean (central moment)

[compact pewter]

@lofty mountain Has your question been resolved?

[compact pewter]

Channel closed

Closed due to timeout

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[compact pewter]

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[tranquil mist]

Hi! 👋🏻, I want to prove that the min(n)=1008² where n nonzero natural number.
Such that n(2017+n) is a perfect square

[sleek meadow]

!status

[compact pewter]

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

[tranquil mist]

I'm 5

Find $n \in \mathbb{N}^*$ such that $n(2017+n) = m^2$, $m \in \mathbb{N}$. Prove that $\min(n) = 1008^2$.

[boreal girder]

William James Moriarty

[sleek meadow]

Could you send their working?

[tranquil mist]

gcd(n,2017+n)=1 or 2017
I suppose that n not multiple of 2017
So n is a perfect square also 2017+n

I'm not sure about that

I do decomposition of factors and I conclude that

[sleek meadow]

n is ofc not a multiple of 2017

[next tide]

i got it

william

[sleek meadow]

Well ill give you a hint! first try rearraning your org equation and trey making perfect squares!

[next tide]

brotatoskitodorito

n^2 + 2017 n = m^2

take n^2 to other side

2017n= (m-n)(m+n)

now, we know that one of them is a multiple of 2017

so let one of them be 2017k( k is a nat number)

hence other is n/k

but m+n >n

hence m-n = n/k

and m+n = 2017k

now from them, we get n=km/(k+1)

[upper bane]

!nosols as a gentle reminder if you are going to complete the working for them!

[compact pewter]

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.

[next tide]

oh ok mb

then do rest

[tranquil mist]

But it can be
In this case we need the min
If we don't get anything with n≠2017k
that mean n is a multiple

[tranquil mist]

2017n=(m-n)(m+n)

[sleek meadow]

It can be !
But again i think in the question it was asked to find minm!

However, thats not going to alter in out working process

[tranquil mist]

n=2017k-m
km/(k+1)=2017k-m
km+(k+1)m=2017k(k+1)
m(2k+1)=2017k(k+1)
I think that's don't work

[next tide]

it does u are rly close, try to relate m and n using n/k

u are halfway there

remember if u have 2 variables, u need atleast 2 equations to solve so always try to hunt for that

[tranquil mist]

Yes
Let's continue
a=product p_i ^α_i
b=product q_i ^β_i
With q_i,p_i prime numbers
if ab=c²
and we have gcd(a,b)=1
So p_i≠q_i
The power will be even cause
In c² the power is even
So the power in a must be even because it's the same apear in c
Same thing with b

so a=x² and b=y²

[next tide]

broda.... u were halfway the answer, why changed approach?

[tranquil mist]

I don't have any clear idea

[sleek meadow]

Should i propose a different approach?

[next tide]

m-n=n/k use that

[tranquil mist]

I try to find the second equation

[tranquil mist]

Yes

[sleek meadow]

Ah okay

[sleek meadow]

Can you express the original statement as x * y = 2017^2?

The reason why we do this is so that we can compare the factors on both sides since 2017 is a prime number!

[tranquil mist]

Let a=n/k
m-n=a
m=a(k+1)

[next tide]

nice

so u get n=mk/(K+1), now use that and the other eqn to find an equation of n and k, and then use logic

[tranquil mist]

We have n²+2017n=m²
2017n²+2017²n=2017m²
2017²=2017m²/n-2017n
2017²=2017n(t²-1)
With t=m/n

[sleek meadow]

Nope you are not allowed to introduce another variable

i need it to be solely in the form of m and n

and no there can not be m or n in the denominator also!!

[tranquil mist]

I get it

We have (m-n)(m+n)=2017n

m-n=n/k

m=n(1+1/k)

So (n/k)(n(2+1/k))=2017n

n×(2k+1)/k²=2017

n=2017k²/(2k+1)

[sleek meadow]

Okay you are going a little off track

Lemme use a latex software

[tranquil mist]

A minute

[sleek meadow]

try adding 2017^2 to both sides

does this make sense?

Now before i give away the next step,

try adding 2017^2 to both sides

[next tide]

now tell me why 1008^2 is n(min)

[tranquil mist]

$$
m - n = \frac{n}{k} \
\Rightarrow m = n \left(1 + \frac{1}{k}\right) \[2mm]
(m-n)(m+n) = \frac{n}{k} \cdot n \left(2 + \frac{1}{k}\right) = 2017 n \[1mm]
\Rightarrow n \cdot \frac{2k+1}{k^2} = 2017 \[1mm]
\Rightarrow n = \frac{2017 k^2}{2k+1}
$$

[boreal girder]

William James Moriarty

[tranquil mist]

gcd(k²,2k+1)=1

[sleek meadow]

Uhm please pick 1 way

and so ill leave you to @next tide

[next tide]

https://tenor.com/view/7-crore-gif-16763107112638466597

there u go mu guys

hence proved (dont forget to mention, it holds 1 mark)

[quartz yoke]

indian spotted

[sleek meadow]

bhaichara

[tranquil mist]

Yes
2k+1=2017
k=1008
k is one of divisors of n
minn→n=k
Min n=1008

[next tide]

min n=k^2 , 1 mark lost bro...

but u got logic so

https://tenor.com/view/social-credit-social-credit-score-meme-john-xina-gif-23649557

[tranquil mist]

.solved

[compact pewter]

Channel closed

Closed by @tranquil mist

Use .reopen if this was a mistake.

[tranquil mist]

Yeah i forgot

Some things got mixed up in my mind while I was solving it, so my focus got scattered

[compact pewter]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[broken lantern]

Hey girlies
So Ik that I have to use AM-GM inequality here but I'm confused as to why

[rotund sphinx]

$$4^{1-x}=4 \cdot 4^{-x}=4 \cdot \frac{1}{4^x}$$

[boreal girder]

Civil Service Pigeon

[rotund sphinx]

tldr rewrite the negative exponent

[broken lantern]

Damn bro got hacked

[gloomy olive]

fr the theo guy

[broken lantern]

I am, unfortunately, still confused

[rotund sphinx]

Use this rewrite and set up AM-GM

What do you get

[broken lantern]

Huh

Okay

I mean I kind of got that just...yk...why does that happen?

[rotund sphinx]

Whenever you have a thing and the reciprocal of that thing, that’s a classic use case for AM-GM because of the cancellation. Obviously multiplying by constants doesn’t change the fact that you get cancellation, which is what you have here.

[broken lantern]

Ouuhhh

I see

Thanks!!

.solved

[compact pewter]

Channel closed

Closed by @broken lantern

Use .reopen if this was a mistake.

[rotund sphinx]

[compact pewter]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[exotic maple]

How do i solve these ones?

I AM a very BIG noob at math

so if anyone knows how to solve it, if you dont mind telling me and simplyfing it

would HELP SO much

[rancid crypt]

Do you have any ideas?

[exotic maple]

no im really lost i dont know math 1 bit

[winter fern]

Just try normal multiplication knowing that i² = -1

[rancid crypt]

Yeah all is normal only you need to use that.

[exotic maple]

whats the answer for the first one maybe i can see whats missing.

[rancid crypt]

We can't give you the answer, we can help you to get it.

[exotic maple]

well it seems like im gonna be stuck here for a very long time it seems challenging

im very shit at maths what do i multiply here

it says "find z1 x z2"

[rancid crypt]

Don't worry start with the first one, we can help you.

[rancid crypt]

You need to multiply the functions.

[winter fern]

Think of it as if it was (x+3)(x-5) just do it as you usually do

Just when you see i² you can replace it with (-1)

[exotic maple]

hold on

but there is no i² only i

or its the same

[winter fern]

If there's just i, you keep it as it is

[exotic maple]

i has to do something with imaginary right?

[winter fern]

But keep in mind that for example the first one, you have an i in z1 and an i in z2, so you will eventually get i² somewhere

[winter fern]

Yes i is an imaginary number, but complex numbers pretty much hold the same algebraic properties

[exotic maple]

okay i see

[rancid crypt]

Yes, use that.

[winter fern]

It's really just that i² = (-1), nothing really changes. When u see i² you use -1 instead, when u see i you leave it as it is

[exotic maple]

but when you say multiply the functions.
do i multiply (cos60 + i sin60) * (cos30 + i sin30)

[exotic maple]

okay i understood this part with the i

but im not getting the plot here what do i have to multiply

[wanton patrol]

evaluate cos60

[rancid crypt]

Right!

[exotic maple]

BUT this is the part right WHAT do i multiply there

the whole thing?

[winter fern]

That and the r values 2 and 3

[winter fern]

Yes

[rancid crypt]

All, yes.

[winter fern]

It's just r1(x +iy)*r2(a+ib)
You multiply them as if they were real values

Nothing changes, quite literally

Just give it a try and let us know

[exotic maple]

i dont understand one bit in here everything just got way too confusing now

well sorry guys i know you guys are probably on your nerves but i seiously dont know any formulas or anything

what is x?

what is iy where did that even come from

[wanton patrol]

it's just:

simplifying 2 expressions and multiplying them

you simplify 1 part and then move onto another

[exotic maple]

how do i simply it?

simplify

[wanton patrol]

well

what is cos60

[exotic maple]

uhh

[wanton patrol]

u gotta memorize these btw. or u just draw the right triangle

,w cos60

[boreal girder]

Results provided by WolframAlpha

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

Requested by 200.300

[wanton patrol]

ur gonna start to recall them after u do a lot of questions

don't worry about that

[rancid crypt]

Or use the calculator.

[exotic maple]

well you see the problem here is that i can understand things once i know the meaning behind them because if i look at that first solution

i dont understand anything, only i know is that i is a imaginary and now i know that i2 = -1

[wanton patrol]

what are you not understanding exactly

[exotic maple]

its hard to explain, i just dont understand what do i need to solve, like what do i need to simply or multiply?

i should multiply (cos60 + i sin60) * (cos30 + i sin30)?

[wanton patrol]

well, it tells you to find z_1 and z_2

[exotic maple]

yes

[wanton patrol]

z_1 and x_2 are given

i meant z_2

so u just multiply them

[exotic maple]

okay yes but this is the part that gets me stuck

[wanton patrol]

i'd start with simplifying first though

[exotic maple]

like what numbers do i simplify

[wanton patrol]

you would distribute first

the 3

what'd you get

[wanton patrol]

cant forget the 3

[exotic maple]

oh wait

when you say multiply

it means the whole thing

i thought you needed to multiply just the numbers in the ()

[wanton patrol]

everything that comes after the = is part of z_1

so you gotta do that

[exotic maple]

what do you mean by distribute?

[wanton patrol]

so

people tend to call it that

it's throwing the "3" in the brackets basically

like

number (something + something_1) = number * something + number * something_1

you're just throwing what's outside the brackets into the party

[exotic maple]

can you give me an example with real numbers?

[wanton patrol]

$3(5 + 8) = 3 * 5 + 3 * 8$

[boreal girder]

mango

[exotic maple]

ohhhh

i got it now

i think i get this part okay

[wanton patrol]

it's the same thing here

[exotic maple]

okay wait hold on

but this has cos60 so?

if 3(cos60 + i sin60) = 3 * cos60 + i sin60?

[wanton patrol]

partly correct

the isin60 needs 3 as well

[exotic maple]

oh yeah i forgot that cus i know it sorry

so basically 3(cos60 + i sin60) = 3 * cos60 + 3 * i sin60

now what?

[wanton patrol]

so now

sin and cos are nothing special

they're real numbers as well

,w cos60

[boreal girder]

Results provided by WolframAlpha

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

Requested by 200.300

[wanton patrol]

,w sin60

[boreal girder]

Results provided by WolframAlpha

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

Requested by 200.300

[wanton patrol]

just gotta put those in place now

[exotic maple]

so cos is techinaclly = 1/2?

[wanton patrol]

cos60 yes

[exotic maple]

okay but what do i do with the 60?

[wanton patrol]

60 is just the angle

u can just

remove it and put 1/2 in stead

like remove the entire cos60

because cos60 is the same thing as 1/2

[exotic maple]

so just 3 * 1/2 + 3* v3/2?

[wanton patrol]

pretty much

just don't forget the i

[exotic maple]

3 * 1/2 + 3* i v3/2?

[wanton patrol]

yea

[exotic maple]

okay

now?

[wanton patrol]

now u gotta do the same exact thing

for z_2

[exotic maple]

okay let me try on my own here

[wanton patrol]

after you're done u just multiply them

[wanton patrol]

alright

[exotic maple]

okayokay let me see if i can do anything

but before i try this shit

is this a hard like math thing

or am i just tripping

the way you guys explain it seems easy

[winter fern]

Ur just tripping

It's pretty basic

[exotic maple]

whatt?

[winter fern]

You just have to focus

[wanton patrol]

it's extremely easy

[exotic maple]

damn

[wanton patrol]

you just tend to see it as hard

math is not an alien invention, people made it up by themselves

[winter fern]

People usually are intimidated by cos and sin and i, mainly because they're more abstract

[wanton patrol]

and you're part of people too, so you won't have an issue understanding it

[exotic maple]

i love maths its just that

1. I didnt focus till the start now i have to learn everything new
2. the languange barrier

english is not my first languange so im trying my best to understand

but thanks to yyou guys

im actually starting to learn some stuff

[wanton patrol]

it's alright, it just needs time

[exotic maple]

okay let me try this now

[rancid crypt]

You are welcome.

[exotic maple]

z1= 3(cos60 + i sin60) z2 = 2(cos30 + i sin30)
z1= 3 * cos60 + 3 * i sin60
we simplify

z1= 3 * 1/2 + 3 * i v3/2

z2= 2(cos30 + i sin 30)
z2= 2 * cos30 + 2 * i sin30
we simplify

z2= 2 * 1/2 + 2 * i v3/2

z1= 3 * 1/2 + 3 * iv3/2 + z2= 2 * 1/2 + 2 * i v3/2

but now here

to the multiplication

[clever copper]

This isn't an actual test is it?

[rancid crypt]

Check cos 30.

[exotic maple]

this is lol

[wanton patrol]

u mixed up cos30 and sin30

[clever copper]

Getting help for this is considered cheating and is against server rules

[wanton patrol]

buddy is trying to learn, leave him alone lol

we're not giving him answers

[exotic maple]

no the test is finished

[clever copper]

oh got it

[rancid crypt]

Oh, ok.

[exotic maple]

yeah but everyone in the class failed

like everyone

[clever copper]

sorry for the distrubance go on

[exotic maple]

so she gave us anotherr chance

no problem

[exotic maple]

hold on let me draw it because its so complicated that i dont know how to use those commands

[wanton patrol]

so sin30 = 1/2

[exotic maple]

cos basically v3/2

[wanton patrol]

cos30 = v3/2

yea

you put 1/2 for cos30

[exotic maple]

ohhh

[winter fern]

Unit circle

[rancid crypt]

Use that and check the cos.

[exotic maple]

let me draw the math equazion its complicted for me to type it with 1/2

and lets see

if i have it correct this time

but before i draw it

what do i multiply then?

i simpified z1 and z2 for example

[wanton patrol]

so you got z_1 right

well

they're asking for

"what is z_1 multiplied z_2"

so u just multiply them

[rancid crypt]

Get z2 and multiply.

[exotic maple]

hold on

but wait

dont mind that handwriting

took me hours

but what was wrong here again?

Z1 = 3 * 1/2 + 3 * v3/2

Z2= 2 * 1/2 + 2 * v3/2

[wanton patrol]

cos30 is v3/2

[exotic maple]

wait

what about sin 30?

[wanton patrol]

it's 1/2

u swapped them lol

[exotic maple]

cos60 is 1/2 no

[wanton patrol]

yea

but you've got cos30

[exotic maple]

so it does not matter that sin 30 is also 1/2?

[wanton patrol]

so

here's the thing

they're kind of flipped

cos30 is v3/2, and sin60 is v3/2

pay attention to this part

[exotic maple]

okay so once again

z1= 3 * 1/2 + 3 * i v3/2

z2= 3 * v3/2 + 3 * i 1/2

[wanton patrol]

yea

[exotic maple]

finally

[wanton patrol]

so now

since you've got common denominators

[wanton patrol]

i believe they had i as well

[exotic maple]

yes ohmy do i keep forgetting

[wanton patrol]

$z2= 3 * v3/2 + 3 * 1/2 => \frac{3 * \sqrt{3}}{2} + \frac{3}{2}$

[boreal girder]

mango

[exotic maple]

thats the finalscore?

[wanton patrol]

since they have common denominators, they can just share the bar

so $z_2 = \frac{3 \cdot \sqrt{3} + 3}{2}$

[boreal girder]

mango

[exotic maple]

it says to find both z1 and z2 tho no

[wanton patrol]

so we've found z2

[wanton patrol]

but this has an i

hold on

it was 3 * 1/2 right

so it'd be 3 * 1/2 * i

$\frac{3i}{2}$

[boreal girder]

mango

[wanton patrol]

ok so $z_2 = \frac{3 \cdot \sqrt{3} + 3 \cdot i}{2}$

[boreal girder]

mango

[wanton patrol]

now we gotta get z_1

[exotic maple]

okay but

before we get to z1

how the hell do we know what do put in top and bottom

like how do we just put 3 * v3 + 3 * i on the top?

how does that work

[wanton patrol]

well

it just depends

hold on

[wanton patrol]

this is wrong btw

what'd u say z2 was

[exotic maple]

z2= 3 * v3/2 + 3 * i 1/2

[wanton patrol]

Z1 = 3 * 1/2 + 3 * v3/2

Z2= 2 * 1/2 + 2 * v3/2

ok no it's not wrong

the way u have it written

it's a mistake btw

$\frac{\sqrt{3}}{2} \neq \sqrt{\frac{3}{2}}$

[boreal girder]

mango

[wanton patrol]

those are not the same thing

[exotic maple]

ohhh

[wanton patrol]

your teacher's gonna deduct points for that

[exotic maple]

so which one is the correct one

the left one or the right one

[wanton patrol]

the left one

cos30 = v3/2

not v3/v2

[exotic maple]

yup got it

im still confused on the

[wanton patrol]

now about this one

[exotic maple]

part where we put those at the top and 2 at the bottom

yes

[wanton patrol]

$z_2 = \frac{2 * \sqrt{3}}{2} + 2 \cdot \frac{1}{2} \cdot i$

[boreal girder]

mango

[wanton patrol]

so this would equate to

$z_2 = \sqrt{3} + i$ right

[boreal girder]

mango

[exotic maple]

yeah im pretty much stuck here now i wont even lie

[wanton patrol]

alright which part

[exotic maple]

look

like how do we get this like

[wanton patrol]

that was actually

incorrect

🤣

i messed it up

[exotic maple]

well i cant een tell if it is or not LOL

[wanton patrol]

are u like wondering how we got to that part

or where it came from

[exotic maple]

yes like how

[wanton patrol]

because i don't know where it came from

i just made it yp

[exotic maple]

you know when i see this

im gonna be wondering

why do we have 3 * v3 at the top?

[wanton patrol]

so you don't get how they just get under 1 bar?

[exotic maple]

yes like how do we just know where to place those numbers?

[wanton patrol]

well, this was a mistake but

if you're asking for like

in general

then

$3 * \frac{sqrt{2}}{2}$

[boreal girder]

mango

[wanton patrol]

the 3

would go up

because this is the same thing as

$\frac{3}{1} \cdot \frac{sqrt{3}}{2}$

[boreal girder]

mango

[wanton patrol]

im sorry

there should be 3

sqrt3

that's why the 3 goes up

so it becomes $\frac{3 \cdot \sqrt{3}}{2}$

[boreal girder]

mango

[exotic maple]

okay whats the full solvment of z2 now like the correcto ne

[wanton patrol]

you've got this one

as answer

[exotic maple]

just that?

[wanton patrol]

i think you messed it up somewhere

[wanton patrol]

yea

u could try and solve z_2 again

[exotic maple]

on the v

[wanton patrol]

yea u messed up by square rooting 3 and 2

[exotic maple]

yes should've made only 3

[wanton patrol]

so that's the answer for z_2

[exotic maple]

v3/2

[wanton patrol]

the 2's cancel eachother

nothing confusing in that part right

[exotic maple]

not that i know of right now to be honest

so just to make sure again

Z2 answer is 2 * 1/2 + 2 * v3/2

im guessing we found Z2

[wanton patrol]

yea

[exotic maple]

we dont have to mess with z2 anymore?

[wanton patrol]

tbh u could have just left it alone at the first place

it's just easier to

simplify it and then multiply

[wanton patrol]

yes

nothing more to do

[exotic maple]

okay good

now what do we do in z1

[wanton patrol]

you can't simplify $\sqrt{3} + i$ further

[boreal girder]

mango

[wanton patrol]

what'd u get for z_1

[exotic maple]

Z1 = 3 * 1/2 + 3 * i V3/2

[wanton patrol]

$z_1 = \frac{3}{2} + \frac{3i \cdot \frac{\sqrt{3}}{{2}}$ right

[boreal girder]

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

[wanton patrol]

oh well

$z_1 = \frac{3}{2} + \frac{3 \cdot i \cdot \sqrt{3}}{2}$

[boreal girder]

mango

[wanton patrol]

let me check rq

[exotic maple]

okay take your time dw

[wanton patrol]

eya thats correct

the only thing that u can do

is to just get them under 1 bar

[exotic maple]

okay. now for the final

[wanton patrol]

$z_1 = \frac{3 + 3i \cdot \sqrt{3}}{2}$

[boreal girder]

mango

[wanton patrol]

so now you just multiply

[exotic maple]

what?