help-0

[proven robin]

and replacing it with the integrals

[mellow grail]

yeah

[proven robin]

but that doesnt seem to work

i dont know

do you know how?

[mellow grail]

Well i am thinking

wait if you write it as a product of 3 terms

(-1)^n * integral 1 * integral 2

[proven robin]

yes

and a sum of this product

now what?

[mellow grail]

uhh

1 = 1/n * n right?

[proven robin]

yeah

[mellow grail]

so if u let x = 1/n, y = n in third identity

u have a double integral

double integral * third identity

bruh that doesnt work

since 1 is not < 1

[proven robin]

hmmm

[mellow grail]

xy = 1 in our case

[proven robin]

<@&286206848099549185>

there's an answer here but it is limited for subscribed users

https://www.chegg.com/homework-help/questions-and-answers/problem-5-find-exact-value-convergent-infinite-series-1-n-n-1-rewriting-sum-double-integra-q42190500

[alpine sable]

'member slader?

yeah i 'member slader

[proven robin]

?

wdym

[mellow grail]

i think this converges to phi

[proven robin]

no

log means ln

[mellow grail]

bruh i used another calculator it recognised my input wrong

[mellow grail]

yeah i got that finally on the calc

[proven robin]

which calc?

wolfram?

[proven robin]

?

[mellow grail]

no

.....

[proven robin]

well

do you have any idea?

[mellow grail]

Nope

[proven robin]

do you know anyone who can help?

[mellow grail]

Ann, MethsIsAlwaysRight,NEONPerseus

[proven robin]

can i mention them?

[mellow grail]

I think so

Fun fact, u can write the double integral as double integral of another sum divided by (-1)^n * x

[proven robin]

@vale wigeon

could you please help us?

[vale wigeon]

augh i literally got pinged thrice in three different channels

[proven robin]

sorry

[mellow grail]

$$\sum_{n=1}^{∞} \int_{0}^{1} \int_{0}^{1} \frac { \frac{-xy}{1+xy}}{(-1)^n*x} dxdy$$

Brruh why did everything go to hell

[proven robin]

that's kinda strange isnt it

[mellow grail]

Yeah but its smthing ig

[proven robin]

the sum in the nominator is with respect to n

but theres no n there

[mellow grail]

Oh shi-

that wasnt supposed to be there

it should've simply been this

[ocean seal]

ItzKraken2

[lone heart]

@proven robin Has your question been resolved?

[lone heart]

@proven robin Has your question been resolved?

[lone heart]

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[near rose]

Hi, What is the value of (a-b)/(a+b) when 9^a=2^(1/b), (a+b)^2=6*log2/log3 for both positive numbers a,b?

[lone heart]

@near rose Has your question been resolved?

[wet nest]

ablog9=log2
ab= log2/log9= log2/2log3
a^2-b^2 = (a+b)^2-4ab = 6log2/log3 - 2log2/log3= 4log2/log3
(a-b)/(a+b)= a^2-b^2/(a+b)^2= 4(log2/log3) / (6 log2/log3)=2/3

[near rose]

wow thank you so much

[lone heart]

@near rose Has your question been resolved?

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[alpine sable]

can anyone help with this

[glacial rover]

The question isn’t phrased great but I’m assuming that the % of willing does not include those who are also willing to pay all and most of the fee

I’m guessing it would say at least half for it to include the other data points

I mean it’s clearly not B or D

I’m leaning towards C as it makes sense it just isn’t exactly correct

But it still could be A, it seems to depend on the extremity of exclusion for mistakes in the statement

As I would argue you can infer that it is correct but it should be induction not certainty

[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

sorry i have brain damage

it helps when people help me

[lone heart]

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[glossy hill]

Hi, is that true ?

[vale wigeon]

@lunar fog please open your own channel. the other person got here first. #❓how-to-get-help

@glossy hill yes

it's just a^2 b^2 = (ab)^2

[lunar fog]

?

[glossy hill]

and a^2 + b^2 = (ab)^2 FALSE ?

[limpid spade]

a²+b²=a²b² looks right to you?

[glossy hill]

i think no

ok thanks you !

.close

[lone heart]

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[floral knoll]

Hi, im currently trying to understand the equation for radioactive decay and activity and i came across the word "decay constant". its calculated like that: λ = ln(2) / T1/2. But does someone know, why I have to calculate it with the ln(2)

[tacit arch]

What kind of answer are you looking for

You want the math derivation?

Or you want to know something else?

[floral knoll]

yes

[stark yew]

Do you know integration, @floral knoll?

[floral knoll]

nope

[stark yew]

Then you won't be able to follow the mathematical derivation.

But here's an overview anyway:

Suppose N is the size of a population of radioactive atoms at a given time t, and dN is the amount by which the population decreases in time dt; then the rate of change is given by the equation dN/dt = −λN, where λ is the decay constant. Integration of this equation yields N = N_0 e^(−λt), where N_0 is the size of an initial population of radioactive atoms at time t = 0.

The half-life, is then given by N(t_half) = N_0/2

Solving this for lambda yields $\lambda = \frac{\ln 2}{t_{\frac12}}$

Suppose $N$ is the size of a population of radioactive atoms at a given time $t$, and $dN$ is the amount by which the population decreases in time $dt$; then the rate of change is given by the equation $\frac{dN}{dt} = −\lambda N$, where $\lambda$ is the decay constant. Integration of this equation yields $N = N_0 e^{-\lambda t}$, where $N_0$ is the size of an initial population of radioactive atoms at time t = 0. The half-life, is then given by $N(t_\frac12) = \frac{N_0}2$. Solving this for lambda yields $\lambda = \frac{\ln 2}{t_{\frac12}}$

[floral knoll]

thank you, ill screenshot it and try to understand

thank you very much

.close

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[tepid knot]

im confused

[ocean seal]

waris

[tepid knot]

can someone tell me how to do this problem

end of year trig killin me

[tacit arch]

,tex .sum diff trig

[ocean seal]

riemann

[tacit arch]

Second row

[tepid knot]

thx

[proven leaf]

you can also figure that sin(x) shifted pi/2 units to the left looks like another curve as well :))

[tepid knot]

im just forgetting sum diff rules

[wet nest]

Cos(90+x)=-sinx

[lone heart]

@tepid knot Has your question been resolved?

[lone heart]

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[late moss]

hello

[late moss]

can i ask a physics related doubt here?

related to relative motion>

[gilded magnet]

Ok

[late moss]

this is the ques

[gilded magnet]

Btw there's a physics server in #old-network but still ok

[late moss]

if we do it wrt truck

[gilded magnet]

Amit bijarnia ?

[late moss]

then why do we take velocity towards west and 10-vsin theta
and now towards east. vsintheta-10
lets say we need to study relative motion of b wrt a
then we just take inverse of velo or acclrn of a and add it to b right?
then why didnt we do vsintheta-10 here

[late moss]

ye

lol

how did u recognize

he is such an awesome teacher.. well better than my teacher at fiitjee atleast

but i couldnt understand this concept

[gilded magnet]

I remember handwriting....i learnt relative from him.....the way he explained was just...

[late moss]

yup

can u explain why he took 10-vsin theta

instead of vsin theta - 10

??

[queen cove]

He's measuring relative to the truck

Not to the start position of the boy

[late moss]

yes

ik

[queen cove]

And so the boy would be getting closer, not further

so the negative

[late moss]

why closer

:/

[queen cove]

Its assuming the truck is moving faster than the boy

[late moss]

he said that to judge a motion of lets say a wrt to b we will take velocity or accln of a and inverse it and add it to velo or acclrn of b

[queen cove]

If the boy was moving faster the 10-Vsin theta would be negatie

and so further away

[late moss]

i still cant understand :(

[queen cove]

both are equally correct, just take into account the negative at the end

[late moss]

which -ve?????

[queen cove]

If you use 10 - Vsin theta, if the boy was getting closer to the truck 10 - Vsin theta would be negative
If you use Vsin theta - 10, if the boy was getting closer to the truck Vsin theta - 10 would be positive

They're both valid, just check whether the end sign makes sense

[late moss]

if we take vsin theta-10.. then to get to the truck the v should be negative.. so vsin theta - 10 would be negative .. shudnt it be?

[gilded magnet]

If vsintheta were greater than 10 then he wud definitely cross?...

Since he's moving rightwards at greater speed

[queen cove]

Yes

[gilded magnet]

So it's obviously the other case

[queen cove]

Yes

but you can use either in your equations

[gilded magnet]

Along with vertical component

[late moss]

well isnt that what we want?

to cross?

[gilded magnet]

Ya observer ko rest me karva do

[late moss]

ye kb bola sir ne?

:(

ha rest mei krwaya to

truck ko

[gilded magnet]

Aur uske liye Jo Kiya vo object ko karo

[late moss]

krdiya

[gilded magnet]

Idk bruh i learnt relative 2 yrs ago

[late moss]

to fir vsin theta + 10m/s west hoga na

[gilded magnet]

Mum burnt Mohit sirs notes

[late moss]

LOL

u prepared for jee too

?

[late moss]

where v sin theta is towards east

[late moss]

yes

[gilded magnet]

Also there's vertical comp up

[late moss]

yup

[gilded magnet]

If vsintheta is greater

[late moss]

yes

[gilded magnet]

Then it's quite straightforward

[late moss]

yup

bhai ek baar kya voice call pe aa skte ho? mai kaafi der se ye question soche jaa rha.. it would be easier for me to explain my doubt @gilded magnet

[queen cove]

You can (with reference to the start point of the boy) figure out his end point. If the truck passes that endpoint before he reached it, he did not cross.

[gilded magnet]

Mera vocal chords thoda damage hua hai

[late moss]

ohhkkkkk npp

so u mean to say i can solve with vsin theta - 10 too?

[gilded magnet]

The question says 'just crosses the road'

[gilded magnet]

In that case he wudnt 'just cross the road'

[gilded magnet]

Which means he shd die or get hurt if he was an instant late

Like 0.01 sec late

[late moss]

why :(

[gilded magnet]

In first case he wud never die

[late moss]

yes

so its good isnt it?

thats what we want

[gilded magnet]

Not just that

He wud always be at a distance from the truck while having just crossed

[late moss]

s o what else do we want

[gilded magnet]

In case2... increment 10-vsintheta a bit and he dies

[late moss]

yup

[gilded magnet]

So here he wud just cross feeling the truck scratch Lil bit off his dress behind smoothly as he just crosses

'JUST CROSSES'≠'CROSSES'

[gilded magnet]

And immense relief,heartbeat

[late moss]

ahhhh

btwww

is vsin theta east - 10 m/s east= 10m/s west - v sin theta west

[gilded magnet]

Btw join the physics server.waha pe petschge hai.petschge Amit sir ka baap hai.

[late moss]

?/?

[late moss]

usse bhi baat kr rha hu abhi

lol

i thought vo foreign ka hai

[gilded magnet]

Ha plasma physics karta hai

[late moss]

o

[gilded magnet]

Max planks pe bhi tha

Solar winds vagera karte te

1 sec

[late moss]

kk

[late moss]

max planks kya hai

[gilded magnet]

Write the vectors on paper you'll get it

[late moss]

ha its equal

but if they are equal then vsin theta -10 se bhi krne ka kuch to way hoga

[gilded magnet]

But you won't 'just cross' you'll always be 8m away from truck

During motion

[late moss]

yup

acha

truck rest pe hhai

isiliye??

cos vsin theta -10 lene se hum aur dur jayege

and just cross nhu krege

?

[gilded magnet]

Actually motion...par Haan in our last image yes

[late moss]

hum to last imsge ke ref mein kr rhe na

solve

[gilded magnet]

Haan

I'm dumb

[late moss]

nhi

i m in 11th

u have much more exp

just started my 11th journey

btw is abj sir enough for jee

[gilded magnet]

Bruh i messed up everything

[late moss]

whyy

[gilded magnet]

I'm not explaining here

[late moss]

dm

?

[gilded magnet]

More than enough

[late moss]

okkk

[gilded magnet]

You can also try nikhil dabai sir for some units

[late moss]

kk

thnx alottt

[gilded magnet]

Wlc broski all the best and don't take things too seriously there r soo many colgs lol

[late moss]

i started my jee prep in 10th

dont wanna wastw all the efforts ff

[gilded magnet]

Bits manipal Amrita lpu SRM many more...

Yea ryt but if u feel like ur trapped or ever feel mentally low...u need reminders

[late moss]

well everything is going smooth rn

[gilded magnet]

Cool

Like my dp then

[late moss]

but chem is such a disgusting subject

[late moss]

yup

[gilded magnet]

Agreed

[late moss]

too much rote learning

and phy ke abhi udhar teacher ache nhi hai

so abj sir se pdh rha

[gilded magnet]

Yea btw u hearda aanand srinivas stayqrious?

[late moss]

yes

[gilded magnet]

Floatheadphysics?

[late moss]

yup

[late moss]

his vids are quite fun to watchh

[gilded magnet]

Edupoint?

[late moss]

nope

[gilded magnet]

Yea actually I'm relearning all the physics now

[late moss]

ahhhh

[gilded magnet]

Starting from motion in straight line

[late moss]

OH damn why tho?

[gilded magnet]

Reached projectile

[late moss]

studying from abj sir???

[gilded magnet]

Academic pressure didn't let me learn stuff from first principles

Now physics is not part of academics as I'm taking btech cse

12th just over

[late moss]

hey btw i have another doubt... when there is no wind and car is moving forward.. then the flag flutters in opposite direcn..but when there is wind and car is moving forward.. we take the velocity of flag same as the velo of car and ins ame direcn

why is that so

[late moss]

u just gave ur jee advanced?

[gilded magnet]

Kinematics nikhil dabai sir

Bro igtg

[late moss]

kkk

[gilded magnet]

Mom asking to sleep

[late moss]

okk cyaaaa

[gilded magnet]

Bye broski all the best

[late moss]

thnx

.close

[lone heart]

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[pallid creek]

The downfall of proof by algebra? 👀

[tacit arch]

What is your question

[pallid creek]

Why is it not working

4^1 is not -4

[north canopy]

then x isnt 1

[pseudo ice]

[pallid creek]

Yes

What did I do wrong

I feel like I have made a dumb error and it's staring right at me

[pseudo ice]

wonder where

[pallid creek]

I don't see it

Square both sides

[pseudo ice]

That introduces extra solutions when you do so

[pallid creek]

What?

[pseudo ice]

e.g. take x = -1, squaring both sides would get you that x^2 = 1, and solving that gets you that x = ±1 (a more silly example)

[north canopy]

yep

[pseudo ice]

This here already has no solutions (for real choices of x, 4^x is always strictly positive)

[pallid creek]

Oh I know

I just wanted to see what would drop out

[north canopy]

,w 4^x=-4

[pallid creek]

Well

[ocean seal]

Results provided by WolframAlpha

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

Requested by [ ]#5399

[pallid creek]

Doesn't that mean

This also has 2 solutions??

[north canopy]

u can rewrite 2 as sqrt4

[pallid creek]

Knew it had i, thought it would come out from the algebra

[north canopy]

¯_(ツ)_/¯

[pallid creek]

But I did the exact same thing

[north canopy]

yea but 4^x can be 2

4^x cant be -4

[pseudo ice]

You squared something that was negative in the first one, which created the solution that didn't exist

[pallid creek]

But -4² is always 16

[north canopy]

yea but u forced it

[pallid creek]

Bro I know.

[north canopy]

u made new solutions

if the base thing is fundamentally wrong, u wont rly go anywhere

[pallid creek]

But that is how people found out what 4^½ was no? They could've believed that 4^1 = -4 and 4 bc that's what the algebra says

[north canopy]

its wrong tho

like 4^x will always be positive

ur forcin ur solution to be a thing, while it isnt correct

[pallid creek]

But assume u didn't know that

Then what?

[north canopy]

plug back in and u see it isnt possible

[pallid creek]

It shows clearly and arithmetically that 4^1 can equal -4. I know this is wrong but I want to know why, and not bc I'm forcing it bc maths doesn't just do that from what I've seen

If I am "forcing it", then all proof is unreliable, bc I haven't forced anything. I have just done algebra

Gonna go out for a bit, I'll read it if it closes automatically or smth

[lone heart]

@pallid creek Has your question been resolved?

[lone heart]

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[little ledge]

someone ples explain the theorems

[little ledge]

bro

u need to open ur own channel

[quiet vector]

!help

[lone heart]

Please read #❓how-to-get-help

[royal plank]

which one specifically?

[little ledge]

all four

i mean nvm 3. a

evyerhing except 3 a

[quiet vector]

for 3 b

u know the inscribed angle relation with central angle right?

[little ledge]

no i wasnt paying attention

[quiet vector]

https://caddellprep.com/subjects/common-core-geometry/central-inscribed-angles/

finish this vid and come back

[little ledge]

ok

[quiet vector]

@little ledge

https://www.youtube.com/watch?v=MyzGVbCHh5M

this might be better

the last one probly doest have a proof

[little ledge]

i get it

[quiet vector]

ok so look into 3 b then

break the required angle into arcs that u know abt

[little ledge]

whatt is teh intersection meant to be the center

there is no center

[quiet vector]

the intersection isnt the center

imagine or draw a center O

[little ledge]

ok

[quiet vector]

the required angle can be broken down to these arcs; this should be enough help for u to figure out the answer

[little ledge]

ohh i see it now

thank u

.close

[lone heart]

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[indigo harbor]

I was asked to differentiate that function

How to find the value of F'(x)

[placid zinc]

I'll call the inside function g(t), and say its anti-derivative is G(t). Then:

F(x) = G(inf) - G(x)

[quiet vector]

if u ignore the bounds...

[placid zinc]

What's the derivative of that equation?

(Where inf is really a lazily applied limit)

[indigo harbor]

But that wouldn't give you what G equals to

[placid zinc]

Sorry I am back

If I take the derivative of both sides, I get:
F'(x) = -g(x)

= -e^(-x)/x

This is a useful method for taking the derivative of any integral with the variable in the bound

[lone heart]

@indigo harbor Has your question been resolved?

[indigo harbor]

Thank you so much

[lone heart]

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[amber tulip]

need help with 2.1.1

[alpine sable]

Try to write it in a way that decomposes the exponent into easier forms

And successively take those

[amber tulip]

what

[alpine sable]

eg. $4^{\frac{2}{3}} = (4^2)^{\frac{1}{3}} = 16^{\frac{1}{3}}$

[trail remnant]

4th root of 81 cubed
sqrt of 81 twice is 3
3 cubed is 27

fractional powers mean roots, where the denominator is the root

the coefficient is then the integer power, in this case 3.

[ocean seal]

seymourdavison22

[proven leaf]

start with this property $\frac{1}{m^{n}}=m^{-n}$

[ocean seal]

XxMrFancyu2xX

[proven leaf]

see if you can figure out the rest from there :))

[amber tulip]

so 1 over the thing = the thing negatively?

then I just do it normally cos it's positive now?

[proven leaf]

just simplify $81^{\frac{3}{4}}$ like you do normally and keep it in the denominator:))

[ocean seal]

XxMrFancyu2xX

[amber tulip]

thank youuu

[amber tulip]

how is it done again?

[proven leaf]

use $(a^n)^{m}=a^{n\cdot m}$

[ocean seal]

XxMrFancyu2xX

[lone heart]

@amber tulip Has your question been resolved?

[alpine sable]

@mathelpers

[lone heart]

@amber tulip Has your question been resolved?

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[wheat isle]

Is this how you would do a question like this

[queen cove]

If it's an evaluate then yes the process is correct, but you forget to make it -1 + h

[wheat isle]

Oh shoot

Alright

Thanks

[queen cove]

All good?

[wheat isle]

I’m a little confused for this one as well

Should I be using m= (y2-y1)/(x2-x1)

[queen cove]

For which part

[wheat isle]

Part a and b

[placid zinc]

Yes

[queen cove]

Yes it seems so

[placid zinc]

That's the slope of a secant always.

[wheat isle]

Okay

So I know my x1 and y1 will be (1,2)

To find my y do I sub in 1.01?

[placid zinc]

x = 1.01, into the equation y = x⁴ + 1? Yeah

[wheat isle]

So we get

=
2.040604

Is that our y2 and y1 is x2=1.02

[placid zinc]

y1 is x2 = 1.02?

[placid zinc]

.

[wheat isle]

Oops

I meant to say

x2 = 1.01

[queen cove]

For a go through (1.01, y1)
For b go through (0.99, y2)
For c find the mean of these values and use that for x = 1

[wheat isle]

So is x2 going to be 1.01

[queen cove]

Sure

[wheat isle]

Is that correct for part A

m=4

[queen cove]

I think so

[placid zinc]

Why is that exact

[queen cove]

Cus the examiners are nice

[placid zinc]

That's really nice

[wheat isle]

No way

[wheat isle]

This is for part B

[wheat isle]

Find the average of both y and x values ? For part c

[queen cove]

I think actually get the gradient of the line though x1 and y1

[wheat isle]

How?

[lone heart]

@wheat isle Has your question been resolved?

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[heady plank]

1. How should i find expression for the apparent nth term a_n of the sequence?
   3,10,29,66,127....

2. Why following expression is wrong for following sequence?
   1,-1,1,-1,1...

My Answer: a_n = -(-1)^n
Actual Answer: a_n = (-1)^n+1

[queen cove]

Those answers are equivalent

[heady plank]

i know, but latter seems to be "preferred" by professor and book

not sure why

[wanton valley]

More compact way of writing it

Essentially its the same as 2*(2)^n

Makes no sense to write it that way

[heady plank]

ouch

[wanton valley]

So more compact would be 2^(n+1)

You get what i mean?

[heady plank]

honestly no, sorry ;P would be great if you can elaborate some more.

[wanton valley]

In mathemathics we want to write terms in a series as compact as possible

So instead of 2*(2^n)->2^(n+1)

Likewise (-1)*(-1)^n=(-1)^(n+1)

[queen cove]

When multiplying two exponentials you add the powers, so $-(-1)^n = (-1) \times (-1)^n =(-1)^1 \times (-1)^n = (-1)^{n+1}$

[ocean seal]

WeAreIngram

[heady plank]

oh god, you are right. can't belive i forgot that rule

thanks, now i understand that.

[queen cove]

Great

[heady plank]

.close

[lone heart]

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[bronze cloak]

i need help

[queen cove]

For (2)

[bronze cloak]

yes

[queen cove]

If the equation has two repeated roots then the discriminant (b^2 - 4ac) is equal to 0

So write out the full discriminant and set it equal to 0

[bronze cloak]

thx

[queen cove]

np

[bronze cloak]

how would i get the value of k

[queen cove]

Solve that new equation

[bronze cloak]

how

idk how to solve when there is a

power

nvm i got it

thx

[queen cove]

np

[bronze cloak]

can u tell me if this is correct

[queen cove]

You also have to consider 2x - 4 = -17

[bronze cloak]

how can i solve this using algebra

[queen cove]

Your value is correct

Another value is x = -6.5

So the sum of all solutions is 4

[bronze cloak]

thx

[solid sonnet]

You can also square both sides and solve

[queen cove]

Yep

Although doing that all the time may lose solutions

[bronze cloak]

i just calculated both and got 4

[queen cove]

Yay

[bronze cloak]

as sum of both

[solid sonnet]

Nice

[bronze cloak]

and for number 2 i got -2

is that correct

[queen cove]

Yep

[bronze cloak]

thx

can u help with q3 if possible

[queen cove]

It would take too long to type out on my phone

Basically you use their hint to simplify it to a quadratic over a quadratic = 73/3

Then you expand out those quadratics and multiply up both sides to get to quadratics equal to one another's which you can then combine into one quadratic and then solve

[bronze cloak]

how do i do that

[queen cove]

Ask someone who's on pc it would take me a week to type on my phone

[bronze cloak]

ok

.close

[lone heart]

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[chrome depot]

For what values of k will the following equation always be negative?

[chrome depot]

The equation is always negative when D < 0

[tacit arch]

What's D

[chrome depot]

Determinant

Is it called that way in english

That thing under the square root in quadratic formula

Where D = b^2 - 4ac, or in this case ->
D = (-2k^2)^2 - 4k(k^3 + k^2 - 3k - 4) = 4k^4 - 4k^4 - 4k^3 + 12k^2 - 16k = -4k^3 + 12k^2 + 16k

So, D = -4k^3 + 12k^2 + 16k

[tacit arch]

This is discriminant, not determinant

[chrome depot]

Then, we follow that D < 0:

-4k^3 + 12k^2 + 16k < 0
divide everything by k
-4k^2 + 12k + 16 < 0
divide everything by 4
-k^2 + 3k + 4 < 0
Solve the equation: -k^3 + 3k + 4k = 0 (roots are 4 and -1)

[chrome depot]

Yeah, that thing. Exchanged them XD

[chrome depot]

And then, logically, for (-inf, -1) U (4, +inf), the function is negative

And that should be the solution

But it isn't

According to the book

[tacit arch]

Show the question and answer from the book

[chrome depot]

It's in Serbian

But here is the image if you want

@tacit arch (in case you are waiting for the ping)

Another observation: k must be negative and D < 0, for this to be always negative

[tacit arch]

,rotate

[chrome depot]

I have translated the problem up there, there's nothing more to it

[tacit arch]

when you divide everything by k, how do you know the inequality sign doesn't change

[chrome depot]

OOOOOOOH

I don't know

That's a good observation

So that's an illegal move?

Wait wait wait

Let me try and solve now!

[tacit arch]

you just have to treat k>0 and k<0 separately

[chrome depot]

Yes!

[chrome depot]

And

In fact

Only k < 0

Because that's what exercise wants

Like, k = a in this problem, and when D < 0 and a < 0

Then the function is completely negative

Solution found!

Thank you very much @tacit arch !

[lone heart]

@chrome depot Has your question been resolved?

[lone heart]

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[brazen berry]

Linear Algebra PQ ive been stuck on

[brazen berry]

does it have something to do with the e_1, e_2, and e_3 vectors?

1
0 = au + bv + cw
0

looking through my notes i cant really find much related to this

or should i put the 4x1 vectors in a system and reduce to ref or rref?

[brazen berry]

i think it's this

[naive valley]

if you stack the three equations together columnwise, you get something of the form TA = B, which you can solve as T = BA^-1

[brazen berry]

I'm not sure i understand that, sorry

any way you can simplify or reword the explanation?

 3 0 2
 3 0 0
 3 2 2
-1 4 2
```?

like that?

[naive valley]

that's B

[river blaze]

You just multiply botch sides with the inverse of A

[naive valley]

A is ```
0 1 -3
-3 -4 4
-2 -2 1

your three equations together are the same as TA = B

[brazen berry]

oh okay, thats making a bit more sense to me

how do i find the inverse of A again?

[naive valley]

you can start with the augmented matrix [A | I] and row reduce it until you get [I | A^-1]

[brazen berry]

okay, i think that makes sense!

ty both!

.close

[lone heart]

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[silver portal]

I need help

[shy dove]

@silver portal when they don't do it for you, labeling sides usually helps.

And putting a variable for the sides you don't know

[silver portal]

So a b c and h

[shy dove]

yeah I suppose. I'll draw some things and send the picture in a second

let's take this for example

can you tell me what side CD is?

[silver portal]

21?

[shy dove]

21 is the whole thing

[silver portal]

10.5

[shy dove]

you don't know if it's split halfway

hint: it's not a number alone, it contains another side

[silver portal]

thats the only side cd can be aint it

[shy dove]

i'm not sure I get you. But does this make sense to you?
CD = 21 - AD

[silver portal]

Ye

Why - ad

I never done this before so i dk

[shy dove]

because the side on the right = whole thing - side on the left

(on the right and left of the dotted line that is)

another way to see it is:

whole thing = left + right
(do - left on both sides)
whole thing - left = right

does this make more sense?

[silver portal]

Ye

whole thing is 21 tho

left isnt given

[shy dove]

yeah that's the thing, you have to work around that, but I'm getting there

using CD you just found (21-AD), BD, and BC, can you think of a way to solve for AD?

[silver portal]

Bc - bd

[shy dove]

I mistyped it sorry 😅

I meant the two sides with known lenghts plus the one with 21-AD

[silver portal]

So 8 and 17

[shy dove]

yes

essentially that bigger triangle on the right

a right triangle might I add, might help you see it

[silver portal]

Weird question but u know what an isoceles triangle is

[shy dove]

two sides have same measures

[silver portal]

If those 2 triangles were split the left would be isoceles and the right would be a right triangle

If i solve those like that would i still get the right answer

[shy dove]

I'm not sure. Honestly I'm seeing one way, but there probably are more. Here I'll send what I mean, maybe I'm just not clear

with the three sides I drew, you can find AD. and with AD, the question mark side

bleh, I need to leave, very sorry about that, but if you use pythagora's theorem with what I've shown you will be able to do it. I believe in you :)

[lone heart]

@silver portal Has your question been resolved?

[silver portal]

<@&286206848099549185>

[lone heart]

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[wheat isle]

With questions like these

[wheat isle]

Where it requires a phase shift

Would this be considered correct or incorrect

Do I have to complete the curve that appears before pi/2 or no

Because it starts at pi/2

[vale wigeon]

sinusoids do not have any inherent start point

you, however, are asked to graph yours over the interval [0, 2pi]

and right now, you are missing the part from 0 to pi/2

so yes you have to complete that too

[wheat isle]

Oh wait

It should be going down to -3

[vale wigeon]

sure should...

[wheat isle]

also I need help with solving tan2(theta)=1

I have a lot of trouble with solving equations over a restriction

so I have tan^-1(1)= 45 degrees for base angle

And tan is pos in Q1,Q3

But the restrictions say [-180,180]

@vale wigeon does this mean Q3 won’t be part of the solution?

[vale wigeon]

i think it's not very helpful to think of quadrants here simply because your equation is tan(2x) = 1

you are best off writing down the general solution, i.e. in degrees, 2x = 45 + 180n

wait, did you mean tan(2θ)=1 or tan^2(θ)=1?

tan2(theta) feels like deliberate ambiguity

@wheat isle?

[wheat isle]

the first one

[vale wigeon]

ok

well you should NEVER write as tan2(theta)

anyway

[wheat isle]

What I thought was I can use quadrants and then divide the angles by 2 afterwards

[vale wigeon]

yeah, that approach makes the domain restriction tricky to account for.

[wheat isle]

But quadrant 3 and the restriction -180,180 confuses me

yep

[vale wigeon]

that's why i am suggesting a different approach that makes the domain restriction straightforward to deal with.

first solve the equation over all real numbers
then filter out only those solutions that belong to your desired interval

[wheat isle]

How do I solve tan(2)theta = 1

I don’t think there’s anything to simplify

[vale wigeon]

this is deliberate and malicious on your part...

there's no other explanation for why you keep putting parentheses in exactly the wrong places

[wheat isle]

tan(2theta)

[vale wigeon]

from tan(2theta) = 1, go to 2theta = arctan(1) + 180n

where n ranges over the integers of course

[wheat isle]

45+180n?

[vale wigeon]

yes

[lone heart]

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[celest stump]

how do i do this?

[lone heart]

@celest stump Has your question been resolved?

[celest stump]

<@&286206848099549185>

[weary widget]

Well...

first, find the intersection point of $y = 3x+2$ and $y=5x+3$.

[ocean seal]

1048576Prog

[celest stump]

3x+2=5x+3
-2x+2=3
-2x=1
x=1/-2

oops

[weary widget]

OK that's ok

[celest stump]

i know you break it down to parametric equations(i think) but idk how to deal with the 3 variables

[weary widget]

expand $(0,-1,-2)+k(1,2,3)$

[ocean seal]

1048576Prog

[weary widget]

like $(0,-1,-2)+k(1,2,3)=(\cdot,\cdot,\cdot)$

[ocean seal]

1048576Prog

[celest stump]

oh

(0+k,-1+2k,-2+3k)

[weary widget]

simillarly expand $(1,2,0)+m(-1,1,0)+n(0,1,-1)$

[ocean seal]

1048576Prog

[celest stump]

(1-m,2+m+n,-n)

[weary widget]

So, (0+k,-1+2k,-2+3k) = (1-m,2+m+n,-n)

[celest stump]

right

[weary widget]

that is
$$

[celest stump]

(k=1-m,-1+2k=2+m,-2+3k=-n)

[weary widget]

ok

It's a multi-variable equation

[celest stump]

yeah..

[weary widget]

well

put the numbers right and the variables left.

[celest stump]

$(1=-m-k,1=m-2k,-2=-n-3k)$

[ocean seal]

Joshii

[weary widget]

How did you get $k=1-m$ to $1=-m-k$？

￥

[ocean seal]

1048576Prog

[celest stump]

oops

$(1=m+k,1=m-2k,-2=-n-3k)$

[weary widget]

ok

[ocean seal]

Joshii

[weary widget]

$([A]1=m+k,[B]1=m-2k,[C]-2=-n-3k)$

[ocean seal]

1048576Prog

[weary widget]

just a label

what will have if $A-B$

[ocean seal]

1048576Prog

[weary widget]

for both side of the equation

[celest stump]

$0=-3k$

[ocean seal]

Joshii

[weary widget]

$\implies$?

[ocean seal]

1048576Prog

[celest stump]

k=0?

[weary widget]

Yeah.

[celest stump]

can i sub that in everywhere?

[weary widget]

replace $k$ to $0$ for $[A],[B],[C]$.

[ocean seal]

1048576Prog

[weary widget]

Then it is a bi-variable equation.

and it can be solved.

[celest stump]

$(1=m, 1=m,-2=-n)$

[ocean seal]

Joshii

[weary widget]

ok. so we can solve $(k,m,n)=?$

[ocean seal]

1048576Prog

[celest stump]

$(1,1,2)$

[ocean seal]

Joshii

[weary widget]

k=1???

k=1???

[celest stump]

oh oops sorry

(0,1,2)

[weary widget]

just, check it. check if $(0,-1,-2)+k(1,2,3) = (1,2,0)+m(-1,1,0)+n(0,1,-1)$.

[ocean seal]

1048576Prog

[weary widget]

Oh it is just a sketch, maybe some calculation in it is wrong. check it. But you know how to solve multi-variable equation.

[celest stump]

yes now i do

$(k=1-m,-1+2k=2+m+n,-2+3k=-n)$

[ocean seal]

Joshii

[celest stump]

$(1=k+m,-3=m+n-2k,-2=-n-3k)$

[ocean seal]

Joshii

[celest stump]

can i do B + C? @weary widget

mv,

nvm

[weary widget]

Yeah.

you can do anything algebraic

because $P=Q,R=S \implies P \circ R = Q \circ S$

[ocean seal]

1048576Prog

[celest stump]

but now i cant get rid of 2 variables

[weary widget]

?

[celest stump]

like if i subtract/add i'll be left with 2 variables

[celest stump]

i re-did the calculation it should be this instead

[weary widget]

OK

what's the equation

[celest stump]

$(1=k+m,-3=m+n-2k,-2=-n-3k)$

[ocean seal]

Joshii

[weary widget]

B+C, then?

[celest stump]

$(-5=m-5k)$

[ocean seal]

Joshii

[weary widget]

so solve the equation with $[A]$ and $[B]+[C]$.

[ocean seal]

1048576Prog

[celest stump]

oh i see

$(6=-4k-3)$

[ocean seal]

Joshii

[celest stump]

i must of done something wrong

oh i think i did wrong order

$(1--5=m-m+k--5)$

[ocean seal]

Joshii

[celest stump]

$(6=0+6k)$

[ocean seal]

Joshii

[celest stump]

$k=1$

[ocean seal]

Joshii

[celest stump]

$(k=1,m=0,n=-1)$

[ocean seal]

Joshii

[celest stump]

thats not the answer tho

:(

<@&286206848099549185>

[sly spade]

What's your question again?

[lone heart]

@celest stump Has your question been resolved?

[odd edge]

I feel like just graphing it would help

Or at least help you see how to figure out how to solve with math

[celest stump]

the way @weary widget said to do it is right im just messing up somewhere

[lone heart]

@celest stump Has your question been resolved?

[lone heart]

@celest stump Has your question been resolved?

[lone heart]

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[minor aspen]

i really dont understand how this happened, why did the + change to - after moving j to the numerator

[minor aspen]

and how can we even move it up there

what rule is that haha

[visual stream]

everyone

The latest definition of time and space
four-dimensional
is achievable

Light distorts space, but time does not, time is like a circle

I do not like China
You can use translation software to translate Chinese
Thanks------

[minor aspen]

sry what

[visual stream]

Discovering patterns can make powerful bombs

[minor aspen]

are you real? @visual stream

[visual stream]

the light just like this

you can think this no real

[alpine sable]

You sound like you tried to Google Translate your messages but accidentally translated them back and forth 20 times

[minor aspen]

hahahah

fr

[visual stream]

but you know the light and time and space very like

[minor aspen]

yes yes very

.close

[lone heart]

Channel closed

Closed by @minor aspen

Use .reopen if this was a mistake.

[visual stream]

yes

[visual stream]

I am china people but I don't like my country

[alpine sable]

hello china people

[visual stream]

you can get important thing in the picture

if the time and the space like this picture

what mean ?

all of the science is a wrong,maybe

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[visual stream]

four-dimensional can give us a turth

[lone heart]

Please don't occupy multiple help channels.

[ornate leaf]

等等

我懂中文

i can have a shot at translating

[visual stream]

可以的

[ornate leaf]

等等

兄弟

你是不是走錯路了

[visual stream]

好

[ornate leaf]

這裏是求助channel

要討論到 #discussion 討論

[visual stream]

我不像你这么爱这个国家（中国）

[ornate leaf]

誰説我愛國

只不過懂中文而已

跟愛國拉不上丁點關係

[visual stream]

我希望美国将来可以统治世界

[ornate leaf]

嗯

個人有保留

[visual stream]

我希望有人可以掌握时间

[ornate leaf]

美國與中國也不是很好啦

[visual stream]

很空间

[ornate leaf]

哦

也合理

欸

其實最好的結果一定是世界和平啦

等等等等

你怎麽上discord

[visual stream]

科技最好的结果是造福世界

[subtle birch]

wow

[ornate leaf]

大陸不是把discord禁了嗎

vpn？

[visual stream]

拜拜了

[ornate leaf]

嗯

好啦好啦

拜拜

.close

[visual stream]

有缘再见

[ornate leaf]

有緣再見

[keen plinth]

[minor aspen]

is that a bot

[keen plinth]

no

[subtle birch]

檢測到中國玩家

[ornate leaf]

who

who's a bot

[minor aspen]

even in chinese he doesnt make sense ?

[ornate leaf]

wdym

me?

me or the other person

[minor aspen]

no haha

[keen plinth]

idk what you're talking about

[ornate leaf]

no i think he makes sense

[keen plinth]

yeah

[ornate leaf]

hes just very against china

[keen plinth]

definitely not bot

lol

[ornate leaf]

yeah

its more like

bots arent allowed to have that kind of opinion

[keen plinth]

[ornate leaf]

it would be abolished by china otherwise

oop

[keen plinth]

i suppose he left

[minor aspen]

why is it in the math channel tho

[ornate leaf]

it started out as math but then like

[keen plinth]

close or not

[ornate leaf]

not yet

[ornate leaf]

i offered to discuss with him in chinese cuz like

his native language is probably chinese

but then he said stuff like

nah i dont like china and shit

[subtle birch]

you know chinses or like you use translator

[minor aspen]

oh wait you actually thought he's not a bot?

[ornate leaf]

but then it devolved into some political conversation lmao

[thorn tapir]

[keen plinth]

that was so random

[ornate leaf]

im bilingual native

the thing in my pfp are my chinese exams

reading and writing

[keen plinth]

dude isnt a bot

[ornate leaf]

hes probably not a bot

no bot speaks like that

no its more like

no bots are allowed to speak like that

or else china's gonna come in do shit

[keen plinth]

no bot just leaves half way through a conversation

[minor aspen]

ahaha true

[ornate leaf]

yeah

[minor aspen]

he just came here to discuss the world state real quick

and peaced out

[ornate leaf]

sanest chinese citizen

[subtle birch]

fr

[keen plinth]

kinda random

.close

[lone heart]

Channel closed

Closed by @keen plinth

Use .reopen if this was a mistake.

[visual stream]

my mind is like a robot
like CHAT GPT
Or CHAT GPT like me?

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[minor aspen]

do you still need help with science

[ornate leaf]

他們認爲你那種行爲好像預設的bot

[alpine sable]

asking the real questions here my man

[vale wigeon]

chatGPT bad.

[long axle]

I think chatgpt is like you

[visual stream]

What is the CHAT GPT neural network element like?

[alpine sable]

hmmmmmmge

how many hmmges do you have

[keen plinth]

5

thats it

im out

[alpine sable]

ran out of hmmges 😔

[keen plinth]

well

.close

[lone heart]

Channel closed

Closed by @keen plinth

Use .reopen if this was a mistake.

[visual stream]

Free neurons are better than CAHT GPT

[alpine sable]

inb4 this gets reopened for like another 3 times

[keen plinth]

[alpine sable]

so true bestie

pop off !! 😎

[minor aspen]

i think china chat gpt will rule

you know?