math-help

[left knoll]

I need to think about it

Somehow it is still confusing me

Yes pls

Gimme an example

You are a teacher?

You just took a random number?

Yeah but from where comes the 3

Yeah but I could add 4 too?

Instead of 3?

So you just added 2/2 there cause you can?

Can I add 3/3 instead of 2/2?

WTH

Yes?

So i have to take the 2

Ahhhh

Wait I understand. Let me tell you. And you can tell me if it is good or not?

I have x^2+2 and if I derive it. It is 2x. So I take this 2 and multiply it with both sides

Wait I will write it down

Is that right? I mean the meaning behind it?

I know. It is just for me. I always have steps in math 😂

Even this little step

See you 🙂 and thank you

Will be someone else here to help?

[left knoll]

does anyone here know how to solve logarithmic inequalities? ive never seen a logarithm in an exponent in these kind of exercises

[cinder yacht]

it's in the form x^a > 8 with a = log10(2) ( a known constant)

to solve it u do ^ (1/a) on both sides
so [x^a]^(1/a) > 8^(1/a)
ie [x] ^ (a * 1/a) > 8^(1/a)
ie x > 8^(1/a)

and u replace "a" with "log10(2)"

[left knoll]

can someone explain this please

[sullen verge]

u got x = ksqrt(y)
if y is increased by 44% u got
x = k * sqrt(1. 44y)
x = k * sqrt(1.44)*sqrt(y)
x = 1.20 * k * sqrt(y)
so x in increased by 20%

[true steeple]

What does the + sign in (b) mean? Union?

[true steeple]

Got it

[ember kernel]

Yep that's the set of all possible linear combinations between vectors from V1 and V2

[true steeple]

Thank you

[glacial dust]

can someone PLZ explain this to me u dont have to like go into practical solving

just give me a ran through everything

mention me if u have time to help ty

[left knoll]

got that, but how do i get rid of the a in x > 8^(1/a) now?

[feral wedge]

is this right or it must be 3/2?

[cinder yacht]

u don't get rid of it, that's just ur answer

[left knoll]

nah man thats not the answer, the answer in my textbook is ]1000; +∞[

[left knoll]

If u look closely the 2/2 wasnt added it makes up 1 (2÷2 =1) it was multiplied after that 3 was taken common in denominator then it was multiplied by 2 of denominator so it became 6 an the 2 of numerator was multiplied by x so next line u got 1/6 × rest of that ( hope u understand and don't mind the spelling mitsakes if there is any

[left knoll]

Oh yes now I understand it. Thank you

Why are the messages deleted?

[left knoll]

Welcome buddy glad u understood

[left knoll]

I dont know

[green flicker]

Marked one please

[dense fog]

@glacial dust

[glacial dust]

aight aight ty

[frozen meadow]

any good websites to practice calculus. college level. just started studying calculus. a website which gives progress features or topic specific questions would be helpful

[cyan night]

khan academy calculus ab is not bad

o college level?

[frozen meadow]

yep

[cyan night]

i have been using pauls online math notes and its really good for revision

https://tutorial.math.lamar.edu/

[frozen meadow]

oh okay. tanx

[cyan night]

"tan x"

[frozen meadow]

do they practice questions

[cyan night]

yeah

[frozen meadow]

alright. and noice math joke btw

[cyan night]

kk

[frozen meadow]

damn this is very easy to use. thanks so much

[cinder yacht]

damn i feel stupid then, but let's see :
a= log10(2) = ln(2)/ln(10)
so 1/a = ln(10)/ln(2)
and we know 8^(1/a) = exp( ln(8) * 1/a)
therefore 8^(1/a) = exp( ln(8) * ln(10) / ln(2))
= exp( ln(10) * ln(2^3) / ln(2))
= exp( ln(10) * 3 * ln(2)/ln(2))
= exp ( ln(10^3))
= 10^3

to not get confused, I was taught to always view log_a(b) as ln(b) / ln(a)
and to view x^c as exp(c * ln(x)) especially when c isn't an integer*

there's probably a shorter way than what I did, but what I did is what I do when stuck with logarithms and exponents

[frozen meadow]

is csc short form for cosec

[cinder yacht]

a faster way to see it would be, since (1/a) is an exponent of 8, to write it as a single logarithm of base 8
so 1/a = ln(10)/ln(2) = 3 * ln(10) / [3 * ln(2)]
= ln(1000) / ln(8)
= log_8(1000)
so 8^(1/a) = 8^(log_8(1000))= 1000

[cyan night]

yep

[left knoll]

nah man its all good!! thank you for taking time out of your day to help me. just a question: how would you solve it if you were to add a log on both sides like this? i havent seen "In" yet so it wouldnt be smart for me to solve it like that

[cinder yacht]

oh yea, log10( x¨ [log10(2) ] ) becomes log10(2) * log10(x)
and log10(8) = log10(2^3) = 3 * log10(2)

so the inequality becomes log10(2) * log10(x) > 3 * log10(2)
you divide both sides of the inequality by log10(2) (knowing that log10(2) > 0, it doesn't change the sign of the equality) which gives :
log10(x) > 3
which is equivalent to : x > 10^3

[atomic dagger]

"If A is a 8x5 matrix such as dim(Ker(A)) = 2, then the dimension of the vector subspace generated by the lines of A is equal to 3. True/false" The answer is true, could someone explain to me why ?

[left knoll]

OH WOW, thank you so much!!!!

[cinder yacht]

théorème du rang

dim(Ker(A)) + dim(Vect(A)) = nombreDeColonnesDeA

2 + dim(Vect(A)) = 5

[atomic dagger]

merci beaucoup!

[cinder yacht]

de rien

[simple bough]

Algebra 2 how do u define always sometimes never?

[ember kernel]

I don't understand the question sorry

[simple bough]

Looks like this sorry to inconvenience ur day

[ember kernel]

Always means the equation holds for any value for x (or a)

Sometimes means the equation only holds for some values for x (or a)

Never means the equation holds for no value for x (or a)

@simple bough

[simple bough]

Ohhh ok thank you have a good day then thank you again

[ember kernel]

Np, you too

[left knoll]

does this have answers to the problems?

[left knoll]

Not all but mostly yes. I recommend it too.

also does anyone know what dominance means in terms of calculus

am asking in terms of this

[ember kernel]

Well the definition is literally in the screenshot

[cinder yacht]

the definition is in the screenshot
the unofficial meaning is that e^x is strong as f**k
like how e^666 is soooo much bigger than 666

and e^(-666) is wayyyy too close to 0 ( (compared to 1/666 for example)

since e^x is so strong, e^x is the big boss that determines where a limit is going if you have a multiplication or division of e^x with something relatively weak like x^k

when x->+infinity or x->-infinity

[left knoll]

tysm ur explanation was perfect!!!!! 😆 ♥️

[left knoll]

i didnt understand what it was trying to say

[cinder yacht]

ur welcome :3

[ember kernel]

In common language that means that e^x diverges "faster" than x^k (e^x gets very big much much quicker than x^k)

[left knoll]

lmao part d, my brain isnt working

pls help twill be greatly appreciated

[ember kernel]

Point of inflection : y'' = 0

That is, (2+x) e^x = 0
Since e^x is never 0, you can divide both sides by e^x to simplify the equation to 2+x = 0

Solved

[left knoll]

ahh ok ok ty

[pulsar turret]

for a relation to be a function, each x-value must correspond with only one y-value but in the textbook I'm using ...

Is there an explanation for this? Or is it only an exception from the general rule applied for a relation to be a function?

[glacial dust]

can someone help me

for c and d

[frozen meadow]

https://tutorial.math.lamar.edu/Solutions/CalcI/ChainRule/Prob9.aspx
can someone explain me this solution

[supple island]

imagine it as a^x, yoy write it as e^ln(a^x)=e(x*ln(a))

and then you do the derivative

[fervent tide]

In the table, you'll notice that each input x has only one corresponding output y, which makes it a function. It doesn't matter that the outputs are the same
This relation could not be a function

 x |  y
-5 |  2
 0 |  2, 3
 5 |  2

because as you see, it is not clear what output we would get for the input 0

[steady solstice]

help me solve these questions...😖

[supple island]

can you post a bigger pic?

[ember kernel]

Hi
I'm always enthusiastic about helping people understand difficult concepts and helping them solve particular questions, but you might not realise that you're asking us to do your whole assignment, no questions asked.

Is there something you don't understand in a specific question ? Are you stuck at some step and you don't know to get past it ? These types of questions I will answer to. But "please do my whole assignment" is too much to ask in my opinion.

I'm saying that because that may (or may not idk) be how other helpers feel about your question

[cinder yacht]

no it's " y''=0 and changes its sign"

[ember kernel]

I had forgotten about the 2nd condition, thanks for putting me right

[ornate plank]

hello people of the internet; I'm supposed to decide wether this function has or doesn't have one single asymptote.
And I have no idea how to even start?

Am I supposed to determine the domain first? Because I don't know how to do that without knowing the value of b

[pulsar turret]

Can someone help me solve this question?

[left knoll]

please help

[left knoll]

can someone help me with this question

I don't know where to go from here

[cinder yacht]

there are 2 types of asymptotes :
-asymptotes when x-> +infinity or - inf, those are often horizontal (sometimes inclined)
for the horizontal, there's a horizontal asymptotes when lim f(x)=a constant when x->+inf or - inf
so that's what you gotta compute

-vertical asymptotes : those can happen when there are points of "disconnection" ("c" for example) in the domain, and when lim f(x) = +inf or -inf when x->c+ or x->c-
so you gotta find the domain, and ur right it does depend on b here. because when computing "delta" of the denominator :
if delta<0 : x^2+2bx+1>0 all the time so D = R
if delta = 0 : x^2 + 2bx + 1 = (x-x0)^2 where x0 is the unique solution of x^2+2bx+1=0, so D =(-inf, x0[) U (x, +inf) and x0 is indeed a point of "disconnection"
if delta>0 : x^2+2bx+1 = (x-x1) (x-x2) : two points of "disconnection" where u gotta compute the limit of f(x)

[willow basin]

hiya! thanks. but i don't understand why x^4-x is not x^3. x^4-x^1 = x^4-1 = x^3. I didn't mean that if u differentiate x^4-x you'll get 3. I wanted to simplify x^4-x^1 before differentiating. why does that not work if u dont mind me asking?

x = cube root of -8. = -2. but this is still wrong. idk maybe i did smth wrong or...idk.

[cinder yacht]

you went from x^4 - x^1 to x^4 - 1

what happened there

[willow basin]

to the power of 4-1

[cinder yacht]

np

no

[willow basin]

i.e 4-1 = 3

?

[cinder yacht]

what you seem to mean is that x^4 - x^ 1 = x^(4-1)

[willow basin]

isn't that the indice rule?

[cinder yacht]

right?

[willow basin]

yeah

[cinder yacht]

like you can "factor in x"

it's not really possible because 4 and 1 are in a power

[willow basin]

yeah exactly

wdym

[cinder yacht]

the rules that you do have are : x * 4 - x * 1 = x * (4-1)
and : x^4 / x^1 = x^(4 - 1)

the rule you used is some mix between the two, which is wrong

[willow basin]

how did i use a mix sorry

i belive i used only the first one

[cinder yacht]

x**^4 - x^1 = x^**(4-1) is false (it concerns POWERS)
x * 4 - x * 1 = x * (4-1) is correct (it concerns MULTIPLICATIONS)

power and multiplication are not hte same

[willow basin]

oh mb i was thinking of x^4 divided x^1 = x^(4-1).

[willow basin]

thank you

[willow basin]

also thank you

have a nice day

[left knoll]

let me know if you have any questions with differentiation

it can be a confusing topic

[willow basin]

cries please i just started it

thanks again tho

[edgy hinge]

find a,b,c (a,b,c is a prime number)

can anyone help me please

[ember kernel]

Trial and error

There's no other way to solve it

[cinder yacht]

when it's so hard u can try special cases
like what if a=b=c? well u find out that would mean 8a^2 = 4a^3 <=> a=2 so a=b=c=2 works

[left knoll]

idk why but I get really messy answers for the x bit. would appreciate it if someone can show me the working out to get the answer for x.

[candid rivet]

Is [ exp(ixn) ]^2 equal to 1 ?

[ember kernel]

Only if x is a multiple of pi (0, pi, 2pi, 3pi,...)

Because exp(i 2pi m) is equal to 1

Posting your question in the appropriate channel #science-help might help you get an answer faster

[ember kernel]

• Start off by isolating x in the first equation to express it in terms of y
• Then replace the x in the second equation by the expression you found in the first one. Carefully simplify the expression to be left with one solid fraction (the one in the answer). That's your final solution for y, as it is not expressed in terms of x
• Go back to the first equation and replace the y by the final value you got for it in the previous step. Carefully simplify the expression and you'll get the final solution for x as well

It's very normal if the equation get messy after the substitutions, that's why it's important to simplify them as soon as you can everytime to make future calculations lighter

[drifting siren]

how do I find the average rate of change for h(x)? I've gotten to this work but I can't figure out how to simplify that huge chunk on the top

Here’s my work:

[shut violet]

Recall the expression a-b^2 = (a-b)(a+b), so that 9-(x-2)^2 = (3-(x-2))(3+(x-2))

[tropic mist]

Could I please get some help with question a. I understand why you would translate it 1 place to the right, but how are the y-axis intercepts and asymptotes found? Also, how is the domain R? The first image is the question and the second one is the answer from the textbook. Thank you.

[vestal kite]

m

[gritty cloud]

Can you help me guys with this?

And what YouTube channel is good at teaching Calculus 1 ?

[left knoll]

1. gradient is negative reciprocal

2. use point gradient formula to find equation of line: y-y1=m(x-x1)

[gritty cloud]

Good

[glad sierra]

https://www.youtube.com/channel/UC6AVa0vSrCpuskzGDDKz_EQ
found this amazing channel
worth it!
enjoy 🙂

[left knoll]

good?

so i guess u get it then?

[gritty cloud]

Yes

Do you know where i can get some more calculus questions to practice with?

A book

Or a channel

[left knoll]

@gritty cloud a good book is schaums calculus

james stewart is another good book

michael spivak is also a good book

3blue1brown seris on youtube called essence of calculus is a good way to introduce calculus to yourself

he goes into a lot of depth

[dense fog]

@gritty cloud fighting

[left knoll]

?

[random flame]

Can someone help me with Q8a? I got an ans that is close to the book’s ans. I’m not sure whether it’s just me or the book having a stroke qwq

Ping me if ya do help and thanks in advance!!

This thing is due tmr and there’s still more questions I haven’t done-

This was the working I did (yep I cancelled it but pretend the line wasn’t there-)

[cinder yacht]

why did you factor x^3-x+6 before computing dy/dx

but other than the fact that it made u waste time, the answer seems correct to me

an even faster way to do it would also be to say that the tangent at P's equation is y = mx+p
since it's parallel to the y=2x-5 line, then m=2 (they have the same m coefficient begause they're parallel)
so y = 2x + p
plugging in y=6 and x=-1
we get 6 = -2 + p so p = 8

[random flame]

Ohh

Okay thanks a lot!!

It makes sense

Am kinda ded at Q8b now- idk what’s the question asking for

[gritty cloud]

Thanks 💜🌹

[weary arrow]

If you draw a coordinate system and mark off the points P and Q, and then draw normals from both those points, they will intersect at point R and form a triangle, which you are supposed to find the area of

A normal from the circle

I dont remember enough geometry to tell you how to do that right off the bat

[random flame]

Ehh it's not Q10b

It's Q8b

[weary arrow]

OH lol

[random flame]

Bruh I wanna legit give up, this is due tmr

It's 9:16pm where I am

And I'm already tired ;;-;;

[sullen verge]

ive been typing the answer for 10 minutes and it just vanished f

[weary arrow]

RIP that sucks

[sullen verge]

yeah

f

[random flame]

Rip ;w;

[sullen verge]

for a you got y=2x+8

then

the tangent to point Q is parallel to that of point P

so bothe have the same slope

[weary arrow]

Do you have geogebra by any chance? I would've honestly just used a math program to solve these im too lazy lol

[sullen verge]

that is 2

so then you have the formula for the derivative of the function that is
f'(x) = 3x²-1

you solve for that equals to 2

[random flame]

Wait where did 3x^2-1 come from-

[sullen verge]

so 3x²-3=0

[sullen verge]

its the derivative of
x³-x+6

[random flame]

Oh-

[sullen verge]

u got it?

[random flame]

Yeah

I rmb smtg abt second derivative of a function in class

But the teacher wasn't emphasising on it so my mind decide to throw that away ;-;

[sullen verge]

then youll get x=1 ou x=-1 (you already used x=-1 for P)
so you find f(1)
youll get 6
so Q(1 ; 6)

[sullen verge]

that not second derivative its the first one

second derivative is when you derive the function a second time

[random flame]

Wait I got f(1)=6 -?

[random flame]

Oh-

[sullen verge]

yeah yeah i made a mistak

[random flame]

Oop issokay

Wait so that's it?

Oh okay it's correct-

Bruh I spent almost one week on this

H o w

Thanks man

I’m sorry for crowding this channel but I’m stuck at Q9-

It’s not even worth staring at my working, I just followed the textbook’s examples

[left knoll]

@random flame still need help?

[random flame]

Yes please qwq

[left knoll]

k

so

derivative of original function

u know how?

to do that?

[random flame]

Yeah

It's the f(x) prime thing right

[left knoll]

yep but the above equation requires chain rule

do u know that?

[random flame]

Yeah

That I know

[left knoll]

ok so chain rule that mf and then use point p and q to find their respective equations

keeping in mind what ever gradient u find at each point

negative reciprocal it cause its a normal

perpendicular

[random flame]

What's negative reciprocal-

[left knoll]

oh ok

so lets say u have x

[random flame]

Is it like m1 * m2 = -1?

[left knoll]

negative reciprocal is -1/x

[random flame]

Ohh

[left knoll]

if x is orginal

[random flame]

Okay

[left knoll]

negative then fraction it basically

ok then

once you have an equation for each point

p and q

make them equal to each other cause they interseact

i cant spell, jesus

[random flame]

LMAO-

Okay hold on lemme do the chain rule thing first

[left knoll]

and then you can find x and y corordiantes

boom point r

get it?

[random flame]

Hol' on

I'll ping ya if I find any troubles

[left knoll]

alg

oh and send photo of where ur stuck

[random flame]

Alright

Thanks for being patient man qwq

[left knoll]

bro all good, u got this

[random flame]

I don't think I can find the equation for point P

If I do 0^-1 it gave me maths error

Which technically means no solution for IGCSE

[left knoll]

0^-1

is 1/0

which is not possible

hold on

let me do it

what is ur derivative

[random flame]

Yeah I did that cuz negative reciprocal

[left knoll]

for gradient?

how did you work out the gradient?

[random flame]

After chain rule, it's 4(x-1)^3

I'll send my working

[left knoll]

yea

[random flame]

Ik it's abit all over the place but that's cuz I wanted to save space -^-"

[left knoll]

u never found the original gradient

[random flame]

Huh-

[left knoll]

oh lmao its on the side

sorry my b

[random flame]

Oop issokay 😂

[dense fog]

Write wrong @random flame

[left knoll]

?

the gradient being zero is weird

@random flame

[left knoll]

what does write wrong mean

[random flame]

Eh

Hmm

[left knoll]

zero gradient is just a point

a dot

[random flame]

If I sub in x=1

[left knoll]

no ur maths is correct

[random flame]

It's gonna be 4(1-1)^3

Aaa-

[left knoll]

one sec

[random flame]

Okay uh

[dense fog]

wrong all @random flame

[left knoll]

how is it all wrong?

[random flame]

Aight sure

[left knoll]

his derivative is correct

his substitution is perfect @dense fog

where is his mistake

Genuinely asking since i may be missing something

[random flame]

I think man's tryna drag you off course

[left knoll]

maybe

[random flame]

Do what ya wanted to do first

[left knoll]

ok ok

[random flame]

Then later I'll check the book ans

Yea

[left knoll]

sorry give me some time to figure this out

[random flame]

Issokay

[left knoll]

when are you gonna slee

p

[random flame]

Depends on when I can finish this

Ima be here till 1am if I have to

[left knoll]

ok so just up to q 10 right?

[random flame]

I got a whole lot of other questions to do

I'm far behind the class cuz I got busy organising school camps

Then now I'm just ded

[left knoll]

i think the question is literally broken

since everything u have done is correct @random flame

[random flame]

Impossible

[left knoll]

check the answer

we will both learn today

if that is ok with you @random flame

[random flame]

That's completely fine ^^

Okay so the ans is (1, 5.25)

[left knoll]

does the answer give working?

[random flame]

Nope-

Sometimes this book's ans gets strokes and doesn't give the correct ans

That's what literally happened while I was doing Q8

I was solving a solved question for a week

[left knoll]

ok one sec

i think i found something

i could be wrong

sorry again for this taking so much time

[random flame]

It's okay

I'm used to it

Add Maths takes me ages to complete

[left knoll]

ok can i ask how you found ur equation

for the second one at least

since i think that may be one of the issues

[random flame]

Using the (y-y1) = m(x-x1) formula

[left knoll]

so lets take point

q

y-5=4(x-2)

y-5=4x-8

y=4x-3

correct?

[random flame]

But then the gradient supposed to be smtg like 1/x?

[left knoll]

no ur correct

i was just checking all possibilities

[random flame]

Ah okay

Yeah

[left knoll]

i think either i have missed something

or the book has messed up

what book are you using?

[random flame]

Hold on

This

@left knoll

[left knoll]

ok there are no worked solution pdf available

let me try the q from scratch

sorry once again

[random flame]

Nono it's okay

I can wait

I've accepted my fate to sleep later than 12am

This is the last yr of high school after all-

[slate dune]

🙏

[left knoll]

i found our issue!

[random flame]

:0

[left knoll]

I AM AN IDIOT

[random flame]

BAHAHA AREN'T WE ALL IDIOTS/j

[left knoll]

i forgot that 0

[dense fog]

You should test basic math @random flame

[left knoll]

does not have a fking reciprocal

[random flame]

LMAOOO

[left knoll]

so use 0 as the gradient

[random flame]

Okay that's interesting

Noice noice

So that means y=4 and x=-2

[left knoll]

for point p

x=0

[random flame]

Wait wha-

[left knoll]

cause if gradient is 0

then 0(x-x1) = 0

[random flame]

Ah yeah

[left knoll]

yep

[dense fog]

Haha

[left knoll]

and then ur other equation is perfect

[random flame]

So that's all I gotta do?

Or is there more-

[left knoll]

i mean solving simultaneously should have done the trick

look i think i have definitely missed something here (my excuse is that i am tired and it is 1am where i am)

[random flame]

o-o

It's okay man

[left knoll]

can i redirect u to someone better

[dense fog]

Ok

[left knoll]

like he is a god @random flame

[random flame]

LMAO

Sure

[left knoll]

@qu\

[supple island]

is this the issue?

[left knoll]

@ember kernel

@random flame

[random flame]

Yeap-

[left knoll]

he has helped me a ton

[random flame]

Aight thanks man

[left knoll]

a genuine sait

[random flame]

Go get some rest

[left knoll]

sorry once again @random flame

[left knoll]

ill try

[random flame]

Nono it's fine

Yeah sleep well aight

[dense fog]

[random flame]

Anyways back to this hellhole-

[left knoll]

@random flame phys student is also really good

@supple island

[random flame]

Ooo aight

Big thank

[left knoll]

he can really help

once again man, sorry i couldnt help u out

[random flame]

Legit thank you man

[supple island]

ok haven't done this in a while but i'll give it a go

[random flame]

You've been a legit huge help

[random flame]

Aight sure thanks

[dense fog]

😊🙃😉🙂

[left knoll]

nah man i think i confused u more

[random flame]

Nah fam you gud

[dense fog]

Goodnight

[random flame]

OH SHT WAIT I FOUND THE ANS ADY @supple island

I decided to ask my teacher

[left knoll]

WHAT WHAT

[random flame]

He said use x=1 as an equation

Y'know how x=1 is part of the coordinate P right

Yeah so apparently that can be used

[left knoll]

OHHH OFCCCC

BROOO IM SO DUMBBB

TYYYYYY @random flame

and ty to ur teacher

[random flame]

OH WAIT NEGATIVE RECIPROCAL IS -1/x

I THOUGHT 1/x LMAO

NO NEEDA THANK ME MAN I'M JUST IN PURE SHOCK LMAO-

We all sleep deprived ppl who still somehow function 👍

Anyways THANKS AGAIN GUYS

SRSLY

Y'ALL AMAZING

[left knoll]

truuuu

k gtg

[random flame]

Baii

[left knoll]

james started stream 4/10

byyyeee @random flame tyyyyyyyy

[cinder heath]

anyone this question please

idk the answer so

it's matrix question

[weary arrow]

Idk if this is how you're supposed to do it but I found a valid solution

@cinder heath

[cinder heath]

oh thanks

[thick steeple]

3 equations? I only know how to solve systems whit 2

[ember kernel]

That's exactly the same methods as systems of 2 variables, but with one extra variable & equation

[sullen verge]

anytime!

[left knoll]

Hi! Is there someone here who is engaged in Olympiad mathematics (or an advanced school course)?

[ember kernel]

Yes

[left knoll]

@ember kernel In my Internet space it is difficult to find decent lectures on functions, functional equations, functional inequalities and the like. Do you have something to share (lectures, sources, books)? I'd appreciate it

[coral sierra]

I've managed the first part of part a where it says to show a product of numbers of the form 3k+1 gives a number of the same form, but I'm not sure how to deduce? I think I can do it by cases by considering what remainders mod 3 the factors of such a number would have but I don't think that would be deducing 😅

any help on how to do this?

[sullen verge]

anyone knows a great math/math-related course available on Coursera?

[ember kernel]

Hi ! Sorry, everything I know about calculus was taught at school/uni. I have no pdf/book/lectures that could help you

[steel crystal]

yo i need some help

could someone help me write a recursive formula for part b of this problem
i have
(0,0), (1,1), (6,36), (35,1225), (204,41616)
and the problem is
Part (b): Find a recursive relationship expressing (xn,yn) in terms of (xn-1,yn-1) for any n≥1.
??

[fast stream]

write a matrix with the coefficients of all the variables

so thatll give you some matrix M

so M(x y z) = (8 1 7)

multiply by inverse matrix of the 3x3

on both sides

then you will get sols for all 3 variables

[ember kernel]

If it can help you :
6 = 6 * 1 - 0
35 = 6 * 6 - 1
204 = 6 * 35 - 6

And yn = (xn)²

[steel crystal]

is there any mathematical way to find that

[ember kernel]

Trial & error + intuition

I checked two dozens of recursive formulas to get this one

Starting by realising that 36 = 6² - 1
And that 204 was close to 6*35

[steel crystal]

hmmm

and then for the y-coordinates?

[ember kernel]

Easy

The equalities 1 = 1² and 36 = 6² seemed obvious to me

And I just needed to verify that 1225 = 35², which is the case

[steel crystal]

ye

so the first one would be

x(n) = 6x(n-1) - x(n-2)

and the second one

[ember kernel]

(x(n))²

[glossy latch]

Can you show the whole problem?

[coral sierra]

That is the whole problem

There’s nothing above that its just another question above it

[glossy latch]

what do they mean when they say "this form"\

(same as 3k+1?)

[coral sierra]

same as 3k+1 yes

[glossy latch]

ah kk

okay, let's start with the assumption that there are a finite amount of ks, such that 3k-1 is prime

stuff like (1,2 and all other values thya are prime with that form)

[cinder heath]

3,4 from matrix

[glossy latch]

if we were to treat kz as the last final viable number such that (3kz-1) is the last prime possible in that form. Funnily enough, this problem uses a lot of part a, which it looks like youve solved. by creating a variable c, which is equal to k-2/3. Substituting this for every k in the set, you end up with the form 3c+1. Taking the product of every possible values of 3c+1 that are prime, add a particular value that guaremtees that the oroginal value of 3k-1 is preserved and then use proof by contradiction

@coral sierra

[left knoll]

The name of the book is “A textbook”?

[cinder heath]

a textbook of engineering mathematics 👀

sem1

author name is important

[left knoll]

Like “a”

Never though a title would be so generic

Lol

[cinder heath]

it's like that so i

@left knoll btw chole bhature are tasty lol 😆 😋

[left knoll]

Ay no homo broooo

Kinda sus ngl

But yes they are delicious

[left knoll]

Q9d where did I make the stupid mistake

[supple island]

x=y²-1 isn't y=sqrt(x+1)

it's y=sqrt(x+1) combined with y=-sqrt(x+1)

[left knoll]

Oh shoot ur right tysm I knew it was dumb mistake lol

[supple island]

also can't you just do the integral in dy?

[left knoll]

I wanted to try this way

[supple island]

ok

[left knoll]

I did dx/dy

But then I wanted to try this way

@supple island

[supple island]

?

[left knoll]

Ignore that lmao

[left knoll]

Being dumb

[ember kernel]

At the end of the left column, there's also a u² that comes from nowhere

After inegrating a function with regards to a variable, you shouldn't have tha variable in the expression of the function anymore

[left knoll]

The u^2 just means unit squared

It something we have to do

[ember kernel]

Oh ok

Pointless to me

[left knoll]

yea its just a rule that you have to do it for exams otherwise we dont get the mark

its stupid

[ember kernel]

It is

Plus it's confusing as you're also taught to use u for the substitution variable

[left knoll]

exactly

[left knoll]

i think im making another stupid mistake, just cant find where

[left knoll]

Thanks anyway for your responsiveness.

[supple island]

what's the question?

[ruby needle]

?

what question , i jusst lrean about it

[ember kernel]

Dude

You computed 24 + 16 = 30

^^

[left knoll]

find shaded area

[left knoll]

jesus christ

my excuse is that its 3 am

and i am tired

[robust oak]

can anyone confirm this is right?

[ember kernel]

I'll answer in 2h, to make sure your test has ended by then

._.

[sullen verge]

isn't that right?

oh fuck

[ember kernel]

x)

That's okay

[sullen verge]

i just read what you wrotr

lets delete

[ember kernel]

Okay then

[soft lintel]

Hey, everyone Can anyone suggest any useful resources to understand linear algebra I'm struggling quite a lot 😅

[ember kernel]

G. Strang's textbook on linear algebra

This book is so well-written, and very comprehensible

If you can't afford it, Strang's whole lectures (he worked at MIT) are available on YouTube for free
Check them out, they're great too. Whenever I didn't understand something from the book, I would go watch the lecture of the difficult chapter to have it explained to me. Here's the link :
https://www.youtube.com/watch?v=7UJ4CFRGd-U&list=PL221E2BBF13BECF6C&ab_channel=MITOpenCourseWare

I can't stress enough how good this source is

[soft lintel]

Thank you soo much 😆 That was so helpful

[teal karma]

Can someone tell me what I should look up on yt to learn this topic

Literally have no idea what this is

[sullen verge]

complex numbers

[soft lintel]

I think it's related to the domain of the function like for example 1/x is not defined on 0 . Maybe this video can help https://youtu.be/djT6-YamHaA

[winter dove]

can someone help me with this, it says show your solution also, thankyou in advance♥️

[weary arrow]

Mean = sum of all numbers divided by amount of numbers
Median = middle number when the numbers are sorted in ascending order
Standard deviation is the formula in the pic

[winter dove]

i need help with this

[winter dove]

i need help with this, can someone help me? thankyou

[left knoll]

@winter dove what does set 1-4 years mean?

[winter dove]

i think its the year like first year, second year, third year, fourth year

[left knoll]

ah ok that makes sense, the wording of discrete distributions is so bad imo

[winter dove]

im so sorry

[winter dove]

can u help me with it?

[left knoll]

i can but im kind of busy rn with some homework 😬😬

sorry

im not free for the next 5 hours so

if someone doesnt help you by then, i will, but most likley someone will do it before me

if not ill do it

[winter dove]

oh its okay, the due is still on the next day

thanks for helping me, im glad

you can take your time first☺️

[left knoll]

thanks girl*, ill try to help if i can 😊

[winter dove]

im not a man😁 im a girl

thankyousomuch in advance

[cinder yacht]

in my country it was called "Sets and maps" smth like that lol

try looking for "bijective injective surjective functions"

[left knoll]

i think i fixed the issue 😁

[dusky orbit]

Help me pls !!! i have a problem : prove that f(x)= x^6 +x^5+x^4+x^3+x^2+x+1 is Irreducible polynomial in F2[x] . How can i know that polynomial is Irreducible or reducible ?? .I have seen some video about that but in this case the deg(f(x)) > 2 , i dont know how to solve this .Thank you very much .:(

[charred mist]

this might sound stupid but isnt this literally impossible?

I found dy/dx which is =12(2-x)^-2

then I equated that to 0 since it says that it crosses x-axis since y = 0 on the x axis

I think I messed up there but Im not sure

[left knoll]

@shrewd swan

[left knoll]

what is the ques?

[shrewd swan]

16th one

[left knoll]

mark ab = 5

[shrewd swan]

Ik

But tan 90 is undefined

[left knoll]

why do you need tan90?

[shrewd swan]

Tan90-x

[left knoll]

its supposed to be tan(90-x)

[shrewd swan]

Oh

[left knoll]

got it?

hehe this chapter

[shrewd swan]

Yeah

[left knoll]

Cute chapter

[shrewd swan]

Tan90-x is cot

[left knoll]

Ye

[left knoll]

yes

[shrewd swan]

They didn't tell that in school smh

[left knoll]

???

[quick comet]

yes

[shrewd swan]

Deleted i think

[left knoll]

They did..you were absent

sin(90-x)=cosx
cos(90-x)=sinx
tan(90-x)=cotx
cot(90-x)=tanx
cosec(90-x)=secx
sec(90-x)=cosecx

[shrewd swan]

This type of question is deleted i did anyways

[shrewd swan]

Yasho big brein

[left knoll]

I challenge you to write all the formulas of trigonometry

[left knoll]

that is defo out of his syllabus

[shrewd swan]

BROO IM IN 10th

[left knoll]

all of this is in 11th ya

[quick comet]

ye just put it there so he can refer

[shrewd swan]

Chilll

[quick comet]

ik

[left knoll]

copy pasting the picture just sent

[frozen meadow]

What...

[left knoll]

Very nice

[shrewd swan]

This part is deleted

[frozen meadow]

U mean that topic is deleted?

[left knoll]

i think complementary pairs is deleted

[frozen meadow]

So cbse is cutting parts

[left knoll]

oh yea since last year

[torpid root]

I did this question wrong in an exam. Is the answer d?

because a and c are just the means I think?

[normal schooner]

can anyone help me understand triangle proportionality? like AA~

[teal karma]

Help with the three exercises at the bottom

[teal karma]

#9

[ember kernel]

Express sinh in its exponential form

[steady tide]

Im not sure how you even simplify the negative 5 in the radicand with 6 cube roots of 5

it doesn't seem possible to me without using imaginary numbers

WAIT NVM

-1 cubed is -1, so i can just factor it out of -5

[idle remnant]

need help with i) please

[sullen verge]

still do?

[sullen verge]

and could you please send me the rest of the exercices, i have a test on complex numbers on Wednesday 😅

[winter dove]

i need help with this thankyou

[sullen verge]

for a its number of individuals in an age group/number of individuals

for b its the same except its in percentile

the last one u cumulate as you go down the age groups

[winter dove]

how to like show the solution? or how to solve it?

like can you show it in a table way?

[sullen verge]

yeahh you can

it is easier that way actually

[winter dove]

in A you just divide it right?

the frequency is 3 divided by the total of frequency?

when you sum up all the frequency, its 191 so, 3 divided by 191?

am i right?

in B, you just do the same, like 3 divided by 191 times 100?

3
__ * 100 ?
191

[winter dove]

i'm somewhat confused on C, how to solve it?

[sullen verge]

yes

[sullen verge]

you just add them so for
11-20
it will be the frequency of 10 and below + the frequency of 11-20

[stiff solstice]

can someone help me with this sum?
sum:
two angles are supplementaire .four times one angle is equal to twice the other angle .Calculate the size of the angles by means of a system of equations and using the substitution method

??

[sullen verge]

x +y =180(cause they re supplémentaires)
and 4x = 2y

then u solve that

[stiff solstice]

sorry bro i mean supplementaire

[sullen verge]

i edited ir

[stiff solstice]

ok

[stiff solstice]

@sullen verge then it becomes

   x+y=180

{
4x-2y=180

right??

[sullen verge]

noo
4x-2y=0

[stiff solstice]

ohh ok

[sullen verge]

(along with the firdt equation)

[stiff solstice]

you mean this ?
x+y=0
{
4x-2y=0

[sullen verge]

noo the first one is still equal to 180

[stiff solstice]

ohh ok

[sullen verge]

x+y=180
{
4x-2y=0

[stiff solstice]

@sullen verge ohh thanx

[sullen verge]

happy to help

then to solve it you find y according to x

or x according to y

then you substitute

[winter dove]

in a percent form?

[winter dove]

is this right?

[coral flint]

The area of a circle is halved when its radius is decreased by y.
Find its radius.

2y + r2
(2-r2)y
2 + r2y
(2+r2)y

[left knoll]

(2+r2)y

[left knoll]

Since the radius is decreased by y, so the new radius is r-y.
We know area of a circle is pir^2, so the new area is pi(r-y)^2
Now it's given that the area of the circle gets halved when reduced by y, so pi(r-y)^2 = (pir^2)/2

[minor cairn]

The answer is 2 I think

[analog stirrup]

anyone good at calc 2

Need help asap please

🥲

@coral flint can you help me?

[coral flint]

Ques

?

[analog stirrup]

have you took cal 2?

Or are taking it?

[coral flint]

Nope

[analog stirrup]

oops, nvm. Thank you

[coral flint]

Yea

[analog stirrup]

@left knoll

How about you?

[frozen meadow]

send ur questions here. there are alot who can help u

[frozen meadow]

if i remember it correctly its like subtracting. f'(2) from f'(5).

[safe ember]

You just draw rectangles and calculate the areas. The rectangles must be one unit wide and their heights must be f(2), f(3) and f(4)

So yeah, it's basically just f(2)+f(3)+f(4)

[left knoll]

does anyone know why its D

[onyx hollow]

Read off the values at f(x=0) and f(x=2) and add then together, what do you get?

[left knoll]

ahh -2 + 2 thank you

[onyx hollow]

Does anybody here understand the process of taking the derivative of the norm_2 of a vector with respect to that vector itself? Having a hard time googling myself to an answer.

[supple island]

the directional derivative?

[onyx hollow]

d/dw(a*|w|_2) is the expression

I'm not sure if I can just do it as 2aw as the final expression

[supple island]

wdym norm_2

is it ^2?

[onyx hollow]

I'm pretty sure that in this case norm_2 is just norm so discard that part

[limber gull]

what this mean lol

[left knoll]

Does anyone know why this is 65 and not 55
For a 2€ coupon, a restaurant allows you to pick your ice cream dessert, out of 10 distinct flavors. The
dessert is served as ice cream balls, with each ball costing 1€. How many different ice cream
combinations can you make?
Note: you don’t have to spend all 2 euros.
Note: the option of choosing no ice cream balls (no dessert) is not considered and doesn’t count.

[supple island]

and a is just a scalar?

[onyx hollow]

correct

[supple island]

can you provide the whole text?

[fervent tide]

It's combinations, often denoted nCr. It means how many different samples of size r you can make from n different elements.

And the order of the elements within the samples is not relevant.

[limber gull]

thanks alot bro

[onyx hollow]

It's just a small part of a larger assignment, but the part encircled is what I want to take the derivative of:

Which I've identified as the norm (they hint about it as well)

[supple island]

well no, it's not the norm

it's quite different, wj is the j-th component of the vector w

[onyx hollow]

But isn't the norm of a vector it squared with itself, which ends up being that expression like this:

[supple island]

oh ok that's waht you meant

ok so you don't actually derive by w to find the minimum

you need to find the gradient=0

[sullen verge]

i think it is (sorry for the late reply)

[peak pollen]

please tell me this isn't in higher IB Math

[sullen verge]

did you try to multiply and divide by the conjugate?

[lament dagger]

I need help with this question, I got to an equation
n = m+2 - 30/m+1 which was given as hint, idk what to do after that

[cyan night]

okay, so they have given that n is a natural number, which means n is a whole number

what does that imply?

m^2 + 3m - 28/m+1 must be a whole number then

which means?

the key thing is to realize, for a fraction to be a whole number, the denominator must divide the numerator

or in other words, m^2 + 3m - 28 is divisible by m+1

you can continue to work from here

also dont forget that it n is a natural number

that means it must be positive

[lament dagger]

oh yeah this clears everything, I got the solution, a lot of thanks mate

[cyan night]

just wondering, (A) is false and (B) is true right?

no problem btw

i got m could be ||5,9,14,29|| because m > ||4||

[lament dagger]

correct options are A and D

[cyan night]

rlly?

hmm i maybe messed up

[cyan night]

what values did u get for m

or a i mean

[lament dagger]

wait you have got a point B is correct and A is wrong

but here for some reason

[cyan night]

if you havent got it i can explain to u how i got B

[lament dagger]

correct options are given A and D

[cyan night]

and C is correct henceforth

[lament dagger]

I got it

[cyan night]

B and C should be true tho

[lament dagger]

umm how about we reread question

which of following are FALSE

lol

🤣

[cyan night]

you got a point xd

[lament dagger]

that was a trap lol

[cyan night]

yepp

[small canyon]

x+1 = A(x^2+1) + (Bx+c)(x), how do i find b and c in this?

got the value of A, but after that i am a bit lost

[ember kernel]

Use the fact that each power of x can have its own equation

Because obviously x^m and x^n are different if m and n are different

That means that ax² + bx + c = dx² + ex + f is only true if a = d, b = e and c = f

Do you know what I mean ?

[small canyon]

ah i get it thanks alot

[silk pond]

someone help me im so bad at logs and exponentials, where do i start

[ember kernel]

Express eveything in the same base

Here, you can see that 27 = 3³

Now, using the property 3^a * 3^b = 3^(a+b), re-express the equation as 3^x = 3^y (you need to find the value of x and y by yourself)

Since exponential functions are injective, 3^x = 3^y implies that x = y

You can solve for n from there

[rustic marten]

hi, I was given the following situation: n = 60, f(A) = 0.3, f(B|A) = 0.5, f(A,B) = 0.15 (f(X) = relative frequencies). Is it possible to find f(Not A, B) and f(Not A, Not B) with the information available? I would say no, because there is no information given about the distribution of the remaining 0.85 = 42 entities in the "Not A"-row. Is this right?

[rancid folio]

I just can't

[onyx hollow]

Try to rewrite the 2 to something and look at the expression again

[rancid folio]

should I turn it into a surd?

etc sqrt4?

[onyx hollow]

Correct, not really sure how this is a proof, but you can try with different neighboring integers that the higher the numbers are the lower this difference will be

So sqrt(10)-sqrt(9) will be smaller than sqrt(9)-sqrt(8) etc...

[rancid folio]

ohhhhh

[rancid folio]

bro Ive been stuck on this for a solid hour thx so muhc

[junior bear]

i need the full demonstration

[left knoll]

guys i think this isnt math but is it a problem when your start smoking weed and then start watching lord of the rings

[weary arrow]

No lord of the rings is great

[zenith sparrow]

which q

[shrewd swan]

4th

[zenith sparrow]

ok gimme few secs

[shrewd swan]

Ye ye

[red spruce]

find the gen term

and then find n

[shrewd swan]

These are my calculations, wrong obv

[shrewd swan]

D i got as 2/5

[red spruce]

3+(n-1)5 = 78

[solemn sphinx]

What happens when -4 goes to the other side -_-

[shrewd swan]

Sorry lol

[solemn sphinx]

Instead of skipping a step, add four to both sides

[zenith sparrow]

which sum soln is this ??
if fourth then application is wrong

[solemn sphinx]

You tenth kids be more careful with signs😭 🫂

[red spruce]

idt u even neeed to write it down tho

mind me ho jaata

[shrewd swan]

Yeah fourth

[zenith sparrow]

i wanted to show step by step

[solemn sphinx]

You can correct it now and see if you're able to reach the ans

[zenith sparrow]

be careful with how you read the question , cause you have solved it wrong

[red spruce]

10thies solve ncert

[shrewd swan]

Yeah

[red spruce]

rd sharma lvl nhi aate jyada

i solved the whole rs and rd still ncert se aaye the

[left knoll]

Rsa best

[shrewd swan]

Rd sharma that guy 🤬

[zenith sparrow]

but concepts are best with rd

and explanation is really good
if you put time into it you can really understand chap in rd without teacher

[red spruce]

it gives u many questions of same type tho

[zenith sparrow]

ye time thoda waste hoga but still concepts will be on point

[red spruce]

i did the whole book , helped me but like for marks ncert hi padhna pada

[zenith sparrow]

zamn

[solemn sphinx]

You have a decent handwriting✨

[zenith sparrow]

for 2 mins itll be okay then itll go to hieroglyphics mode

[shrewd swan]

So I got d as 2

I referred from a website

How does A6 turn into 6

[zenith sparrow]

which sum

[shrewd swan]

A6 just randomly turned into 6

How doe

[paper obsidian]

here's your answer

[zenith sparrow]

i need the question to give ye your answer bro

[shrewd swan]

The same question man

[left knoll]

what's the question

[zenith sparrow]

i sent the solution right

[shrewd swan]

Wait NVM i got it

Lol I forgot to change signs when I sent it to the other side

[paper obsidian]

75 divided by 5 isn't 12 bro

[shrewd swan]

That was literally all 😂

[zenith sparrow]

ye just realized bro

[paper obsidian]

haha

[zenith sparrow]

i was thinking bout that for a long time lmao

[paper obsidian]

anyways you know how to solve it now...

[shrewd swan]

I do 😼

[paper obsidian]

great

continue

[shrewd swan]

Man I'm too slow with math

[paper obsidian]

good luck

[zenith sparrow]

isnt common difference given in the question

[red spruce]

keep ur phone down , u dont neeed it

[zenith sparrow]

common difference d is 5

you have to find n

[shrewd swan]

Bhai 4th sub question I meant

[zenith sparrow]

ohh

[shrewd swan]

You went and did the 4th question lol

I'll make it more clear next time sorry lol

[zenith sparrow]

im a bit slow today

i wrote 80 /5 as 12

[shrewd swan]

It's not the phone I get stuck on simple things

[zenith sparrow]

and only meow noticed it

[red spruce]

6th term is 6 pehla term is -4 d nikao what the isssue

[shrewd swan]

Now I got it no issue

[red spruce]

5d=10

d=2

ho gya

[shrewd swan]

Bhai Han samaj gaya

[red spruce]

oki

[zenith sparrow]

ye i think he got it , i went and did the forth sum thats where the entire confusion began plus i did wrong division

i wanna punish myself so bad

[red spruce]

u dont punish urself for bad maths , the regret is enough boyo

[zenith sparrow]

but this is 4th grade math 💀

[untold patio]

Hello there, wanted to ask how i "reduce" this
x*(x+4)-(x+2)^2
thanks in advance

[supple island]

well

try to develop it

then quite a bit of terms cancel out

[ember kernel]

What I would do is try to get a polynomial of the form ax²+bx+c, and then factorise it using the quadratic formula (or an identity)

[untold patio]

okay thank you very much i'll have a go at it

[crude crag]

anyone here who is good at probability?

i'm struggling with this problem

basically

given 100 candies: 50 milk-flavored candies, and 50 chocolate-flavored candies

and 2 boxes

[tough elk]

I simply have no idea what to do

[crude crag]

now you have to put the candies into the two boxes

then you'll get blindfolded, the candies will be mixed up randomly, and the two boxes will be swapped randomly multiple times

when the blindfold is removed, you will choose a candy from the box

if it's a chocolate one, you win

if it's a milk one, you lose

question: how should you put the candies into the box so that the probability that you pick the chocolate candy is maximized?

[silk pond]

does anyone know how to do 3k please i am confusion

[left knoll]

try to reduce the bases to have only 3^something x 3^something then you can simplify that to 3^(something +something). then when the bases of LHS and RHS are both 3, you can solve for the equation with only the powers.

[ivory minnow]

and if you reduce the bases to have only 3^ then you will get => 3^[(n^2)+3n] = 3^[3n+(1/4)] and if bases are the same then you can just get this equation => n^2 + 3n = 3n + (1/4) and 3n's cancelled each other then => n^2 = 1/4 and you will get the result as n = +1/2, n = -1/2

[solemn sphinx]

With the given conditions, it looks like a uniform distribution in my opinion,so any method of putting candies won't change the probability of picking up either one of these candies. But I'm curious to know the solution.

[solemn sphinx]

Use substitution.
ln_w_ = z
Hence w = e^z
Hend dw = e^z dz

This becomes an integration in z instead of w now. Therefore the limits will change from w to ln_w_
ln1=0, ln_e_ =1

[gritty cloud]

Idk how but im struggling with this

[gritty cloud]

I need the soultion please 🌹

[left knoll]

For second image replace all the h with 0

Wait is there more to the second image?

Looks like first principles

1st image:

Y = (X-4)^1/3

Dy/dx = 1/3(x-4)^-2/3

Sub in a for part i)

And sub in 1 for part ii)

Sub into derivative i mean

[gritty cloud]

Is that the correct answer for image 1

[left knoll]

which part are you asking