help-49

[copper ginkgo]

bo

no

this is a unit of coding + maths

topics not that complicated

[midnight plank]

@copper ginkgo Has your question been resolved?

[copper ginkgo]

<@&286206848099549185>

[midnight plank]

@copper ginkgo Has your question been resolved?

[midnight plank]

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[bronze rover]

does anybody know about the math behind RSA encryption

[bronze rover]

pls

[chilly cobalt]

is this the right one? under "operation"
https://en.wikipedia.org/wiki/RSA_cryptosystem

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[verbal pumice]

The answer is probably gonna be of the form $\frac{f(x)}{1 + x + \sqrt{x}}$

[grand pond]

dyxn

dyxn

[verbal pumice]

I mean this sort of leads back to solving the integral so it's not very useful

[grand pond]

dyxn

[verbal pumice]

but is there a more direct solution to it

[grand pond]

dyxn

[verbal pumice]

¯_(ツ)_/¯

[floral apex]

i didnt solve it

calculator says ostrogadskys method

ive never heard of that

[floral apex]

on this

[midnight plank]

@verbal pumice Has your question been resolved?

[verbal pumice]

lmao

[midnight plank]

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[twilit field]

FInd an example of a function and a convergent seqeunce $(x_n) \to x$such , $f(x_n)$ converges, but not to $f(x)$

[midnight plank]

Please don't occupy multiple help channels.

[grand pond]

What a wonderful world !

[twilit field]

oops, closed

Nothing comes to mund

*mind

[wary trail]

nothing comes to mund either

I'll have a think :p

[twilit field]

oh

[lyric charm]

what if your function returned 420 on some inputs, 69 on others, and your sequence managed to hit only one of those

[wary trail]

$x_n = 1/n, x = 0, f(x) = \delta(x)$

[grand pond]

Xetrov

[twilit field]

would $x_n =\frac{1}{n}$ and $f(x)=\frac{1}{x}$ work

[wary trail]

wouldn't converge

[grand pond]

What a wonderful world !

[lyric charm]

x_n = 1/x is sussy

[wary trail]

that just generates your natural numbers no?

[lyric charm]

ok and what does your f(x_n) converge to

[twilit field]

what's \delta (x) here

[twilit field]

It doesn't converges, right

[wary trail]

1 if x=0, 0 elsewhere

[twilit field]

hmm

Does that actually converge though

[wary trail]

yup, it converges to 1

because anything in x_n is non zero, so all elements are 1

[twilit field]

hmm, okay

thanks

.close

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[last slate]

how can I do this

[midnight plank]

@last slate Has your question been resolved?

[ornate glacier]

Think from the inside the cube

Like think of two faces the front and the back

They touch at the center

Therefore 2*radius =side of cube

By using the given information , radius =1 cm

Therefore diameter =2

[obsidian torrent]

But wouldn't that mean the hemispheres would intersect with each other

[prime rapids]

Here's a top-down view

[ornate glacier]

Why?

Oh ok

I misunderstood as if all the hemispheres touch at only one point

[midnight plank]

@last slate Has your question been resolved?

[last slate]

how did you get square root of 2

[split coral]

pythagoras

[last slate]

oh right

!close

.close

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[zenith mirage]

i qas told to find solutions to a system of equations hich is hat i have ritten don to find and i need to anser the question on the screenshot
can someone help alk me through? i forgot ho to do a system of equations and it feelsdifferent noe that im qorking ith other variables and i already have x ans y

[zenith mirage]

!occupied

[midnight plank]

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[zenith mirage]

@winged phoenix

pleasedelete your message

[cedar mason]

you can just place the second equation in the first one

[zenith mirage]

Hoq

I dont knoq hoq

Or qhat to put qhere

theyre the same format

so idk hat to do

[cedar mason]

$ab^4 \equiv ab^3 b$, yeah?

[grand pond]

Percy

[zenith mirage]

yes

[cedar mason]

so

[zenith mirage]

i dont get qhat youre tryingto get me todo

its not clear to me

[cedar mason]

whats the second eq

[zenith mirage]

5=ab^3

but

theyre both

have b valies

so putting b ouldnt do anything but

makeit more b

[cedar mason]

...

[zenith mirage]

you dont have to act cocky dude

get out if youregonna be like this

[cedar mason]

$ab^4 \equiv \mathbf{ab^3} b$, yeah???

[grand pond]

Percy

[zenith mirage]

.close

[midnight plank]

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[zenith mirage]

ffs dude

be respectful and not fucking rude

andacting like i knoq everything

i came here for help

[cedar mason]

[zenith mirage]

and youre sitting herebeing rude as fuck

this happens every fucking time dude

i ask for help

and people are mad that i dont e=understandsomething

dont fucking garner for any sympathy fromme

[fossil seal]

Fr

What happend to ur W button

[novel flax]

brain rot kid

[zenith mirage]

.close

.reopen

[midnight plank]

✅

[zenith mirage]

i qas told to find solutions to a system of equations hich is hat i have ritten don to find and i need to anser the question on the screenshot
can someone help alk me through? i forgot ho to do a system of equations and it feelsdifferent noe that im qorking ith other variables and i already have x ans y
and please adhere to the rules

● Respect that other people might be at a different stage in their education than you, what is obvious to you might not be obvious to them.
● Be rational and exercise critical thinking. If someone makes a mistake in their reasoning, politely point it out.
‌
I am sloq. And im sorry. Pleasedont be rude to me. Please be patient and understanding and knoq that i struggle qith some seemingly common-sense things.

this is qhat i have

[sudden yacht]

Where did all the w go??

[tawny apex]

uhh divide both the equations and youll get your b

then plug in the value of b in any of the two equations and youll have a

[zenith mirage]

Please dontgo off topic

[zenith mirage]

I dont knoq hoq

Read up above all of this

and see that i really have no clue

like genuinely

Aria

Do not

Please

Alberto

Leave

[tawny apex]

uhh so you dont know how to divide?

or you dont know how to solve?

[zenith mirage]

Solve

ell

idk qhat to divide

either

[sudden yacht]

Ok, I was going to help you but if you don't want... alright 🤷‍♂️

[tawny apex]

okay so you have 2 equations

[zenith mirage]

Mhm

[tawny apex]

one is: $5=ab^3$

[grand pond]

Aria

[tawny apex]

and the other is: $10=ab^4$ right?

[grand pond]

Aria

[zenith mirage]

can i include the dot for multiplication just so its not confusing bc even a little differnce can thro me off im sorry

but yes

[tawny apex]

ok

so dividing the LHS we get: $10/5$

[grand pond]

Aria

[tawny apex]

which is 2

[zenith mirage]

ait ait ait

ait a mnute

[tawny apex]

yeah?

[zenith mirage]

i dont ant an anser or for someone to ork it out for me

i ant tolearn ho to and apply it

like
telling me astep and i can try it

[tawny apex]

ok ok

[zenith mirage]

and see if i did it good sorry

[tawny apex]

but there are various different steps for different system of equations

so like if a system of equation has variables that are getting subtracted or added we will perform different opertaions than what we are doing rn

[zenith mirage]

thats fine

right?

[tawny apex]

here, the system of equations have variables which are multiplied together

so think like this, we need to find the variables

[cedar mason]

<@&268886789983436800>

[sudden yacht]

<@&268886789983436800>

[tawny apex]

uhh <@&268886789983436800>

[sudden yacht]

Whoops thanks Percy

[zenith mirage]

huh

[tawny apex]

yeah back to the question

[zenith mirage]

idek hat happenedlol

[tawny apex]

so we need to find the variables right?

[sudden yacht]

A spammer, he sent some dangerous link

[zenith mirage]

yes

ait

im confused no

hich is a and hich is b

[tawny apex]

we need to find a and b

they both are the variables

[zenith mirage]

of the (number, number)

in my anser choics is it (a,b)

[tawny apex]

no they are not the (number,number)

[zenith mirage]

no like

[tawny apex]

its for (x,y)

[zenith mirage]

in the anser format

[tawny apex]

the mcqs?

[zenith mirage]

okay no im really confused

[tawny apex]

they are also for x and y

[zenith mirage]

so hy am i finding a and b

[tawny apex]

listen, so you have to find the values of a and b

then plug those values into the given equation of $y=a*b^x$

[grand pond]

Aria

[zenith mirage]

so finding a and b and then graphing the lines and seeing hich lines up ith the other 2?

[tawny apex]

then by trial and error method, we need to find the value of x and y from the mcq that satisfy the eqn

[tawny apex]

no no we dont need to graph it

[zenith mirage]

itd make it easier

if i did

[tawny apex]

no no it wont

[zenith mirage]

but itd sho me

right there

[cedar mason]

$\cdot$

[grand pond]

Percy

[tawny apex]

oh thanks

[cedar mason]

asterisks is crazy

trump level shit

[tawny apex]

lmaoo frfr

[zenith mirage]

i like astericks tho :(

[tawny apex]

nah discord makes it crazy

[sudden yacht]

Huh? Do you usually write asterisks for the multiplication (on paper)?

[zenith mirage]

no i just like hoq they look

[sudden yacht]

That would be a very bad practice, honestly

[tawny apex]

he likes asterisks in general lol

[sudden yacht]

Ah alright alright

[zenith mirage]

i qant totake a break but itd be selfloathing and i really am stressed right noq

but

i cant because im on the train of thought

and if i do ill lose everything

and getmore stressed

[sudden yacht]

If you're stressed, don't do maths. Otherwise that's quite obvious you can't understand fully

[tawny apex]

yeahh

[zenith mirage]

irealy really need to get this done

[tawny apex]

just relax for a while

[zenith mirage]

this is 2 eeks overdue

[sudden yacht]

Take breaks during the study. They're fundamental to restore the brain and make it relax and get back to work

[zenith mirage]

ive only been on this for like

20 minutes

at the most

today

[tawny apex]

uhh maybe take a 5 min break and then come back to it later

[zenith mirage]

but i cant

i already forgot it

[sudden yacht]

Then you can't also understand (and do) the exercise

[zenith mirage]

its not funny to me,i alqays forget things

qhni try tolearn them

itsokay guys thank you for trying i really appreciate you

.close

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[dusty portal]

What have you tried?

[regal barn]

no idea

[dusty portal]

A general rule of thumb would be to multiply the top and bottom by sec^2(x).

[regal barn]

i did tanx=sinx/cosx

then lcm

[dusty portal]

Eh...

[cedar mason]

id try writing tan as sin/cos and doing a+b-x thing

oh pi/6

[regal barn]

i see

[cedar mason]

not sure

[dusty portal]

Yeah, doesn’t work out too nicely for Kings Rule.

[regal barn]

and then?

[dusty portal]

u-subsitution.

[regal barn]

can you tell me what will bve u?

tanx.secx/(sec^2x+tanx.sec^2x)

[dusty portal]

What derivative do you have?

[regal barn]

to what?

( \int_0^{\pi/6} \frac{\tan x \sec x}{\sec^2 x + \tan x \sec^2 x} , dx )

[grand pond]

Tom

[regal barn]

ohh wait

if i take tanxsec=u

then du=sec^3x+sec^2x.tanx

soo

let me take secx common

i am stuck at choosing

please help

[dusty portal]

Choosing?

[regal barn]

1/du seems bad

[dusty portal]

Why not try u=sec(x)?

[regal barn]

sqrt(u^2-1)/(u+du)

still bad form

[dusty portal]

You could always convert tangent into secant.

[regal barn]

yes i did

tan^2x=sec^2-1

[dusty portal]

You would have ||du/(u^2+sqrt(u^2-1)u^2)||

[regal barn]

sqrt(u^2-1)

[dusty portal]

Notice the derivative of sec(x).

[regal barn]

hmm got it

$$
\int_1^{\frac{2}{\sqrt{3}}} \frac{du}{u^2 + \sqrt{u^2 - 1}u^2}
$$

[grand pond]

Tom

[dusty portal]

<@&268886789983436800> Here too

[regal barn]

$$
\int_1^{\frac{2}{\sqrt{3}}} \frac{du}{u^2(1+\sqrt{u^2-1})}
$$

[grand pond]

Tom

[regal barn]

u there?

[dusty portal]

Yes.

Another subsitution.

[regal barn]

and what it will?

[dusty portal]

What derivative do you have, again?

[regal barn]

this time no idea

[dusty portal]

||1/u||

[regal barn]

( \sqrt{u^2 - 1} )

[grand pond]

Tom

[dusty portal]

I gotta go to class.

[regal barn]

ahhhhhhhhhhhhh

i will ask you tomorrow

need to rest

gn

.close

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[lean ibex]

for number 7 is A" = (-6,24), helping a friend out and i wanna make sure my work is correct!

[lean ibex]

nevermind! i got it solved

.close

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[dull yoke]

not too sure how to start

[unreal plank]

I need help with these angles "Angle Relationships & Parallel Lines"

[buoyant yoke]

!occupied

[midnight plank]

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

[dull yoke]

firstly, does 'compute numerically' mean i have to do this by hand, or i should do some sort of approximation method with python

[robust isle]

well if you want to compute shit with 100x100 matrices by hand you do you

they prolly mean with a computer yes

[dull yoke]

ah ok

thanks @robust isle

.close

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[mystic salmon]

where did i go wrong?

[buoyant yoke]

lol

[mystic salmon]

yeah i noticed

ok so let me explain my thought process

i graphed out the curve on a rectangular plane

from [0, 2pi]

and i got the points (0, 6), (pi/2, 2), (pi, -2), (3pi/2, 2), & (2pi, 6)

and i noticed that [pi/2, pi] would be in quadrant 2 but would have a negative r value, so would go in quadrant 3

and [pi, 3pi/2] would be in quadrant 3, reflected to quadrant 1 with the negative r value

but C is wrong

i don't get why, or how i would arrive at the right answer

figured it out

.close

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[molten bay]

In this given figure o is incenter of ABC triangle AO/OE=5/4,CO/OD=3/2 then find BO/OF?

[midnight plank]

@molten bay Has your question been resolved?

[midnight plank]

@molten bay Has your question been resolved?

[violet rivet]

do yk what's incenter?

[versed sierra]

DO=OF=OE

[midnight plank]

@molten bay Has your question been resolved?

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[ashen glade]

@everyone can somebody help me with this worded simultaneous equation

[lyric charm]

stop doing these everyone pings that don't work

[ashen glade]

sorry can you still help with my work please

[lyric charm]

ok, what have you tried?

[ashen glade]

nothing I am a bit stuck on how to start it

[lyric charm]

do you know what it means for a triangle to be equilateral?

[ashen glade]

the sides are equal in length

[lyric charm]

ok

can you write down an equation that says the left and right sides are equal

[ashen glade]

(b+6) = (2a+b)

[lyric charm]

ok

can you simplify this equation in any way

[ashen glade]

(b-b+6) = (2a+b-b) then equal to 6 = 2a divide by 2 a = 3

[lyric charm]

ok you have just found the value of a

you are halfway there

[ashen glade]

(2a + b) = (4a - 1)

(2(3)+b) = (4(3) - 1)

6 + b = 11

6 - 6 + b = 11 - 6

b = 5

a = 3 and b = 5

@lyric charm thank you

[midnight plank]

@ashen glade Has your question been resolved?

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[steady trail]

$lim_{x\to 0} \frac{\sin(\frac{1}{x})}{\frac{1}{x}}=1?$

[grand pond]

Task Bot

[solid iris]

if $\lim_{x\to a}g(x)=b$ and $g(x)\ne b$ for all $x$ "near $a$" then $$\lim_{x\to a}f(g(x))=\lim_{y\to b}f(y)$$

[grand pond]

ロケットジャンプ

[solid iris]

this is what ur thinking about. carefully check that all conditions hold

[steady trail]

The limit of 1/x does not exist

[solid iris]

so the theorem doesnt apply and we must find another way to compute the limit

[sudden yacht]

True, but luckily sin(1/x) / (1/x) is an even function, so you can just focus on x → 0+

[solid iris]

u dont have to do that using what im thinking

[sudden yacht]

Ah sorry I thought of another thing

[steady trail]

Since sin (1/x) is limited and x tend to 0, then the limit is 0?

[sudden yacht]

Yes

[steady trail]

👍

Was there another way?

[solid iris]

no thats the easiest way

[steady trail]

Okay!

thank you all

.close

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[solid iris]

np!

[midnight plank]

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[strong stump]

How can I expand $\cos(kz)$ around $z = 2$?

[grand pond]

jandro0103

[grim vector]

By expend you mean taylor series or something else ?

[strong stump]

taylor series

maybe i need to write z-2 = 0 ?

[grim vector]

You can let z = 2+h with h tends to 0

[strong stump]

then sub t = z-2?

[grim vector]

Yeah same idea

[grim vector]

I think this is risky

[grim vector]

This sounds better

[strong stump]

ok ive got cos(2k + kt)

now x = 2k + kt ?

[grim vector]

Hm

The thing is that your x won't be near 0

Id do addition formula of cos

[strong stump]

ye its wrong

whats taylor in t = 0 ?

[grim vector]

The same as if x = 0

[grim vector]

Here

But for t instead of x

[strong stump]

but in my case if i sub x = 2k + kt

when t -> 0 its not valid anymore

[grim vector]

I think you don't

You do cos(2k+kt) = cos(2k)cos(kt) - sin(2k)sin(kt)

And develop for cos(kt) and sin(kt)

[strong stump]

maybe i can use $f(t) = f(0) + f'(0)t + \dots$

[grand pond]

jandro0103

[strong stump]

so its $\cos(2k + kt) = \cos(2k) - k\sin(2k)t + \dots$

[grand pond]

jandro0103

[midnight plank]

@strong stump Has your question been resolved?

[grim vector]

Ah maybe

[strong stump]

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[twilit field]

Trying to solve Q15

[twilit field]

I was thinking $f= xyz+ 4z^2$

[grand pond]

What a wonderful world !

[gaunt nimbus]

?

[twilit field]

part(a)

And then using FTC

[gaunt nimbus]

oh

[twilit field]

(72+36 -0 = 108

[gaunt nimbus]

doesnt align for the z component?

$f = xyz + z^2 ?$

[grand pond]

k

[gaunt nimbus]

@twilit field

[twilit field]

oops

yea

my bad

[gaunt nimbus]

yea

and ftc

and u're done

[twilit field]

Couple more questions

in 19 f = xtan(y)

[gaunt nimbus]

mhm

[twilit field]

so the integral is 1

[gaunt nimbus]

tan(0) = 0?

[twilit field]

mmm

yea

so 2

[gaunt nimbus]

${f(2, \pi/4) - f(1, 0) = 2(1) - 1(0) = 2 - 0 = 2?}$

[grand pond]

k

[twilit field]

hmm

yea

As for the second one

$x+ ye^{-x}$

[grand pond]

What a wonderful world !

[twilit field]

so

$1- ( 1 + 2e^{-1})$

[grand pond]

What a wonderful world !

[gaunt nimbus]

other way

[twilit field]

idts

just realised the order is opposite for multivar FTC

[gaunt nimbus]

?

[twilit field]

hmm

okie

my bad

so $2e^{-1}$

[grand pond]

What a wonderful world !

[gaunt nimbus]

i think os

[twilit field]

okie

thanks

[gaunt nimbus]

👍

[midnight plank]

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[twilit field]

To start, we parameterize the circle by $x = 4 \cos(t), y = 4 \sin(t); 0≤t≤2π$
\
we then have $\int_{0}^{ 2 \pi} 4^5 \sin^4(t) \cos(t) dt - \int_{\pi /2}^{ 3 \pi/2} 4^5 \sin^4(t) \cos(t) dt $

[grand pond]

What a wonderful world !

[zealous schooner]

Why not just integrate from -pi/2 to pi/2 instead?

[quiet hinge]

just use the natural bounds you get

[twilit field]

so I paramatrize it from -π/2 to 3π/2

[quiet hinge]

what

[quiet hinge]

.

-pi/2 and 3pi/2 are the same mod 2pi

[twilit field]

no, so like when I write the paramtrization of the cirlce, this is what I write, right

oof, this isn't pretty

[gaunt nimbus]

How about ||r’(t)||?

[twilit field]

$4^6 ( \frac{ sin^{5}(t)}{5})$ is the integral

[gaunt nimbus]

Multiplied by Speed

[twilit field]

oh right

[gaunt nimbus]

No?

[grand pond]

What a wonderful world !

[twilit field]

yeah, norm of the derivative

[twilit field]

so that's just giving me 4^6 \frac{2}{5}

,w 4^6* (2/5)

[grand pond]

Results provided by WolframAlpha

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

Requested by math_rocks

[gaunt nimbus]

Why?

Let’s slow down

Ye

U correct

[twilit field]

Thanks!

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[wind oxide]

helo

[wind oxide]

If atleast one root of the equation$x^2-2mx+m^2-1$ lies in $\left(-2,4 \right)$ then set of complete values of $m$ is given by-

[grand pond]

rak³en

rak³en

rak³en

rak³en

[wind oxide]

Then lastly take the union

no help

[midnight plank]

@wind oxide Has your question been resolved?

[wind oxide]

<@&286206848099549185>

[runic hamlet]

well for m=3 then f(4)=0 so the inequality in your first case doesnt work

so f(a)=0 or f(b)=0 are the cases you missed

[midnight plank]

@wind oxide Has your question been resolved?

[wind oxide]

o thats what happens at those points

okay okay thank u

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[dusty portal]

I just want to get this over with

[dusty portal]

I don't even know where to start

.close

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[wicked sun]

this is the solution.

don't we miss a "r" in the right side integral ?

[gaunt nimbus]

yes

[wicked sun]

thanks

[midnight plank]

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[flat spire]

How can I convert this to RREF I’m struggling to think what would the next best move be

[flat spire]

So I guess the only thing I can do is change R1 into something but idk what

[cedar mason]

RREF what

[flat spire]

What

What do you mean what

[cedar mason]

I'm not familiar I guess

[flat spire]

Huh?

Reduced row echelon form

[runic hamlet]

eliminate the 3 above the -1

multiply second row by -1

then you have rref

[flat spire]

How would I get rid of the 3? That’s what I thought I should do but I don’t know how to do that without drastically changing anything else

[runic hamlet]

add the second row to the first

[flat spire]

But if I do that then 3-1 = 2 no?

[runic hamlet]

well obviously multiply it by 3 first

[flat spire]

o

ohh byeah ok

[cedar mason]

oh gau'ssian elimination?

[runic hamlet]

yes

[cedar mason]

lmao makes sense

[flat spire]

After doing R1 + 3R2 I get
1 0 4 5 | 10 im not sure if that’s right

[runic hamlet]

it is

[flat spire]

Okay

Wait how does this get me RREF though don’t the 4 and 5 have to be a zero

[runic hamlet]

no

[flat spire]

I’m confused

[runic hamlet]

rref means row echelon form, all pivots =1 and only zeros above the 1s

[flat spire]

But above the 1s for my second row I have 4 and 5 which aren’t zeros

[runic hamlet]

only the pivots

not all 1s that appear randomly somewhere

(also in that second row there is only the one 1, the other two are -1)

[flat spire]

Wait so like this?

[runic hamlet]

yes

[flat spire]

ohhhhhh ok

thank you

.close

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[umbral scroll]

how can i find the variations of the f function

[tough shale]

Is that f'(x)?

[umbral scroll]

yes

[tough shale]

Are those steps related?

[umbral scroll]

yeah

[tough shale]

So when you took the LCM, e^-x should be multiplied with (x+1)² right?

[umbral scroll]

ohhh yeah right right

[tough shale]

And I think doing it without taking the LCM would be much better

[umbral scroll]

LCM ?

[tough shale]

Of the denominator

That's what we call it here anyways... (in my country)

When you take the common denominator

[umbral scroll]

oh ok

yeah

so how can i do this

[tough shale]

The 2/(x+1)² part

It's in 2 u^(-2) form

Apply the power rule for integration

[umbral scroll]

uhm yeah

wdym integration

[tough shale]

To find f(x)

Is that not the question?

[umbral scroll]

direction of variation of the function

i need to find the direction of variation of the function

[tough shale]

Oh

[umbral scroll]

so for that i need to solve f'(x) =0

[tough shale]

What does the ' mean then?

Derivative?

[umbral scroll]

yes

df/dx

[tough shale]

I thought you were supposed to find f(x) lol

[umbral scroll]

sorry in france we use the ' for derivative

[tough shale]

As in increasing/decreasing?

[umbral scroll]

yes in R+

[tough shale]

No, that's what is used everywhere

[umbral scroll]

ok so do you know how to find the direction of variation of the function of f here ?

[tough shale]

Okay, so taking the common denominator again (sorry, mb)

Yes

[umbral scroll]

yes

[tough shale]

The denominator is always positive right?

[umbral scroll]

yes as it is a square

[tough shale]

Yes

[umbral scroll]

oh yeah i know now

as it is in R+

[tough shale]

And for the numerator, it's $2 + (x+1)^2e^{-x}$

[umbral scroll]

it's always increasing

yeah well its 2 + instead of 1 +

i found the solution thank you

🙂

[tough shale]

Okay

[grand pond]

Suika

[midnight plank]

@umbral scroll Has your question been resolved?

[midnight plank]

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[wraith cosmos]

Help

[wraith cosmos]

I need help know how to write them in the forms it’s asking me I see there is test ones but still not good enough for me

[wary ridge]

js put them in the exponential form

and then solve it

do u want help solving a particular part

[wraith cosmos]

Yes

How do I make them like the examples she’s asking

[wary ridge]

have u studied exponents?

[wary ridge]

bro dont disappear

[wraith cosmos]

No

I haven’t

[wary ridge]

ok

whats $2*2$

[grand pond]

Shashwat

[wraith cosmos]

4

[wary ridge]

dyk u can even write it as $2^2$

[grand pond]

Shashwat

[wraith cosmos]

yes

thats 2 to the power of 2

[wary ridge]

thats exponents

[wary ridge]

u just gotta put powers in here

[wraith cosmos]

yea buy see how shes asking 1

1-4 = 1-256 or .003

or that dont matter

[wary ridge]

she put it in exponents

[wraith cosmos]

i can just do 4 to the power of 4

[wary ridge]

then she multiplied it

(1/4)^4 = 1^4/4^4 = 1/256

$(1/4)^4 = 1^4/4^4 = 1/256$

[grand pond]

Shashwat

[wraith cosmos]

what abt bed mas

[wary ridge]

e here means exponents

so ur doing exponents over here

b-brackets
e-exponents
d-division
m-multiplication
a-addition
s-subtraction

[wraith cosmos]

So I can do them like this

[wary ridge]

it wud be 4

not 5

[wraith cosmos]

oh yea sorry

[wary ridge]

$(-5)^4 = 5^4 = 625$

[grand pond]

Shashwat

[wraith cosmos]

is it not in the -

tho

it would not be - 625?

[wary ridge]

if u multiply - * - u get +

havent u studied that

[midnight plank]

@wraith cosmos Has your question been resolved?

[wraith cosmos]

Oh ok how can I get bedmas th

Tho

[wraith cosmos]

Well when I use in a calculator it’s -625

[last slate]

-(5⁴) and (-5⁴) are different

Even number gives you +ve sign

odd number gives you -ve sign

[midnight plank]

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[wraith cosmos]

Well I’m just confused because this is on my explanation sheet

[last slate]

if the bracket isn't specified in the que

then first find the power

then add the same sign

@wraith cosmos if sign is in the bracket then follow the rule which I said

[wraith cosmos]

ok

[last slate]

what is -2³

[last slate]

c and f are incorrect

(-3)²= -3*-3 =9 (two -ve multiplies make+ve)

[wraith cosmos]

Would it be positive numbers

[vague remnant]

the answer on f is correct

[last slate]

-5²= -(5*5)= -25

[last slate]

his step is incorrect

[vague remnant]

(-3)²

[wraith cosmos]

I’m confused am I right or wrong

[last slate]

on C your answer was incorrect and on f your step was wrong

[last slate]

huh?what did I do incorrectly?

oh I get it

Thanks

@wraith cosmos ?are you still confused?

[wraith cosmos]

sort of

[last slate]

on what part

[wraith cosmos]

on what to know whats a - power and whats not

like with the brackets etc

[last slate]

if the exponent (power) is even then the answer would be +ve

Like 2,4,6,8,...

if the exponent (power) is odd then answer would be -vr

this is valid for all -ve numbers

for +ve number the answer would be +ve irrespective of exponent

@wraith cosmos okay?

[wraith cosmos]

ok

i see

[midnight plank]

@wraith cosmos Has your question been resolved?

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[steady trail]

$\lim_{z\to 0} |\frac{e^\frac{1}{z}}{z-1}|$

[grand pond]

Task Bot

[steady trail]

How to calculate this ?

[vocal inlet]

was about to say some stupid shit...just wanna ask
is it some kind of complex limit ?

since you used z

[steady trail]

Yes

[vocal inlet]

well then I'M OUT

[golden comet]

I feel like it should be infinity

but yeah dont know how to do complex limits

[steady trail]

I think it does not exist because e^1/z for 0^+ and 0^- has 2 different limits

[vocal inlet]

now that you pointed it out

[midnight plank]

@steady trail Has your question been resolved?

[summer geode]

yes

It does not exist for that reason of e^1/z

being divided by z-1 near z = 0 would not affect that

[steady trail]

Thanks!

.close

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[wraith cosmos]

Help

[midnight plank]

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[dusty portal]

Does $\mathrm{lcm}(a, b)=2^n$ for $a, b, n\in\bZ^+$ force either $a$ or $b$ equal to $2^n$, and can this be generalized for any non-negative integer base?

[grand pond]

;(

[sharp coral]

yes, and only for prime bases

i think it helps to think of it in terms of the prime factorization

[dusty portal]

Ok

[midnight plank]

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[rustic tartan]

$\int_0^{\frac{\pi}{4}} \frac{\cos(x) + x \sin(x)}{\cos^2(x)}$ dx

[grand pond]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[grim sierra]

rewrite as cosx/cos^2x + xsinx/cos^2x

= secx + xsecxtanx

[rustic tartan]

I can't use sec

[grim sierra]

wdym you "cant use sec"

[rustic tartan]

Didn't learn it yet

[grim sierra]

i find this very hard to believe

you're doing integrals and apparently dont know what sec is

[rustic tartan]

Exactly

[grim sierra]

okay ive got the solution

[rustic tartan]

I tried to use {(e^ix + e^-ix)/2}

[grim sierra]

let g(x) = 1/cosx

then use g(x) instead of secx

[grim sierra]

are u doin git without sec as a challenge..?

[rustic tartan]

As i said, i didn't learn sec(x)

I had learned sin, cos and tan

[grim sierra]

then replace all your secants with 1/cosx

it will still work 💀

[grim sierra]

or do this

[grim sierra]

and theres absolutely know way youre allowed to use this and youre not allowed to use secx

[rustic tartan]

$\int_0^{\frac{\pi}{4}} \frac{1}{\cos(x)}$ dx + $\int_0^{\frac{\pi}{4}} \frac{x \sin(x)}{\cos^2(x)}$ dx

[grand pond]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[rustic tartan]

$\int_0^{\frac{\pi}{4}} g(x) + x \cdot \tan(x) \cdot g(x)$ dx

[grand pond]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[rustic tartan]

@grim sierra

What now?

[grim sierra]

rewrite g(x) = 1/cosx

[rustic tartan]

.

[grim sierra]

∫ 1/cosx dx = ∫ cosx/cos^2x dx = ∫ cosx/(1 - sin^2x) dx
u = sinx, du = cosx
= ∫ 1/(1 - u^2) then proceed with pfd

then do the same thing for ∫ xtanx/cosx dx when ibping

[rustic tartan]

Using integral by parts?

[grim sierra]

yes

[rustic tartan]

Hmm...

[rustic tartan]

∫ 1/(1 - u^2) = ?

Arctan?

[rustic tartan]

<@&286206848099549185>

[midnight plank]

@rustic tartan Has your question been resolved?

[rustic tartan]

<@&286206848099549185>

[rustic tartan]

...

$\int_0^{\frac{\pi}{4}} \frac{\cos(x) + x \sin(x)}{\cos^2(x)}$ dx

[grand pond]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[grim sierra]

∫ 1/(1 + u)(1 - u)

use partial fractions

google how to do pfd integration

[midnight plank]

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[midnight plank]

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[regal barn]

how can I proceed it? should I open the derterminant which will be long process? i can't think of any smart approch

[silver bluff]

you can do row column operations

since that doesnt change the value of the determinant

addition

you can try any way you want across a row or a column

since each row/column has x+1,w and w^2 it doesnt really matter how you add

[regal barn]

thank you very much

i got it now

1+x+w+w^2 and it will be x take common out of each row

so it will be divisible by x

[midnight plank]

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[molten bay]

O is incenter of this triangle and we are given some information which is

AO/OE=5/4

CO/OD=3/2

BO/OF

[tawdry laurel]

@molten bay

Hello

Are you aware of this angle bisector theorem

Note AO, BO, CO are angle bisectors

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

Hello

Yes you are right

[tawdry laurel]

Ok

[tawdry laurel]

@molten bay

[molten bay]

What can I use of it?

[tawdry laurel]

Okay

[molten bay]

Actually let me google

[tawdry laurel]

We have AO / OE right

[molten bay]

I see

[tawdry laurel]

Since BO is bisector and CO is bisector

AB / BE is AO / OE which is 5/4

Can you do similar for the other angles and side ratios?

(the diagram is very misleading)

[molten bay]

Ahh why?

[tawdry laurel]

Because AO looks like more than 5/4 OE

[molten bay]

AB/AC=AD/AF

[tawdry laurel]

Or it could just be my perception

[tawdry laurel]

uhh how did you get that

[molten bay]

I didn't understand how you compared

In one triangle

[tawdry laurel]

Ok

[molten bay]

Bisector theorem?

[tawdry laurel]

So the angle bisector theorem in AEB

Since BO is bisector right

So by the theorem

[molten bay]

AB/BE

[tawdry laurel]

AB/BE is AO/OE

[molten bay]

Ohh got it

Thanks

[tawdry laurel]

Which is 5/4 yay

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

So what should I do next?@craze.cv

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[crude totem]

a is from R

i have to find a so that the matrix is of type Cramer

what should i do next or is it not this the best approach?

.close

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[sweet shuttle]

can these be multiplied together ?

[sweet shuttle]

I thought the number of rows in a has to match the number of columns in b

a is 3x2 and b is 2x2

[subtle blaze]

Do you think yes or no

[sweet shuttle]

what ? yes or no to what ?

if it can be multiplied ?

[subtle blaze]

Yeah

[sweet shuttle]

nah but it says to multiply them so I assume yes

[subtle blaze]

Do you know that AB is not the same as BA?

[sweet shuttle]

yeah

[subtle blaze]

So which are you looking to calculate

[sweet shuttle]

both

[subtle blaze]

Try one first

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[tough shale]

If n is even, it is given that the function is either strictly decreasing or strictly increasing at x = c

But f'(c) = 0 right?

How is it possible?

Shouldn't it be a tangent to the graph at that point?

Can anyone pls give me any example which the given statement supports but not my statement.?

[fallow scarab]

Can you fix this sentence

What is "it"

[tough shale]

The line with slope 0 at that point

f'(x) is slope right

[fallow scarab]

Yes tangent lines at local max and min have slope zero for differentiable functions

[tough shale]

??

[fallow scarab]

f(x) = x^2

The first derivative at x=0 is zero but f'' = 2 > 0

[tough shale]

n = 1 which is odd. Hence there is a local minimum at x = 0

[fallow scarab]

Oh then x^3

[tough shale]

Yes, so we get a tangent at x = 0 and that the graph has a point of inflection

But wiki also said that the function would be strictly increasing at x = 0 which I don't see the case in the graph (and it also contradicts the tangent)

Is what my doubt is

[midnight plank]

@tough shale Has your question been resolved?

[fallow scarab]

x^3 is strictly increasing around 0

[tough shale]

But slope is 0 at x = 0

[fallow scarab]

Do you know the definition of strictly increasing

[tough shale]

Doesn't strictly increasing mean f'(x)>0 in that region?

[fallow scarab]

No

[tough shale]

Oh mb

Then what is it?

[fallow scarab]

https://mathworld.wolfram.com/StrictlyIncreasingFunction.html

[tough shale]

Oh yeah

Mb

Thanks

Another query:

Why did i get no inflection points?

[zealous schooner]

,w derivative of (x-1)(x-2)^2(x-3)

[grand pond]

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[zealous schooner]

,w factorise 4x^3-24x^2+46x-28

[grand pond]

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[zealous schooner]

looks like your first derivative is correct

,w double derivative of (x-1)(x-2)^2(x-3)

[grand pond]

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[zealous schooner]

you do have inflection points, those are the points where f''(x) changes sign

i.e. f''(x) = 0

[tough shale]

But shouldn't f' and f'' be both 0?

[tough shale]

As this

[zealous schooner]

no

f' being 0 doesn't affect any value of x being an inflection point

an inflection point is a point across which the concavity of the function changes

[midnight plank]

@tough shale Has your question been resolved?

[tough shale]

So is it wrong in wiki?

[midnight plank]

@tough shale Has your question been resolved?

[solemn gazelle]

Hello! How are you? I don't know what it says on the wiki, but in general, if you have a function f and need to calculate the inflection points, you follow this procedure:

1. Find f''(x)

2. Solve f''(x) = 0

3. The sign of "f" is studied (Maybe you say it in another way, but it is a diagram that indicates for which values of x the images f"(x) are positive, negative and zero)

4. As kheerii said before, you have to look at the roots of f" (the zeros you put in the diagram above) and see if there is a change of sign before and after that root.

By change of sign I mean "first it's negative and after the square root it's positive" or "before the square root it's positive and after the square root it's negative." In those two cases, the point is a inflection point of f.

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[last slate]

how do I find the domain of g

[solemn gazelle]

You need the rational expression to give you numbers between -1 and 3. Observe which values of x satisfy that (you can solve an inequality, graph it)

[last slate]

I got 1/2 < x and 3.5 < x

what do I do now

<@&286206848099549185>

<@&286206848099549185> \

bro i need help

[midnight plank]

@last slate Has your question been resolved?

[last slate]

<@&286206848099549185>

[fervent ember]

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[last slate]

@fervent ember

[fervent ember]

assume x = 3, calculate (x+1)/(x-2)

[last slate]

4/1?

4