help-17

[royal tide]

I forgot to distribute the -

[reef agate]

Also it seems like you just turned the - into +

[royal tide]

yeah cause the - wasnt distribture

can you explain how they got 1/2 and where the 1 next to (y’)^2 came from

[reef agate]

No you shouldnt distribute before you even multiply

You should just say $\qty(y\prime)^2 = \qty(x^4 - \frac 1{4x^4}) \qty(x^4 - \frac 1{4x^4})$

[twin meteor]

King Leo

[royal tide]

yeah

then you foil right

[reef agate]

And only then you must distribute

[royal tide]

then thats the y prime to the second power

can you explain my friends works pls

[reef agate]

Which step of your friend's work do you not understand

[royal tide]

Like there’s a 1 next to y prime squared

And it’s 1/2 for her

[reef agate]

Your friend solved for $1 + \qty(y')^2$ to conform with $\int_a^b \sqrt{\textcolor{cyan}{1 + \qty(\dv{y}{x})^2}} \dd{x}$

And $-\frac 14 - \frac 14 = -\frac 24 = -\frac 12$

[twin meteor]

King Leo

King Leo

[vocal sleet]

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[wintry arrow]

working on a sequence excerise, everything makes sense except part 4 how did we get the equation of wn

[paper depot]

what do you mean?

$w_n = 2u_n + 3v_n$ is the \textbf{definition} of the previously unseen sequence $(w_n)$.

[twin meteor]

Ann

[wintry arrow]

so basically Wn=2Un+3Vn both which ive workedon in other parts

i dont understand how we solved part 4

this is the answers frommy teacher

[flat whale]

which step do you get lost first

[wintry arrow]

okay so we replaced with the equation we got before for u in2b and v in 3b, but then after it became Wn+1 i got lost

[flat whale]

where exactly is this step

your handwriting and color choice is hard to read/follow

[wintry arrow]

sadly its my teachers i dont understand it either

[flat whale]

screenshot the exact line you get lost

or two lines

[wintry arrow]

where did the 9 come from

[vocal sleet]

@wintry arrow Has your question been resolved?

[flat whale]

9(n+1) = 9n + 9

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@wintry arrow Has your question been resolved?

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[peak matrix]

thats too many terms imo

btw you could also avoid taylor

$\frac{\ln\left(\cos\left(x\right)\right)}{\cos\left(x\right)-1}\cdot\frac{\cos\left(x\right)-1}{x^{2}}$

[twin meteor]

MathIsAlwaysRight

[peak matrix]

you can rewrite it like this

Remember this limit?

lim ln(x+1) / x = 1

oh

you can use that

but i'd do that on the log

1 + cos(x) - 1 would work

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[hard hound]

can u do five without fermats little theorem

[paper depot]

sure, you can just rawdog the first few powers of 5 and 17

might help that 17 ≡ -5 (mod 11) so this whole thing is equal to 2 * 5^22

so really you just need to rawdog the first few powers of 5

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[lyric wigeon]

Ive gotten this far but im confused how which f(x) to put first in the intergral

[hollow sundial]

well currently your bounds don't work sadly

remember that the line $2-x$ intersects with $\sqrt{x}$ only once

[twin meteor]

BuilderDolphin

[hollow sundial]

so only one of the solutions in $(x-4)(x-1)=0$ would work

[twin meteor]

BuilderDolphin

[hollow sundial]

(they intersect only once since both have a constant sign slope at any point)

[lone scarab]

they intersect at x=1

[hollow sundial]

once you get the intersection point you can split the integral and evaluate the volumes depending on which function is bounding on each side

[lone scarab]

yep, y=sqrt(x) is bounding till 1, y=2-x afterwards

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[royal cipher]

So far I just tried expanding the whole thing but its looking like that isnt the solution here because i have this equation that looks impossible to work with. My prof said to recall amgm but im still lost here, so am I even on the right path?

[paper depot]

first note that if any letter is 0 then the entire thing becomes (LHS) ≥ 0 which is trivial

so assume they are all strictly positive

then you can divide both sides by klm and write:

(k/m + l/m)(k/l + m/l)(l/k + m/k) ≥? 8

and i think expanding that and applying the ineq t + 1/t ≥ 2 should help somehow

[royal cipher]

where does this inequality come from?

[paper depot]

various sources

the assumption underlying it is t > 0

t + 1/t - 2 = (t^2 - 2t + 1)/t = (t-1)^2/t

[neat rose]

@royal cipher what did u try?

[royal cipher]

expended left side of inequality

rn im trying to work with what ann suggested

[neat rose]

aah u can use AMGM

$a+b \geq 2\sqrt{ab}$

[twin meteor]

Goëtia

[neat rose]

rewrite this for k+l & l+m & m+k

[royal cipher]

im still a bit lost here im sorry, but I rewrote it

[neat rose]

show me

[royal cipher]

k+l ≥ 2sqrt(kl)
k+m ≥ 2sqrt(km)
l+m ≥ 2sqrt(lm)

[neat rose]

multiply them

[royal cipher]

ok wait i see the vision

oh ok so you just end up with the proposition

ok yeah he said this would be easy once its like there in front of me I see now

ty

[neat rose]

close channel if ur good to go @royal cipher

[royal cipher]

oh wait how to i close

nvm

.close

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[zenith copper]

hi

[zenith copper]

can someone help me w a binomial expansion q please

[hushed pewter]

!da2a

[vocal sleet]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[zenith copper]

srry if its hard to read

ik i need to multiply

but idk how

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@zenith copper Has your question been resolved?

[zenith copper]

no

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@zenith copper Has your question been resolved?

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[lusty crow]

If i have limits in summations is there a way to calculate it out heres an example

[vocal sleet]

@lusty crow Has your question been resolved?

[regal bane]

Infinite limits outside of summations can sometimes be recast as Riemann sums and turned into an integral

[lusty crow]

Does that work in this case

[steady plover]

Try subbing n/k as x and 1/k as dx

limit of the integral: 0 to 1

It’s what kaynex said

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[round ether]

what is being describe here

[round ether]

This is for regular expressions

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[round ether]

<@&286206848099549185>

.close

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[bitter delta]

what's the answer for this?

part c

I believe a and b are right

[tidal umbra]

if $f(g(x))=g(f(x))=x$ what kind of function are $f(x)$ and $g(x)$?

[twin meteor]

yajat

[bitter delta]

inverse?

idk

[tidal umbra]

yes

[bitter delta]

thanks

[tidal umbra]

that should be your asnwer

[bitter delta]

.close

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[mighty adder]

How do I go about solving this step by step ?

[reef agate]

Do you know how to integrate the h'(x)

This will get you some form of h(x) = ... + C

[vocal sleet]

@mighty adder Has your question been resolved?

[wary hull]

Do the subsitution u=1-x^2 when integrating

$\int x\sqrt{1-x^2}dx$ with $u=1-x^2\implies du=-2xdx\implies -\frac{du}{2}=xdx$

[twin meteor]

mathisfun

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[shadow spire]

Could somebody help me with this geometry problem A regular triangular pyramid is intersected by an oblique plane that contains one base edge and is perpendicular to the opposite lateral edge. This plane, together with the base plane, forms a dihedral angle of α. The truncated part of the pyramid, between the base and the oblique plane, has a volume V. What is the height of the pyramid? This is the solution that I have to get

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[trim prawn]

i hate series

[trim prawn]

series should not exist

[reef agate]

Do you have a question regarding series?

[wary hull]

!da2a

[vocal sleet]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[trim prawn]

no I'm just venting

[atomic jasper]

this is not the place to do it

perhaps another channel will fit better, look around

[trim prawn]

It is the place if I say so

[atomic jasper]

thats not how this works

[wary hull]

You would not be victorious

[rugged orchid]

Theres #discussion

[light jay]

its not its a help channel for people who need help with questions

[reef agate]

💀

!done

[vocal sleet]

If you are done with this channel, please mark your problem as solved by typing .close

[light jay]

you having a vendetta against series isnt a question though

[trim prawn]

.close

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[balmy plaza]

uwu

[balmy plaza]

need help

[wet halo]

@balmy plaza

[balmy plaza]

i mean.

uhhh

suz

[edgy kayak]

???

[pale perch]

[wet halo]

😔

[balmy plaza]

c.lose

[wet halo]

😔

[balmy plaza]

why sad broooo

[reef agate]

Because typo

😔

[balmy plaza]

how to close it

.close

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[wet halo]

😌

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[dull geode]

hey i dont really get this proof

[dull geode]

this is the question, what i dont get is othr implication

if $b$ has a prime faccorization of the forma blah bkah then b divides a

[twin meteor]

SushiMan

[hybrid flicker]

This is so weird as we absolutely need to suppose a and b in N and not in Z

[dull geode]

idk this is the teachers sol

[hybrid flicker]

since there's no +-

[dull geode]

this was one of the two sols

[hybrid flicker]

so you need help with this implication?

[dull geode]

well i mean i tried doing it on my own

lemme take a photo

[hybrid flicker]

let $a = p_1^{a_1}p_2^{a_2}\ldots p_n^{a_n}$ and $b = p_1^{d_1}p_2^{d_2}\ldots p_n^{d_n}$ with $0\leq d_i \leq a_i$

[twin meteor]

rafilou is not not born in 2003

[dull geode]

ok so i dont get how to do the reverse implication

[hybrid flicker]

$p_i^{d_i}\mid p_i^{a_i}$

[twin meteor]

rafilou is not not born in 2003

[dull geode]

yes i agree with this

[hybrid flicker]

so

multiply all of those

you know a|b and c|d implies ac|bd

[dull geode]

ts is what i have so far

ignore the last two lines

[hybrid flicker]

oh wow you don't need to go that far

no need to implicate the gcd at all

[dull geode]

oh yeah dont worry about tha

thas for the other question

[hybrid flicker]

you only need that

[hybrid flicker]

and this property

of course you can also see it directly with:

[dull geode]

what im so confused

[hybrid flicker]

$a = p_1^{a_1}p_2^{a_2}\ldots p_n^{a_n} = p_1^{d_1}p_2^{d_2}\ldots p_n^{d_n} \times p_1^{a_1-d_1}p_2^{a_2-d_2}\ldots p_n^{a_n-d_n}$

[twin meteor]

rafilou is not not born in 2003

[dull geode]

so

the two implications are practically teh same thing

[hybrid flicker]

oh no not at all

[dull geode]

cause thats what i did for the first diection

here

[hybrid flicker]

where does the "b = p^..." come from?

and how do you know the same primes are involved in b's decomposition?

[dull geode]

from the question

[hybrid flicker]

if you're only supposing b|a

[dull geode]

oh yeah i was confused about that

so im just assuming stuff here

ok wait

[hybrid flicker]

yeah you're assuming too much

remember that if you only suppose b|a

you don't even have access to this

for all we know more primes could be involved

(and also if b is not a natural integer, that's not necessarily true)

(as we could have b = - ... or b = 0)

[dull geode]

Is this better

[dull geode]

ok so i got where you wanted to m with the reverse implication, i just dont get what you mean by this

[vocal sleet]

@dull geode Has your question been resolved?

[dull geode]

<@&286206848099549185>

[dull geode]

.close

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[vocal sleet]

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[valid path]

can someone walk me through a proof of mathematical induction, lets just say for this question: $3^{n}>3 \times 2^{n}, n>=3$

[twin meteor]

linden

[valid path]

$3^{n}>3 \times 2^{n}, n \geq 3$

[twin meteor]

linden

[valid path]

base case, n=3
3^3 > 3*2^3
27 > 24

base case true

i forgot the name of this step

$3^{k}>3 \times 2^{k}$

[twin meteor]

linden

[valid path]

and now we assume if P(k) is true then P(k+1) must be true

$3^{k+1}>3 \times 2^{k+1}, n \geq 3$

[twin meteor]

linden

[valid path]

and now idk what to do

get P(k) inside P(k+1)??

$3 \times 3^{k}>2 \times 3 \times 2^{k}$

YO ignore the n>=3 i forgot to remove that

[twin meteor]

linden

[wary hull]

This is the inductive step

[wary hull]

Get P(k+1) in some form of P(k) to show it is true

We want $3^{k+1}>3\cdot 2^{k+1}$

[twin meteor]

mathisfun

[wary hull]

And get it to $3^k>3\cdot 2^k$

[twin meteor]

mathisfun

[valid path]

is it not this?

[wary hull]

It is indeed so

[valid path]

and now what do i do from here

[wary hull]

We have $3^k>2^{k+1}$

[twin meteor]

mathisfun

[valid path]

oh we do too

[wary hull]

Yes

[valid path]

what can we use this for

[wary hull]

Now we know that $3^k>3\cdot 2^k$ by the inductive hypothesis

[twin meteor]

mathisfun

[wary hull]

Therefore, since $3^k>3\cdot 2^k>2\cdot 2^k>2^{k+1}$, our inductive hypothesis is proved

[twin meteor]

mathisfun

[valid path]

and then we just finish with the

conclusion

or whatever its called

[wary hull]

Yes

[soft walrus]

QED?

[wary hull]

[valid path]

okok

lets see if ive got that down

with a super easy example

2^n > 2n, n is natural number greater than 2
base case:
2^3 > 2(3)
8 > 6

inductive step:
P(k) => 2^k > 2k

assume P(k+1) is true

P(k+1) => $2^{k+1} > 2(k+1)$

[twin meteor]

linden

[wary hull]

$P(k+1)\implies 2^{k+1}>2(k+1)$

[twin meteor]

mathisfun

[valid path]

thank you

$\implies 2\codt2^{k}>2k+2$

$\implies 2\cdot2^{k}>2k+2$

[twin meteor]

linden

[valid path]

i forgot what to again but we persevere

due to the inductive hypothesis we know that $2^k>2k$ therefore, $2*2k>4k>2k+2$

[twin meteor]

linden

[valid path]

chat im lost again 😭

[wary hull]

a

$2^k>k+1$

[twin meteor]

mathisfun

[valid path]

but

2k>k+1

[wary hull]

We know $2^k>2k$ by inductive hypothesis

[twin meteor]

mathisfun

[wary hull]

So $2^k>2k>k+1$ is $k>1$, which is true here

[twin meteor]

mathisfun

[valid path]

no way its that simple

thank you so much mathisfun

again

i think i got it

you'll see me back in like a day if im not

.close

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[arctic timber]

I've got this thing thats practice for a quiz coming up and I cant figure out what I did wrong. It was something deriving it at the start but idk where it is. Can someone help?

[wise drift]

product rule for 2nd term on the left and also for the right

[wary hull]

Finding the $\frac{dy}{dx}$?

[twin meteor]

mathisfun

[arctic timber]

oh youre right

[wary hull]

Note that $\dv{x}xy=(x)'y+x(y')$

[twin meteor]

mathisfun

[arctic timber]

thanks

nice thank you guys

[wary hull]

Welcome

Let me see if I can cook up a sol

[arctic timber]

wait i think i just did something wrong again

is the derivative of xy^2 = x(2y*y') +y^2?

[wary hull]

Yes

[arctic timber]

so then am i going to be dividing by y' at some point

idk what i did im in basically the same position but with different numbers now

[wary hull]

No

[wise drift]

combine the like terms now

[wary hull]

$2x+xy'+y-3y^2y'=y^2+2xyy'\implies y'(x-2xy-3y^2)=y^2-2x-y$

[twin meteor]

mathisfun

[wise drift]

like bring em to one side

[arctic timber]

where did the other y' go

[arctic timber]

okok

but how do you get rid of the y' on the left

[wary hull]

Where

point it out

[arctic timber]

2xyy'

the y' disappears

[wary hull]

Nope

[wary hull]

It's the second term on the left

[arctic timber]

do I facor it with the other side?

[wary hull]

Yes

[arctic timber]

OH I GET IT

[arctic timber]

but one last thing why is x(2y +y') not 2xy+ xy'

i get everything else but can you just not distribute like that?

[wary hull]

Huh

it is same thing

The thing inside is just multiplied though

Not added

[arctic timber]

oh youre right

mb

thank you so much

[wary hull]

Should I post sol

[arctic timber]

sure I think i got it but itd be nice to double check\

[wary hull]

$2x+xy'+y-3y^2y'=y^2+2xyy'\implies y'(x-2xy-3y^2)=y^2-2x-y\implies y'=\frac{y^2-2x-y}{x-2xy-3y^2}$

[twin meteor]

mathisfun

[arctic timber]

yup perfect

thank you

[vocal sleet]

@arctic timber Has your question been resolved?

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[vapid kiln]

Quick question here: Does this equation have 1 or 2 solutions?

[outer warren]

did you try solving it?

how many solutions did you get?

[vapid kiln]

Got 2

-28 and 22

[outer warren]

can you show your work

[wary hull]

Plug it back in

Both of them

[vapid kiln]

Uhh square both sides and then we have
| x+3 | = 25

x+3=25 x=22
x+3=-25 x=-28

[outer warren]

you're conflating the abs val property

[wary hull]

Huh

[outer warren]

$\sqrt{m^2} = |m|$ \
however $(\sqrt{n})^2 = n$

[twin meteor]

ℝαμOmeganato5

[wary hull]

No need for abs

We are taking principle root here

Therefore the argument inside the radical (radicand) must be positive

[vapid kiln]

Ahh I didn't notice that

[last agate]

imagine the intersection of the graph of sqrt(x) and the horizontal line y = 5

sqrt(x + 3) is a transformation away from its parent function sqrt(x)

in particular a horizontal translation

i think it should be pretty obvious now how many solutions there is to that equation

not that u can’t plug in your answers back to get rid of the extraneous solutions u introduced by squaring both sides to but yeah

[wary hull]

^

[outer warren]

it wasn't really an extraneous solution

[last agate]

yeah okay, there is none

they just solved a completely different question lol

[wary hull]

Should just be:\
$x+3=25$\
$x=22$

[twin meteor]

mathisfun

[vocal sleet]

@vapid kiln Has your question been resolved?

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[vague torrent]

$a \times b \times c = d \times e \times f \times g \times h \times i$

i need to prove no solution exists if we use the natural numbers 1 through 9 (each number must be distinct)

[vague torrent]

distinct as in a neq b neq c neq d and so o

for example 123=45678*9

[twin meteor]

polar

[novel cliff]

well that is easier than you think

think about the digit 7

[vague torrent]

yeah?

[novel cliff]

if you have a number

let me put it thid way

if you multiply some numbers and a 7, you would agree that the product is a multiple of 7 right?

[vague torrent]

indeed

[novel cliff]

therefore if the equation need to work, then both sides need to be multiple of 7 cause at least one 7 is used

but we only have 1 of 7, therefore only one product can be a multiple of 7 as 7 is a prime number

[vague torrent]

okay i got it already

[novel cliff]

ok :>

[vague torrent]

thank you very much sur

sir

[novel cliff]

np

[vocal sleet]

@vague torrent Has your question been resolved?

[vocal sleet]

Channel closed

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[vocal sleet]

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[waxen storm]

Hey,can someone help me with solving this problem

[fresh ivy]

need some help

[paper depot]

<@&268886789983436800> test cheater + channel occupied

[paper depot]

which subpart

[waxen storm]

The first one

[waxen storm]

I don’t think he’s cheating

Cause

It’s march 2

This quiz was in February 27

[paper depot]

well he should still take his own channel lmao

[waxen storm]

True

[paper depot]

but ok point taken

[waxen storm]

Okay so basically

She wants p(x) in terms of x

Since we have that 0<x<2

I don’t know how to solve it like that

I got that MCN is 90 degrees since it’s a rectangle given and angles in a rectangle are 90

[waxen storm]

Are they both x?

[paper depot]

DN = BM = x

it says so in your question

[waxen storm]

Shouldn’t it be only BM

oh

So when I try to get the area it should have an x in it

[paper depot]

of course it should

that's what "in terms of x" means

[waxen storm]

Oh

Then do I just put everything

Its basexheight over 2?

Shouldn’t I get MN since it’s the hypotenuse to get the base

And then I can take MC as a height since it’s a 90 degree angle

[paper depot]

don't use the letter x as a multiplication symbol

the line of attack for finding the shaded area is to do it as [ABCD] - [ADN] - [MNC]

[waxen storm]

Oh I have to get the shaded area

[waxen storm]

What do you mean by that

Nvm

I got it

Do I subtract the whole rectangle from the 2 triangles as well

Or do I say ABCD=ADN-MNC

And we are subtracting the area right

[paper depot]

other way around

you are subtracting the two white triangles from the rectangle

[waxen storm]

Ohhh so I put the rectangle lastly

[paper depot]

...

[ABCD] - [ADN] - [MNC]

i was unambiguous here

rectangle minus triangle minus other triangle

[waxen storm]

OH OKAY

But I don’t know how to get the area for MNC

Since I don’t know the base

[paper depot]

yes you do

it's 4 - x

[waxen storm]

NC is the base?

[paper depot]

well it's the horizontal side

[waxen storm]

I thought the base is the hypotenuse

[paper depot]

that can sometimes help but not here.

[waxen storm]

Oh okay

[paper depot]

you have a right triangle in which both legs are known

those are perpendicular to each other and so can play the roles of base and height

[waxen storm]

Oh I get it now

That’s a rule

So does it become
MNC= (4-x)(2-x)/2

It’s -11/2?

Nvm it’s 3/2

I dont underatsnd how (2)(x)/2 gives a 5

[fresh ivy]

what number? i forgot to tell, so hard for me the no.1

[waxen storm]

That’s for the area of ADN

[fresh ivy]

that is not a cheat, just a assessment. will pass tomorrow, just delayed becuz of lacking time. pls help on number 1, so hard. that's calculus

[paper depot]

!occupied

[vocal sleet]

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).

[paper depot]

you're still in somebody else's channel

[waxen storm]

Uh I don’t know what I’m solving

[paper depot]

neither do i

[waxen storm]

Erm

[paper depot]

didnt you just want the area of the thing in terms of x

[waxen storm]

Yes

[paper depot]

why are you trying to substitute random shit into x now

[waxen storm]

I don’t know…

I failed math

I don’t know what I’m doing

..

[paper depot]

you got the area of MNC correctly as (4-x)(2-x)/2

[paper depot]

2x/2, or just x, is the aera of ADN

now subtract both of these from the area of ABCD, which is 8

you will get an expression in terms of x

[waxen storm]

I don’t know I got 13-2x over 2

[paper depot]

you are not supposed to plug anything into x at this stage. x is just x.

[paper depot]

show ALL your work.

[waxen storm]

But my calculator gave 3 over 2

[paper depot]

...what did you put into your calculator

[waxen storm]

Look

[paper depot]

i have this exact same model of calculator

it does not have a way to simplify expressions

x is probably something left over from when you last used a mode or function that required a variable

it is a garbage value and has nothing to do with this problem

you should be doing this BY HAND.

...

[waxen storm]

.

[paper depot]

oh for crying out loud omg

[waxen storm]

Why would I

I can use a calculator

[paper depot]

it only hit me now that i jumped straight to like part 3 of the problem

[waxen storm]

Erm

[paper depot]

part 1 asks for ONLY the area of MCN

[waxen storm]

Yes…

It’s 3/2

[paper depot]

no it's not just 3/2.

it's going to depend on x.

anyway using a calculator ≠ just chucking your entire problem into it wholesale, yes?

[waxen storm]

I dont underatsnd anything ur saying

.

Look

[paper depot]

let's start from the beginning, then?

[waxen storm]

Okay

[waxen storm]

My calculator is tweaking

[paper depot]

can you show me what happens when you print just x

[waxen storm]

Okay

[paper depot]

ok so this 5 ended up here from some previous calculation you did.

so no, you just aren't using it correctly.

again,

this calculator cannot do algebraic simplification for you.

a calculator is not a magic "throw problem in, get answer out" box.

[waxen storm]

I was using it for proportionality’s I never used it for algebraic expressions

Oh

[paper depot]

yet here you are, trying to use it to do algebra.

[waxen storm]

Well this is gonna be hard

[paper depot]

how old are you and/or what grade are you in

[waxen storm]

9th

I’m 15

[paper depot]

ok so this is approximately the right difficulty level for you then

yes, you are going to need to simplify this stuff by hand. expand carefully, collect like terms, all that jazz...

[waxen storm]

I just have difficulty with algebra normally I can solve roots tho I just don’t get what they want in the question

[paper depot]

they want you to simplify the expression (4-x)(2-x)/2

thats it

no more no less

[waxen storm]

How do I do that

Do I expand and reduce (4-x)(2-x)

[paper depot]

dunno what you mean by "reducing" but expanding is the right thing to do.

[waxen storm]

Uh I don’t know that’s what my teacher says

[paper depot]

ok well can you do it and show me your steps and answer

[waxen storm]

Wait it’s + xsquared

[paper depot]

yes, the x^2 should have a plus sign before it

[waxen storm]

I added that

[paper depot]

is combining -4x - 2x into -6x the thing your teacher calls "reducing"?

[waxen storm]

It looks like an identity

[waxen storm]

Kinda

It’s when you like break down the parentheses

She calls it expand and reduce

[paper depot]

oh so as a fixed phrase..

ok whatever

[waxen storm]

But when we factorize we add parentheses

[paper depot]

(8 - 6x + x^2)/2 is correct

[waxen storm]

That looks like an identity tho

Is it

Yk (a-b) squared

I’m not sure if it’s right tho

Cause a squared and nothing squared gives 8

Do I take a common thingy which is 2 or I can’t since the x^2 doesn’t have a 2

[vocal sleet]

@waxen storm Has your question been resolved?

[waxen storm]

.close

[vocal sleet]

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[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[frigid snow]

math isnt doing its thing

its supposed to look like this

[paper depot]

forgor the minus sign

in the exponent

in front of the fraction

[frigid snow]

oh im so silly

[paper depot]

btw on desmos you can do normaldist(mu, sigma)

[frigid snow]

thats very useful

just checking if i got the minus in the right place

[paper depot]

yup

[frigid snow]

cool

however

flat line

[paper depot]

you are zoomed way out

what are your mu and sigma?

zoom in to near whatever mu is

[frigid snow]

zoomed in on it

[paper depot]

adjust x axis scale to something like... 0 to 15 i guess?

go into settings

[frigid snow]

that is a very small bump

[paper depot]

i mean youve got a pretty spread out curve

[frigid snow]

so this is fine?

[paper depot]

i plotted your curve with your mu and sigma

[idle cedar]

general equation for bell curve / SD curve [if found helpful]

e^{-x^{2}}

[twin meteor]

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

[idle cedar]

oh wait it has to be with mu and sigma

[paper depot]

OP was already using this

[idle cedar]

a

[frigid snow]

im wondering why im not getting the same results as you

[paper depot]

it is just a matter of having a poorly scaled viewing window

nothing more

[idle cedar]

i was about to say that

im pretty sure one can scale y-axis by holding shift and dragging the y-axis down or up on PC, or pressing and holding then dragging the y-axis down or up on mobile

[frigid snow]

i'll probably do this on mobile

[paper depot]

wait you can do that on desktop??

[frigid snow]

i just tested

yes you can

okay yeah that makes this so much easier

[idle cedar]

yes

[idle cedar]

you can scale it to match this original scale

[frigid snow]

i roughly got it

im happy with this though

thank you so much

.close

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@jade vale Has your question been resolved?

[jade vale]

.close

[vocal sleet]

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[fading portal]

I answered IN=a.CE now im stuck on EC=c.AB
can someone explain how we start answering

[shadow spire]

No

[fading portal]

open a new help

geometry ppl?

[queen root]

hi again

u proved that vector DC = 10/7 EC

and u have vector DC = 10/3 AB

thus, 10/7 EC = 10/3 AB

and js find the coefficient c

[vocal sleet]

@fading portal Has your question been resolved?

[fading portal]

yeah i found that C=7/3

Sorry i was offline
when i was offline i answered it

but now im thinking about b.

[queen root]

b

seeing ABED is a parallelogram, then vector EB = DA

[fading portal]

yeah

[queen root]

mb i went afk

u also have ABHJ a parallelogram so vector JA = HB

now u can go work it out

[fading portal]

dw

yup

HQ=JQ-AB?

cuz JH=AB

[queen root]

yes

[fading portal]

or that wont work cuz well have 0=0

lol

[queen root]

oh

i suggest first off prove that AD / AJ = BC / BQ = 1/4

[fading portal]

okkay

[queen root]

do yk how

[fading portal]

u mean AD/AJ

[queen root]

yes mb

[fading portal]

dw

yup thales

[queen root]

yes

and in the start u proved that AJ = 4AD to build point J

donc le rapport est 1/4

then, apply another thales in triangle HBQ, u have E on (HB) and C on (BQ)

and (EC) II (HQ)

[fading portal]

Yeahh

[queen root]

now the other thales

u wanna find EC/HQ

[fading portal]

EC/HQ?

I want to ask something

[queen root]

yes

[fading portal]

how do these ideas come to ur mind

like how

[queen root]

uhhhhh

[fading portal]

XD

[queen root]

like

[fading portal]

i wanna know the secret\

[queen root]

idk

[fading portal]

😆

[queen root]

if u get good at math

[fading portal]

yah

[queen root]

u get better eyesight

[fading portal]

hm

[queen root]

better eyesight

better visualisation to the problem

[fading portal]

i wanna do sciences maths next year inshaallah

[queen root]

yes, cause u wanna EC since u have it in terms of AB

[queen root]

ur 2ème?

and inshallah

[fading portal]

2eme of what

[queen root]

année lycée

[fading portal]

1ere annee lycee, 5eme wlla tronc commun

next year 2eme

[queen root]

oh

[fading portal]

yes

[queen root]

different bac programs ig

[fading portal]

im not in bac

[queen root]

no ik that

[fading portal]

ye

[fading portal]

ohh true

[queen root]

if u do thales in BHQ

u get EC/HQ = BC/BQ

which is 1/4, we js proved it

[fading portal]

ye

[queen root]

so EC = 1/4 HQ

and EC = 7/3 AB

[fading portal]

yeah

[queen root]

so 7/3 AB = 1/4 HQ

so whats b

[fading portal]

b=7

[queen root]

what

[fading portal]

21/3

ah nah

[queen root]

7 × 4 is not 21

[fading portal]

28/3

[queen root]

yep

[fading portal]

wut lol

i messed up

yeah anyway

9.3333333333333333333333333333333333333333333333333333333333333333333333

yeah

yey i got it

[queen root]

crazy

[fading portal]

Thank you so much for helping me!

[queen root]

js a reminder tho

[fading portal]

yeah?

[queen root]

when u have distance BC = 2MC for example

that doesnt instantly imply that the vectors are the same too

[fading portal]

hm

yes

true

[queen root]

u have to also add (BC) II (MC) and that (BC) et (MC) ONT LE MEME SENS

[fading portal]

ill keep that in mind

does that apply to this exercise?

[queen root]

yes

u have to say it

[fading portal]

where

[queen root]

caracteristiques d'un vecteur

[queen root]

after every thales we have done

[fading portal]

ohhh

[queen root]

and yeah, exercice done

[fading portal]

done

splendid

[fading portal]

🥳

Crazy how we both speak french, the exercise IS originally in french, but we keep talking in english

[queen root]

yeah i guess

[queen root]

though i like to answer it in french better

[fading portal]

me too

i just transalted it in english cuz i thought no one understood french here

Allah i3awn and tysm again @queen root

[queen root]

np

[fading portal]

.close

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[round belfry]

i cant manage to find F.

[scenic ravine]

Hmm?

$F(x) = \int_{1}^{x} (1+ \frac{1}{2t^2}) e^{-t^2}$

[twin meteor]

What a wonderful world !

[vocal sleet]

@round belfry Has your question been resolved?

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[vague vessel]

Help idk what to do

[gentle sleet]

is it supposed to be $\int\frac{x^2}{\qty(4-x^2)^\frac32}$ or $\int\frac{x^2}{\qty(4-x^2)^3}$

[twin meteor]

Sepdron

[vague vessel]

3/2

[gentle sleet]

oh didn't see the sqrt

[gentle sleet]

use exponent rules, it'll be $(cos\theta)^{2\cdot\frac32}$ in the denominator

[twin meteor]

Sepdron

[vague vessel]

Ok now I have tan(theta)-d(theta)

[gentle sleet]

can you show the process?
it should be (sec²(theta)-1)d(theta)

[vague vessel]

Yes I got the answer now

Thanks

[gentle sleet]

oh nice, np

[vocal sleet]

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[sturdy jungle]

Hiya, have i done this correct? :) Sorry if it's a bit confusing, was running out of space in my notebook.

[frozen bobcat]

that works, yes

[digital brook]

Its correct , another way you could solve it is to first multiply both sides by 3

[sturdy jungle]

thanks :D

.close

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[vague vessel]

How am I supposed to move the e^-xcos(2x) on the right to the left? Wouldn’t that remove it

[wary hull]

Subtract it

[vague vessel]

Doesn’t that remove it

[frozen bobcat]

there's probably a sign mistake somewhere

[wary hull]

From?

Subtract on both sides

[lilac pebble]

what is the symbol that looks like a colon?

[halcyon ice]

Not nessecarily
sometimes when trying to do this trick it just cancels out so you have to try to repeat it a different way

You got here by ibp right?

[vague vessel]

Multiply

Nah I probably did do a sign mistake this is what I did

[halcyon ice]

Try letting e^-x be u instead of dv

itll make the last part a -(the integral) instead of a (plus the integral) so youll be able to solve for it without cancelling it out

Btw fyi your du is wrong it should be -2sinx in the first part and 2cos2x in the second part

that isnt a sign mistake but still itll mess up the final answer

[vague vessel]

Oh yeah i was doing the integral of u instead by accident

I’m cooked

[halcyon ice]

It happens dw

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[terse berry]

Hi I have a question. I am stuck in an exercise because I know the answer logically but I don’t know if I should solve it in another way

[reef agate]

,rccw

[terse berry]

I am having trouble with the second question

The b)

[reef agate]

D'abord, vous devais utiliser $()$ comme $1^2 + \qty(-\sqrt 3)^2 = r^2$

[twin meteor]

King Leo

[terse berry]

Ensuite ?

[reef agate]

Quelle est le radius correcte?

[terse berry]

I found 2

[reef agate]

✅

Its asking for the tangent line perpendicular to the given tangent line

@terse berry

Did you simply find the other tangent line?

[terse berry]

So I solved the first thing which is finding the tangent line that was shown of the point P in the circle but then I just got confused about the rest and got stuck about how I’m supposed to find this perpendicular line. Is it just geometry because when I look at the circle I feel like I can say what the line is ?

[reef agate]

Perpendicular lines have negative reciprocal slopes

$$y_1 = m_1x + b_1$$
$$y_2 = m_2x + b_2$$
$y_1 \perp y_2$ if $m_1 = - \frac 1{m_2}$ and $m_2 = -\frac 1{m_1}$

[terse berry]

Ohhhh

Ok I think I got it

[twin meteor]

King Leo

[terse berry]

I also have another question how can I find the points of the curve of my tangent line when it is vertical ?

[reef agate]

Honestly, the only way ik how to explain it is with parametrics

But ig its just $\lim_{x \to c} \qty[\dv{y}{x}]_{x = c} = \pm \infty$, assuming $y$ is continuous at $x = c$

[twin meteor]

King Leo

[terse berry]

For the horizontal I did this

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[gray fox]

i stared but now im stuck

[gray fox]

here is my work so far

[wary hull]

Good

Now note that your sum is $\frac12\qty(1-\frac13+\frac12-\frac14+\frac13-\frac15...)$

[twin meteor]

mathisfun

[wary hull]

So what terms cancel out?

Hint: rearrange the terms in the sum

[gray fox]

how did you get this?

[wary hull]

I just wrote each term out

Like plugging in n=1, n=2...etc

Remember that $\sum_{n=1}^{\infty}a_n=\lim_{n\to\infty}\sum_{k=1}^n a_k$

[twin meteor]

mathisfun

[atomic jasper]

(telescoping series)

[wary hull]

Yes yes

Better to explain concept though

[atomic jasper]

no need to if they already know it, just cant recognise it

[gray fox]

okay so the only thing that would stay is 1 right?

[wary hull]

Not only one

There is one more term that is not subtracted

What is the first negative term in sequence

[gray fox]

1/3? but there is a positive 1/3 too so wouldnt it cancle?

[wary hull]

Well what are the positive terms I listed

[gray fox]

1/2 then 1, 1/2, 1/3

[wary hull]

Ok, so does 1/2 get cancelled

[gray fox]

no i guess it wouldnt

[gray fox]

so would i have 1/2(1+1/2)?

[wary hull]

Yep

[gray fox]

so is it .75?

[wary hull]

Check it

[gray fox]

ya it was right

.close

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[desert estuary]

is 1/tanx restrictions (pi * n)/2?

[simple mason]

If you are talking about 1/tan(x) specifically then yes
But if you are talking about cot(x) then it is only n*pi

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[grim sail]

How i get these wrong

<@&286206848099549185>

Im short on time

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[trail mesa]

this is correct

perhaps they are looking for 0, 0, 1, 2 tho

[grim sail]

Yea ur right

Das so dum

But ty

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[viscid bridge]

Guys, why is the answer k = 8 instead of k=>8?

[last agate]

[last agate]

you've solved for all values of k for which gf(x) >= 39

but you also want the "min TP" to 39 and not something else

for example, if you set k = 9, then your min TP wouldn't be 39 but rather something > 39

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[viscid bridge]

Yes

[viscid bridge]

sorry for the late respond

[hybrid flicker]

the range of the function is gf(x) >= 39

so your minimum, 5k-1, is not just bigger or equal to 39

since the range of the function contains 39, 5k-1 = 39

[viscid bridge]

I dont really get it

[hybrid flicker]

<@&268886789983436800> ad

[viscid bridge]

I feel like i get it but im not sure if i really get it

[hybrid flicker]

let me try to put it another way

you have a function gf

you're given that the range is [39,infinity)

[viscid bridge]

Yes

[hybrid flicker]

you found that the minimum also equals 5k-1

[viscid bridge]

Okk

[hybrid flicker]

but you're given the range here, so you already knew the minimum was 39

you found two expressions for the same value

they must be equal

[viscid bridge]

Yes

Ohh

[hybrid flicker]

39 = 5k-1

[viscid bridge]

Thats why it is =

Thankuuu sm