help-36

[lime summit]

Look at the part where there is blue and green text, there is n

You see it

?

In the end of that line they multiply by n

[thin urchin]

Ohhhhh

I seeeee

It s cuz i havent simplifed

By n/n^2

It gives n right

[lime summit]

The solution the does lim (n) × lim (the other term)
So you get infinity×3 wich is infinity

[lime summit]

Yes

[thin urchin]

Okii and when i factorise i always tell myself the sign by the the thing in front of my brackets but so is that wrong

Should i simplify

Just in case

[tranquil pine]

the thing infront your brackets?

[final saddle]

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[shy nacelle]

Prove that, for every natural number n, there are n distinct natural numbers a1, a2,...,an such that a1^2+a2^2+...+an^2 is a perfect square.

[shy nacelle]

I did the base case and inductive hypothesis, but I am stuck on the inductive step

[gritty zodiac]

Show it

[shy nacelle]

Here it is

[gritty zodiac]

Hmm is d necessairly a perfect square

Sure you assumed you can have a perfect square from k squares

But that shouldnt necessairly mean that when you take k+1 the first k should make a perfect square

[shy nacelle]

Why not?

[gritty zodiac]

Well for starters, you assume that there exist k squares such that they will add up to a perfect square, not for every k squares they will make a perfect square

I wont lie i dont remember the solution but id suggest trying the route of making the sum of k+1 squares a sum of two different sums whise total number of...numbers, fk English aint my native tongue, add up to k+1

Id suggest

Make your c squared

A sum of two or more squares

Which are perfect squares

And both can be achieved with less than k sqares

Which is what your hypothesis allows

Hmm no actually i think thats the wrong way

Actually nah

You can reduce the sum of k+1 squares into a sum of k or less than k squares by having some that make a perfect sqaure

Then they will make a perfect square based on your hypothesis

[shy nacelle]

Wait so was my original a_1^2+...+a_k^2= c^2 right?

[gritty zodiac]

I wouldnt immediately write that c is a perfect square

First id say in your hypothesis that k is >=2

Then at the end id ad

Lemme look at your paper

D squared and a_k+1 squared ste distinct squares etc and we can find a perfect square for k=2

It would go Like that but i dont rly Like how its built...

[gritty zodiac]

They dont have to make a perfect square just cuz its a sum of squares

[shy nacelle]

Right

[gritty zodiac]

And you should say beforehand that you selected a1 a2...ak so that they make a perfect square, not heres k of them so they make a perfect square

Cuz not all a1 ....ak will make a perfect square

But you can select them

Heres how id do it

Select a1 and a2 which make a perfect square

Which you can do because of your hypothesis

Lets call their perfect square a b

Then select a3 a4...ak + 1 so that b plus all of those ( which is k natural numbers total) make a perfect square

Itd be over then except im not sure we can state Here that for one square there are other squares so they add up to a perfect square

[shy nacelle]

Just so that I can understand, we can make this step because there are less than ak squares

[gritty zodiac]

There should be ak squares total which is ok, the only issue is can you Pick exact number of squares for a specific square so that they all add up to a perfect square

[shy nacelle]

I see

[gritty zodiac]

All you know is that for some number k you can take k specific distinct natural numbers and their squares will add up to a perfect square

[shy nacelle]

I see, so what can we do from here?

[final saddle]

@shy nacelle Has your question been resolved?

[shy nacelle]

Well there is a hint that I received from the textbook if that helps. Use Induction. The inductive hypothesis will be a_1^2+⋯+a_k^2=b^2. Then, consider a_1^2+⋯+ak^2+b^2. This sum is not guaranteed to be a perfect square, but a related sum is. It uses one more idea: That 3^2+4^2=5^2.

[gritty zodiac]

Sorry net died

Still need help?

[shy nacelle]

Yes please

[gritty zodiac]

I think you can just select b squared + a_3 squared plus ...+ a_k squared such that b squared is a sum of a_1 squared plus a_2 squared and b plus a_3 plus ... a_k are a perfect square

[shy nacelle]

Why?

[gritty zodiac]

You have a sum of k squares at most in each case

But all together are a sum of k+1 squares

[shy nacelle]

But then how does this tie to the k+1 case (I have to go to pick someone up so I won't be able to repond, but I'll come back ASAP)

[gritty zodiac]

Cuz you have k+1 numbers in total

The trick was to just write one square as sum of 2 swuares

Thus reducing the problem into one that your hypothesis covers

[final saddle]

@shy nacelle Has your question been resolved?

[shy nacelle]

No

I have returned

[shy nacelle]

For the last part of the sentence do you you mean b squared is a sum of a_1 squared plus a_2 squared and b squared plus a_3 squared plus ... a_k sqaured are a perfect square?

[final saddle]

@shy nacelle Has your question been resolved?

[gritty zodiac]

Yes

[final saddle]

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[forest rose]

do I just jump right into L'Hopital's rule from here?

\lim_{x\to5^-} \frac{\sqrt{25-x^2}}{x-5}

[forest rose]

,align
\lim_{x\to5^-} \frac{\sqrt{25-x^2}}{x-5}

[soft zealot]

Ratatosk

[forest rose]

I can't think of any way to get rid of the sqrt, but i mean, it stays even after l'hopital and it's undefined afterward, so idk

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[wooden gorge]

heyyyyy... im back

[wooden gorge]

i need help with just the first part i thought this was math not science

i know the height, its 6

i think

but a...

freaking A

question a is so c

con

confusing

if someone wants to help ill send a piece of my hair /j

[ornate solstice]

@wooden gorge For Question A you just plug in the values given for volume and base into the equation given which is v = 1/3(b)(h)

[ornate solstice]

Then you solve for h

[wooden gorge]

h

huh

!!!

[true oak]

@wooden gorge just subtitue the values you have for the given variabkes

[wooden gorge]

im very bad at volume so i have no idea what any of this or that means

i only know to divide the volume and the base

for the second one i think

?

for question b

yeah that one

[true oak]

the question tells you the equation you need to use

[wooden gorge]

mhm

1/3=bh

right?

[true oak]

B

[wooden gorge]

uh huh

[true oak]

you know the volume

so you subtitute the variable

[wooden gorge]

the 1/3?

[true oak]

you also know the height

no

[wooden gorge]

oh

v

?

or wait

[wooden gorge]

OH

[ornate solstice]

volume is given as 216

[wooden gorge]

mhm

yes those

[true oak]

so you put the numbers where the letters are

[wooden gorge]

huh

36 = b

?

is that what youre saying

[true oak]

yes

[wooden gorge]

ohh

[true oak]

are you new to variables?

[wooden gorge]

not new, just terrible

im a 9th grader and my 8th grade teacher kind was bad at his job

[true oak]

It's ok, just understand that the letter is just a placeholder essentially

[wooden gorge]

oooh

[true oak]

once you know what the letter is, you can put in the number value

[wooden gorge]

writing this down rn

[true oak]

ok ok, can you show the equation you have now?

[wooden gorge]

so it should be
216 = 1/3 (36) 6

if im not wrong (which i probably am)

[true oak]

It's ok to be wrong

but you don't know the height of the shape

so that would be your unkown or x value

[wooden gorge]

ohh

so
216 = 1/3 (36) x

?

[true oak]

yes

then you isolate the x value

on one of the side

then it's just calculator work :)

[wooden gorge]

sorry to ask but how do i do that

[true oak]

first, get rid of the fraction

[wooden gorge]

and how would it be calculator work

[true oak]

then

[wooden gorge]

alright

so like
just get rid of it completely?

[true oak]

no

[wooden gorge]

oh

[true oak]

do you know how to multiply 1/3 to = 1?

[wooden gorge]

uhm

by 3?

[true oak]

yes, but since yo

multiply 1 side

you have to it to the other side

[wooden gorge]

alr

how would that be written out in an equation form

[wooden gorge]

ooo desmos

i know that app

[true oak]

Yes so to get rid of a number

you can to do the oposite of what it is doing

[wooden gorge]

so i write dhat equation down? (Minus the numbers slashed out ofc)

[true oak]

yeah, then you get rid of the 36 and voila

[wooden gorge]

216-36?

[true oak]

no, since it's multiplication

the oposite would be division

think of the pais of operators (+ and -) (x+ ÷) (^2 + sqrt)

[final saddle]

@wooden gorge Has your question been resolved?

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[dense crescent]

first image gives original function and its derivative; second image shows what i am stuck on.
i'm aware that second derivative is just taking the derivative of the first derivative, but how would i apply it to the point (1,2)? do i apply it to the point after getting the second derivative or do i apply it on the first derivative before solving for second derivative?

[dense crescent]

looking through it again i would presume applying it after as it would equal to zero if i plugged them into the first derivative

i think

now i cannot find the derivative, i get stuck when i do the quotient rule

[vital crag]

!show

[final saddle]

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

[dense crescent]

the one on the very bottom is

where i am stuck at atm

[final saddle]

@dense crescent Has your question been resolved?

[jaunty portal]

mmm in the line just before the last one, why do you have dy/dx twice in the end? 🤔

(2xy)' = 2 y + 2 x y'

[dense crescent]

do i not multiply the numerator by derivative of denominator making it (6x^2 - y^2) * 2 y' ?

oh

wait

oooh

[jaunty portal]

after that, it's ok to factor dy/dx

substitute what you have for dy/dx, and that should be it 🤔

[dense crescent]

could you elaborate? sorry its just not clicking in my head right now

[jaunty portal]

I mean after you do that correction, it's ok to factor dy/dx as you did. Remember that you already have an expression for dy/dx, so just plug that in

[dense crescent]

wait oh my god its all coming together

[final saddle]

@dense crescent Has your question been resolved?

[dense crescent]

.close

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[fathom stag]

I'm a bit confused with how I should proceed

[fathom stag]

how do I rewrite it in that form

[cosmic warren]

Have u tried plugging in (x-3) for f yet

[fathom stag]

No

I don't think we are meant to use a graphing calculator for this

[cosmic warren]

no like

It says y = 1/x

That's ur f

f is y

x is x-3

If u just plug it in then the rest is algebra

Like adding fractions with common denominatora and shizz

[fathom stag]

OOOOHHH

[fathom stag]

So it'd be 1/x-3

Thanks

.close

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[finite blaze]

25a⁵×5a⁴÷a³

[tranquil pine]

ok what can you rewrite 25 as

it looks like a square doesnt it

[finite blaze]

5⁶

[tired walrus]

no, 25 isn't equal to 5^6.

[finite blaze]

5a⁶

[tired walrus]

do you mean "I think the answer is 5a^6. Can someone tell me if I'm right?"

[finite blaze]

No, I dont know how to calculate

[tranquil pine]

ok well first what can 25 be rewritten to

[finite blaze]

5²

[tranquil pine]

Yes

so you have (5^2 a^5 x 5a^4)/a^3

do you think you can combine on the numerator

[finite blaze]

5a¹⁴/a³?

[tranquil pine]

no

first what is 5^2 times 5

[finite blaze]

25²?

Or 5¹⁰?

[tranquil pine]

No

5^2 is just 5 times 5 remember

What is 5 times 5 times 5

written as 5 to an exponent

[finite blaze]

5³

[tranquil pine]

Correct

so you have 5^3 times a^5 times a^4 on the numerator

Can you combine a^5 and a^4

[finite blaze]

a⁹

[tranquil pine]

Yes

now you have

5^3 times a^9 over a^3

do you think you can get rid of the a^3 at the bottom to make the entire thing not a fraction

[finite blaze]

5³×a⁶

[tranquil pine]

Yep

Thats it

Nice work

[finite blaze]

Thank you for helping me

[tranquil pine]

no problem

[finite blaze]

.close

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[sacred inlet]

each day john leaves home at the same time and cycles to same route to work

on monday, his avg spd was 18km/h and he arrives 4mins late
on tuesday, his avg spd was 24km/h and he arrives 6 minutes early

time taken to travel on monday is t minutes, form an eqn in minutes for the time taken to travel to work on tuesday

[sacred inlet]

why is ans t-10?

dont understand

[tired walrus]

well

imagine this

let's say john's work day begins at 10:00

on monday, at what time did he arrive?

and on tuesday, at what time did he arrive?

[sacred inlet]

monday he would arrive at 10:06
tuesday he would arrive at 9:54

[tired walrus]

10:06?

he's four minutes late.

[sacred inlet]

oh my bad

10:04

[tired walrus]

yeah

so you see how the time difference between these is 10 minutes right

[sacred inlet]

ahhhhh

i see

yup that makes sense to me

thanks 😄

.close

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[mint orbit]

I need some help with this guy. I've been assured that it's possible to derive a clean condition on z such that the inequality holds (it's a line segment in the complex plane): $$\qty| 1 - \frac{ z^2 \pm z\sqrt{z^2-4} }{ 2 } | < 1$$

[soft zealot]

jan Niku

[mint orbit]

IDK if there is some kind of obvious algebra trick im missing

[charred chasm]

[mint orbit]

trivially $4 > z^2 \pm z \sqrt{ z^2 - 4} > 0$

[soft zealot]

jan Niku

[charred chasm]

i was working on this last night

[mint orbit]

but what the actual hell do you do from here

[gaunt solstice]

this is just the pq formula?

or almost

nvm

[mint orbit]

its challenging

[magic sparrow]

yes, it is possible

[mint orbit]

i said ah fuck it ill just plot it

but matlab hates it

i cant get the line to show up for any mesh size i pick

[charred chasm]

so my understanding is that it's stable if the discriminant is >= 0

is that correct?

[mint orbit]

hmm? how do you get that

[charred chasm]

Well this is supposed to be a solution to $x'' + \lambda^2 x = 0$

[soft zealot]

TooManyCooks

[charred chasm]

My guess was that you don't want complex roots

because you can have exponentially increasing functions

[mint orbit]

complex is fine? are we talking about two different polynomials

Youd have to explain more which discriminant you are talking about

[mint orbit]

hrm?

[charred chasm]

So you did the substitution right? The one with linear recurrence relation?

[mint orbit]

yea

[charred chasm]

That's how you got the equation you have

[mint orbit]

i mean i can track through it or send a pic, if youd like

[charred chasm]

the quadratic

[mint orbit]

correct

[charred chasm]

No that's fine, i got it on my notes

I was thinking of what you meant by "stability" last night

Since the problem was adamant about certain values < 1

[mint orbit]

well the roots of the polynomial depend on a parameter

since the roots pop out into the z^n they clearly have to be < 1

so you require |z(b)| < 1 and solve for b

[charred chasm]

wait how is that clear

all you have is a quadratic equation in z

[mint orbit]

because we aren't interested in solutions that just explode to infinity

[charred chasm]

z?

or x

[mint orbit]

right

[charred chasm]

or does it even matter

[mint orbit]

x?

i dont have x's

[charred chasm]

right. so x will blow up to infinity if lambda is complex

[mint orbit]

the original recurrence is $A^n_k = z^n$

[soft zealot]

jan Niku

[mint orbit]

n is some time step

it blows up if you dont constrain the roots

[charred chasm]

can you derive that condition strictly from the quadratic equation or no?

[mint orbit]

here

[charred chasm]

because that doesn't seem to follow for me

[mint orbit]

does this help?

ive used z twice

[charred chasm]

|z|< 1 came out of nowhere

[mint orbit]

well for the same number

[mint orbit]

it just doesnt seem to me like all of this additional context is helping

the problemt can be entirely divorced from everything here

[charred chasm]

yeah nvm

[mint orbit]

up to the inequality

so i dont know how much of this drudging is going to assist in moving forward unless im really really misguided

[mint orbit]

[uncut mulch]

Do you want to do a poll, ask a friend or eliminate 50% of the options

[mint orbit]

i just want help lol

[mint orbit]

I'm starting to think now that this is wrong, isnt it

this doesnt make sense because b isnt real

i have to do it some other way

[final saddle]

@mint orbit Has your question been resolved?

[final saddle]

@mint orbit Has your question been resolved?

[mint orbit]

No

[final saddle]

@mint orbit Has your question been resolved?

[final saddle]

@mint orbit Has your question been resolved?

[mint orbit]

https://tenor.com/view/hmm-sure-catbruh-gif-21408636

[dry light]

sin^2(x) + cos^2(x) = 6

[stiff belfry]

@mint orbit could you remind me of the equation from which this inequality arises

[mint orbit]

this is probably all of the context you could desire

[stiff belfry]

yes i remember

lemme take another crack

[mint orbit]

my teacher gave me a hint but im on a mental verge so i may be misremembering

that we really want to find the boundary where |z| = 1

which is something sometning z^n = 1 or something?

so you have radius 1?

i cant remember

[charred chasm]

isn't it 1 if $\lambda \Delta t = 0$?

[soft zealot]

TooManyCooks

[mint orbit]

this wont be true

[charred chasm]

i'm just looking at the quadratic you had last night

[mint orbit]

i think it corresponds to either a string with no tension or a non-numerical scheme

[charred chasm]

yeah i think so too

[stiff belfry]

from where did you get this

when I do the problem I wind up with the condition $$\left|z + \sqrt{z^2 + 1}\right| < 2$$

[soft zealot]

rat says

[mint orbit]

work here

how illegible is it

you end up with $z^2 + (b^2-2)z+1=0$

[soft zealot]

jan Niku

[vital crag]

handwriting's fine. the photo's just bad

lighting probably

[mint orbit]

,w z^2 + (b^2-2)z + 1 = 0

[vital crag]

and it's small

[soft zealot]

Results provided by WolframAlpha

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

Requested by janniku

[charred chasm]

b is lambda delta t, right?

[mint orbit]

it is

[stiff belfry]

oh I see

[mint orbit]

i think this is wrong, unless youre using some kind of nice substitution

[stiff belfry]

yes I am using a different substitution

all i did was collect the coefficient of $z$ into a single variable

[soft zealot]

rat says

[mint orbit]

you mean the b^2-2

[stiff belfry]

yes

i had hoped this would make resolving the sqrt easier

it has not

there will be some substitution trick that makes this doable, I suspect

[mint orbit]

maybe we can do this by just solving for the region or whatever my teacher was saying

aka just solving $\qty| b^2 \pm b\sqrt{b^2-4} | = 2$

[soft zealot]

jan Niku

[mint orbit]

although idk how to solve this either

[bleak granite]

would squaring both sides not work? sorry if I'm like just not braining here

[mint orbit]

idk if im being really honest im not even sure what nice way exists to express the magnitude of a complex expression like that

[mint orbit]

.

[bleak granite]

yeah ok idk half of those words, think this math may be a bit out of my sphere of math knowledge sorry man

[mint orbit]

i mean im not sure how to talk about the size of the expression absent some trick

some method might exist but idk it

[fierce orchid]

take cases on whether b^2>=4 or b^2<4

Assuming b is real

[mint orbit]

b is complex

[fierce orchid]

It isn’t, nevermind

Yeah I realzied

[mint orbit]

infact i know the solution

it should be entirely imaginary

[fierce orchid]

I read the rest of the convo

[mint orbit]

but

i appreciate it

[fierce orchid]

Oh nice

[mint orbit]

i know the answer i just cant get there

[fierce orchid]

Is it an easy trick or very hard

Oh I thought yiu knew the steps

[mint orbit]

idk im having mental issues today my teacher told me a good hint but i couldnt gather it

i dont know the steps

he said somethign about how we only want the boundary

so you can solve this as an equality with one

and then its z^n = 1

which means something?

because we know its e^itheta

i have no idea what he was talking about

[fierce orchid]

How z^n=1?

[mint orbit]

im not sure

[fierce orchid]

Hmm

[mint orbit]

i was hoping itd make sense to someone else

[fierce orchid]

x^2-z^2x+z^2=0

If the solutions to above is x, then we have |1-x|=1

And only z^2 is involved so I think we can just let z^2=a

But how do we use this

[vital crag]

@mint orbit not that I think a numerical analysis class would cover this, but have you learned Rouche's theorem
https://en.m.wikipedia.org/wiki/Rouché's_theorem

[mint orbit]

i have not, ill take a look

im gonna close this because i think im going to bed for several hours

but i appreciate all the thoughts yall

.close

[final saddle]

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[final saddle]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[cerulean igloo]

I was given a hint that im supposed to factorise this and conclude something from there, i arrived at f(x)=(x^2+ax/2)^2 +(3-a^2/4)x^2 +bx +1
not sure how to take it from there. what i can see is that the left term is always positive

but im not sure whether making the (3-a^2)... part >=0 will suffice

because its possible perhaps that it is negative but less in magnitude

[brittle cave]

wait

[rustic wedge]

?

[brittle cave]

if all the terms are positive

and perfect squares are positive

isn't there no maximum

[rustic wedge]

*bx may be negative btw

[brittle cave]

one sec

[rustic wedge]

Hmm might be a little overboard but u can try calculus, if you know how to find maxima and minima

[brittle cave]

second derivative test 💀

am i dumb or is there no maximum though

[cerulean igloo]

i want its derivative to have 3 real roots, and the points where f' is 0 needs to be above x axis. but doing that is harder than it sounds especially with these variables, i got that a^2>8 for f' to have 3 real roots

[cosmic warren]

Could factor as x^2(quadratic) + (another quadratic) maybe, and completing the square

[rustic wedge]

To f(x), there isn't, but a^2 + b^2 should have a maxima

[cerulean igloo]

yes that is what i had attempted and arrived at the expression i sent above

[brittle cave]

why

isn't the only restriction that f(x)>=0

can't a and b increase forever

[cerulean igloo]

no

[brittle cave]

which would be a higher a^2+b^2 value

[cerulean igloo]

hop into desmos and put random values

[brittle cave]

why not

[cerulean igloo]

youll see the graph can go below

the x axis

[brittle cave]

just keep them positive?

[cerulean igloo]

there has to be a certain set of values of a and b

for which f(x) is never negative

[rustic wedge]

Sometimes certain values give complex roots too

[cosmic warren]

well it looks annoying ngl but you could try completing the square on the 2nd quadratic also

[brittle cave]

please elaborate

don't get

wait am dumb

[rustic wedge]

You could set f(x) = product of two monic quadratics and then solve for their coefficients, and find the roots, and then set them all > 0. But that does seem like a metric crap ton of work

@cerulean igloo ?

[cosmic warren]

By which I mean (3-a^2/4)(x + b/(2(3 - a^2/4))^2 +(1 - b^2/(4(3 - a^2/4)^2)) . Then on the constant term we want 4(3 - a^2/4)^2>= b^2, then 4(9 - 6a^2/4 + a^4/16), then 36 >= 6a^2 + b^2 -a^4/4, 12>=a^2, maybe write as 0>= -a^4 + 24a^2 +(b^2 - 144) and solve for a^2 in terms of b^2

this is messy af but it's going somewhere maybe

[final saddle]

@cerulean igloo Has your question been resolved?

[cosmic warren]

a^2 = 24 +- sqrt(24^2 + 4(4b^2 - 144))/2 then can set discriminant equal to 0 i think?

Wait don't need that

[final saddle]

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[zinc bridge]

im confused on how i can find the area of bdc

[zinc bridge]

can anyone help me

or am i stupid

yea im stupid

i just didnt see it

.close

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[errant sigil]

Quick question

[errant sigil]

The triple scalar product

u · (v x w)

Its a scalar, right? First you compute the cross product, then the dot?

Oh, derp, I guess that's right in the name. There's a typo in my teacher's notes which threw me a bit

[tired walrus]

yes

[errant sigil]

Thanks!

.close

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[sour sentinel]

need help with an integral

[sour sentinel]

$\int\frac{dx}{x^2+ax+1}$

[soft zealot]

skittle

[sour sentinel]

anyone?

<@&286206848099549185>

[fresh quest]

hi

[sour sentinel]

hi'

[fresh quest]

1s

let me try

ok

[sour sentinel]

yes?

[fresh quest]

(x+1/2a)^2 +1-(a^2)/4

[sour sentinel]

$(x+1/2a)^2+1-a^2/4$

[soft zealot]

skittle

[sour sentinel]

thats wolfram's way isnt it

[fresh quest]

1 s me send photo

[sour sentinel]

i dont quite understand

[fresh quest]

last line without ^2

at t

and sgrt

[sour sentinel]

i dont understand the handwriting

can you try latex?

[fresh quest]

$(x+\frac{a}{2})^{2} = x^{2} + ax+ \frac{a^{2}}{4}$

it u understand?

[sour sentinel]

yes

so the denominator is equal to: $\left(x+\frac{1}{2a}\right)^2-\frac{a^2}{4}+1$

[soft zealot]

skittle

[fresh quest]

ok in ur integral u dont have $\frac{a^{2}}{4}$

[soft zealot]

binibini

[fresh quest]

yes

[sour sentinel]

now what

[fresh quest]

mistake

a in nominator

[sour sentinel]

?

[soft zealot]

binibini

[sour sentinel]

so the denominator is equal to: $\left(x+\frac{a}{2}\right)^2-\frac{a^2}{4}+1$

[soft zealot]

skittle

[fresh quest]

yup

[sour sentinel]

now what

[fresh quest]

then u take $(x+\frac{a}{2})=t*\sqrt{\frac{-a^{2}}{4] +1)}$

how make sqrt?

[sour sentinel]

\sqrt{}

what is t

[soft zealot]

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

[fresh quest]

wait

[sour sentinel]

$\left(x+\frac{a}{2}\right)^2=\sqrt{-\frac{a^2}{4}+1}$

something like that?

[soft zealot]

skittle

[fresh quest]

(x+\frac{a}{2})=t* \sqrt{\frac{-a^{2}}{4} +1}$

me back

[sour sentinel]

what is t???

[fresh quest]

1 s

[fresh quest]

this is a substitution method

[sour sentinel]

but what did you substitute

[fresh quest]

so you will replace thex with some "t".

[sour sentinel]

i dont really understand

[fresh quest]

$(x+\frac{a}{2})=t* \sqrt{\frac{-a^{2}}{4} +1}$

[soft zealot]

binibini

[fresh quest]

where

[sour sentinel]

this thing

what exactly did u do

[fresh quest]

u have it in denominator of ur integral, and we want in denominator $x^{2}+1$

[soft zealot]

binibini

[sour sentinel]

ok

so what did u do

??

[fresh quest]

1 s

[sour sentinel]

?

[fresh quest]

do you understand more now?

[sour sentinel]

can you just expain what do i do after that step?

[sour sentinel]

nope

[fresh quest]

ok

u need $z^{2} +1$

[soft zealot]

binibini

[fresh quest]

$t^{2} +4$

[soft zealot]

binibini

[sour sentinel]

WHAT THE HELL IS T

[fresh quest]

some example

to solve

what x will you substitute to get $x^{2} +1$

[soft zealot]

binibini

[sour sentinel]

in what expresssion

[fresh quest]

u have intergral of $\frac{dt}{t^{2} +4}$

[soft zealot]

binibini

[sour sentinel]

from when exactly

i dont understand u at all

sorry

[fresh quest]

this is just an example

[fresh quest]

can you solve such an integral?

[sour sentinel]

i think so

[fresh quest]

show ur work if u solve

[fresh quest]

this is your new unknown that you substitute for x

[fresh quest]

yup , good

[fresh quest]

u see , u took 4 before t, in ur task u can too

$(\frac{-a^{2}}{4} +1)$

[soft zealot]

binibini

[fresh quest]

u can take it before x

[fresh quest]

like u did here

[sour sentinel]

factor it out?

[fresh quest]

yup

[sour sentinel]

how?

[fresh quest]

it=P (for easier writing )

[sour sentinel]

this is what i get

[fresh quest]

yup

[fresh quest]

out it , before ur intergral

[sour sentinel]

??

[fresh quest]

the same as here

but ur $t=x+\frac{2}{a}$

[soft zealot]

binibini

[sour sentinel]

i rewrote it

[fresh quest]

ok

without ^2 ,is ur t

[sour sentinel]

how do u know

[fresh quest]

look, here u had $\frac{t^{2}}{4}$

[soft zealot]

binibini

[fresh quest]

sqrt of it

is $\frac{t}{2}$

[fresh quest]

t/2 is ur "u" here

[soft zealot]

binibini

[sour sentinel]

ok

??

<@&286206848099549185>

we dont seem to progress

[sour sentinel]

this is the question

[sour sentinel]

and this is the factored denomionator

[final saddle]

@sour sentinel Has your question been resolved?

[final saddle]

@sour sentinel Has your question been resolved?

[final saddle]

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[hasty cipher]

Im not sure where I am going wrong

if anyone can help guide me

[tiny gorge]

there's a way way way easier way to approach this

do you know the formula for u x v involving sin(theta) and the formula for <u,v> involving cos(theta)?

[hasty cipher]

yes

[tiny gorge]

i recommend using those here

[hasty cipher]

hmm, okay. Is there a reason im failing at this method?

I assume I simply cant add then subtract the equation in blue to try and solve

[dusky fjord]

oh hell nah

[final saddle]

@hasty cipher Has your question been resolved?

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[tidal granite]

hello could someone please help me graph this?

[tidal granite]

im unsure

would i graph y=-7 as a point?

and then x<1 as a line from 1 to infinity?

<@&286206848099549185>

[final saddle]

@tidal granite Has your question been resolved?

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[tidal granite]

plz <@&286206848099549185>

[tidal granite]

https://cdn.discordapp.com/attachments/908076393198419988/1152774051933204480/image.png

i dont know how to graph this

[marsh temple]

!15m

[final saddle]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[warm ether]

isnt it just the graph of y=-7 on the domain of x<1

[tidal granite]

wym

how would i draw that

<@&286206848099549185>

someone plz 😭

[tranquil pine]

yes

what is it?

[tidal granite]

plz

this

[tranquil pine]

yes

[tidal granite]

https://cdn.discordapp.com/attachments/908076393198419988/1152774051933204480/image.png

[tranquil pine]

graph it

[tidal granite]

ive tried

i dont know how to graph this

how do i graph smth like that in general

idk how to graph (x,-7)

[tranquil pine]

ah its a set rooster

[tidal granite]

or x<1

[tranquil pine]

set rooster bro

[tidal granite]

whats that

[tranquil pine]

set rooster method in set theory

[tidal granite]

ive never heard of that

[tranquil pine]

oops

i meant roster

[tidal granite]

how do i do that

[tranquil pine]

well for such element of x tending to y you have to include its functions and its graphical representation including its limits, asymptotes, domain, range, and y - intercepts

[tidal granite]

what should i start with

i have no info on this how am i supposed to graph with only this information

[tranquil pine]

just put x = -7

and color the area behind x = 1

[tidal granite]

im doing it online

its an online graphing tool for hw

but ill try

[tidal granite]

why x=-7?

the -7 is a y value

[tranquil pine]

um

make equation

[tidal granite]

its not working

idk what to do

cajn someone plz help

[tidal granite]

how do i make an equation out of this

<@&286206848099549185>

.close

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[wary parcel]

this is the hint from the book

[final saddle]

@wary parcel Has your question been resolved?

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[livid meteor]

So question

Would this be the proper way to write out my equation? My biggest concern is 4yt as to whether or not that is the correct way to input it

[final saddle]

@livid meteor Has your question been resolved?

[steep sequoia]

I think you should write y instead of y(t) as your question specifies.

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[forest topaz]

Anyone know what the line in the middle means??

[tired walrus]

@forest topaz do you mean |

[forest topaz]

yea

[tired walrus]

if so: that means "divides"

or "is a factor of"

[forest topaz]

So b is a factor of a?

a/b = remainder 0?

[tired walrus]

no, other way around. a|b means a is a factor of b

[forest topaz]

thanks

[forest topaz]

Do you know

What that symbol with three dots mean?

Like the inverse three dots

[tired walrus]

??

show me the symbol in context please

[forest topaz]

Also universal statements can be disproved by counterexample right?

But that cannot be proved by example

[tired walrus]

i said in context,

not a tiny image that contains only the symbol itself and nothing else...

right now it looks like a slightly fucked up letter y

[forest topaz]

=

[tired walrus]

can i have the entire page, please

uncrop

i THINK it might be $\because$, meaning ``because''

[soft zealot]

Ann

[forest topaz]

exactly

Yeah that symbol

And that turned upside down 180 degrees is therefore right?

[tired walrus]

yes

i have to say im not really happy with how it took 2 tries before you could properly show me what you were asking about

[forest topaz]

Sorry my first answer was definitely vague, I only had a very low quality screenshot of my professor's lecture notes so its my bad

[tired walrus]

well you could've sent the entire thing still

[forest topaz]

At the time I though that the image might be good enough but I apologize if it wasn't clear

[final saddle]

@forest topaz Has your question been resolved?

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[wary parcel]

can someone please explain this solution? Especially the underlined part

[hidden kernel]

it says that, you can use 2005x1 instead of 2005x2005 cause every hxw rectangle in the 2005x2005 can be put in 2005x1 as 1xw

[wary parcel]

yup

[hidden kernel]

The question now is about how many rectangles which contain the middle square over the total amount of rectangles

[wary parcel]

right

[hidden kernel]

i don't understand why 2005 + 2C2005

oh wait i got it

it is 2005*2005/2

...

i mean

2005 squares + 2005*2004/2

so 2005*2006/2

for rectangles 1 - 1002, the amount of rectangles which include middle square is 1003

so for 1004 - 2005 too

[wary parcel]

wait a second. why can these be the same in 2005 ways, and different in 2c2005 ways?

[hidden kernel]

what do yo mean by 2005 ways?

[wary parcel]

look at the underlined sentence in the solution

[hidden kernel]

you mean when transitioning from 2005x2005 to 2005x1?

[wary parcel]

no

[hidden kernel]

so for 1003 there are 2005 rectangles

oh

cause for 1x1 there are 2005 ways to put in a line

and for othere there are 2C2005

[wary parcel]

why 2c2005 for others

oh wait choose a beginning and an end in a line?

[hidden kernel]

right

[wary parcel]

okay

[hidden kernel]

so 1003*2004 + 2005

or, as it says 1003^2

so 1003^2/(2C2006)

[wary parcel]

1003 squares to choose from the left side, 1003 squares to choose from the right side? (that forms a beginning and an end

[hidden kernel]

is that clear?

(sorry, ban inet)

[hidden kernel]

right

[wary parcel]

yup, thank you so much:))

[hidden kernel]

no problem

[wary parcel]

.close

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[spark nova]

Rational Equation

[spark nova]

I'm confused on where should I start multiplying

[hidden kernel]

multiply both side by 2x(x-2)

[spark nova]

so x-2 x 2x then 2x x 2x?

or am i wrong?

[hidden kernel]

you have to get $3x*2x - 4x(x-2) = 5(2x)(x-2)$

[soft zealot]

TimK

[spark nova]

is this correct?

[hidden kernel]

no

[spark nova]

where did i do wrong?

should i find the lcd?

[hidden kernel]

Use this

[spark nova]

i have to multiply 5x twice right?

[hidden kernel]

right

[spark nova]

10x^2-20x?

or 10x(x−2)?

[hidden kernel]

both

[spark nova]

so i could just use any of them?

[hidden kernel]

right

[spark nova]

my answer is 6x - 4x - 8x = 10x^2 - 20x

but i still have to find the value of x

im still confused cuz this is how my teacher taught us

[hidden kernel]

ok, what is $3x*2x - 4x(x-2) = 5(2x)(x-2)$ simplified?

[soft zealot]

TimK

[hidden kernel]

without any brackets?

isn't it $6x^2 - 4x^2 + 8x = 10x^2-20x$?

[soft zealot]

TimK

[hidden kernel]

simplify: $28x = 8x^2$

[soft zealot]

TimK

[hidden kernel]

then $x(8x-28) = 0$

[soft zealot]

TimK

[hidden kernel]

@spark nova

[spark nova]

honestly i dont know what to do cuz i cant keep up with the lesson so i have no idea if it should be simplified or not, all i know is that we have to find the value of x

[hidden kernel]

look at the steps i sent

[spark nova]

how did it became 4x^2 + 8x? why did the sign change?

[hidden kernel]

-(-) = +

[spark nova]

oh right

[spark nova]

how did it came to this?

[hidden kernel]

-28x from both sides

[spark nova]

where did the -28x come from?

[hidden kernel]

it is to cancel out 28x

[spark nova]

ohh okay i get it now

[final saddle]

@spark nova Has your question been resolved?

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[green quail]

very dumb question but why is the domain of 1/|x|-x (-∞,0)

[green quail]

OH

nvm

im braindead

.close

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[undone owl]

Let ABCD be a parallelogram and M a point inside it, if angle MBA is congruent to angle MDA, demonstrate that angle MAB is congruent to angle MCB.

[final saddle]

@undone owl Has your question been resolved?

[undone owl]

<@&286206848099549185>

anyone

<@&286206848099549185>

<@&286206848099549185>

[undone owl]

<@&286206848099549185>

[undone owl]

.close

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[undone owl]

.reopen

[final saddle]

✅

[rugged zodiac]

Ha

Yess

Stef

@undone owl haaa

[undone owl]

can you help me?

@rugged zodiac ?

[rugged zodiac]

With what @undone owl

[undone owl]

with the problem

[rugged zodiac]

Iam the helpless unhelpful helper

Which prob

[undone owl]

Let ABCD be a parallelogram and M a point inside it, if angle MBA is congruent to angle MDA, demonstrate that angle MAB is congruent to angle MCB.

[rugged zodiac]

No

I still haven't read that chapter

It's 11 one iam on 4

[undone owl]

What?

[rugged zodiac]

YouTube is there or chat gpt

[rugged zodiac]

Iam a highschool student

[undone owl]

I tried both

[rugged zodiac]

Oh

[undone owl]

9th?

[rugged zodiac]

<@&268886789983436800>

[split acorn]

what the

[rugged zodiac]

<@&268886789983436800>

[rugged zodiac]

Yeas

[split acorn]

why are you pinging the mods

[undone owl]

ok

[rugged zodiac]

He is struggling

Help him

[split acorn]

so?

[sonic umbra]

?

[agile nacelle]

What's going on here?

[rugged zodiac]

How did you know

[split acorn]

bro just decided to call mods to help

[undone owl]

I guessd smh?

[split acorn]

with this punchline

[rugged zodiac]

Pak?

[agile nacelle]

It doesn't look like any moderator action is needed here. The OP was a bit trigger happy with the helper ping, but we usually don't punish that.

[undone owl]

trigger happy? what does it mean?

[agile nacelle]

You're only supposed to ping helpers once.

[undone owl]

ok thx ill remember in the future

[agile nacelle]

As for your question, the first step is to draw a diagram. Do you have one?

[undone owl]

yes

[agile nacelle]

You should show it then; that will make it easier for helpers to see if they have any ideas.

[undone owl]

i need to log on my phone

a sec

[agile nacelle]

(I've drawn my own diagram now, and I don't immediately get any ideas, sorry. Interesting property if it holds.)

[agile nacelle]

(You'll probably want to move M a bit out of the way so it doesn't look like it lies on AC. But, as I said, I don't really have any ideas).

[undone owl]

Thank you anyway

[agile nacelle]

Well, I suppose you can't just move M willy-nilly. Perhaps if you make the side lengths of ABCD more different, you can get M to fall more clearly off the diagonal.

[undone owl]

ok lemme try

[agile nacelle]

Idea: If you select a coordinate system where the slopes of AB and AD are minus each other, then "angle MBA = angle MDA" simply means that the slopes of MB and MD are also minus each other, and "angle MAB = angle MCB" means that the slopes of MA and MC are minus each other.

and "angle MAB = angle MCB" means that the slopes of MA and MC are minus each other.
No, that is nonsense -- it doesn't hold :-(

Yes it does!

(Tropo doddering around and arguing with himself).

Since we're now talking about slopes rather than angles, the assumptions and conclusion all stay the same if we scale the y-direction independently of the x-direction. So without loss of generality we can assume that the slopes of AB and AD are ±1, in other words that ABCD is a rectangle!

[final saddle]

@undone owl Has your question been resolved?

[agile nacelle]

And we know we can assume ABCD is a rectangle, we can pick a much nicer coordinate system for it: one where the coordinates are A(0,0), B(0,h), C(w,h), D(w,0).

In this setting "angle MBA = angle MDA" becomes x/(h-y) = y/(w-x) and we need to prove that is the same condition as x/y = (h-y)/(w-x). But that is clearly the case.

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[tranquil pine]

Need help to prove Q2b with induction

[tranquil pine]

.close

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[spare light]

Hi, I was doing an exercise and I found myself stuck at the second step of the exercise:

Ex:
U0=0 and for any natural number n, Un+1=1/2-Un

1. calculate U1, U2, U3,U4
   I found U1=0.5, U2=1/1.5, U3=0.75, U4=0.8

2. Deduce a conjecture of the explicit form of Un as a function of n.

I've tried several things here, but nothing seems good to me.

Could you help me and explain to me how to do it please?

[spare light]

here is the exercise (written in French) if it can help you

<@&286206848099549185>

[winter sleet]

1/2 2/3 3/4 4/5

[spare light]

?

[winter sleet]

n/(n+1)

0.5=1/2 1/1.5=2/3 0.75=3/4 etc

[spare light]

Oh okay thanks

didn't see that

[winter sleet]

don't forget to .close if there are no more questions

[spare light]

no thanks

.close

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[subtle lava]

.close

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