help-26

[wise salmon]

Yea

[night halo]

The lim of x as x approached 2

Would be 2

[wise salmon]

This case it would be negative infi?

[night halo]

X is approaching negative infinity in this case

[wise salmon]

And y would go to infinite?

[night halo]

Yea

[wise salmon]

ur so fucking smart I love you

[night halo]

😭

I didn’t get the help I need but glad I could help u ig

[wise salmon]

So is option 1 true or

[night halo]

Not necessarily

It could be true

But the question is asking MUST

Which is only option 3

Now that this is done I’m closing this to get another case to get my help 😭

[wise salmon]

dis the options

[night halo]

Option D

III must be true

And it’s the only one which has 3

[wise salmon]

wat grade r u in

[twin heath]

3 can't ve true if p(x) < 0 at x> 2

[night halo]

Started undergrad just now 😭

[night halo]

It says p(x) < 0 when x>=2

[twin heath]

Infinity is greater than 2

Oh shit that's negative infinity mb

[night halo]

No wait

[wise salmon]

can u be my friend

[night halo]

Am I tripping

[wise salmon]

🥺

[night halo]

I might have autism I can’t lie

[twin heath]

No it's negative infinity you were right

[night halo]

Ok, hold on can u help me with mine sir 😭

Did u do calc 2

[night halo]

Bruh

I need help with these serphic

[wise salmon]

🫤

He

Pre uni math

[night halo]

This is calc 2 😭

I took it in hs

It’s AP calc BC

I just have autism and forgot how to do them

[twin heath]

Oh lord those are disgusting

[night halo]

Indeed

[wise salmon]

Am taking that next yr

Should I

Or go ab first

[night halo]

R u a 10th grader 😭

[wise salmon]

Am a freshman rn

🥲

[night halo]

Take BC, p sure they cover AB and BC if u take BC

Atleast at my school they did

[twin heath]

Give me a bit to try these out yeah

[night halo]

Appreciate it brother 🫡

[topaz sinew]

@night halo Has your question been resolved?

[night halo]

@twin heath u get anywhere?

[twin heath]

Kind of but I got no papet

Paper

[night halo]

can u type it out? or is it too hard

cause i been struggling on it since last night. I tried photomathing and shit but I can't figure it out

[twin heath]

I know latex but gimme a sec to verify if I'm cooking or if the kitchen has burnt down

[night halo]

lmfao ight

[twin heath]

Verified: I cooked

So

Pull the first question back up

So

First of all, we split it into two integrals, right?

[twin heath]

Yeah I've got it my method works

[night halo]

yuh

we add the 2 integrals or wtvr

[twin heath]

Do you notice anything about the coefficient of the integral on the left?

How is it related to the exponent

[night halo]

oh shoot

derivative?

[twin heath]

Close to it

I'm gonna be fr I forgot how to do u sub here so my method is a little jank but

We have something like

$(g'(x)-7)e^{g(x)}$

[thorny flame]

Serphic

[night halo]

ok

[twin heath]

So we get smth like

Let me cook rq

[night halo]

🔥

[twin heath]

$\int_{5a^3+7a}^{5b^3+7a} (u'-7)e^u du$

[thorny flame]

Serphic

[twin heath]

It looks worse except for one thing

If we just distribute the 7 out, we get a pretty obvious derivative

[night halo]

I see

[twin heath]

So we get at least one free integral.. except

[night halo]

im sorry for dumping this ass problem on u bruh, this hurt my brain trying it earlier

[twin heath]

No it's okay I'm so close to being done I just need to change bounds properly

[night halo]

ight

can you tag me once u figure it out, im gonna be doing something so I can't see the screen. Ill hear the ping though

also @twin heath I figured out the second question so no need to work on that

WAIT @twin heath i think i mightve figured it out, gimme a sec. ill put it on clean paper and send so u can double check

[twin heath]

I'm sorry my brain fried on the bounds 😔

[night halo]

Check this

We didn’t need to do all that

It’s a bit messy

But factor and then u sub

Leads us to just having to integrate e^u

Which gets us what I circled

ignore the graph and stuff cause thats a different question

[twin heath]

Okay so roughly what I did

Lemme just SS desmos

[wise salmon]

I have to do that for calc 2

[night halo]

nah its much easier in hs, i just didn't get credit for it in college so I have to take it again but its harder in college

so im cooked

Concepts are the same tho

[wise salmon]

Wat ur major

[night halo]

Aerospace Eng lmfao

@twin heath add me back, imma need your ass next week when I run into another pothole like this

ill just dm u next time

imma close this for now

.close

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[dapper sparrow]

Hi im learning error in taylors theorem

[dapper sparrow]

What happened to the (n+1) above the f(c)

am I missing something?

is it becasue f'''(e^c) is just e^c?

[humble grotto]

I really dont know honestly

[vernal matrix]

(well, f(x) = e^x, and f'''(c) is the e^c, but basically )

[topaz sinew]

@dapper sparrow Has your question been resolved?

[dapper sparrow]

@vernal matrix ok bet thabk you lol

Just so I understand, I basicly use this formula

Set c = a and x for lower and upper bounds

and thats it

[vernal matrix]

Upper and lower bounds for the remainder term as per here?

[dapper sparrow]

Yes like

in my homework it states :Use the error formula in Taylor’s Theorem to find an upper bound for the error |some func|. Compute the actual error.

So that would be just taking a or x for c (whichever gives me the higher value)

[vernal matrix]

In this case, they can use the endpoints because that third derivative, f'''(x) = e^x is an increasing function, so because c is between a and b, then f'''(c) = e^c is (strictly) less than e^b, as c is (strictly) less than b

[dapper sparrow]

Ok wait here is another problem where the taylor series is not centered around 0:

[vernal matrix]

If you have the corresponding function for | f^{(n + 1)} | being either increasing or decreasing, you can estimate the error above with | f^{(n + 1)}(b) | in the increasing case, and | f^{(n + 1)}(a) | in the decreasing case, but in general it may not be either of those

[dapper sparrow]

Oh

so i cant pick either a or x?

Becuase c is in between x and a, so to get the max bound I should pick whatever gives me the highest f(c) function no?

[vernal matrix]

Whichever gets you the highest value for $\abs{f^{(n + 1)}}$, yes, but that may not be either $a$ nor $x$

[thorny flame]

@vernal matrix

[dapper sparrow]

oh, like it can be a number in between?

[vernal matrix]

Yep, it potentially could be, there are probably some examples for those (but I can't do examples to save my life )

[dapper sparrow]

Lmao makes sense

But in the vast majority I should be ok picking either or?

[topaz sinew]

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[vernal matrix]

In many cases you may get given, I think you should be (but I hold no liability if you find one where it isn't there's bound to be one you get given where it's not )

.reopen

[topaz sinew]

✅

[raw owl]

Can I have help with like, all of this

[vernal matrix]

!occupied

[topaz sinew]

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).

[dapper sparrow]

lmaoo

well

thanks for the heads up :D

One last clarifying question

If I have a problem such as : Find the third Taylor polynomial

Is it asking 3rd order or just for a trinomial?

[vernal matrix]

third order, do the Taylor polynomial up to (x - a)^3

[dapper sparrow]

@vernal matrix ok the reason im asking is because the next part says "Find the fourth Taylor polynomial"

And it comes out to 0

so do I just show i did the work and put 0 :D

[vernal matrix]

Pretty much, I'd think do you have the question?

[dapper sparrow]

Yes, not with me though it was in lectures

I can send it in like 15 when I get back

[vernal matrix]

Sure thing, but yeah, if, say, the 4th derivative evaluates to zero, then I'd think you're to leave it as you find it

[dapper sparrow]

Ok just wanted to make sure because it also has calculate error parts which really cuts down the work

[vernal matrix]

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[topaz sinew]

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[hard terrace]

Can someone please help me out understanding how to do this problem?

[hard terrace]

from what I understand p is a ratinal function and q is an integer 0 or higher

p-q is a decimal value

so that would mean this is true right?

[vernal matrix]

p is any rational number ("a fraction"), q can be any integer, positive, negative or zero

"take any rational number, you can always find some integer which has distance from said rational being at most 1"

[hard terrace]

I thought Z was only things greater than 0

oh nvm

that's N

mb

[vernal matrix]

Yep

[hard terrace]

Wait

this is a dumb question

but if q is negative and is within a ||

does that still mean the negative sign can change

it can right

the end value

is still positive

[vernal matrix]

It could potentially, remember you're finding |p - q|, so if p - q ends up negative, you can e.g. multiply that by -1 to make it positive

[hard terrace]

oh wait i get it now kinda

but 1 shouldn't be greater than p-q each time right

what happens if p is 7/3 and q is 0

or are they trying to say when 1 is greater than p-q those conditions are true

[vernal matrix]

What they're saying is that I give you a p, and then you have to find some q such that |p - q| ends up being at most 1

So, you gave me p = 7/3, you can't choose q = 0, because that leaves you with |p - q| = 7/3 > 1

[hard terrace]

so that means it's false then

[vernal matrix]

But, I can then give you q = 2

Then |p - q| = 1/3, which is at most 1

[vernal matrix]

It's not, it's true

Give me another choice of rational number

[hard terrace]

1/3 and 2

[vernal matrix]

If you gave me p = 1/3, I'll take q = 0 (and |p - q| = 1/3)
if you give me p = 2, I take q = 2 (and |p - q| = 0)

[hard terrace]

I thought the whole point was to check all possible cases

[vernal matrix]

nooo, it's basically like giving you a game

[hard terrace]

so if I can find a value that is true in this case then it'll be true?

[vernal matrix]

The statement is effectively saying "pick any rational number - can you find some integer such that the distance between them is 1?"

[hard terrace]

oh ok

that's the part I'm a little

[vernal matrix]

If, instead, they said $\forall p \in \bQ, \forall q\in \bZ, \abs{p - q} \leq 1$, you'd be right here - that's saying "the distance between all pairs of rationals and integers is at most 1" (which is not true)

[hard terrace]

confused about

[thorny flame]

@vernal matrix

[vernal matrix]

But basically, this is what you're meant to do: for any rational I give you, you need to find some corresponding integer such that the distance between them is at most 1

[hard terrace]

wait I didn't search up what this meant ∃

I think that's why I was confused

I didn't realize it meant there exists

ok now I understand what you're saying

[vernal matrix]

Yeaa, $\exists$ is "there exists..."

[hard terrace]

if there is a possiblilty that it exists then this case is true

[thorny flame]

@vernal matrix

[vernal matrix]

And yep wanna give me more rational numbers, and make me choose q's?

[hard terrace]

sure but I think I understand it if you want I can give you more

[vernal matrix]

As long as you have an idea of showing it, then it's all good

[hard terrace]

thanks for all the help

[topaz sinew]

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[covert current]

when I'm solving "prove that x = x " problems, am I allowed to manipulate both sides of the equation

[covert current]

or like as an easy example

uhhh

wait what is an easy proof

[verbal shadow]

there's no problem with manipulating both from a logical standpoint

but it's not usually the intention of the person asking the question

it's usually expected you make one into the other

[covert current]

uhhhh

can you check my work and see if it's too much

[verbal shadow]

sure?

[verbal shadow]

i suck at combos, lol

[covert current]

it's just binomial theorem

what i did at the end was

divide by x on both sides

is this a proper proof

[rose sierra]

If you're trying to prove a=b, you can't use a=b anywhere in your proof

[covert current]

i didn't did i

[rose sierra]

idk

[covert current]

I started with the expension of (1+x)^m+n

[rose sierra]

What does this notation mean

looks like binomial coefficients from your work but I haven't seen it written this way before

[covert current]

which part of the notation

yes they are combinations

[rose sierra]

Is $^{m+n}C_1 = \binom{m+n}{1}$?

[thorny flame]

Ari

[covert current]

yes

@knotty finch

<@&286206848099549185>

sorry wrog pimg lol

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[topaz sinew]

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[onyx anvil]

Express the area of an equilateral triangle as a function of the length of a side.

[vernal matrix]

Any ideas of how to start?

[onyx anvil]

well im just not sure if the answer is 1/2xBxH or a= sqrt3/4 s^2

i just dont know if either of them are right to be honest

[vernal matrix]

The second one (how did you get that?)

(the first one is in general, but involves both one side of a triangle and its height, you want to express it only in terms of the equilateral triangle's sides)

[onyx anvil]

well in my notes from lecture those both were mentioned but I was just confused on which one was the correct one

im very confused in this homework i need to go to office hours

A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 16 in by 28 in
by cutting out equal squares of side x at each corner and then folding up the sides. Express the volume V of the
box as a function of x.

this one im also confusedo n

[vernal matrix]

Most likely, they said that they used the first to get the second, I would think? (the base is s, the height you work out to be sqrt{3}s/2)

As for the second one, it might help to draw a picture of the situation, and what you end up getting...

[onyx anvil]

okay im working on it right now, im trying to figure it out

[onyx anvil]

ok i got it, i just didnt understand how that could make a box

v(x)=(28-2x)(16-2x)(x)

[vernal matrix]

Yep, there you go sometimes easier to visualise with pictures

[onyx anvil]

What do all members of the family of linear functions f(x) = c + x have in common? Sketch several members of
the family.

for this one I just said they ahve slope parallel lines and the y intercept in common

and im working on the sketches

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[ivory sorrel]

prove that $\lim_{x\to 0} \frac{\sqrt{1+x}-1}{x}$ is 1/2

[thorny flame]

Veni, vidi, perii

[ivory sorrel]

so

$\frac{\sqrt{1+x}-1}{x} - \frac{1}{2} <\varepsilon$

[thorny flame]

Veni, vidi, perii

[ivory sorrel]

or $\frac{\sqrt{1+x}-2-x}{2x}<\varepsilon$

[thorny flame]

Veni, vidi, perii

[ivory sorrel]

hmm

I think I'd be better off starting from $|x-1|<\delta$

[thorny flame]

Veni, vidi, perii

[junior fable]

do we know at this point that limit evaluation is a ring homomorphism?

[ivory sorrel]

what

this is calc 1

[junior fable]

pfff lmao

ok my b

that $\lim_{x\to c}\left(f(x)g(x)\right)=\left(\lim_{x\to c}f(x)\right)\left(\lim_{x\to c}g(x)\right)$ and $\lim_{x\to c}\left(f(x)+g(x)\right)=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)$

[shrewd horizon]

tbf that is a fairly basic property, just uh, "a ring homomorphism" sure is a way of phrasing it

[ivory sorrel]

I can't use the binomial approxaimation, can I

[thorny flame]

Flip

[ivory sorrel]

I know this

BUT

I want to do it from first principals

[junior fable]

cool

[ivory sorrel]

Or is that a bad idea?

[junior fable]

I know I just think it's funny

[thorny remnant]

shouldnt it be $\frac{2\sqrt{1+x}-2-x}{2x}<\varepsilon$

[thorny flame]

convergence

[ivory sorrel]

oops yesh

so , I now square both sides

or ywk

yeah this is a horrible idea

provides no insighst whatsoever

Should probably just use basic properties of limiys

.close

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[ivory sorrel]

thanks

[topaz sinew]

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[steep wren]

im trying to find out what the domain of this graph would be with the transformations but i dont get how to with this problem

[steep wren]

<@&286206848099549185>

i can always reexplain by the way

[twin heath]

what was the original function?

[steep wren]

here let me send a picture

[twin heath]

that is... a "fun" looking graph

one moment

[steep wren]

lol okay

i do have the answer key but i just didnt understand how they got there either

[twin heath]

okay so

pardon me splitting the function up into three for this (desmos doesn't have notation for one piecewise function)

but

[steep wren]

yeah thats fine

[twin heath]

b realistically doesnt matter since it's in the middle, right?

we care about the two endpoints

[steep wren]

maaybe let me look at rhis

i dont get it😔

[twin heath]

hold on if i show you the graph it might be

less shitty

[steep wren]

okok

i just dont get what the translations i need to do are if that makes sense

[twin heath]

yeah no that's valid

[steep wren]

thats the original graph right

[twin heath]

yep

[steep wren]

okok

so how do i follow the translations the equation is telling me

[twin heath]

so

two ways you could do this

right

[steep wren]

alr

[twin heath]

you basically want to make it so that (x+1)/2 -1 stops at the exact last value in the range

very shitty way of explaining it but

for exampel

the last portion of the graph stops at

x = 5

[steep wren]

ya ya

[twin heath]

the function is only defined up to when x is equal to 5

so

our last point in the domain is

when

$\frac{x+1}{2}-1 = 5$

man.

DGSJKAOSDKJGIS

fuck latex

[steep wren]

😭😭

[thorny flame]

Serphic

[twin heath]

there we go

[trim ermine]

how do you even write that up

[steep wren]

do i just solve for x here

[twin heath]

yep

[twin heath]

wdym

oh there's like

[trim ermine]

command like this

[twin heath]

there's a bot (the TeXit bot)

[trim ermine]

I would rather write in ona paper and send it

[twin heath]

counterpoint: im too lazy to write

[steep wren]

is there an easier way to do this like for the other problems

not the fraction ones

i just follow the transformations and do it for the x values

[twin heath]

im gonna say i dont think so

because

what this is basically doing is averaging the value of x between x and 1, and the subtracting one

it's.... not very fun to imagine what that would even look like

[steep wren]

LOL okay thanks for ur help

[twin heath]

i mean

wait

there is a way

[steep wren]

okkay

[twin heath]

if you convert it to

a(x/2 - 1/2)

you instead have something like

"horizontally stretch it by a factor of two, and then shift to the right by a half"

wait

no you wanna do it the other way around hold on 💀

[steep wren]

i tried to do that and i just didntt get the same answer as the worksheet

[twin heath]

thats because for that, x itself has to be shifted

like

(x+5)/2 means shift 5 to the left and then stretch by two

[steep wren]

ohh

[twin heath]

but x/2 + 5 doesnt have a interpretable meaning

[steep wren]

i was confused because of the -1 at the end if the thing

[twin heath]

yeah you want to try and move everything into the same coefficient as x

and then you should be able to interpret it the same way

[steep wren]

okay okay

[twin heath]

this shit gave me a panick attack like 5 months ago too dw 😭

[steep wren]

LMAO thanks sm for ur help

[twin heath]

np!

[steep wren]

.close

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[rain sluice]

In N=B+Ae^(kt) is B or A the initial value

[woeful drift]

!original

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[rain sluice]

for exponential growth questions

for N=Ae^(kt) A is the initial value, so then what is B

[opal vault]

if the "initial value" you're talking about the value at t = 0, the initial value is A + B

[rain sluice]

in this question u gotta use dN/dt = k(N-B) to see the growth equation then use that, why is B 10 if the initial is A + B

[topaz sinew]

@rain sluice Has your question been resolved?

[vestal sigil]

because B is the final mass

[rain sluice]

the question says the initial is 10g, and the answer says that B = 10, so ur saying B is both the initial and final mass??

[vestal sigil]

your N is the transformed mass, logically at the start it's 0

and at the end it is the initial

the transformed mass grows exponentially from 0 to 10

[rain sluice]

wdym, these are the graphs

[vestal sigil]

k<0

[rain sluice]

why does that matter

[vestal sigil]

its written in your solution e^(-kt) with k positive

so, the constant being multiplied by t must be negative

otherwise, the graphs will go to infinity

[vestal sigil]

ohhhh, 3/5 not 3/2

you made a mistake in desmos lol

you confused me haha

[rain sluice]

truee so the red is the amount tranformated and the blue is the amount remaining?

[vestal sigil]

exactly

[rain sluice]

ok ty

.close

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[snow gyro]

Guys

[snow gyro]

How will I graph this manually?

<@&286206848099549185>

[neon iron]

by taking values

lol

ofc values >=0

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[hexed root]

anyone can help?

some numbers are Fibonacci sequence and some are harmonic sequence

[rigid geyser]

Try changing the fractions to have the same numerator. For example, 3/2 = 6/4

[hexed root]

okay done, then???

[rigid geyser]

Did you find the patterns? Which one are you stuck on?

[hexed root]

1😅

[rigid geyser]

So what are all the terms?

[hexed root]

terms?

[rigid geyser]

I mean, have you found the missing part of the sequence?

[hexed root]

not yet

[rigid geyser]

So we have 6/4, 6/7, _, 6/13. We need to figure out the _. We can sort of guess that it will be the pattern 6/_.

The denominators are 4, 7, _, 13

[hexed root]

should be 10

[rigid geyser]

Okay, now do you know the whole sequence?

[hexed root]

so 6/10

[rigid geyser]

Yes.

[hexed root]

nice

so that's the answer?

[rigid geyser]

You can just add them up.

[hexed root]

oh okay thanks a lot

.'close

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[quartz igloo]

😎😎😎😎

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[vapid dagger]

Could somebody explain how I would prove the sequence will always be arithmetic? Maybe give me a hint or soemthing 😭

[arctic briar]

wait why are u doing 10-4 and 18-10

does the qu tell u to do that?

[vapid dagger]

It just said find the first 3 terms

And it gives sum

[neon iron]

are you an IB kid, car?

[vapid dagger]

So I’ll just minus the sum from previous

[vapid dagger]

Unfortunately yes

[neon iron]

to show it's arithmetic you must show there's a constant difference

[vapid dagger]

OkY

Lemme brainstorm for a sec

[arctic briar]

oh right it was sum

[neon iron]

in other words, $S_{n} - S_{n-1} = d$ for all $n$.

[thorny flame]

45

[vapid dagger]

What😭

OHH

[neon iron]

see it now?

[vapid dagger]

Yes

Do I do it like 2 times to prove it

Wait nvm

[arctic briar]

u need to prove sadly

so u need to show for all n it works

[neon iron]

all you have to show is that $S_n - S_{n-1}$ is a CONSTANT; if it is a constant value, this implies the sequence is arithmetic.

[thorny flame]

45

[opal vault]

well not Sn but un

[vapid dagger]

Okay

Lemme try it

[neon iron]

S_1 = u_1, S_2 = u_1 + u_1 + d

S_2 - S_1 = d

this holds for all pairs

and S_n is given in the question

[opal vault]

read what you just wrote

[arctic briar]

[neon iron]

what?

[opal vault]

S2 - S1 = (u1+u1+d)-u1?

is that equal to d

[neon iron]

Oh

[opal vault]

I highly doubt something as Sn = n^2 + 3n is arithmetic

[trim ermine]

You just have to prove that they have common difference ig

[opal vault]

It's Sn-Sn-1 that is arithmetic

[neon iron]

that's a good point

[opal vault]

not Sn

let un = Sn - S(n-1)

[neon iron]

yeah you gotta derive u_n from S_n then

[opal vault]

show that un - u(n-1) = d

[neon iron]

my bad @vapid dagger you gotta derive u_n from S_n

[vapid dagger]

I think I flopped

[neon iron]

which is a bit more work

no you didn't

I made a mistake

[arctic briar]

u can also prove Sn is 2S(n-1)+d

[vapid dagger]

What does this mean😭

[neon iron]

find S_1 and S_2 by plugging in n = 1 and n = 2 into the equation

[arctic briar]

un is the value at the nth place in the sequence

[neon iron]

then use the fact that S_1 = u_1 and S_2 = 2u_1 + d

[arctic briar]

mb

[vapid dagger]

Wait why is s2 2u1+d

[neon iron]

Because $S_2 = u_1 + u_2$

[thorny flame]

45

[neon iron]

and $u_2 = u_1 + d$

[thorny flame]

45

[arctic briar]

*u1 is 1st term, u2 is 2nd term etc

[vapid dagger]

OHH

[arctic briar]

mm just remember ppl might not use the same letters as u guys to depict maths, like i think over here we use t, not u

[vapid dagger]

I see

Okay ima try the question

[trim ermine]

And we use a

[arctic briar]

exactly why car was confused

[cold mortar]

@vapid dagger simply do Sn-Sn-1 and it will give you Tn that is the nth term of the sequence and then do Tn-Tn-1 that will be the common difference d and this must be constant then it is an AP sequence

[vapid dagger]

Is S or T the sum?

[cold mortar]

S is the sum and Tn denotes nth term

[vapid dagger]

Oh okay

I feel like I should be getting this but I can’t sequences and series makes me wanna cry😭😭😭

[cold mortar]

its the easiest if u do logically

[vapid dagger]

I’m low on logic☹️

[cold mortar]

here Sn-Sn-1 that you have calculated gives Tn i.e nth term of sequence not the d

[vapid dagger]

Oh okay

[cold mortar]

so Tn=2n+2 this is the nth term so d=Tn-Tn-1 that diff. b/w n th term and n-1th term

d=Tn-Tn-1=2n+2-(2(n-1)+2))=2 i.e constant so its an AP

[vapid dagger]

OHHH

I FINALLY GET IT

THANK YOU SO MUCH YALL😭😭

[arctic briar]

WWWWWW

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[vapid dagger]

I’m lost again ☹️

[vapid dagger]

How do you do the sum of the sequence if n=0??

Wont the formula not work

Also it’s formatted weirdly with n being the power 😭

It sgeopmetric right

[neon iron]

,rccw

[neon iron]

yes its geometric series

[vapid dagger]

🤌

[golden blade]

Do you have a question on a or b?

[vapid dagger]

A, and probably b because I can’t do it without a

[golden blade]

10^-0 is what

[vapid dagger]

1

[golden blade]

so whats the problem with n = 0

[vapid dagger]

Oh so do I just add 1 to my answer

[golden blade]

,, 10^{-n} = \frac{1}{10^n} = \left ( \frac{1}{10} \right )^n

[thorny flame]

bacc

[vapid dagger]

OHH

[golden blade]

The first term is 1

and r = 1/10

[vapid dagger]

that felt so obvious 😭 :(

[golden blade]

like a car crash

[vapid dagger]

Wouldn’t it be 1/10

[golden blade]

n = 0

[vapid dagger]

Zeroth is 1

Oh

[golden blade]

yea

[vapid dagger]

Yay it worked out

Thank you

[golden blade]

For b)

,, \log ( \left ( \frac{1}{10} \right )^n ) = n \cdot \log ( \frac{1}{10} )

[thorny flame]

bacc

[golden blade]

It's a gaussian sum

[vapid dagger]

Is n=x??

[golden blade]

no

,, -\log(10) \cdot \sum_{n=1}^x n = -91

[thorny flame]

bacc

[golden blade]

use the gaussian sum formula

[vapid dagger]

I have no idea what that is

[golden blade]

you are about to

[vapid dagger]

This is all I got

Mahbe that’s called the Gaussian sum??? Idk

[golden blade]

So here

x(x+1)/2

[vapid dagger]

Thank you

I just gotta check if I learnt this before

MYbe I forgot

[golden blade]

maybe you forgor

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[mellow arrow]

is that you in your pfp

anyways

!status

[topaz sinew]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[quiet flax]

2

we know that LHS = (1 + cos(2x))/2 + (1+cos(2x+2pi/3)/2+(1+cos(2x-2pi/3)/2

but what now

we can factor a 1/2

maybe we can try using cos(x+y) = cosxcosy - sinx siny

[mellow arrow]

use this on the latter two

[quiet flax]

hm

= 1/2(3+cos(2x)-cos(2x))

ahhh yees

[mellow arrow]

as simple as that

[quiet flax]

now we can use cos(2x) = 1-tan^x/1+tan^2x

alriht

thankks!

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[mellow arrow]

wut

[quiet flax]

tan^2 x

[mellow arrow]

why

[quiet flax]

or cos^2x - sin^2 x whichever

they are the same identity

cos(2x) = cos^2 x - sin^2 x

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[novel yoke]

Sorry, i know i've been asking the same question again and again
But now, i have the time to wait

[novel yoke]

Please check my answers and please give me hints..

What did i do wrong..

The answer is supposed to be -4

[cunning kayak]

yo

can send ques

[novel yoke]

oh

wait

[cunning kayak]

what have you done in denominator?

[novel yoke]

oh, someone told me i should change sin^2 2x -1, so i change it to -cos^2 2x and they tell me that i am correct

[novel yoke]

This

[cunning kayak]

BTW cos²x ≠ cosx²

[novel yoke]

Oh?..

[novel yoke]

Hmmm

[novel yoke]

What should i've done then?

[cunning kayak]

wait

[novel yoke]

I turn x to y + pi/4 so that lim y goes to 0 ..
Hm...
Mmmm
If i think about it again..
Why did i do that

[novel yoke]

nvm, i still dont know what to do 😅 ahah

[cunning kayak]

i need pen paper

wait lemme finish my dinner

[novel yoke]

oh okay

no worry

[cunning kayak]

sin²2x = 1 - cos²2x

and then you replace cos²2x with (2x)² cuz ending to 0

have they taught you abt it^

[novel yoke]

but you can only do that to sin and tan?

not cos..

cuz thats an exception

[cunning kayak]

maybe I am mistaken

i actually need pen paper

[novel yoke]

oh mm, no worry, take your time

i understand that

[cunning kayak]

do you know L Hospital?

[novel yoke]

no, not yet

😅 at the end its easier to use L hopital

dang

[cunning kayak]

yeah ok solved

so we can rewrite numerator as (π-4x)² right?

and denominator as -cos²2x

[novel yoke]

yeah

[novel yoke]

nice

[cunning kayak]

now

h = x - pi/4
which gives x = h + pi/4

correct?

[novel yoke]

yeah

[cunning kayak]

okay

now plug in values

which would give something like this

[patent cipher]

nice

vgood

[novel yoke]

😅 hmm

[cunning kayak]

yeah dw

he is my friend

was telling me 💀

[cunning kayak]

also

right?

[novel yoke]

right

[cunning kayak]

and what is cos(pi/2 + x) ?

[novel yoke]

-sinx

[cunning kayak]

and then what is

(cos(pi/2 + x))^2

[novel yoke]

-sinx^2?

[cunning kayak]

nop

it would be only sin^2 x

[novel yoke]

ohh

[cunning kayak]

(-1)^2 = 1

-ve will become +ve after squarin

[novel yoke]

ohhh

[cunning kayak]

so in denominator we have
-sin^2 h

right?

-ve was already there

[novel yoke]

yea

[cunning kayak]

sin^2 h = h^2

[novel yoke]

ohh

[cunning kayak]

yeahh

so in denominator we will have: -4h^2

[novel yoke]

yeah

so its -4

[cunning kayak]

now can u simplify numerator

[novel yoke]

cuz the numerator is 4h^2

[cunning kayak]

yup

[novel yoke]

yeah

[cunning kayak]

16h^2

[cunning kayak]

got it or have some doubt?

[novel yoke]

hmm im gonna redo it

try it again myself

[cunning kayak]

sure

i think you get confused with square?

[novel yoke]

no, maybe its still with the cos in the denominator

[cunning kayak]

wait

lets rewite

[novel yoke]

pi/2 is because pi/4 * 2 right?

[novel yoke]

oh right

[cunning kayak]

i just saw

you sent message that you got it

what went wrong

[novel yoke]

the emoji

but i dont wanna bother others by telling them again and again so

yeah

[cunning kayak]

oh yeah i remember

dw

wont happen again

i will tc

[novel yoke]

oh, okay thanks

[cunning kayak]

👍

[novel yoke]

anyway, thankyou

[cunning kayak]

[novel yoke]

i guess imma close now

.close

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[grand trellis]

is there any other way to do this other than manually counting?

[hearty turret]

you can probably get a better handle on what numbers are "good"

[grand trellis]

done that and the basic steps

[hearty turret]

what did u characterise good numbers as?

[grand trellis]

rn + (1+2+3....+m)

wait

[hearty turret]

as in you can probably get a better characterisation

[grand trellis]

yea

m is r-1

[hearty turret]

see which numbers you can/can't write as a sum of consecutive positive integers, then use that to make conjectures/prove ur conjectures

[grand trellis]

hmmm alr

the number "n" should be divisible by r-1?

no it doesnt work for all mb

wait nvm i got it

r(2m + r -1) = 2n

from here i think i can just take values of r

and we have the set too so we dont need to find limiting cases ig

tysm

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[gritty orchid]

how to expand from this to this.white paper to black paper soln

[charred pecan]

idk i'm on class rn but u can use this (wait for upload)

∗sin²𝛼+cos²𝛼=1 sin2𝛼=2sin𝛼cos𝛼 −1≤sin𝛼≤1 & −1≤cos𝛼≤1
∗ Si : α + β = 90° →cos𝛼=sin𝛽 & sin𝛼=cos𝛽
In other words : cos(90°−𝛼)=sin𝛼;sin(90°−𝛼)=cos𝛼
Ex : cos60°=sin30°=1/2

; cos30°=sin60°=√3/2

; cos90°=sin0°=0 ; cos45°=sin45°=√2/2

∗ Si : α + β = 180° →sin𝛼=sin𝛽 & cos𝛼= −cos𝛽
In other words: sin(180°−𝛼)=sin𝛼;cos(180°−𝛼)=−cos𝛼
Ex : sin150°=sin30°=1/2

; cos150°=−cos30°=−√3/2

∗sin(−𝛼)=sin𝛼;cos(−𝛼)=cos𝛼

really sorry i cannot help u more man if u don't get it by developp the factorise form

maybe tag helpers

[gritty orchid]

no i just want to know if we can change the 5h-15..... to (h-3)^2+k^2=16

[charred pecan]

i think so

[gritty orchid]

i wanna see the steps for that sorry if it sounds lazy but i genuinely cant simplify it

i was hoping to understand it by looking at someone solving it

[charred pecan]

yes ik

ok i start it

[hearty tartan]

u know that sin^2 x + cos^2 x =1 right?

[gritty orchid]

are questions like this fine here?

[gritty orchid]

yes

[charred pecan]

man i WRITE ALL OF IT

[gritty orchid]

i wanted the expansion from 5h-.... to the one provided in soln before it changes into x and y

[hearty tartan]

ok give me a min

[charred pecan]

30s for me

[hearty tartan]

i had to get my phone to click pic 🙂

[gritty orchid]

i wanted something different but this is way simpler ,so thanks for the trick

.close

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[halcyon willow]

solved it but wanna check

[halcyon willow]

if its correct

i got 2 sqrt((x+1)^3)/3 + 2sqrt(x^3)/3 +C

[vernal matrix]

,w diff 2 sqrt((x+1)^3)/3 + 2sqrt(x^3)/3

[thorny flame]

Results provided by WolframAlpha

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Requested by chartbit

[halcyon willow]

so uh am i wrong

[vernal matrix]

So much for laziness

[halcyon willow]

😭

[vernal matrix]

How did you get your answer?

[halcyon willow]

x+1 =t

dx= dt

after rationalisation

ofc

dr becomes 1

and we get sqrt x+1 and sqrt x

use x+1 =t and evaluate the other with standard formula

am i wrong?

[vernal matrix]

Seems a bit sus why x + 1 for t?

[halcyon willow]

dx= dt?

and we get sqrt of t

[vernal matrix]

From that sqrt{x^2 + 1} term?

[halcyon willow]

yea

[vernal matrix]

You can't do that x^2 + 1 is different to x + 1

[halcyon willow]

no

wait

i meant

sqrt x+1

dx=dt

[finite storm]

Where did u get the x+1 from?

[halcyon willow]

rationalise the nr

dr becomes 1

we get sqrt x+1 + sqrt of x in nr

[finite storm]

sqrt(x^2 +1) + sqrt(x), no?

[halcyon willow]

nope

sqrt x

shit

i wrote it wrong

its sqrt x+1

yeah

[finite storm]

The split into two integrals

[halcyon willow]

yeah did

[finite storm]

Long division prolly safest way to proceed

But if u want to do it the other way

[halcyon willow]

hmm i guess

id did it right

i just wanted to check ans

nvm

[finite storm]

That’s fine

Alrighty

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[lilac fern]

I NEED HELP CONVERTING PARABOLA GENERAL FORM INTO STANDARD FORM. Because when I put my answer in desmos, and the original equation of general form of that problem. It doesn't matches. I don't know what am I doing wrong.

[lilac fern]

But mathway said the answer is

(y + 2/7)^2 = 8 (1/7) (7x - 13)

And it matches in desmos, it matches the general form. And my question is WHERE DID THE (1/7) CAME FROM???

My final answer is almost the same as what should be the answer, but I'm confuse on where the 1/7 came from?

my answer is

(y + 2/7)^2 = 8(7x - 13)

Mathaway's answer, and it actually matches on desmos compared to mine is

(y + 2/7)^2 = 8 (1/7) (7x - 13)

<@&286206848099549185>

[compact zodiac]

what's the question?

[random yew]

yup whats the question

[lilac fern]

where did (1/7) came from

[random yew]

the whole exercise what does it require

[lilac fern]

this is the answer

(y + 2/7)^2 = 8 (1/7) (7x - 13)

my answer was

(y + 2/7)^2 = 8(7x - 13)

[compact zodiac]

well i see x in one side and y on the other

[random yew]

bro whats the prime equation that you need to convert

gimme a starting point i cant find where the 1/7 came from like this

[lilac fern]

Converting general form to standard form of a parabola

[compact zodiac]

and what is the equation of the parabola you are converting?

[random yew]

can we see miss parabola

[lilac fern]

So the general form starts off with 49y^2 - 392x + 28y + 732 = 0

After that you will put similar terms

[compact zodiac]

yeah that's what we needed from the start

[lilac fern]

49y^2 + 28y = 392x - 732

After that I isolate the 49 from 49y^2

49(y^2 + 28y)

which I need to divide the 49 to 28

28/49 becomes 4/7

With that, I can finally do the completing the square

49(y^2 + 4/7 +

the formula of completing the square is (n/2)^2

so 4/7 divided by 2 would become 2/7 then I squared it which became 4/49

49(y^2 + 4/7 + 4/49) = 392x - 732

now we need to also bring 4/49 to the other side

49(y^2 + 4/7 + 4/49) = 392x - 732 - 49(4/49)

I canceled it

49(y^2 + 4/7 + 4/49) = 392x - 732 - 4

49(y^2 + 4/7 + 4/49) = 392x - 728

uhm

[neon iron]

oh you are missing a y: It should be 4/7 y.

[lilac fern]

forgot about that one, but even if I add it. It wouldn't change the answer

[neon iron]

ok let us go step by step from here

[lilac fern]

49(y^2 + 4/7y + 4/49) = 392x - 728

[neon iron]

49(y+4/7)^2 = 392 x -728

(Also, it is + 4 on the right, not -4)

[random yew]

2/7

not 4/7

[neon iron]

oh thx

[lilac fern]

49(y+2/7)^2 = 392 x -72

because I was doing a perfect square, to satisfy the trinomial it'll be y+2/7

[neon iron]

yea stupid of me thx

(-728 in your last reply)

Ok, then bring the 49 to the right

[random yew]

8x-728/49

[neon iron]

(y-2/7)^2 = (392/49) x - (728/49) = 8 x - 104/7

This fits with the solution with the 1/7

[lilac fern]

49(y^2 + 4/7y + 4/49) = 392x - 728

what I did to the right side is I factored it

their lcd is 56

so

396(7x - 13)

and after that

I need to remove the 49 in

49(y+2/7)^2

So I divide both sides by 49 to cancel

49(y+2/7)^2 = 396(7x - 13)
49 49

which is how I got my answer

(y + 2/7)^2 = 8(7x - 13)

This is the standard formula I need to achieve, so that I can graph it

[lilac fern]

help

[neon iron]

hey

Let us look at the right side. It is 392 x -728 = 56 (7x -13)

then divide 56 by 49

which gives you 8/7

thats it.

[lilac fern]

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

HOLYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

[neon iron]

I know the feeling

[lilac fern]

@neon iron THANK YOU

THANK YOU SO MUCH

YOU DIDNT KNOW, HOW LONG IVE BEEN CONTEMPLATIN

THANKY OU

CAN I ADD YOU

[neon iron]

😂 you're welcome

sure

[lilac fern]

OGM

GEPGNKESGNRSGw

THANK YOU FRFR

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[unique tendon]

Can somebody explain to me what the concept of the component of a vector is?

[neon iron]

What is a vector?

I mean, to you?

[unique tendon]

Something that has both magnitude and direction?

[neon iron]

okay. that's good. Do you want to do 3 dimensions or 2 dimensions? it doesn't really matter, actually

[unique tendon]

2D

[neon iron]

okay. Do you know how vectors are added, I mean visually?

[unique tendon]

Yes

[neon iron]

okay, then imagine in 2D, two vectors which are pointing in different directions. not exactly opposite, but different directions. Call them A and B

[unique tendon]

Okay

[neon iron]

Now, you can get any other vector you want by combining A and B by multiplying them with numbers and adding these: I mean things like 5A - 2B or A+3.5B or any other combination k A + l B

[unique tendon]

Got it

[neon iron]

Okay, now say you draw a grid in your 2D space, with an x-Axis and a y-Axis.

Then you can take X to be the vector which goes one unit in positive x-direction

and Y is the vector that goes one unit in positive y-direction

Then every other point can be written as a combination k X + l Y

[unique tendon]

Understood

[neon iron]

And this vector k X + l Y is represented by (k, l).

[unique tendon]

Yes

[neon iron]

Another reason one does this is that on the grid, you have the point (k, l). And if you take the vector (k, l) and let it begin at (0,0), then the tip of the arrow will be at (k,l)

So points in 2D and vectors in 2D are kinda the same thing, but the vectors as a description make more sense when thinking about adding vectors, or multiplying them with a number.

So the components of a vector are the number of units you need to go from the origin (0,0) in x-direction (first component) and y-direction (second component), so that you get the vector.

The end.

[unique tendon]

Wait,
Wdym by "the vectors as a description make more sense when thinking about adding vectors, or multiplying them with a number."

[trim ermine]

[unique tendon]

Yes, but sometimes i am given two vectors and told to find the component of one vector along the other vector, what does that mean?

[neon iron]

Like, if you have two points (3,4) amd (2,1), then you can just add them by (3+2, 4+1) = (5,5), but what does it mean? But when you draw the vectors (3,4) and (2,1), the addition visually makes sense.

[trim ermine]

@unique tendon vectors is a topic as a whole itself, I would recommend you to watch lecture on yt related to it. You would understand it fast there as here we can only write and explain you

[unique tendon]

Could you suggest any?

[trim ermine]

Do you understand Hindi ?

[neon iron]

It might mean this: If you are supposed to find the component of A along P, then you choose some vector perpendicular to P, call it Q. Then A can be represented as a combination of P and Q, say A = 3 P + 5 Q. Then the component asked for is "3". I think this is the most likely what is meant.

[trim ermine]

I can suggest some in that language

I don't know about English ones

But the content is mostly the same

Gautam explained it the best way

If you still can't understand, go and search up a lecture

[unique tendon]

Like this?

[neon iron]

yea

[unique tendon]

What if the angle between P and A was an awkward angle such that you can't draw a perpendicular line from the tip of A onto P?

[neon iron]

you extend P to a line.

[unique tendon]

And if they are in 180° ?

[neon iron]

If A is perpendicular (90 degrees) wrt. P, then the component would be zero. If it is 180 degrees, then it would be negative.

[unique tendon]

What does negative component indicate?

[neon iron]

if you have vector P, then 2P is double length as P, but same direction.

-P is same length as P, but opposite direction

-2P is double length, but opposite direction

[unique tendon]

Yes

[neon iron]

so negative component means that when you project A onto the line which you get by extending P, it lands on the other side of P. This happens when the angle between A and P is bigger than 90 degrees. Just draw a picture and you will understand