help-33

[lucid hinge]

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[marsh citrus]

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[meager prawn]

2n^2 + (4n - 2) + (4(n + 1) - 2)
2n^2 + (4n - 2) + (4n + 4 - 2)
2n^2 + (4n - 2) + (4n + 2)
2n^2 + 8n

Right?

My textbook says the answer is

2n^2 + 4n + 2

What's going on here?

[proven timber]

Show the original question

Like, a picture or screenshot

[meager prawn]

Okay one sec

[proven timber]

2 + 6 + 10 + ... + (4n - 2) = 2n²

So 2 + 6 + 10 + ... + (4n - 2) + (4(n + 1) - 2) = 2n² + (4(n + 1) - 2)

You added 4n - 2

[meager prawn]

Hmm?

I only posted LHS btw

[proven timber]

I know

4n - 2 is included in 2n²

[meager prawn]

Is it?

[proven timber]

Maybe I'm mistaken but that looks like a 4n - 2 to me

[meager prawn]

Sure I agree. But I thought the idea was to substitute the green part with 2n^2 and simplify the rest? Am I mistaken?

[proven timber]

That green part is not 2n²

2n² has 4n - 2

If I said x = 1 + 2, and I had 1 + 2 + 3, I couldn't say that's = x + 2 + 3

Cuz x = 1 + 2, not just 1

[meager prawn]

Ah okay, that makes sense, thank you. So I should be substituting the green here with 2n^2 ?

[proven timber]

Yes

[meager prawn]

Makes sense, thank you 🙂

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[supple flower]

Is this a textbook typo?

[marsh citrus]

@supple flower Has your question been resolved?

[marsh citrus]

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[tiny cape]

Yes, ish. https://books.google.com/books?id=_341EAAAQBAJ&pg=PA18

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[still temple]

god i need help

[proven timber]

What have you tried

[still temple]

nothing yet

[proven timber]

What does 10 ft 3/s mean

Ah

Okay so you know dV/dt = 10 ft³/s, right?

You want dr/dt

Do you know any formulae to relate the radius of the cone and the volume?

[still temple]

yes

[still temple]

hmmmm

no

[proven timber]

What's the volume of a cone?

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

V=1/3hπr²

[proven timber]

And what is h in terms of r?

[still temple]

height

right?

[proven timber]

But how does it relate to r

"whose height is twice its radius"

[still temple]

2h=r

[proven timber]

No

That would give a height that's half the radius

[still temple]

h = 2r

[proven timber]

Yes

Substitute that into the formula

[still temple]

v = (1/3)2r*pi*r^2

@proven timber

[proven timber]

Yep

[still temple]

and now i have to derive?

[proven timber]

Simplify that

Then differentiate

[proven timber]

r * r² is not r²

[still temple]

2r^3 then?

[proven timber]

Yes

Now differentiate

[proven timber]

Don't forget to use chain rule on r

[still temple]

i did forget?

[proven timber]

Yes

You're supposed to differentiate with respect to t, not r

[still temple]

to t?

[proven timber]

Yes. You're trying to find dr/dt

[still temple]

hmm

[still temple]

yes

so

i ahve to

@proven timber i dont understand bro 😦 could you graph it?

[proven timber]

There's nothing to graph

r is a function of t

So what's the derivative of the formula with respect to t

Using chain rule

[proper cove]

https://tutorial.math.lamar.edu/classes/calci/relatedrates.aspx

[still temple]

and what is the way

to add that

to this

[proven timber]

Read this

[still temple]

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[thick elk]

Q, find the coordinates of the stationary points on the curve with equation y=x^2\left(6\sqrt[3]{x}-7\right)

[thick elk]

oops that didnt paste well

[trim quest]

$y=x^2\left(6\sqrt[3]{x}-7\right)$

[elfin berry]

tatpoj

[thick elk]

ah thanks

[trim quest]

👍

Where are you stuck?

[thick elk]

each time i try solve it i end up way off

idk where i go wrong

[trim quest]

Can you show what you did?

[thick elk]

one sec

In this attempt I got the answer 0,0 which is right but I didn't get 1,-1 which is the second answer

[trim quest]

$\frac{d}{dx}[f(x)g(x)]$ is not equal to $f'(x)g'(x)$ in general

[elfin berry]

tatpoj

[trim quest]

Looks like you just took the derivative of each factor separately, but that doesn't quite work

You either need to use the product rule, or distribute the x^2 first before you differentiate

[thick elk]

ah ok let me try rq, thanks

$ 0=x^{-\frac{1}{3}}-\frac{7}{2}x$

thats what im up to

oops

$0=x^{-\frac{1}{3}}-\frac{7}{2}x$

[elfin berry]

NightHorn

[thick elk]

@trim quest

[trim quest]

The x^-1/3 isn't quite right

$x^2 \cdot x^{\frac{1}{3}} = x^\frac{7}{3}$

[elfin berry]

tatpoj

[thick elk]

oh of course its ADD

[trim quest]

yep

[thick elk]

$0=x^{\frac{4}{3}}-x$

[elfin berry]

NightHorn

[thick elk]

is that right?

[trim quest]

The exponents are right but your coefficients are gone

Before differentiating, what do you have?

[thick elk]

$y=6x^{\frac{7}{3}}-7x^2$

[elfin berry]

NightHorn

[thick elk]

then differenciating

and div 14

[trim quest]

OH

I didn't realize you divided by 14

yeah, you were right

[thick elk]

ah haha thanks

so what do i do now?

[trim quest]

This is good. Do you know how to solve it?

for x

[thick elk]

$x^{\frac{4}{3}}=x$

?

[elfin berry]

NightHorn

[thick elk]

or no

one sec

$x^4=x^3$

[elfin berry]

NightHorn

[trim quest]

I think what you want to is keep the right side as 0 and factor the left

[thick elk]

how, the methods i can think of wont work with that power

[trim quest]

From here, factor out x

[trim quest]

Oh, I mixed up the left and right when I said this

[thick elk]

haha np i understood

[trim quest]

$$0 = x^\frac{4}{3} - x$$
$$0 = x(x^\frac{1}{3}-1)$$

[thick elk]

$0=x\left(x^{\frac{1}{3}}-1\right)$

[trim quest]

Ooh the tex bot is struggling again

but that looks right

[thick elk]

haha lol

[trim quest]

x ( x^1/3 - 1)

= 0

[thick elk]

yeah

ah okay I get it, thanks!

[trim quest]

Np 👍

[thick elk]

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[merry shale]

Hey all, how would I find the solutions to Cos(x/2 - 1) = 0?

[merry shale]

I know π/2 and 3π/2 make Cos into 0 but how would I account for the other numbers inside the parentheses?

[cursive echo]

set those numbers equal to the pi/2 and 3pi/2 @merry shale

[merry shale]

My god I can't believe I forgot that

Thank you for that I was blanking super hard

[marsh citrus]

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[pliant junco]

I do not understand c and d

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.reopen

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✅

[marsh citrus]

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[stone cobalt]

stuck on b onward

[stone cobalt]

-3 and 3 would be critical values because they are endpoints

idk what else past that

oh wait i got b

its -2, 0, 2 because f'(x) = 0

okay so c onward

i also dont know where to go with this

okay update i've figured out 4 and just need to evaluate how to do 5

i'll ping now

<@&286206848099549185>

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[still temple]

Well first thing

[worthy trellis]

Have you tried anything?

[lunar herald]

ngl idk where to start xD

would u cube root it first to make it 1/17?

[still temple]

don't mind me, give me a sec

Oops

[lunar herald]

huh?

so n is the number right

[still temple]

$\frac{1}{\sqrt[3]{17}} = (\sqrt[3]{17})^{-1} = (17^{\frac{1}{3}})^{-1} = 17^{\frac{-1}{3}}$

[paper karma]

$\sqrt[n]{x^m}=x^{m/n}$

[still temple]

bruh bad formatting

[elfin berry]

Mohammad

[still temple]

hold on

There u go

aaaaa

oops

[elfin berry]

trololol !

[paper karma]

$\frac{1}{\sqrt[3]{17}}=\frac{1}{17^{\frac{1}{3}}}=17^{-\frac{1}{3}}$

[elfin berry]

Mohammad

[lunar herald]

so thats the answer?

[paper karma]

Is that all you care about?

[lunar herald]

no but abt 50 ppl put 50 different things so its kinda confusing to keep up.

[paper karma]

It's all the same under different masks

[lunar herald]

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[crimson frost]

👋 I need help with this problem

[crimson frost]

There's a snek 🐍.
It can move in two directions, left (L) and right (R).
In the beginning, the snek is looking towards north.
The snek's tail stays at the origin, but the snek's body makes a step in the given direction.
When the snek moves, it's left and right change ("it's looking at another direction", so it's left is relative to the snek's last move).

So for example, this is what the beginning looks like

And this is what it looks like, when the snek goes to the right

[sleek lake]

completely fair

[crimson frost]

This is what RR looks like

[proper zodiac]

Youre playing snake 🐍

[crimson frost]

Now, here's the problem I've been trying to solve.
Given a sequence of moves (for example, RLLR, RLLLLR, RRR, etc), how can we predict when the snek crashes into itself?

[crimson frost]

Yes basically yes, but there are apples everywhere so at each step the snake gets longer

Now, for some reason, this problem turned out to be extremely difficult for me lol

[sleek lake]

so forward is forbidden

[proper zodiac]

What kind of answer are you looking for here?

[crimson frost]

Yup!

My question is: I made a function, it takes two arguments

It takes the direction the snake wants to take, so R or L, and the snake's previous move

This way, I can convert a RLLRL kind of string to a "South, East, North, East" kind of string

[still temple]

interesting question, by predict do you mean "snek is forced to crash" or "snek has already crashed"

[crimson frost]

No, just whether it crashes or not, it doesn't matter at what step

[crimson frost]

Can this be an operation for a group, where the group's set is made up of all the possible states that a snek can be in?
I haven't done much stuff with groups, but everything I tried failed in some way

[proper zodiac]

I don't see groups being very helpful here

I feel like the soliton to this is going to be an algorithm rather than some formula or operation on the string

[crimson frost]

Yeah, it's probably like 10 lines of python (pun intended ), but I wanted to solve it in some cooler way

[still temple]

every other point has to be one of vertical or horizontal - say start is (0,0), then if the snake is at (a,b), then it can only travel north/south iff a+b mod 2=1

[proper zodiac]

You ideally want to keep track of which direction the snake is facing so you can at each step in the string give (x,y) coordinates of the snake and then it becomes a problem of determining if/when the snake reaches the same coordinates

Perhaps something mod 4 can do that

[still temple]

lol

great minds think alike

[proper zodiac]

Lol

I'm thinking something like L is -1 and R is +1 and you mod 4 each step and the number you have tells you the direction

Once you know direction you know which of (x±1,y±1) you'll have as next coordinate

[crimson frost]

Wow this works super well

[proper zodiac]

Yeah what makes this difficult is the change of reference frame each time the snake moves, bc you decided L and R are relative to where the snake is facing, so first thing u wanna do is move to absolute positioning

[crimson frost]

Yes that's the first thing I did, now I use north, south, east and west

Is there a mathematical entity or object that keeps track of "what has happened in the past"?
For example, in this case, can I have something that uniquely defines each snek's state?

Oh, I'm dumb, the string of moves already does uniquely define a single snek

[proper zodiac]

Hm, possibly markov chains

This seems markov chainy

But that's kind of more probabilistic

[still temple]

I think the position in the chain already determines the n/s or e/w directions

[proper zodiac]

Eh markov chains only care abojt what just happened

[crimson frost]

Ok people
Thank you so much for your help!

I'll get back to the problem, it has been really fun to work on
I suggest to try to mess around with this system of relative left and right, there are some nice and somewhat hidden patterns

Thanks again @proper zodiac @still temple!

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[clear plume]

Are these the same?

[still temple]

no

[clear plume]

well for 9) i have 2f(x)+3. How would I flip that to make 10)

or rather what's goin on that i'd have to represent in a diff way

i remember there being some sort of order of operations when it came specifically to describing functions with h. shrinking and v. stretching etc.

but not the specifics

[marsh citrus]

@clear plume Has your question been resolved?

[clear plume]

<@&286206848099549185> pinging cause 19 min later

[tiny cape]

Shifting is addition, stretching is multiplication. If you keep track of what is being shifted and what is being stretched (use parentheses, they can be useful!), you'll see why they're different.

Your answer for 9 is correct, but don't try to "flip that" to make 10, just do what it says to construct it instead.

[clear plume]

2f(x+3) would that be it??

including it in the paranthesis puts it ahead of PEMDAS of the stretch?

[tiny cape]

Not quite. f(x) is the thing you want to be shifting, not x.

[clear plume]

so i'd still need to do f(x) + 3

but then stretch f(x) + 3 by a factor of 2

[tiny cape]

Indeed.

[clear plume]

2(f(x) + 3) seems clunky

but is that it??

[tiny cape]

It is!

You can even multiply it out, if you wish.

[clear plume]

would that be 2f(x) + 6?

[tiny cape]

Yep!

[clear plume]

Thanks for the help 🙂 have a good one! i understand now why they're not the same

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[vapid nexus]

is it 3(2x^(3) - 1)^(2) * 3x^(2) - 3(2x^(3) - 1)^(2) * 5x^(4) instead?

[cursive echo]

@vapid nexus not quite

[vapid nexus]

so what am i getting wrong

[vapid nexus]

yeah

but still incorrect, unless i mistyped

oh i did mistype

makes sense now

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[lucid hinge]

For 6

It’s just 2025 right?

For each element, the numerator is different

[teal hearth]

yes

[lucid hinge]

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[rancid wolf]

Hey, I’m supposed to find whether following sets are sub spaces of P2 and if yes, what are its basis.
S1={p€P2; p(0)=1}
S2={p€P2; p(1)=p(2)}
I know I’m supposed to check whether it contains 0, and is closed under addition and scalar multiplication, for the basis I’m not 100pr sure, I understand this for R2, but polynomials and P2 is making me struggle…

[marsh citrus]

@rancid wolf Has your question been resolved?

[rancid wolf]

<@&286206848099549185>

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@rancid wolf Has your question been resolved?

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@rancid wolf Has your question been resolved?

[marsh citrus]

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[smoky marsh]

why does this not work? isnt this just power rule of logs?

[smoky marsh]

*reverse order

hey im back

[proper zodiac]

The 1/x power would need to be inside the log

Applied to x not the whole log(x)

[smoky marsh]

ohhhhh, so just my notation is wrong

hng

arrrrr, this dont be making sense!

[main idol]

Wot

[proper zodiac]

Well what is even the question

[main idol]

Oh are you supposed to give it a wrong answer

[smoky marsh]

no, the answer is supposed to be 0

[proper zodiac]

Well yeah apply log like you were doing

[smoky marsh]

so ln(x)/x is right? cause i dont see why it would be wrong

[proper zodiac]

Yeah?

No one said it was wrong

[smoky marsh]

alright, i'll report back in a min

ok so i got 1/x but when the limit is 0, that would equal to infinity

ln(x)/x=(1/x)/1=1/x

[still temple]

💀💀

[smoky marsh]

i think he meant to prove it wrong lmao

[proper zodiac]

ln(x)/x is not 0/0 or inf/inf

[smoky marsh]

well using l'hospital method, i got the 1/x, if i didnt and sticked with ln(x)/x with x approached 0 then i would get 1/0, still a problem

how do i evalute the limit of ln(x)/x

[proper zodiac]

It's -inf*inf = -inf

It's not an indeterminate form and it's a result you can prove in general

[smoky marsh]

alright thanks

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[desert cloud]

<@&286206848099549185>

.help

[marsh citrus]

Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[desert cloud]

.tag

[old dawn]

hey

[desert cloud]

hi

[old dawn]

yd? is that yards? xd

[desert cloud]

can you help me?

yeah

[old dawn]

i think it tells you that the circumference of the bottom circle is6yd

weird drawing

and height of the cone is 7yd

[desert cloud]

yeah find the volume i guess?

[old dawn]

turn the circumference of that circle into radius

and just calculate

[desert cloud]

okk

[marsh citrus]

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[wary lichen]

oh wow math

[wary lichen]

what

.close

[marsh citrus]

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[woven harness]

how do you do this

[marsh citrus]

@woven harness Has your question been resolved?

[marsh citrus]

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[torpid anvil]

x^2+ax+b the function passes through (3,0) and (11,0). How do I find out what a and b are?

[dry prawn]

Plug in your values of x and y, you'll have 2 equations with 2 unknowns (a and b)

[torpid anvil]

Huh?

[dry prawn]

What about what I said doesn't make sense?

[torpid anvil]

Is there a simpler way other than brute force

[dry prawn]

This way is simple enough

[still temple]

Steakanator didn't even mentioned anything like that.

It's not bruteforce.

[torpid anvil]

bruh

[dry prawn]

If you want an alternate method, you can note that x=3 and x=11 are both zeros of the function

[torpid anvil]

alright then. I guess I have to plug in everything

thanks

.close

[marsh citrus]

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[steady gale]

discriminant?

[lunar marlin]

hello?

[steady gale]

or maybe not. srr i only guessed that u can calculate discrimant

[lunar marlin]

Im not sure what discriminant is but I'm working on intermediate algebra and I'm completely lost on the factor section

are we able to do calls?

<@&286206848099549185>

[steady gale]

i can tell you all process and answer using photomath 😄

[lunar marlin]

Is that the best way for explanation

[steady gale]

its the fastest way

[lunar marlin]

is there someone i can talk to directly?

[still temple]

have u heard of completing the square

[lunar marlin]

like factoring out the exponents?

[still temple]

hm?

like first dividing by 81 to get the normal form

[lunar marlin]

I got (9y2+18y)2

[still temple]

huh

[still temple]

actually

[merry terrace]

81y^2+9y+9y+1

Factor it

[lunar marlin]

9y+1^2

.close

[marsh citrus]

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[marsh citrus]

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[onyx hemlock]

Hi! I need some assistance with finding a formula for this word problem. I’m a bit confused on where to start

[marsh citrus]

@onyx hemlock Has your question been resolved?

[onyx hemlock]

<@&286206848099549185>

[hazy lion]

their suggestion is good

define some variables

see if you can identify the dependent and independent variables

[marsh citrus]

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[marble glacier]

anyone can help me with l'hopitals rule?

[slim solstice]

wats ur qn buddyboyo

🚑

[marble glacier]

limit

of

Nln(1+a/N)

[slim solstice]

send pic

[marble glacier]

wrt N ofc

[slim solstice]

just send a pic

[marble glacier]

https://gyazo.com/2cba325e177efd145019fb4830e17007

[slim solstice]

ok wat does lhopital state

fx over f'x?

[marble glacier]

oof all i know u keep dy/dx'ing until u get a number

Ya

[slim solstice]

aight so did u try to differentiate it

wat expression did u get

[marble glacier]

Ya

[slim solstice]

wats the fraction u got

fx / f'x, what does it look like

send pics

[marble glacier]

-a/N(N+a)

oh

[slim solstice]

ok

[slim solstice]

ok so u got this

[marble glacier]

Yea

[slim solstice]

-a is the numerator?

[marble glacier]

YEa

[slim solstice]

and N(N+a) is denominator

[marble glacier]

ya

[slim solstice]

ok

just divide ontop and below by N^2

[marble glacier]

Ok

[slim solstice]

then the top n bottom will converge

[marble glacier]

(a/n^2) / a/n

[slim solstice]

wait

tf

is N + a in the denominator or not

send a pic

of the expression dude

typing math in text is super unclear

[marble glacier]

ok now ill divide by n^2?

[slim solstice]

yea

wait

[marble glacier]

was i supposed

to diff it

[slim solstice]

ok nvm i think that makes numerator and denominator both converge to zero

[marble glacier]

with the N

too

[slim solstice]

yes wtf

[marble glacier]

....

Ok

that shall be done

[slim solstice]

dude thats d/dN

[hidden plaza]

$$\lim_{N \to \infty} N\log{\left(1+\frac{a}{N}\right)} = \lim_{N \to \infty} \frac{\log{\left(1+\frac{a}{N}\right)}}{\frac{1}{N}}$$
Candidate for Lhopital rule as we have $\frac{0}{0}$ situation
$$\lim_{N \to \infty} \frac{\frac{-a}{N^2 \left( \frac{a}{x} + 1 \right)}}{\frac{-1}{N^2}} = \lim_{N \to \infty} \frac{aN}{N+a}$$ from here evaluate the limit as normal

this took me way to long to way to long to write

[elfin berry]

Pure

[marble glacier]

AHH

so the N in front

u divide by N^-1

make it a fraction

and apply l'hopital

Hm

thanks

.close

[marsh citrus]

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[stray crater]

how would I find the horizontal and vertical asymptotes of xln(1+1/x)? do I first have to rewrite and use L'H?

[hushed burrow]

As x -> 0: ln(1+1/x) - ln(1) goes to 0. 1/x goes to positive infinity. So you use L’Hopitals rule: Take the derivate of ln(1+1/x) - ln(1)/1/x to find the limit. Simplify the derivative, cancel terms, and voila, you have found the limit.

Correct me if I'm wrong 🙂

The limit as x->0 of xln(1 + 1/x) is undefined - so use L’Hopitals rule on the limit as x->0 of (ln(1 + 1/x) - ln(1)) / (1/x) to achieve the value 0.

[marsh citrus]

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[solemn vessel]

https://media.discordapp.net/attachments/610130794953834496/1044697699833872414/image.png

[solemn vessel]

just the graph question

ill try figure out question 2 myself

im not sure how to breakdown the statements given

number 2 aswell actually i wanna compare my answer

[scarlet ingot]

@solemn vessel for the graph question, f'(x) represents the gradient of the curve, so its asking if its negative or positive in the region a<x<b

and the second derivative, f''(x), represents the rate of change of the gradient - it should be known that at local maximum points, the second derivative is negative. and at local minimum points, the second derivative is positive

[solemn vessel]

so its c and d

since it goes from negative to positive

from a and b

[scarlet ingot]

yeah exactly

[solemn vessel]

you got an answer for number 2 or na?

jus wanna make sure if im on the right track

[scarlet ingot]

well now u need to use the second condition for each to figure out if its c or d but yeah its between those

[scarlet ingot]

what was your method?

i could give the answer but I feel that ruins the fun

:)

[solemn vessel]

bet

ima try

[solemn vessel]

just differentation then subbed in e and 0

ln became 1/x

and uhh

[scarlet ingot]

differentiating f(x) is a good first step

chain rule will help there

let $u=ln(x)$, $\frac{d}{dx}ln(ln(x)) = \frac{d}{du}ln(u) \cdot \frac{d}{dx}ln(x)$

[elfin berry]

Syrenate

[scarlet ingot]

yeah?

[marsh citrus]

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[solemn vessel]

ye i had that i was jus watching a chain rule video

thx

[scarlet ingot]

aight cool nw

[marsh citrus]

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[rotund violet]

@cobalt yacht do you know cauchy-schwarz inequality?

@cobalt yacht are you there?

[cobalt yacht]

Yes

I know

[rotund violet]

so we do this with cauchy schwarz

wait just gimme a minute let me make sure

actually i dont think cauchy schwarz will neccesarilly work

these are olympiad problems so like these will take time

yeah ask this in #competition-math

ill think about this

[marsh citrus]

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[rotund violet]

.close

[marsh citrus]

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[pearl zealot]

quick one

[pearl zealot]

a^2-a equals what?

Like if im using the quadratic formula

And i have like root : a^2-a

How am i gonna simplify that yk

[cloud iron]

$a^2 - a$ or $a^{2-a}$?

[elfin berry]

lexitorius

[pearl zealot]

The first one

[cloud iron]

That would be a difference of squares

Uh maybe

[pearl zealot]

So i cant simplify that?

[still temple]

you can take out common factor

[cloud iron]

Yeah you can factor it but it wouldn't be diff of squares I was wrong

You can factor an a out of both terms

[still temple]

yea

[pearl zealot]

Like a(a-1)

or what

[still temple]

yep

[pearl zealot]

And then what?

If i have to solve the root of that

[still temple]

thats just a monomial

[pearl zealot]

So just take it like that?

[still temple]

you mean $a(a-1)=0$ ?

[pearl zealot]

Nooo

[elfin berry]

13part

[still temple]

ah

[pearl zealot]

Like a ( a -1 ) under a root

And before that root

a+-

[still temple]

square root??

i dont understand

[pearl zealot]

So like a +- root (-2a/2 ) - a

Quadratic formula as i said

[still temple]

ok so you do mean this

[pearl zealot]

Yea just that under a root yk

[still temple]

you dont really need a quadratic for this one

[pearl zealot]

Like its already in a quadratic formula

Like my equation is x - 2a +a/x

What i did was multiply with x

so i get x^2-2ax+a

[still temple]

ahh ok

[pearl zealot]

And i can use the quadratic formula out of that

Yk

You know the topic group/sets of functions?

[still temple]

ah i dont

[pearl zealot]

All good it doesnt matter lmao

So what do i do after that?

[still temple]

hahah ok ok

[pearl zealot]

Like if its under the root

[still temple]

with the quadratic?

[pearl zealot]

Yea

[still temple]

ok ok

so

[pearl zealot]

now its like

a +- root (-2a/2 ) - a

[still temple]

im kinda new to latex math so hold on

[pearl zealot]

All good lmao

[still temple]

if you have $x^2-2ax+a, x=\frac{-(-2a)\pm \sqrt{(-2a)^2-4a}}{2}$

well if i didnt simplify so much

[elfin berry]

13part

[still temple]

there we go

you can simplify it: $x=\frac{2a\pm \sqrt{4a^2-4a}}{2}$

[elfin berry]

13part

[still temple]

$=\frac{2a\pm \sqrt{4a(a-1)}}{2}$ (we took common factor out)

[elfin berry]

13part

[still temple]

you can simplify it a bit more but ill leave it up to you :)

[pearl zealot]

Wait whatttttt

My question is to show wich functions have like 1 or 2 or none zeros

And i think i can just leave it like that

And see what i can set in for a to get some zeros

Related to the task

I think thats what i have to do

[still temple]

well do what you have to do! if its an equation, just use whatever is necessary

[pearl zealot]

But wait

Whats 0 rooted?

Like root ( 1^2-1

So i get 0 under the root

@still temple

Thats equal to 0 right?

[still temple]

ok first of all what do you mean by under the root

ah yes yes

$\sqrt{0}=0$

[elfin berry]

13part

[marsh citrus]

@pearl zealot Has your question been resolved?

[marsh citrus]

@pearl zealot Has your question been resolved?

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[marsh citrus]

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[cinder ether]

how do i get d and e

[marsh citrus]

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[cinder ether]

.close

[marsh citrus]

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[dim dust]

Hiii how do I solve this

The wheels of Wouter's bicycle have a radius of 30 cm. If Wouter moves his bicycle 75 cm, what angle are the wheels turned over? Calculate the magnitude of that angle in radians and sixty degrees.

[brisk scarab]

calculate in sixty degrees means?

[dim dust]

In the ° way

I just google translated the question 💀

So sorry for some weird wordings

[brisk scarab]

ah okay

so basically

if the radius is 30cm

then you can find circumference

and we know that the circumference is basically an arc of 360°

so you need to find out how many degrees is 75cm, IF circumference (whatever it is) is 360°

its abt comparing both

you'll get degree measure. then you can find the radian meaure through a formula which i fogor
I'll send the formula here

[dim dust]

Yea I got the formulas but the math was just not mathing haha thanks a lot

<<33

.close

[marsh citrus]

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[peak viper]

Whata the next step, can you give hint

[wind light]

It’s a famous number!

[peak viper]

i dont understand the 2 formulas, can u give me the name of it so i can go on yt ?

if u dont mind

[wind light]

Well as x approaches infinity of (1+1/x)^x what does it approach?

[peak viper]

e

[wind light]

yea

[peak viper]

and whats the next step, im blocked there

if i replace i get 1 power infinity

its an indetermined form

[wind light]

Why are you trying to use lHopital on it?

[peak viper]

i need to simplify it

the answer is e power 12

but idk how we got it, so im trying that

[wind light]

Oh the first one

[peak viper]

oh, i get it why u asked this, we know its = e but the limit is not on its propore form

this is the proper form and we have to make the previous limit on this form and we have to solve it

idk if you got it lol

[wind light]

Yea okay so pretty sure you can use log rules to simplify it

[peak viper]

how to use it?

[wind light]

So do you know the log power rule?

[peak viper]

i mean i havent used log in the whole excercise

[peak viper]

ik but i have never used log for this certain excercise

do you want me to show what i have done ?

[wind light]

no so

[peak viper]

i dont understand the whole method

[wind light]

Okay

I can’t tell what’s happening there

[peak viper]

can you ping another helper?

[marsh citrus]

@peak viper Has your question been resolved?

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[rancid tendon]

how can i solve this problem?

[distant peak]

this problem... makes no fcking sense or I'm dumb

[rancid tendon]

the same think that i said

[rancid tendon]

I think the teacher was wrong and wrote absolute magnitude instead of mass

because mass is written with "m" and absolute magnitude with "M", but i still dosent know how to solve that

[sleek lake]

this was invented 2000 years ago

vintage math

[rancid tendon]

do you know how to solve it?

[sleek lake]

you just need the origin, what luminosity is 0 M or 1 M

i don't know the reference point

[rancid tendon]

i dont know the luminosity, is the point of the problem

[placid oak]

,w luminosity formula

[elfin berry]

Results provided by WolframAlpha

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

Requested by hazysleet#7757

[placid oak]

[sleek lake]

right there we go

[rancid tendon]

oh thx!

[sleek lake]

,calc 100^((4.83+0.8)/5)

[elfin berry]

Result:

178.6487574852

[sleek lake]

times the sun luminosity

,calc 178.65 * 3.848 * 10^26

[elfin berry]

Result:

6.874452e+28

[sleek lake]

that many

[rancid tendon]

thx for everything

.solved

.close

[marsh citrus]

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[mint plinth]

Hey. Would this make sense?
There is no commutative linkage, but a and b can be linked (because the result is again a rational number)

[devout mauve]

are you translating some terms from another language?

[mint plinth]

Yes exactly

Which terms are wrong?

[devout mauve]

linkage is pretty weird

[mint plinth]

Yea its what the translator gave me

Is there a good translator for it? I only know the german term

Kommutativitätsverknüpfung ist the german word.

[devout mauve]

I would just call it commutative operation

[mint plinth]

Okay, thanks

And does what I wrote there make sense?

[devout mauve]

now with that out of the way

[mint plinth]

Im trying to self-learn it right now

[devout mauve]

first we need to remove the element 0

cause we can't divide by it

[mint plinth]

oh the fraction is wrong? wow

[devout mauve]

but then yes, dividing is indeed a non-commutative operation over Q\{0}

[mint plinth]

Okay, so how I wrote it it does make sense? Are there some example questions I can do on that topic?

[devout mauve]

for all nonzero rationals a,b we have that a/b is again a nonzero rational number

and in general a/b != b/a

what topic exactly. group theory in general?

[mint plinth]

Yes :)

I will try asking my teacher at school too, but we have a complete different topic there

[devout mauve]

well yes group theory is not usually taught in school in germany lul

or in most countries I assume

[mint plinth]

Mhm yes sadly

[devout mauve]

I mean most people wouldn't understand it anyway. sadly some teachers probably included

anyway, what kind of questions do you want. more examples of groups/non-groups?

[mint plinth]

Yes, maybe example questions that I can try solving but I have the right answers to compare

As i only find university material when I google for it

[devout mauve]

well group theory is university stuff

that really shouldn't be surprising

[mint plinth]

Yes, haha. Im trying to solve a question we got in class, but in 3 dimensions instead of 2. That's what I need that all for to understand the next parts.

[devout mauve]

what question

[mint plinth]

Well there was a golf course. A golfer plays the ball. Given was the highest point of the ball and the start. Also the slope of the terrain.

So we had to calculate the height where it stops.

But I found it boring

So I'm trying to solve it how he described it. He said it flew 5m to the right. He mentioned it as example but I would like to calculate it

[devout mauve]

and how is group theory relevant here?

[mint plinth]

From my research I'll have to do vector areas. And a YTber "Daniel Jung" did a playlist on it. And he said that we should know a few terms, which included group theory, rings, objects, etc.

@devout mauve This playlist was the only resource I found that didn't confuse me

[devout mauve]

english term is vector space

and ehh

I wouldn't really say that you need that explicitly

[mint plinth]

My bad, my english is horrible

[devout mauve]

at least not for that question

[mint plinth]

The vector space, or group theory

[devout mauve]

if anything that probably makes the question harder

[mint plinth]

mhm okay

[devout mauve]

sounds more like you want multivariable calculus

[mint plinth]

Will that be covered in the next few months of highschool?

Or would I have to learn that myself, and start right now?

[devout mauve]

might depend on your state. we didn't cover it

[mint plinth]

Okay, thanks! I will have a look at it

Thanks for the help 👍

[devout mauve]

youre welcome

[marsh citrus]

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[torn bloom]

Need help

[torn bloom]

Topic: Half Angles and Double Angles formulas

Anyone free?

I am trying, but didn’t find any clue yet.

<@&286206848099549185>

[spark berry]

j

Use this

Then square both sides

Then equate

[torn bloom]

Okay

Sin 0/2 = 3/5

[spark berry]

At the end, you will only have cos = the number you had got

Yes

[torn bloom]

Wait a sec

Which one should i use

[spark berry]

Sin theta/2

That one

[torn bloom]

sin^2 theta/2 = 9/25

What next?

[spark berry]

Noo

[spark berry]

Use this

Sin theta/2 = the first part of the picture

[torn bloom]

9/25 = +- 1-cos/2

[spark berry]

Yes but fix your left side

That came from your mistake earlier

[torn bloom]

Isnt it 3/5)^2

[spark berry]

Ah yeah

Yes you are correct

Just erase the +-

Anything that would came out of square would always be positive

9/25 = 1-cos/2

[torn bloom]

Yeah

[spark berry]

multiply both sides by 2

[torn bloom]

18/25

[spark berry]

Yeah

Then fix the equation to make cos (x) = the number

[torn bloom]

18/5 - 1 = cosx

[spark berry]

Yeah

[torn bloom]

Which is 13/5

So means 13 is adjacent

[spark berry]

Then find inverse of cosine

Noh, just pull out your calculator

It sometimes do the trick

[torn bloom]

How to find inverse

We didn’t learn

[spark berry]

Just use a calculator

It's the cos^-1 with it

And don't forget to configure your calculators to degrees

[torn bloom]

👀 sorry

Don’t get it

[spark berry]

It might be in radians. You will get a different answer and it seems your lesson is all about degrees

[torn bloom]

Answer in book is 24/25

[torn bloom]

Yes

[spark berry]

Do you have calculator like this?

Ok

[torn bloom]

Yes

[spark berry]

So at sin cos and tan

At the top

There's a sin^-1 cos^-1 and tan ^-1

Press shift then cos

[spark berry]

Then put this

There's also a typo with your answer

13/25

[torn bloom]

Wait

Not working

It says error

Cos^-1 13/25

[spark berry]

Wait I just realized you are wrong here

You got it the other way around

[torn bloom]

Why

[spark berry]

1- cosx =18/25

1 = cosx +18/25

1 - 18/25 = cos x

[spark berry]

And if the error persists, you're probably still at radian mode

[torn bloom]

cos^-1 = 7/25

Lemme try

[torn bloom]

Wym

[torn bloom]

Error

[spark berry]

at the right top corner of your screen, can you see a D or an R in it?

[torn bloom]

Oj

Should i make it D

[spark berry]

Yeah

[torn bloom]

I see a D

[spark berry]

It should be in D

[torn bloom]

Still error

[spark berry]

BRUH, I'm getting an answer

[torn bloom]

Is there any other day

[spark berry]

I'm getting an answer here

[torn bloom]

How?

It says

Match: Error

[spark berry]

Can I see your calculator or something

[torn bloom]

Ye

[spark berry]

Are you sure you are pressing shift first before you press cos

[torn bloom]

Yep

[spark berry]

Cause I'm getting an answer here

[torn bloom]

Im 100% sure

Its showing it

cos^-1 of 7/25

[spark berry]

Well I'm getting an answer here

[torn bloom]

Idk why didn’t work

I can show u

And what do we do now?

[spark berry]

Yeah that's it

[torn bloom]

Oh the book shows 24/25 lol

[spark berry]

Checking here it says this

I dunno, can you show me the picture of you're book

Like where the key to correct is

[spark berry]

,rotate

[spark berry]

There's something wrong here, I can feel it in my balls
Is that the entirety of the activity?

[torn bloom]

Wym

Maybe there is a simple way

Second one, I found just by using the cos2x formula

[spark berry]

OOOH god damnit, I wasn't reading the instruction properly

Find sin theta

[torn bloom]

Yeaaaaa

[spark berry]

Use this

[torn bloom]

73.73

[spark berry]

Yes at sin theta

Or maybe just the whole cos^-1 inside the sin

[torn bloom]

Bro wym?

[spark berry]

Yeah mission success

I got 24/25

[torn bloom]

? Lol

K

May i wonder how? 🤣

[spark berry]

sin (cos^-1 (7/25))

[torn bloom]

Yeo

[spark berry]

K dio, you can go do your own thing now. I think we are finished here

[torn bloom]

Okay thenks

[marsh citrus]

@torn bloom Has your question been resolved?

[marsh citrus]

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[astral coral]

Hi, I have a list of straight lines (co-ordinates) and I must check if all the lines are connected, what would be a good way to do this?

[astral coral]

I was thinking to compare each line to see if it has an intersection (common co-ordinate) with another line and if it does then that line intersects something but then there could be two different separated structures with lines that intersect

[marsh citrus]

@astral coral Has your question been resolved?

[cobalt sentinel]

Wdym by connected?

And separated structures?

[viral lily]

There can't be two disconnected structures connected inside, only parallel lines never intersect

[astral coral]

There is a list of lines, each list of lines has the co-ordinates of the points on the line e.g (0,0),(1,0),(2,0),(3,0),(4,0),(5,0)

so for example, if a = (0,0),(1,0),(2,0),(3,0),(4,0),(5,0) and b = (2,0),(2,1),(2,2),(2,3),(2,4) would be intersecting since (2,0) is common so this would class as connected

I am fine understanding when 2 are connected (have an intersection somewhere) but when there are more lines I am unsure of what to do

[marsh citrus]

@astral coral Has your question been resolved?

[marsh citrus]

@astral coral Has your question been resolved?

[next ravine]

You can just describe those lines with equations

[marsh citrus]

@astral coral Has your question been resolved?

[still temple]

Are they lines or line segments?

[marsh citrus]

@astral coral Has your question been resolved?

[marsh citrus]

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• Show your work, and if possible, explain where you are stuck.
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[maiden seal]

Good morning/evening. I am kind of blocked on this problem. what's the approach to solve it?

[proper zodiac]

Notice that x²+x-1 = x²-x-2 +(2x+1)

[exotic meteor]

Can't you just take 4th root of all sides and get x^2 - x - 2 + 2x + 1 = x^2 +x - 1

[maiden seal]

Right... I am still not sure where to start though

[proper zodiac]

You have an equation of the form a⁴+b⁴ = (a+b)⁴, think about for what values of a and b this can be true

[exotic meteor]

Could use the abc-formula to get 3 parenteses on the left side, or am I wrong again?

[proper zodiac]

No idea what you mean

[exotic meteor]

You get a new equation by doing so:
((x+1)(x-2))^4 + (2x+1)^4 = (x^2+x-1)^4

[proper zodiac]

It's the same thing, you would end up factoring later

[proper zodiac]

Factoring from the start in fact makes it less likely to notice this

[exotic meteor]

Isn't that the same as i wrote? Except the parantese

[proper zodiac]

You arrived there by "taking the 4th root of both sides" which is not how 4th roots work

Youre viewing it as an equation I am saying it's an identity independent of this question

[marsh citrus]

@maiden seal Has your question been resolved?

[marsh citrus]

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