help-28

[fathom cairn]

Solve it

You can divide by 3

5/3 * 3^x+1 = 2^3x-1

[short siren]

Small hint

The 3^{x+1} and 1/3 share a common base

[fathom cairn]

that means you can write: (1/3)^x?

[short siren]

No

$a^{b} \cdot \frac 1a = a^{b-1}$

[glossy valve]

VulcanOne

[fathom cairn]

So that means 3^x * 1/3?

[fathom cairn]

.

Yes

Checked

[short siren]

$3^{x+1} \cdot \frac 13 = 3^{x+1 -1} = 3^{x}$

[glossy valve]

VulcanOne

[fathom cairn]

Why -1?

[short siren]

$3^{x+1} \cdot \frac 13 = 3^{x+1} \cdot 3^{-1}= 3^{x+1 -1} = 3^{x}$

[glossy valve]

VulcanOne

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@fathom cairn Has your question been resolved?

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[flat lion]

How are they getting that ratio

cc’/0.6 = 0.84/1.44

They mention similar triangles so im guessing its something like

This tells me that AA’ = 0.6. Not sure if thats right, but if it is, how did they get that

[short siren]

Triangle εcc' is similar to triangle εAA'

[flat lion]

Alright i think that makes sense but where does 0.6 come from

[short siren]

So you have 1.8 as a total width

And above they mention that the middle width is 0.6

[flat lion]

Oh

Oh i see

Ok last question is why do they know cc’ = dd’

One side of the triangles have size = 0.84. They are both right triangles. Is that enough to know cc’ = dd’

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[echo summit]

Hii umm

[echo summit]

Anyone know how they got this

[gritty rose]

translate

or ask in physics server in #old-network

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@echo summit Has your question been resolved?

[thin flint]

Theyre talking about laying a needle down on water

Which then is in equilibrium due to surface tension gamma and gravity

The question they are asking is how do they get 2 L gamma cos theta = g

l is length of needle M is mass of needle gamma_w is surface tension of water, the other gamma is that of soapy water theta is the angle

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[obsidian radish]

So uh

What is a general term

[wintry grotto]

Google ur

It

[dry light]

it’s simply the general term, as it says

like the general term for a quadratic expression would be, ax^2+bx+c

[obsidian radish]

Ohhh

why was every explaination so confusing

[dry light]

that’s maths for you

[dry light]

do you understand how to do the question?

[obsidian radish]

ill gonna give it a shot

i assume we do

$$
\binom{13}{k} x^{13 - k} \cdot y^k
$$

[glossy valve]

dannylewastaken

[obsidian radish]

yeah?

or would i also need to add the summutation too

[dry light]

i’m ngl i was going to use pascal’s triangle

[obsidian radish]

💀

pascal 13 high

lmaooooo

but i assume that is what it meant by general term, yes?

[dry light]

yeah

[obsidian radish]

oh waiti should add summutaiton

or is that excluded

[simple totem]

that is the term

if you add summation that is the whole polynomial

[obsidian radish]

Ohh that makes sense

[gilded hinge]

nned that summation

[obsidian radish]

what

ok now i got someone saying i do

and someone saying i dont

[simple totem]

this is the general term

[obsidian radish]

[simple totem]

if you include a summation, then you get all the terms

[obsidian radish]

yeah okay

[gilded hinge]

oh ok

[obsidian radish]

yeah that makes sense

[gilded hinge]

general term

[obsidian radish]

alr thank you

[gilded hinge]

yeah

[obsidian radish]

.close

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[runic bloom]

isn't it B and C?

[gilded hinge]

yes

D converges because of the alternate series rule

A obviously converges

B's term is equivalent to 1/3n so it diverges

C is equivalent to 1/3 + n² which diverges

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@runic bloom Has your question been resolved?

[runic bloom]

okay ty

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[undone mural]

I need to express logk(1/4) in terms of k

I dont even know where to begin other than rewriting the equation different ways that I can't see helping me

[meager dew]

it's already written in terms of k

[undone mural]

I mean k = ...

now I know what k equals, it says it

i just need to prove it ig

[meager dew]

what does log_k(1/4) equal

[undone mural]

nothing, just needs to be rewritten as k = ...

all I'm given is log4(a) = k

[meager dew]

what is your original problem

[undone mural]

for example, the answer to (i) would just be 2k

[gilded hinge]

idk why they use log4()

if i were you i'd just go back to ln()

and log4() = 4*ln()

and use the basic ln properties

[undone mural]

ln()

?

[simple totem]

this is not true

[undone mural]

yeah I was thinking that

4*ln() = ln(⁴), but there's nothing to power to 4 so it doesn't work

[gilded hinge]

sorry i meant

[undone mural]

yes but then I'm left with another unknown variable, b

[gilded hinge]

but 1/4 = 4^(-1)

[undone mural]

yes 1/4 = 4^(-1) but I don't see how that might help me progress further to solving it

[simple totem]

im confused at what they're asking for

is this not already in terms of k?

[undone mural]

no, it's not really in terms of anything at the moment. They want you to find k = ?

[simple totem]

no

it is in terms of k

finding k = would put it not in terms of k

[undone mural]

for example, 2^a = 8 in terms of a would be a = log2(8)

that's my understanding of it anyway at least based off the other parts of the entire question

the answer is -(1/k) but I have no idea how they get that

[simple totem]

no

[simple totem]

this is solving for a

if you're given k = ln(a), then ln(a^2) in terms of k is ln(a^2) = 2ln(a) = 2k

[undone mural]

okay, that's my bad. The question is asking "If log4(a)=k, express logk(1/4) in terms of k"

[simple totem]

yes and im saying they already expressed it in terms of k

[undone mural]

the answer is -(1/k), idk what to tell you

I just need to figure out how they get that answer

maybe it means with k being the only variable?

cuz log is technically a variable, so it's probably just asking to express it without the log

[simple totem]

log is not a variable

and that answer doesn't make sense as far as i can tell

[undone mural]

I mean function

Well the book isn't wrong, so idk what to say

[simple totem]

,w log_4 10

[glossy valve]

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Requested by maximo#5992

[simple totem]

that's k

then ,w log -(1/1.660964)

[undone mural]

where did that 10 come from

[simple totem]

a wasn't specified

i'll just show it doesn't work for some a

,w -(1/1.660964)

[glossy valve]

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[simple totem]

,w log_1.660964 (1/4)

[glossy valve]

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Requested by maximo#5992

[simple totem]

see how they're different?

[undone mural]

but letting a equal anything will give a different answer every time?

[simple totem]

well if you had $\log_4(a^2)$

[glossy valve]

maximo

[simple totem]

then you get 2k

no matter what a is

[undone mural]

yes

[simple totem]

so it doesn't matter what the value of a is

so they are indeed equal

[undone mural]

because u can sub in log4(a) for k

[simple totem]

but in this final question

if you change the value of a

then you get a different value for log_k(1/4) and -1/k

so clearly the answer is incorrect

[undone mural]

you're 110% certain the answer in the book is wrong?

[simple totem]

we just saw the proof

i would hope you would be certain too

[undone mural]

there has to he some rule in logarithmic that allows the answer to be -(1/k)

[simple totem]

did you not see the two numbers differed?

books make mistakes

[undone mural]

but the fact this book gets all proof checked would make it very unlikely its wrong

[simple totem]

it still happens, often

[undone mural]

not saying it can't be, just very very unlikely

[simple totem]

if you choose to believe the book that's ok, but you should also believe the numbers you just saw

computed not by me, but by wolfram alpha

which show the book is incorrect

it's up to you to decide what to do with that information

[undone mural]

wait so you said k = log4(10), giving you that set of numbers as the result

obviously here you're just letting a = 10, but if it was let's say 8, the answers would be totally different

[simple totem]

the point is

for part a

or part i

where you're asked to find log_4(a^2)

it's independent of a

a can be anything and the answer is correct

log_4(a^2) = 2k

[undone mural]

yes

[simple totem]

but in this last one

[undone mural]

but part v is dependant?

[simple totem]

yes. in fact it's incorrect for most values of a

[undone mural]

in the change of base rule, what if I let the base equal a?

I'll send pic to show what I mean

[simple totem]

i know what you mean

you can try but im done with this specific question, i think i've shown enough already

good luck 🫡

[undone mural]

thank you

@simple totem I know u just said ur done, but I think solved it. Does that look right?

[simple totem]

no

[undone mural]

why not

[simple totem]

i don't see how the last 2 equations hold

[undone mural]

-loga(4) = -(1/k)?

[simple totem]

how you got from -k^-1 / log_a(k) to that

[undone mural]

the bottom left equation

cross multiplication

[simple totem]

bottom right

but then you're not talking about log_k(4^-1) anymore

you're talking about a completely different expression

[undone mural]

what

which equation are you talking about? the (-logk(4))(loga(k)) one?

[simple totem]

if you cross multiply here you're not talking about -log_k(4) anymore

[undone mural]

yes

[simple totem]

your result is

$-\log_a(k) \cdot \log_k (4) = -k^{-1}$

[glossy valve]

maximo

[simple totem]

which is not what we're trying to show

[undone mural]

yes but look at the very bottom right then

those 2 multiplied give -loga(4)

which, if flipped / powered to -1 gives -(1/log4(a))

and log4(a) = k

so we can sub k in, giving us -(1/k)

[simple totem]

log4(a) = k is given

if you multiply them you get

$-\log_a(4) = -k^{-1}$

[glossy valve]

maximo

[undone mural]

yes, now flip the loga(4)

and u get -(1/k)

[simple totem]

what

no k^-1 = log_a(4)

k = 1/loga(4) = log_4(a)

you're back to the start

https://www.desmos.com/calculator/qpsrxqwaz0
here's a graph with k = log_4(x), and then log_k(1/4) and -1/k

you'll see the graphs don't match

for any x

[undone mural]

this looks right to me

[simple totem]

yes, but did you see what the question wants you to write?

it's log_k(1/4)

not -1/log_4(a)

i feel like you're not reading what im sending

[undone mural]

the answer in the book says -(1/k) which I know u say is wrong, but if its possible to get that as an answer, then it must be the answer

[simple totem]

it is not an answer

you are not representing the original expression

i showed you 3 different times why they are not equivalent

you can get to -k^2/3! if you wanted to

but that doesn't mean it's the answer to the question

[undone mural]

I didn't mean it's an answer to the equation, I mean it's an answer as in, a final judgement of what the question is asking

[simple totem]

and i am saying it is not

again, read what i am sending

i said that -1/k is not a way to write the original expression

and i've shown you why that is the case

if you still believe those 2 forms are equivalent then i can't help you

[undone mural]

okay wait, what exactly is the error in the pic where I showed me getting -(1/k)

[simple totem]

you aren't expressing log_k(1/4)

you're now expressing a completely different thing from what the question asked you to

[undone mural]

yes but at what point did I go off track

[simple totem]

i've pointed this out multiple times if you scroll up

when you cross multiplied

i've said this before

[undone mural]

there's no error there as far as I know

[simple totem]

there isn't, but you're no longer talking about the same expression

im done here, it seems you're not reading what im sending

[undone mural]

I am reading what you're saying, I just feel like we're talking about 2 different things. If there's no errors in my workings and I was only working with logk(1/4), then how could've I got the expression to something completely different

[simple totem]

your equality doesn't go all the way through

you haven't expressed log_k(1/4) in terms of k

you wrote a completely different expression in terms of k

you're not doing what the question is asking you to do

you get to the point where you have this

you are no longer writing log_k(1/4) in terms of k

you are now talking about -log_a(k) * log_k(4)

the question wanted you to write:
log_k(1/4) in terms of k

you wrote:
-log_a(k) * log_k(4) in terms of k

-1/k is not an answer to what was asked of you

[undone mural]

ah yes, I see now. I've just let a = 16, making k = 2 and if you fill that into the answer the book gave and the original expression, u get -2 = 1/2

[simple totem]

this shows you they're not equal

[simple totem]

these show you they're not equal

i don't know why you were so hesitant about believing hard empirical proof

[undone mural]

the first proof you gave I just wasn't comprehending well enough

[simple totem]

it is exactly what you did here

[undone mural]

just couldn't wrap my head around it for some reason

[simple totem]

im just saying it doesn't hold for some values

which is exactly what you just did

[undone mural]

yeah I'm honestly not sure what I was even thinking when I was looking at it. Idk is it because I'm just tired or what

anyways, at least I got to a conclusion... eventually... thank you

[simple totem]

it's because you saw the answer and chose to believe the book could not make a mistake, then sought out that answer

there's tons and tons of typos all over the place in these books

[undone mural]

no no I knew the book could make a mistake, just whatever I was thinking for some reason thought something else

it's almost 3am, I'm just not functioning at all ig

anyways, I'm going to close this, thank you again

.close

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[rose finch]

How do I assess the validity of a factored polynomial?

[simple totem]

by validity do you mean if it's the correct factorization?

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@rose finch Has your question been resolved?

[rose finch]

i got it already, thank you

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[dusty patio]

can someone verify for me that i got an answer correct? ill send the picture in one second

[dusty patio]

"Find each integral. As part of your work explain which formula from the basic integral you
are using "

[torn jolt]

,w integral of 7 * (4th root of x)

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[torn jolt]

yes

[dusty patio]

alright thanks!

.close

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[hushed idol]

Hello, return of a familiar problem:

[hushed idol]

https://cdn.discordapp.com/attachments/1079554318698561537/1079600750633701426/IMG_4541.jpg

I have the base cases down and I assume that it's true for all values up to k.

How do I go about using this?

I think I need to prove that $3 \cdot 2^n + (-1)^(n+1) \cdot 2$ is equal to something

but I'm not sure how I construct that something

[glossy valve]

Солдат удачи

[hushed idol]

wait it would be that term

added to the original an

I'm messing around on a blackboard on the other side of the room, please ping me for the noise if you contribute

[sonic quiver]

Assuming a_n holds, show that a_(n+1) also holds. If you use the identity given for a_(k+1) then it follows directly from there

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@hushed idol Has your question been resolved?

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[ocean swallow]

is this not A?????

if the limit is finite

why is it not converges

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@ocean swallow Has your question been resolved?

[gritty rose]

That's not enough for convergence

Or what do you mean by "the limit"?

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[dense edge]

which of these lines would be an acceptable answer? "find the derivative"

only the last one? or professors should not be too strict about it

let's say you don't spot the common factor, and hand it in before that

[glass crystal]

ask your teacher what they would accept

[dense edge]

OK

.close

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[brisk parrot]

Can someone check these? I just wanna make sure I'm understanding limits and one-sided limits correctly

[brisk parrot]

Whoops forgot b.) but I'm thinking it's 1?

Oh I think a.) Is supposed to be 0

[sharp vine]

yeah

rest is correct I think

[brisk parrot]

yay!! thanks

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[hot herald]

stop deleting, make a new channel, again

[pine merlin]

I'm not deleting though!

I swear I'm not
Every time I send the question it keeps disappearing and idk why

I tried sending it in the probability and statistics channel and it disappeared there too

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[torn jolt]

Can someone help me with #4

[iron wraith]

Draw the line y=x and visualize what happens to one of the triangles when it is reflected through the line.

[torn jolt]

Oh ok

you’ve answered 3 of the 4 questions

theres only one answer choice left

I just realized that lol

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[bitter nymph]

Is this not just a)?

[limber flicker]

What makes you think a

[bitter nymph]

cuz the tangent line equation is as y = f(a) + f`(a)(x-a)

[limber flicker]

sure, it's correct, I was just thinking maybe giving reasoning relating y-b=m(x-a) , which is of course what they've done above in the photo you've sent

[bitter nymph]

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[brave mural]

I’m bad at rounding but if I were to round 2.5 to the nearest tenth would it stay the same or would it be 3 or 2.6?

[warm abyss]

It would stay the same

[brave blaze]

you cant round it to 2.6

neither to 3

[brave mural]

Why?

[gritty rose]

https://www.splashlearn.com/math-vocabulary/wp-content/uploads/2023/01/Thousandths-1.png

[trail oasis]

Yes it would stay the same because it's already a tenth

[brave mural]

Ok thanks guys

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[eager locust]

how to convert from minutes to seconds?

[eager locust]

/60 or *60?

[void widget]

Minutes

Remember
1min=60s

[eager locust]

yes

[void widget]

min/60=sex

sec*

💀

[eager locust]

oh alright

thanks

[void widget]

Welcome

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[fallow meteor]

can someone help me with maths?

[lime jackal]

sure

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[idle bison]

I was doing eigen values of matrices

[idle bison]

and came across cubic polynomial

need help how to find the value sof x

of cubic polynomial

please help, I have exam in 2 days time and im losing hope... :c

[torpid stag]

Please show your work

[idle bison]

Alright

I was trying to solve it

The equation on top is the equation i need to get lambda values for

If someone can help me get those please, it will be much appreciated

?

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[fierce gate]

How can one figure our the height of a cylinder given the base area and the radius

[atomic blade]

Can't without knowing the volume

[fierce gate]

That's what I thought lol

My math book is losing it

[atomic blade]

Burn it

[fierce gate]

calculate the desired(the one you're searching for) size of the cylinder

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[torn jolt]

Yo

[full forum]

Please don't occupy multiple help channels.

[torn jolt]

I am done, but how do i manipulate this equation

I need to isolate for c

[gritty rose]

closed your other ch

start by cross multiplying

[torn jolt]

emh

there is only one

[gritty rose]

or just multiply by 1+c

[torn jolt]

would this be correct

is my question i forgot to type it

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@torn jolt Has your question been resolved?

[gritty rose]

you can verify by plugging into your equation

one of these

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[dire flame]

guys how do i do this :(

I don't understand at all

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[marble meteor]

can someone show me how to complete the square of this equation : x^2+y^2+6x+2y+6=0

[marble meteor]

i havent done this type of math in a while so i forgot what to even do for this

[rustic hamlet]

Typing this out will be painful - give me a moment to write.

[light sonnet]

https://youtu.be/fWTEGeEtttI?t=193

[rustic hamlet]

Oh - or the video lol woops

[marble meteor]

wait but whats in the empty spaces

[rustic hamlet]

I was about to ask you - so I'm showing and not just doing.

Do you know how to complete the square?

[marble meteor]

i forgot , i haven't done this for like 2 years

[light sonnet]

It's the values that you plug in when you complete the square

[light sonnet]

Google is your friend

[rustic hamlet]

No worries - so, for the first blank - take half of 6, then square it.

[light sonnet]

Hence the video

[rustic hamlet]

What would that =?

[light sonnet]

Watch it

[rustic hamlet]

Yeah, that's a good idea. To brush up.

[marble meteor]

i found a video prior to asking but i couldnt figure out how they got the numbers in the blanks

[marble meteor]

9?

[rustic hamlet]

Then we'll go through it this way so you have a reference if that's helpful.

Right!

[light sonnet]

There are hundreds of videos

Try watching the one I linked, and try the math on your own

Post your work, and someone can check it

[rustic hamlet]

Ok, now for the second blank - you're going to take half of 2 and square it, and put it in the blank. Just like before.

What would go in that blank?

[marble meteor]

1

[rustic hamlet]

Great job.

[marble meteor]

thanks for the video

[rustic hamlet]

Math is balanced, and we just added a 9 and a 1 to the left side of this equation, right?

So, to keep it balanced, we must add a 9 and a 1 to the right side of the equation as well. That's what those blanks are for.

[marble meteor]

i see

[light sonnet]

I wouldn't phrase it like this tbh. Because one, you're spoonfeeding and two, the coefficients of x are always going to be 6 and 2, and three, you should let them try

Not do it for them

[marble meteor]

my brudda he is helping good

[rustic hamlet]

Okie doke - I'll back off. I just thought showing and then when they watch the video they'd have a reference to follow to know what steps to take.

[marble meteor]

man

[marble meteor]

PLEASE let him do what hes trynna do

[light sonnet]

Well, as you stated, you haven't done that math in like two years, hence the video, so watching that video could jog your memory on how to do it

[marble meteor]

i know but if i just have an example like this i can just refer to it

[light sonnet]

Then you can attempt it yourself with that video as assistance

The video has an example

[marble meteor]

aint no way

ok so is there anyway for you to let twentelise show me or is that off the table

[light sonnet]

Well, the purpose isn't to spoonfeed you and give you answers. You ought to try it on your own, and as I stated, that's what the video is for. If you watch it, it goes through that exact same process that twentelise is showing

[marble meteor]

right so the fact that i want twentelise to show me rather than the video has no effect on this

[light sonnet]

So you mentioned before, you haven't done that math in a while, that video is a way to help kickstart and help you recall that info. And as mentioned, the video and twentelise process is the same

[marble meteor]

why are you so good at avoiding my questions

[light sonnet]

You know, instead of the back and forth, you could have finished watching that part of the video

[marble meteor]

buddy

[light sonnet]

And tried it on your own by now

[marble meteor]

twentelise could have finished this whole problem with me by now

[light sonnet]

And you could have watched that video and attempted it on your own and finished it by now

[marble meteor]

why dont you want him to help me

[light sonnet]

the purpose isn't to spoonfeed you and give you answers. You ought to try it on your own

[marble meteor]

im not trying to get answers im just trying to understand the process

[light sonnet]

Look up resources if you forget certain processes

[marble meteor]

why would i be able to ask a question here if im instantly going to get shut down and told to watch a video

i dont like videos because i dont usually understand very well from them

so people like twentelise are my saviors

im tired now

[light sonnet]

So you're telling me that if you forgot the process of pythagorean theorem, you're going to go straight to a server instead of looking up some resources to jog your memory on pythagorean theorem?

[marble meteor]

yeah

thats why the math help section exists

doesnt matter what the question is

people are here to help and to answer

[light sonnet]

And google is there to help and answer too

[brisk obsidian]

Mind if I take over. This seems to have devolved into an argument. 😉

[marble meteor]

im just trying to understand why the person doesnt want me to ask a question here😭

its fine

im going to go now

[rustic hamlet]

I messaged you, potato. I didn't mean for it to go all nuclear here.

[marble meteor]

.close

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[elder sphinx]

Hi, i need help understanding market based analysis mathematically

[elder sphinx]

i don't understand why this is equal

This is from this chapter

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[green lily]

Can i get some help with b)ii)

[green lily]

i know that it is 2x+3y-z= ...

but i dont know how to find that constant

nvm i got it

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[tired perch]

Hello, when drawing a phase diagram of a linear system of differential equations, are we free to chose any variables we want for both axes ?

[fast peak]

well it should be relevant to the system

[tired perch]

It’s tripping me out

I’m studying the case of a system X’ = AX, A being a 2x2 matrix

Usually we construct the phase diagram using u1 and u2 as axes, both vectors being distinct eigenvectors of both eigenvalues

But in the case of A not being diagonalisable, we chose different axes

We chose z1 and z2 as axes, both of which being the coordinates of a vector Z that we considered being : Z = P-1 * X

My question is could we have drawn the phase diagram using u1 and u2 as axes ? U1 being an eigenvector and u2 a vector we whose such as u1 and u2 are linearly independent and P = (u1,u2) ?@fast peak

[fast peak]

well try it

see what you get

[tired perch]

I’m stuck haha

I get this if i impose conditions on the constants

So i tried to deferentiate it and the derivative is positive if we have this :

But i can’t quite visualise how to represent -u2/u1 on my diagram

[fast peak]

I thought u1 and u2 were supposed to be vectors

why are you dividing them

[tired perch]

True

Good point

Any tips ? @fast peak

[fast peak]

I am not exactly sure what you are trying to do here

drawing phase diagrams and trying to solve the system are two different things

[tired perch]

But instead of z1 and z2 using u1 and u2

[tired perch]

Yep my system is solved, we get this :

Never mind language barrier is killing me

Maybe you can help me with another question, here we are approximating our solution by a function

My question is why don’t we have ln(C3 * s) at the very end there ? @fast peak

[fast peak]

well I have no idea what is going on there

that said, remember log laws

ln(c*s)=ln(c)+ln(s) and then it would get merged with c1

[tired perch]

We’re simply approching w by a function

So w = f(s) and s = z

[tired perch]

I tried but surely if you use that you should end up with what we have plus a constant ? Lets say C3

[fast peak]

you would get s c1+s c2 ln(c3) + s c2 ln(s)

the first two terms are just s*constants

so just another constant c4

so s c4 + s c2 ln(s)

[tired perch]

Ohhh shit

Thank you man

Just couldn’t see that last part idk why

Ofc !

Thanks man take care ❤️ @fast peak

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[crude jackal]

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[tired perch]

Hello, in this scenario for a linear system of differential equations, does the value of a impact the phase diagram ?

[tired perch]

I find this for a = -2

And this for a = 2

(Axes are the same on both diagrams)

Can anyone check if my diagrams are valid ?

This would be the general solution to my particular problem fyi

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[tired perch]

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[slow cave]

List the side lengths of △FGH in order from smallest to largest, given that m∠H=58°, m∠F=54°, and m∠G=68°.

[scenic yew]

We can use the hinge theorem that states that the bigger the angle, the longer the length of the opposite side to the angle

Basically think of it like an actual door. The angle is the angle on how u open the door, and opposite side is the length of the opening in the door.

As you open the door larger, the angle opposite to the opening also get's larger

And because of how we use notation on triangles, we usually name the angle by using the name of the opposite side

that's why mF opposite side is also F just on how we usually name our sides and angles on a triangle

There you have it, now you can answer your question now

[slow cave]

Ok thanks

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[calm field]

is there a faster way of converting decimals to binary without using a table? like smth that could be quick to do on paper

[split sphinx]

there is

you divide the keep dividing the number by 2, and the remainder you get each time would be binary number. For example you have 57. 57/2 = 28 r1, 28/2 = 14 r0, 14/2 = 7 r0, 7/2 = 3 ** r1**, 3/2 = 1 r1, 1/2 = 0 r1

so you have the binary number of: 111001

hoping this explanation is understandable

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[manic nebula]

am I write in thinking the least upper bound here is n=3?

[torn jolt]

Yes

[manic nebula]

or is there no l.u.b because sqrt(10) is not a positive integer

[torn jolt]

Root 10 is not an integer

3 is

[manic nebula]

ok

I was kind of thinking of this example

you can't say sqrt(2) is the l.u.b because obviously it's not rational

[heavy wolf]

Yes the difference is the domain

[manic nebula]

but there's no l.u.b at all because we keep approaching sqrt(2) and thus can't define the l.u.b

[heavy wolf]

Yep the domain is key here

[manic nebula]

ok, thanks. get it

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[full forum]

@fluid prism Has your question been resolved?

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@fluid prism Has your question been resolved?

[onyx inlet]

@fluid prism

What on earth are you doing

Use quadratic formula not factorize

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[full forum]

Please don't occupy multiple help channels.

[gaunt nest]

Sum of interior angles = (n-2) x 180

So for heptagon exterior angles would be 7 x 360 - (7-2) x 180

[dense egret]

thank u

[warm abyss]

Don't give out answers

[dense egret]

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[torn jolt]

I'm confused here. Shouldn't they have found g'(x) and then g'(fx)?

[gritty rose]

why do you think that?

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[inner mango]

Any hints on how I can proof that for arbitrary polynomials p1,p2,p3 that (1-1/p1(n))^p2(n)<1/p3(n) for n large enough?

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[void arch]

I was wondering if n=1 would this still hold?

[ivory cairn]

if n =1 then S_n is the trivial group. There's only one element in it

[void arch]

Thanks for verifying haha, I got doubtful for a second

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[wanton mesa]

how do i know to use the sum of arithmatic sequence formula here instead of geometric?

[polar valve]

how would you use a "geometric" thing?

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@wanton mesa Has your question been resolved?

[wanton mesa]

Sum of geometric series

[polar valve]

do you have a geometric series?

[wanton mesa]

Yes?

Wait no

Because p =1

But if p is for example =2 it would be ageometric series wouldnt it

[polar valve]

write the first 5 or 10 a_n and decide if it is or not. but your question was for 4,1,4,1,4,1,4, ...

But it isnt relevant if it is or not. you should calculate the sum 4.1.4.1.... and i think it doesnt matter which way you use to get this sum.

You should choose your prefered way.

[wanton mesa]

Ah ok thanks

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[warped gazelle]

If 5 persons can do 5 jobs in 5 days then 50 persons can do 50 jobs in how many days?

[warped gazelle]

I don't have a question about the problem

But what topic are these types of question considered in?

<@&286206848099549185>

[opaque quartz]

Ratio and Proportion

[warped gazelle]

oh ok ty

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[hollow moth]

is this a sufficient proof:

Base case: If there is only 1 container, it must contain all m items, and m > 1.

Assumption: Assume that if m items are put into k containers with m > k, k = 1,2,3,...n, at least one container must contain more than one item.

Induction step: consider n+1 containers. Case 1: the (n+1)th container contains no items:
using the assumption, we know that one of the n containers must contain more than 1 item.

Case 2, the (n+1) container contains a ball:
if we have m > n+1, we also have that m-1 > n. Thus we can apply the inductive hypothesis and say that one of the containers belonging to n must contain more than 1 item.

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@hollow moth Has your question been resolved?

[hollow moth]

<@&286206848099549185>

[hollow moth]

<@&286206848099549185>

[hollow moth]

<@&286206848099549185>

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@hollow moth Has your question been resolved?

[hollow moth]

<@&286206848099549185>

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@hollow moth Has your question been resolved?

[velvet marlin]

ngl my induction is shit, but are you not supposed to prove n first?

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[hoary hatch]

Having quite a bit of trouble with these questions
If anyone could help me, that would be great

[onyx glen]

!status

[full forum]

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

[hoary hatch]

Oh

Need confirmation on how to do the first one
No idea how to do the second one

For the first one, is the root (x-6i)?
And is -6i expanded like any other variable?

Wait
Don't I also need the conjugate of (x-6i) to be one of the other 2 roots?

So (x+6i)(x-6i)(2x+3) are all the roots?

And I still don't know how to do the second one

[full forum]

@hoary hatch Has your question been resolved?

[hoary hatch]

i think i solved the first one
but i'm still yet to figure this one out, and i'm quite lost

[earnest perch]

look at the discriminant

if the discriminant is negative, it only has complex roots

[hoary hatch]

b²-4ac in relation to 0

[hoary hatch]

oh
i didn't know that

i thought it only applied to where the parabola was on the cartesian plane giving you positive/negative

oh

[earnest perch]

it determines how many real roots the polynomial has

[hoary hatch]

the standard cartesian plane is replaced with the real/imaginary plane

wait
but that doesn't??

[hoary hatch]

why is this the case?

[earnest perch]

were not really doing that

[hoary hatch]

no?

okay

[earnest perch]

so when u have a second degree polynomial and you want to get its roots, you use the formula

the formula has a sqrt(b^2-4ac)

[hoary hatch]

you mean x=-b+-...?

[earnest perch]

so if b^2-4ac happens to be negative, the roots must be complex numbers

[hoary hatch]

🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🫣 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 🍌 <- Tap

[earnest perch]

yes

[earnest perch]

lol

[elfin stream]

Btw quick question, k is 72 in the first one?

[hoary hatch]

i don't think it matters what k is, does it?

[elfin stream]

Probably doesnt, but I was wondering, so asked

[earnest perch]

wait did your third root involve k?

[hoary hatch]

i thought that the first 2 roots would be conjugates of one another, giving me (x∓6i), and using the last value, which is 108, i thought it would be like (x+3) or something

was that incorrect?

[earnest perch]

yea so (x-6i) and (x+6i) are roots

but to figure out the third one

you gotta divide the whole thing by (x-6i)(x+6i) = x^2 + 36

[hoary hatch]

d in a standard cubic will always be the product of the last term in all 3 roots

and d is 108

[earnest perch]

o

[hoary hatch]

so the final one has to be 3, right?

or -3, rather

because i^2=-1

[earnest perch]

yeah that makes sense

[hoary hatch]

or
wait

-i*i

-i^2

1

[earnest perch]

3 because (x-6i)(x+6i) = x^2 + 36

[hoary hatch]

so 108/6^2

which is 3

[elfin stream]

-36iz = 108
iz = -3
z = -3/sqrt(-1)
z = 3i

No?

(z is the last root)

[hoary hatch]

you squared the 6, but not the i??

[earnest perch]

but the leading coefficient is 2 so (x-6i)(x+6i)(x-3) doesn’t work

[elfin stream]

Oh oops 😅

[hoary hatch]

oh
right
i did think of that

so i just put 2 before (x+3)

so (2x+3)

[hoary hatch]

is that just a rule, or is there an explanation to it?

[earnest perch]

the formula for solving second degree equations has a sqrt(b^2-4ac),
so if b^2-4ac happens to be negative, the roots must be complex numbers

[hoary hatch]

oh

i see

but that's the quadratic equation

why does it apply to a cubic in that way?

oh

[earnest perch]

the question 2 isnt a cubic

[hoary hatch]

that's not a cubic

[elfin stream]

Because a cubic can be factorized in the form of two expressions:
[x-(-x)][x^2 + bx + c]

[hoary hatch]

yeah, i just came to that conclusion myself

[elfin stream]

Wait which r we on?

[earnest perch]

#2 right now

[elfin stream]

Oh

Oops i was on 1

[hoary hatch]

i made an oopsie

[earnest perch]

but i have a feeling #1 is missing something

where did that k vanish

[hoary hatch]

we can always ping helper to see if we're missing something

or just expand it

[earnest perch]

ima try doing long division

[hoary hatch]

we have 3 potential roots and an equation
i'm going to expand and see if i got it right

the first and last terms are right

and the second one, too

i did it

and x=72

@earnest perch@elfin stream

thanks for your help today

i'll be doing physics now

bye bye

.close

[earnest perch]

i think the coefficient of x isn’t necessarily 72

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[hoary hatch]

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✅

[earnest perch]

like the last root has to be in terms of k

k

[hoary hatch]

why's that?

[earnest perch]

otherwise we get 72x, not kx

and k isnt necessarily equal to 72

[hoary hatch]

what

but i have all the roots

and i expanded the roots

what is left to be there but the full cubic?

@earnest perch

and if part of the cubic, like the x term, is missing, then is it not whatever was given in the cubic equation?

did i miss something?

[earnest perch]

like what if k was different

wouldnt the third root be different too

[hoary hatch]

it can't be anything else, though

[earnest perch]

oh how do you know

[hoary hatch]

because i have all 3 roots

and expanding them gives the coefficient, 72, for x

[earnest perch]

o yea actually you’re right

ye sorry about that

[hoary hatch]

that's fine

we all make oopsies

imma gonna do physics now

bye bye

[earnest perch]

aight cya

[hoary hatch]

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[wild owl]

Hello, does anyone know how to solve the 3rd one ? Thanks before

[earnest perch]

to calculate it, get the point where the line and the curve contact (which looks like its at about x=1.6)
then the shaded area is equal to the integral from 0 to 1.6 of the blue line’s function minus the integral from 1 to 1.6 of f(x)

does it make sense why that is?

[wild owl]

I see

[wild owl]

Thanks for the explanation. i'll try that

[earnest perch]

alr, and 1.6 is not the actual value

u gotta find it too

you’re welcome

[wild owl]

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[lone lake]

What are the prerequisites for learning calculus efficiently?

[hot herald]

algebra
trig / geometry
functions

exponents/logs

[lone lake]

thanks.

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[opaque quartz]

This question is about operators in Quantum Mechanics and linearity:

Are there theorems that deduce whether certain combinations of operators are linear or nonlinear?

e.g, consider operators Ai and Bi where all Ai are linear, all Bi are nonlinear

Can you say irregardless of what A and B are
-If A+B Is Linear/Nonlinear?
-If A-B is Linear/Nonlinear?

-If A1+A2 Is Linear/Nonlinear?
-If A1-A2 is Linear/Nonlinear?

-If B1+B2 is Linear/Nonlinear?
-If B1-B2 is Linear/Nonlinear?

[opaque quartz]

Or will it always be a case-by-case, depending on which linear and nonlinear operators you have?

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@opaque quartz Has your question been resolved?

[opaque quartz]

furthermore, actually, what actually -is- the combination of operators A + B

e.g if A=dy/dx and B=(d^2y/dx^2)

what is (A+B)(f(x))

[full forum]

@opaque quartz Has your question been resolved?

[coarse patio]

the linearity of a combination of operators depends on the specific operators involved and how they are combined

[opaque quartz]

so there is no generalization of
linear + linear = linear

and so on so forth

[coarse patio]

first, the sum or difference of two linear operators is always linear. This means that if Ai is linear and Bi is linear, then (Ai ± Bi) is also linear

also

the sum or difference of a linear and a nonlinear operator is generally nonlinear. This means that if Ai is linear and Bi is nonlinear, then (Ai ± Bi) is generally nonlinear

example

[opaque quartz]

generally, meaning this cannot be used as definitive proof?

[coarse patio]

consider the case where A = dy/dx (a linear operator) and B = d^2y/dx^2 (a nonlinear operator). Then, (A+B)(f(x)) = dy/dx (f(x)) + d^2y/dx^2 (f(x)) is a nonlinear combination

[coarse patio]

yes

[opaque quartz]

👍 i did figure nonlinear + linear would tend to be nonlinear and linear + linear be linear

i was more wondering if there was anything definitive to that

[coarse patio]

There is no definitive rule or theorem that can be used to determine the linearity of a combination of operators without analyzing the specific operators involved and how they are combined. The linearity of a combination of operators depends on the individual operators and their algebraic properties, so it is not possible to make a general statement that applies to all cases

but

there are some general principles, such as the linearity of the sum or difference of linear operators, that can be helpful in determining the linearity of certain combinations of operators. so though, determining the linearity of a combination of operators requires careful analysis of the individual operators and how they are combined.

[opaque quartz]

that makes sense, checking for specific cases doesn't seem too hard anyway

that aside, if possible can i ask a question in the same field, about wavefunctions and probability?

[coarse patio]

what

[opaque quartz]

what as in 'what you have said is just incomprehensible to me' or are you asking for the question

[coarse patio]

whats the question

[opaque quartz]

this defines the probability of a tise in the region between a and b

does psi have to be normalised in order to compute this

or does it not matter

[coarse patio]

the wave function psi(x) does not necessarily need to be normalized

[opaque quartz]

so normalisation is only needed for if we're computing expected values?

[coarse patio]

its necessary for any computation involving probabilities or expected values in a quantum system

[opaque quartz]

is this for psi(x) specifically?

as in, what you are saying would not apply to a tdse, psi(x,t)?

[coarse patio]

normalization condition applies to any wave function in quantum mechanics, whether it is time-independent (psi(x)) or time-dependent (psi(x,t))

for a time-independent wave function psi(x), the normalization condition is expressed as ∫|psi(x)|^2 dx = 1

this means that the probability density function |psi(x)|^2 must integrate to 1 over the entire range of x

[opaque quartz]

ah, i see my confusion

i read this as 'unnecessary' and blanked into 'okay, so what is it for'

that makes more sense

[coarse patio]

👍

[opaque quartz]

thanks for the help! should be good for now

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[spark valve]

what is geometric reaosn why this has rank 2?
2 −4 10
2 3 −4
4 2 0

[gritty rose]

find the dimension of the image

[spark valve]

last 2 columns can create the first one

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[outer pasture]

could someone help me with trigonometry identities?

[outer pasture]

$\frac{cos(A)}{1+sin(A)}+\frac{1+sin(A)}{cos(A)}=2sec(A)$

[glossy valve]

Astassa

[sweet tangle]

Do you need to prove that this is true?

[outer pasture]

yeah

[sweet tangle]

start off by turning the left hand side into just one fraction

[outer pasture]

i got this
$\frac{1+cos^{2}(A)+2sin(A)+sin^{2}(A)}{cos(A)+sin(A)cos(A)}$

[glossy valve]

Astassa

[sweet tangle]

yeah thats right, but for now leave the denominator as cosA(1+sinA)

then then top can be simplified as cos^2(A)+sin^2(A)=1

[outer pasture]

im not sure i get it

that would leave 2+2sin(A) on top right?

[sweet tangle]

yeah exactly

$\frac{2+2sin(A)}{cos(A)(1+sin(A))}$

[glossy valve]

SVRNHead

[sweet tangle]

and can you see a common factor in the top and bottom that can be cancelled?

[outer pasture]

1+sin(A)?

2

[sweet tangle]

yeah exactly

which gives 2/cosA

which is by 2secA

[outer pasture]

awesome, thank you!

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[hollow moth]

i just dont understand this problem at all

(1+sqrt(5))/2 is 1.618, so am i proving that any fibonacci number is greater than 1.618?

[torn jolt]

you are proving the nth Fibonacci number is greater than phi^(n-2)

but yes, phi is just a number that's approximately 1.618

[hollow moth]

and n is greater than or equal to 2