help-19

[latent scaffold]

You're missing a minus sign in there

[steep hare]

hmmm

4-3000/x^2?

now im sensing we set this to 0

[latent scaffold]

Indeed.

[steep hare]

soo
4-3000/x^2=0

4x^2=3000

x^2=750

x ≈27.39

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[nova ore]

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✅

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[hollow lantern]

Hi so I'm working through vellemans how to prove it, and I've hit a snag

[hollow lantern]

#4

I try and do what weve done so far and expand out from the outside in

So I go off and find this identity

[boreal crag]

that doesn't look like a self-contained statement

[hollow lantern]

But already by doing so I'm doing it wrong

Apperntly the union of let's call it p is oh onope don't use the definition we gave you, you should know, despite this is us teaching you set theory from the ground up its the power series

So is there a way to get from the original through definition of union through definition of power series etc, cause I seem to have excess sets I don't know how to get rid of

My working out so far

Sorry your right

[odd edge]

@hollow lantern Has your question been resolved?

[hollow lantern]

Newby to set theory, any help apreciated

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[sick mulch]

hi can someone help with this question?

[lunar hedge]

page 73

[sick mulch]

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[white glen]

Hi! I'm just wondering if I'm on the right path for this exercise? I ended up with 1/2 being the distribution for L but I'm not really sure if I'm even on the right path

[odd edge]

@white glen Has your question been resolved?

[white glen]

<@&286206848099549185>

[hardy panther]

...

I am pretty sure

that you need to find the volume of this solid

[white glen]

Why is it in R^3?? Isn't it R^2?

[hardy panther]

f is a function of two variables, that's 3 dimensions

what you are calculating is the volume of this solid

Harry can arrive at any time from 0 to 60 minutes.
Ron can arrive at any time from 0 to 60 minutes

hmm

actually I think I may be doing something wrong here because I was acting like there were only 3600 combinations of H and R

but there are infinitely many

[white glen]

because its continuous right?

[hardy panther]

yes.

[white glen]

but youre sure my initial calculation was wrong right?

[hardy panther]

I didn't really scrutinize it

lemme see

[white glen]

ok!! thanks

[hardy panther]

R-H looks right because Ron has to arrive later than Harry

However, I think Ron's lower limit of integration should be Harry's time

since we clamp the function f(R,H) at zero when R<H anyway

moreover, it looks like you are integrating the first integral w.r.t. Ron and your upper limit is...Ron.

[white glen]

ok so u suppose we should have the integral for ron be from h to 60 and harrys interval from 0 to r?

Like this?

[white glen]

oops😔

[hardy panther]

I think it should be this

$\int_0^{60} \int_h^{60} r-h \quad \text{d} r\text{d} h$

[clever fjord]

Melvin Eugene Punymier

[hardy panther]

now...what I would expect though

is multiplying each outcome by its probability

but every outcome is equally likely

and we have infinitely many points

and I'm just rusty with this

so how do we take care of that

[white glen]

we keep getting 1/2 😭

[hardy panther]

what do your notes say

[white glen]

we intergrate the joint pdf right? not r-h cause we need to find the distrubution when l = r-h

[hardy panther]

oh...

I think we just have to divide by the area

because that's the full range of H times the full range of R?

...which would be division by 3600

but does that come out wrong?

do you have the answer?

[white glen]

So we tried that, which resulted in it equal to 1/2

[white glen]

No😔

But if most of our steps were right, we might think our answer might be correct. Just checking if our steps werent wrong

[hardy panther]

I got 20 minutes

$\frac{1}{3600}\int_0^{60} \int_h^{60} r-h \quad \text{d} r\text{d} h$

[clever fjord]

Melvin Eugene Punymier

[white glen]

Oh like that

No we calculated $f_{x,y}(x,y)$, which was $\frac{1}{3600}$ and put that in the space you originally put r-h in

[clever fjord]

aer

[hardy panther]

Mrm...

[white glen]

^^because of this

[hardy panther]

so you think the pdf is constant?

[white glen]

Yea we kinda thought that didnt make sense either but its a constant if r-h is between 0 and 60 anythjng before is 0 and after is 1

[hardy panther]

before what and after what

I think we can write this in a way such that the probability of any time for Harry and Ron is constant, but we need the outcomes involved (R-H)

99% sure

[white glen]

So everything lower than 0 is 0, and everything after 60 is 1

[hardy panther]

the hint says that the probabilty that L = R-H is is zero, is nonzero

in fact, since we clamped all negative values to 0, that would make P(L=0) = 1/2

[white glen]

Yes

So then the probability for everything else is also 1/2???

[hardy panther]

the probability that L>0 is 1/2

I may not have even calculated what they were asking you for

pft! They didn't even MENTION "expected value"

ok so you do have a uniform distribution 😅

[white glen]

Yes😭😭Expected values are for our next chaoter so its nothing to do with that

[white glen]

For the L?

[hardy panther]

what should your limits of integration be then

🤷‍♂️

you know what would be really cool?

an example.

[white glen]

😭

[hardy panther]

you got formulas but no examples?

hang on I got a stats book right here

ok I agree with you

because double-integrating the joint pdf should get you 1

and that would happen if you were doing

$\int_0^{60} \int_0^{60} \frac{1}{3600} \text{d}r \text{d}h$

[clever fjord]

Melvin Eugene Punymier

[hardy panther]

so your joint pdf is just 1/3600

integrating the joint pdf across the infinite range of values for each variable is supposed to always get you 1

so you would say that

$f_{RH}(r,h) = \frac{1}{3600}$

@white glen

[clever fjord]

Melvin Eugene Punymier

[odd edge]

@white glen Has your question been resolved?

[white glen]

Hi im sorry im in a lecture rn😭i'll reply later if youre still there

[hardy panther]

confirm with your professor and let me know what they say about this one 👍

[white glen]

Yes!! We also had this.

[white glen]

Im not in a lecture for the class for this question 😔😔the answer to this question will be revealed next week unfortunately. I'll update you on it if you want I guess

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[junior leaf]

I have a grid that grows with this pattern and, given two numbers, I want to sum all the numbers inside the rectangle that is formed by them

[junior leaf]

As an example, 13 51 would return 358

Restrictions are

1 <= A <= B <= 10^9 Where A and B are the corners of the rectangle
1 <= Rows <= Cols <= 10000
1 <= Answer <= 10^18

don't know how to go about this because of the big number restrictions
it needs to run in under a second too

tried finding a formula but came up with nothing good

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[steel raven]

.reopen

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[steel raven]

this seems fun

you're programming this i assume?

oeis might be useful. for the row starting with 4 you can use this sequence (which has a formula) https://oeis.org/search?q=4%2C+8%2C+13%2C+19%2C+26%2C+34%2C+43&language=english&go=Search

similarly https://oeis.org/search?q=7%2C+12%2C+18%2C+25%2C+33%2C+42%2C+52%2C+63&language=english&go=Search

the references include a diagram here (supposedly might have errors). the formula for each row seems to have a pattern you can derive in general.

oh it's transposed but that shouldn't matter

[ornate belfry]

make a 2d array where you calc the cumsum for ind i,j

then for any rectangle with upper corner (x_1, y_1) and lower corner (x_2, y_2)

do cumsum(x_2, y_2) -[ cumsum(x_2, y_1) + cumsum(x_1,y_2) - cumsum(x_1, y_1)]

[junior leaf]

I plugged the next line in and got something very similiar, but that -2 got me really confused

[steel raven]

which -2 ?

[junior leaf]

I don't know, making -in the worst case- a 10000x10000 array sounds very inneficient

[junior leaf]

(n^2 + 5*n - 2)/2.

I was looking at the
n(n + 5) / 2 pattern that emmerges, but the -2 got rid of it

[ornate belfry]

Ye I assumed for any element in array

But if u have this specific configuration

U can prob do o(1)

[odd edge]

@junior leaf Has your question been resolved?

[steel raven]

i see what you mean now. this should continue the pattern tho
(n^2 + 5*n - 2)/2 = n*(n+5)/2 -1

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[vast sigil]

!help

[odd edge]

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[vast sigil]

How would I solve this, I have worked out that the hypotenuse of triangle is 9

<@&286206848099549185>

[proven harbor]

Just named the figure O, A, B, D

can you find OA?

@vast sigil

[vast sigil]

Yeah, OA is 2.4 cm right

[soft solstice]

You mean 3?

[proven harbor]

Now do you know that all radius are equal?

[vast sigil]

I did square root of 1.8^2 + 2.4^2

[soft solstice]

That is 3

[vast sigil]

is oa a radius

[proven harbor]

1.8² + 2.4² = 2.4² ?

[proven harbor]

Tell me what radius is?

[vast sigil]

Radius is distance from center of a circle to any point on the circumference

but i dont know where the center is

[proven harbor]

The center is O

I think they forgot to mention it

[vast sigil]

yeah i was wondering too

so if center is o then radius is 3 cm

which means d is 3 cm?

[proven harbor]

Yup

[vast sigil]

Sounds good

And for this one.

I'll try solving it then coulld you tell me if i went wrong anywhere

[vast sigil]

So I got that 12 cm was the missing side of the triangle using pythagoras, then if the radius is 20cm the diameter of the circle is 40cm, and if we're subtracting 40 from the radius + the missing side which is 12cm thats 40 - 32 = 8 so d is = 8

[proven harbor]

Correct

Nice

[vast sigil]

Thanks bro means a lot

[proven harbor]

Btw you don't need diameter

Just do OC - OD

To get CD

[vast sigil]

How would I know what OC is though

wait nvm

[proven harbor]

OC is radius lol

[vast sigil]

20 - 12 *

Yeah fair enough

Alright thats me finished thanks for helping take care 🫡

[proven harbor]

Wlcmm

Same bro

! Done

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[thick fiber]

K)

am I on the right track

that equation I got just seems really ugly

[pastel dew]

hey marito!

[thick fiber]

hi!

[pastel dew]

I'm working out currently , so I m unable to help you 😭

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[junior ether]

guys im trying to do part d

the plot for part c is this and i got the volume to be 17.2794 about

the plot for part d is this and idk what is the right bounds for t to make it so it doesnt double count any volume

like i tried the upper limit of t as 2pi and it gave a very small value like .61

which isnt right cuz its almost the same shape as part c so it should have a volume of near 17

[odd edge]

@junior ether Has your question been resolved?

[olive needle]

The shapes shouldn't be the same size. The latter should be smaller.

[odd edge]

@junior ether Has your question been resolved?

[junior ether]

wait why does the plot extend to like -3 to 2 on the y axis tho then

like the t values range from 0 to 4pi

[olive needle]

My bad. I didn't change the interval to 4pi.

[junior ether]

so why tf am i getting such a small volume

hold on

ima show u my code for it

this is for part c\

[junior ether]

this is for part d

there is a tip that says
Watch out, because as t runs from 0 to 4 π, this shell wraps around twice.
If you integrate with respect to t from t = 0 to t = 4π, your answer will be off.

the only differnece between part c and d is that the denominator in radial[s] is 16 not 8

and then the bounds are from 0 to 4pi

instead of 2pi

[junior ether]

dyk whats wrong

[junior ether]

.close

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[outer wadi]

Bit confused on this problem, not sure why my simulation is getting -269

[outer wadi]

If we have mean = 2

then the rate is 1/2 right

oh

wrong variable

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[timber elk]

when finding the area between two curves how do i know hich function to subtract by which in the definite integral?

[amber schooner]

upper - lower

[timber elk]

so I need to graph them?

[patent hazel]

not necessarily

[amber schooner]

always a good idea

[patent hazel]

u can just use FTOC

[timber elk]

what is that?

[patent hazel]

fundamental theorem of calc

calc is short for calculator

[timber elk]

how so?

[amber schooner]

joseph, i’m not so sure you have any idea what you’re talking about

[timber elk]

😭

[patent hazel]

no i know it

math is just my 5th language

[timber elk]

what?

[amber schooner]

we use the fundamental theorem of calculus to evaluate definite integrals regardless of whether or not the problem refers to finding the area between two curves

[amber schooner]

ignore him, he’s clueless

[timber elk]

ok, so i should graph them just to be sure

[amber schooner]

yea

and find the intersection of the curves

[patent hazel]

sorta what i said

just in simpler terms ig

graph, find what function is on top and which is on the botton in the interval (ie. top function has larger y value)

[bitter folio]

W calculator

FTC is fundamental theorem of calculus

[amber schooner]

bro you’re yapping

[bitter folio]

fr

[patent hazel]

and then do the integral from a to b of top function - bottom function

[amber schooner]

man didn’t say this shit

😹😹

[timber elk]

i have no idea how FTC has to do with this bro

[patent hazel]

math majors try not to be condescending challenge

[bitter folio]

thats how to integrate

[patent hazel]

it doesnt lol

[amber schooner]

$\int_a^b f(x) dx = F(b) - F(a)$

[clever fjord]

knief

[patent hazel]

just use a calc

[amber schooner]

bro you’re trolling in a help channel?

[bitter folio]

💀

[bitter folio]

im not even in college

[amber schooner]

they’re trying to learn sir

[patent hazel]

im teaching them

[amber schooner]

💀

saying just use a calculator isn’t teaching them bruh

[patent hazel]

btw if the functions cross, u split the integral at the intersection and then recalculate for each segment

o also for horizontal slicing, do the indefinite integral of the right function - left function instead

[rotund nebula]

yo can someone help me next

[patent hazel]

@amber schooner got u

[timber elk]

ok thanks @amber schooner

[bitter folio]

advertising frfr

[rotund nebula]

been waiting 5 hours

[amber schooner]

you’re welcome

[timber elk]

5 hours is crazy

[rotund nebula]

everyone else would rather solve x+1

or 5/5

[patent hazel]

ok what is ur question

💀

[mystic saffron]

https://tenor.com/view/skull-gif-25709972

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[odd edge]

Please don't occupy multiple help channels.

@kindred solar Has your question been resolved?

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[leaden arrow]

idk if this is the right place to ask but is there any easier way to remember these that’s maybe more intuitive or do i just need to make a quizlet deck lol

[leaden arrow]

they’re all just really similar and it confuses me

[wanton bison]

maybe it helps if you derive them on your own

[forest sky]

well you can note that the "co" functions have the same derivative as their corresponding functions up to a sign difference

[leaden arrow]

yeah i noticed that

i guess i just don’t understand where the absolute values and square roots and stuff actually come from

heavy on the absolute values

the other ones make pretty ok sense

[leaden arrow]

i will try this

what would i set arcsinu equal to?

[forest sky]

also the u' here is just from the chain rule so you don't really need to include it

[leaden arrow]

i might just write them all out like 20 times i think they’d be stuck in my brain then

seems like i just need to study more lol thanks for the advice

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[mystic saffron]

not sure where to start this problem.

[warped glacier]

SOHCAHTOA

you want to find the (downward) height ofc

you need one more side

the speed of the airplane and the time give you the distance
what kind of distance is this? (which direction)

[mystic saffron]

oh wait i think i understand

would i do the 400 mph times sin(3) first?

[warped glacier]

cool, when you're ready write out the equation

[warped glacier]

no, you're not using 400

distance = what * what?

or if you're stuck, rearrange speed = distance/time

also don't get tricked, 15 mins is 1/4 hour

[lapis terrace]

I feel like <ABE is wrong, can someone try?

[warped glacier]

!help

[odd edge]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[mystic saffron]

this is the work I had done im not sure if its correct though

[warped glacier]

weird, I don't see 400 * 1/4 = 100 miles in your work

[mystic saffron]

i actually just tried solving it in a different way using 400 times 1/4 and i ended up getting the same answer

[warped glacier]

basically opp/adj = tan(3 deg)

so opp = 100 tan(3 deg)

,w 6.8 - 100 tan(3 deg)

[clever fjord]

Results provided by WolframAlpha

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

Requested by sooouth.

[warped glacier]

yeah you seem to have done that

I just couldn't tell cause your work looks super confusing

without any mention of the horizontal distance or a calculation of that

without seeing tan 3 degrees anywhere

without seeing tan = opp/adj

you get the idea

[mystic saffron]

what makes it confusing do you think? Do you think it would be better to rework the equation similar to how you did it instead so my instructor would have a better understanding of how I got my answer?

[warped glacier]

yes

it's about how you communicate using mathematics

you should write down things so that they relate to parts in the question

so that other people can understand what you are doing

you don't have to write particularly much

$d = st = 400 \cdot \frac{1}{4} = 100$

horizontal distance = adj = 100

$\tan 3^{\circ} = \frac{opp}{adj} \implies opp = 100 tan 3^{\circ}$

hence height = $6.8 - 100 \tan 3^{\circ} = 1.559 \cdots \approx 1.6 \text{ mi}$

[clever fjord]

south's secret twin brother

[warped glacier]

this is much much clearer

you don't need to write every single detail and clarification

but see how this is much more understandable?

notice how I used words such as horizontal distance and height to signpost the calculations I'm doing

and abbreviations such as d = st and adj are understandable

given the context of speed/distance/time and trig

actually altitude instead of height would be slightly better, as height could be the height of the right triangle

it's not 100% perfect but miles better than what you wrote

[mystic saffron]

okay yes that makes alot more sense thanks alot

I can see how my work could have been a bit unclear for sure.

I have a couple more different questions if thats alright

[odd edge]

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[ocean lantern]

isnt it supposed to be r cuz r^3-r^2 should give me r

[feral scaffold]

on the top, it's r^3 * r^3, not just r^3

[ocean lantern]

yes but if I factor out r^3

[feral scaffold]

can you show what that'll look like?

[feral scaffold]

you can't do that
they're multiplied with each other

[ocean lantern]

well if they're multiplied to each other shouldnt I be able to take it out

actually if thats not possible then ig they did it this way

[feral scaffold]

if you aren't convinced, let's replace them with numbers
so you're kinda saying 2*3*2*4 = 2*(3*4) which isn't true

op discord formatting, hold on

[ocean lantern]

ok

[feral scaffold]

yea that one's right

you can only factor things out this way if it's addition
so if it's r³sin³θ + r³cos³θ, then you're able to factor it to r³(sin³θ+cos³θ)
since it's being multiplied here, you can't do this

[ocean lantern]

hmm ok

thanks

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[feral scaffold]

np

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[broken oyster]

Hello for part c) i only found 6 roots which are

[broken oyster]

these

idk whr the last two come from

[quasi sparrow]

What last two are you talking about

,w roots z^8-(5/2)z^4+1

[clever fjord]

Results provided by WolframAlpha

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

Requested by rie.mann

[quasi sparrow]

Oh you just use part a) again

[broken oyster]

Huh? I already used all of part a?

[quasi sparrow]

Use it on another root

[broken oyster]

ohh I see

[broken oyster]

also for part b) i only knew that one of the roots was 4sqrt 2 cus i got the answers but when i solved it i got P(z)=1/2(2z^4-1)(z^4-2)=0 and I'm wondering what happened to the 2z^4-1 part

[quasi sparrow]

Did you typo your factoring

[broken oyster]

uh probably lemme check

[broken oyster]

nope

ok wait

ye idk

oh crap i did

theres meant to be a 1/2 at the front

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[broken oyster]

<@&286206848099549185>

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[rose cedar]

this task was done in my lecture.. and i have no clue how we passed the first line... i wrote it down exactly how it was written on the board. id appreciate if someone could explain how im even supposed to start solving stuff like this:

[rose cedar]

<@&286206848099549185>

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@rose cedar Has your question been resolved?

[river spindle]

$\lim_{t\to\infty} (1+\frac{1}{t})^t = e$

[clever fjord]

viiviiiix

[rose cedar]

ye i get how the last part works... idk how to get to it

[river spindle]

We convert the problem in such a way that we can use this limit.

We write denominator in the numerator as is then we add remaining terms in numerator and substract what wasn't there.
like
$\frac{x^3+2x^2+1}{x^3+x-7}$
We write
$\frac{(x^3+x-7)+(2x^2-x+1+7)}{x^3+x-7}$

[clever fjord]

viiviiiix

[river spindle]

then we seperate the terms to get 1 + .. form

[rose cedar]

i see

okay thanks!

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[hollow frigate]

Hi I'm stuck on question b)-b), I have proved the inequality in b)-a) and plugged p/q into it. From that I found that:

q*U_n < p < q*V_n

But that's pretty much it, I don't really know how to move forward from there.

[hollow frigate]

(Btw I could find that because I proved that
n!U_n + 1/n = n!V_n)

[odd edge]

@hollow frigate Has your question been resolved?

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@hollow frigate Has your question been resolved?

[hollow frigate]

<@&286206848099549185>

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@hollow frigate Has your question been resolved?

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@hollow frigate Has your question been resolved?

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@hollow frigate Has your question been resolved?

[mystic saffron]

What is n.n!

[mystic saffron]

As you increase n the inequality stays true right?

However the lhs and rhs of the inequality are equal when n is large enough as per question a?

What can you say about this as relates to contradictions

[odd edge]

@hollow frigate Has your question been resolved?

[hollow frigate]

I'm not sure if that's what you're asking but it's because
1/n = n!/(n!*n)

[hollow frigate]

Yeah when n tends to infinity but with limits strict inequalities become non strict inequalities right?

[odd edge]

@hollow frigate Has your question been resolved?

[mystic saffron]

Yes my bad sorry to mislead you

[hollow frigate]

It's okay I also thought about it

[prime basalt]

Note that in b)-a) when n is large enough n!e becomes an integer. For example when n=q, we have n!e = (q-1)!p.
On the other hand, n!u_n = integer + 1/n. Try reaching a contradiction from here using b)-a)

[hollow frigate]

What do you mean by
"n!u_n = integer + 1/n"?

Oh wait

[prime basalt]

n!u_n = n!(1 + 1/2! + 1/3! + ... + 1/n!) + 1/n

[hollow frigate]

n!U_n is an integer wow I did not recognize that

[hollow frigate]

Oh you mean V_n right?

Anyway I think I got it, thank you @prime basalt, note to self: true for all n means you can choose any value of n that you want to find a contradiction

[prime basalt]

Oh I misread the task

Yeah n!u_n is an integer

[hollow frigate]

Yeah it's fine, thanks again

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[cobalt heath]

hi

[shadow berry]

Hello

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[rich tusk]

Hey hey, I need some help attemtping to connect the following equality to L functions:
\sum_{k=-\infty}^\infty \frac{1}{(a + kq)^{2s}} = \sum_{\substack{n \equiv a \pmod{q}}} \frac{1}{n^{2s}}. where n runs through the integers

[rich tusk]

$\sum_{k=-\infty}^\infty \frac{1}{(a + kq)^{2s}} = \sum_{\substack{n \equiv a \pmod{q}}} \frac{1}{n^{2s}}$

[clever fjord]

M47R1XAdV#9685

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@rich tusk Has your question been resolved?

[rich tusk]

no

<@&286206848099549185>

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[rich tusk]

.reopen

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✅

[quasi sparrow]

Do you know what $n \equiv a$ mod $q$ means?

[clever fjord]

riemann

[rich tusk]

yes

[quasi sparrow]

Can you write out what it means

[rich tusk]

yh n = kq + a such that k is an integer

[quasi sparrow]

When the sum is over all all such n, can you write what that sum means in terms of k

[rich tusk]

wait wdym?

[quasi sparrow]

Rewrite this summation

[quasi sparrow]

In terms of k

[rich tusk]

ah yh yh it would be :
$\sum _{k = -\infty}^\infty} \frac{1}{(a+kq)^{2s}}$ right? that would be the equality right? or do you mean something else?

[clever fjord]

M47R1XAdV#9685
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[odd edge]

@rich tusk Has your question been resolved?

[quasi sparrow]

Yes so what's your question

[rich tusk]

how can i conncet the sum to a dirichlet L function, L(2s, chi)

where chi has conductor q and is real

[quasi sparrow]

What are you actually trying to do

If you're working on a problem just show it

[rich tusk]

its not a specific problem, im attemtping to find a way to get an L(2s, chi) out of the following expression :
$\sum_{a = 1}^{q-1}\chi(a)\sum_{k=-\infty}^\infty \frac{1}{(a+kq)^{2s}}$ so i was trying to work with the second sum to see what to do

[clever fjord]

M47R1XAdV#9685

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@rich tusk Has your question been resolved?

[rich tusk]

<@&286206848099549185>

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@rich tusk Has your question been resolved?

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[scarlet pond]

Am I missing something? What am I doing wrong

[signal trail]

You have flipped the values, sin(0)=0 and cos(0)=1

and etc.

It looks right when the values are flipped though

[scarlet pond]

ohhh bruh

[scarlet pond]

So I just gotta flip my sin and cos right?

[signal trail]

yeah

just flip the values

👍

[scarlet pond]

tyy

[signal trail]

yw

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[outer quail]

Ok so I'm not exactly looking for an answer but i want to know what i have to learn to get here. i can do limits with conjugates. like specifically i know i have to learn trig identities but i m just having a real hard time with it

[narrow crypt]

try

facotring

cos^2 x = (1+sinx)(1-sinx)

[signal trail]

remember tan^2theta= sin^2/cos^2theta

[cursive meadow]

b/c cos^2x = 1-sin^2x
difference of squares

[cursive meadow]

r u ment to cancel out ?

[narrow crypt]

yes

[cursive meadow]

k yea it makes sense

[outer quail]

(sin^2(theta))/cos^2(theta) (1-sin(theta) = tan^2(theta)-tan^2(theta)sin(theta) ?

[cursive meadow]

where'd you get the second tangent squared?

oh

you distributed into it

[outer quail]

yeah

[cursive meadow]

usually the strat is to put everything in terms of sin and cosine and work from there

this'll work here b/c of the identity + factorization "k" mentioned

if u want a hint feel free to ask

but i think u got this

[outer quail]

what exaclty is the bracket doing

[cursive meadow]

it's saying

multiply tan^2theta by quantity 1-sintheta

nothing but defining ur terms

[outer quail]

so i do (sin^2/cos^2theta)* (1-sin(theta)

[cursive meadow]

yup

and what did we say cos2^theta is equal to?

[stuck matrix]

Can y'all help me with this

Im just an intern

[cursive meadow]

mate post it in an unused help chat

[stuck matrix]

Ok

[cursive meadow]

alr @outer quail where were we 😅

ok so we have this

now lets use identities + manipulation

[outer quail]

(sin^2/(1-sin(theta)^2)* (1-sin(theta)

[cursive meadow]

almost

cos^2x is 1 - sin^2x

that's one of the identities from manipulating pythogrean formula

difference of squares now ?

[outer quail]

so the 1-sin^2(theta) goes were the cos^2theta is

[cursive meadow]

yup

and what can we do with 1-sin^2(theta)?

[outer quail]

1-sinx * 1+sinx

and you get sin^2theta/1+sin

can we now use pi/2 as theta?

[cursive meadow]

yes

[outer quail]

so the lim is 1/2

[cursive meadow]

yup

[outer quail]

so step by step turn all into sin and cos then whether you have sin or cos then change it in either respectively. or with these types of problems you will just end up with just sin or cos or could you end a problem with both or no?

[cursive meadow]

just see what's making the function undefined and try to get rid of it

tools are:

setting everything into sine / cosine if you're completely lost
noting the trig indentities
realizing you can modify some trig identities like the factoring shown above

[outer quail]

alright :) patience very appreciated and thanks for the math help

[cursive meadow]

nw
if you need help memorizing identities or want more practice/help ask away

[outer quail]

i just need a good place to find identities rather than google images or just a place that has like a structure learning how the identities work

[odd edge]

@outer quail Has your question been resolved?

[outer quail]

im gonna add you i can ask for identites and that quesion

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[odd edge]

Please don't occupy multiple help channels.

[hard hill]

Can anyone see if this is right and possibly correct my work

[velvet heart]

f(g(9)) ?

[odd edge]

@hard hill Has your question been resolved?

[hard hill]

wdym

[velvet heart]

oops, got the wrong example, f(g(9)) is defined, but f(g(2)) isn't

[hard hill]

so everything is correct besides f(g(2))

[velvet heart]

idk, i didn't check them all

do you know how to figure out whenever the composition is defined or not?

[hard hill]

not really

[velvet heart]

if you have f(g(x)), first find g(x) then check if you can input it in f

for f(g(9)), you have that g(9) = 1, so f(g(9)) = f(1) which is defined as 3

just replaced g(9) with 1

[hard hill]

oh ok

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@heady plover Has your question been resolved?

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@heady plover Has your question been resolved?

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[warm ore]

what can i google to find how to solve these types of examples

[clever fjord]

M47R1XAdV#9685

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[burnt holly]

what transformation is c)?

[hushed dock]

ivnerse

inverse

[burnt holly]

ohhh thanks

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[burnt holly]

explain and answer 8. a,b,c,d please!

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[burnt holly]

.

[mild fjord]

please help I tried transposing dividing multiplying didnt work

[orchid sphinx]

a. Even, b. Odd, c. Even, d. Odd

[bitter folio]

ZX is diameter

So what angle XAZ

[mild fjord]

XAZ as a whole is 180 right?

[bitter folio]

Mhm

So 11w+5w+4 = 180

Then find w

[mild fjord]

wait

I keep finding W as 2

[bitter folio]

Wat

[mild fjord]

earlier

[bitter folio]

16w = 180 - 4

[mild fjord]

ohhhhhh

[bitter folio]

W knief reactions

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[mild fjord]

it didnt 😦

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[blazing quiver]

Hi

[odd edge]

Please don't occupy multiple help channels.

[blazing quiver]

<@&286206848099549185>

[tender sigil]

heads up ur school is shown on the paper, might want to edit it out

[brisk urchin]

provide a clearer image of the question

[vagrant stirrup]

I'm assuming you need help with part ii?

[vagrant stirrup]

!15m

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[winged nimbus]

I don't know how to start

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[pseudo mountain]

if A squared to the power of b to the root of 7, find b+a

pls help

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@winged nimbus Has your question been resolved?

[winged nimbus]

bruh

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[lusty night]

How do we reach the implication with tan and sec in the explanation here?

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[compact steppe]

hi, can someone help me understand how did we get cos alpha and sin alpha from yn^2 and xn^2 respectively

$$ x_n^2 + y_n^2 = 1 - \frac{2}{1+4n}$$

$$ x_n = \sqrt{1- \frac{2}{1+4n}} \cdot \cos \alpha$$

[clever fjord]

atif

[compact steppe]

@winged vortex

yea yea just with sin

somehow one of the variables dissapears but the corresponding trig func appears as a factor on RHS

yea i dont think this is incorrect, but i want to know how we created the relation you mention

like what are the missing steps algebraically

or is there an explicit substitution statement missing

@winged vortex

mm no it's analysis 3 i.e. multivariable calculus, it's a small part in an exercise where we test for uniform convergence of another function

myb i got it, let me just try to check quickly

yea you're missing context for the motivation of that

the function we are testing for u. cont. is sin(pi/1-x^2-y^2)

but i don't think it's relevant so i omitted

are we just defining x_n and y_n here?

okok

(listening)

yea i know the equation

not so much w the parametric one but i see where you are going

just a sec let me check

yeah that's it

many thanks

im wondering if x_n = ... was derivable without knowing the parametric eq of circle?

makes sense

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[grand wind]

How do i find vertical / horizontal asymtote

[quasi sparrow]

https://www.cuemath.com/calculus/asymptotes/

[grand wind]

The number if 3sqrt -4/3, how’d i graph it

Its undefined

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@grand wind Has your question been resolved?

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[wanton snow]

Hey

[mystic saffron]

id eyeball aswell

[candid canyon]

mb

[limber glade]

channel opened by mistake or what?

[wanton snow]

hello

can anyone help me with a Problem of surface area and volumes

[limber glade]

just post the question

[wanton snow]

bruh don't joke on it ik it's ez just I'm so weak at math

And it's a 10th standard edition 💀lol

@limber glade

bro you there

@limber glade

@limber glade

[limber glade]

yeah yeah

[wanton snow]

@limber glade

[limber glade]

i was reading

[wanton snow]

help me out bro i got my exam the next day I'm cooked

[limber glade]

GODDAMIT DON'T SPAM

[wanton snow]

yeah sorry for that

[limber glade]

fine

[wanton snow]

thx a lot

[limber glade]

so you know the formula for the surface areas (both total and curved) of the cone and cylinder?

[wanton snow]

wait yeah I wrote then out

I can't understand the formula just rote learned 😭

[limber glade]

fine for now

[wanton snow]

cone is πrl

[limber glade]

just write them once

[wanton snow]

And cylinder is 2πrh+2πr²

[wanton snow]

Ok

Done

@limber glade

what should I do next

[limber glade]

ok so what do you think the question is wanting

[wanton snow]

So a tent is surmounted to a conical top height and dimater is given also the slant height we need to find the cost of canvas at the rate of 500 per m²

while the base will not be covered

[limber glade]

yeah

so basically the CSA of cone + CSA of the cylinder

you can get the radius from diameter

[wanton snow]

oh but how do we know it's CSA or TSA that confuses me

[wanton snow]

Ye

[limber glade]

you don't need base = CSA

imagine the structure of the tent

[wanton snow]

oh and if it's Base it's TSA

yupp got it

[limber glade]

you don't set up a tent where there is a base in the conical top part

[wanton snow]

Oh

btw got the radius so now should I just apply the formulas

how do I get the canvas cost should I multiply it with the value i get through CSA of coen + Cylinder

@limber glade

[limber glade]

ok so what did you get as the total area?

[wanton snow]

i haven't started yet

i was waiting for the answer that is that what I should do

[limber glade]

since the rate is per m cube, you can just multiply the rate with the total area, that's it

[wanton snow]

Ok

[limber glade]

and as for the answer, good luck

[wanton snow]

thx

[limber glade]

if you're done then close the channel

[wanton snow]

Alright

Thx for your help

[limber glade]

yw

[wanton snow]

got the answer finally 💀🔥

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[hoary flower]

For this question I don’t think any of the statements are true… because doesn’t it depend on how it is partitioned?

[timid fiber]

Lets see I will help you with this one but I’m slow typer and a high schooler so if you don’t mind

[signal yacht]

'partioned'?

[tepid pelican]

If you dont think they're true, produce a counterexample for each of these

[timid fiber]

let a in A such that A is a subset of P(A), and b in B such that B is a subset of P(B) then, and for the RHS its like a or b in A unions B such that P(A union B) this one is clearly true right sincethe power set of A unions B must contain the power set of A and power set of B let me see the other one of a

But not equal ofc

Because you always have elements that is a subset of A unions B but not A union B

[signal yacht]

it says proper subset though

[tepid pelican]

does it?

i see $\subseteq$

[clever fjord]

MæthIsAlwaysRight

[timid fiber]

But isn’t that just subset?

[signal yacht]

oh no it doesn't lmao

yes it is i misread

[timid fiber]

I think B is true though

[tepid pelican]

I'd be careful with "always" here

[timid fiber]

Let me try to prove it, let x be in P(A) intersects P(B)

Then x must be a subset of A and a subset of B

[signal yacht]

must it?

[timid fiber]

For the RHS, that x is a subset of P(A intersects B) then x must be a subset of P (A intersects B) which implies that it must be a subsets of A and a subset of B

[signal yacht]

it's just any powerset no

oh it's powerset

man im tired ill see myself out

you're doing welll

lmao

[timid fiber]

It must

If it’s not in either of them they won’t have intersection

[signal yacht]

i somehow read powerset and thought superset until i said powerset myself

[timid fiber]

I think B is true though

a is false

But I am working on my own problem, I think it’s not hard to prove just pick arbitrary elements

And try to prove by contradiction

[tepid pelican]

in simple words, find a counterexample

[timid fiber]

Since X is a subsets of A Union B then let u be an element of X blablabla

a the counter example is trivial is that the subset of a union B must not equal to the union of P(A) and P(B)

Which is relatively easy to come up with I think

For b i genuinely think it is correct

[tepid pelican]

b is indeed correct

as you proved

[timid fiber]

Anyone help me with puzzle I am in help - 37 😭😭

[odd edge]

@hoary flower Has your question been resolved?

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[indigo mist]

Can someone help me please with the f) one?

[olive needle]

,rotate

[olive needle]

Are you solving for x?

[indigo mist]

Yeah

[olive needle]

What have you tried so far?

[indigo mist]

I tried doing 5-2x and thag gave me 3x but the equation won't gave me the correct answer

The teacher didn't explained how to solve this kind of equations so I was guessing

[tacit wasp]

5 - 2x ≠ **(5 - 2)**x

[indigo mist]

What is the difference?

[tacit wasp]

7•(5 - 2x) should ring you a bell about distributing

[tacit wasp]

The parentheses!
(5 - 2)•x = 3•x

5 - 2•x can't be simplified, instead

[indigo mist]

Ooh, I see

[tacit wasp]

If the teacher didn't tell anything about parentheses it's quite a big problem, because otherwise you can't solve this

[indigo mist]

I see

I think I will just ask her tommorow then

Thanks anyway!

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[hoary flower]

I don’t think b is correct though? Take A ={1,2} and B ={2,3}. Then partition A as {{1,2}} and B as {{2},{3}}. Then the Union is {{1,2},{2},{3}}

[tepid pelican]

B is intersection

[hoary flower]

But the intersection of A U B is {2}

[hoary flower]

Could you explain?

Oh wait

[tepid pelican]

You dont get to partitiom A

P means power set

[hoary flower]

Oh wait hmm

[tepid pelican]

P(A) is the set of all subsets of A

[hoary flower]

Yes

I want to prove first statement of part b then?

Because second state,ent is not true

[tepid pelican]

Both are true

[hoary flower]

How so?

Take the example I gave but with intersection

Then P(A) intersect P(B) = empty set

But empty set not equal to P(A intersect B)

[tepid pelican]

Not true

[hoary flower]

Why?

[tepid pelican]

Both P(A) and P(B) contain {2}

[hoary flower]

Because I can partition however I want right

[tepid pelican]

And both of them contain the empty set

[hoary flower]

No, P(A) contains {1,2}

Not {2}

[tepid pelican]

Ypu dont get to partition stuff

P(A) contains ALL the subsets of A

[hoary flower]

Wait…

[tepid pelican]

So it contains {2} as well

[hoary flower]

OHHH, is that defintion of partition?

[hoary flower]

Wait… but is the set {2} the same object as 2?

[tepid pelican]

Does your book define that symbol P somewhere?

[tepid pelican]

No

P usually means powerset, not partition

Unless specified otgerwise

[hoary flower]

It defines as partition.

And says that is set of subsets of X (P(x))

[tepid pelican]

Partition is wrong word to use

[tepid pelican]

Yes, this is the proper definition

So P(A) contains ALL subsets of A

[hoary flower]

No, but it must satisfy these conditions

[tepid pelican]

Including {} and {2}

[tepid pelican]

Which conditions

[hoary flower]

these ones i believ

[timid fiber]

What are all subsets of B?

[hoary flower]

i don tthink it is all subsets

[tepid pelican]

Huh okay

[hoary flower]

yeah sry it wasnt the most clear

[timid fiber]

Partition

[hoary flower]

using these deifnitons i mean

[tepid pelican]

I dont see how the notation P(A) is justified then

[timid fiber]

P is conventionally power set

[hoary flower]

sry I meant the weird P

[tepid pelican]

This isnt a unique set

Did your book define a power set earlier?

[timid fiber]

In set theory partition is like X\ {R} where R is equivalent classes

[hoary flower]

no

[timid fiber]

This is a freaking proper partition

[tepid pelican]

What book do you use?

[timid fiber]

P is conventionally for power set

[hoary flower]

these are just lecture notes

would it be alright to transition to question

[timid fiber]

And the question you proposed is classic I think

[hoary flower]

because using the defintion above idk

[tepid pelican]

The definition above defines a partition

[hoary flower]

because using this, I dont beleive that any of the statments are true

[tepid pelican]

But it doesnt define P(A)

[hoary flower]

yes i am talking about partition

i mean partition

when i say P

but P

[timid fiber]

Usually partitions are defined rigorously like X / ~

[tepid pelican]

Using that, none of the statements even makes sensr

[timid fiber]

Or simply X \ {R} R is any equivalent classes

I have never seen using P

[tepid pelican]

Its just a symbol

And there are different kind of partitions, distinguished based on the context

[hoary flower]

ok

[tepid pelican]

And this is also the main problem

You couls have a partition C of A

[timid fiber]

But it doesn’t show equivalent classes for the previous set equal problem

Which are the equivalent classes

[tepid pelican]

P is just a specific name

[timid fiber]

Alright

[tepid pelican]

And stuff like P(A) makes no senae

[timid fiber]

Oh I see you switched the question

I thought you were saying previous question which confused me

[tepid pelican]

Because P is a set, it's not a function and it's not something thats supposed to have an input

[tepid pelican]

Its the same question

[hoary flower]

i am still talking about this question

[tepid pelican]

The pic is just a relevant defn

[timid fiber]

This is easy the second part is basically definition of equivalent classes

Again you can draw a map to prove this I think

But the previous question, which I believe it’s a power set notation for sure

[tepid pelican]

But ye, my point is that the defn 6.1 defines what "partition" means and it uses a symbol P to denote some partition. This however has nothing to do with the usage of P in P(A)

[timid fiber]

Because the second pic it gives equivalent classes which make it defined..

[tepid pelican]

I also suspect that the question is talking about powersets

[timid fiber]

Agree

[tepid pelican]

Its a quite common exercise in the context of powersets

[hoary flower]

oh okay

[timid fiber]

You need equivalent classes for partitions? If my memory is right

[hoary flower]

perhaps i will read a bit more

buut if we were to use the defintion of a parition to solve the question

[tepid pelican]

Depends on the context

[hoary flower]

then none of the statment would be true right

[timid fiber]

Is partition even defined without mentioning equivalent classes? I don’t think so right?

[tepid pelican]

Partition can mean different things

[tepid pelican]

Here it was defined like this

[timid fiber]

Partitions are fundamentally equivalent classes

[tepid pelican]

It was defined

[timid fiber]

That one is classic though

[tepid pelican]

There are different kinds of partitions

[raw burrow]

2^7x + 17$ = 2^7 + 34$

I need help

[timid fiber]

Despite unconventional notations the second question is basically asking definition of equivalent classes

But that partition on image is defo the one I have been mentioning

I mean I am from high school maybe my brain isn’t good 😭😭

[hoary flower]

wait so

am asking just based upon defintion of partition

none of the statements in the problem are true then?

since it is an arbitrary partiion

[tepid pelican]

None of them makes sense

[tepid pelican]

If you assume that it can be an arbitrary partition, then none of them, yeah

[timid fiber]

It’s but it’s not arbitrary it sin P

[tepid pelican]

Wdym

[hoary flower]

hmm i see

[timid fiber]

Since its in P

P is basically X \ ~ with weird notation

So basically