help-33

[red dagger]

shit cancels immediately

amyways igtg sleep

goodnight!

[proper kite]

You too oomfie

[potent token]

oomfie

[obtuse inlet]

AS "BRUHHHHHH" SAID, ITS BETTER TO UNDERSTAND THEM THAN TO USE THE MNEMONIC.

BUT THE MNEMONIC SAVED ME A FEW TIMES SO HERE:
"SOHCAHTOA"

[potent token]

why all caps

rude

[obtuse inlet]

SOH CAH TOA
S=O/H , C =A/H , T=O/A

[obtuse inlet]

MY HUMBLEST APOLOGIES =(

[proper kite]

Yeah ik the trig rules and whatnot
I was speaking on moreorless those identities used in general equations

[potent token]

also idk what you mean by its better to understand: unless you know the specific nomenclature and basis for the names, then its memorization for sin vs cos vs tan

[proper kite]

But he did solve my question tho, I think I’ll just drop having to memorize em

[potent token]

do you want identities?

[proper kite]

Yeah sure, above the 3 main ones

[potent token]

what three main ones

[elfin berry]

! extremum

[proper kite]

The reciprocal, quotient and pythagorean identities

[potent token]

double angle

um

[proper kite]

Negative angles right?

Or smth like that

[potent token]

use unit circle for negative angles, reflected angles, e.t.c

$\sin{a + b} = \sin{a}\cos{b} + \sin{b}\cos{a}$

oop

[elfin berry]

! extremum

[potent token]

$\cos{a + b} = \cos{a}\cos{b} - \sin{a}\sin{b}$

[elfin berry]

! extremum

[obtuse inlet]

THIS IS WHAT I USED IN SCHOOL

[proper kite]

Ah I see
What are some suggestions to make my arithmetics faster, since I’m still awfully slow with rational expressions and so on
(I’m not particularly the best with dealing with operations, even then I still get confuse with applying pemdas fully)

[modern falcon]

hi

yall

drink water

stay hydrated

[proper kite]

Very interesting, will take note

[potent token]

you're getting reported

no but like dont butt in with random info

[obtuse inlet]

BUT WATERS IMPORTANT

[potent token]

um

so remember

cotan is the reciprocal of tan

tan is sin/cos

[proper kite]

Yeah huh

I think I got those down in my brain

[potent token]

sec is the reciprocal of cos

csc is the reciprocal of sin

aare you god at evaluating trig functions at easy angles?

are you familiar with radians

do you know inverse trig

do you know the derivative of all the trig functions?

[proper kite]

I don’t think I have touched there yet

[potent token]

do you know the taylor series for them

oh my bad

[proper kite]

That no,
I think the only function I rlly know for trig

[potent token]

ignore everything past radians

[proper kite]

Is literally just the basic euler one

Ie

Etheta=costheta+isintheta or whatever

[potent token]

ya

then you can solve for cos

i think

i forget how

no thats not perfect

let me show

[proper kite]

Amazing

Wait hol up just one question before you demonstrate further @potent token

[potent token]

$e^{i\theta} = \cos{\theta} + i\sin{\theta}$

yeah what

no typo mb

[elfin berry]

! extremum

[proper kite]

I think the only core issues I rlly have dealing with some equations is that my operational skills aren’t the best especially when I have like fairly bad dyslexia and adhd which makes it ever much slower

You think practice is the only way to make it any better?

I think its more or less because of my very bad planning skills and making out the steps I have to do to map everything out

I do know someone did suggest me to learn some number theory to improve it as, it’ll make it easier for my brain to make out those connections

[potent token]

sry i was eating

yeah

um

to memorize

just write it down a bunch of times

taking notes on cumpter doesnt improve retention

wdym @proper kite

im confused what you saying

what is hard?

[proper kite]

Well its not rlly because its hard, its more or less because idk, I just have a very hard time organizing and planning things in general, and its no better than when doing mathematics
Even smth as simple as Pemdas I trip up on the some of the most trivial bs

[potent token]

yes

write down stuff that you forget

it will come to you easier

thats what i do

when my teacher gives me a google classroom code or smth

i write it down like 10 times

then i have no problem remembering

oh shit

i forgot it

😅

oops

it was like

i3ij2j or smth

i forgor

there was an n somewhere in there

ok back to you

🇺

[proper kite]

Oh I’ve been doing it, especially when I had an oversight on things
(Tbh I shouldn’t had been lazy and finished my last year math with a D) (I didn’t pay attention in class and only on the last minute I rushed in with a 2 As and a B- on my exam

[potent token]

so tell me

do you make too many silly mistakes?

for silly mistakes

i always read what im saying out loud

[proper kite]

Because lets just say it transferred over to this year and I have a D+ rn since I failed my first two tests

[potent token]

although that might not be possible

its important

[proper kite]

A lot

Even writing numbers wrong

[potent token]

but do you know the course material

[proper kite]

I once misread 45.8 as 65.8

[potent token]

was that bc of your dyslexia? sorry abt that

[proper kite]

Yeah I do know, pre calc

[proper kite]

Oh yea definitely

[potent token]

you mentioned you're dyslexic, right?

yeah

thats tough

idk how to help with that

but

for silly mistakes

i recommend

for preparation

[proper kite]

Yeah and I also have some fairly bad adhd as well
So my executive functioning isn’t the best

[potent token]

writing your notes down many times

and saying them out loud

[proper kite]

Definitely will

[potent token]

and during the test

[proper kite]

I’ve noticed thats also where I notice my mistake after tryn fix em for like 5 minutes

[potent token]

talk out loud

yah

ok

i gtg

to the gym

bye

good night

well for me

idk where you live

[proper kite]

Oh right cya

[potent token]

good night

[proper kite]

Canada, but yeah gl hf thank you

[potent token]

ok bye

[marsh citrus]

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[dull dawn]

does anyone know how to find C when taking the integral of a vector

[iron meadow]

!original

[marsh citrus]

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

@dull dawn Has your question been resolved?

[dull dawn]

.close

[marsh citrus]

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[vestal mist]

im stuck on how to do this

[vestal mist]

using what ive learned

i get
dy/3=y^2dx
then i integrate both sides getting
y/3=xy^2+C
i solve for C by plugging in y(4)=3 to get C=-140/3
so now do i solve for y in the above equation and plug in C?

[hollow sparrow]

ah there you go

move all factors of y to one side

so you have $\frac{dy}{dx} = 3y^2$. I would do the following: $$y^{-2}dy = 3dx$$ and then try again

[elfin berry]

MellowDramaLlama

[hollow sparrow]

$\int y^{-2} dy = 3 \int dx$

[elfin berry]

MellowDramaLlama

[vestal mist]

oh

1/(-3y^3)=3x?

@hollow sparrow

[hollow sparrow]

hi please don't ping I"m helping other people

[vestal mist]

oh sorry i wasnt aware of that

[hollow sparrow]

check your integral on the left hand side again.

remember the power rule for integrals is this

(also don't forget your + C on the right side)

[vestal mist]

-1/y=3x+C

[hollow sparrow]

yep bingo, now just solve for y and you should be gucci

[vestal mist]

thanks!

[hollow sparrow]

yep np

[vestal mist]

i get C=-37/4

doesnt seem to be right

[hollow sparrow]

ok so it's kind of weird thing but that + C doesn't get carried over into the denominator

so here's what I'm getting

$$ -\frac{1}{y} = 3x + C$$
$$ -y = \frac{1}{3x} + C$$
$$ y = -\frac{1}{3x} + C$$

[elfin berry]

MellowDramaLlama

[vestal mist]

why does the +C not get carried into the denom

[hollow sparrow]

remember the constant C is just a veritical translation of your graph since an integral is a family of functions

[vestal mist]

oh

i see

[hollow sparrow]

ok now try and solve with that

(you also didn't negate the -1/3x which could've also been an issue)

[paper raptor]

hey do I know you?

[vestal mist]

um,

[hollow sparrow]

huh

[vestal mist]

i plugged in y(3)=4

[hollow sparrow]

oh

oh wait I"m dumb

I'm sorry

the +C does carry over into the denom

butyou didn't negate the function

try this: $$ y = -\frac{1}{3x + C}$$

[elfin berry]

MellowDramaLlama

[vestal mist]

ooh

yay!

thanks!

[hollow sparrow]

my bad (it's been a minute since I"ve done this lol

[vestal mist]

no worries lol

.close

[marsh citrus]

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[marsh citrus]

@tulip folio Has your question been resolved?

[rare dock]

Hey man,what do you need help with?

[marsh citrus]

@tulip folio Has your question been resolved?

[marsh citrus]

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[fathom nest]

I am in calculus 2 and we are doing maclaurin expansion. I am now lost at what do so after I expand the sin(x^3) out. The -x^3 and the division by x^9 is there I am confused

[fathom nest]

Thanks for the help in advance

Here is my professors work for some reference

<@&286206848099549185>

.close

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[trim pulsar]

could someone give me a hint as to what im doing wrong?

[trim pulsar]

.close

[marsh citrus]

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[upbeat oar]

ik how to solve this

but

my teacher got 0.16 ft/ sec and

idk how she did that. bc she said one of the sides = 23 bc of pythagorean thereom. but when i do it i get 8

like for the side

one of the sides . cuz we’re given 20 and 12

it’s a right triangle

-0.3 ft/sec btw

and i keep getting 8 but my teacher got 23.. for B

c is the ladder btw

[marsh citrus]

@upbeat oar Has your question been resolved?

[upbeat oar]

it’s related rates..

[marsh citrus]

@upbeat oar Has your question been resolved?

[marsh citrus]

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[stiff dove]

Can someone guide me through this?

[stiff dove]

,rotate

[gleaming pecan]

put x = 1

[stiff dove]

Why

[stiff dove]

Is there a more algebraic method?

[gleaming pecan]

look, it is written on your photo, that, x is any real number right ?

so

you can subsitute whatever

in palce of x

if you put x = 1

then you got required sum

multipleid by 2

and on left side also 2

so you can eaisly see, that sum of coefficents = 1

that si algebraic method, i jsut wrote it

[marsh citrus]

@stiff dove Has your question been resolved?

[stiff dove]

What if I sub in x=0?

Wait that doesn't work

.close

[marsh citrus]

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[left mango]

How do i do all of this

[sage onyx]

Do you know what the discriminant is?

[left mango]

yes

b^2-4ac

[sage onyx]

Ok so you should be able to do the first question, yeah?

[left mango]

i just not sure about the positive definate and indefinate mean

[left mango]

im not sure

[sage onyx]

Positive definite means it’s greater than zero if you put in any number other than 0

[left mango]

oh

alr

but

do i justt sub somehting in

or do i do b^2-4ac

[sage onyx]

so what does the a,b and c mean in that formula?

[left mango]

a =3 b = (12-k) c= 12

[sage onyx]

Exactly

So with that you’ll get a function in K

[left mango]

huh

why

[sage onyx]

Well you substitute that into the discriminate formula

[left mango]

how

oh

nvm

yra

what do i do after

[sage onyx]

Then youre asked to find the values of K such that it’s strictly positive

[left mango]

i got

144-24k+k^2-144

= 24k-k^2

[sage onyx]

You’ve flipped your signs there

Just in the 2nd step

[left mango]

huh

oh

oops

24k+k^2

what di i do here

[sage onyx]

K^2-24k

[left mango]

yes

[sage onyx]

You now need to work out for what values of K this is strictly positive

[left mango]

how?

[sage onyx]

So you just solve k^2-24k>0

[left mango]

how do i do that?

[sage onyx]

Well you’re looking to solve for k so what do you think you would do

[left mango]

move 24k over

and root

but i dont think thats right

[sage onyx]

Well taking the root wouldn’t help because you’d still have a k on both sides

But the first step is correct

[left mango]

yea

[sage onyx]

How could you isolate a factor of k?

[left mango]

uh

k(k>24)

?

[sage onyx]

Remember that k^2=k*k

So you have k x k>24 x k

[left mango]

yes

[left mango]

oh

What do u so tho

[sage onyx]

Just think about it for a while it’s probably easier than you’re expecting

[left mango]

divide

k over?

k>24

no

k<24

[sage onyx]

Yeah you divide, if you’re unsure about the direction of the inequality what could you do to check?

[left mango]

uh

idk

im pretty sure if you divide or times over u have to switch it tho

@sage onyx is that right?

sry for ping

someone please help me

How do i do d) and q3

My brain cant

[still temple]

oh shit

i used to solve this shit

[left mango]

Please help

My brain

Cant brain

My math isnt mathing

[still temple]

for the d one

u need like the equations dont u ?

initial velocity ?

[left mango]

which eqauation

velocity as in time in air

[still temple]

d

[left mango]

?

[still temple]

no like

intial velocity

when it was fired

[left mango]

doesnt say

[still temple]

also the angle of inclination..

[left mango]

it just says

[still temple]

uhm

what chapter is this from

is this conics

[left mango]

a shell is fired from level ground after t seconds, its height h in meters is givin by h=50t-5t^2

[still temple]

might refer to a parabolic curve

[left mango]

it is

its quadratcic functions

yay!

[still temple]

h = 50t - 5t^2 hm

whats the derivate of height

[left mango]

what

[still temple]

dh/dt would be useless..

[still temple]

bro what grade is this from

[left mango]

10th

[still temple]

wait..

[still temple]

all u have to do it equate height in this and then solve

for t

[left mango]

i dud

did

[still temple]

is the answer wrong?

[left mango]

no

answer says 15 s

[still temple]

uhm so?

[left mango]

idk how to get htat

[still temple]

just solve

wait

375= 50 t - 5t^2

so

[left mango]

i litterally did

i divided by 5

[still temple]

5t^2 - 50 t+375

[still temple]

yes exactly

[left mango]

and then t^2 - 10t + 75

you cant factorise this no more

[still temple]

my brother..

[left mango]

my bad

[still temple]

yes

and i am sure u can

let me think

[left mango]

gl

[still temple]

t^2 -10 t + 75

hm

it has complex roots

[left mango]

it

should be

not complex roots

which is why i think its wrong

<3

[still temple]

yeah

probably is

[left mango]

im very smort

real

[still temple]

u need -375 for it to be right

[left mango]

i need

to get smarter

[still temple]

😭

[left mango]

-5t^2+50t-375

would that work

sir

[still temple]

no

[left mango]

im gonna commit

[still temple]

5t2 -50 t - 375

so

t2 - 10 t - 75

[left mango]

that doesnt work

[still temple]

so t2 - 15t + 5t - 75

[left mango]

you cant just whip out a - 375

[still temple]

t (t - 15) + 5 (t - 15 )

[still temple]

which is why the question is wrong

[left mango]

your wrong

idk

i hte math

[still temple]

youre*

[left mango]

https://tenor.com/view/plink-cat-plink-cat-gif-1794292671885121408

[still temple]

stupid maths*

[left mango]

rel

well well well

help me with the one below

u got this

[still temple]

recheck the question

[left mango]

💪

[still temple]

no i have my own doubts

[left mango]

no

i legit sent a photo

[left mango]

right here

i need this last question to be done

i need to hop on val

and then

derank

[still temple]

solve it yourself

if u have time to play val

u can spend time on maths

[left mango]

no

i cant solve it myself

cause i dont know how

thats why im here

which is also why i cant play val

[left mango]

so please help me 🙏

[marsh citrus]

@left mango Has your question been resolved?

[left mango]

.clsoe

[marsh citrus]

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[marsh citrus]

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[still temple]

hi guys

[still temple]

would the answer be A and C

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

i was thinking a) cuz binomial and c) cuz multinomial, just need someone to confirm

[tight furnace]

Looks correct

So the answer is D)

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

.close

[marsh citrus]

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[still temple]

thanks

[marsh citrus]

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[pale peak]

not sure how to start the question

[pale peak]

is the force right on the labeled diagram between floor and the ladder

even if i had it not sure where to go from there though

<@&286206848099549185>

[marsh citrus]

@pale peak Has your question been resolved?

[pale peak]

<@&286206848099549185> is anyone able to help? i think it wants to shut down this channel

[fluid mica]

Hellooo

@pale peak

[pale peak]

hi

[pale peak]

can you help?

[fluid mica]

Alrighty

Mechanics

Yes

So we know F1 is friction yes?

[pale peak]

yes

[fluid mica]

F1=F2 yes?

[pale peak]

i think so foir horizontal equilibrium

[fluid mica]

Mhm

So we need to find out what the friction is

To do that

We use F=ur

[pale peak]

yes

[fluid mica]

F being mew

Yea that’s it

[pale peak]

and u=tan60 right?

[fluid mica]

No mew is the coefficient of friction

0.2 in this case

[pale peak]

ooh ok

how do i work out r though

[fluid mica]

R is the reaction force which acts on the ground

Just like how horizontal forces equal, so do vertical

[pale peak]

so its just 20g N

[fluid mica]

Yah

Now we have R and we have the mew

[pale peak]

so its just 0.2 x 20g

[fluid mica]

Mhm

Btw don’t just assume horizontal forces equal each other and that vertical forces equal each other

We only do this in 2 cases

When it says that the system is in equilibrium

[pale peak]

it says its wrong

[fluid mica]

Or when it asks the minimum or maximum force required for smth

[fluid mica]

Hm

[pale peak]

does r act vertically or is it at an angle and just some component vertical

so then the 60 degrees actually matters

[fluid mica]

Vertically

Give me a sec let me see what’s going on

[pale peak]

ok

[fluid mica]

Should be 39.2?

[pale peak]

it says that is wrong

[fluid mica]

Is there anything else to the question?

[pale peak]

just that

its from some questions that a few needed to resolve moments as well so that could be something?

[fluid mica]

No moments here as doesn’t give distance

[pale peak]

but because its a uniform ladder the weight acts at centre so just use a variable value ?

[fluid mica]

Oh no

I’ve misunderstood the question

Apologies

[pale peak]

its ok

[fluid mica]

Was thinking I was going crazy 😂

[pale peak]

lol

how do you do it?

[fluid mica]

So in this case

It’s talking about another external force

Which would be applied to the ladder

To keep it in equilibrium

So when horizontal forces equal

[pale peak]

so friction + external force = force from wall

[fluid mica]

Exactly

We’ve worked out friction

But we don’t know what the force from the wall is

And to work that out we would need to use moments 😭😭😭😭

[pale peak]

lot of effort but sure

gotta love moments questions

[fluid mica]

Lol

Take moments from base of ladder

[pale peak]

ok

[fluid mica]

So take moments from A

[fluid mica]

Everything good?

[pale peak]

not sure how to resolve the 20g force to be perpendicular to the bar for the moment

[fluid mica]

Take a look at this diagram

The forces acting from A are cancelled

So we don’t need to worry about S (external force), the friction and the reaction at the base of the ladder RA

[pale peak]

how do you know the angle acting from 20g is 60 degrees

[fluid mica]

Where?

[pale peak]

between 20g and the ladder

[fluid mica]

Woah woah woah

[fluid mica]

Fixed it

Also that angle isn’t necessary here

[pale peak]

surely need resolve the forces to be perpendicular to the ladder though

[fluid mica]

But it’s because the dotted line is perpendicular meaning its at a 90 degree angle

[fluid mica]

We can see that inside the triangle we have a 60 degrees, also have a 90 degrees meaning we are left with 30 degrees

Thus if it’s perpendicular where one side is 30

Other must be 60

Instead of having to think of all this always just remember that when there’s a ladder like this or anything at an angle

And there’s a force acting down

[pale peak]

ok ill give it a go

[fluid mica]

The angle next to it will always be the same as the one inside

[pale peak]

so would this be right?

just did it and got it right

answer was 17.4 N

Ty alot for helping @fluid mica

.close

[marsh citrus]

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[toxic flame]

how can I solve it since I dont have the value in the y axes?

[marsh citrus]

@toxic flame Has your question been resolved?

[marsh citrus]

@toxic flame Has your question been resolved?

[marsh citrus]

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[silver quarry]

How do I make a sign chart with this equation? f(x) = (6(x+3)(x-1)^2 ))/(x-3)^3

How do I make a sign chart with this equation? f(x) = (6(x+3)(x-1)^2 ))/(x-3)^3

[lean flame]

Damn the bot is really broken

[silver quarry]

what do I do

[lean flame]

Where you stuck

[quaint arrow]

is this help channel open?

[silver quarry]

i have the equation simplifed but I am not sure what to do with that information

(x^2 - 6x + 9)(x-3)

The line on the bottom right is supposed to be the sign chart

??

can i have help please?

[silver quarry]

How do I make a sign chart with this equation? f(x) = (6(x+3)(x-1)^2 ))/(x-3)^3

[eternal edge]

Sign chart is finding when is the function positive or negative, right?

[silver quarry]

yes

i understand the multiplicty law to know when signs stay the same or flip

but i dont know how to tell what it starts off as

[eternal edge]

We know that a function can change sign whenever it's either discontinuous or zero

Right?

[silver quarry]

i learned that it changes sign when the multiplicity is odd

so please enlighten me

i sent a picture up above

should i just plug -4 in?

[eternal edge]

That may not hold for non-rational functions

Consider sign(x)*(x^2)

Anyway, did you get what I said before?

[silver quarry]

ok

[eternal edge]

This

[silver quarry]

can you show me what that would look like'

its hard for my small brain to understand

[eternal edge]

Apart from some examples, there is nothing to show. The thing is that a continuous function won't change sign unless it passes through 0 at some point

So a possible approach is to find all points that the function is 0 or discontinuous at

Because that is going to tell us the only places the function is allowed to change sign on

[silver quarry]

yeah i was gettin at examples but i think i get it now, could you show me how that would work

I plugged in -4 and it gave me a negative number which also agrees with the graph

[marsh citrus]

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[eternal edge]

Okay now that you got the places at which the function may change sign, you have broken the real number line into a few pieces

[silver quarry]

so what should i do

let me send you a picture of the actual question

[eternal edge]

Hmm they ask you to do it in a different way than what I'm saying

[silver quarry]

yeahhhh

and if i do it a different wey i get penalized ive tried

?

[eternal edge]

Yeah I don't think I can explain it clearly because I'm not familiar with that way of solving this

Sorry :/

[silver quarry]

could you at keast tell me if my answre is right@eternal edge

so i can kinda have a feel for if im doing it somewhat right

[marsh citrus]

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[still temple]

how is -sin(300) not w

[eternal edge]

Nothing a graph calc can't tell you already, but the answer is not right. Although all signs are flipped so you might have missed a sign

[silver quarry]

ok

[marsh citrus]

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[rough fulcrum]

whats the formula for finding the surface area of a right sqaure-based pyramid

[rough fulcrum]

?

[rough fulcrum]

how do i find the slant height

[serene knoll]

Draw a right triangle with the height as a side and half the length of the side of the base as the other side

then use pythagoras

[rough fulcrum]

I have this, is 13 the slant height?

[serene knoll]

It's not the slant heigh if it is referring to the black line, no

It might be easier to visualize here

[rough fulcrum]

ohhh

[marsh citrus]

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[marsh citrus]

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[celest birch]

someone please help me with this

[celest birch]

i dunno where to start

i need to simplify it

[austere yarrow]

you can start by distributing 2(x+3)

[stark sandal]

you should start by distributing (x+3)(x-3)

because identity

[austere yarrow]

oh yeah

[marsh citrus]

@celest birch Has your question been resolved?

[austere yarrow]

i think u have to close it

oh wait

x and check?

[marsh citrus]

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[cobalt yacht]

I already got the proof that it works for odd numbers

[cobalt yacht]

But how can I prove it doesn’t work for even numbers?

[glacial hedge]

Why would it not ?

[cobalt yacht]

Well I tried some even numbers and it doesn’t work

[glacial hedge]

How did you prove it for odd numbers

[cobalt yacht]

Looking at n|(n-k)^n + k^n for k = 1,2,…,(n-1)/2

[glacial hedge]

That's much more obscure than what I had in mind

[cobalt yacht]

It’s just mod n

[glacial hedge]

Yeah ofc it works when you expand because they nicely simplify the k^n
And there's no middle term

So in general mod n you'd be left with (n/2)^n

If n is even

[cobalt yacht]

Yeah but idk how to go about the case where n is even

[glacial hedge]

Does n divide (n/2)^n

Evaluate (n/2)^n mod n

[cobalt yacht]

Why is this relevant?

[glacial hedge]

n divides a number iff it's 0 mod n

[cobalt yacht]

No where does (n/2)^n come from

[glacial hedge]

This argument pairs the terms together

Except that if n is even, you have an odd number of terms

[cobalt yacht]

This argument doesn’t work for n even

[glacial hedge]

So the middle one is leftover

[cobalt yacht]

Ud get a 2k^n hanging

[glacial hedge]

Oh right

That too

Ok change of method

Evaluate the entire sum mod n

[cobalt yacht]

U mean the original sum or which sum?

[glacial hedge]

Original

[cobalt yacht]

How do I do that?

[glacial hedge]

Wouldn't you happen to know a theorem about the value of a^n mod n ?

[cobalt yacht]

Like Euler theorem?

[glacial hedge]

I don't know 2 such theorems

[cobalt yacht]

Not sure what theorem you’re referring to

[glacial hedge]

It's a mod n

,w Fermat's little theorem

[elfin berry]

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[cobalt yacht]

N has to be prime for that

[glacial hedge]

Chinese remainder theorem says it doesn't

Well it's a bit more complicated actually

But it still holds

a^n = a mod n

[cobalt yacht]

Not sure what you’re on about take a=3 n=8

[glacial hedge]

,w 3^8

[elfin berry]

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[glacial hedge]

It's 1 mod n

Right

[cobalt yacht]

Well that’s not a mod n

[glacial hedge]

Had a bit of a moment

[cobalt yacht]

It’s aight 😂

[glacial hedge]

I haven't done any arithmetic in too long

[glacial hedge]

Gonna invent a proof for this, brb

Cause then the sum would be n-1 mod n which is just about never 0

[cobalt yacht]

It’s not gonna always be 1 mod n I don’t think

6^8 is 0 mod 8

Well yeah if it’s even

[hidden plaza]

consider n = 2^(p)q where q is odd and do some modulo sheninegan with mod 2^p

youd need euler's theorem for this

[cobalt yacht]

What’s the goal here I’m not seeing it

[hidden plaza]

Suppose n = 2^(p)q where q is odd and its true that n|1^n + 2^n +...+ (n-1)^n [insert some modulo shenanigans] deduce that 2^(p+1) | n - a contradiction.

[cobalt yacht]

Okay ill try …

[marsh citrus]

@cobalt yacht Has your question been resolved?

[marsh citrus]

@cobalt yacht Has your question been resolved?

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[cobalt yacht]

.reopen

[marsh citrus]

✅

[marsh citrus]

@cobalt yacht Has your question been resolved?

[marsh citrus]

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[still temple]

I know how to do it with things such as sinacos2a+cosasin2a

but how would i do it with this

[summer trench]

,, \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \cdot \tan B}

[elfin berry]

यजतलमाओ

[still temple]

hm

would tan(a+b) =$\frac{tanA+tanB}{1+tanA\cdot tanB}$?

[elfin berry]

RecRio

[still temple]

also where are you getting this from

[summer trench]

yes

[summer trench]

those r identities, we proved them

[still temple]

is it from this?

[marsh citrus]

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[marsh citrus]

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[tacit solar]

I think it's 28 cuz the allude to that twice but is it from 30-12 or 58-30

[marsh citrus]

@tacit solar Has your question been resolved?

[cunning fiber]

what is 28

[tacit solar]

Width

[cunning fiber]

of the opening?

[tacit solar]

Yes

[cunning fiber]

half of the width is 28

[tacit solar]

But w should be shorter than l

[cunning fiber]

eh that's more of a convention than anything

[tacit solar]

So w is 56 ?

[cunning fiber]

yes

[tacit solar]

And they tell you to find area but before finding width which is weird because I don't think you'd do second question before first

Then the last question says go to find BD

[marsh citrus]

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[marsh citrus]

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[hidden plaza]

U tried smth?

[rapid verge]

well

the indicators are bernoulli right

sum of bernoulli is binomial

..?

[marsh citrus]

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[odd sparrow]

Simple question, whats happening to get from first to second line

[lone heart]

sin^2(x) = sin(x) * sin(x)

So you can just separate the fraction using the above

[odd sparrow]

ohh right I see

Do you also perhaps know waht is happening here?

Does the cos^2x become the 1-sin^2x

[lone heart]

It didn't

cos^2(x) + sin^2(x) = 1

Applying associative property, you can group the terms so cos^2 + sin^2 - sin^2cos^2 is the same as (cos^2 + sin^2) - sin^2cos^2

[odd sparrow]

ohhhh thank you

I see that now

makes senes, cheers

.close

[marsh citrus]

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[marsh citrus]

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[vernal forge]

what's the exercise

[marsh citrus]

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

[vernal forge]

so, simplify?

$a^{3n}b^{2n+1} = a^{2n}a^n b^{2n}b$

[elfin berry]

artemetra

[clever loom]

Isn't it just

[vernal forge]

well this is the first step

i'm not giving you the answer right away lol

[vernal forge]

@frozen geode do you see how that's true?

$a^{n}b\sqrt{a^{n}b}$

[elfin berry]

artemetra

[vernal forge]

just type out regularly

[elfin berry]

Amin

Amin

[vernal forge]

uh

something went wrong lol

[vernal forge]

look here

$\sqrt{a^{2n}a^n b^{2n}b}=\sqrt{a^{2n}b^{2n}a^n b}$

[elfin berry]

artemetra

[vernal forge]

how's that for a hint

the last step is literally 1 operation away from the final solution

put what

yeah i was showing without it

it's the same thing

$\sqrt{a^{3n}b^{2n+1}} = \sqrt{a^{2n}a^n b^{2n}b}$

[elfin berry]

artemetra

[vernal forge]

if you are done then type ".close" here

[marsh citrus]

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[rare sail]

I have been thinking about this for a while and I can't figure it out.
Imagine you toss 2 fair coins without looking at the output. The sample space is {HH, HT, TH, TT}.
You have a parter that has seen the output of the 2 coins and tells you that one of them is Heads. Now, you calculate the probability that the other is also Heads. Intuitively, because coin tosses are independent, you'd say it's 1/2, it doesn't matter what the other coin is. But, looking at the sample space, if you know one of them is heads, you have the event {HH, HT, TH}. Every outcome has a probability of 1/4 so the probability that the other coin is tails is 1/2 but the probability of being also heads is 1/4? I know this is wrong but I don't really understand where the reasoning is failing.

[wheat osprey]

Intuitively, because coin tosses are independent, you'd say it's 1/2
not quite. I think it's similar to the monty hall problem

because...informally, the knowledge that one is heads sort of carries over all the events

the only event that has no heads is the all tails event

[rare sail]

but how is the output of one of the coins affecting the probabilities of the other?

[wheat osprey]

oh I missed the part where you're wanting the probability of the other coin being heads

P(2 heads|one is heads) = P(2 heads and at least one is heads)/P(at least one is heads)
= (1/4)/(3/4)
= 1/3

it's because the events "2 heads" and "at least one is heads" are not independent

[rare sail]

now that's more unintuitive

but it doesn't make sense how you toss a coin and it turns heads, you go to another galaxy a million years later you toss another coin and its probability to be heads is 1/3?

how's that possible?

[wheat osprey]

no - the chance that second coin is heads is still 1/2.

[rare sail]

I think I lack basic understanding of conditioned probabilities

[wheat osprey]

the 1/3rd here is given that you will get at least one heads, you will get 2heads

so you know that you're not getting 2 tails

that 1/3rd chance also includes the possibility that your first flip is tails

[rare sail]

oh I think I get it now

[wheat osprey]

P(2 heads|the first flip was heads) is 1/2

[rare sail]

yeah yeah I get it

[wheat osprey]

conditional probability is hell to understand, yeah

[rare sail]

omg you solved an entire night of nightmares

that was haunting me lmao

🙏

Thanks

.close

[marsh citrus]

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[flint reef]

The question is as follows: Given tan 4pi/9 = 5.6713, determine the following (a and b part) to 4 decimal places

Is my process for b) correct?

Answer in book

[marsh citrus]

@flint reef Has your question been resolved?

[flint reef]

<@&286206848099549185>

[marsh citrus]

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[marsh citrus]

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[hasty ruin]

seems a little vague
but i guess i can claim lim{x -> inf} 1/x has limit -100 in that definition

as x goes to inf 1/x gets closer and closer to -100 because it gets smaller and smaller

[glass silo]

"gets closer"

[marsh citrus]

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[marsh citrus]

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[celest shadow]

Im completely lost can anyone walk me step by step ive tried watching a vid but got the wrong answer twice

[boreal cloak]

Find the average o...?

[boreal cloak]

both questions?

[forest light]

how do you do division

[boreal cloak]

8 hours ago...

[loud kernel]

https://discord.com/channels/268882317391429632/488120190538743810

[celest shadow]

Yeah im completely stuck ive watched 3 diff videos i keep ending up with wrong answer

I got one last try or else they mark me down on hw

[boreal cloak]

can you take a picture without the text being cut off?

[boreal cloak]

you know how to find the average slope?

[celest shadow]

Kinda yeah

I jist tried it again

[boreal cloak]

what did you get?

[celest shadow]

I got 1.0080

[boreal cloak]

are you looking for exact or decimal answers?

[celest shadow]

Yeah decimal answers

[boreal cloak]

1.008 is too low

how did you get that?

[celest shadow]

I did f(1)-f(6)/1-6

[boreal cloak]

ah

$\frac{f(b)-f(a)}{b-a}$

[elfin berry]

mud ツ

[boreal cloak]

it should be $\frac{f(6)-f(1)}{6-1}$

[elfin berry]

mud ツ

[celest shadow]

Ohhh

Lemme try it real quick

[boreal cloak]

it should be the same though...

[celest shadow]

6.899?

Idk wtf i did above i mustve messed up somewhere

[boreal cloak]

somewhere...

[celest shadow]

Wait

I fucked up with my calculator

Its 2.8989

[boreal cloak]

yes

[celest shadow]

Thank you i see

Now can you show me how to approach the second one if you dont mind?

[boreal cloak]

do you know what the mean value theorem states?

[celest shadow]

Like f(b)-f(a)/b-a?

[boreal cloak]

kind of

if f is differentiable on (a, b) and continuous on [a, b], then there is at least one number c in (a, b) such that $f'(c) = \frac{f(b)-f(a)}{b-a}$

[elfin berry]

mud ツ

[celest shadow]

So i need to differentiate f?

[boreal cloak]

yes

[celest shadow]

So just 5/x^1/2

[boreal cloak]

yes

[celest shadow]

Then i plug in c?

[boreal cloak]

you're trying to find c

but yes

[celest shadow]

So how do i take 5/x^1/2 to find c?

[boreal cloak]

plug c into the derivative function, and equate that to what you found in the first part

[celest shadow]

Ok so 5/x^1/2=2.8989

[boreal cloak]

since by the mean value theorem,

yes

[celest shadow]

Ohhhhh

[boreal cloak]

to be honest, it's probably easier if you have the exact answer

[celest shadow]

exact answe?

[boreal cloak]

of the average slope

[celest shadow]

So i shouldnt round my answer for the slope?

[boreal cloak]

that's right

[celest shadow]

Alright ill try it again

i got 2.9747 for the second

[boreal cloak]

hm

[celest shadow]

2.97474488 for the full answer

[boreal cloak]

i was looking for an answer with radicals

does this make sense?

[celest shadow]

Yeah thats a lot easier

[boreal cloak]

alright, then set that equal to f'(c)

and solve for c

[celest shadow]

10sqrt6-5/x^1/2?

[boreal cloak]

there shouldn't be x anywhere...

remember that you plugged in c for x?

so $2\sqrt{6}-2 = \frac{5}{\sqrt{c}}$

[elfin berry]

mud ツ

[boreal cloak]

i would multiply by $\sqrt{c}$ on both sides, so you would get $\sqrt{c}(2\sqrt{6} - 2) = 5$

[elfin berry]

mud ツ

[boreal cloak]

then use algebra to isolate c

how is it going along?

[marsh citrus]

@celest shadow Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[brittle widget]

15cos2θ + 20sin2θ + 7 = 0 for 0<=θ<=180 degrees

[brittle widget]

<@&286206848099549185>

[marsh citrus]

@brittle widget Has your question been resolved?

[remote barn]

uh, I don't know how to explain this
5(3cos2θ+4sin2θ)=-7
3cos2θ+4sin2θ=-7/5
3cos2θ+4sin2θ+4=-7/5+4
3cos2θ+4(sin2θ+1)=-7/5+4
3(cosθ^2-sinθ^2)+4(2sinθcosθ+sinθ^2+cosθ^2)=-7/5+4
3(cosθ+sinθ)(cosθ-sinθ)+4(sinθ+cosθ)^2=-7/5+4
(sinθ+cosθ)(3(cosθ-sinθ)+4(sinθ+cosθ))=13/5

wait I replied too early

[brittle widget]

oh yea i see what you did there

do i have to turn cos and sin to tan somehow?

[remote barn]

Not necessarilY

[remote barn]

(sinθ+cosθ)(3(cosθ-sinθ)+4(sinθ+cosθ))=(sinθ+cosθ)(7cosθ+sinθ)=13/5 and I actually have genuinely no idea how to proceed from here I feel like what I did was useless

[remote barn]

expanding, we get 6cosθ^2+8sinθcosθ=8/5

[remote barn]

I feel like you find the answer through discriminants

b^2-4ac__>__0

So like
sinθ^2-3/5__>__0 which is weird

[marsh citrus]

@brittle widget Has your question been resolved?

[brittle widget]

eh just whatever at this point

.close

[marsh citrus]

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[marsh citrus]

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• Be polite and have a nice day!

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[lucid hinge]

i dont understand how to solve this question altogether

[marsh citrus]

@lucid hinge Has your question been resolved?

[lucid hinge]

<@&286206848099549185>

[marsh citrus]

@lucid hinge Has your question been resolved?

[lucid hinge]

.close

[marsh citrus]

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