help-13

[brisk nova]

idk

[lyric narwhal]

what do you mean you aren't allowed to use trig

it's literally a trig question

[brisk nova]

thats what was told

[wicked mantle]

Is sine law allowed?

[brisk nova]

idk

no

pure geometry

[wicked mantle]

no?

[muted timber]

wait, do they expect you to solve it just with pythagoras?

[lyric narwhal]

I doubt it's possible that way

maybe using some weird construction

[wanton isle]

divide it into right triangles

[lyric narwhal]

yeah exactly we have practically no information here

[brisk nova]

idk

[muted timber]

OOF

[brisk nova]

they just mention pure geo

[wicked mantle]

Bruh

I did solve it on my own. In my method, sine law is required

[lyric narwhal]

yeah one of either sine or cosine law

will be used

probably sine law is better

[wicked mantle]

True

[brisk nova]

soo

what should i do?

[lyric narwhal]

ask your teacher if you're allowed to use trigonometry

maybe you misunderstood

[brisk nova]

kk

[wanton isle]

even right triangles uses trig

[wicked mantle]

Here’s my method
ABC is similar to ADC

[brisk nova]

yeah

[wicked mantle]

I guess it’s enough for you to do the rest

[lyric narwhal]

$\Delta ABC\sim\Delta DBA$

$AC=5, BC=AB+2$

[wraith dagger]

kheerii

[lyric narwhal]

hmm, maybe similarity might be enough to solve it

creating equations for AB, BD and DC

[crystal raptor]

(\sim) 🙂

[wraith dagger]

kheerii

[lyric narwhal]

thanks

[wicked mantle]

Alr, if you have any problem with the progress, just open a new channel

I just wanna inform you of an easier approach

.close

[cedar kiln]

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[cedar kiln]

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[past sonnet]

all i have written down is

a + 1, a + 2

I really dont know what to do after

[undone halo]

well you can simplify it to

a and a+1

[past sonnet]

yes

[undone halo]

because those are already consecutive

[past sonnet]

yep

[undone halo]

ok so i have no idea what to do but lets just see what happens

so the square of a is

[past sonnet]

a^2

[undone halo]

and we have (a+1)^2

[past sonnet]

a^2 + 2a + 1

[undone halo]

ok so we have

(a+1)^2 and (a)^2

[past sonnet]

yep

[undone halo]

subtract them

to get the difference

[past sonnet]

2a + 1

[undone halo]

ok

whats a + (a + 1)

[past sonnet]

ah! I see

[past sonnet]

2a + 1

ty

quick follow up q

[past sonnet]

when it says "squares of ..."

whatever comes after, should i square it?

[undone halo]

Probably

It might change from problem to problem so you just need to analyze it carefully

[past sonnet]

i see

thanks!

[undone halo]

close the channel when done

[past sonnet]

.solved

[cedar kiln]

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[tawny drum]

im clueless

i tried dividing then u sub

it didnt work

[mortal hemlock]

i think using trig identities would work

[tawny drum]

nvm i figured it out

u can just divide out a -1/2

then u can easily integrate it to form of Ln(n)

.close

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[dense walrus]

Having a hard time with understanding question 13, that's it's answer tho, ping if you can help

[uneven quartz]

having a hard time reading the question on my low-resolution pc.

[inland wigeon]

How do I make a channel ?

[undone halo]

Lets take it step by step

You know that one dozen of "articles" is 3060

[dense walrus]

Yes...

[uneven quartz]

go here and type or paste your quesion.

[undone halo]

So how much is 1 article

Seeing that 12 is 3060

[dense walrus]

3060 divided by 12

[undone halo]

= 255

[dense walrus]

Yes

[undone halo]

Now Sarah sold it for 20% profit

Meaning they increased the price by 20%

So whats 20% of 255

[dense walrus]

51

[undone halo]

Now add that to 255

[dense walrus]

306

[undone halo]

So for each garment they sold it for 306

[dense walrus]

Mhm

Meaning 255 is the selling price?

[undone halo]

Actually wait ill look at the question again

No

[dense walrus]

Yeah that's what's got me confused rip

Ohh got it got it

306 is the selling price

[undone halo]

Ok I see

306 is the already discounted price

[dense walrus]

Oh

[undone halo]

As in they already gave the 10% discount

[dense walrus]

Ok

[undone halo]

The discount of an item is as the problem said

(Marked Price - Discounted Price)/Marked Price

[dense walrus]

Mhm

[undone halo]

Let x be the Marked Price since its unknown

The discount is 10%, so its 1/10

so its

(x - 306)/x = 1/10

[dense walrus]

Yeah that's what got me lost

[undone halo]

Ok

Multiply both sides by x

x[(x-306)/x] = x[1/10]

The left fraction's x gets cancelled out

x-306 = x/10

Still following?

[dense walrus]

Yeah

[undone halo]

Now multiply both sides by 10

[dense walrus]

Got a question in mind but finish first

[undone halo]

To cancel out the right fraction

10(x-306) = x

10x - 3060 = x

This part should be simple enough for you to solve

[dense walrus]

Yea

So I was looking at the question and I think I somewhat understand

I was wondering why they would equate the mp-sp/mp to 10/100

I'm used to (mp-sp/mp)× 100=90

So you said 306 is after the discount right?

[undone halo]

Yes

[dense walrus]

Ok I think I got it

Thank you so so much

[undone halo]

You can check yourself

Whats 10% of 340

[dense walrus]

34 oop

[undone halo]

Subtract that by 340

And youll find it perfectly matches the discounted price of 306

[dense walrus]

AHH yeah yeah

Alright thank youu

[undone halo]

Close the channel when done

[dense walrus]

Alr

.close

[cedar kiln]

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[neat raven]

hi

[cedar kiln]

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[lavish quarry]

Whats help you need

[neat raven]

.close

🙂

!original

[cedar kiln]

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.

[dusk finch]

.close

[cedar kiln]

Channel closed

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[dusk finch]

(you have another channel open, use that)

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[cedar kiln]

@umbral linden Has your question been resolved?

[cedar kiln]

@umbral linden Has your question been resolved?

[nova glacier]

@umbral linden hi

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[umbral linden]

1

[nova glacier]

Do you know how such path must look like?

Did you draw it?

[umbral linden]

ok actually i lied about being at 1

i got to a point where I did a bionomial of (16, 8) and got the total paths

12870

then idk what t odo

[nova glacier]

Did you try to draw it?

It might help

[umbral linden]

yes

i drew it

[nova glacier]

Ok so either you go below or above the square right...

[umbral linden]

yea

should i just subtract all the paths taht go through the square then

[nova glacier]

Or count directly

But you could also do that...

Let's choose one

[umbral linden]

counting directly would take to long

12k+ different routes

unless i did that horribly wrong

[nova glacier]

No,

I'm thinking about splitting into paths that go below and above

If it's below, than you must not go above the point (-2,-4)

That is, in the first 8 steps, you must have at most two up

[umbral linden]

(-2,-4)?

[nova glacier]

I mean 2

[umbral linden]

ok

[nova glacier]

(2,-2)

[umbral linden]

yea i get that

[nova glacier]

So how many are there?

that go below

[umbral linden]

uhh i manually count it?

[nova glacier]

You can

you want to find another way?

[umbral linden]

yea

lemme see

[nova glacier]

I think it would be annoying to find the paths that go through the square too

[umbral linden]

yea

[nova glacier]

but lets do that

[umbral linden]

gimme a sec lemme grab my paper

[nova glacier]

Let's do both ways, whatever

Let's finnish what we started with goes below

The ones that go above is the same by symmetry

So you must get to x=2 without getting to y=-1

[umbral linden]

it cant even be y=-2

[nova glacier]

It can

[umbral linden]

it can?

[nova glacier]

They say with boundaries right?

[umbral linden]

oh

i misread that part

[nova glacier]

so you either need to get to (2,-2), (2,-3) or (2,-4) and from there to (4,4)

That's simple calculation

[umbral linden]

(2,-1) is in the square thogh

so is (2,0)

[nova glacier]

OOPS

(2,-3) or (2,-4)

i meant

My coordinates are off

[umbral linden]

I don't know how to add up all the different methods though

I would assume its a 2^n system

but When we get into the square, we limit it so

I don't know what to do

[nova glacier]

just count the number of ways you can get from (-4,-4) to (2,-4) and multiply by the number of ways to get from (2,-4) to (4,4)

Then repeat for (2,-3) and (2,-4)

Add them up

multiply by two for the paths that go up

and you are done

[umbral linden]

theres gotta be abetter wy lmao

[nova glacier]

That's very simple man, but I'm with you

We can try to find a simpler way

[umbral linden]

I know its simple but its tedious

[nova glacier]

But let's finish this one its very simple

[umbral linden]

ok

[nova glacier]

(-4,-4) to (2,-2) -> (2,-4) to (4,4) is 8C2 * 8C2

[umbral linden]

what does the C mean

[nova glacier]

Choose

[umbral linden]

I just started learning combinatorics sorry

[nova glacier]

$\binom{8}{2}$

[wraith dagger]

Cain

[umbral linden]

ok

yea that was what I was gonna do

[nova glacier]

(-4,-4) to (2,-3) -> (2,-3) to (4,4) is 7C1 * 9C2

(-4,-4) to (2,-4) -> (2,-4) to (4,4) is 1 * 10C2

[umbral linden]

why is it 7 choose 1

and 9 choose 2

I know is 7 steps

but why choose 1

[nova glacier]

7 steps you need to choose the one step up

[umbral linden]

ok

[nova glacier]

Then 9 steps you need to choose the 2 steps right

[umbral linden]

oh I see

Ok

alright gimme a minute to process the information

Why isn't it 7c6

because of the 6 going right

oh wait

those are the same thing

lol mb

[nova glacier]

yes

same

[umbral linden]

I keep forgetting these thing lol

[nova glacier]

nCk = nC(n-k)

[umbral linden]

ok yea I get that

so 252 and 45

wait what why is it 1

1x10c2

oh

nvm

[nova glacier]

because you have one way to get to (2,-4)

[umbral linden]

yes

sorry I am slow lol

ok yea

these are the valid pathways

[nova glacier]

but it someway doesn't give a valid answer

[umbral linden]

what did I forget then

or are all the answers wrong

[nova glacier]

What did you get?

oh

[umbral linden]

correct answer is d apparently

idk how to get that

[nova glacier]

This doesn't seem to get us there

Not sure why

Let's try a different way and figure this out

[umbral linden]

alright

[nova glacier]

Let's try to count the paths that goes through the square

[umbral linden]

ok

[nova glacier]

oh I know why

[umbral linden]

so starting from (-2, -2) to (2, 2)

[nova glacier]

We double counted

because from (2,-3) you can get to (2,-2)

and we counted this path already

[umbral linden]

?????

[nova glacier]

I mean, when we counted the paths that reach (2,-2)

and added the paths that reach (2,-3)->(4,4)

we counted some paths twice

[umbral linden]

oh

yea

we did

[nova glacier]

We need a different way to avoid that

basically we need to remove the paths that go up

on the next step

after reaching (2,-3)

or after reaching (2,-4)

so

Basically we need to get (3,-3)

and (4,-4)

[umbral linden]

starting from (-4, -4)?

[nova glacier]

yes so to fix this

we need to replace (2,-3) with (3,-3)

and (2,-4) with (4,-4)

To avoid double counting

[umbral linden]

so instead of
(-4,-4) to (2,-3) -> (2,-3) to (4,4) is 7C1 * 9C2
(-4,-4) to (2,-4) -> (2,-4) to (4,4) is 1 * 10C2

we should do
(-4,-4) to (3,-3) -> (3,-3) to (4,4) is 8C1 * 8C1
(-4,-4) to (4,-4) -> (4,-4) to (4,4) is 1* 1

one sec lemme edit the bionomials

[nova glacier]

(-4,-4) to (3,-3) -> (3,-3) to (4,4) is 8C1 * 8C1
(-4,-4) to (4,-4) -> (4,-4) to (4,4) is 1 * 1

[umbral linden]

yea

ok

[nova glacier]

, 2*((8 choose 2) * (8 choose 2) + (8 choose 1) * (8 choose 1) + 1)

[umbral linden]

so then we get 64 +1

[nova glacier]

,w 2*((8 choose 2) * (8 choose 2) + (8 choose 1) * (8 choose 1) + 1) //N

[umbral linden]

huh??

[nova glacier]

1698

[umbral linden]

gimme a sec to think this through rq

[nova glacier]

I multiplied by 2 to add the paths that go above the square

[umbral linden]

oh we forgot that

the first time

[wraith dagger]

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Requested by cain0196

[nova glacier]

I didn't

[umbral linden]

I did

lol

[nova glacier]

But that's the right answer apparently

We got there eventually

lol

[umbral linden]

lol one sec i'm very slow

[nova glacier]

It's just putting everything into a calculator

[umbral linden]

why did u do 2(8c2)

like where did you get the 8c2 from

[nova glacier]

(-4,-4) to (2,-2) -> (2,-2) to (4,4) is 8C2 * 8C2

(-4,-4) to (3,-3) -> (3,-3) to (4,4) is 8C1 * 8C1
(-4,-4) to (4,-4) -> (4,-4) to (4,4) is 1 * 1

[umbral linden]

ok

[nova glacier]

That's the 3 options we had

[umbral linden]

yep

ok

[umbral linden]

why is it from (2,-4)

instead of (2,-2)

[nova glacier]

typo

[umbral linden]

oh ok

why did we choose (3, -3)?

[nova glacier]

Our original idea was (2,-3)

But then we saw that if we go up after (2,-3), we already counted this path

so we must go right

so instead of (2,-3), we switched to (3,-3)

To avoid counting those that go from (2,-3) to (2,-2)

[nova glacier]

we already counted them here

[umbral linden]

oh

[nova glacier]

A cleaner way to see this

Is that you must go through the diagonal

(z,z)

But the only places you can go through are (2,-2), (3,-3), (4,-4)

Now we can use this observation to get a cleaner solution

With the paths that go through the square (the way you wanted in the first place)

Can you see how?

[umbral linden]

one sec sorry I'm redrwaing this all online to better visualize it

[nova glacier]

K

Digest it afterwards lol

[umbral linden]

ok

I get it

[nova glacier]

Now do the second way

see if you got it

The paths that go through the square must cross the diagonal in which points?

[umbral linden]

wait...

wait what diagonal r u talkign about

sorry

if we start from (-4, -4)

we go from there to lets say (-2, 2)

then we go from (-2, 2) to (4, 4) right?

[nova glacier]

Yes, but we want to move on to the second solution

Where you count the paths that go through the square

[umbral linden]

ok

[nova glacier]

instead of those going around

[umbral linden]

so we start from (-4, -4)

we go to (-2, -2)

and we get to (2, 2)?

[nova glacier]

This isn't good enough

[umbral linden]

all the solutions wehre -2>x>2 and -2>y>2

[nova glacier]

if you go through the square

draw a line from (0,4) to (4,-4)

[umbral linden]

ok

[nova glacier]

That's the diagonal I'm talking about

I'm saying that any path must cross this diag

Path that goes from (-4,-4) to (4,4)

and if it gets into the square

it must cross it where?

[umbral linden]

(2, 0) or (0, 2)?

[nova glacier]

no

This is not the diag

[nova glacier]

the diag is (x, -x)

oopss

I messed up with the coordinates

I meant (-4,4) to (4,-4)

for the diag

[umbral linden]

ok

[nova glacier]

you can see that this is not the case in the picture because you could have go through the square without crossing the diag there

[nova glacier]

yes

now if you get into the square

you cannot get out without crossing the diag inside the square

[umbral linden]

wdym get out

[nova glacier]

you need to finish in (4,4) eventually

so the path needs to get outside the square

somewhere

there's an exit point

[umbral linden]

how can you do taht with the first diagonal?

I mean

like it still must cross it right?

[nova glacier]

not inside the square

you could go to (-1,-4) and then up to (-1,4) and then to (4,4)

[umbral linden]

oh

[nova glacier]

You must cross it but not inside the square ...

[umbral linden]

ok i see

[nova glacier]

so now if you get in the square you must cross in (-1,1), (0,0) or (1,-1)

you can calculate the number of paths that go through these as we did

and then subtract from the total number of paths

To get the solution

try that

[umbral linden]

does (-2, 2) count as in the square?

[nova glacier]

no because you can cross there without being inside the square

the "bad" paths are those that are getting in

not on the boundaries

[umbral linden]

ok

[nova glacier]

Let me know if it works out

I g2g now

[umbral linden]

:c

ok well uve been a lotta help ty

[nova glacier]

yw

[umbral linden]

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[mighty shuttle]

if you have -1, how would you make it 1

[crimson sedge]

,w graph |x|

[wraith dagger]

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Requested by firstisomorphism

[mighty shuttle]

exactly

which is what you're doing here

[cerulean linden]

This is just how absolute value is defined. Here’s a convenient way to think about it using vectors: let’s say you have a vector to the point -1 from the origin on the real number line, then the length of this vector would be 1. The absolute value will always represent the length of the vector on this line.

[cedar kiln]

Channel closed

Closed by @cunning spoke

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[neat raven]

https://tenor.com/view/nice-gif-21458880

[neat raven]

yes. the extent of the vector.

[cedar kiln]

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[tidal prism]

I only got one question, where did this five come from

[restive shard]

f(-1) = 5

[tidal prism]

Is it because we have the x and y values and since f(c) equals Y we substituted the value?@restive shard

Of y into it

[restive shard]

it’s more that you have a function with an input
simplifying it to a number lets you do your basic operations like adding and subtracting easier

substituting is the only way we can get an actual numerical answer without variables

[tidal prism]

Isn't that basically what I just said

[restive shard]

idk why you mentioned y

there’s no y on the paper

[tidal prism]

The (-1,5)

[restive shard]

the second value isn’t necessarily always the y

it could just be represented as f(x) instead

which tends to be the norm when talking about functions from now on

[tidal prism]

Yea

In this section f(c) and f(x) are the same

And y equals f(x)

[restive shard]

i wouldn’t say that
x is a variable, c is a constant

[tidal prism]

Yea you're right

So my question is, if we substituted 5 into the right side of the equation since f(-1)= 5 why didn't we substitute it into the right side where it says f(c+x)

[restive shard]

we did basically did, but we showed more steps

technically you could write the 5 as 4-(-1) to make it make more sense

but it’s all the same thing

[tidal prism]

Isn't it 4-(-1) because we substitute the original equation into it

4-x

[restive shard]

no cuz you also have the delta x

so it becomes 4-(-1+ delta x)

which becomes 5-delta x

[tidal prism]

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[quartz salmon]

Determine the equation of the normal line (in parametric form) to the surface given by $z=x^2+y^2$ at the point $(1,1,2)$. The normal line is the line that passes through the point $(1,1,2)$ and is perpendicular to the surface at that point.

[wraith dagger]

dghf

[quartz salmon]

I don't know where to begin. Partial derivative?

And then?

[cedar kiln]

@quartz salmon Has your question been resolved?

[quartz salmon]

Since it’s z = x^2 + y^2, then the gradient of f is (2x, 2y, -1).

Evaluated at the point (1,1,2), it’s (2, 2, -1)?

In parametric form, we write r(t) = r_0 + td, where r_0 is (1,1,2), and d is (-2,-2,1). Good so far?

So the parametric equation becomes $$\begin{cases} x = 1 - 2t \
y = 1 - 2t \
z = 2 + t
\end{cases}$$

[wraith dagger]

dghf

[quartz salmon]

Can someone check my work?

2nd time a mentally disabled moron interrupts my help channel today

.close

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[floral arrow]

Yes, though you could just write it (x,y,z) = (1,1,2) + (2,2,-1)t
or (x,y,z) = (1,1,2) + (-2,-2,1)t, same thing

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[undone star]

Scope of question is high school math so it would be preferable if your answers aren’t too over the top

[undone star]

To paraphrase the question, it is essentially asking why can we treat the differential operators as fractions in most cases

And if so

Why is the second derivative not written as something like

(I typed it up on overleaf because idk how to use the discord bot)

I’m not really sure how to justify this as really, we have just been taught in school that the chain rule works and u can just cancel things

The question does come with the hint that (f+g)’ = f’ + g’ where f and g are some arbitrary functions

Again, I’m not really sure what this is meant to allude to

[cedar kiln]

@undone star Has your question been resolved?

[undone star]

<@&286206848099549185>

[near mango]

damn

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@undone star Has your question been resolved?

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[deep kite]

hello, im not sure how to start

[cedar kiln]

@deep kite Has your question been resolved?

[deep kite]

<@&286206848099549185>

[clear umbra]

have you learnt the sum and difference identities for sin and cos?

[deep kite]

yeah

[clear umbra]

then heres a start: connect D and B together

[deep kite]

Pythagoras theorem?

[clear umbra]

yes

[deep kite]

idt it works coz it will break up the θ

[clear umbra]

thats why i asked you if you learnt the sum and difference identities before

connecting those two points break the angle into two parts

[deep kite]

but then i would have to break up BD too and the working would get very complicated

also, it’s a 3 mark question, the working shouldn’t be so complicated

[clear umbra]

not really?

[clear umbra]

it really isnt

[deep kite]

im thinking of toa cah soh

[clear umbra]

ok this is a good start

you dont really need to know the exact value of theta

[deep kite]

θ can vary

[clear umbra]

?

[deep kite]

like u can change θ to change the perimeter

[clear umbra]

oh yeah

but for now it doesnt matter

because we're trying to find the equation for the perimeter

[deep kite]

am i supposed to do some simultaneous trigo by finding 2 expressions for the height

[clear umbra]

nope

you see how there is two angles after connecting D and B

[deep kite]

yea

[clear umbra]

BDC and ADB right?

[deep kite]

yep

[clear umbra]

BDC is a smaller angle

lets call it something like k

so whats the remaining angle

[deep kite]

θ-k

[clear umbra]

and the hypotenuse is common for both right triangles yes?

[deep kite]

yes

[clear umbra]

ok so what is the side AB

using trig ratios

[deep kite]

ok i get it now

lemme try it out

[deep kite]

ok but that doesn’t look anything like 125 or 175

[clear umbra]

well can you expand sin and cos of theta - k?

[deep kite]

Oh missed out a coefficient

for the sin θ sin k

[clear umbra]

do you know the values of sin k and cos k?

[deep kite]

yeah

[clear umbra]

then put them in

[deep kite]

Woah!

it works

[clear umbra]

wasnt so complicated after all

[deep kite]

eh

Seeing it is tough

well thx

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[gentle cape]

,w normcdf(-(1-0.99)/2)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by electrophon

[gentle cape]

i tried to use normcdf(-(1-0.99)/2) on matlab but ithe output doesn't yield the correct5 answer

[neat raven]

,w 1*234

[wraith dagger]

Results provided by WolframAlpha

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Requested by anshulyolo121

[cedar kiln]

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[oak knoll]

Please, help me with the proof of this statement:

[potent echo]

Hi!

@oak knoll

Let me know when you come back

[azure swift]

If Adult Einstein needs help with this, then I’d be surprised if Baby Einstein will be able to help

[azure swift]

Didn’t you ask this a while back but for L(f,D) or am I tweaking?

[cedar kiln]

@oak knoll Has your question been resolved?

[olive wind]

Did that get answered?

This is a somewhat cool problem, I had something similar back when I took analysis

Not exactly this

But similar

[azure swift]

I don’t recall, but if it did; would maybe be beneficial to use

But it maybe doesn’t matter

[olive wind]

If it did this would be kind of pointless

The proof is symmetric

[azure swift]

Yea, like I’m guessing they didn’t just understand it rather

[olive wind]

This needs very careful manipulation of inequalities even if I give the proof outline

It's hard to rigorously write everything

So understandings is pretty important

[azure swift]

The notation is slightly different so im unsure if this is the same proof im thinking I had

Like U here is the upper staircase integral or whatever it’s called? Tried translating it lmao

[olive wind]

Upper Reimann integral

[azure swift]

Ah ok

[olive wind]

It's inf over all partitions of the upper sum

[azure swift]

While the integral on the right is the inf of all such integrals?

[olive wind]

@oak knoll Do you have that the Reimann integrator allows convergence to the integral with equally spaced partitions

[olive wind]

Inf on all the sums

[azure swift]

Oh thought u said it was the left one so was confused

[olive wind]

Left one is an upper sum with respect to partition D

Right one is the upper integral

Which is the inf over all partitions of the upper sum

[azure swift]

Oh lmao

[olive wind]

Not necessary

But nice

[azure swift]

Hm so maybe im not recalling the correct one, but is one way to argue via a choose of delta with respect to some diagonal length of each subset of the partition?

Oh nvm this is one variable lmao

I should just stop interfering at this point

[olive wind]

The way I would do it is to get a random upper sum within eps/2 of the integral

[cedar kiln]

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[olive wind]

RIP

[azure swift]

@oak knoll !!!

[cedar kiln]

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[harsh raven]

how to get involved in advanced channels? thanks.

[undone halo]

#get-advanced-access

[harsh raven]

thank you

i click the roles of advanced, but still do not have the permission

ok, I think its been reviewed

.close

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[brisk nova]

idk where to begin

[dawn flume]

In any triangle, the median, altitude, and angle bisector intersect at specific points: the centroid (median), orthocenter (altitude), and incenter (angle bisector). When these three are concurrent, it usually implies that the triangle has special properties.

[brisk nova]

how do we know that this is only possible when the centiod oethocenter and circumcenter ore concurent

i thought of thats

[compact junco]

I would think about where those centers could possibly be

[brisk nova]

but couldnt find a way to prove it

[compact junco]

is this math oly

[brisk nova]

hmm

[compact junco]

istg ive seen this somewhere but i cant quite put my finger on iti

[dawn flume]

For these three lines (median, altitude, and angle bisector) to be concurrent in a single point, the triangle must be an isosceles triangle. This means two sides of the triangle are equal

[compact junco]

wait huh why

[brisk nova]

yeah

we cant prove that

the point dont need to concurrent

[compact junco]

I might consider dropping perpendiculars from A and O onto BC

(where O is the point where the 3 lines intersect)

If we call those perpendiculars A' and O'

theres quite a few more pieces of info we get

e.g. AA' and BE intersect at the orthocenter

[compact junco]

geogebra strat fr

[brisk nova]

yeah

ill try that

would OO' bisect BA'

or as per the diagram

QQ'

[compact junco]

Sorry Im not sure what you mean by that

[compact junco]

this problem is quite frustrating

[brisk nova]

is QQ' = A'J/2

[compact junco]

Not necessarily

[brisk nova]

okay

[compact junco]

I mean in this diagram it certainly looks like A'J < 2QQ'

[cedar kiln]

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[slim vigil]

need to prove this using

eulers formula

e^itheta = costheta +isintheta

[mortal hemlock]

try writing every term seperately

[slim vigil]

can we try

that

summation but

multiploication

the pie thingy

[mortal hemlock]

write out 2cos(x)-1 first

[slim vigil]

ok

[mortal hemlock]

using eulers identity

[slim vigil]

how

:c

[mortal hemlock]

e^(ix)=cos(x)+isin(x)
e^(-ix)=cos(-x)+isin(-x)

what happens when we sum these two equations?

[slim vigil]

uhhh

e

2cosx

=

e + 1/e

power

thing

[mortal hemlock]

just keep it as e^(ix) and e^(-ix)

[slim vigil]

yeah

'ok

damn alright

now - 1 huh

so e^(ix) + e^(-ix) - 1

= 2cosA

[mortal hemlock]

\begin{align*}e^{ix}&=\cos(x)+i\sin(x)\+\left(e^{-ix}&=\cos(-x)+i\sin(-x)\right)\\hline\end{align*}

[wraith dagger]

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

[slim vigil]

yeah i got that

now what

we got 2cosA value

[mortal hemlock]

indeed

[slim vigil]

-1 value

[mortal hemlock]

e^(ix)+e^(-ix)=2cos(x)

[slim vigil]

now same for next term?

[mortal hemlock]

yes

[slim vigil]

ok

so

e^(ix) + e^(-ix) = 2cos2a

[mortal hemlock]

almost

[slim vigil]

wait wat

how is it same

for all angles

is it now?

[mortal hemlock]

e^(iA)+e^(-iA)=2cos(A)

[slim vigil]

oh

yes right

[mortal hemlock]

e^(i2A)+e^(-i2A)=2cos(2A)

[slim vigil]

bet bet

so now we take pie?

product notation

ig lets try

[mortal hemlock]

sure

[slim vigil]

doing

[crimson sedge]

hi

[mortal hemlock]

hi

[crimson sedge]

need some help?

[mortal hemlock]

!help

[cedar kiln]

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

[slim vigil]

yeah i mean

sure

ok

i got

n-1 pie k = 0 e^(i(2^k)A) + e^(-I(2^k)A) - 1

is this correct

idk latex sorry

[mortal hemlock]

i dont understand what you tried to type

[slim vigil]

there

uhh

dont you get that

[mortal hemlock]

no

what is the equal sign for?

[slim vigil]

here

this

everything in product notation

[mortal hemlock]

$\prod_{k=0}^{n-1} e^{i2^kA}+e^{-i2^kA}-1$

[wraith dagger]

Flappie

[slim vigil]

yes that

now what

use the formula?

that would give 2cos(2^k *(A) - 1)

brrrrr

[mortal hemlock]

thats back where we were

[slim vigil]

oh ok so now?

[mortal hemlock]

hmm, maybe something with induction

[slim vigil]

yeah that

i dont get that

[mortal hemlock]

try (2cos(x)-1)(2cos(2x)-1)

just multiply try9ng

that

but in the e power form

[slim vigil]

this thing

maybe

[mortal hemlock]

this was given?

[slim vigil]

lets just say it is generated

[crimson sedge]

cant you just multiply 2cosa +1

[mortal hemlock]

.....

[slim vigil]

dont hope for it to be correct

[mortal hemlock]

!nogpt

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[slim vigil]

yeah that is why

multiply?

uhh would u use euler in that method?

i need an euler method

[crimson sedge]

this yields (4cos²a -1)(2cos2a-1)... which is (2cos2a + 1)(2cos2a-1)...

[slim vigil]

i know that method

i need a euler formula method

what my teach needs...

[crimson sedge]

aight then

[slim vigil]

man i need to get after this

<@&286206848099549185>

.close

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[lone burrow]

What’s my next step here I’m lost

[mortal hemlock]

you can factor that as a square

y^4+1/2+1/(16y^4) = (a+b)^2
a=? b=?

[lone burrow]

How

There’s no solutions

[mortal hemlock]

$(a+b)^2=a^2+2ab+b^2=y^4+1/2+1/(16y^4)$

[wraith dagger]

Flappie

[lone burrow]

Did you have one in mind

[mortal hemlock]

i wonder what this could be

[lone burrow]

y^2

[mortal hemlock]

🤔

[lone burrow]

How does that get you 1/2

[mortal hemlock]

haha sorry this is a bit too passive aggressive

[lone burrow]

I tried that half an hour ago

[mortal hemlock]

a=y^2
b=1/(4y^2)
a*b=y^2/(4y^2)=1/4
2ab=1/2

[mortal hemlock]

theny ou calculated wrong

[lone burrow]

I think you’re onto something here

Alright got it thanks

.close

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[mortal hemlock]

👍

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[crimson sedge]

anyone know how to solve this?(where x is worth 2)

[mental trail]

are you trying to get a or x?

[crimson sedge]

a problably

its for my math test

[mental trail]

so

[crimson sedge]

idk what my teacher want on this

[mental trail]

get rid of denominators (meaning multiply both sides by same quantity)

then put everything with "a" on one side

the rest on the other side

[crimson sedge]

k, i go try

i get 19a = 11a

what i should do now?

[agile vortex]

Try again

@crimson sedge A must = 2?

[crimson sedge]

my teacher said that x = 2

[agile vortex]

Show your working I thing you went wrong somewhere

[crimson sedge]

wait

i make it wrong

srry

[agile vortex]

Its fine

[crimson sedge]

get 36a = 18a

what i should do? 2a = 18 - 36?

[agile vortex]

I think you’ve messed up with getting rid of the denominators

How did you do it?

@crimson sedge

[crimson sedge]

u multiplied the numeratos by the denominators

4(2a . 2+1) = 2(2+a . 2+3)

[leaden shore]

move 4 to the right side multiplying the binomial 2ax+1, smae with 2 but to the left side multiplying x+ax+3

[leaden shore]

kk you already did it, now expand it

after expanding it try to factorize the letter of your interest, either x or a, whatever your teacher asked you to get the value of

just play with algebra

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[coarse geyser]

just so you know these are differential equations, basically I want to know if he just redefined the constant c as ln|c| so he can use the logarithmic rules

[coarse geyser]

cause -ln(x) = ln(1/x)

and ln(1/x) + ln(c) = ln(c/x)

right?

[split viper]

yeah

what is your doubt though? that's pretty straightforward

[coarse geyser]

just to be sure

.closed

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[cedar kiln]

Please don't occupy multiple help channels.

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[quartz salmon]

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[warped cliff]

why does the blue graph contain a limit but the yellow graph doesn't?

[warped cliff]

for the yellow graph because the limit from the left and right side are different, there isn't a limit

shouldn't that be the same for the blue?

why is it that the limit for the blue graph as x approaches 4 is 6.7

[hearty plank]

maybe it was just a bad drawing? it's true that the limit doesn't exist if the left and right limits are different. The blue graph has both sides approach a value ~ 6.7, which is why it's the limit.

[restive shard]

the yellow one should have equivalent left and right limits

at least from the visuals

[warped cliff]

yes but if you approach from the left the slope would approach negative infinity and vice versa

oh wait

is it because the yellow graph is referring to the limit of the slopes

[hearty plank]

I think I understand what you're trying to ask, but could you maybe rephrase your question?

[warped cliff]

you know how a derivative of some point can be defined as the limit of the secant lines at points a distance away from that point as the distance approaches 0

that's what the yellow graph is referring to

[hearty plank]

yup, but I don't think that would relate to the limit not existing. if you're trying to find the limit of f(x) as it approaches x=c, then it will have a limit.

[warped cliff]

yeah that's true but khan said specifically that the limit as x approaches that point for the yellow graph doesnt exist

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-4/v/differentiability

you can check it out if you want

time stamp is 6:10

nevermind

im tripping

i guess i mixed "the limit of this expression" with "the limit of this graph"

thanks though

.close

[cedar kiln]

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[cedar kiln]

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[elder night]

Question number 2 in this textbook. I just need to know how to solve it, not the answer please. What information do I need in order to get the prior and posterior probabilities for the female military recruits for example. I have just learned probability and bayes theorem and I'm still confused. And also what is the question asking when it says XX as a function of XX on a graph.

[elder night]

This is the work that I have done for Q1 using Bayes Theorem

My issue is that I feel that either I dont understand there's already information given, or there isnt. And am I expected to assume? I'll attached the full background brief after this which does not say anything about the specific groups - female recruits in NJ, gay men in Providence, RI, or IV drug users in NYC

[cedar kiln]

@elder night Has your question been resolved?

[cedar kiln]

@elder night Has your question been resolved?

[cedar kiln]

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[acoustic quartz]

∫sin^4(x) * cos^4(x)dx

[acoustic quartz]

is this equal to this

i used the power reduction formua

[upper laurel]

graph the derivative of f(x) in desmos by typing d/dx f(x) and see if it lines up with sin(x)^4 cos(x)^4

[acoustic quartz]

it doesnt match up

but im not sure waht my mistake is

[upper laurel]

can you show the power reduction formula

[acoustic quartz]

Here’s my work too

[cedar kiln]

@acoustic quartz Has your question been resolved?

[acoustic quartz]

<@&286206848099549185>

[wanton sail]

Please don't post in other people's help channels if you're not helping

[azure trench]

@acoustic quartz For this one, you can do int (sin^2x)^2*(cos^2x)^2 dx

= int ((1-cos 2x )/2)^2 * ((1+cos 2x) / 2)^2 dx

= 1/16 int (1-cos2x)^2*(1+cos2x)^2 dx

= 1/16 int (1-cos^2 2x)^2 dx

{ a^2 - b^2 = (a+b)(a-b)}

= 1/16 int (sin^2 2x)^2 dx

[acoustic quartz]

r u using half angle identity?

[azure trench]

1- cos^2 2x = sin^2 2x, pythagorean

Then I'll use power reduction

[acoustic quartz]

wait where did the /2 come from?

[azure trench]

Power reduction for sin^2 x

[acoustic quartz]

Wait for here

does that use half angle?

[azure trench]

That uses power reduction formula:
sin^2 x = 1/2 * ( 1- cos2x)
cos^2 x = 1/2 * (1+cos2x)

[acoustic quartz]

Wait whats this formula called?

[azure trench]

I'm not sure for this one, haven't used it nor heard of it

[acoustic quartz]

ok i see

i understand now thanks so much

[azure trench]

np

[acoustic quartz]

.close

[cedar kiln]

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[crimson sedge]

yo

[cedar kiln]

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[chilly bolt]

are these 2 thigns the same?

what is the significance of the upper case T in the projection matrix formula

or does u^T essentially just mean v

[upper laurel]

do you know matrices

[chilly bolt]

i just started learning them

[upper laurel]

the ^T means to flip the matrix diagonally like this

this is "transposing" the matrix

A^T is the transpose of the matrix A

[chilly bolt]

is it cause u can only do dot product that way

[upper laurel]

you dont have to