help-4

[clever needle]

i should use it untill the remainder is 0?

like continue with the the divisors

[toxic ridge]

You can then find the other zeros via quadratic

[toxic ridge]

So you found 1 right!

There's no guarantee that the other roots are rational

[clever needle]

wait so i have to find the zero of the polynomial

[toxic ridge]

Yes

[clever needle]

wait so if x =1 then x-1 would be its factor

right?

[toxic ridge]

Yes

Now that's called factor theorem

[clever needle]

then whats the next step

[toxic ridge]

Now you divide the polynomial with x-1

You'll get a quadratic

[clever needle]

yeah i know that

i mean the

divison

[toxic ridge]

Then see if the quadratic is factorisable or not

[toxic ridge]

You don't know how to divide one poly with other?

[clever needle]

i do

but im not 100% confident with it

[toxic ridge]

Show me your work after doing it

I'll tell you if its correct or not

[clever needle]

idk i can im on pcc

but ill try to explain what i did

[toxic ridge]

Ok atleast tell me the quadratic you obtain

[clever needle]

i gotta divide this with x-1

right?

[toxic ridge]

Yes

[clever needle]

okay

[fringe lava]

If you know (x - 1) is a factor, you don't need to divide
You can express it as (x - 1)(ax² + bx + c)
a must be 1, c must be 120
(x - 1)(x² + bx + 120)
Equating x coefficients,
-bx + 120x = 142x
b = -22
(x - 1)(x² - 22x + 120)

= (x - 1)(x - 10)(x - 12)

[clever needle]

But am I allowed to do that on my exam if we havnt been introduced to that method?

[toxic ridge]

It's more of a logical intuition than a method

And i don't think any math teacher would not allow intuitive methods to fasten your calculations

[clever needle]

yeah

[pallid oasis]

I added up the coefficients. Sure rrt finds all rational roots, but in this case its overkill

[toxic ridge]

Again, adding up the coefficients is essentially putting p(1). Rrt is just a word to describe what you're doing

[clever needle]

@toxic ridge

x² - 22x + 120

x - 1 | x³ - 23x² + 142x - 120
-(x³ - x²)
----------
-22x² + 142x
-(-22x² + 22x)
------------
120x - 120
-(120x - 120)
------------
0

[toxic ridge]

Yep

Now see if you can factorise the quadratic

[clever needle]

middle term splitup?

[toxic ridge]

Yep

[clever needle]

x sqaure - 10x - 12x + 120

[toxic ridge]

Yep

[clever needle]

now we group thm

x(x - 10) -12(x-10

[toxic ridge]

Yep

[clever needle]

(x-10)(x-12)

[toxic ridge]

Yes

And the question is done

[clever needle]

but why is there (x-1) here

(x-1)(x-10)(x-12)

[fringe lava]

cuz it's a factor

lol

[clever needle]

is it neccassory

to put it

aanyways

thank you @toxic ridge , @fringe lava, @pallid oasis

.close

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[jade canopy]

What is the definition of a long memory time series

[sleek nebula]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

@jade canopy Has your question been resolved?

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@jade canopy Has your question been resolved?

[low olive]

iam bad in math

[lyric sundial]

<@&268886789983436800>

[limpid raft]

Do not use random peoples help channel to troll.

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[sullen tendon]

!statua

!status

[vale dock]

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

[sullen tendon]

2

[sullen tendon]

Over here

[boreal orchid]

Part a, b or c

[sullen tendon]

All of it

[boreal orchid]

Okay, well, do you know what the condition for two lines to intersect are

[sullen tendon]

When the gradient isn’t the same

@boreal orchid

[boreal orchid]

That's true technically, but not exactly what we need

[sullen tendon]

is there a certain condition?

Is it the rule
y-y1 = m(x-x1)

[boreal orchid]

Yes, more to do with the whole equation as opposed to the gradient

[sullen tendon]

Or smt?

[boreal orchid]

That's used to get the equation of a line when you have 2 co-ords and the grad

[sullen tendon]

Ohh

So could you please explain this question for me

[boreal orchid]

Well, if the two lines intersect, what that must mean about their solutions

[sullen tendon]

Uhh

Idk

[boreal orchid]

Okay, let's dial it back, when you have an equation for a line, like, let's say y = 4x + 2

To find the co-ordinates of it, you plug in your x value and get your y value correct ?

[sullen tendon]

Yeah

[boreal orchid]

And if they intersect, that means at some point, they must have the same co-ordinates right

[sullen tendon]

Yep

So I can set them to be equal?

[boreal orchid]

Precisely

[sullen tendon]

Once I rearrange

K I’ll give it a shot

[boreal orchid]

Coolios

[sullen tendon]

Ok so

I got

Y= -x/2 + 3.5
Y = -x/2 -k/4

and do I set them equal?

@boreal orchid

[boreal orchid]

Er sure, that's a little unconventional but it works

[sullen tendon]

Ok

@boreal orchid I got k = -15

Do I put that back into the equation before

,w -x/2 + 3.5 = -x/2 -k/4 solve for k

[rocky lotus]

Results provided by WolframAlpha

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Requested by clqvicle

[sullen tendon]

Oh

[boreal orchid]

Yeah I was gonna say

[sullen tendon]

Mb was doing mental addition

I mean multiplication

[sullen tendon]

Anyways what do I do now to find the intersect

[boreal orchid]

Well you have your values of k for which the graphs intersect

[sullen tendon]

I mean not intersect

[boreal orchid]

So, you can think about it graphically, surely there's either below or above that value, they don't intersect

[sullen tendon]

I’m confused

It says for the value of k not intersect

But if they didn’t intersect it would be a matter of the gradient

[boreal orchid]

Yes, but keep in mind it says "values"

[sullen tendon]

And the gradient is -1/2 for both of them

[boreal orchid]

Implying there's multiple

[sullen tendon]

What does that have to do with stuff tho

How can I find the multiple values?

[boreal orchid]

Well, think about it this way, if it intersects at k = 14

[sullen tendon]

-14

[boreal orchid]

Both me and my calculator got k as 14

[sullen tendon]

Oh

[boreal orchid]

I'm unsure where you got -k/4 from

I assume it's a simple sign error

Regardless

[sullen tendon]

No it’s -k/4

[boreal orchid]

Hm

Am I stupid chat

[sullen tendon]

It’s probably me

I’m tweaking

[boreal orchid]

Yeah no your line 2

Is messed up

Happens to the best of us don't worry

[sullen tendon]

Nvm it’s my fault

[boreal orchid]

But back to the question, if the lines intersect at k = 14, what happens when k isn't 14

[sullen tendon]

They are Parallele

Parallele

Parallel

[boreal orchid]

More importantly

[sullen tendon]

They never intersect

[boreal orchid]

When k is not 14

So for what values of k, do the lines not intersect

[sullen tendon]

Infinite?

why do linear equations have to have another variable 💀😭😭😭

[boreal orchid]

Well, close

Infinte apart from 1 value

[sullen tendon]

When k ≠14

[boreal orchid]

Splendid

[sullen tendon]

ohhhhh

So that’s the answer of a.?

[boreal orchid]

Yeps

[sullen tendon]

Yep Ty

[boreal orchid]

And I assume B and C are obvious from that

[sullen tendon]

B. K = 14

Right?

@boreal orchid

[boreal orchid]

Yes

[sullen tendon]

And for C.?

[boreal orchid]

Think about it

[sullen tendon]

The lines will. Only intersect at k = 14

[boreal orchid]

And for all other values

[sullen tendon]

Wdym

How is it all other values

Ohhh

At k = 14, the lines are the same

Bc we know that y = -x/2 + 3.5

Y = -x/2 + k/4

Sub k = 14

And we get 3.5

tysm @boreal orchid

Thanks for helping me understand

[boreal orchid]

No problem

[sullen tendon]

.close

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[warm mesa]

3+5x6x(4-5)

[stark wedge]

<@&268886789983436800> potential troll

[warm mesa]

?

[unreal blade]

3+5*6x(4-5)
3+30x(-1)
3-30x

[warm mesa]

yes

ty

[shut lion]

Please don't post troll questions here.

[pallid oasis]

Why is one x being treated as * and the other as presumably a variable??

[warm mesa]

ok

[shut lion]

If you have an actual question it's fine. It just wastes peoples time and takes up channels if not.

.close

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[unreal blade]

i just guessed

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[north scarab]

could somoene answer this

[stark wedge]

numerator can be whatever tf it wants to be

[north scarab]

sorry i meant zero in numerator

[stark wedge]

guh.

do you mean like "still 0/0" or "numerator is 0 but denominator isn't"

[north scarab]

numerator is 0 but denominator isn't

[stark wedge]

what's 0/a

with a not equal to 0

[north scarab]

so zero divided by a non zero value

will equal 0

forgot about that

[stark wedge]

you forgor 💀

[north scarab]

then why is it only refering to zero in the denominator

[stark wedge]

wdym

[north scarab]

why hasnt it said if the sub of x=a results in a zero denominator AND numerator

[stark wedge]

ah..

dunno, guess the slide was poorly made?

[north scarab]

maybe, nots ure

sure

[noble anvil]

Yeah, they should have said the numerator tends to 0 as well.

[stark wedge]

really like

if ☆ stands for anything NONZERO, then:

• ☆/☆ = limit exists
• 0/☆ = limit exists and is equal to 0
• ☆/0 = limit doesn't exist and is infinite
• 0/0 = more work required

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[safe fulcrum]

@worldly stirrup yes, so X means you choose the X from that bracket, and O means you don't choose the X

[safe fulcrum]

so O means you choose the Y from that bracket

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[vale dock]

@safe fulcrum Has your question been resolved?

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[dusk silo]

so for this question the forces P experiences : friction of A (upwards, parallel to plane), reaction of A (perpendicular to plane), friction between P and plane, reaction of P

why doesn't P experience the tension of the string (downwards, parallel to the plane)?

[gritty leaf]

If P isn't connected to the string then why would it experience tension?

[dusk silo]

why does it experience the friction of A (upwards, parallel to plane)?

[gritty leaf]

P wants to slide downwards, so the friction between it and the plane would act in the opposite direction i.e. upwards

Sorry no thats not right

[dusk silo]

it's okay take your time

[gritty leaf]

is this an further mechanics question

[dusk silo]

normal mechanics

[gritty leaf]

Okay, B is moving downwards, so A is accelerating up, P is stationary, A is trying to slide up P so the friction force acted upon A due to P is down the incline.

You see anything wrong with my chain of thought? I don't see how it could be upwards parallel to the plane

[dusk silo]

so you see how A moves upwards

the friction between A and P moves down

therefore using N3L

that friction is acted upon P upwards

i'll send the markscheme

[gritty leaf]

Yes I see

So what's the issue?

[dusk silo]

yeah so i understand it the friction of A pulls P UP the plane

now if we're doing N3L

why doesn't the T go downwards (on P)

[vale dock]

@dusk silo Has your question been resolved?

[vale dock]

@dusk silo Has your question been resolved?

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[vale dock]

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[civic scarab]

.close

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[frozen hare]

I need help just with the figure

this was the way I thought

pink is the tangent

not exactly

I tried draw in a point, no in intersection

in the green point

so the tangent is on the intersection

?

hmm

so how would be the figure

can you draw for me?

pls

like this?

[slate folio]

yea

[frozen hare]

hmm

dont make sense for me

[slate folio]

now you can just solve for the angle of the radii

the angle of those two purple segments

[frozen hare]

but idk this angle

I'm confuse

kkkkk

if the purple segment with pink segment is 90

how the pink segment is 180 with your complementary angle

?

[slate folio]

wait a minute

[slate folio]

this surely doesnt have enough information

nvm I misunderstood the question

[frozen hare]

yeah

[frozen hare]

what is nvm?

[slate folio]

you have to express d in terms of r

@frozen hare

What is the total sum of the four angles I marked

[frozen hare]

360

[slate folio]

yep

[slate folio]

and if you know the angle between the pink lines, the angle between the tangents and the radius, what is the angle between the purple segments

[frozen hare]

idk

because betwwen the pink is 120

bot and top

but the purple has a space

[slate folio]

and what is the angle between the tangent and the radius

the angles you marked

[frozen hare]

90?

[slate folio]

yep

[frozen hare]

90 +120

210

but 120 is a segment

and the rest

should be 60

not 90

no?

I thought other way to draw

see

[slate folio]

wait lemme draw

[frozen hare]

and like this?

[frozen hare]

here r>1

[frozen hare]

here r<1

[slate folio]

focus on these angles

[frozen hare]

ok

[slate folio]

the sum of those angles is 360 right?

[frozen hare]

y

[slate folio]

^ same as here

[frozen hare]

x?

[slate folio]

you only need to care about the upper 120° angle

ignore the other one

[frozen hare]

why

[slate folio]

you want to focus on the angles between the purple segments

[frozen hare]

ok ok ok

[slate folio]

as I said those angles adds to 360

because the angles around a point always add to 360

[frozen hare]

is 300

the angle between the purple segments

?

120 + 90 + 90

[slate folio]

the angle you just calculated is this

(I accidentally deleted a green line)

[slate folio]

I need the smaller angle however

[frozen hare]

so is 60

sorry

[slate folio]

yep

and do you know the cosine theorem in triangles

[frozen hare]

a² = b² + c² - 2bc * cosX

[slate folio]

so the lengths of the purple segments are given

the length can't be negative right?

[frozen hare]

yeah

it is rong

sorry

1 second

[slate folio]

so you're only left with the positive answer

the algebra you did seems correct

[frozen hare]

no because, i forget the r²

[slate folio]

oh yeah I missed the r^2

[frozen hare]

kkkkkkkkk

that is right

so

[slate folio]

nice

[frozen hare]

for 2)

the minimum is when the derivate is 0

here the question

[slate folio]

what's your problem here

[frozen hare]

I started with the figure problem

and now I just want to understand the other parts

3. is easy, I just need the valur of r

but 2) idk

i think in 2, the r is minimum when the derivate is 0

[slate folio]

yes

[frozen hare]

so

that its

ty

man

[slate folio]

no problem

[frozen hare]

you're so smart

.finish

.end

[slate folio]

.close

[frozen hare]

ty

.close

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[brave saffron]

can someone explain this solution to me I don't get it

[trail patrol]

The first part is obvious I guess?

I mean 24 * (2 + 4 + 6 + 8)

[brave saffron]

yeah

but the next part doesn't

[trail patrol]

we simply take 2 times each hour (because there is am and pm) and 12 only once, because we start at 12:05

wait no, 12 was taken twice as well

oh yes, because there is 12:00 pm and 12:00 am in the 24 hour-period

so it chimes at 1 am once, at 1 pm once, at 2 am twice, at 2 pm twice, and so on

(counting only chimes that are equal to the hour)

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[upbeat wind]

I'd like some help understanding why the "a" in the parenthesis doesn't become a negative once the minus outside becomes flipped to positive. I was under the impression that all the values need to be "flipped"

[uncut pollen]

"a"?

[noble anvil]

What a

[uncut pollen]

wdym

[upbeat wind]

oh sorry wrong image

[uncut pollen]

ah

[polar oxide]

what do you want it to be?

[noble anvil]

rajel your active gone

RIP

[upbeat wind]

The proper answer is a+2b, but I came to the conclusion it was a-2b which is wrong according to my book

[uncut pollen]

cause the negative is being carried by the 2

-2(a-b)

(-2)(a-b)

(-2)(a)+(-2)(-b)

lmk if it makes sense or not

[upbeat wind]

Yeah I think I get it now, thank you very much

[uncut pollen]

np

[upbeat wind]

.close

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[halcyon estuary]

Hi

[halcyon estuary]

Hello

This is the question

Pls help what transformation is this talking about

Like a veritable compression by factor of 4/5

Then what about in the middle

Do I facto 15x plus 6

Oh also a reflection in the y axis from the begitave sign

<@&286206848099549185>

[vale dock]

@halcyon estuary Has your question been resolved?

[torn fox]

Yes, factor that out. You want the coefficient on x to equal 1
Remember that this transfofmation will do the opposite of what you expect bc of its location
Are there any translations vertically or horizontally?

[halcyon estuary]

Hello

Yes vertically up 3 units

[halcyon estuary]

But you can't factor it out perfectly

[torn fox]

Fractions are acceptable. Remember to simply

@halcyon estuary does the factoring make sense? Do you need an example on what i mean about fraction?

[halcyon estuary]

Yes please

I tried on my own but I still don't know how to make x on its own

Like -3(45x-18)

Since -3 would be 1/3

[torn fox]

$4x+1 = 4(x+\frac{1}{4})$

[rocky lotus]

Pai

[halcyon estuary]

Oh

[torn fox]

You would divide out the 3. You were multiplying while factoring 😅

Thatbwould be -3(5x-2) instead

[halcyon estuary]

But doesn't the c need to be alone

[torn fox]

Did you mean x?

[halcyon estuary]

Yes

Sorry

[torn fox]

Yes it does
I was just correcting what it should look like after factoring the -3 you were trying to use

Did you try factoring out the -15 yet? Or still working on it?

[halcyon estuary]

Is that what we are doing right now?

So the final simplest answer is -3(5x-2)

But how do I describe that

[torn fox]

Thats just for common factors that are integers
You still need to factor out the 5

[halcyon estuary]

The five x is just like rise over run

Uhh

How

[torn fox]

.

Look at that again
Thats what i meant about fractions
You can go from -3(5x-2) or from -15x+6
You'll get to the same place either way

You're basically diving by whatever number you're trying to factor out
So in my example i was factoring out a 4
So 4x got divided by 4 AND the 1 got divded by 4
1÷4 = 1/4

[halcyon estuary]

So would the answer be

-15(x-6/15)

Or like 0.4

How would I even describe that

[torn fox]

Yes 👍
Reduced fraction is preferred but youre fine with 0.4 too

[torn fox]

Well, you know outside of f(x) are vertical transformations
So that means inside would be...?

[halcyon estuary]

But it's not a porabola it's a line

Because it's not squares

Or wait nvm

[torn fox]

You're right, its not a parabola
However its not a line either
F(x) is just a function of some sorts right now

[halcyon estuary]

Oh no wait

[torn fox]

We can even call it a parent function of some type

[halcyon estuary]

So the same rules apply?

So it would be an expansion of 15 moves right 0.4

Units

Ok thank you

[noble anvil]

I was about to react to that whyd you remove it lol

[halcyon estuary]

lol that was accident 😭

[noble anvil]

ik

[torn fox]

Expansion as in opposite of compression right? Then yes perfect
Make sure you say its horizontal as well

[halcyon estuary]

Yes ok thank you so much!

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[tranquil zephyr]

Mind's going blank

[tranquil zephyr]

how do i do this

[fair wing]

I can't digest the starting two lines of the question, is the questions language is weird or I cant understand?

[tranquil zephyr]

..

occupied

[opal pendant]

take a general matrix

of 3x3

a,b,c
d,e,f
g,h,i

[tranquil zephyr]

sum of their squares = 6 right

[opal pendant]

yes correct

that is the only important part

[tranquil zephyr]

after that?

[opal pendant]

what is the problem here then

its simple pnc

[tranquil zephyr]

im horrendous at pnc 🙏

[opal pendant]

a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=6

now

notice that they each have to be from the set {-1,0,1}

by each i mean a,b,c,d,e,f,g,h,i

[tranquil zephyr]

yes

[opal pendant]

now notice that they are all squared

so anything that u give will be positive

[tranquil zephyr]

mhm

[opal pendant]

now 6 of them can be 1,3 can be 0

[tranquil zephyr]

yes

[opal pendant]

6 can be -1, 3 can be 0

now in those 6 we just have to choose between 1 and -1

because the output is the same

do you understand?

[tranquil zephyr]

one sec

i understood those two cases

wt abt 3 -1, 3 1 and 3 0

[opal pendant]

yes that is what im saying

[tranquil zephyr]

and all other cases

[opal pendant]

^^

[tranquil zephyr]

oh so it would be covered by that

kk

[opal pendant]

so now, first we have to choose the 3 entries we want to be 0

how will we do this

[tranquil zephyr]

9c3

[opal pendant]

yes

now for the other 6 entries

[tranquil zephyr]

2 choices

[opal pendant]

we have 2 options, 1 or -1

yes

[tranquil zephyr]

so 2^6?

[opal pendant]

yes exactly

[tranquil zephyr]

k

lemme calculate

[opal pendant]

now we just multiply these 2

[tranquil zephyr]

so that would just be the answer?

9C3 * 2^6

[opal pendant]

yes

do u understand why?

[tranquil zephyr]

yes

[opal pendant]

yeah then you should get ur answer

[tranquil zephyr]

ty

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[glass kelp]

"lie on the diameter"

are they on the edge

[dusky pulsar]

incomplete question no?

[glass kelp]

i believe ye

[small vine]

idk

[opal pendant]

lets use some common sense they obviously lie on the circle

[noble anvil]

Just do it considering they are lol.

[small vine]

should i just assume that its on the end?

alright

so just find the midpoint and it will be fine?

[glass kelp]

if they lie on the edge, yes

the midpoint is the center

use 2r = D

and plug into ur usual circle equation

[small vine]

okay

ty

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[ornate dragon]

So my question isn't regarding this entire thing

it's more about this

i can solve this given enough time

but i'm afraif i might mess up during an exam

i saw a method where people "broke up" IBP

but i was wondering if there was any way i could make this simpler to calculate

[warped heart]

Wdym by "broke up"? Is it integrating 3x*log and also 4*log?

Cuz no, the way to proceed is indeed an IBP

[ornate dragon]

no, i meant this

hold on

i'm not sure what it's called but i've seen it before and i was wondering whether or not it was useful or not

or worth learning

[cobalt crow]

di method?

[warped heart]

I have no idea what this is; probably best to stick to the class's material, isn't it?

Ultimately, if you gotta do an IBP, you gotta do an IBP

[ornate dragon]

that is ibp

thats the thing

[autumn whale]

thats supposed to be a "faster" way of doing IBP but it really doesnt affect much

[ornate dragon]

i suppose

[autumn whale]

mainly when you need repeated applications of it

its supposed to be more organized

but you wont go wrong doing ur conventional way

no harm in doing it the normal way

[warped heart]

IBP for me is:
$$\int_a^bu(t)v(t)\d t=\Bigl[u(t)V(t)\Bigr]_{t=a}^{t=b}-\int_a^bu'(t)V(t)\d t$$

[rocky lotus]

import matplotlib.pyplot as plt

[ornate dragon]

yup we agree on that

but this is ibp just more organized i guess

[warped heart]

It's best to avoid mistakes to do each step one by one with the usual one

[ornate dragon]

from what satvik said

[warped heart]

Even if you gotta do 2 or more IBPs, do them one by one

[ornate dragon]

i could do this

[warped heart]

It avoids confisions and issues

[ornate dragon]

then i'll have poly divisions

and i can integrate that easily

[warped heart]

You can already do one IBP with u(x)=3x+4 and v(x)=log(x+2) though

[autumn whale]

yea

but me personally i just do conventional IBP

its not like its gonna go wrong anyway you just have to go one by one

[ornate dragon]

alright

[ornate dragon]

well i mean

if there's a 10% chance i make a mistake

and i have to integrate like 4-5 times total

then

[warped heart]

You know what?

As a teacher, I can tell you that even when students make computational mistakes, we give'em good grades

[ornate dragon]

not at my uni lol

[warped heart]

It's okay to make a mistake, and what matters most is that you get the method right

When you make a mistake because you went an orthogonal direction to what was required to do, then yes you're gonna get a bad grade

[ornate dragon]

they are known for being extremely rigorous with their corrections
to the point where missing a single dx can make the prof. cross out your entire exercise and grade you 0

it's stupid but

[warped heart]

[x] Doubt

[autumn whale]

oh wow

well

[ornate dragon]

i've literally seen it in person lol

[autumn whale]

i would probably say

[warped heart]

I mean yeah it is stupid, but I doubt they're gonna try that for real, especially if you can go complain to the uni

[autumn whale]

you can do conventional IBP just keep it organized so you dont lose track or anything

but like

if it gets painstakingly annoying

[warped heart]

Anyway, I'm digressing; got any more question regarding the IBP and the diff eqn you had to solve?

[ornate dragon]

no that's alright

[autumn whale]

learn DI real quick and go for it

[ornate dragon]

i was just wondering if there was anything easier beyond regular IBP

but since i'm working with ODEs

there's quite a few "long" integrals to solve

[warped heart]

Idk, if the uni crosses out something for a missing dx, they won't enjoy an unconventional untaught method

[ornate dragon]

they're somehow fine with that as long as your work makes mathematical sense

it's weird

[warped heart]

Sometimes you gotta do the computations and sometimes you won't be able to find a way to circumvent them!

[ornate dragon]

but my uni is known for being strict

like complaining won't do much because you kind of knew what you were getting into when you applied, lol

[autumn whale]

well in essence its still IBP

but i guess if they're pretty strict about it then you can do DI

it's not like DI is hard to do especially when you already know conventional IBP

so

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[vale dock]

Please don't occupy multiple help channels.

@drowsy olive Has your question been resolved?

[sweet remnant]

try a higher level math channel

[wraith heart]

#help-46

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[vivid fiber]

What is ur ques?

[lunar compass]

how is the equation for this 3 circle?

[vivid fiber]

For the yellow area?

[lunar compass]

yeah

[vivid fiber]

Oh ok

do u know what is "compliment", "union", and "intersection"?

[lunar compass]

yea i know

[vivid fiber]

ok, so why r u stuck?

where are u stuck at*

[lunar compass]

in the placement

[vivid fiber]

huh

wdym

[lunar compass]

my answer is (A intersect B) union (A intersect C)

is it correct?

[vivid fiber]

wait let me see

oh so the anmswer doesnt have to be in the multiple choice thing?

[sweet remnant]

that is an accurate way to describe the region but its not the only way

[vivid fiber]

yeah

[lunar compass]

im confused at the part the 3 circle intersect

[sweet remnant]

one of the answer choices also accurately describes the region

[lunar compass]

yeah

wrong multiple i think

that made me confused too

[vivid fiber]

u mean the answer isnt in the choices?

[lunar compass]

yeah

or im wrong?

[vivid fiber]

the answer is there

its just really not obvius

[lunar compass]

then, how is other way?

[sweet remnant]

its there but its tricky to visualize

[sweet remnant]

^

[lunar compass]

hmm

[vivid fiber]

the one in the options 💀

[vivid fiber]

yeah

the way its notated makes our intuition think that it must be wrong

[sweet remnant]

yeah lowkey I did like 2 passes

[vivid fiber]

lol same

[lunar compass]

hmm, im still confused, which one in the choice was right?

[vivid fiber]

try to visualize each one of them slowly

just close the picture, then look at the options

take pencil and paper and then draw it

[lunar compass]

B?

[vivid fiber]

yes

[lunar compass]

wow

[vivid fiber]

you just need to be careful in like visualizing

[sweet remnant]

another way to approach is it try to invalidate some options

[vivid fiber]

some options are obviously wrong

[sweet remnant]

you can stop early in some cases instead of visualizing the whole thing for each choice

[lunar compass]

ahhh

[sweet remnant]

for example in e, we have something union B^c

[vivid fiber]

u can easily sign off a few choices here

[lunar compass]

E obv, wrong

[sweet remnant]

Which means the stuff in A but not B should be there too, but its not so E can't be right

for D, you have something union B

which means all of B definitely should be in the region, but its not, so D can't be right

[lunar compass]

and C, its outside the C , didnt include the inside C

[sweet remnant]

for C, since you intersect with the set C^c, you can be certain nothing in the set C can be in the region, but there is, so option C can't be right

and options A and B have like compound regions on both sides so if you aren't too comfortable with invalidating, its probably easier to just draw them out

but i think learning how to invalidate options is definitely a better way to absorb the intuition

behind like all math problems at large

[lunar compass]

well, i didnt think that way.... cuz this thing isnt common here, thankyou so much @sweet remnant@vivid fiber

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[slate folio]

I need help with this randomly difficult problem that appears on my TSA practice test

[slate folio]

S.ABC is a tetraheron. D, E and F lies on the medians SM, SN and SP of triangles SBC, SAC, SAB so that SD/SM=1/2, SE/SN=2/3, SF/SP=3/5. The plane (DEF) cuts the tetrahedron into 2 parts. Find the ratio of the volume of the big part/small part.

[vale dock]

@slate folio Has your question been resolved?

[slate folio]

<@&286206848099549185>

[modest sinew]

Placing S at the origin and expressing D=(b+c)/4, E=(a+c)/3, F=(3/10)(a+b) in terms of the basis (a,b,c) yields a 3x3 matrix with determinant 1/20 so Vol(SDEF)=1/20 of Vol(SABC) and the plane DEF therefore splits the tetrahedron in the ratio big:small = 19:1

[slate folio]

uhh I got the answer but didnt know how to do it

correct answer is actually around 3,63

[modest sinew]

huh?

[slate folio]

I did see your original solution briefly

and the answer seems correct

[modest sinew]

Vector check gives Vol(SDEF)=1/20 of Vol(SABC) so the plane removes a 1/20 sized wedge at S; the rest is 19/20, hence big:small = 19:1

[slate folio]

also I'm haven't gotten into uni yet so I dont know anything about barycentric coordinates or matrixes

but this is supposed to be solveable for a 12th grader like me

[modest sinew]

That is definitely a hard question

that seems like an AIME training question ngl

[slate folio]

this question is way way harder than what appears on TSA test realistically

my teacher just put it in for some reason

[modest sinew]

yea i had to double check it because my original answer seemed wrong

definitely way harder than what would appear

[slate folio]

gonna close this and ask my teacher tomorrow

.close

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[dull vortex]

Hi can someone help me with this question

[modest sinew]

set 3p + 4q = 10(p – q). This gives 3p + 4q = 10p – 10q so 7p = 14q and p = 2q. substitute p = 2q into (p^2 + 2q^2)/(p q): the numerator is 4q^2 + 2q^2 = 6q^2 and the denominator is 2q*q = 2q^2, giving 6q^2 / 2q^2 = 3

[safe fulcrum]

just a note that you shouldn't give the full solution if the person hasn't shown any working

[modest sinew]

thanks, will do so going forward

[dull vortex]

yea im a bit confused on the first step tho

[modest sinew]

whats the confusion

[dull vortex]

why do you set the other side 10(p-q)

[modest sinew]

Because the original equation is

(3p + 4q) / [2(p - q)] = 5

multiplying both sides by the denominator 2(p - q) clears the fractions

[sleek nebula]

Cross multiplying the original statement

[dull vortex]

oh i get it

[modest sinew]

3p + 4q = 5 * 2(p - q) = 10(p - q)

[dull vortex]

you bring the denominator to the other sied

alr thanks for the help

[modest sinew]

np

[dull vortex]

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[mental skiff]

Hello i have a doubt regarding riemann's explicit prime counting function how do we calculate the non trivial zeros of riemann zetas function computationally

[mental skiff]

This is the definition riemann gave

i am like building a prime counter computationally for that for approximation i am planning to use 200-500 non trivial zeros of like riemann zeta function

how do we calculate them computationally

[cloud coral]

Hmm... That im not completely sure. It can be proven all nontrivial zeros have real part in (0, 1) and we know they have to come in conjugate pairs

[mental skiff]

right

[ebon glade]

just take a precomputed list?

[modest sinew]

Compute Z(t)=exp(i*theta(t))*zeta(0.5+i*t) with the Riemann Siegel O(sqrt(t)) sum, scan t for sign changes of Z, polish each bracket with Newton/Brent, and the resulting heights t give the zeros rho=0.5+i*t you need

[mental skiff]

Thats good but like the precomputed list would also have been found by an algo right?

[mental skiff]

Noice lemme look

[cloud coral]

That computation assumes RH for correctness... then again practically all the nontrivial zeros we found so far satisfy RH so a pet project assuming it can find the zeros

[mental skiff]

Are there non trivial zeros of riemann zeta function that have Real part as integer or simple rational numbers like 3/2?

We need riemann zeta zeros right order or what zeros matter does it matter too

[cloud coral]

Provable that real part in (0, 1)

[mental skiff]

i know we need to take non trivial

[mental skiff]

ok

[modest sinew]

the Riemann Siegel series itself is unconditional; what does use RH is the naive “find every zero by scanning only the critical line” inpractice you bracket sign changes on the line then run Turings test to confirm no zeros off the line were missed so the computation is valid even if RH were false

[mental skiff]

still like 1/3?

[ebon glade]

well all of them have (conjecture) the simple real part 1/2

[mental skiff]

lol sorry i got mistake

i meant imaginary part

a+bi b is like 1/2 3/2 or some

Riemann conjecture itself is that real part is 1/2 of a non trivial zero i meant like the imaginary part

cause i checked here

http://www.plouffe.fr/simon/constants/zeta100.html

[ebon glade]

well feels unlikely but who knows

[mental skiff]

Non trivial 100 zeros

[cloud coral]

Hmm i see lemme google

[mental skiff]

these are all messed up

so this means like non trivial zeros with imaginary part like 1/2 or 3/2 have not been found

i mean like simple rational numbers?

This means that every complex number with real part 1/2 is not a non trivial zero of riemann zeta function but every complex number which is a zero has Re(s) 1/2 right?

so like there are infinitely many zeros and infinitely many non zeros too right?

hello?

[cloud coral]

There are obviously infinitely many nonzeros

[mental skiff]

Have zeros like i told above not been found

In the video i was watching

https://www.youtube.com/watch?v=dktH8hJadyU

If one proves that the error in Li(X) approximation to Pi(x) is 0(root(x)logx) the riemann hypothesis gets proved cause the 0(rootxlogx) derivation depends on that riemann hypothesis is true

deriving an exact error function would indeed prove riemann hypothesis

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[pliant palm]

can someone explain this in simple terms

[stark wedge]

explain the what or explain the why?

as in, what exactly would you like explained here:
(A) "WTF is this even saying?"
(B) "I get what this is saying but why is it true?"
(C) "Secret third thing."

[pliant palm]

a

i don't understand shi

[stark wedge]

have you dealt with logarithms before at all

[pliant palm]

noo

first timer

[stark wedge]

ok hm well maybe you should (re)watch the video you got this from

[pliant palm]

its been 3 years since I last studied math

[stark wedge]

or we could go over the basics right now

[pliant palm]

he just started blabbering about this law of log

[pliant palm]

suree

[stark wedge]

are you familiar with exponentiation

[pliant palm]

no

[pliant palm]

like i said-

[stark wedge]

im trying to figure out where to start from lmao

3 years erodes memory at different rates for different ppl

[pliant palm]

yeah there is an urgent need that I need to understand logs out of nowhere

[stark wedge]

do any of the words "power", "indices" or "exponent" ring any bells

[pliant palm]

i know power and exponents

aren't they the same ?

[stark wedge]

yes basically

so like if i asked you to work out something like 2^5 would you be able to do it

[pliant palm]

yeah

we have to multiply 2 to itself 5 times

40 ig

[stark wedge]

we multiply 5 copies of 2 is how i like phrasing it

but yes

no it's not 40

how did you get 40

[pliant palm]

wait

my bad

its

32

[stark wedge]

what's that urgent need btw

yes 2^5 = 32

[pliant palm]

I need to learn big o and for that I need to understand logs and expo

[stark wedge]

bruh

well ig you will need to brush up on like

roughly all of precalculus i guess

anyway...

ok

now lets go over some more stuff related to exponentiation

[pliant palm]

sure

[stark wedge]

lets say you had the equation $x^3 = 729$ (and $x$ is positive here, for simplicity's sake). how would you solve for $x$?

[rocky lotus]

Ann

[pliant palm]

9 ?

if I'm not wrong

[stark wedge]

how'd you get 9

[pliant palm]

i remember cubes and squares of some numbers I studied in middle school

lol

[stark wedge]

okay so table recall

alright lets change it up a bit

[pliant palm]

yeahh

[stark wedge]

$x^3 = 20$

[rocky lotus]

Ann

[stark wedge]

how would you solve this one

(it's no longer any nice number)

[pliant palm]

yeah

its not perfect square or cube

idk

[stark wedge]

do you know what a root is

[pliant palm]

yes

[stark wedge]

ok right

the value of x here is the cube root of 20.

$x = \sqrt[3]{20}$ as it's written

[rocky lotus]

Ann

[pliant palm]

wait

why did 3 went up there

[stark wedge]

this is the notation for cube roots

[pliant palm]

oh

[stark wedge]

have you seen it before?

[pliant palm]

alr

not the cube one

but the square one

but i understand

[stark wedge]

oh dear.

[pliant palm]

go ahead

[stark wedge]

the square root answers the question "What^2 gives this number?"
the cube root answers the question "What^3 gives this number?"
...
the n'th root answers the question "What^n gives this number?"

(we don't have any fancy names for roots of degree higher than 3; they're just called 4th root, 5th root, 6th root etc.)

[pliant palm]

alr

[stark wedge]

so anyway this was kinda warmup

and a bit of context-setting

meaning that if we want to find an unknown base in a power, then we will end up using roots in some way or another.

[pliant palm]

yeah

[stark wedge]

if you're even slightly curious you might ask how to find an unknown exponent instead

[pliant palm]

using logs ?

[stark wedge]

exactly. that's their like... origin story

[pliant palm]

mhm

alr

[stark wedge]

the logarithm answers the question "(base)^what equals this number?"

in notation, the base is written as a subscript

[pliant palm]

so

in

log^2

2 is the base ?

[stark wedge]

subscripts are written with _ not ^

[pliant palm]

oh

dorry

[stark wedge]

for example, as we've discussed earlier, $2^5 = 32$. thus, $$\log_2(32) = 5$$

[pliant palm]

mhm

[rocky lotus]

Ann

[pliant palm]

so

in log_2, 2 is the base

right ?

[stark wedge]

yes

[pliant palm]

and

what if

we say log^2

what's the diff

[stark wedge]

mmm

ok so like

if you write just log without an explicit base, you need to be able to infer it from context

for example in an engineering context a missing base might imply log_10

in math it might imply log_e

in computer science log_2 because of shit like binary search trees and the like

$\log^2(x)$ would then mean the same thing as $(\log(x))^2$

[rocky lotus]

Ann

[stark wedge]

it might come up but it is slightly esoteric

[pliant palm]

that completely went over my head

why does x gets squared here

[stark wedge]

no

the x does not get squared

the logarithm gets squared

[pliant palm]

oh

x becomes a part of log

[stark wedge]

if i wanted to square the x i would have put the ^2 inside the parentheses right next to the x

[pliant palm]

but how

[stark wedge]

don't worry about it just now i'd say

anyway point is you should treat the base as a key part of the log notation that you should not drop

[pliant palm]

how's this the same thing

[stark wedge]

definitionally

[pliant palm]

alr

I was just curious

[pliant palm]

okay so base is important

[stark wedge]

it's sometimes convenient not to write a dozen parentheses, it's nothing more than that -- just a convenience

yes

mmmm i'd say we are still kind of a long way from like big-O stuff

but in any case

[pliant palm]

I'm cooked

anyways go ahead

[stark wedge]

you know exponent laws, yes?

stuff like a^(p+q) = a^p * a^q

[pliant palm]

def not

[stark wedge]

bruh

[pliant palm]

i only know that

[stark wedge]

well i guess you have to relearn those as well

[pliant palm]

a^0 = 1

[stark wedge]

otherwise you really are cooked 💀

[pliant palm]

shit man

[stark wedge]

sautéed and served with garnish

baked at 350°F for a full hour

[pliant palm]

hahaha

don't brag ur cooking skills now

back to the topic now

[rocky lotus]

MOOJY

[pliant palm]

this goes wayy beyond my comprehension

[stark wedge]

it's some kind of config option idr where it is but if you type ,help you should figure out how to navigate it

btw don't put plain text inside dollar signs\

[pliant palm]

yeah

yess @stark wedge c explained it

[stark wedge]

texit will activate even if you put a formula inside dollar signs somewhere in the middle of your msg like this $x^2+y^2=z^2$ and it'll render everything outside the dollars as plain text without $this ugly italicized no-spacing shit$

[rocky lotus]

Ann

MOOJY

MOOJY

[stark wedge]

it's not a latex thing it's a bot thing