help-33

[still temple]

3^x+5=2?

no

1

[sinful thistle]

no

yes

[still temple]

3^x+5=1

but you can divide them?

[sinful thistle]

do you know how to write 1 as a power of three?

[still temple]

1^3?

[sinful thistle]

no...

what i mean is

can you write 1 = 3^a for some a

[still temple]

oh

x+5

has to be 0

in order to make 1

like a has to be 0

[sinful thistle]

yes

[still temple]

ahhh

thank you

[sinful thistle]

so 3^(x+5) = 3^0

nw

[still temple]

can you also do it in a way of

like just multiplying

54 to 3?

[sinful thistle]

??

no not exactly

you'd need logarithms for that

[still temple]

oh rlly

im goin to learn log rn

idk this lesson is so messy

[sinful thistle]

ah

honestly this way is much easier

[still temple]

i should try to be more familiar with expnent equation

before i start right?

[sinful thistle]

logs are too complicated

[still temple]

start learning about logs?

okok

ty smh

so much

[sinful thistle]

np

[marsh citrus]

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[still temple]

.close

[marsh citrus]

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[marsh citrus]

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[small stream]

A^TA is a "check" on whether the columns of A are orthonormal. If they are, the matrix behaves like a set of independent axes in space, and A^TA=I verifies this property.

Is this true?

[marsh citrus]

@small stream Has your question been resolved?

[static quarry]

yes, A^t A = I means precisely that the dot product of each column of A with itself is 1, and the dot product of two different columns is 0

i.e. the columns are orthonormal if and only if A^t A = I

(this is assuming A is real, if it's complex then you need to use the conjugate transpose instead)

[small stream]

Then can we say that if A^TA =I, then A is symmetry?

[static quarry]

no, not necessarily

try this one

     0     0     1
     1     0     0
     0     1     0

[small stream]

How did you come up with an example so quick

[static quarry]

started with the identity matrix and shuffled the columns around

we know its columns are orthonormal so any shuffled version also has that property

some of the shuffled versions are symmetric, like

     0     1     0
     1     0     0
     0     0     1

but others like the one above are not

[small stream]

But you perform matrix multiplication very quickly. I first write A ^ T and then multiply the result element by element, which is time costing

[static quarry]

no i didn't multiply it at all

the columns are orthonormal in I, that's obvious

and if i move the columns around they stay orthonormal, i don't have to compute A^T A to confirm that

so just look for a shuffle that is not symmetric

[small stream]

Ooh

Tysm!!

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[marsh citrus]

@vivid hatch Has your question been resolved?

[marsh citrus]

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[sour cliff]

Hey! I need help with this question, the answer is C, but i don't understand how the 10 get up there, how the 5 goes infront of the 7 and how the 10 got p when p is supposed to be with 2. I hope my explanation on my trouble wasn't confusing!

[eager plank]

ok..

It is i

Right..

lets go step by step..

[sour cliff]

HELPP IS MY EXPLAINATION TOO CONFUSING

[delicate berry]

you should start by putting all terms other than s on one side, here on the right

so

5t-2s=7/p - 10

5t-2s=(7-10p)/p

-2s=(7-10p)/p - 5t

-2s=(7-10p-5pt)/p

multiply by -1 on both sides

then:

2s = (5pt + 10p - 7)/p

divide by 2 on both sides

s = (5pt + 10p - 7)/(2p)

hence c

@sour cliff

[sour cliff]

Hold on I'm trying to understand them

[sour cliff]

How did the 10 got p?

[delicate berry]

the original equation was:

5t-2s=7/p - 10

i wanted 10 to have p as a denominator

so i multiplied by p and divided too

5t-2s=7/p - (10p)/p

then as denominator was same i could put the numerators together

hence: (7-10p)/p

[sour cliff]

Was 10 was multipled by p to match 2p then?

[delicate berry]

no

if i write
$$10= 10$$
would that be correct?

[elfin berry]

AnshumanNeon

[sour cliff]

What

[delicate berry]

look, 10 is same thing as 10 right?

yes or no

bear with me for a moment

[sour cliff]

Yes

[delicate berry]

if on the left hand side i multiply by 2, but then divide by two immediately, would i still get 10?

$$\frac{10 \times 2}{2}=10$$

[elfin berry]

AnshumanNeon

[sour cliff]

Yesss

[delicate berry]

same thing if i do it with 3 or anything other number

so if i have 10

and i want to write it in some way such that i have its denominator as p

i just multiply and divide it by p

as $$\frac{10 \times p}{p}=10$$

[elfin berry]

AnshumanNeon

[sour cliff]

So we just add p because of 2p?

And it won't change anything cause 10 was immediately divided by p after multiply p?

[delicate berry]

no

[sour cliff]

God

[delicate berry]

don't think about 2p like that

[delicate berry]

you're right here

2p was made a littler later

on the left hand side i had "2s", but i only wanted "s"

so i divided both sides by 2

so "2s" became "s"

and on the right hand side, we went from "p" to "2p"

that 10p/p has nothing to do with it

[sour cliff]

I think i could grasp it somewhat

[delicate berry]

i don't want you to think of doing 10p/p for 2p

because that's not what was done

2p came up on its own because we wanted it to be independently of 10p/p

[sour cliff]

If i didn't know the context I would think you're talking about a romance between 10p who's with p but 10p used to like 2p and even change for them

That aside

[sour cliff]

Why do we need to put 5t on the bottom? I thought we need to put it on top

[delicate berry]

no its not on bottom

my bad

its like this:

$$-2s = \frac{7-10p}{p} - 5t$$

[elfin berry]

AnshumanNeon

[sour cliff]

Oohhhh

[delicate berry]

then we do the same thing like we did with 10 to get its (5t) 's denominator to be "p"

$$-2s = \frac{7-10p-5tp}{p}$$

[elfin berry]

AnshumanNeon

[sour cliff]

Why di we need to multiply by -1? How do i know when I need to do so?

[delicate berry]

because we just want "s" not "-s" or "-2s"

-1 x -1 = 1

so we multiply by -1

tbh multiplying by -1 is same thing as dividing by it

so you can even think of it as we divide both sides by -1 which makes more intuitive sense

[sour cliff]

Oh so the (-) is for both 2 and s, not just s?

[delicate berry]

yes

-2s = -1 x 2 x s

we just want s

so we divide by 2

and multiply/divide by -1 (because its the same thing for -1)

[sour cliff]

Why is the 10 not affected by -1 multiplication?

[delicate berry]

it is

look there, earlier it was "-10p"

after -1 multiplication it became "+10p"

[sour cliff]

Ahhhh

Am I doing smth wrong here

[delicate berry]

no

-5x-1 is 5

it is

[sour cliff]

I thought we should be getting -7..

[delicate berry]

why?

[sour cliff]

Because we got -7 here

After multiply them all by -1

[delicate berry]

after you multiply by -1, the signs switch

-5pt becomes 5pt

-10p becomes 10p

7 becomes -7

[sour cliff]

So the 5 and 7 just switched side?

[delicate berry]

no

it is just convention to write whatever is in plus first

if you write it like:
$$s= \frac{-7+10p+5tp}{2p}$$

[elfin berry]

AnshumanNeon

[delicate berry]

thats still correct

[sour cliff]

Ahhhh I get now

Unrelated question to the current topic, how can we know if we should put a number on the top or bottom of the /?

[delicate berry]

hmm

in this case if you see when i divided by 2 on both sides, the 2 was transferred to below the /

that's because
$$\frac{1}{x} \times \frac{1}{y} = \frac{1}{xy}$$

[elfin berry]

AnshumanNeon

[sour cliff]

I get that, but I don't get how it's relevant

[delicate berry]

so
$$\frac{-7+10p+5tp}{p} \times \frac{1}{2} = \frac{-7+10p+5tp}{2p}$$

[elfin berry]

AnshumanNeon

[delicate berry]

well i didn't get your question clearly to be honest.

its more complicated

to explain that

and i have no idea where to start

if you multiply some fraction, then the multiplation will always come on top

if for some number has same denominator, then the number will always come on top

[sour cliff]

I

Don't get it

Since when was 2 a fraction or denominator number?

[delicate berry]

we divided by two

division by two is sasme as multiplication by 1/2

[sour cliff]

Ahhhh

Okok i think I get it

Thank you so much! I should close it now right?

[delicate berry]

if you have no other questions

then yes

[sour cliff]

.close

[marsh citrus]

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[shy pumice]

Hello, i got this awful ass proof, can anyone validate that it is correct or not? or what the flaw in it is? (the induction party feels very shaky to me, but i am not sure how to do it. tao's hint was to use induction)

[hollow glen]

where did you make use of the induction hypothesis?

[shy pumice]

to be clear, i did want to, but i tried to focus on the hint if it would work. however, he says the induction is "somewhat degenerate in the sense that it does not use its inductive hypothesis", which is also why i skipped it.

the entire induction feels a bit "shaky" to me

[hollow glen]

oh yeah right, if you use induction, you need to show that for any a++ there exists a b, such that b++ = a++. The induction hypothesis would be that a has a predecessor b' with b'++ = a

but you don't need this induction hypothesis

[hollow glen]

also did you mean to say "a++'s predecessor is a"?

[shy pumice]

this is how the solution manual did it, but i felt like a proof by contradiction is (saner?) easier to understand. but it also doesent feel very correct.

[shy pumice]

oopsie, yes. i will fix it later

[hollow glen]

I think the proof is fine this way, you just don't have as many tools available yet, later in maths this kind of induction shouldn't occur anymore, at least I've never seen it

[shy pumice]

alright, thank you!

[hollow glen]

to make it more clear, I would explicitely write that b=0 in the first case and b=a in the second

[shy pumice]

.close

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[rough kraken]

Can someone help me out with this problem

[rough kraken]

Anyone here?

[faint narwhal]

U must have drawn something wrong

[rough kraken]

No this is all correct

[faint narwhal]

Is the base a straight line?

[rough kraken]

Yeah

[unreal oxide]

what are u required to do here?

find x and y?

[faint narwhal]

and two scratch mean they have same length?

[rough kraken]

Yep I have to find x and y

No two scratch mean it is equal

I think it is an isoceles triangle

[faint narwhal]

Ok, then fine

[rough kraken]

U think u can help me out?

<@&286206848099549185>

[unreal oxide]

well y is quite obvious isnt it

[rough kraken]

What do u mean?

How is it obvious

[unreal oxide]

what triangle theorems about angles are u familiar with

[rough kraken]

You mean the properties of a triangle?

[unreal oxide]

yeah okay what properties do u know

[delicate berry]

x=105, y= 70?

is this correct?

[glossy flint]

I think so

[unreal oxide]

yes but help them figure it out

[rough kraken]

Yeah it is

But can you help me do it

[glossy flint]

Yeah, do you recognize any particular type of triangle there?

[rough kraken]

Yeah I do

It is isoceles triangle

[glossy flint]

Awesome

So what's the main property about the angles in an isosceles triangle?

[rough kraken]

Two sides of the triangle is equal

[unreal oxide]

what about angles?

what can u say about angles in an isosceles triangle

[rough kraken]

I didn’t get you?

Sum of all angles is 180

[unreal oxide]

that is true of any triangle

anything particular that u know about the angles of isosceles triangles?

[rough kraken]

All that I know is that they are equal

[delicate berry]

@rough kraken try to use all properties of triangle and isoceles triangles to find as many angles as you can. just go on and on from there. you will eventually reach it

@unreal oxide is already explaining it, so i won't interrupt

[rough kraken]

Ok thanks neon I’ll try that

[delicate berry]

follow silver

[rough kraken]

Ok

[delicate berry]

i just gave ideas

[unreal oxide]

actually u can take over neon if u dont mind

coz i might have to leave

[rough kraken]

Ok sure

[delicate berry]

alr then if you're busy

[rough kraken]

So neon did you figure out the answer by using the properties of the triangle and isosceles triangle ?

[unreal oxide]

thanks ❤️

[delicate berry]

yes

[rough kraken]

Ok I’ll try it

[delicate berry]

would you like to try or should i explain the process?

[rough kraken]

If u can explain it pls

[delicate berry]

well this property is not of isoceles triangles but of equilateral

[rough kraken]

Equilateral triangle?

[delicate berry]

in isoceles triangles, angles adjacent to equal sides are equal

[delicate berry]

where all sides are equal in a triangle.

[rough kraken]

I don’t get it

[delicate berry]

here in this isoceles triangle. side AB= side AC

then, angle ABC is also equal to angle ACB

which has its own proof

[rough kraken]

Oh because of the two lines scratch?

[delicate berry]

are you talking about BC and BC ?

this is true for any isoceles triangle

[rough kraken]

No like what does this mean

[delicate berry]

this means that these two lines marked with double croses are equal in length. but not not equal to those lines marked with single crosses

[rough kraken]

So single means not equal and double mean equal?

If you can, can you show me how did you do it step by step on paper

I don’t understand anything on words

[delicate berry]

any number of lines crossed means whichever lines that are crosses are equal to each other.

so all lines crossed with single line means all of those are equal separately

all lines crossed with double line means all of those are equal separately

all lines crossed with triple line means all of those are equal separately

[rough kraken]

Ok yeah I get that

[delicate berry]

and so on

basically in an isoceles triangle, angles opposite to equal sides are equal

[rough kraken]

Can you show me how did u do it on paper?

[delicate berry]

hmm. ok

[rough kraken]

thanks that would help a lot!

Did u get it

[delicate berry]

lemme send it. wait a min

[rough kraken]

I’ll review it

Thanks so much I understood I will, do it by my own

Can you add me as friends

So that if I have more questions I can have help?

[delicate berry]

done

@rough kraken close this if no more questions are left

[rough kraken]

+close

!close

If the angles of a triangle are in the ratio 1:2:1, find all the angles of the triangle. Classify the triangles in two different ways

.close

[marsh citrus]

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[pliant willow]

how do you prove that a function is strictly increasing on an interval ?

[devout mauve]

depends on the function

you can either show the relevant inequality holds

or another classic is to show the derivative is always >= 0 with only =0 at isolated points

[pliant willow]

oh

i used the derivative to find the direction of variation

so if there is no extremum in a given interval the function is strictly increasing ?

[devout mauve]

well it could also be decreasing

[pliant willow]

yes ok

tysm i love you

[devout mauve]

but extremums are where those two change

[pliant willow]

i never realised it was that simple

[devout mauve]

well and there is saddle points obviously

[pliant willow]

yes i mean like the place in my variation table where there is a bar

[devout mauve]

like x=0 at x^3

[pliant willow]

for example here

in ]4; +oo[ it is strictly increasing

but in ]x2 ; +OO [ it isnt

right ?

[devout mauve]

give me a sec, I need a minute to understand your table

[pliant willow]

it's a random one i found on the internet

i picked it to illustrate the thingy

[devout mauve]

ok but which of those is the function and which is the derivative

I dont understand why the 4 is even mentioned

who cares about roots

[pliant willow]

ig the first part is the derivative

and the actual function isnt given

as it is of no use in variation study

[devout mauve]

I mean it looks like its just strictly increasing from x2 to infty

single points dont matter

its only not strictly increasing if its constant for a bit

[pliant willow]

oh but there is a forbidden value

[pliant willow]

ohhh

ok

uh

[devout mauve]

x^3 is strictly increasing

even tho it has that point at x=0

[pliant willow]

you got one that aint srictly increasing ?

[devout mauve]

https://i.imgur.com/uG2R8uU.png

[pliant willow]

👍

function not curve lol

[devout mauve]

you mean "formula" ?

I can only come up with piecewise stuff

[pliant willow]

uh ig

[devout mauve]

generally most nice formulas you can write down wont be constant for some time and non constant for other times

[pliant willow]

they be rare asf right ?

functions that aint strictly increasing or decreasing on intervals

[devout mauve]

most functions you will come across are just strictly increasing or strictly decreasing or switch between those

[pliant willow]

tysm

i luv you

you gave me wise help

.close

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[little prism]

how can i do this?

[little prism]

what is long division

sorry I am not very acquainted with english math terms

okay I have yet to study polynomials

[cunning fiber]

Nvm just do $u=xe^{\arctan x}$ then

[elfin berry]

Civil Service Pigeon

[little prism]

I just saw I could write tan^2u+1 as tanx derived

so basically

i think i have f' * g + f * g'

yea and then e^u stays the same

so I just get tangent of u times e ^ u

thanks

.close

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[ancient thicket]

How do I calculate the inverse of following equation.

[ancient thicket]

Shit

There

Please just explain to me how I do it, no asking me questions about whether I know this or that.

[quaint elm]

equations don't really have inverses

[ancient thicket]

Bruh

My book literally told me to find the inverse

More precisely it says

In Exercises 13-20, f is a one-to-one function with inverse f^(-1). Calculate the inverse of the given function in terms of f^(-1).

[quaint elm]

the inverse of the given function

not equation

[ancient thicket]

Ah, my bad, I did not mean equation

[quaint elm]

right

take f^-1 of both sides

[ancient thicket]

Just that?

[quaint elm]

well

or rather uh

ok ignore the h(x) part

you want to find the inverse of f(2x)

simplest way to do that is to let y = f(2x)

and then solve for x

[ancient thicket]

Then I would have

h(x)=y

[quaint elm]

i just said to ignore the h(x) part

[ancient thicket]

Then

Okay

[quaint elm]

good?

[ancient thicket]

If I calculated it ccorrectly

Correct?

[quaint elm]

yep

[marsh citrus]

@ancient thicket Has your question been resolved?

[marsh citrus]

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[velvet pike]

Hello!
I have this table and want to calculate di/dE.
This would be the right solution but i have no idea how to get there

[marsh citrus]

@velvet pike Has your question been resolved?

[winter anchor]

<@&286206848099549185>

[marsh citrus]

@velvet pike Has your question been resolved?

[marsh citrus]

@velvet pike Has your question been resolved?

[marsh citrus]

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[wise oxide]

How r u suppose to know how much times ur supposed to deriv when ur finding power series by differentiation??

[fervent rampart]

in what situation? like finding the taylor series for a function?

[wise oxide]

Like it just says find the power series

[fervent rampart]

could you post the problem you are working on?

[wise oxide]

Like for 1/(1+x)^3 why they derive twice

But for 1/(1+x) it’s just once

[main idol]

the word is differentiate

[main idol]

show the whole thing

[wise oxide]

Actually idk how much times they even do it but sometimes random videos do 1 and sometimes 2

[fervent rampart]

in general to find the taylor series you have to have a formula for the nth derivative, so you take the derivative as many times as it takes to find the pattern

[wise oxide]

I don’t get it

This is Taylor series?

What

[proud ice]

yes

[proud ice]

Are they showing examples? Maybe they're stopping when the pattern is apparent. A taylor series required infinite differentiations

[shrewd dove]

Looks like they’re using substitution

You manage to get the series into a form like 1/(1-x) but replace x with whatever you found

So for a you’d just put -x instead of x in the sum

B and c id have to think more closely about bc I haven’t done this stuff in forever

[proud ice]

I think OP's question was more about what are taylor series in general

[shrewd dove]

Oh my bad

[marsh citrus]

@wise oxide Has your question been resolved?

[marsh citrus]

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[wary kite]

word

[shrewd dove]

Agreed

[marsh citrus]

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[proud ice]

https://tenor.com/view/cat-crazy-cat-keyboard-cat-fast-gif-5753018

[marsh citrus]

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[umbral rain]

hello i'm not sure how to draw a diagram of this cus i'm confused by the wording

[marsh citrus]

@umbral rain Has your question been resolved?

[umbral rain]

<@&286206848099549185>

[marsh citrus]

@umbral rain Has your question been resolved?

[umbral rain]

<@&286206848099549185>

[outer lodge]

You are given a magnitude and a direction. Sketch a coordinate system such that each unit is simply a metre

[umbral rain]

Alright thank you

.close

[marsh citrus]

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[marsh citrus]

@chrome falcon Has your question been resolved?

[chrome falcon]

.close

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[wild crystal]

Please help

[marsh citrus]

@wild crystal Has your question been resolved?

[wild crystal]

i cant calculate the data of the graduates from english, history and geo

i could only calculate the no. of ppl pursuing french german and spanish which is 15k, 15, and 10k

.close

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[marsh citrus]

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[near elbow]

how many different arrangements are there of the 11 letters in the word REQUIREMENT in which the two Rs are together and the three Es are together?

[near elbow]

guys, i tried doing 2 x 7! x 3 but i think it's wrong

[proud basin]

imagine the EEERR as a single letter as they have to be together, so now you have (EEERR)QUIMNT which would be 7!, then you can permute EEERR

[near elbow]

ohhhhhh

for EEERR is 5!/(3!2!) rightt?

[near elbow]

thank you so much!

.close

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[molten wyvern]

hi, i don't really understand the wording of this problem

[molten wyvern]

is this the graph for all of them?? or is it just f'

i kinda just winged part b 😭

[wary kite]

it says "the curve is the graph of.."

at the start of each question sir

for part a it’s f, for part b it’s f’, and for part c it’s f’’

[molten wyvern]

so its the same graph for everything

im slow nvm got it

do not do math at 3 am

[wary kite]

true

you should’ve been watching jon jones and charles oliveira

[marsh citrus]

@molten wyvern Has your question been resolved?

[marsh citrus]

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[woeful shadow]

have i done something wrong in question ‘h’?

excuse my small handwriting

i think the equation’s been rationalised in the answer key

[fathom ridge]

Looks alright

[woeful shadow]

in the answer key there was a root 2 in the numerator and a root x in the denominator, i think mine’s just not fully simplified

[fervent rampart]

you can split up the denominator as [\sqrt{2x} = \sqrt{2}\sqrt{x}] then use the fact that [ \frac{2}{\sqrt2} = \frac{\sqrt 2 \cdot \sqrt 2}{\sqrt 2} ]

[elfin berry]

cloud

[woeful shadow]

yesss this is what i was talking about

i forgot about it

thank you though

ima head out

👋

.close

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[restive silo]

Hello

[restive silo]

In linear algebra there's a question giving 3 vectors and asking if they do span R³

What does that mean?

I understood we should search for consistency but if our system does then what should I do next ?

[fervent rampart]

so the span of a set of vectors is all possible linear combinations of those vectors

so "do these vectors span R^3" is asking whether you can express any vector in R^3 as a linear combination of the given vectors

[restive silo]

This is still kinda confusing, sorry

So we take 3 vectors and make them in echleon form right?

So that we can see if they're consistent or not

[fervent rampart]

so if we have three vectors $\vb v_1$, $\vb v_2$, and $\vb v_3$, a linear combination of those vectors is
[ k_1 \vb v_1 + k_2 \vb v_2 + k_3 \vb v_3 ] where $k_1, k_2, k_3$ are scalars

[elfin berry]

cloud

[restive silo]

Cool

[fervent rampart]

so asking ``do these vectors span \R[3]'' means asking, does the linear system of equations [ k_1 \vb v_1 + k_2 \vb v_2 + k_3 \vb v_3 = \vb b ] for every vector $\vb b$ in \R[3]?'

[elfin berry]

cloud

[fervent rampart]

using the definition of matrix multiplication, we can rewrite this as [ \begin{bmatrix} \vb v_1 & \vb v_2 & \vb v_3 \end{bmatrix} \begin{bmatrix} k_1 \ k_2 \ k_3 \end{bmatrix} = \vb b ] where we have taken $\vb v_1, \vb v_2, \vb v_3$ as the columns of a matrix

[elfin berry]

cloud

[restive silo]

Ah I see

[fervent rampart]

so we do the rref of the matrix and ask ourselves, "does the equation Ax = b have solutions for every possible vector b?"

[restive silo]

so bcz we have 3 linear combinations it's R³ and we should search if we have a consistent system for b or not. Am I correct?

[fervent rampart]

yes, we are asking whether it is consistent for every possible vector b

[restive silo]

But the solution set will be R¹, right?

[fervent rampart]

what do you mean?

[restive silo]

I mean when we make the system in echelon and we figure out it is consistent

My teacher puts for vector b. x y z for example and after we reach into echelon they substitute for them general numbers

And then they solve the system to get the solution set if there any. To get values of x y z

So those will be always like this
{[]}

Just one matrix so R¹

[fervent rampart]

the solution set to a system of linear equations is generally a vector / set of vectors

[restive silo]

I see. This thing has nothing to do with the question itself

[fervent rampart]

because you have individually solutions for x, y, and z, but you package them all together as a vector

[restive silo]

Yep right

Can I ask one more thing

To determine linear combinations are independent we should have all pivots elements when that matrix is turned into echelon form

Is that correct?

[fervent rampart]

we determine if vectors (specifically the column vectors of a matrix) are linearly independent that way

[restive silo]

Okay that's great

Thanks alot

.close

[marsh citrus]

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[marsh citrus]

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[abstract hull]

What is the substitution

[abstract hull]

Or write denominator as 1- root and make it - tan^-1

[amber birch]

you want to differentiate this or integrate this?

[abstract hull]

Differentiation

By substitution

[amber birch]

you mean the chain rule then

so yes substitute u as the bracket

$\frac{dy}{du} = \frac{1}{1 + u^2}$

[elfin berry]

south, just south

[amber birch]

now you just need to find du/dx

[abstract hull]

No

[proud arch]

Hey @abstract hull

Again

Lol

[abstract hull]

Yooo

[proud arch]

Put e^ x = tan 2 theta

And do as we did the other day

[amber birch]

oh you mean that

but like

this way is perfectly fine

[proud arch]

1+ cos2x = 2cosx ^2

And what not

You will get simple theta at the end

[abstract hull]

After this what? I'm dogshit when substitution

[abstract hull]

Yeah but the question is solve by substitution

[proud arch]

1 + e^2x = ( sec 2 theta )^2

This Darn AI bruh

Converting my symbols to emojis

Lol

[abstract hull]

Lol

[proud arch]

@abstract hull

Got something

?

[abstract hull]

So tan 2 theta / sec2theta-1

[proud arch]

Yep

Sin 2theta / 1- cos 2 theta

And you know what to do from that

[marsh citrus]

@abstract hull Has your question been resolved?

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[abstract hull]

How did u get there

[marsh citrus]

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[umbral rain]

i got up to BH.CA = |a|^2 + 2a.b - 2b.c - |c|^2 and idk what to do from there

[tough igloo]

could u show full working out?

is thag hsc 2022

omg that is

[tough igloo]

there shouldnt be a b in there

[tough igloo]

what was ur BH and CA ?

[bitter terrace]

NOOOO NOT HSC

[tough igloo]

aussie too?

[bitter terrace]

ye man

[tough igloo]

yoooo

[bitter terrace]

i got a yr left :,)

[tough igloo]

2025 graduate?

[bitter terrace]

yep

[tough igloo]

AY ME TOO

u do 3u?

[bitter terrace]

anyway

[umbral rain]

i got BH = BO + OH = a + 2b + c

and

CA = a-c

[tough igloo]

um

r u sure

check again

[umbral rain]

wait which one

[tough igloo]

just BH

[tough igloo]

it is BO + OH

[umbral rain]

ohh wait is it just a + c

[tough igloo]

yuh

try applying dot product again

!

r u also 2025 grad?

[umbral rain]

ok that makes it a lot simpler thx!

[umbral rain]

ye lmao

[tough igloo]

nwsss

[bitter terrace]

technically CA would be -c+a

be careful with that

[tough igloo]

hello fellow soldier

[tough igloo]

no

its a-c or -c+a

[umbral rain]

hello there

[bitter terrace]

OH SHIT

LOL

[tough igloo]

LMAO

[bitter terrace]

did not see that arrow correct

[tough igloo]

youll notice its diff of 2 sqaures

and youll get |a|^2 - |c|^c

notice how this is a circle

[umbral rain]

ohh and then radius is the same

so its just zero

[tough igloo]

ie magntiude of OA or OB or OC are all the same lengths (all r radius)

[bitter terrace]

ye

exactly

[tough igloo]

mhmm

[umbral rain]

rightt ok thanks guys

[tough igloo]

thus proven

[umbral rain]

yep thanks!

[bitter terrace]

thats crazy tho

[tough igloo]

nwwws

[umbral rain]

ikr

[tough igloo]

i wish ylal goodluck!

[umbral rain]

u too!

[bitter terrace]

gl for the next yr guys

[umbral rain]

last yr ;-;

[tough igloo]

im not doing 3u hsc next year!

[tough igloo]

thanks mate

[bitter terrace]

accelerated??

[tough igloo]

yuh

[bitter terrace]

BRUH SAVE ME TAKE ME WITH YOU

[tough igloo]

but i wish u ugys goodluck

yall got this

[umbral rain]

yess gl

[tough igloo]

keep practicing! WERE GONNA E4 THIS

[umbral rain]

yess hopefully

[tough igloo]

I SITLL HAVE TO SIT HSC FOR THE REST OF MY SUBJECTS 😭

[bitter terrace]

oooft pls not the asian 5

[tough igloo]

the what

[tough igloo]

beleive in urself!

[umbral rain]

its like phys chem eng ext math ext 2 or smth

[umbral rain]

:)

[tough igloo]

take ur time reading the question carefully and dont rush or else youll make silly mistakes

[umbral rain]

yepp

[bitter terrace]

phys maths 4u like eco or software and bio

[umbral rain]

and eco?

[bitter terrace]

yeye

[tough igloo]

u predicted me

sorta

[tough igloo]

u guys gonna pick up 4u?

i odnt do chem

replace chem for bus n eco n bio

[bitter terrace]

im dropping that

[tough igloo]

and u got my subejcts

[tough igloo]

ahhh

[umbral rain]

already doing it ;-;

[tough igloo]

whyyy

[bitter terrace]

bro i have too many units

[tough igloo]

u hsould keep it

[tough igloo]

same 😭

[bitter terrace]

lolllll

[tough igloo]

im on 14...

[bitter terrace]

NO WAY

[umbral rain]

damn why

[tough igloo]

4u math eng adv eco bus phy bio

MY SCHOOL WONT LET ME DROP

SFNDVKFDBV

i cna drop next term...

[bitter terrace]

THATS CRAZY

[bitter terrace]

oh saving grace

[umbral rain]

bruhh thats dumb

[tough igloo]

imgonna run away from school i have 6 back to bakcs in 4 days

[tough igloo]

ikr

[tough igloo]

in wk8

[umbral rain]

jeez thats tough

[tough igloo]

welp

[bitter terrace]

jp ext aint required an exam so hehe

[tough igloo]

i hope u guys have a better time than me

[bitter terrace]

man i hope you get that subject drop fast

gl guys

👍🫡

[tough igloo]

turn to understand the concept before going into qns blind, thats my advice

[umbral rain]

lmao idk abt that i chose rly difficult subjects

[tough igloo]

thanks bro

[umbral rain]

fr

[tough igloo]

u tooo!

[tough igloo]

me too 😭

[umbral rain]

ok but tbh i cant believe its already wk 6

[tough igloo]

DONT SAY THAT

[umbral rain]

LOL

[tough igloo]

im gonna kms before wk8 starts

[umbral rain]

u too

[umbral rain]

lmao im joining u i feel so underprepared for everything

[tough igloo]

twinning 😭🤝

2022 qn14b is also vectors it sso goofy

[umbral rain]

😭

[tough igloo]

im actually so cooked for everything

[umbral rain]

kinda crazy how did u know the exact question from the exact paper

[umbral rain]

lmao me too dw

[tough igloo]

i sat the hsc this year 😭

[umbral rain]

OH

RIGHT

[tough igloo]

i already idd all the apst appers lmao

[umbral rain]

ohhhhh ok that makes sense

[tough igloo]

anyways gotta go do 4u hw

goodluck!

[umbral rain]

lmaoo ok good luck!

thanks again

:)

[tough igloo]

np!

gotta help a fellow soldkier out

[umbral rain]

have fun with yr 12 :D

[tough igloo]

u tooo!!

[umbral rain]

lmaoo

.close

[marsh citrus]

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[marsh citrus]

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[grizzled monolith]

In a 10×10 square grid,
22 cells are marked. Centurion counted the number of side-adjacent marked cells for each of the 100 cells of the grid, while Millennial did the same but only for each of the
36 boundary cells. It turned out that there is exactly one count of 4 among Centurion’s results, and the sum of all Centurion’s counts is 64. What is the sum of Millennial’s counts?

[outer lodge]

What do you mean by one count of 4?

[grizzled monolith]

I think it means there is exactly one cell in the grid that has 4 side-adjacent marked cells.

[outer lodge]

Is a side-adjacent marked cell not just a cell that is marked and touching the bounds of the grid?

[grizzled monolith]

Correct

[outer lodge]

If that’s the case then you can’t have a cell with 4 side-adjacent marked cells next to it right?

[grizzled monolith]

Slrr, ahhh right, oki I'll try that one

I got 32, 24, 23, 34, 36, 64..they're all wrong

[outer lodge]

How are you getting these? Im still a bit confused on the question

[marsh citrus]

@grizzled monolith Has your question been resolved?

[marsh citrus]

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[loud jolt]

one sec

after re arranging

[ancient acorn]

what is the answer

for this

??

[loud jolt]

wrong channel

[ancient acorn]

sorry

which is the channel

to answer this question

[loud jolt]

#❓how-to-get-help

oh wait

it’s #help-32

this is yours

[ancient acorn]

no one is answering

I guess I need to wait

[loud jolt]

so just wait

also someone is answering

[ancient acorn]

sorry

[loud jolt]

.close

[marsh citrus]

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[marsh citrus]

@median stag Has your question been resolved?

[median stag]

<@&286206848099549185>

[hidden dawn]

if i understood well

actually i didnt understand very well

[marsh citrus]

@median stag Has your question been resolved?

[brittle shuttle]

A hint is that the number of total line segments is C(n,2) and the number of blue line segments is exactly n. Now you should set an equation and solve for n

[marsh citrus]

@median stag Has your question been resolved?

[marsh citrus]

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[ancient acorn]

hi

[ancient acorn]

can I have some answers to some questions please

[marsh citrus]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[ancient acorn]

can you help me

[marsh citrus]

@ancient acorn Has your question been resolved?

[marsh citrus]

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[ancient acorn]

could you help me

with this

please

is anyone online

so we go a triangle is 180 degrees

would I have to minus the sides to get the answer

[dire pulsar]

Use the rule sine which is opposite divided by the hypotenuse. This means

sin 42 = 6.5/x

Solve for x use a calculator

@ancient acorn

[ancient acorn]

hi

what do I have to do

sorry

[dire pulsar]

Solve the equation because u are solving for the sine ratio

[ancient acorn]

so I need to write

sin 42

on the calculator

is the answer 6.5

x

??

are you afk

or not

[ionic cradle]

Hey, do you have any idea how it could be solved? What tactics is that work for?

[ancient acorn]

erm

we know a triangle is 180 degrees

would I have to minus the lines to get the answer

[ionic cradle]

Have you learnt anything with trigonometry

[ancient acorn]

no

I dont think so

[ionic cradle]

I’ll try to solve it

[ancient acorn]

thanks

do you know how to solve it or not

[ionic cradle]

So this is the basic and you can prove that ABH and AHC are similar triangles because they have the same angles and idk what to do from there

[ancient acorn]

ohh can I give you a different question for you to answer

[ionic cradle]

I can try

[desert agate]

It’s better to use SOHCAHTOA

[ionic cradle]

I think you just have to add 180 to 149

[ancient acorn]

to get the answer

[ionic cradle]

Yea, I personally haven’t studied these but I asked a friend

[ancient acorn]

could you do this one for me

can I add you on dis

[ionic cradle]

Sure

[ancient acorn]

there I sent

[north wadi]

3 * 4

three ways to choose a starter * four ways to choose your main course

[ionic cradle]

That’s just the total

[north wadi]

is it not asking for all possible combinations

ah

thats kinda tedious

[ionic cradle]

Oh wait ya

[desert agate]

FB,FC,FL,FR,PV,PC,PL,PR,SB,SC,SL,SR

[north wadi]

lol

[ionic cradle]

I was thinking about combinations but it’s easier

[north wadi]

start from F(all letters on the right row)

and iterate to the next eltter

a bit annoying but manageable

[marsh citrus]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

@ancient acorn Has your question been resolved?

[marsh citrus]

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[neat elbow]

hello, i need help on how to do the fourth one please

[astral kite]

what have you tried

[limber hearth]

Index form ?

[wary kite]

^m/n

[astral kite]

a^b

[neat elbow]

i don't know where to start

[limber hearth]

Ok ok

[astral kite]

ok

roots are exponential powers

how can we use this to help us

[limber hearth]

10 = 10^1
Sqrt(10) = 10^1/2

[astral kite]

yeah that

now

[neat elbow]

i genuinely have no idea since this is a topic i'm really bad at

[astral kite]

oh

well square root of x is bsaically x^2=10

right?

[neat elbow]

yeah

[astral kite]

so basically, if we were to isolate x,

x^2^1/2=10^1/2

since x^a^b=x^ab

that means that x=10^1/2

thus, sqrt10=10^1/2

now

what is the number 10 in an exponential form

it's 10^1

since the numbers 10 and sqrt 10 are multiplied

10^1*10^1/2

is our number in exponential form

[wary kite]

what

[astral kite]

I'm trying to visualize it

[wary kite]

🤔

[astral kite]

(I know theres a negative component)

[wary kite]

sqrt(x) = x^2 = 10??

[astral kite]

no it's situational

sqrt10 is x^2=10 for positive

[wary kite]

i think it’s clearer to start from

$x^{a+b} = x^a \cdot x^b$

[elfin berry]

knief

[astral kite]

alr

you can teach him your ways

[wary kite]

to understand why roots can be expressed in terms of powers we should use this fact

consider the sqrt(x)

if we wanted to express sqrt(x) in terms of an exponent let’s see what happens when we multiply sqrt(x) * sqrt(x)

can you tell me what sqrt(x) * sqrt(x) is @neat elbow

[neat elbow]

would it just be x

[wary kite]

yes and x is just x^1

so we can say if sqrt(x) were to be expressed in terms of some power, then we know that

$\sqrt{x} * \sqrt{x} = x^1$

[elfin berry]

knief

[wary kite]

but we know from this that x^a * x^b = x^(a+b)

so for now we don’t know what sqrt(x) is as a power but let’s say it’s x^a for some a

and we will solve for that value of a

so

$\sqrt{x} \cdot \sqrt{x} = x^1 \iff x^a \cdot x^a = x^1$

[elfin berry]

knief

[wary kite]

yes?

[neat elbow]

yeah

[wary kite]

so we can now use

$x^a \cdot x^a = x^{a+a} = x^{2a} = x^1$

[elfin berry]

knief

[wary kite]

hence 2a = 1

thus a = 1/2

which means that sqrt(x) = x^(1/2)

we can extend this idea for roots other than the sqrt such as the cube root or 420th root etc

and we will find that

$\sqrt[n]{x} = x^{\frac{1}{n}}$

[elfin berry]

knief

[wary kite]

note that with the sqrt it’s just when n = 2

we just don’t write 2 there by convention

[wary kite]

does this make sense to you

[neat elbow]

yeah

[wary kite]

so you have

$10\sqrt{10}$

[elfin berry]

knief

[wary kite]

given that sqrt(x) = x^1/2

how can we express sqrt(10)

[neat elbow]

sqrt(10) = 10^1/2

[wary kite]

mhm perfect

now 10 is just 10^1

so

[wary kite]

we can use this fact

$10^1 \cdot 10^{1/2} = ?$

[elfin berry]

knief

[neat elbow]

10^3/2

[wary kite]

mhm

that’s it

[neat elbow]

oh damn wth

thank you so much for your help

[wary kite]

yep

you’re welcome

[neat elbow]

my exam is actually in less than 3 hours and this genuinely helped me a lot, thank you 🙏

[wary kite]

mhm no worries

good luck

[neat elbow]

thank youu, have a good day/night

.close