help-39

[prisma kernel]

I mean, you already know the area of the lower part yeah

so Area of figure - 1

[prisma kernel]

And for area of figure divide it into two rectangles and add their area

[green meteor]

This?

[prisma kernel]

Yup, looks correct

[green meteor]

Ok 👍

Thanks🙏

.close

[pearl pond]

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[cobalt apex]

Can I get help with c I don’t know how to begin

.close

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[fallow narwhal]

yo could somone pls provide some assistance with this q pls in the box is what ive got so far

[tiny egret]

theres a couple ways to do this; you can navigate the board mentally using only those moves and write down the moves you make as you go, or turn it into a system of two equations

which would you be more comfortable doing?

[fallow narwhal]

i would like to do it as a system of two equations as that is what they recommended

like ik with vector math its js 3 down and 6 right but like we dont have those vectors defined js these topsy turvy x's and y's which is annyoing icl

[tiny egret]

so with that, we have unknowns a,b, and we have ax + by = (6,-3)

can you see that?

[fallow narwhal]

hmm wait so

yeah yeah makes sense so thats the first eq

[tiny egret]

thats actually the full equation

[fallow narwhal]

oh what-

[tiny egret]

since x,y are vectors

[fallow narwhal]

oh wait true

[tiny egret]

$x=\begin{pmatrix} x_1 \ x_2 \end{pmatrix}, y=\begin{pmatrix} y_1 \ y_2 \end{pmatrix} $

[jolly parrot]

lyric

[tiny egret]

and we know those values

so go from there

[fallow narwhal]

so from that eq im solving for a and b

correct?

[tiny egret]

ye

[fallow narwhal]

i js cant visualise how i can get to 3,6 from there by multiplying to the x and y vectors

coz ive enviosned how to get there like step by step using the addition of those vectors

and if i js sum them up i get

-2x+y

[tiny egret]

we can see that's not correct

[fallow narwhal]

ohhh wait yeah ok i was like thinking of each move being individual not vectors adding up my fault

i still dont get how im meant to figure out a and b tho

i genuienly cant visualise it

[tiny egret]

solving the equation would avoid having to visualise it

or, you can use some drawing software to track moves and try figure it out there

[fallow narwhal]

yeah but how can i solve for two variables with one eq ):

[tiny egret]

well, $$ax+by=\begin{pmatrix} 6 \ -3 \end{pmatrix} \quad\implies\quad a\begin{pmatrix} x_1 \ x_2 \end{pmatrix} + b\begin{pmatrix} y_1 \ y_2 \end{pmatrix} = \begin{pmatrix}6\ -3 \end{pmatrix}$$ $$\therefore \begin{pmatrix} ax_1+by_1 \ ax_2+bx_2 \end{pmatrix} = \begin{pmatrix} 6 \ -3 \end{pmatrix}$$

[jolly parrot]

lyric

[tiny egret]

this gives two equations

[fallow narwhal]

hmmmm

gimmme a sec

[tiny egret]

nw

[fallow narwhal]

ohhhhhhhhhhhhhhhhhhhhhhhh

i got it gng

i realised u can js find x and y

by finding the distance of each one

ty for ur help

[tiny egret]

also though you might be constrained to the chess board

so might not work

[fallow narwhal]

oh damn

i mean i got it right

ty tho

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[sharp smelt]

oops, lemme make the soln and question 1 message

.clsoe

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[shell brook]

try using math

[cinder flower]

[shell brook]

this looks so condescending

[unborn abyss]

[shell brook]

why am i getting stared at..

[cinder flower]

messages deserving get ’d

it’s the circle of life

[still hamlet]

[shell brook]

:AA_Sus:

imagine that the emoji worked

please

[shell brook]

okay i feel bad for OP

[vestal tapir]

because of the wondeful things he does?

[shell brook]

because he opened this channel for seeking professional help

[sharp smelt]

I think I know how to do this, just need my work checked

The issue is typing all this out

formatting matrices in TeX Is NOT FUN

[feral sedge]

the channel is closed btw

just a reminder

[unborn abyss]

.use an online generator for that

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[short orchid]

guys how to find critical region by binomial distribution?

[short orchid]

i tried converting binomial to normal distribution and got x<=5.342 but i think it's wrong

[pastel umbra]

Why're you converting it though?

[quick star]

it says "use a binomial distribution"

[pastel umbra]

You can directly just calculate those P(X <= [number]) straight out of the Binomial

[pastel umbra]

This too

[cinder flower]

[quick star]

ilyy

[short orchid]

im trying to use the z = (x̄ - μ)/ (s/ sqrt(n)) formula to find x̄

bcs thats the only way i know

[short orchid]

ooooo

wait

[cinder flower]

kuromi necklace

[pastel umbra]

This is an A Level paper, ...right?

Judging from the source

[short orchid]

i thought critical region is like the point in the distribution where it becomees rejected so we have to find that point to find P(X>=critical region) for the type 1 error?

[short orchid]

yes

[pastel umbra]

yes

[pastel umbra]

Right, so I would imagine you have a Casio scientific calculator for this

[quick star]

omggg, that's so cool!! more kuromii~

[cinder flower]

so cute

[pastel umbra]

@quick star @cinder flower Elsewhere, please

[tranquil bolt]

Yes

[pastel umbra]

Whether the typical Classwiz one or the CG50/100 one, both of those have the distribution calculation option (that you should be familiar with from when you started the binomial distribution chapter earlier on)

[sterile python]

Can you get the casio tho?

[pastel umbra]

If you're asking about the direction of the inequality though, look at your hypotheses from (a) - it's one-tailed, and the H1 here is less than 0.2, so our CR has to look at being less than some value

[pastel umbra]

I can't say for sure that OP is in the UK, but if they are [guessing again from the paper being an A Level which is typically a UK qualification#], they're a) sold in quite a few places, and b) bulk-bought in by a lot of secondary schools (esp. the graphing calculators)

idk where @short orchid's gone, though

[sterile python]

Yeah true, anyways we mostly do statistics by hand in asia

[pastel umbra]

Good for you

But maths exams at this stage don't test your computational skills; though you can do them by hand (and in some cases you are taught to, because some questions are algebraic in nature so a calculator wouldn't work), that's not really the point of the paper

[pearl pond]

@short orchid Has your question been resolved?

[short orchid]

Ok guys im back

[short orchid]

Yes

[short orchid]

Yes rejection region is on the left

[short orchid]

Yes we dont use calculator for the distribution but it’s nice to know that theres a function about it in my calculator

I dont understand why we can do P(X<number) directly

[pastel umbra]

Binomial CD if you're using the Classwiz

https://youtu.be/QEFZHNaufxY

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[sharp smelt]

Define $f(x)= \begin{cases} x & x \in \Q \cap [0,1] \ -x & x \in (\Q)^c \cap [0,1] \end{cases} ; g(x)=x^3$

[jolly parrot]

wai

[sharp smelt]

oops

this won't work

.close

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[sharp smelt]

.reopem

.reopen

[pearl pond]

✅ Original question: #help-39 message

[sharp smelt]

$f= \begin{cases} 1& x≥0 \ -1& x<0 \end{cases}; g(x)= x \cos\left( \frac{1}{x} \right)$

oh it might

Oh yeah, it does , the upper and lower sums will diverge in opposite directions

[jolly parrot]

wai

[tropic saddle]

whats R[0,1]

[sharp smelt]

riemann intwegrable on [0,1[

[pearl pond]

@sharp smelt Has your question been resolved?

[ivory basin]

Yeah I think that one's too discontinuous

Or in this case, discontinuous enough

[pearl pond]

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[sharp smelt]

thanks

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[royal galleon]

from what I understand it wants us to find the biggest cylinder contained in the frustum. Wouldn't that always be the height of the frustum and the radius of the top part. I don't see where h is playing a role

[gray tundra]

hey

[pearl pond]

@royal galleon Has your question been resolved?

[pastel lagoon]

h is part of the dimension of the shape.

you plug a number in for the shape, you get the volume of the cylinder that can be cut from the log of that shape with h in its parameter.

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[arctic tinsel]

need someone patient to help cause I'm slow at algebra atm

[arctic tinsel]

idk where to start

[proper nova]

start with the exponent rules

turn $8^x$ into $2^{\text{something}}$

[jolly parrot]

1 divided by 0 equals Infinity

[arctic tinsel]

2^3(x)?

[proper nova]

yes

$2^{3x}$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

so on the numerator, you got that

on the denominator, you got $2^y$

[arctic tinsel]

yep

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

or the fraction would turn into: $\frac{2^{3x}}{2^y}$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

what exponent rules could be applied here

[arctic tinsel]

since its being divided maybe subtract the exponents?

[proper nova]

yes!

[arctic tinsel]

ohh

[proper nova]

damn you're not slow at all

[arctic tinsel]

so we have 2^3x—y

[proper nova]

yes

[arctic tinsel]

OH AND 3-Y IS 12

[proper nova]

🔥

[arctic tinsel]

2^12?

OHH

[proper nova]

yep

[arctic tinsel]

I GET IT

[arctic tinsel]

THANK YOU SO MUCH

[proper nova]

yeah you're not slow at all

!done

[pearl pond]

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[arctic tinsel]

God bless you man

[proper nova]

np

[arctic tinsel]

ty

.close

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[proper nova]

i think ts SAT kind of question

[pearl pond]

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[daring bay]

yo i plugged 3t into this function what did i do wrong

[west sapphire]

(3t)^3 is not equal to 3t^3

[plush bramble]

,tex .exp rules

[jolly parrot]

riemann

[daring bay]

oh my god

shit

i forgot that

[daring bay]

is there another rule thing like this for fractions?

like 3t/2 woudl be the same as t * 3/2 right

[west sapphire]

yes 3t/2 = t * 3/2, but it has nothing to do with exponents

just commutativity of multiplication

[pearl pond]

@daring bay Has your question been resolved?

[daring bay]

ty

[pearl pond]

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[gilded girder]

What part did I mess up on?? Sorry if the work is messy

[plush bramble]

,rccw

[plush bramble]

x^3 - x does not equal x^2

,tex .int rules

[jolly parrot]

riemann

[plush bramble]

Sum rule

[gilded girder]

oh yea

ty

i missed the entire unit 7 in class so im trying to cram everything in rn for a test tmr and im like losing my mind

so could i rewrite it as 8 integral x^3 - 8 integral x??

[pearl pond]

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[narrow breach]

I'm not sure how to get started with this one

[chrome lintel]

Suppose the stone started with velocity 0 ft/sec

[chrome lintel]

Did it help?

[narrow breach]

let me write it down real quick

[chrome lintel]

ok

[narrow breach]

am I heading in the right direction with this?

[chrome lintel]

Yeah

[narrow breach]

so from here how do I find S0? or is this just it? (do I divide by 32??)

[chrome lintel]

Look, you have to use the equations S(t) and S'(t), in order to solve the problem.

I'm still writting the complete solution

[narrow breach]

ok, can I show you what I have come up with?

[chrome lintel]

Yes

[chrome lintel]

checking

Your only mistake was setting S0 at t, it should be S0 at t0.

It makes a lot of difference

If you substitute S0 as 32t, in S(t), it's describing an initial height which changes with time.

It's not the physical situation

[chrome lintel]

And we should have -(185+S0)/32=-t1

so the negatives will cancel out

@narrow breach

[narrow breach]

gotcha, I'm gonna try re-writing that so I can understand the process a little bit better

Did I do something wrong? when I was doing t1 I got the fraction as a positive

[chrome lintel]

nothing wrong in this picture

[narrow breach]

am I missing something? I dont see how your fraction is negative

[chrome lintel]

I've made a mistake in the photo and I corrected it here.

[narrow breach]

ohh my apologies I over looked that

[chrome lintel]

-(185+S0)/32= - t1

[narrow breach]

I'm having a hard time understanding these next few steps you have done

[chrome lintel]

I'll explain

it looks like it's not an easy exercise

The equations seem to be describing a building next to a cliff.

a building which has a height S0=64 ft

and the cliff's height is over 600 ft

[chrome lintel]

because when you are at the time t0=2, your velocity is 0, right?

If your initial speed was 0 ft/sec, and your final speed is -185 ft/sec, considering the original equation S(t)=-16t²+S0, that is the conclusion.

[chrome lintel]

Unless the exercises gives you other information.

@narrow breach

[narrow breach]

thats all the info I am given

I'm not sure where this section came from

[chrome lintel]

Yes, that's right

[narrow breach]

ohh is this a factoring situation?

[chrome lintel]

The negative altitude means that the rock fell under the building's floor level.

[chrome lintel]

indeed

[chrome lintel]

you just missed a t0

it should be -16t0(t0-2)

[chrome lintel]

Without more context info, that's a difficult math/physics question for beginners.

[narrow breach]

yea my prof does this quite a bit from what I've heard

[chrome lintel]

If you've got more questions, feel free to ask for help.

[narrow breach]

is this an Sn ?

[chrome lintel]

S0

[narrow breach]

gotcha

[chrome lintel]

my handwritting is a bit odd today

[narrow breach]

is there a name for this formula?

this to be more specific

[chrome lintel]

the S(t) is just one of the Torricelli's equations

S(t1) is the value of Torricelli's formula at t=t1

[narrow breach]

how did you get the value for the 735.76ft?

[chrome lintel]

I'm terribly sorry, I've made a stupid mistake at the first solution. We need to use S'(t)=-32t, instead of S(t)=-32t+S0

the solution is much easier

it's about 535 feet tall building

[chrome lintel]

The derivative of the constant S0, should be 0

There's no cliff in the problem, the building is really tall

@narrow breach

I double checked this time

[narrow breach]

it's alright, thank you for checking. I'll just read through this real quick

let me make sure I know the proper steps real quick. So first I set v(t) = s'(t) to get t1 then I plug that into the original formula and set it equal to zero to find the height?

[cinder flower]

decent handwriting

[chrome lintel]

Yes, and when you substitute t1 in the original formula, you have to consider that S(t1)=0.

[narrow breach]

gotcha, thanks for the help 😄

.close

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[left vector]

hi

[left vector]

where did i go wrong

the ans in book is 29.2

i have tried like 3-4 times still making some mistake

[minor cloud]

both your sum of f_i and your sum of f_i \cdot d_i are wrong

your f_i is especially noticable. 8+9+10 already sum to a number higher than 25, let alone the other ones in the column

[left vector]

ahh i dont know how did i mess up so bad lmao

[left vector]

i did not see that 10 mb

[minor cloud]

nps

[left vector]

thanks i got my ans have a nice day : )

[minor cloud]

you too!

!done if you have nth else, and see you around!

[pearl pond]

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[left vector]

.close

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[frozen shadow]

hi,,, i dont understand where 2xy came from

[frozen shadow]

got the same answer as the final one but without - 2xy

[cinder flower]

,rotate

[cinder flower]

good handwriting

[frozen shadow]

thank you

[cinder flower]

this is what you're looking at?

[frozen shadow]

yup

that was one of the examples

i redid it today but lkke

[cinder flower]

there is a y*2x in the middle of the top line

[frozen shadow]

no hyeah but the thing im supposed to differentiate is

x²y + 2y³ = 3x + 2y

my signal is butt but im trying to send what i did today

[cinder flower]

is this what you're wondering about? (ignoring the middle 2 scribbled out lines)

[frozen shadow]

hyeah im mostly wondering where + y • 2x came from ??

like am i tripping

[cinder flower]

product rule on x^2y

[frozen shadow]

ehat x^2y

[cinder flower]

the x^2y in the top left corner here

[frozen shadow]

wouldnt it just be

2xdy/dx

oh wait hold on

ohhhhh

i see now

okay thank you lemme redo it

.close

[pearl pond]

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[cinder flower]

no problem

[frozen shadow]

ehy didnt the y in 2x • y get dy/dx attached to it ?

[frozen shadow]

hello again

[frozen shadow]

the question sent first sorry

[pearl pond]

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[naive dirge]

hello

[naive dirge]

i am trying to find the right term

adding, minus, multiply, and divide

what is the term describing these things

[heavy onyx]

arithmetic

[naive dirge]

arithmetic computation

[feral sedge]

operators?

[naive dirge]

is that it?

[feral sedge]

it's not a commonly used term as far as I know

[naive dirge]

if I were doing some equation solving, I could not avoid arithmetic computation

[naive dirge]

what is the more common term

[feral sedge]

I don't know

I would just call them the four fundamental operations

[feral sedge]

I mean this sounds fine, though I don't have all the context

[naive dirge]

I am writing an essay which revolves around how repetitive tasks can lead to cognitive bias.

One example is constantly doing this "fundamental operations" can lead to a biased way of thinking.

one may confuse a dash sign in an essay with the minus sign in an equation

[feral sedge]

Ah ok yeah in that case maybe the arithmetic operations or something

Or you might just need to write them out idk any universally recognized name for them

[grave mirage]

Basic arithmetic operations

[warped violet]

They are called operators

The symbols themselves are not computation

+, -, / or whatever doesn't do anything

It just tells you what to do

And the process of "doing" is called computation

[pearl pond]

@naive dirge Has your question been resolved?

[vestal tapir]

they are just functions

we call them operators because they don't look like functions with ()

and cuz they are so basic

[pearl pond]

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[pearl pond]

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[boreal slate]

any hint plz ? i aint sure if my approach is ok

[tropic saddle]

are you forbidden from using a calculator to compute 1.04^2 ?

[boreal slate]

yep

[tropic saddle]

are you sure?

[boreal slate]

no

but ig thats the whole point of Rn lol

[tropic saddle]

no

computing 1.04^2 is possible entirely by hand

at that point a calculator is just a shortcut

by replacing 1.04 with 1.1 you are making your bounds much worse

[boreal slate]

ok , but choosing 1.04 gives 0.5408

[tropic saddle]

which is certainly better

continue like you did

you should see what equation you ultimately have to solve

[boreal slate]

you mean it goes (1.04)^3 then 4 then 5 until i get what i need ..

[tropic saddle]

yes but dont forget the factorial thats building up in the front

[boreal slate]

ofc

[boreal slate]

btw is using ln(1+x) could maybe let me avoid using calculator?

[tropic saddle]

oh whoops I didnt even notice that you arent using ln(1+x)

you should definitely use ln(1+x)

[boreal slate]

oh

[tropic saddle]

but the computation is essentially the same

[boreal slate]

yeh

thx

< 3

.close

[pearl pond]

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[pearl pond]

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[proud vessel]

The question a is to find the area of the triangle

[proud vessel]

This is my working

According to the textbook my answer is wrong, it should be +96

I am unsure where my error is

[warped violet]

And what is your answer

At least you labeled some things this time

[proud vessel]

My answer was x^2-12x-96

[warped violet]

... And how did you get to that

Literally just one line written on the bottom right

[proud vessel]

The area of the square is 12 x 16 = 192

[tardy reef]

so what happened to that 192 in the end?

[proud vessel]

I added together the area of the smaller triangles , then subtracted them from the total

[warped violet]

Well incorrectly it seems

Would be nice if we would see the work where you did the adding and subtracting

[tardy reef]

I dont see that 192 used later anywhere

[proud vessel]

The triangles are labeled as A2 A3 and A4

[warped violet]

I am aware

And their areas are calculated correctly

[proud vessel]

Oh, I see, I just forgot to subtract -96 from 192

Easy

.close

[pearl pond]

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[warped violet]

¯_(ツ)_/¯

Not even a thank you or bye

[proud vessel]

I mean, saying “well incorrectly it seems” isn’t exactly helpful so like

[warped violet]

Unfortunately I can't see what isn't shown

If you wish not to show what you need help with, I can't help

Or what you did incorrectly

And to be honest I'd get it if this was the first time. But I've participated in your channels a couple of times now and it seems like you are simply ignorant in this regard ...

Not to mention I willingly spent 10 minutes at least trying to help you here (not trying to make it sound like that's the problem, it was my choice). I think we also thank people for at least trying to help us, even if unsuccessfully.

[pearl pond]

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[low matrix]

Alr so this is f_x and f_y

idk how to find the critical point through the denominator

how do i get a point from this weird equation

[calm wing]

$\frac{a}{b}=0 \implies a=0$

[jolly parrot]

artemetra

[low matrix]

I also have to tackle the case of partial derivative being undefined

[calm wing]

why

[low matrix]

The denominator will be zero in that case

[low matrix]

Read my question please

[calm wing]

why would an undefined partial derivative mean a critical point?

[low matrix]

my textbook says so

[calm wing]

interesting ok

[calm wing]

complete the squares for x and y

[low matrix]

this is the equation

the same one

[calm wing]

are there any restrictions on the domain

cuz this has infinitely many solutions

[low matrix]

idk man

[low matrix]

how so

[calm wing]

oh actually! important point: they say "either or does not exist"

so exactly one must exist

[calm wing]

but if this is zero then both partial derivatives do not exist

[summer imp]

I don't think this is how it's supposed to be interpreted

It's quite common to have the critical points be those points where derivatives aren't defined

As soon as at least one of the partials isn't defined you can classify that point as a critical point

[low matrix]

nah both can be undef too

[summer imp]

In this case they just get a whole hyperbola worth of critical points (that come from the undefined partials)

[low matrix]

how are you guys innterpreting this 😭

i don understand

[calm wing]

hm it's weird wording then

[summer imp]

This is the equation of a hyperbola

[low matrix]

yeah

[calm wing]

yes that's the only way

[low matrix]

but how are there infinite points

[summer imp]

It just means any point on this hyperbola is a critical point

You can still check for those where the partials vanish

[calm wing]

all of these points satisfy the equation of the hyperbola

[summer imp]

Then critical points will be those points on the hyperbola where your partial derivatives don't exist, and those points where they are both 0

[low matrix]

but all points on the hyperbola make this term zero right?

[summer imp]

Yes

Hence why they are critical points.

[low matrix]

so infinite critical points huh

tf

[summer imp]

This isn't really surprising though this is just where your f(x,y) = sqrt(blahblah) vanishes

But yes, there are infinitely many critical points

[low matrix]

so how do i find the critical points for this function

how to write the answer i mean

im getting (2, -3) from the numerators btw

[summer imp]

You just say the critical points are (2,-3) and those points satisfying the hyperbola equation

[low matrix]

what if i am also asked to determine the extrema

how can i check infinite points then

[summer imp]

Well your function is nonnegative

And it’s 0 on this hyperbola

So all of the hyperbola would have minimums

So you would just do your analysis as always to classify the points where the derivatives vanish and go back to the points where they don’t exist to classify those too using other means

[low matrix]

ermm im lost

[low matrix]

how did you determine this btw

[summer imp]

Everything is under a square root

Whatever is spits out is at the very least 0

[low matrix]

and what abt this

what does dis mean

[summer imp]

The denominator in your partials

It’s exactly f

So the critical points on the hyperbola satisfy f(x,y)=0

[low matrix]

oh so all the points will give absolute minima = 0

i see

[summer imp]

Well = 0

But yeah

[low matrix]

yuh mb

and i can also find the det of hessian matrix for (-2,3)

[summer imp]

Yes

[pearl pond]

@low matrix Has your question been resolved?

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[daring bay]

do i have to mutliply -2 with everythng here?

[minor cloud]

only one of the brackets.

[daring bay]

and for the rest i can just multiply -2b with 1/2b and 2 with -1/2?

[minor cloud]

you're missing a couple of multiplications there

remember, distributive property

or FOIL, if you've learnt of that

[daring bay]

hold up

[daring bay]

can i just do(a^2-b^2)

[minor cloud]

how does that help here

[proper nova]

this simplifies nicely im not going to lie

[minor cloud]

you don't have a + sign and you don't have the right numbers to do so

that being said I see another helper eager to take over, so in the interests of not flooding you with extra text I'll let them take over

[daring bay]

okay so how do i solve (-2b+2)(1/2b-1/2)

[proper nova]

can you see what you can factor in the second bracket?

[daring bay]

yeah

1/2

[proper nova]

cool

so factor that out

[daring bay]

1/2(b-1)? or

can i just replace it with one

[proper nova]

the 1/2 and -2 cancels out to -1

[violet knot]

what is infinity+1 bc me and my friend are arguing abt this

[minor cloud]

!occupied

[pearl pond]

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).

[proper nova]

!nosols

[pearl pond]

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.

[proper nova]

and for your question, it's undefined because infinity is not a number

[daring bay]

how is -2 relevant tho in this factoring

[violet knot]

Thank you

[left sequoia]

Yea sorry

[proper nova]

you had a -2 as a factor of your original question

[daring bay]

yes i already multiplied that wth my first bracket

[proper nova]

oh, i was talking about the original question

[left sequoia]

Dont multiply with first bracket , try to make the expression more simple by getting rid of ugly stuff like fractionw, multiply with second bracket

[daring bay]

so leave the -2 out for now?

[proper nova]

or you can factor 1/2 out as i said

would still give the same result

either way is fine

[daring bay]

but wait there gotta be a way to solve (-2b+2)(1/2b-1/2)

just by multiplying with each other?

or am i stupid

[rich charm]

Guys....

[proper nova]

technically that would give you the same result, you aren't stupid

the thing is that it would get messy pretty quick

so we tend to look for cleaner methods

[rich charm]

is it possible that x+x⅔(±√5^4.7) = 573^√56

[proper nova]

!occupied

[pearl pond]

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).

[daring bay]

oh okay

so then we have (-2b+2)1/2(b-1)

[proper nova]

for a clean method, don't multiply the -2 with the first bracket

[daring bay]

oh

okay so -2(b-1)1/2(b-1)

[proper nova]

and do notice that -2 * 1/2 = -1

[daring bay]

but where are we doing that even

idk wym by that

ohhh

[daring bay]

i got you now

but do i have to multiply the -2 with that? its after the bracket

[proper nova]

uhh no

your -2 got disappeared after multiplying with 1/2

[daring bay]

yes but you can do that? wait

[daring bay]

you can just multiply like this?

[proper nova]

yes

[daring bay]

damn ok i didnt know that

and then you have -1(b-1)(b-1)

and then you can do (a^2-b^2)

?

[proper nova]

no

no need for a^2 - b^2

do you see the (b-1)(b-1)?

[pearl pond]

@daring bay Has your question been resolved?

[daring bay]

yeah

[proper nova]

well

that's 2 same thing multiplied with each other

[daring bay]

i can just do b^2-1^2

[proper nova]

so what does that become?

[proper nova]

well no you can't

because that would be (b-1)(b+1)

[daring bay]

oh

[daring bay]

im not sure

[proper nova]

$a \cdot a = a^2$

[daring bay]

b^2+1

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

so $(b - 1)(b - 1) = \dots$

[proper nova]

no again

[jolly parrot]

1 divided by 0 equals Infinity

[daring bay]

yeah b^2 + 1

b * b = b^2

nah

[daring bay]

wait you do everything in the bracket times everything in the other bracket right

[proper nova]

in this case, imagine your $b - 1$ as $a$ in the case above

[jolly parrot]

1 divided by 0 equals Infinity

[pearl pond]

@daring bay Has your question been resolved?

[pearl pond]

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[crimson siren]

what do the stupid letters mean in math

[shell brook]

what are the "stupid letters" in question?

[cinder flower]

what stupid letters

[crimson siren]

x y and z

allat crap

[shell brook]

NOOO

boog

NOOO

[crimson siren]

math is numbers

[shell brook]

they can mean what you want them to mean

[crimson siren]

not positions

[shell brook]

usually they're variables

[crimson siren]

chicken?

like x chickens and y chickens

[shell brook]

YEAH!

[shell brook]

backflips

[shell brook]

slayla, you're typing a lot

please have mercy on him

[crimson siren]

reciting the whole bible 💔

einstein last words: me equals MC SCARED

[verbal whale]

!redir

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[verbal whale]

@crimson siren

[shell brook]

was this a troll help channel btw?

[crimson siren]

no i actually didnt know

💔

[versed mica]

have you considered a career in comedy?

[crimson siren]

no

[kindred plover]

good, don't

[shell brook]

slayla stopped typing.

[crimson siren]

good now i can play geometry dash

thanks for helping me nerds i will be back when i need to know more

[cinder flower]

when you are telling a story to someone you might name a man in the story bob just to give him a name to refer to. the listener won’t know who exactly it is, just that it’s a man. when you are ‘telling a story’ about a number you might name it x. it’s some number

[shell brook]

"nerds" is crazy

[shell brook]

dont know what you cooked but you cooked regardless

[pearl pond]

@crimson siren Has your question been resolved?

[grave mirage]

TPAB cover btw

K dot really fell off

[shell brook]

no self advert

[plush bramble]

<@&268886789983436800> spam

[shell brook]

wow mods work fast

good job mods

good to know that all of my messages are being moderated

[lilac jackal]

[shell brook]

your name reminds me i got hiragana to learn

[knotty schooner]

mods in this server r great. always happens in my help chats as well

[cinder flower]

they’re alright

[knotty schooner]

[shell brook]

are you saying that cuz you arent a mod

i will give up anything to be a mod

and bring revolutionary changes to this server

[cinder flower]

no i don’t want to be a mod

[knotty schooner]

[pearl pond]

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[arctic tinsel]

Q7. Prove Pascal's Identity with logical steps.

[arctic tinsel]

i searched it up on yt and google

but honestly

i cant understand anything

:(

[cinder flower]

i have an extremely simple proof for ts

[arctic tinsel]

go on

[steep saddle]

dont you just add the fractions together do some manipulation and it becomes nCk

can you write pascals identity at least

[arctic tinsel]

i can write it but idk how to use latex

[steep saddle]

then expand it into the definition

[steep saddle]

js do it on paper

[arctic tinsel]

ok

one sec

[cinder flower]

${n\choose k}$ is the number of size $k$ subsets of ${1,2,\ldots,n}$. a subset either includes $1$ or it does not. there are ${n-1\choose k}$ subsets that don't include $1$, and ${n-1\choose k-1}$ subsets that include $1$. hence $${n\choose k} = {n-1\choose k} + {n-1\choose k-1}$$

[jolly parrot]

slayla

[cinder flower]

qed

[arctic tinsel]

OH OKAY

that makes alot more sense

[cinder flower]

combinatorial proofs once again showing their power

[arctic tinsel]

thank you for helping i been stuck here for a minute

.close

[pearl pond]

Channel closed

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[grave mirage]

W proof

[cinder flower]

[pearl pond]

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[quick venture]

I just need these arguments clarified

Are they correct becuase to me clockwise is negative and anticlockwise is positive

[steep saddle]

you can always add 2 pi to make a negative angle positive

[quick venture]

I was just thinking for the last one in the fourth quadrant

Wouldn't it be -tan^-1(-3/3)

Because it's going clockwise

[steep saddle]

no negative sign

[quick venture]

I don't understand why

[steep saddle]

what is arctan(-1)

the range is -pi/2 to pi/2

[quick venture]

-45

[steep saddle]

ok so -45 degrees or -pi/4 radians

[quick venture]

Yes

[steep saddle]

that's clockwise 45 degrees

so it's already negative

why do you need another negative sign

[quick venture]

Hmm

Okay

Thanks

.close

[pearl pond]

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[pearl pond]

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[median shadow]

Is Q3 fine so far

[west sapphire]

@cinder flower what's your verdict on the handwriting here

[median shadow]

I know the writing is a bit ambiguous but

This current step feels so unelegant that my spidey math senses are ticking off

Something doesnt seem quite right

[plush bramble]

show how you got dy/dtheta and dx/dtheta

[median shadow]

i just tookt he derivative

the*

d/dxsinx = cosx

d/dxcosx = -sinx

[shell brook]

Good handwriting

[median shadow]

thanks

[steep saddle]

@ slayla

[shell brook]

I feel like we should focus at the question on hand

[median shadow]

hopefully ms. slaya approves of the handwiritng

[median shadow]

yeah

so yeah like is this ok

the step

then i just do a u sub

and call it there

[plush bramble]

what is this step

[median shadow]

i added the 1s

then

factored out the 2

took it out the sq root

[plush bramble]

oh that's a + 1

[median shadow]

thats about it

[plush bramble]

,tex .half angle

[jolly parrot]

riemann

[plush bramble]

try using the sin half angle identity now

[median shadow]

but how

wait

[plush bramble]

compare

and

[median shadow]

OHHH

thats just 2sinhalf

ill keep at it and see what i get

[cinder flower]

the curvy calligraphy is almost decent but the messiness really detracts from it. i think the messiness is partially caused by the pencil lead size being too big for the size of the writing, but also just some carelessness and poor strokes. like those equal signs are so careless. overall, mid

[shell brook]

It's good

[median shadow]

i was rushing lol

wait i have a better one

[shell brook]

Hey, you're writing is fine 🫂

Positive vibes all around

[median shadow]

@cinder flower

now what

[median shadow]

im not satisfied until slayla approves

[shell brook]

I'm gonna be the head of handwriting soon

[median shadow]

this changes things

[steep saddle]

i cant read any of that

[cinder flower]

you would still benefit from a smaller pencil lead i think. i might call it beautiful had it been written with that. but it's better than the other one

[steep saddle]

i do everything digitally

[cinder flower]

good calligraphy but not always the most legible

like this. what is the word on the left?

[median shadow]

maybe 0.5 will be better

[median shadow]

find

[cinder flower]

is it find?

[median shadow]

my F's are like that

my Ts are similar too

[cinder flower]

well it's not just the f

[median shadow]

its the same except without the little - at the middle

[cinder flower]

the i and n and d are also slurred

[median shadow]

oh well

wait i should focus on my question

[cinder flower]

maybe

[median shadow]

i got 8r

Wait one wait

This has to be it

[plush bramble]

looks right: https://proofwiki.org/wiki/Length_of_Arc_of_Cycloid

[median shadow]

thanks

ill try to memorize the half angle ones, though i dont like them

.close

[pearl pond]

Channel closed

Closed by @median shadow

Use .reopen if this was a mistake.

[plush bramble]

,tex .rocket trig

[jolly parrot]

riemann

[plush bramble]

some more for you to memorize

[median shadow]

i think the power reduction formulas should be added to the rocket trig as well honestly

hopefully they add that in

but thanks!

[pearl pond]

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[empty rover]

can somebody teach me this?

[past basin]

NH3 is neutral I’m pretty sure

It’s ammonia

So the Cu is what gives the whole thing its +2 charge

So if we take NH3 we know its charge is 0

[empty rover]

I thought NH3 has a charge of -1?

[past basin]

Does it?

Isn’t it NH4+

Then when u lose the H it’s just NH3

Based on my assumption of it being neutral tho, each H has an oxidation number of +1

So the N must have a number of -3

For NH3 to be 0

So my guess is A

Not a chemistry expert tho

[empty rover]

yeah the answer key says A

but I'm kinda confused how oxidation works here

[past basin]

For neutral molecules, the oxidation number is 0

And H has an oxidation number of +1

[plush bramble]

https://discord.gg/eexdsFw

[empty rover]

bro this server is fucking useless

[empty rover]

these charges comes from the periodic table right?

[past basin]

The +1 charge of H?

Yeah ig

[empty rover]

basically for all the rules of oxidation

[past basin]

H in general always has a +1 charge in these problems and O has a -2 charge

[empty rover]

like for O, halogens etc

[past basin]

There is some exception where H can have a -1 charge but I forgot what that is

[viscid shale]

Metal Hydrides

[past basin]

Oops I meant -1 oxidation number

Maybe it’s the same case tho

[viscid shale]

Yeah, i understood

[empty rover]

ok guys

eh can someone explain to me

this qn

I'm still quite confused

how we apply the rules of oxidation here

[past basin]

One rule of oxidation is that if the molecule is neutral, its net oxidation number is 0

NH3 is neutral

So it has an oxidation number of 0

[ivory basin]

And some things have a set oxidation number, check if your book has a table for this

[past basin]

Another rule is that H has an oxidation number of 1 expect in the case specified above

[ivory basin]

For example hydrogen is usually +1 except in metal hydrides

[past basin]

Yeah

[ivory basin]

Oxygen is usually -2 except in peroxide and shit

[past basin]

N+3*1=0

N=-3

[empty rover]

oh so like what other compounds are neutral?

[iron basin]

obvious ones

[ivory basin]

Most compounds

Ions are the ones that aren't zero

[past basin]

That’s not something u rlly need to memorize

NH3 is the main one to memorize is neutral ig

U don’t even rlly need to memorize is because u alr know NH4+

[empty rover]

my book gave me Na+ = +1 oxidation, Cu2+ = +2 oxidation state, Br- = -1 oxidation state and O2- = -2 oxidation state

[viscid shale]

You prob need to understand the idea of lewis acid-bases and what a coordination complex is

[past basin]

Coordination complex?

[viscid shale]

yeah, this is a metallic coordination complex

[past basin]

I dont believe I’ve learned about it yet

[viscid shale]

Cause bonds here are coordination bonds

[ivory basin]

What grade are you in @empty rover

[past basin]

U mostly just need to know some basic oxidation values and the few simple rules for these imo

And some basic algebra ig

[ivory basin]

From your username I'm guessing you're Indian. And if so, coordination complexes are grade 11-12 stuff

[viscid shale]

The problem is Nitrogen gets pretty messy because the oxidation values depend on the kind of compound its forming

plus minus 123, plus 45 iirc

[empty rover]

I'm in polytechnic

[empty rover]

yeah but I'm not from india

[past basin]

Uni?

[empty rover]

The N is 3- right from the periodic table?

[empty rover]

one level lower than Uni

[past basin]

It doesn’t always work like that tho

Which is why u can only trust a few oxidation values (H, O, and whatever else ur book gives u)

[past basin]

Interesting

Haven’t heard of it

Before

[empty rover]

ok so let me get this straight

as long as theres a neutral compound in the formula, the total oxidation state is 0?

[past basin]

The total oxidation state of that compound is 0

[empty rover]

so it makes the whole thing an oxidation state of 0 right?

[past basin]

For ur example for ex

[tardy reef]

Ligands of the complex in this particular case forms a coordination bond, and is neutral. This need not be a general case.

[viscid shale]

Imma try to show why NH3 can be neutral here and still bond to Cu

[past basin]

The whole molecule has an oxidation number of 2+ still

But NH3 has an oxidation number of 0

That’s what I meant by that compound

The neutral compound

[empty rover]

so the 2+ comes from Cu?

[past basin]

Yeah

[viscid shale]

NH3 has 5 valency electrons, with a oxidation value of -3 it can take 3 +1 Hydrogen atoms

And is left with a free pair of electrons

[empty rover]

so is it the case that since H has a charge of +1, we take 1 x 3?

[past basin]

Yeah

H always has an oxidation number of +1 except for metal hydrides

[viscid shale]

The fact that nitrogen is left with a free pair of electrons allows it to act as a "base"

Which means it can donate them

And for our case, that means that it can bond to the copper ion

Without actually increasing its own electron count

Since the NH3 remains neutral, the overall count remains at +2

Meanwhile the copper ion gets 8 extra electrons (with 2 free bonds still possible based on the original +2)

[empty rover]

where it gets the 8 extra electrons from tho?

[viscid shale]

The electrons the 4 molecules of ammonia "donated"

Since you get 2 from each

Still, all this implies you know that ammonia has a free electron pair

[pearl pond]

@empty rover Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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[dim parrot]

K is an odd integer greater than 1 then k^33-k is divisible by?

[dim parrot]

I'm new on discord

[signal mango]

dont worry

[dim parrot]

k(k^32-1)

k(k^16-)(k^16+1)

k(k^8-1)(k^16+1)(k^8+1)

[signal mango]

$$k^{33} - k = k(k^{32} - 1)$$Using the difference of squares repeatedly:$$k(k^{16} - 1)(k^{16} + 1)$$$$k(k^8 - 1)(k^8 + 1)(k^{16} + 1)$$$$k(k^4 - 1)(k^4 + 1)(k^8 + 1)(k^{16} + 1)$$$$k(k^2 - 1)(k^2 + 1)(k^4 + 1)(k^8 + 1)(k^{16} + 1)$$$$(k-1)k(k+1)(k^2 + 1)(k^4 + 1)(k^8 + 1)(k^{16} + 1)$$

[jolly parrot]

papaop

[dim parrot]

k(k^2-1)(k^2+1)(k^4+1)(k^8+1)(k^16+1)

[signal mango]

put $$ at the start and end

$$k(k^2-1)(k^2+1)(k^4+1)(k^8+1)(k^16+1)$$

[jolly parrot]

papaop

[signal mango]

like this @dim parrot

[dim parrot]

Yeah

Soooo

[versed mica]

{}

[signal mango]

you wnated the solution or every passage?

[dim parrot]

What will we do next?

Thought?

[signal mango]

we need even terms since k is odd

[dim parrot]

Why we need even terms?

I didn't get your logic

[signal mango]

You need those even terms—like $(k^2+1)$, $(k^4+1)$, etc.—because they appear when you factor the expression. Since $k$ is odd, every one of those terms is an even number, adding extra factors of 2 to the total.

[jolly parrot]

papaop

[signal mango]

you understand?

[dim parrot]

But why are we looking specifically even terms?

[signal mango]

Because those terms are the only way to find the highest common divisor.When you factor $k^{33} - k$, the algebra naturally produces $(k^2 + 1)$, $(k^4 + 1)$, and so on. Since $k$ is odd, every one of those terms is even (odd + 1 = even).Each of those "even terms" adds another factor of 2 to the result. Without counting them, you'd miss out on a huge chunk of the total divisor ($2^7 = 128$).

[dim parrot]

What is wrong if I look at odd terms?

[jolly parrot]

papaop

[signal mango]

Nothing is "wrong" with looking at the odd terms (like $k$ or $k^2+1$ if $k$ were even), but in this specific problem, there aren't any.Since $k$ is odd:$k$ is odd.$k \pm 1$ are both even.$k^2+1, k^4+1,$ etc., are all even (odd + 1 = even).

[jolly parrot]

papaop

[dim parrot]

I see

[signal mango]

is it clear?

[dim parrot]

Hmm

So it will be always divisible by 128

[signal mango]

exactly

[dim parrot]

Thanks

[signal mango]

no probs do .close

[dim parrot]

.close

[pearl pond]

Channel closed

Closed by @dim parrot

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[boreal slate]

hey i would like some help solving this integral

[boreal slate]

any (small) hint will be cool

[patent vessel]

$$ \int \dfrac{dx}{x^2 \sqrt{1-\frac{1}{x^2}}}$$

[jolly parrot]

robin.dabanc_

[patent vessel]

This?

[boreal slate]

yep

[patent vessel]

Ok got it

Can you simplify the root by clearing the denominator first?

[boreal slate]

..

[patent vessel]

Ok you have the integral on the extreme right