help-4

[versed granite]

this is the wrong channel for your question @midnight pier

[midnight pier]

He send me here

[wet granite]

😓

[versed granite]

use one of these

[cunning sky]

B thì cái 2/sqrt(x)+3 ông nhân với sqrt(x)-3 ý
xong r cái trên (đm quên tên r) thì có 2sqrt(x) - 6 - sqrt(x) + 5 là ra sqrt(x) - 1 và ykyk

với cả (a-b)(a+b) = a^2 + b^2 nhé

[wet granite]

😮 Vietnamese

[cunning sky]

câu c) thì ông đưa B = 1/(sqrt(x) - 3)
thì ta có A - B = căn x trên căn x trừ 3

[jovial edge]

2/9 mà vẫn học bài

chăm nhỉ

[wet granite]

wait how do I close the thing…

[jovial edge]

.close

[vale dock]

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[jovial edge]

like that

.reopen

[vale dock]

✅

[wet granite]

Thanks

.close

[vale dock]

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[sullen isle]

#1409222558007689328 message

[oak crane]

what is unknown

[sullen isle]

someone pls help

[west cloud]

what have you tried?

if you haven't drawn a diagram, do that rn

[vale dock]

@sullen isle Has your question been resolved?

[vale dock]

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[dire helm]

Do the letters look right?

[runic scroll]

are you allowed to write out the names of the Greek letters just like that?

[dire helm]

For now, yes, i don't have access to special characters on the typing platform I use

[runic scroll]

if you are, it might confuse readers as to why you did it for delta and theta, but not the omega, which I presume is an omega because of the angular frequency context.

not the w*

[dire helm]

It'll be fixed at the end, did I get the variables right

[runic scroll]

without checking the actual values, those variables look like the standard way to describe the linear acceleration, time, phase angle, and angular acceleration (w), force (f) and linear displacement (x) respectively.

if my interpretation matches yours, you might be good to go.

[dire helm]

It's the values I'm having trouble with

[tidal swift]

reminder that your initial question asked to check the letters

but if you want your values to be checked, maybe show your working too

[dire helm]

This is just interpretation of the word problem that colors my approach to the rest of the problem

[runic scroll]

then, where did a come from?

plus, I assume this is physics. you might want to include the units of each quantity here.

[dire helm]

A's acceleration
T's time
Wi is initial angular velocity
wf is final angular velocity
Delta theta's angula

[runic scroll]

I know a is linear acceleration, but where did the value come from? might want to show that.

,calc 637/12.5

[rocky lotus]

Result:

50.96

[runic scroll]

is this how you got a?

[dire helm]

yes

[runic scroll]

do you then know what this quantity actually represents?

you took the total angle of revolution and divided it by the amount of time taken.

what you found is not the angular acceleration or even linear acceleration.
what you found has units of radians per second, otherwise known as angular frequency/velocity.

[dire helm]

ok, good to know

so that's wi, I assume delta theta's a or did I get that part right

[vale dock]

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[dire helm]

Do the values look right?

[vale dock]

@dire helm Has your question been resolved?

[vale dock]

@dire helm Has your question been resolved?

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[wheat spire]

The Answer should be C I guess but my approach is giving me A

[wheat spire]

Help me please as this question is taking a lot of my study time

I have an exam in like 3 months

[hardy coral]

There is no answer C

Read the question carefully; it's not about how much actual product he is selling, it's about how much profit he's making

[wheat spire]

The question was really bugging me so I found an non-confusing way

Let's suppose the dealer is selling 10kg for 10$ but using counterfeit weight he sells 8kg for 10$ but before selling he adds 20% impurities taking his content to 12Kg now if he was selling 8Kg for 10$ he would sell 12 Kg for 15$. If he was honest he would have got 10$ but now he got 15$ which is 50% more

[rocky lotus]

Finerkit

[wheat spire]

And what do u mean by "There is no answer C" please Elaborate

[hardy coral]

The answers are 1, 2, 3, 4

[wheat spire]

That makes no sense lol

What do u think about his approach?

[hardy coral]

?? Look at the picture you sent

[wheat spire]

I already told what I did is incorrect

Read the follow up

[hardy coral]

I feel like I'm speaking to a deaf person

[wheat spire]

It's maybe I'm not understanding what u trying to communicate to me

[hardy coral]

Where is answer C?

[wheat spire]

Oh you were talking about that?

I thought you were talking about the answers

[hardy coral]

... these are the answers

[wheat spire]

No I mean...

I didn't expect you were correcting me about 1,2,3,4 to A,B,C,D confusion

I thought you were correcting me on my approach of answering the question

Hence the misunderstanding

[hardy coral]

I'm not correcting, I'm trying to understand what you mean by

The Answer should be C I guess but my approach is giving me A
which is the first thing you typed

[wheat spire]

Yeah bcuz I didn't focus on what's not relevant

My bad I guess

[hardy coral]

Seems pretty relevant to me; I still don't know which answer is supposed to be correct and which you came to

[wheat spire]

Thanks for nothing I guess?

[hardy coral]

I suppose you don't want my help then; feel free to ping helpers and good luck

[vale dock]

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[silent lichen]

hello i just started a pre-calculus class and I'm already confused lol. I've tried solving them and using chat gpt to help explain it but It just doesn't make sense, and on another math app it says the square root of 9a^2 is equal to 3a so now I'm even more confused.

how i go on about solving these two is that I find the square root of both 9 and a^2, which just leaves me with 3a.
and for the second one I understand slightly more I believe the rule i used for the first question doesn't apply when you're doing an addition problem under a square root? but I would still appreciate it if someone explained why neither of these are correct.

[hardy coral]

the rule i used for the first question doesn't apply when you're doing an addition problem under a square root?
Yeah that'd be right

[keen tundra]

note that (a+3)^2 = a^2 + 6a + 9 which is not equal to a^2 + 9

[manic igloo]

sqrt(9a^2) = 3*|a|

[silent lichen]

so, was the question with sqrt(9a^2) incorrect because they didnt put 3(|a|)?

[manic igloo]

Yes, for negative values of a, sqrt(9a^2) = -3a.

[silent lichen]

ohh i see..

[silent lichen]

i dont really understand lol is this for the first or second question

[keen tundra]

this for the one with addition

this shows why they are not equal

[hardy coral]

It means sqrt(a^2 + 6a + 9) = a + 3 (if a >= -3)

[keen tundra]

if they were equal if we take the square of both sides and get (a+3)^2 = a^2+9 which as I said is not true

[silent lichen]

sorry im still here im just trying to think lol

[hardy coral]

• sqrt(a^2 + 6a + 9) = |a + 3|
• sqrt(a^2 + 9) =/= a + 3

sqrt(a^2 + 9) can't be simplified in the domain of real numbers

[cunning sky]

this

[silent lichen]

okay i think i get it what you guys are doing is you reversed the problem to check? by doing the square root of a^2+9 you get a+3, but I get lost when you guys put the 6a in there, where is that from?

[cunning sky]

uhhh

[silent lichen]

i get it now that it simply doesn't work but I just wanna understand how you guys like mathematically proved it

[cunning sky]

(a+b)^2 = a^2 + 2ab + b^2

[silent lichen]

oh this is some formula i need to memorize

[hardy coral]

$\sqrt{pq} = \sqrt{p} \cdot \sqrt{q}$

[rocky lotus]

Nel

[hardy coral]

This is pretty much the only thing you can do with square roots

If p is a square, you can then simplify

$\sqrt{x^2} = |x|$

[rocky lotus]

Nel

[hardy coral]

So for your original question, $\sqrt{9a^2} = \sqrt{9} \cdot \sqrt{a^2} = |3| \cdot |a| = |3a|$

Of course if you realize that 9a^2 is the square of 3a, you don't need to split and recombine

[rocky lotus]

Nel

[hardy coral]

(and |3| is just 3, so you can also write 3|a| if you want)

$\sqrt{p + q} \neq \sqrt{p} + \sqrt{q}$, so it doesn't work for $\sqrt{a^2 + 9}$

[rocky lotus]

Nel

[silent lichen]

okay.. i think i understand it now, would I be safe to just leave the question : sqrt a^2 + 9 to be unsolvable because addition just doesn't work like that under square roots, and you'd only be able to simplify it any further if it were to be sqrt a^2 (9) ?

[magic orbit]

have you heard about imaginary numbers

[silent lichen]

yes

[magic orbit]

so you know that i^2=-1

[silent lichen]

yes

[magic orbit]

can you apply that to simplify a^2 + 9

[keen tundra]

it is still not going to be a square

if you mean a^2 + 9 = (a-3i)(a+3i)

[magic orbit]

youre right my bad

[silent lichen]

i think i'm good for now thank you guys for the help you are all some smart fellers

[vale dock]

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[midnight pier]

Can someone check what I did wrong for these two problems?

[midnight pier]

55 and 57

[vale dock]

@midnight pier Has your question been resolved?

[stark wedge]

immediate: you've still got "5 looks like S" disease

[midnight pier]

Damnit I usually correct it before I post now after what you said

I forgot again

Sorry

[stark wedge]

uhhh ok hold on

your 9.6 km figure looks correct to me and it's what i get

the 11.4 should be 11.5 though

uhhh one sec

i think maybe the book's answer key has a typo here -- or maybe the problem statement

bc the answers in the book imply a distance of 12 km

rather than 15 as was written

in 57 you fell victim to crude rounding

the actual southward distance is like 1.35

[midnight pier]

the answer keys drive me a lil insane this is maybe the 4th error in the book

[stark wedge]

it would have been better to figure out the horizontal dist as 6.5 sin(78)

and not go thru tan at all

[midnight pier]

Cause my answer for the side I put to use in tan won’t always be for certain?

[midnight pier]

Does that mean the text book made another mistake with the 1.4?

[stark wedge]

no it's fine

your approach made you have a rounding error, is what im saying

[midnight pier]

Ohh okay

Thank you, Ann!

.close

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[vale dock]

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[obsidian raft]

Need help

[neat hearth]

Okay send it

[obsidian raft]

I need to figured out how old is tung tung tung tung sahur

[neat hearth]

Bro

[tidal terrace]

.close

[vale dock]

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[obsidian raft]

How do i know

NOO

[neat hearth]

Youtube shorts other way

[obsidian raft]

I asking fr 😭

[wraith heart]

<@&268886789983436800> troll i guess

[neat hearth]

Sure gng

[stark wedge]

not a math question.

[obsidian raft]

Nah its bro im rlly asking

[obsidian raft]

Oh mb

[ivory valley]

technically...

[vale dock]

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[opaque gazelle]

can anyone tell me why it isn't 74/75?

[opaque gazelle]

I did the math and deleted it

I marked the shorter segment of the hypotenuse x

and the diagonal of square S as y

and then I solved, found y as 0.4, found x as 1.8, and then I solved and found the ratio to be 74/75

[distant galleon]

Did you incorrectly assume that these three points are collinear

[opaque gazelle]

bruh

wait

I see now

that is so bs joijfioerjfiweg89jeworij

thank you

.close

[vale dock]

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[vale dock]

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[glass kelp]

Geometric sum?

what’s the issue

[cobalt crow]

I need help

[glass kelp]

Which question and progress?

;3

[cobalt crow]

i)

progress, null

[glass kelp]

Have u dealt with geometric sum

B4

[cobalt crow]

like

long time ago, in intro to precalc

[glass kelp]

Right

So

Let me walk u through the proof again

[ashen prawn]

1+2+2 ^2+...+2^n

[glass kelp]

Consider ${\sum_{i=0}^n ar^i}$ and ${r\sum_{i=0}^{n} ar^{i}}$. Then,
\begin{align*}
r\sum_{i=0}^n ar^i - \sum_{i=0}^{n} ar^i &= (ar + ar^2 + \dots + ar^{n+1}) - (a + ar + \dots + ar^n)\
&= ar^{n+1} - a = a(r^{n+1} - 1)\
\sum_{i=0}^n ar^{i} (r-1) &= a(r^{n+1} - 1)\
\sum_{i=0}^n ar^i &= \boxed{\frac{a(r^{n+1} - 1)}{r-1}}
\end{align*}

Bruh

[ashen prawn]

?

[cobalt crow]

can someone help

[ashen prawn]

.

[glass kelp]

@cobalt crow does this look familiar

[cobalt crow]

someone help me step by step

[ashen prawn]

isnt it this

[glass kelp]

Exactly

[rocky lotus]

k

[cobalt crow]

how did this part happen?

ah there was a typo before

now its fixed

[glass kelp]

But

U get it, right?

[cobalt crow]

kind a

[ashen prawn]

u shd revise it and then solve ig

[cobalt crow]

this is the revision? wtf u mean

[ashen prawn]

im saying that arent u solving gp after long time?

[cobalt crow]

what about it ?

[glass kelp]

Compare it to question 1

Can u see what a and r are

[ashen prawn]

u mustve forgotten some aspects hence to brush up u go yourself or take help from k here to freshen the concept

[cobalt crow]

@glass kelp

a = 1

r = 2

,, \sum_{i=0}^{n} 2^i = \frac{2^{n+1} - 1}{2 - 1} = 2^{n+1} - 1

[rocky lotus]

Renato

[cobalt crow]

@glass kelp @glass kelp @glass kelp

[glass kelp]

Ye

,w sum from i = 0 to n 2^i

[rocky lotus]

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[cobalt crow]

help please dude

@glass kelp

[glass kelp]

Wdym

U did the first question already

2^(n+1) - 1 is the answer

[cobalt crow]

i need help with the other ones aswell

[glass kelp]

It’s just this

[glass kelp]

^

Over and over

[cobalt crow]

i find it hard

[ashen prawn]

2nd one?

[glass kelp]

Think for a sec

[cobalt crow]

dude is not easy

we are assuming |r| <= 1

thats the assumption

[cobalt crow]

q ∈ R dude

@glass kelp

you here?

[glass kelp]

Fair point

[ashen prawn]

here a=q and r=q

[glass kelp]

If ${|q| \geq 1}$, then the series diverges

[rocky lotus]

k

[cobalt crow]

obviouslt dude

you guys are handwaving , shit is not easy

[glass kelp]

What handwaving have i done

[cobalt crow]

q in R my dude

[cobalt crow]

you said 2) is plug and chug

[glass kelp]

Oh nvm

It’s not an ‘infinite sum’

So the formula still holds :/

[cobalt crow]

can you elaborate

[glass kelp]

We are talking abt “finite sum”

[glass kelp]

we talk about |r| when we want to know whether the infinite sum converges or diverges

[cobalt crow]

the sigma in 1) is upto n, n is not infinite

[glass kelp]

However we are doing finite sum rn

[glass kelp]

However the 2)?

[cobalt crow]

ohh

[glass kelp]

Similar thing

[cobalt crow]

since all of them is finite sums

all of them converge

we dont care about the radius

[glass kelp]

Ye

So we can js apply the formula :/

[cobalt crow]

plug and chug, mb

[ashen prawn]

,w sum from i=0 to n x^i

[rocky lotus]

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[cobalt crow]

dude

no wolfram

[ashen prawn]

im js checking

[cobalt crow]

[ashen prawn]

but can u explain hows it diverging?

[cobalt crow]

its not diverging

because its an finite sum

finite sums always converge

dont worry about it. this is not analysis class

[glass kelp]

It diverges if it is an infinite sum

if |r| >= 1

[ashen prawn]

i knew somthing was wrong with what k was saying

earlier

[glass kelp]

Sure

Let’s js continue

[ashen prawn]

yea got it thx

[cobalt crow]

we only care about the radius if its infinite

?

[glass kelp]

Yup

[ashen prawn]

math wasnt mathing

[glass kelp]

Now back to 2)

[cobalt crow]

yep

[ashen prawn]

radius?

[cobalt crow]

r

[ashen prawn]

thts the common ratio

[glass kelp]

Cool

So

What would the answer be for ii)

[ashen prawn]

is this the ans for 2 if x is replaced by q

[cobalt crow]

a = 1

r = q

[glass kelp]

That is for when i=0

Think abt how would the formula change for i=1 instead of i=0

[cobalt crow]

,, \sum_{i = 0} = \sum_{i = 1} - a_0

[ashen prawn]

@glass kelp

[cobalt crow]

just subtract 1

wait, I might have made an oopsie

[glass kelp]

I=1 not i=0

[ashen prawn]

mb then

[glass kelp]

Ok

[ashen prawn]

,w sum from i=1 to n x^i

[rocky lotus]

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[glass kelp]

Mhm

[rocky lotus]

Renato

[ashen prawn]

yea now it fits

[cobalt crow]

dude dont use wolfram wtf

dont be a cheater

[ashen prawn]

i dont feelcomfortable with latex

im a new learner

[cobalt crow]

,, \sum_{i = 0} = a_0 + \sum_{i = 1}

[glass kelp]

What

[rocky lotus]

Renato

[glass kelp]

Cool

[cobalt crow]

this!!

[glass kelp]

And we want ${\sum_{i=1}}$

[rocky lotus]

k

[glass kelp]

So

Isolate that

[cobalt crow]

,, \sum_{i = 0} - a_0 = \sum_{i = 1}

[rocky lotus]

Renato

[cobalt crow]

and a_0 = q^0 = 1

[glass kelp]

Mhm

[cobalt crow]

this is why i said just subtract 1

yeah!!

[glass kelp]

Cool

[ashen prawn]

tru

[cobalt crow]

a = 1 , r = q

,, \frac{q^{n+1} - 1}{q - 1} - 1

[rocky lotus]

Renato

[glass kelp]

Good

[cobalt crow]

3 and 4 is getting icky

but should be the same thing but to the power of 2

XD

[glass kelp]

So what’s a and r

[cobalt crow]

,, \left(\frac{q^{n+1} - 1}{q - 1} - 1\right)^2 = \sum_{i=0}^{n} q^{2i}

[rocky lotus]

Renato

[glass kelp]

Slow down

[cobalt crow]

same as in 2

[glass kelp]

[ \sum_{i=0}^n (q^2)^i]

[rocky lotus]

k

[cobalt crow]

?

[glass kelp]

Consider what a and r are again

[cobalt crow]

but dude

well, ok

r = q^2

[glass kelp]

[ \sum_{i=0} (a)^2 \neq \left(\sum_{i=0} a\right)^2 ]

[cobalt crow]

a = 1

[rocky lotus]

k

[glass kelp]

Yup

[cobalt crow]

ok, fair enough 😂

,, \frac{q^{2n + 2} - 1}{q^2 -1}

well we need to subtract a_0

a_0 = q^0 = 1

[glass kelp]

What abt that -1 at the end

[cobalt crow]

oh start index is 0

im stupid

[rocky lotus]

Renato

[cobalt crow]

this ^

iv) is fucked up dude

any ideas?

ohh I see

,, \sum_{i = 0}^n + \sum_{i = n}^{2n} = \sum_{i =0}^{2n}

[glass kelp]

=?

[rocky lotus]

Renato

[cobalt crow]

@glass kelp @glass kelp

[glass kelp]

Yup

That’s the idea

[cobalt crow]

but...

how do I execute it?

[glass kelp]

Try isolating ${\sum_{i=n}^{2n}}$

[rocky lotus]

k

[cobalt crow]

,, \sum_{I = n}^{2n} = \sum_{i=0}^{2n} - \sum_{i=0}^n

[rocky lotus]

Renato

[cobalt crow]

we dont know sigma 0 to 2n

[glass kelp]

Really

[cobalt crow]

care to elaborate

?

[glass kelp]

There is nothing unique with n here

It can be any number

[cobalt crow]

oh ok i see

ok we got this

,, \sum_{i=0}^{2n} q^i = \sum_{i=0}^{2n} q^i - \sum_{i=0}^n q^i \ \sum_{i=0}^{2n} q^i = \frac{q^{2n+1} - 1}{q - 1} - \frac{q^{n+1} - 1}{q - 1}

@glass kelp ?

you here? dude

[glass kelp]

Why is it 2n+2

[cobalt crow]

2n + 1

😭

[rocky lotus]

Renato

[glass kelp]

Simplifying would be nice

Oh wait

Uhh

[cobalt crow]

,, \sum_{i=n}^{2n} q^i = \sum_{i=0}^{2n} q^i - \sum_{i=0}^{n-1} q^i \ \sum_{i=n}^{2n} q^i = \frac{q^{2n+1} - 1}{q - 1} - \frac{q^{n} - 1}{q - 1} \ \sum_{i=n}^{2n} q^i = \frac{q^{2n+1} - q^{n}}{q-1}

[glass kelp]

[ \sum_{i=0}^{2n} q^i = \sum_{i=0}^{2n} q^i - \sum_{i=0}^{n-1} q^i]

[rocky lotus]

k

[glass kelp]

Mb

I forgot that subtracting the nth term would start it from n+1 not n

Sorry

Regardless, the idea is correct

So it would simplify to

[ \frac{q^{2n+1} - q^n}{q-1} = \frac{q^n(q^{n+1} - 1)}{q-1}]

[rocky lotus]

k

[glass kelp]

N-1 not n

Sorry

[cobalt crow]

yes because we are overcounting the last one twice

in here

[glass kelp]

Yes

[cobalt crow]

we are overcounting

[glass kelp]

Mb, I didn’t notice it

[cobalt crow]

no worries. good catch

[glass kelp]

Did too many integrals 🤦‍♂️

[cobalt crow]

?

[glass kelp]

Anyhow

Simply switch q^{n+1} to q^n

Cuz n-1+1 = n

[cobalt crow]

was this shit easy or no?

[glass kelp]

Little typo

But yes

[cobalt crow]

what?

[glass kelp]

The final answer is correct

[glass kelp]

It’s from i=n

Not i=0

Typo problem

[rocky lotus]

Renato

[glass kelp]

https://tenor.com/view/me-atrapaste-es-cine-its-cinema-cinema-esto-es-cine-gif-12869046600151364058

[cobalt crow]

Scorsese cinema

gif

[cobalt crow]

ok same shit

just factorized

[glass kelp]

Exactly

So we did it right

[cobalt crow]

I appreciate it hard, K

.close

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[cobalt crow]

thanks ^^

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[somber cloud]

im pretty sure my study guide is wrong, but i have no clue how to prove it (just started calculus)

[short solar]

present your case , what do you think vs what was given in guide

[somber cloud]

i intuit that e^(x->infinity) would increase quicker than (x->infinity^(k, not not going to infinity))

but i guess the easiest way to prove/disprove is just plugging in the numbers

[short solar]

is it tho?

are you aware of lhopita's rule

[somber cloud]

im not there yet but thats what chatgpt gave me to prove it goes to 0

[short solar]

also why is the guide wrong

[short solar]

ah ic , tho its nice you are keen to learn , its suggested not to use ai sources as it hinders with learning and potential missinformation

did you write this?

[somber cloud]

i agree in principle but the convenience is oh so delightful

[somber cloud]

thats what was already written

[short solar]

oh nah that looks wrong then

i thought you did that

[somber cloud]

lol yeah

[short solar]

it is not printed so ig whoever had the guide before you fumbled

[somber cloud]

perhaps

both are the answers, i didnt write either

[short solar]

that is correct , howver i rn cant think of a valid written proof except for lopital to help your case

[somber cloud]

thank u

imma mark it resolved since im pretty sure ill get a better grasp of whats happening once ive learned a little bit more of some other stuff

.solved

[vale dock]

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[eager umbra]

can someone js give me the answers and go slow w it

im dumb asf

😭

[stark wedge]

we won't give you answers, sorry!

[verbal badger]

cant give answers but if you show your attempt we can help

[eager umbra]

well can i tell u what i got

[verbal badger]

yes, and tell us how you got it

[eager umbra]

idek what im doing but for a im gussing g(x) is js y so so

g(-2) = 2, g(0) = -2, g(2)=1

idk abt g(3)

[verbal badger]

yeah pretty much, g(x) is the function, so it's like interpreting it as y

i think you can just eyeball it no?

[eager umbra]

i dont get g(3)

is it 0

[verbal badger]

why would it be 0

[eager umbra]

idk

[verbal badger]

how did you get g(-2)?

[eager umbra]

by looking at where the y is

is it 2.5 then

[verbal badger]

so why dont you do that for g(3)?

[eager umbra]

the seperate graphs is confusing me

[verbal badger]

It's the same function (both are g(x), or y, same thing). It just has a discontinuity, or "jump"

But both chunks of the curve make up the same function regardless

[eager umbra]

oh

so its 2.5?

[verbal badger]

yeah

[eager umbra]

okay uh

for b i think its x=-4

right

[verbal badger]

yes thats correct. but why isn't it also x = 2?

[eager umbra]

um

[verbal badger]

(it isn't x = 2, just confirming you know why it isnt)

[eager umbra]

is it bc its open

[verbal badger]

yeah

[eager umbra]

whats the diff between the open one and closed

yo my grammar is so bad rn

sorry

[verbal badger]

the closed point is the literal value of the function at that point, while the open point explicitly indicated that it isn't that value

[eager umbra]

ohh

[verbal badger]

so at x = 2, the closed point is at y = 1 while theres an open point at y = 3 and onwards. This says that "at x = 2, y = 1, but when x is slightly above 2, y becomes very close to, but not equal to, 3"

[eager umbra]

i see

okay for c

its asking for which points is greater than 3

none is greater than 3 tho

[verbal badger]

less than

[eager umbra]

oh

OH

i looked at it wrong

[verbal badger]

and its talking about intervals, not points

[eager umbra]

less than or equal too

[verbal badger]

yeah, kinda a trick question

[eager umbra]

uhh

is it -4 and 2

i dont get this one

[verbal badger]

Well does y ever become more than 3?

[eager umbra]

no

[verbal badger]

so the interval where y is less than or equal to 3 is...

[eager umbra]

is it not (-4,2)

or something

[verbal badger]

why not (-4, 4)?

y is still less than 3 on the chunk between 2 and 4

[eager umbra]

oh

i forgot abt the other one

ty

is the domain also that then

im assuming range would be (-2,3)

i dont get e

at all

😭

is it (-2,1)

[woeful rover]

e asks where the graph is going up

[eager umbra]

oh

uh

(0,2)?

[woeful rover]

Yep that's correct

[eager umbra]

okay tysm

[woeful rover]

What if it asked for the interval where the graph is decreasing?

[eager umbra]

(-4,0)

do i do the other one too

[woeful rover]

Yea

[eager umbra]

the opened one

(2,4)

[woeful rover]

Ye

[eager umbra]

is there supposed to be anything between it

[woeful rover]

Now how do we join those 2 intervals together?

[eager umbra]

like for the answer

[eager umbra]

is it that

U

thing

[woeful rover]

Ye

[eager umbra]

ohh

okay

normally i understand it

its js that the separate ones confuses me

ill js ignore that

[woeful rover]

Ye the more you do these problems, the more you'll get used to it and understand it better :)

[eager umbra]

tysm

[woeful rover]

Glad we helped out

.close

[vale dock]

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[smoky grail]

There are five persons belonging to group A and five persons belonging to group B in how many ways can these persons be seated on a circular table such that no two persons belonging to the same group are seated together?

[lost marlin]

Assuming all the people of each group are the identical

Then how many ways are there ?

[smoky grail]

9! * 9

[lost marlin]

How did you get that

[smoky grail]

I've learned that in circular arrangement one object has to remain fixed and the rest of the process is same as linear arrangement so now we have nine persons and nine seats and there are 9! ways to seat them but the object to be fixed can also be chosen in 9 different ways

[lost marlin]

For now let us assume all people of group A to be identical and Group B to be identical

[lost marlin]

This doesn't work in this case

As it will also count group A people sitting together

[lost marlin]

@smoky grail if this is the case then how many arrangements are there

[smoky grail]

wait before moving further is 9! * 9 correct for the case when all people are identical

[lost marlin]

If people of each group are identical this won't be the case

[smoky grail]

let's call them objects instead of people

if all objects are identical there is 1 way

if all are distinct there are 9! * 9 ways

[lost marlin]

It should be just 9!

Since the first person doesn't matter

[smoky grail]

ooh okay i got it

moving from the distinct case know if we assume that objects in group B were identical then it should be 9! / 5! ways

[smoky grail]

and if objects in group A were to be identical too then it would become 9!/(4! * 5!) ways

[lost marlin]

Oh wait you are considering all cases

I thought the one with the constraints given

[smoky grail]

but was it obvious to you why i placed 5! here instead of 4!

[lost marlin]

?

[lost marlin]

I still don't get this

We do have to keep in mind that they are alternating

But this doesn't ensure that

[smoky grail]

case 1: Alternating arrangement assuming all objects in group A and group B are distinct among themselves 5! * 4!

[smoky grail]

this was my question is it correct

case 2: Alternating arrangement assuming all objects in group A and group B are identical among themselves 1 way

[lost marlin]

Yes it is correct

[smoky grail]

thanks i could have asked ai but i don't believe in it's answers

[lost marlin]

Wait

[lost marlin]

Can you explain why you did this

[lost marlin]

And how you got this from the other expression you told

[smoky grail]

There are 5! different ways in which five distinct objects belonging to group B were arranged among themselves i am just taking away that

[lost marlin]

Alright but how did you use this to get the answer

[lost marlin]

Oh wait I get it you applied this

[lost marlin]

To this

And got 1

[smoky grail]

Only difference is that this is not ensuring alternation

maybe but i'm not so confident about it

[lost marlin]

Instead we can work in the opposite direction

We know that if they are alternating

Then there is only 1 arrangement when they are identical

So we multiply by 4! For A and 5! For B

[lost marlin]

And get this

[smoky grail]

yeah and i was also saying that was it obvious to you that we are using 4! for A and 5! for B

[lost marlin]

Yes

[smoky grail]

it can also go the other way round ? Right

[lost marlin]

Yes if we take B initially

[smoky grail]

I was actually trying to solve this question A Senate committee has $5$ Democrats and $5$ Republicans. In how many ways can they sit around a circular table if each member sits next to two members of the other party? (Two seatings are the same if one is a rotation of the other.)

[lost marlin]

A way to circumvent the 5! and 4! Would be to consider the identical chains as seperate

Let D represent democrats and R as republicans

DRDRDRDRDR ,
RDRDRDRDRD are identical cases but we treat them differently for now
The possible linear permutations will be 5!5! + 5!5!

Now 10 permutations are resulted from rotations

So 5!4!

I am not sure if this is helpful or not

Since I am assuming you wanted some reason as to why the asymmetry existed

[vale dock]

@smoky grail Has your question been resolved?

[smoky grail]

it's confusing but i doubt the possible linear permutations for 5 D and 5 R are just 5!5!

why are you adding another 5!5!

[lost marlin]

Since one of the rotations begins with R and one with D

[smoky grail]

for linear arrangement it think when we say 5!5! it covers all the cases including the one in which D is seated at seat 1 and the one in which R is seated at seat 1

[lost marlin]

From 5! * 4! We can see some asymmetry which depends on which group we take first

But that is because we are fixing the first person to sit to be a D

Instead if we consider both cases together

We get this

@smoky grail i forgot to ask this but where exactly are you facing the problem in the question

[smoky grail]

   b. 2 4 6 8 10
case 1: D's being seated at a and arranged in 5! different ways * B's 5! different ways at b
case 2: B's being seated at a and arranged in 5! different ways * D's 5! different ways at b

[smoky grail]

my original questions is answered pretty much at this point

at this point

i was confused about this addition (5!5! + 5!5!) which i tried to did in circular arrangement too

[smoky grail]

but in circular arrangement a and b are similar

[lost marlin]

Each rotation is identical right

So if we somehow find the total permutations we can divide it by 10

But if we start our permutations with DRDRDRDRDR

We will only cover half of the cases

So we consider RDRDRDRDRD too

But there is another way to see this

[smoky grail]

and this only applies to linear arrangement ? right

[lost marlin]

Let us say there is only one arrangement DRDRDRDRDR now we want to consider how many of these arrangements will be identical when considering circular permutations

[lost marlin]

Yes I was just trying to explain why the 4!5! Comes

[vale dock]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lost marlin]

.reopen

Oh wait I can't do that

[vale dock]

✅

[smoky grail]

okay i understood it

[lost marlin]

If we just consider DRDRDRDRDR the linear permutations are 5!5!

But the identical circular rotations are only 5 instead of 10

So 5!5!/5

We cannot rotate this to all places

So D must go to the third place

Reducing possible rotations

[smoky grail]

We are ensuring alteration

[lost marlin]

Yes

Also while rotating this arrangement

We only have 5 options

I drew the diagram wrong but I hope the idea is clear

[smoky grail]

it's clear pretty much

it's just that i need spend time on solving some problems

without any help and it will fit in

thanks i did understood how circular arrangements differ from linear ones especially when we have constraints like alteration

.close

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[torn plank]

Does anyone know why I need to sub 2 guesses values of x to nSolve? Calc skills

[torn plank]

ion get the difference between nSolve and Polyroots

[tawny moon]

i think the calculator internally uses a numerical approximation method like Newton's method to find roots

these methods start with one guess and produce only one root per guess

[deft canyon]

yes It takes Newtons method

[tawny moon]

so if there's more than one solution, one guess will not reveal the other ones

[deft canyon]

<@&286206848099549185>

[tawny moon]

also ties in to the advice there - you should probably determine the exact number of solutions first, then estimate guesses

[tawny moon]

why are you pinging helpers

[deft canyon]

need help

[tawny moon]

!occupied

[vale dock]

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).

[tawny moon]

please move to another channel, and observe the 15-minute rule before pinging helpers

[torn plank]

Back

Yeah sorry, was letting the channel marinate with answers

[tawny moon]

np

[torn plank]

How do I know when to use nSolve or polyRoots

[deft canyon]

polyRoots(<polynomial>, <variable>)

[tawny moon]

look at the question

[deft canyon]

eg polyRoots(x^3 - 4*x + 2, x)

[tawny moon]

don't think OP asked how to use polyRoots

[deft canyon]

someone asked

[copper stump]

nSolve would give you the first solution value that you find

[deft canyon]

he

nSolve(<equation>, <variable>, <guess or interval>)

[tawny moon]

OP asked for how to know when to use either function, not how to use either function

[copper stump]

but the upside is that nSolve would get you a solution for all kinds of functions (not restricted to polynomials)

so, if you got a non-polynomial function, you should go for nSolve

[torn plank]

Ah, so it will give the closest answer to the approximation okay

[deft canyon]

ye

[torn plank]

So a logarithmic equation is best in nsolve

[green urchin]

dawg

[tawny moon]

wot

[green urchin]

wrong ping

sorry

[green urchin]

dawg

[green urchin]

discord tweaks alot

[deft canyon]

]ye use nsolve

[copper stump]

,rccw

[torn plank]

Question 65

part c

[weary pilot]

fucking bacterial martians

[torn plank]

I got a negative answer for time

[weary pilot]

idt theres a part c

[torn plank]

b

My bad

[deft canyon]

bro theres no pc

[torn plank]

You can tell im failing English lol 😂

[deft canyon]

wht question u nd hp

[deft canyon]

is it the time one?

wheres the question

[torn plank]

yeah

[copper stump]

you are correct

the temperature is in fact increasing with time

you should have noticed that in the part a.ii

[torn plank]

oopsie daisy

[deft canyon]

ok

math is so fun

[torn plank]

I thought t is greater or equal to zero but lowkey forgot it was increasing

Thank you guys

[deft canyon]

are u done

[torn plank]

Oh one more question

Is there a way of doing this without ln? (Haven’t done natural log yet)

[deft canyon]

Wht question is it cant see clear

[torn plank]

or just plug into nsolve

[deft canyon]

Oh ye just Plug nsolve

[torn plank]

Okay that’s all thanks

🙏

.close

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[celest geyser]

Can anyone help me with this please?

[woven cedar]

!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

[tidal turtle]

umm

in 1st

when I put 1, its 1

nd when I put 2 its 13

so the value must be between 1 nd 2

[woven cedar]

!nosols

[vale dock]

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.

[tidal turtle]

?

[vale dock]

@celest geyser Has your question been resolved?

[midnight pier]

I have a question

It's a full exercise

Can you please help me

[stark wedge]

!occupied

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[celest geyser]

Hiii tysm for ur response but im sorry I still don’t understand (this is my second class of add maths😭

[midnight pier]

Ok here are the questions

[tawny moon]

!occupied sir

[vale dock]

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).

[tawny moon]

please move to #help-47

[vale dock]

@celest geyser Has your question been resolved?

[vale dock]

@celest geyser Has your question been resolved?

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[torn sequoia]

I need help guys with law of exponentss

(2x^-1y2 / 3x^2y^-3)^2

I got:

4/9x2y2

[stark wedge]

ok just to confirm

original problem: $\left( \frac{2x^{-1}y^2}{3x^2y^{-3}}\right)^2$ \ \ your answer: $\frac{4}{9} x^2 y^2$

[rocky lotus]

Ann

[stark wedge]

did i read you correctly? @torn sequoia

[torn sequoia]

Wait let me show a picture

We got same answer

But the variables are outside

[stark wedge]

i didn't claim any answer

[torn sequoia]

Oh

[stark wedge]

i was simply trying to read yours and confirm

whether i understood you correctly

[torn sequoia]

The texit and I got the same answer

[stark wedge]

TeXit is just a bot that writes math pretty

it didn't do any of the math it just displayed what i wrote

nothing else

anyway! your answer is wrong.

show your work for how you got it.

[torn sequoia]

Oh

[stark wedge]

i think i have an idea of where you messed up, but i want to see what you did & how so that i can tell you how to fix it.

[torn sequoia]

We just started on law of exponents on 9th grade

[stark wedge]

ok right..

so you know $\frac{y^m}{y^n} = y^{m-n}$, yes?

[rocky lotus]

Ann

[torn sequoia]

Ye

[stark wedge]

yeah so

what's 4 - (-6)

(it isn't -2)

[torn sequoia]

10

So howd texit got the same answer

[tidal swift]

texit is not providing any answers

[stark wedge]

texit is not providing any answers

texit is a bot and it is copying the things i'm writing, but made pretty

[tidal swift]

$how do you think texit is providing any answers if i can use it for shit like this LMAO$

[rocky lotus]

Seia

[torn sequoia]

Im not really that active on this discord server so i dont know ANYTHING about texit

[stark wedge]

ok then thats your lesson 0

texit's job is to format math notation using a markup language called LaTeX

[torn sequoia]

Whats the right answer then?

[stark wedge]

i was trying to guide you to it

[torn sequoia]

I just need to check the answer

[stark wedge]

starting with fixing the y's

y^4 / y^(-6) = y^(4-(-6)) and i asked you to work out yourself what 4 - (-6) is

[torn sequoia]

10

[stark wedge]

right

so that should be the exponent on the y.

[torn sequoia]

4y^10 / 9x^6??

[stark wedge]

use ^ for exponents please.

also the 3 on the bottom should be 9

(and on your paper it is)

[torn sequoia]

Yeye typo

[torn sequoia]

So is this right?

[stark wedge]

yes now it is

[torn sequoia]

Oh thanks

😁😁😁

[tropic mountain]

yes, you could also write it as

$\frac{4}{9}x^{-6} \cdot y^{10}$

[torn sequoia]

What

[tropic mountain]

crap

[rocky lotus]

licentia

[vale dock]

@torn sequoia Has your question been resolved?

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[merry jasper]

Is this correct

[merry jasper]

also can someone help me explain this

[merry jasper]

thanks

yes that

@scenic kettle

[late badge]

?

[merry jasper]

is this correct

way to

use the chain rule

[late badge]

which one?

there's two photos

[merry jasper]

wait

im sending a new one

i learned it from organic chemistry

[late badge]

yeah it's fine

just no -ve in first term

[merry jasper]

-ve?

negative?

[late badge]

as f'(x) of cosec(x) = -cosec(x)cot(x)

yes

-ve -ve cancels out

[merry jasper]

wait im sorry, i think i cropped the wrong one

[late badge]

lmao

[merry jasper]

there

the original question is

(cot3x)^2

[late badge]

yaeah. seems true

it's fine

[merry jasper]

okay one more question

in his videos

he doesnt put the x

but on our book we use this

[ashen prawn]

@merry jasper its the same where we write f' or f'(x)

nothing varies

its just diff notation

[fringe niche]

no it does

f' is the derived function

f'(x) is the derived function when we apply x as a parameter/variable

they're two different objects

[merry jasper]

okayy, but is this correct

can i just write it like this

[fringe niche]

idk what f(x) to know f'(x)

also where are you from you use csc we just keep tan cos and sin

so csc(x) = 1/sin(x) then csc'(x) = -cos(x)/sin²(x)

[merry jasper]

this is f(x)

[fringe niche]

so 1/tan²(3x)

if you have 1/f and f does not become null in the chosen interval where you will derive

then (1/f)' = -f'/f²

so just apply that

-(2 x 3 (1+tan²(3x)) x tan(3x))/tan^4(3x)

[merry jasper]

ohh, so i covnert the cot first

to reciprocal identities

and then

i do the chain rule?

[fringe niche]

if you are confomrtalbe with using csc and cot and what not

go ahead

but i never use them

[merry jasper]

Heres the question, i only need to answer 9 and 10

[fringe niche]

yeah i don't use csc and cot

[merry jasper]

does that mean, i wont have to memorize the derivatives of csc and cot etc

[fringe niche]

i just stick to sin and cos and tan

[fringe niche]

yes

you just need to know what they are

and apply simple derivatives

[merry jasper]

ohhh

damn

thats a big help

[fringe niche]

too much learning is going to slow you down when it's useless

[merry jasper]

they didn't teach that in our discussion, so i didnt know mb

or we did, i just missed it ig

what's that for

[fringe niche]

nah nvm

that's for anti derivatives

forget what i said

[merry jasper]

wait how do i find the derivative of 1/tan

[fringe niche]

apply this

if you don't know how people found that equation just prove it using the definition

[merry jasper]

im sorry, which will become?

which is the f prime and f^2

[fringe niche]

https://www.youtube.com/watch?v=FLAm7Hqm-58