help-23

[lean otter]

Just here

You do it and tell me

If (f(n) > 0), increment (n) by 1 and go back to step 3 (n=1) and compute f(n)

[flat frigate]

weig

[lean otter]

And if (f(n) \leq 0), you got it. The current value of (n) is the solution

[flat frigate]

weig

[lean otter]

The smallest n = 1 btw

[wheat umbra]

So i now calculate everything?

[lean otter]

What is this inequality for btw?

[lean otter]

I mean yeah

[wheat umbra]

School need to make presentation due tomorrow

[wheat umbra]

okay

[lean otter]

i mean bro

you'll try n = 1

it'll satisfy the ineq

[wheat umbra]

so n = 2 is positive?

[wheat umbra]

i only should try this value for n?

but what can i change then?

[lean otter]

ye ye

but its not the smallest

[lean otter]

nah like it satisfies the inequality from the two conditions i gave you

If (f(n) > 0), increment (n) by 1 and go back to step 3 (n=1) and compute f(n)

And if (f(n) \leq 0), you got it. The current value of (n) is the solution

you got it = stop right there

[lean otter]

Change what?

[wheat umbra]

mhh

[wheat umbra]

i don't get this what is step 3

[lean otter]

start with n = 1 and compute f(n)

meaning try a value for n

that isnt the one that didnt work just now

[wheat umbra]

so i approximate n till f(n) is smaller equal 0

[lean otter]

Correct

[wheat umbra]

okay

[lean otter]

Increment n by 1

Until you get sth that corresponds

but as i told you

1 works for this specific inequality

[wheat umbra]

n = 1 => f(n) = -0,83
n = 2 => f(n) = 0,13
n = 3 => f(n) = 23,52

somehow it seems that the numbers are getting bigger and bigger

[lean otter]

why would you keep doing it

[wheat umbra]

Wait...

[lean otter]

first one satisfies the condition

end of the story

[lean otter]

this

[wheat umbra]

That's all i don't try to get near?

[lean otter]

i mean hey it's the only one that applies haha

it's the smallest

you cant go any smaller than this

[wheat umbra]

but i can use non-integers as well right?

[lean otter]

I do not know

Depends on your exercise

Are non-ints allowed

If that's the case you would need a numerical method

[wheat umbra]

...not with Binomial coefficient or?

[lean otter]

Like the bisection or newton's method

[lean otter]

In this context

If

n

represents a count of items or events

then it should be an integer

If n can be a non integer

then you would need to clarify what it represents

and how the binominal coeff is defined for non int n

[wheat umbra]

What is when we are starting from $\fraq{49!}{42,5}\leq(n-1)!(50-n)!

[lean otter]

42.5 makes it unclear

[wheat umbra]

first i thought it would be a simple math project :/

[lean otter]

I need more context and details

about the problem

[lean otter]

I mean this is nothing man

You just need to know clearly what the problems is asking for

[wheat umbra]

We don'z talk about that it is not beautifull

[lean otter]

Damn

That doesn't help haha

Like is there a wording of the problem

[wheat umbra]

so we start again with $$
0.85 \leq \frac{{49 \choose {n-1}}}{50}
$$

[flat frigate]

octodino

[lean otter]

bro keep it simple if you dont have any further details

[wheat umbra]

and then multiply by 50

[lean otter]

why would you want it to be a non int

https://tenor.com/view/monkey-wtf-monkey-gasp-monkey-gif-22578374

[wheat umbra]

it seems to simple idk

[lean otter]

It will only make it more complex

If the problem doesn't clearly state whether it can be a non int or not

It probably means it can't

Is this for highschool?

[wheat umbra]

Yeah kind of

[lean otter]

Damn you're stressed for the presentation

Was this problem given by the teach/prof?

[wheat umbra]

this one

[lean otter]

Oh damn I didn't see this my badd

[wheat umbra]

no problem

[lean otter]

Yeah yeah

Correct

The hypergeometric distribution is appropriate

Smh my bad I didn't see this

[wheat umbra]

ahh okay, no problem

[lean otter]

Ok so

[wheat umbra]

Yeah than i think i try to presentate that then to my teacher today

[lean otter]

so the probability mass function of the hypergeometric distribution is indeed

$$
P(X=k) = \frac{{K \choose k}{{N-K} \choose {n-k}}}{N \choose n}
$$

[flat frigate]

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

[lean otter]

do you understand or did i jump too far

[wheat umbra]

Nono

I understood

[lean otter]

ook

[wheat umbra]

Thats the theme which i presentate

[lean otter]

here you got

• (N) is the size of the population (50 fans)
• (K) is the number of successes in the population (1 muffle sport thing)
• (n) is the number of draws number of fans selected)
• (k) is the number of successes in the sample (we want the muffle sport things to be selected, so (k = 1))

[flat frigate]

weig

[wheat umbra]

Yes and wer're searching n

[lean otter]

and you want to find the smallest (n) such that (P(X = 1) \geq 0.85)

[flat frigate]

weig

[lean otter]

gives you the following ineq

$$
0.85 \leq \frac{{1 \choose 1}{{50-1} \choose {n-1}}}{50 \choose n}
$$

[flat frigate]

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

[lean otter]

no way you would solve this anatically

[wheat umbra]

That means?

[lean otter]

Use a numerical method

Start with n=1 then compute the probability for increasing values of of n until you find an n that makes the probability at least 85

ok?

so again

define the function

find the smallest n

and the same thing we did

[wheat umbra]

ummm so i try till i get the samlles n?

ah okay

is this analytical?

[lean otter]

no this is a numerical method

nah like you cant just guess it

thats what i meant by analytical

and

in this context

as i said earlier

[lean otter]

.

i didnt know it was fans

so now

in this context

to answer your question

no you can't use non integers

like wtf is 2,3 fans

https://tenor.com/view/cta-gif-27652056

[wheat umbra]

never saw one? XD

[lean otter]

fr fr

alr you compute this ok

[wheat umbra]

yup

[lean otter]

1' define function

$$
f(n) = \frac{{1 \choose 1}{{50-1} \choose {n-1}}}{50 \choose n} - 0.85
$$

2' find the smallest integer (n) such that (f(n) \leq 0)

3' start with (n = 1) and compute (f(n))

4' if (f(n) > 0), increment (n) by 1 and go back to step 3
if (f(n) \leq 0) stop there you're good

[flat frigate]

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

[wheat umbra]

but yes this makes more sense to use integers because this already makes no sense if we use a comma number in cobinatorics

[lean otter]

yes absolutely

[wheat umbra]

and then we get n = 1 as the solution

[lean otter]

solution is 43

n = 43

[wheat umbra]

what

[lean otter]

https://tenor.com/view/monkey-gif-19923815

[wheat umbra]

isn't our function this $$
h(50;4;n) := P(X=1)\geq\frac{\binom{1}{1}\binom{50-1}{n-1}}{\binom{50}{1}}
$$

[flat frigate]

octodino

[lean otter]

yes

why tf did my keyboard write 43

😭

i was like why he confused we already mentioned it

n = 1

[wheat umbra]

huh smth wrong here...
$$
\frac{\binom{50-1}{1-1}}{\binom{50}{1}} = \frac{1}{50}
$$

[flat frigate]

octodino

[lean otter]

what about it

[wheat umbra]

ahh no i was dumb

yess we are now have a probability

forgot

[lean otter]

its correct

https://media.discordapp.net/attachments/1037507234382499850/1088906051652440224/caption.gif

[wheat umbra]

did you presave them?

[lean otter]

i mean yeah

[wheat umbra]

hahah

okay really thank you man

[lean otter]

no worries man

https://tenor.com/view/kermit-cat-cat-dribble-cat-ball-jumping-gif-21104303

[wheat umbra]

but they are good tbh

[safe radish]

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[visual maple]

yee i need help someone tryna give me the answer for this rq

[plucky elk]

Server doesn't give out answers. Do your own work

[visual maple]

fuck u bitch

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[quasi bison]

!status

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[turbid wave]

My first step was to rearrange making it 1>= P(a) + p(b) - P (a intersection b)

=> 1>= P (A union B)

[quasi bison]

ok so far so good

you'll need <=> but yes

[turbid wave]

im sorry i didnt understand

[quasi bison]

if you write =>, it will be treating your goal (P(A ∩ B) ≥ P(A) + P(B) - 1) as an assertion

this is a logic thing

[turbid wave]

okay

[quasi bison]

what you really want to say, if you want to make this into a proof and not just rough work, is that the inequality P(A ∩ B) ≥ P(A) + P(B) - 1 is equivalent to P(A ∪ B) ≤ 1.

and what can you say about the inequality P(A ∪ B) ≤ 1?

[turbid wave]

i know it is always true

but how do i prove it

[quasi bison]

how do you know it is always true?

[turbid wave]

cause the maximum probability is 1 right?

[quasi bison]

could say that.

[turbid wave]

but like how do i write it in a formal proof?

[quasi bison]

could also say that the probability of any event always lies between 0 and 1, which is either an axiom of probability or a direct consequence thereof.

[turbid wave]

okay

wait lemme write up the entire thing and then pls let me know if its ccorrect

let me know if i should make any changes

@quasi bison

[quasi bison]

third line is just horrible typographically

and also in terms of wording

also capital Omega and not lowercase omega

also no spaces before any commas

also "the probability of any event lies between 0 and 1" without any omega

[turbid wave]

okay

[turbid wave]

what all changes to make

[turbid wave]

changed this

[turbid wave]

changed this

[turbid wave]

changed this

[quasi bison]

4 pings back to back

[turbid wave]

im sorry

didnt realise

is this fine now?

@quasi bison

[quasi bison]

would prickly also get rid of "As A, B \subseteq \Omega" (but if you insist on keeping it, at least make the B uppercase)

[safe radish]

@turbid wave Has your question been resolved?

[turbid wave]

alright thank you so so much for the help

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[steep fractal]

This is the exercise I was just doing. Everything is correct except a) range and b) ii) domain.
in the book as solution it says range in a) is smaller or equal to 9. same thing for domain in b).

to me that makes no sense, because it is incorrect. if in a) domain is 3 < x < 7 then the range cant be lower than -7.

Could someone explain to me where i went wrong and why my solution isnt same as in book

[steep fractal]

you can see the exercise in picture no1 and my work in pic no2

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@steep fractal Has your question been resolved?

[steep fractal]

<@&286206848099549185>

too bad 😦

.close

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[mint veldt]

Is the example where x=1 and x = -5 a typo? They changed the inequality in these examples. But in these examples, the author said these would be false. But, if x =1, 1<2 is T and 1^2 = 1 < 4 is also T.

[mint veldt]

Specifically, is this inequality not supposed to be flipped?

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[keen tapir]

Hi can somebody help me solve this

[keen tapir]

I heard you can use hospital on this

But idk how

A is a real number btw

[unique bison]

Call 911, you might get help

[keen tapir]

Really ???

Ok

Thx

[plucky elk]

l'hopital, not hospital

[keen tapir]

Same thing

😭😭😭

[unique bison]

On a more serious note, I'd multiply the numerator and denominator by replacing the minus with plus

[plucky elk]

it isn't

[unique bison]

So you get a fraction

[keen tapir]

Regle d hopitale

[plucky elk]

it really, really isn't

[keen tapir]

I tried it

[unique bison]

l'Hôpital's rule

Maybe show ur work

[keen tapir]

K wait

I did this then

Worked with the hopital rule

[unique bison]

,rotate

[unique bison]

You can simplify the top

[keen tapir]

Srry im rewriting everything

Cuz u wont understand

What i alr wrote

@unique bison

[unique bison]

-4x

[keen tapir]

😭😭 oh yeah

Thx

[keen tapir]

Idk what to do from here

[unique bison]

Not sure where u got that last line from but

What do u think u should do

[keen tapir]

I derived the last line

[keen tapir]

Maybe factorization

[unique bison]

Why?

[keen tapir]

Idk

[unique bison]

Just do l'hopital

[keen tapir]

There’s a 2

And a4

Ok ok

[keen tapir]

But the top will become 0

[unique bison]

Why not use it on the previous line?

The last line doesn't seem correct anyway

[keen tapir]

🤷‍♂️

.close

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[noble thicket]

are my answers correct?

[flat frigate]

You must be a bot manager to use this command!

Jun 21, 2023, 6:07 PM

[noble thicket]

!status

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[noble thicket]

4

[quasi bison]

checkbox #3 should be ticked but isn't.

[noble thicket]

right thats also correct

thx

.close

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[noble thicket]

!status

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[noble thicket]

4

[quasi bison]

checks out

[noble thicket]

thx

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[red sandal]

Hey, im stuck on this math probibility question. I dont know where to even start, can some help me out?

[hard crest]

for the first step, go through each of the 7 cards and imagine, "if this card was the first to be turned over, and the rest were still face down, what is the probability that the next card I turn over will be lower than this one?"

[red sandal]

mhm

so 5/6

[hard crest]

well, yes, if it's 5/6 that will tell you that that card was indeed the first one turned over

[red sandal]

🤔

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[noble thicket]

how do i solve for 'c' in this?

[noble thicket]

.closes

.close

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[raw pivot]

how do I solve this differential equation

[raw pivot]

$\dot{x}= ax + by$ and $\dot{y}= cx + dy$ $a,b,c,d \neq 0$

[flat frigate]

tobi.

[raw pivot]

I tried writing this in matrix form and finding the eigenvalues is that a good idea

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[raw pivot]

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[thorny lily]

Could someone help me in this question?-
I have this volume And I have to put it in iterated integrals of this format:

[thorny lily]

Being the answer this :

But I don't understand the limits of the last integral, the dz one

[obsidian oracle]

As you can see from this thing, the "z" boundaries are from the 1st integral

So we're taking 0 <= z <= 1

[thorny lily]

yes, precisely

But I don't know why it is 1

[obsidian oracle]

Since 0 <= z <= 1 - y, y being positive or null

[thorny lily]

Oh, I thought it was the last bc it is the last one I caculate

[obsidian oracle]

Yes you might have read them in the wrong order

[thorny lily]

I'm sorry, but I still don't get it, that only tells me the y range, no?

bc int that y can't be greater than 1, right?

[obsidian oracle]

Actually, you can get that the upper bound for z will be 1

Since 1-y <= 1

Then 0 <= z <= 1-y <= 1

[thorny lily]

Ahhhhh, got it!!

Thx!!!

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[spring oxide]

I did a task where i had to simulate values from an arima model where i choose its order(p, d, q) and then interpret the estimated acf/pacf with the theoretical ones. I got this as response from my teacher "These two paragraphs indicate that it is not clear how the ACF and the PACF of ARIMA models are expected to behave. According to you, an ARIMA(p, i, q) is expected to have p significant lags in the ACF and q significant lags in the PACF. This is not true. Correct this exercise". Any guidance on why im wrong and what is right or maybe point me to a source where i can read about it and learn? I've been taught that acf gives q significant lags and pacf p significant lags but i must be wrong. Any help would be appreciated

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[wraith prism]

So here by comparing our questions what is N and thita?

@sage shore

N=3
Thita=2PI/7

??

Ohh got it

[safe radish]

@wraith prism Has your question been resolved?

[safe radish]

@wraith prism Has your question been resolved?

[safe radish]

@wraith prism Has your question been resolved?

[sage shore]

Hii

Yes, N=3 and Thêta=2pi/7

[wraith prism]

I lost other formulas

[sage shore]

And the result is correct too

Well done!

[wraith prism]

Yeah thanks luna

Does it work only for sina+sin2a type form

Which is only AP?

[sage shore]

AP = ?

[sage shore]

yes, there is another Lagrange formula for sine sum

[wraith prism]

Arithmetic progression

Only two formulas?@sage shore

What about tan?

[sage shore]

Never heard of a tan formula :/

[wraith prism]

Ohh then mystery solved

New question uploading

.ckose

.close

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[wraith prism]

How do we solve it?
Here i tried this 4 ways

Then differentiation which is 2,0,-2 doesn't exist

But here i want to learn how to break it correctly. I gave minus to both but alternatively

[safe radish]

@wraith prism Has your question been resolved?

[final plinth]

What are you not understanding?
Question, can be solved without writing, just remember non-diff+diff-> non-diff
and |x-a| is non-diff at x=a

@wraith prism

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[wraith prism]

Could you explain it please?

[wraith prism]

I didn't understand what you mean?

[final plinth]

What is exactly your doubt, the actual method is given in the solution while I gave the shortcut method

[wraith prism]

Interval is my doubt

How to solve modulas

The solution is unclear

They didn't mention properties for x and -x

I wrote my solution so that i can show what is my knowledge regarding

[final plinth]

You are only interested in derivative at 2

So no need for 4 cases

Think in nbd of 2

[wraith prism]

What if they have asked for any other cases

What will be our approach that time?

4 cases right?

[final plinth]

Yeah

[wraith prism]

Hmm okay

So what is your short trick?

I wanna learn it now

[final plinth]

If a function is made of (differentiable + non-differentiable) functions, then it is non differentiable

And |x-a| is non differentiable at x=a

Your question consists of |x-2| which is non differentiable at 2, so entire function is non differentiable at 2

[wraith prism]

How is it non differentiation?

[final plinth]

I don't remember the proof

[safe radish]

@wraith prism Has your question been resolved?

[wraith prism]

I meant x-a

[final plinth]

Where does |x-a| has sharp corner?

At x=a, a sharp corner is there which means two tangents. So differentiation is not possible

[wraith prism]

X=a

Ohh remembered thanks

[safe radish]

@wraith prism Has your question been resolved?

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[normal raven]

.close

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[craggy trout]

For 23 would I multiply the num and denom by sqrt(2) to get it into the form sinx/x?

[austere spade]

you dont need to, since theta goes to 0 so does sqrt(2)*theta

to see it easier you can let u=root(2)*theta and take lim u-> 0

same expression

[craggy trout]

Okay but what about for 25?

I need to get rid of the 4 in the denom right?

[austere spade]

yeah you'd need 3y in the denom to use that identity

[craggy trout]

Hmm

[austere spade]

multiply top and bottom by some constant you can determine to 'get rid of' the 4

[craggy trout]

So by 4

$\frac{4sin3y}{y}$

[flat frigate]

dopediscorduser

[craggy trout]

$\frac{sin3y}{4y} * \frac{4}{4}$
$\frac{4sin3y}{y}$

[flat frigate]

dopediscorduser

[austere spade]

thats true

but you need a 3y in the denom now

[craggy trout]

So multiply by 3/3 then?

[austere spade]

so you can use that identity

also wait

no thats not true

4 * 4y = 16y

if you multiply top and bottom by 4 you get 4sin(3y)/16y

[craggy trout]

Ah you’re right

I guess by 1/4 then?

[austere spade]

mhm

[craggy trout]

Then multiply by 3?

[austere spade]

yep

[craggy trout]

Could you clear up some confusion I have about this property

Why does it work if you have 3y in your sin arguments and 3y in your denom, but does not work for 3sin(y)/3y?

[austere spade]

it does

the 3s cancel to become sin(y)/y

[craggy trout]

$\frac{sin3y}{3y}$

[flat frigate]

dopediscorduser

[craggy trout]

$\frac{3siny}{3y}$

[flat frigate]

dopediscorduser

[craggy trout]

So these two are identical in regards to the property?

They both are 1 as they approach 0?

[austere spade]

they're equivelant expressions yeah

if you were to take their limit at 0

but that second expression has 3s for no reason really

since they cancel

[craggy trout]

Thanks

[safe radish]

@craggy trout Has your question been resolved?

[safe radish]

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[solemn vault]

Hi

[solemn vault]

Should work??

[flat frigate]

_basudev

[hard crest]

how do you get $n + 2 | 3^z + (n-1)^z$?

[flat frigate]

Hayley

[solemn vault]

.

N-1+3 = n+2

Shit I messed up

[flat frigate]

_basudev

Since we know $ n \,\, \not{|} \,\, a$ and $n \,\, | \,\, b\,\, \implies \,\, n \,\, \not{|} \,\, b+a$

[solemn vault]

n doesn't divides (b+a)

Ahh shot we messed up let me write that again from the beginning

Should work...

[hard crest]

yeah I get why $n + 2;|2^z + n^z$ but how does that imply that $n+2;|;3^z + (n-1)^z$ ?

[flat frigate]

Hayley

[solemn vault]

We have the seires $\varepsilon$ right?? We club 2nd term with the last one 3rd term with the second last one and soo on...

[flat frigate]

_basudev

[solemn vault]

So if we (2+3) | (2³+3³)

Using that god damn property

Get it

?

[hard crest]

... no
we have 2 + 3 | (2^z + 3^z) sure
but how does that help us show n + 2 | 2^z + 3^z ??

[solemn vault]

Where did I wrote thY??

[hard crest]

i thought that was the argument you were making? that we can replace 2^z with 3^z
maybe not
ok i still don't see how we can show n + 2 | 3^z + (n-1)^z

[solemn vault]

We've some small typo

Wait

[hard crest]

yeah i did that in my head lol
but how is this equation justified?

[solemn vault]

It is true .... I don't know what you mean by "how is this Justified"

.close

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[maiden barn]

cos(x) + sin(x/2) +2 = 0

[maiden barn]

How should I approach this ?

[grim plover]

use sin(x/2) identity

[maiden barn]

What does it say ? I have never heard of it ?

[lean otter]

if u don’t know half angle formula, you can also apply double angle formula on cos(x)

[brave wolf]

other way to approach this would be realising that cos(x) + sin(x/2) = -2 if and only if cos(x) and sin(x/2) are both equal to -1.

[maiden barn]

Now that's an approach!

[brave wolf]

This is true because if one of them was greater than -1, it would mean that the other one is even smaller than -1 which is impossible for sin and cos

[vapid anvil]

Great approach

[grim plover]

this would require them to be consistent with the values of sin(x) and cos(x) (and to think critcally enough)

[maiden barn]

But it's a cool idea!

[grim plover]

True

[maiden barn]

Thanks for the answer!

.close

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[obtuse plover]

Nice one

[safe radish]

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[daring edge]

So this is so easy but I have a few core questions that are super important

[daring edge]

so for one, WHERE does the integral start?.. like is it 7 or 8 wtf??

secondly, how do you really set this one up?.. because technically the y = 2 and y = -1 are the boundaries of the actual area??..

[hard crest]

you basically have to turn your head on its side here

[daring edge]

technically you could split it up into two integrals

[hard crest]

y is the new x and x is the new y

[daring edge]

subtract them from eachother since its with respect to y anyways

so is that the process here?...

like you find the area under the y=o line and the area above it and subtract those?..

but like WHAT area is shaded??

[hard crest]

there's only one enclosed area in that graph

[daring edge]

ah so its the area to the right of the parabola

but like how do you begin this

[hard crest]

no, that's not enclosed

[daring edge]

oh WOW

ITS THE ONE TO THE LEFT okay

[hard crest]

the edges of the viewport don't count as boundaries

yeag

again you'll kinda need to turn your head on its side

[daring edge]

okayy so what do we do now?!

[hard crest]

you'll have $\int (\cdots)\dd{y}$

[flat frigate]

Hayley

[daring edge]

what's the upper limit?.. and like how do you put that into integral form

like putting all 3 of those into integral form

[hard crest]

the limits will be the bounds of y, how small and how big can y be?

[daring edge]

oh shoot so it's -1 to 2

[hard crest]

yeah

and the integrand will be the bound on x (as a function of y)

well, higher minus lower

[daring edge]

then the top function is the parabola, and the bottom one is the 0 function

[hard crest]

yep! exactly

[daring edge]

wow okay that makes so much sense

thanks sm!!

.clsoe

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[safe radish]

Show your work, and if possible, explain where you are stuck.

[hasty wagon]

i can think of a way

wait... lemme rethink

hmm finding the discriminant should be the fastest way...

b²-4ac<0 and m-2>0

yea

no problem

please show

you don't have to

just take a pic will do 🙂

I won't

let's see

i see the problem

[grim plover]

I lost the solution on -3m^2 + 8 < 0

Shouldn't it be the other way around?

-3m^2 + 8m > 0

[hasty wagon]

it's -3m²+8m<0

not -3m²+8<0

yea, you forgot that XD

[grim plover]

If it's less than 0 the sqrt becomes complex no?

[grim plover]

Bruh I forgot to point that out

[hasty wagon]

hehe, it's okay

[hasty wagon]

we just consider the real case.

[grim plover]

Huh???

Sqrt less than 0 real..?

[hasty wagon]

right

should be ≤0 instead

i missed that

because the quadratic ≥0

anyways, i think that's it

[grim plover]

Ohhh
I FORGOT AN A (XD) \j

[hasty wagon]

sure

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[keen mural]

Consider: h.x^2 + k.y^2 + z^2 = 25
Find the real values for h and k so that the equation represent a hyperboloid of one sheet which intersects the plane z = 3 as a hyperbola with a focal edge parallel to the y axis

[keen mural]

I don't know how to solve this, I know that since the focal edge is parallel to the y axis that k has to be negative and h has to be positive but I'm not sure what to do beyond that

also that y has to be equal to 0

but then I'm left with h.x^2 = 16

but I don't know how to find the value for h with the available information

[safe radish]

@keen mural Has your question been resolved?

[keen mural]

<@&286206848099549185>

[young nexus]

why?

and if y = 0, why do you know that k has to be negative? if y = 0 then k could be anything, as it doesnt matter.

[safe radish]

@keen mural Has your question been resolved?

[safe radish]

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[viral loom]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

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[grave flint]

sum of exterior angles in a polygon is 360 degrees

I dislike the amount of theorems you need to remember for geometry

there's probably a better approach to start basic and prove every but it's time consuming and I wouldn't know how to go about it

I'm very poor at geometry

[grave flint]

i wasn't even sure about this it just looked right so I searched it up to make sure

Something like that sounds right

[viral loom]

So, did your answer work?

Cause there is something to do there

Nice

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[digital zinc]

Need help plss

Anyone??

.close

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[wraith prism]

,w minimum and maximum for 3x^(2/3)-x^2

[flat frigate]

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[obtuse jackal]

#bots

[wraith prism]

How to solve it?

I guess 1 and 2nd derivative

$2/x^(1/3)-2x$

[flat frigate]

arjunn5589

[wraith prism]

X^4/3 =1

X=1

$-1x^-(2/3)/3×x^2/3-2$

[flat frigate]

arjunn5589

[wraith prism]

So it is decreasing when we put x=1

I guess question answers are wrong

<@&286206848099549185>

[versed leaf]

Max and min of the output

[wraith prism]

My answer is 2 and 1

But they are saying 0 is minimum

[eager marlin]

I’m pretty sure it’s just the first derivative = 0 and solve for x

[wraith prism]

Yes

[eager marlin]

So what’s the issue

[wraith prism]

Answer is C

X=0

But my answer is x=1

,w minimum and maximum for 3×(x^2)^1/3-x^2

[flat frigate]

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[young nexus]

does the first derivative at x = 0 exist? i guess not.

[young nexus]

draw the graph an you will see that near (0/0) all y-values are greater 0, in this sense 0 is a local minimum.

[wraith prism]

What about maximum?

Ohh it's 2

[wraith prism]

Yes. But i am trying to solve it mathematically

[young nexus]

and how will you do this as the first derivative at x = 0 is not defined?

[wraith prism]

Ohh i solved it

Calculation mistakes

[wraith prism]

It is defined

X=0,1

So i got y=2,0

First derivative is
$2x{(1/x^(4/3)-1}$

[young nexus]

what is the first derivative at x=0?

[flat frigate]

arjunn5589

[young nexus]

wolframalpha means:

and in my world is 1 / 0 undefinded, but maybe it is different in yours.

[wraith prism]

In my world I don't use desmos/wolframalfa for every question. So maybe a calculus mistake. Need to check again. Maybe it's allowed to use in your world

[young nexus]

hahaha, i see you do not use wolframalpha. bye.

[wraith prism]

Hehe bye

Only x=1 will be root

@young nexus

But to find the minimum in this question

[young nexus]

it is really simple. it depends what "minimum" means. if "minimum" is defined as a minimal value where the first derivative is zero, then there is no minimum. wolframalpha said this to you.
if "minimum" means a minimal value such that all y-values in the near are bigger you need to know how the function looks like.

first option is answered, second option is answered.

now it is your choice.

[safe radish]

@wraith prism Has your question been resolved?

[wraith prism]

How will you do it without desmos?

I want to know how x=0 is minimum

[young nexus]

draw it by hand?

i mean where is the problem? you know the first derivative everywhere except one point (x = 0). you just need to know how the function behaves in the near of 0.

[safe radish]

@wraith prism Has your question been resolved?

[wraith prism]

How will you draw a curve by hand in the exam l? You have only 2 minutes to dolve the question not 1 hour@young nexus

[young nexus]

hmmm, is this really a problem? you know you need only to test x = 0. plug in x = 0 gives you y = 0. take a value in ]-1,1[ then you know 1 > x^2 > 0, then you know that if a (as x^2) is in ]0,1[ then a^(1/3) > a so you know 3 a^(1/3)-a > 0. so you know in the near of 0 all values > 0 so (0/0) is a local minium.

[wraith prism]

.close

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[pallid lantern]

what am i doing wrong here

[plucky elk]

,w simplify 9^x - 6^x - 4^x, x = log(2/3, (sqrt(5) - 1)/2)

[flat frigate]

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[plucky elk]

your answer's right, wolfram is just rounding

,w simplify 9^x - 6^x - 4^x, x = log(3/2, (sqrt(5) - 1)/2)

[flat frigate]

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[plucky elk]

just a typo in the answer

[pallid lantern]

i dont think so

i checked on the internet too

same ans

[plucky elk]

do you see this?

[plucky elk]

show

[pallid lantern]

@plucky elk

[hard crest]

that's got a different thing in the logarithm

[pallid lantern]

doesn't matter

[pseudo scroll]

yes it does

Look you have a √5 - 1 while they have a 1 + √5

[pallid lantern]

ok then here's the another source

[pallid lantern]

oh hm

[hard crest]

yes, $\frac{1+\sqrt5}{2} = \frac{2}{-1+\sqrt5}$

[pallid lantern]

how

[flat frigate]

Hayley

[pseudo scroll]

Rationalize the numerator and you'll see for yourself

[pallid lantern]

am i doing something wrong here then

[hard crest]

nope your answer is correct

[pallid lantern]

o

[plucky elk]

i already showed you that here

[hard crest]

one of these branches produces a log of a negative number

[safe radish]

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[strong warren]

If you have f(x) being irreducible in a finite field F = Z/pZ[x]. Why is the class of x a zero of f(x) in F/(f(x))[y]?

[peak estuary]

cause f([x]) = [f(x)] = [0]

cause [ ] work nicely with + and *

[strong warren]

ohhh

Thanks

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[peak wraith]

i dont understand the first step?

[lilac patio]

what dont you understand

it says the cone is 2/3rds of the cylinders height and also gives the height of the cylinder has 2r

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@peak wraith Has your question been resolved?

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[storm sage]

How to do part b)

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[storm sage]

I don't know where to begin for part b

Can anyone explain

[hasty wagon]

do you recall how to find median if all the numbers are given?

[storm sage]

Yes, I can recall that

But theres an unknown variable y

[hasty wagon]

true

you know the way to find median differs when there are even numbers of terms vs odd number of terms right?

[storm sage]

Yes

[hasty wagon]

good, let's start

we want the median to be 2

so, it cannot be exactly between 1 and 2, which makes it 1.5

[storm sage]

Yes so its odd number of terms?

[hasty wagon]

yep

but by thinking using the other way round

if we know the y that makes it exactly between 1 and 2

we can just take away one 1 to make it exactly just reaching 2

[storm sage]

If its odd number of terms though, so 12 + 3 + 1 so 1 + y = 16?

[hasty wagon]

I'm not following

[storm sage]

If I have 11 numbers, the 6th number is the median

[hasty wagon]

yea

[storm sage]

5 numbers in front, 6th, 5 numbers at back

1 + y, median = 2, 16

Nvm I think im just confusing myself 😢

[hasty wagon]

i think that's the way for finding the minimum instead of max

[storm sage]

So what do we do?

[hasty wagon]

,rotate

[hasty wagon]

first we find the y for this situation

after that, we take away one 1

then the median will be moved to the right

and becomes 2

[storm sage]

Yes

[hasty wagon]

after that, the new y will be the maximum

[hasty wagon]

Now it's your turn

[storm sage]

Are you saying to move the right corner 1 to the left to make 2 the median?

[hasty wagon]

nah

[storm sage]

How can we find y tho?

Guess n check?

[hasty wagon]

,rotate

[hasty wagon]

when 1+y=6+12+3+1
then the median will be 1.5 (was typo)

[storm sage]

Yes because it becomes even number of terms

So we remove 1 from there

[hasty wagon]

yep

[storm sage]

1 + y = 6 + 12 + 3

y = 20?

[hasty wagon]

careful

1+y=6+12+3+1

[storm sage]

But dont u have to remove 1?

[hasty wagon]

yes

[storm sage]

1 + y = 6 + 12 + 3 + 1 -1

Like that?

[hasty wagon]

yea

[storm sage]

Ohhh

[hasty wagon]

I thought you're doing step one

yea, you did it

👍

[storm sage]

Thank you so much ur great at explaining 👍

[hasty wagon]

cheers!

[storm sage]

.close

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[little sorrel]

this is a right?

[safe radish]

@little sorrel Has your question been resolved?

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[slow fern]

[
(Ax-b)^{T} (Ax-b)=0
]

[slow fern]

How to expand this?

[plucky elk]

do you know dot products

[slow fern]

Yes

[quasi bison]

are you forced to expand

show entire problem exactly as it was stated

[slow fern]

The professor expanded, and I could not understand

[delicate bobcat]

Did they write any entries for the matrix or vectors

[slow fern]

https://youtu.be/ZUU57Q3CFOU?t=1828

[quasi bison]

$((Ax)^T - b^T)(Ax - b)$

[flat frigate]

Ann

[slow fern]

Right

[quasi bison]

$= (x^T A^T - b^T)(Ax - b)$

[flat frigate]

Ann

[plucky elk]

do you know dot products of a sum/difference of vectors?

[quasi bison]

= expand as you normally would since the distributive law still holds with matrices

[slow fern]

$= x^T A^T Ax - x^T A^T b - b^T Ax + b^Tb$

[slow fern]

How does two middle terms simplify to this?

[quasi bison]

$(x^T A^T b)^T = b^T A x$

[flat frigate]

Ann

[quasi bison]

we conflate a scalar with a 1×1 matrix holding said scalar

[slow fern]

Right

.close

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[eager widget]

guys so this is my question

[eager widget]

and this is my working

can someone pls tell me where i went wrong because it shows as math error for me on my calculator....

[atomic geode]

3rd last step

@eager widget

[eager widget]

how did i go wrong?

what was the right way to do it?

[atomic geode]

u mistook multiplication for addition

[eager widget]

what where?

[atomic geode]

..

[eager widget]

yea but i dont understand how

what was the right thing to do then?

[atomic geode]

441 + cos(120) (4x^2) = 5x^2

[eager widget]

yea i wrote that right?....

[atomic geode]

after that

u wrote 441 + cos120 = x^2

it should have been 441 = 5x^2 - 4x^2(cos120)

[eager widget]

yeah thats cuz i took the (4x^2) and minused it from the (5x^2)

[atomic geode]

u cant

[eager widget]

i mean i subtracted (4x^2) from both sides

[atomic geode]

4x^2 is being multiplied with cos120

use BODMAS

[eager widget]

oh

[atomic geode]

multiplication first

[eager widget]

so how to i multiply it tho

i mean its a variable

[atomic geode]

u know trigonometry?

[eager widget]

yea i just learnt it

im in 10th grade so its kinda new for me thats y im kinda confused....

[atomic geode]

u know how to solve cos( 90 + theta) ?

[eager widget]

um no

[atomic geode]

that is needed in this question

cos(90 + theta) = -sintheta

cos( 90 + 30 ) = -sin 30 = -1/2

cos 120 = -1/2

[eager widget]

oh shit

so i should just multiply -1/2 and (4x^2)

[atomic geode]

ya

[eager widget]

oh okay so then i get (-2x^2)

[atomic geode]

yup

[eager widget]

gimme a sec

ill put it on the board

so is this correct?

[eager widget]

is this correct bro?

[atomic geode]

yes

[eager widget]

OMG THANKS SO MUCH UR THE BESTTTT

.close

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[rancid sorrel]

is this the correct way of calculating the area?

[velvet kettle]

No

[rancid sorrel]

So how do I do it?

[velvet kettle]

Try look at the graph again, maybe you switched something 👀

Actually that’s a bit ambiguous sorry

[rancid sorrel]

should it be (x^2+3x-5) - (2x-3)??

[velvet kettle]

Yes

[rancid sorrel]

because y is in - right?

if y was positive it would have been like I did it

[velvet kettle]

Yea since they’re both below y, we would see which one is lower instead of higher

Generally yes

[rancid sorrel]

I did not know about this

thank you

🙂

[velvet kettle]

Wait

Make sure to put a negative though because generally area is positive

Your answer will be negative

[rancid sorrel]

How do I do that?

like at the final answer?

[velvet kettle]

Simply just get your answer and at the very end

Yep

[rancid sorrel]

u sure?

[velvet kettle]

Well depends on the question

[rancid sorrel]

@velvet kettle ?

[velvet kettle]

Huh you want me to do it for you

[rancid sorrel]

photomath says its -9/2

.close

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[bitter oyster]

A square has sides of length 3x cm. A rectangle is 2x cm by (x + 7) cm. The area of the
square is twice the area of the rectangle. Find the dimensions of each figure.

[bitter oyster]

i know its a stupid question but

[lean otter]

have you done anything so far

[bitter oyster]

3x*3x = 4x^2 + 28x

of im not wrong

[lean otter]

seems correct so far

so you can simplify that to $9x^2=4x^2+28x$

[flat frigate]

richard_22

[bitter oyster]

i was about to say ye
5x^2 + 28x = 0

[lean otter]

nope

[bitter oyster]

wait nope

[lean otter]

$5x^2 - 28x = 0$

[bitter oyster]

my bad

[flat frigate]

richard_22

[bitter oyster]

9x^2 = 4x^2 + 28x

hm ok

[lean otter]

do you understand how i reached that

subtracted both sides by 4x^2

and then subtracted both sides by 28x

[bitter oyster]

OH

ye i get that

[lean otter]

then you just want to solve for x

you can factor the x out

[bitter oyster]

hm?

[dry sigil]

x(5x-28)

[lean otter]

x is a common factor in both those terms in the left hand side