help-39

[near brook]

INTEGRALS ARE FUN

.close

[pearl pond]

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[pearl pond]

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[tough lynx]

This is just a general question to those who are good with binary/hex/decimal

What is in the screenshot is an exponent field of e5. Basically where e5 is put to the power of what's to the left and right.
I'm trying to understand what's going on but can't really seem to wrap my head around it.

If I have a look at the first row we have simply (e5)^x where x can go between 0-f in hex.
I've tried working out e5^2 but can't seem to make it 4c.
I've converted e5 into 11100101 (which is 229 and e5 in binary) and then done some binary multiplication but I still don't come up with 4c. I get 217...am I doing something wrong here?

[toxic lichen]

i think those might be exponents modulo 256

[tough lynx]

Yeah. That's what I'm working with as well

[toxic lichen]

let's see

,calc (16 * 14 + 5)^2 mod 256

[jolly parrot]

Result:

217

[toxic lichen]

,w 217 to hex

[jolly parrot]

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[toxic lichen]

huh.

that looks like it should be d9.

[tough lynx]

2 things
1 - Wow that was quick
2 - Crap. Does that mean that this resource is inaccurate?

[toxic lichen]

is any explanation given for this table

so like

[tough lynx]

"When any of these numbers is exponentiated multiple times, the original number is reached again after 255 exponentiations. For example, here is a chart, in hexadecimal notation, of the number 0xe5 being exponentiated 255 times:"

That's the text above the field

[toxic lichen]

huh.

[tough lynx]

This could be a chance for me to actually correct this and then use it as an example for uni.

Should I go through with this or?

[toxic lichen]

the 0e to 01 thing is also suspicious

because if the document is to be trusted then 0e * e5 = 01 mod 256

but it isnt

so something strange is going on.

ok hold up now

i think those might not be plain binary numbers.

did the surrounding text, or the book this is in, mention such a thing as GF(256) or finite fields by any chance

[tough lynx]

Ok I might be trying to work with Galois fields

[toxic lichen]

oh so THATS whats going on.

[tough lynx]

Reason. I'm trying to research AES

[toxic lichen]

okay that makes everything make sense all of a sudden

GF(2^8) is constructed as F2[x]/(f), where f is an irreducible 8th degree polynomial over F2

and these bytes represent elements of GF(256) as (bit0) + (bit1)x + (bit2)x^2 + ... + (bit7)x^7

im not sure if f can be recovered from this table alone but i think it must have been given somewhere in the text

[tough lynx]

Maximum that I've been provided in this is a set of Rijndael generators, those generators expressed in hex. And then the following text and table

[toxic lichen]

can you show those...?

[tough lynx]

I've seen the use of GF(2^8) in other documents covering this before tho

Here's the above text that I'm working with

[toxic lichen]

hmm

but the text does not construct rijndael's field itself?

i for one don't want to go through all 32 possible candidates for the polynomial mod which arithmetic is done

[tough lynx]

Oh yeah. I don't want to force someone to go through that many possibilities.

As for the rijndael field? Nothing.

Most of it has been helpful in my understanding. Few bits of C++ code in order to explain what's going on but it's not actually shown the base Rijndael field

[toxic lichen]

strange. why would they not show it?

[tough lynx]

Maximum that I've found is a text document with a list of all 128 generators slapped into an exponent field and a log field

[toxic lichen]

i mean ok let's try piecing this together...

e5^2 = 4c, they say.

e5 = 1111 0101 = x^7+x^6+x^5+x^4+x^2+1, 4c = 0100 1100 = x^6 + x^3 + x^2

,w (x^7+x^6+x^5+x^4+x^2+1)^2 - (x^6 + x^3 + x^2) mod 2

[tough lynx]

Isn't e5 = 11100101?

[toxic lichen]

ah fuck.

well WA didn't do what i wanted it to do anyway so

[tough lynx]

[toxic lichen]

,w (x^7+x^6+x^5+x^2+1)^2 - (x^6 + x^3 + x^2)

tf

[tough lynx]

Really doesn't like that

[toxic lichen]

it's just a polynomial what is there not to like

yeesh

ugh.

,w simplify (x^7+x^6+x^5+x^2+1)^2

[jolly parrot]

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[toxic lichen]

oh but wait. hold on.

i can use char 2 magicks and simplify this myself

[rustic gate]

try expand rather than simplify

[toxic lichen]

because... in char 2, (a+b)^2 = a^2 + b^2

so we have

x^14 + x^12 + x^10 + x^6 + x^4 + x^3 + x^2 + 1 must be divisible by some degree 8 irreducible

in hindsight this was prob not a good idea

idt i am any closer to figuring out what poly the rijndael field is based upon

[tough lynx]

"A rcon operation that is simply 2 exponentiated in the Galois field."

No idea if this might help. But it's something further back in a different area with the Rijndael Key Schedule

pls don't hurt me if it helps more

[rustic gate]

seems to be x^8 + x^4 + x^3 + x + 1

[toxic lichen]

[tough lynx]

Strange that we're getting a polynomial to the power of 8 while trying to keep everything 8 bits long

[rustic gate]

,w expand (1 + x^2 + x^5 + x^6 + x^7)^2 - (x^6 + x^4 + x + 1) (x^8 + x^4 + x^3 + x + 1)

[jolly parrot]

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[rustic gate]

x^6 + x^3 + x^2

[tough lynx]

001001100
That's 4C alright

[meager trellis]

well i worked out that this is (x+1)(x^13 + x^12 + x^9 + x^8 + x^7 + x^6 + x^3 + x + 1)

but also you seem to have found the answer on wikipedia so i guess that isn't helpful

[rustic gate]

[tough lynx]

And all of this could have been solved if I'd looked into the Rijndael finite field

Bugger

Thanks for all your help tho. Very much appreciated

Ok maybe I've got one more question.
Snow put this polynomial: (1 + x^2 + x^5 + x^6 + x^7)^2 - (x^6 + x^4 + x + 1) (x^8 + x^4 + x^3 + x + 1)
(1 + x^2 + x^5 + x^6 + x^7)^2 makes sense as this was e5 in polynomial form
(x^8 + x^4 + x^3 + x + 1) makes sense now as this is the Rijndael finite field in polynomial.
(x^6 + x^4 + x + 1) My question lies with this. Where has this come from?

[meager trellis]

i think it's... division?

if you expand out (1 + x^2 + x^5 + x^6 + x^7)^2 mod 2

you get some big polynomial of degree 14

but since we're in this finite field

we want to get rid of some multiple of x^8 + x^4 + x^3 + x + 1

because that's equal to zero by the construction of the field

[rustic gate]

i worked it out by doing division

[meager trellis]

so if you just divide that big polynomial by x^8 + x^4 + x^3 + x + 1, you get some quotient, x^6 + x^4 + x + 1, and the remainder, x^6 + x^3 + x^2, which is the actual result that you get in the field

[tough lynx]

So all that's been done is the polynomial that's squared has been divided by the rijndael polynomial?

[meager trellis]

yep

[tough lynx]

Ahhhhhhh

I seeeeeee

That makes more sense now

[rustic gate]

,tex<polynom> \polylongdiv{x^{14} + x^{12} + x^{10} + x^4 + 1}{x^8 + x^4 + x^3 + x + 1}

what

[meager trellis]

wait did you just get the bot to divide the polynomials for you

[rustic gate]

yes

[meager trellis]

huh

[rustic gate]

doesnt support doing it in F2 though annoyingly

[tough lynx]

I guess those negatives in the texit screenshot is from your first typing?

Something not adding up between that and the polynomial you showed before

[rustic gate]

no its doing the division over R

[meager trellis]

the negatives are because it's just- yeah

[rustic gate]

if you mod 2 then all works out

[meager trellis]

it's not doing characteristic 2 polynomial division
that's also why there are 2s in there

[tough lynx]

I mean yeah... -1mod2 is 1 afterall

[pearl pond]

@tough lynx Has your question been resolved?

[tough lynx]

.close

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[void sandal]

whats the question?

[midnight haven]

That as an equation

[opal parcel]

It’s not A that’s for sure

[midnight haven]

R u sure

I have 17 questions

[opal parcel]

Yeah cause the graph represents a quadratic equation

[midnight haven]

So

How do i get the answer

[void sandal]

it cannot be A as it isn't a qudratic nor D as it is negative x. To determine between B or C factorise to find the roots of the equation

[midnight haven]

how do i factor

[waxen cypress]

the roots of the graph is $x = -2$ and $x = 6$ so it must be in the form of $a(x + 2)(x - 6)$.

[jolly parrot]

Nayaka [ꦟꦪꦏ]

[midnight haven]

so am i right

[waxen cypress]

yes

it's facing up so the coefficient of x^2 should be positive

[waxen cypress]

is this from a test?

[midnight haven]

no

he

hw

[waxen cypress]

you should at least do some work or point out what you don't understand.

[midnight haven]

I am

Im halfway through already..

but this is on something i wasnt here for

So i need@help

[waxen cypress]

alright

[jolly parrot]

Nayaka [ꦟꦪꦏ]

[midnight haven]

144

[opal parcel]

Do you know how to factorise?

[midnight haven]

No..

[opal parcel]

Oh ok you need to know that

[midnight haven]

thats why i meed help

[opal parcel]

Maybe watch a YouTube video on how to factorise, it’s difficult to explain thru text

[waxen cypress]

Why would it be 144?

[midnight haven]

Why are there dillar signs

Thats not math lol

So

Who can help

[opal parcel]

Judging by the question, can you guess what the graph would look like?

[midnight haven]

It says how many seconds is it in the air

So

is it 16

32

144

[waxen cypress]

none of them

you should have followed my direction

[waxen cypress]

try to answer this first

if you don't know how, tell us where you stuck

[midnight haven]

IS IT 324

[waxen cypress]

no

[opal parcel]

LMAO

[midnight haven]

is it 4

[waxen cypress]

you can't just randomly guess the answer..

the rocket returns to the ground, so it has the same height as the ground, which is...

[midnight haven]

Well im looking at the graph

[opal parcel]

do you understand quadratics?

[midnight haven]

is it 9

Or

[opal parcel]

The x-axis represents time

[waxen cypress]

alright I guess I'm out

[opal parcel]

The y-axis represents height

When y=0, the height is 0

[opal parcel]

LMAO

[midnight haven]

IT WAS 9

HELLO

[opal parcel]

Yeah

But

Did you just guess?

[waxen cypress]

okay I'm back

or do you plot the graph on geogebra?

That's not how you do it on a test, though. I'm just trying to give you directions so you can do it on your own next time

[opal parcel]

I mean plotting the graph will help them visualise the problem better but they shouldn’t become dependent on it

[midnight haven]

OK SO I GOT IT WRONG

[waxen cypress]

should've answered this first

[waxen cypress]

yes, i agree with you

[opal parcel]

Yeah, you need to learn how to factorise

[midnight haven]

Isk how

jelp

HELP

[opal parcel]

Take out a common factor of 5

[jolly parrot]

Nayaka [ꦟꦪꦏ]

[midnight haven]

u cant its ofd

odd

or 2

?

[opal parcel]

Huh

[midnight haven]

2+3

[waxen cypress]

okay hear me out

[midnight haven]

I keed help

[waxen cypress]

can you divide 5 by 5?

yes i know

[midnight haven]

is it C

[waxen cypress]

I'm trying to help you but you wouldn't listen

[midnight haven]

Ya

[waxen cypress]

alrighy

can you divide 45 by 5?

[midnight haven]

Ya

Its

9

IRS C

CS C HAS A 9

[waxen cypress]

not yet

we're not done

[midnight haven]

is it right tho

[opal parcel]

HAHA

[midnight haven]

Is it right

[waxen cypress]

you got a point tho

[midnight haven]

so im almost

Is it A

[waxen cypress]

45/5 = 9, that's what you meant right?

[midnight haven]

Ya

But c has 9 in the answer

So

[waxen cypress]

okay wait

[midnight haven]

Hurry -

[waxen cypress]

$5x^2 - 45 = 0$

[midnight haven]

I have 5 mims left

To finish

[jolly parrot]

Nayaka [ꦟꦪꦏ]

[midnight haven]

SO ITS 3

AND 0

CS 9 ROOT IS 3

right

or

[waxen cypress]

actually it's B

[midnight haven]

Oh

How

[waxen cypress]

plug in x = 0 here

[waxen cypress]

you'll get -45 = 0 which is wrong

so 0 ain't a solution

[midnight haven]

Id this

[opal parcel]

You’re just rushing thru your homework and not trying to understand it. It’s not going to help you in the long run 😭

[midnight haven]

No but i jabe 2 minutes left

[midnight haven]

this is my last math class

Can someone help

PLEASE

[opal parcel]

x=-1 is a root so (x+1) is a factor

[midnight haven]

so is it right

[opal parcel]

(x-4) is a factor

[midnight haven]

with x+1

[opal parcel]

Yeah but I’m not so sure about the first box

Like idk 💀

But the other two are correct

[opal parcel]

What grade are you in?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[finite crown]

https://gyazo.com/e4c3df3a6723e58ad8ba1c9f3e6318c7

[finite crown]

How does one find all these solutions without using a calculator

Im so lost

I can get the 0,-1 and 0,1

but idk how to get the 0.5's n shi

[leaden cape]

Do we have f(x, y) = 0?

[finite crown]

why would this matter?

[leaden cape]

Nevermind. My bad for asking.

[plush bramble]

can you show the whole page

[finite crown]

https://gyazo.com/d8258d375fa81e0f7aef2e5786bf946f

I can do the Hessian matrix and eigen values

i just dont know how to find all them values

like how would anyone think of 0.5

[plush bramble]

try completing the square for the terms in parentheses

3x^2 + y^2 - 1 = 0

x^2 + 3y^2 - 1 = 0

[finite crown]

hm

[plush bramble]

,w plot x^2 + 3y^2 - 1

[finite crown]

ok

[plush bramble]

okay that plot wasn't very helpful

[finite crown]

How do i complete the square of that

there is no co-eff of x

[plush bramble]

x^2 is a perfect square already

[plush bramble]

actually completing the square is unnecessary

just solve for x or y here
x^2 + 3y^2 - 1 = 0

then plug it into the other equation in parentheses
3x^2 + y^2 - 1 = 0

[finite crown]

Oooo

i can do that

[plush bramble]

or maybe just solve for x^2 or y^2

may may algebra easier

[finite crown]

but is there not infinite values? like x = 0.1 makes y = root(33)/10

[plush bramble]

the solutions have to satisfy both equations

[midnight haven]

hello this was my channe

Im not done

[finite crown]

Since they both equal 0, can i not equate them to eachother. Because when i do i get x = y

[midnight haven]

help my makeup is sweating off

wtff

[finite crown]

now how it works

[plush bramble]

!help

[pearl pond]

Please read #❓how-to-get-help

[plush bramble]

that loses some solutions.

[leaden cape]

It might help to organise the ideas as follows:
From the first equation either y = 0 or y^2 = 1 - 3x^2.
Use these two cases with the second equation.

Each of these two cases will have two cases when used with the second equation but it should work out quickly.

[finite crown]

BIg thanks got both all them half ones

[finite crown]

Thanks ill keep this in mind for my next one

.close

[pearl pond]

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[midnight haven]

hello

[fleet sky]

howdy. What's your question?

[ivory swallow]

goodbye

[fleet sky]

did you have a question?

[pearl pond]

@midnight haven Has your question been resolved?

[fleet sky]

@midnight haven if you have a question re-open the channel

.close

[pearl pond]

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[silent verge]

So i don’t know why

Y=0 is not an symptote

[light helm]

what makes you think that it should be

[silent verge]

Because

Astmptote means

Not reachable

Here

This is why i think y=0 is an asymptote

HA

Horizontal A

Please tell me what im doing wrong tho

[light helm]

Not reachable
misconception

asymptotes can be crossed infinitely many times

asymptotes describe end behaviour

[silent verge]

Oh i see

I had the misconception for some reason

Thanks for correcting

[pearl pond]

@silent verge Has your question been resolved?

[pearl pond]

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[sharp belfry]

Find the volume of the contents of the bowl K, given by x^2 + y^2 ≤ z ≤ 1

[sharp belfry]

I need a clue. I don't know where to begin.

[leaden cape]

Do you know how the sketch the curves z = x^2 + y^2 (= r^2) and z = 1?

[sharp belfry]

I don't get this part "(= r^2) "

[leaden cape]

In view of cylindrical coordinates, x^2 + y^2 = r^2.

[sharp belfry]

yeah

I know that

[leaden cape]

I don't see what is so difficult to understand then.

This questions begs the usage of such a coordinate system so I put r^2 in brackets.

[sharp belfry]

my problem is that I wanted to use this $\int\int\int 1 dxdydz$

[jolly parrot]

afeAlway

[sharp belfry]

But the problem is that I don't have a bound for x and y? So should I bound $-\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}$?

[jolly parrot]

afeAlway

[leaden cape]

Similarly to we talked above before, the result exists here we may use

[leaden cape]

-1 <= x <= 1 will be the bounds as R (for the theorem) would be a circle on the xy-plane of radius 1. For, the soup bowl is defined above such a circle.

[sharp belfry]

Why -1 <= x <= 1 ?

how do you know the radius is 1?

[leaden cape]

$-\sqrt{1 - x^2} \leq y \leq \sqrt{1 - x^2}$ represent a circle.

[sharp belfry]

hm

[jolly parrot]

stabulo

[leaden cape]

After all, these two curves are the explicit functions of the implicit function x^2 + y^2 = 1

[sharp belfry]

yes

but we don't know if x^2+y^2=1 tho do we?

what if x^2+y^2 = 0.5?

[leaden cape]

So to clear things up, we find the two surfaces intersect when z = x^2 + y^2 = 1. From this we can determine that the surfaces intersect on the cylinder x^2 + y^2 = 1.
If z = 0, we have the circle on the xy-plane, if z = 1, we have the curve of intersection.

[sharp belfry]

but the rest of the bounds is correct? I don't need to change anything?

[leaden cape]

You should obtain as one possible order

$V = \int_{-1}^{1} \dd{x} \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \dd{y} \int_{x^2 + y^2}^{1} \dd{z}$.

[jolly parrot]

stabulo

[leaden cape]

Well then you would be answering a different question.

It's easy to imagine this calculation is going to be annoying which is why I suggest using the theorem

The we would have 0 < r <= 1, 0 <= θ <= 2π, r^2 <= z <= 1.

It's nice because R is that circle which can be easily described in such coordinates and the z bounds lend them self (very conveniently) to it too.

[sharp belfry]

Wait so if I understand correctly this theorem is suggesting variable substitution. But is it the normal variable substitution? As in where I calculate the determinant of the jacobian? Does it have a name so I can read more about it. What is the theorem called?

[leaden cape]

You need not use a Jacobian to prove the result.

Theorem 6 is basically Theorem 5 I posted but you evaluate the double integral in polar coordinates instead of rectangular.

Remember like with the double integral and the related iterated integral, the triple integral is not a iterated integral by its definition.
A theorem to help you calculate it is seen in Theorems 5 and 6.
Of course, by our discussion from before, it can be reduced to a triple iterated integral but it's helpful to write it as in Theorem 5 as you then know what region R we are considering integrating over when it reduces to the double integral after the first iterated integral is computed.

I'll include this to complete this particular discussion.

[sharp belfry]

Is there maybe an example you can show me? What book are you using?

[leaden cape]

There is but it might not be easy for you to understand, I'm not sure.

It's Advanced Calculus by David V. Widder.

[sharp belfry]

you're right. This is too hard, is there maybe a website you recommend I should check out since I'm struggling with this concept as you might have already noticed.

[leaden cape]

Paul's online notes.

[sharp belfry]

I will check that out. Again thanks for the Help.

[leaden cape]

It's not too hard tbh but there may be many things you don't see immediately.

Glad to help.

[pearl pond]

@sharp belfry Has your question been resolved?

[pearl pond]

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[earnest wave]

hi pls help idk how to do it w out an average value

[earnest wave]

there is an average but no bounds*

is what i meant

[leaden wadi]

We cannot help on tests.

[pearl pond]

@earnest wave Has your question been resolved?

[earnest wave]

irs not a test bro @leaden wadi

irs my homework 🤪🤪

i got it anyway.

.close

[pearl pond]

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[merry carbon]

Well, it does say test, so forgive our scepticism

[pearl pond]

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[worn lynx]

Very quick question

[worn lynx]

When finding the inverse of a square root for instance sqrt(y-2)

You can’t do +2 until you remove the sqrt root right?

[coarse dawn]

Yes

[worn lynx]

Ok I’m going to solve this problem reslly quick

Then show u what I have

Im pretty sure I did something wrong

[coarse dawn]

,rotate

[coarse dawn]

What's the original problem?

[worn lynx]

This is the original

[coarse dawn]

And the goal is?

[worn lynx]

Finding the inverse, isolating y

A close answer choice I found on the self-checking is $x^2+2$

[jolly parrot]

hej

[worn lynx]

But idk how that will be an answer choice if I have the square root

I got it from when I squared the y-2

Square root on both sides to cancel out the y-2

[coarse dawn]

I'm gonna be honest, your process isn't clear at all

[worn lynx]

Ok I’ll start over

Ohh wait I see why

Im not supposed to square root both terms

To undo a squareroot you just square it again and for the x you’re just doing x^2

So my answer is x^2+2

Thank you both

.close

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[stark panther]

Not sure what I’m supposed to do

[coarse dawn]

,rotate

[pearl pond]

@stark panther Has your question been resolved?

[pearl pond]

@stark panther Has your question been resolved?

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[strong void]

Would this be C?

[warm current]

explain your reasoning

[pearl pond]

@strong void Has your question been resolved?

[strong void]

Because 7x^2 - 21y^2 would be (7x -3) (7x+3)?

[warm current]

But 7*7=49

[strong void]

?

7x3?

[rustic gate]

expand (7x - 3)(7x + 3) and see what you get

[strong void]

49x^2 - 9

So is that it?

[warm current]

Few things, you started with 7x²-21y², and ended up with 49x²-9, so I imagine it's not the correct answer. Also, your y disappeared

[strong void]

So would B be correct? (x+y) (x-y) = x^2 + y^2

[warm current]

Where did you get x²-y² from B?

[strong void]

I meant to put +

[warm current]

❓

[strong void]

?

You put x^2 - y^2

[warm current]

Yeah, before you edited it

[strong void]

Yea

[warm current]

How did you get x²+y² from B?

[strong void]

(x+y) (x-y)

[warm current]

Expand (x+y)(x-y) and see what you get

[strong void]

x^2 - y^2?

[warm current]

!show

[pearl pond]

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

[pearl pond]

@strong void Has your question been resolved?

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[cold minnow]

So hello. Currently I am struggling to solve this math problem

I believe I've misunderstood the method my teacher taught us, and I'm lost. So what I'm looking for what's the method one would use to solve a question like this

[cold minnow]

Ah wait, I made a mistake. Let me repsost the image real quick

My apologies. Here's the correct question

[pearl pond]

@cold minnow Has your question been resolved?

[leaden cape]

What have you been taught?

[cold minnow]

So from what I understand.

1. First sketch the graph to determine if it's negative or positive
2. Find the x intercepts
3. Put them in the integral
4. find the anti-derivative of the equation
5. Slot the higher x-intercept into the anti-derivative then subtract it by the lower anti-derivative with the lower x-intercept

The x intercepts are both 3 and -3 right? So this would be the final equation

[pearl pond]

@cold minnow Has your question been resolved?

[slow glade]

@cold minnow My friend, do you still require help or were you able to solve it ?_?

I got the area as 36sq Unit

[cold minnow]

I'd still like some help if that's possible

[slow glade]

Sure, so were you able to draw the graph ?_?

Or you want me to explain from basics ?_?

[cold minnow]

Let me draw the graph, I had it on my laptop but it went flat

[slow glade]

I am extremely sorry for such a weird drawing. I actually had to draw it with my mouse, and it's sort of hard to draw with the mouse.

[cold minnow]

Literally same lmao

[slow glade]

Now we have to find the area of the shaded region.

[slow glade]

LMAO Yeah

Anyways

So we can do one thing for that: we can subtract the upper area (The one which is created by the line y=0) by the area of the parabola in order to get the shaded area.

So it will be

And -3 will be lower limit and +3 will be the upper limit

Because those are the only two common points, both of the equations have the same points and intersect with each other.

[cold minnow]

I see

[slow glade]

So now, integrating it and solving the equation, we get the area as 36 sq unit.

Again, I am sorry for having such bad drawings, but I hope they are helpful to you in any way.

[cold minnow]

They were very helpful. Thank you very much. I see where my error was now
I forgot the brackets after the integration which caused me to mess up my order of operations. Thank you very much for the help

.close

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[coarse sage]

The ratio of HM and GM between two numbers a and b is 4:5, then find the ratio of the numbers.

[coarse sage]

,, \frac{\frac{2ab}{a+b}}{\sqrt{ab}}=\frac{4}{5}=\frac{2\sqrt{ab}}{a+b}

[coarse sage]

$\frac{2}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{b}}{\sqrt{a}}}=\frac{4}{5}$

[coarse sage]

Consider √a/√b=t.

Then, $$t+\frac1t=\frac52$$

[rustic gate]

are you gonna solve the whole question

[coarse sage]

I'm showing my work so I can know where I went wrong

[toxic lichen]

@rustic gate he's showing his progress

so far this seems to be correct

[coarse sage]

,w 2t^2-5t+2=0

[jolly parrot]

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Requested by Hazysleet#7757

[coarse sage]

So, a:b=4:1 or a:b=1:4

But, 4:1 is given as the correct answer

and 1:4 is also in the options.

Why is that?

[toxic lichen]

because you were fucked over, that's why.

HM and GM are both symmetric -- if a and b were to swap places, neither of these means would change its value.

however the ratio of a to b would have to invert following such a change.

[scenic summit]

it should’ve stated that a>b

[toxic lichen]

!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.

[toxic lichen]

also don't just jump in and ignore the convo that had been going on prior

[coarse sage]

Who asked?

Oops he deleted.

[toxic lichen]

so my point is, you couldn't have distinguished 4:1 from 1:4 based on the data given.

[coarse sage]

Ok

$5^{1+x}+5^{1-x}, \frac{a}{2},5^{2x}+5^{-2x}$ are in A.P

[coarse sage]

I am asked to find the value of k such that a \geq k

[toxic lichen]

not the greatest value of k such that a ≥ k?

[spiral rock]

sorry, wont happen again

[spiral rock]

use the A.M. >= GM property.

[coarse sage]

$a=5^{1+x}+5^{1-x}+5^{2x}+5^{-2x}$

[coarse sage]

Let t=5^{x}

Then, $$a=\left( t+\frac{1}{t}\right)^2+5 \cdot \left( t+ \frac{1}{t} \right)-2$$

[coarse sage]

and for (t+1/t)=k.

a=k²+5k-2

[coarse sage]

I get a=(-33)/4 to be the minimum value of this function

Am I correct?

,w minimum value of k^2+5k-2

[jolly parrot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Hazysleet#7757

[coarse sage]

seems right.

What could be a more faster approach?

[toxic lichen]

a = t^2 + 1/t^2 + 5(t + 1/t)

t^2 + 1/t^2 and t + 1/t are both ≥ 2

(and in fact achieve their minimum at the same value of t but that matters less)

[coarse sage]

Hmm

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[old hemlock]

hi! i dont understand how to explain why there is only one way to score in rounds 4 and 5

and how to score in 6 pin bowling

<@&286206848099549185>

pls help

<@&286206848099549185>

[woven pine]

Hello!

[old hemlock]

hi!

i need help with this

i got part a and b but idk how to do part c and d

<@&286206848099549185>

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[peak lark]

i dont know where to start 😭

[mossy canopy]

Hmm

f(k+1)>f(k) tells us that f is always increasing, but i dont see how we could narrow it down even further

[autumn fossil]

it's f(k+1)>f(f(k))

[mossy canopy]

Ohhh

[peak lark]

ye

[mossy canopy]

If we assume that f is a bijection then we can apply the inverse and get: k+1>f(k)

A trivial solution for this would be f(k)=k

But i dont think we should assume bijectivity

Maybe we can prove that it is a bijection that would be cool

Actually if f is a bijection then f(k)=k is the only solution since f(k)=k-1 is not well defined since f(0)=-1 which is not positive

[pearl pond]

@peak lark Has your question been resolved?

[pearl pond]

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[hollow lantern]

I'm having trouble evaluating this integral. The answer is 9/2 but I am not quite sure how to get there.

[warm current]

What have you tried

[hollow lantern]

this is from a study session i had with a bunch of people last night so we virtually tried everything including seperating the integral, distributing in the negative, etc

we couldn't figure it out

[warm current]

Also, with respect to what variable is this integral being integrated?

[hollow lantern]

x

[warm current]

It needs a dx then

[hollow lantern]

mb i forgot to write it

but that doesnt change the answer

Let me get what i just did now

[warm current]

Your $-1^2$ was the mistake

[jolly parrot]

SWR

[hollow lantern]

wait so it becomes posiitve

oh

wait wha

im confused

it stays negative

[pearl pond]

@hollow lantern Has your question been resolved?

[warm current]

It should be (-1)²

[hollow lantern]

oh

[pearl pond]

@hollow lantern Has your question been resolved?

[pearl pond]

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[dark jay]

After they set the derivative equal to 0, where did the 2 in front of the cos2x go?

[dark jay]

Is it a mistake or am i missing something

[midnight haven]

they divide everything by 2

[dark jay]

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[narrow sandal]

I dont get how these multiplicative ones are recognised

[timber scroll]

Try thinking about them by choosing a number

Let's say 2
2>0 is true
Now if we were to multiply -1 to both sides
It becomes
-2>0 which is not true
And hence we have to flip the sign around to make it true
-2<0 is true

[pearl pond]

@narrow sandal Has your question been resolved?

[narrow sandal]

How will this closure property one be found

[pearl pond]

@narrow sandal Has your question been resolved?

[pearl pond]

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[tepid panther]

Is the period for this graph 2pi or 4pi?

[tepid panther]

and what would the values in-between the pi be?

like from pi to 2pi

would that be 3pi/2?

and for 2pi to 3pi 5pi/2?

nvm I figured it out -- it's 2pi

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[tepid panther]

hello

[warm current]

Did you still have a question?

[pearl pond]

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[tepid panther]

me? @warm current yeah.

[tepid panther]

.reopen.

.reopen

[warm current]

[tepid panther]

is this right?

the amp is 3

it's a sin graph because when y=0 x=0

the period is pi

but I couldn't find the choice

[warm current]

Could you show the whole question, @tepid panther ?

[tepid panther]

sure.

[tepid panther]

here you go.

to find the period, we should do 2pi/b

pi = 2pi/b

cross multiply

[warm current]

2pi/|b| to be technical

[tepid panther]

yeah

piB = 3pi(right?)

or 2pi^2?

[warm current]

where'd you get 3 from?

[tepid panther]

cross multiplication

I didn't know if it would be 3

our original problem is pi = 2pi/b

[warm current]

From P=2pi/b?

[tepid panther]

yeah

2pi * pi

pi * b

[warm current]

cross multiply gives you pi*b=2pi

[tepid panther]

ohhhh

i did pi/pi

not pi/1

[warm current]

oopsie

[tepid panther]

🤦‍♂️

so b = 2 right?

no b =1*

?

how come?

[warm current]

pi=2pi/b, right?

[tepid panther]

yes.

wait no

[warm current]

plug in b=1 and b=2 and see for yourself

[tepid panther]

i got pi * b = 2pi

divide both sides by pi

b = 1pi no?

[warm current]

[tepid panther]

pi * 1 = 2pi

1pi = 2pi

pi * 2 = 2pi

2pi = 2pi

alright B = 2 my bad

so is my answer choice right? @warm current

[warm current]

yes

It was perfect

[tepid panther]

down to the last minute detail

(don't think I didn't catch that reference 😆)

would I be wrong on this question too?

<@&286206848099549185>

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[tepid panther]

am I wrong on this question? (especially the second one)

[warm current]

.yes, they were all correct

[tepid panther]

but doesn't b correspond with the period? SWR.

[warm current]

no

well sorta

"corresponds" yes

but b is irrelevant

It just wants the period

[tepid panther]

alright.

thanks.

when it wants the full equation of the graph that's when the b becomes relevant right?

alright.

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[odd barn]

can anyone help me with dividing polynomials with long division

[odd barn]

.close]

.close

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[odd barn]

.close

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[quiet goblet]

If the diagonal matrix is A=diag{a1,a2,a3,...,an} then is the adjA=diag{a2 * a3 * .. * an, a1 * a3 * a4*... * an,...,a1 * a2 * a3*... * a(n-1)}?

[pearl pond]

@quiet goblet Has your question been resolved?

[pearl pond]

@quiet goblet Has your question been resolved?

[pearl pond]

@quiet goblet Has your question been resolved?

[pearl pond]

@quiet goblet Has your question been resolved?

[plush bramble]

can you prove it?

[quiet goblet]

Noo, these are the shortcuts/tricks my teacher gave me

[plush bramble]

then just take it as a fact. if you don't want to, you should try to prove it

[quiet goblet]

Alright

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[quiet goblet]

Suppose there is a function h(x)=log of f(x) to the base a - log of g(x) to the base a. And I've to find the domain of h(x). Do I seperately find f(x)>0 and g(x)>0 and then take intersection or do I just do f(x)/g(x)>0

[midnight haven]

$h(x)=\log_a f(x) - \log_a g(x)$

[quiet goblet]

Yes

[jolly parrot]

Toby

[midnight haven]

the first - you should do them seperately

[quiet goblet]

Well base doesn't necessarily matters since it's constant but that's my exact question

[midnight haven]

we want log f(x) to be defined, as well as log g(x)

[quiet goblet]

Ok I'll do that

[glacial sequoia]

As long as it's a positive value other than 1

[quiet goblet]

Do I also need to consider the f(x)/g(x)>0 after that or it will do the work?

[midnight haven]

if both f and g are >0, then f/g > 0 automatically

[quiet goblet]

Because f(x) and g(x) can be simultaneously negative in this case

Yes

[quiet goblet]

What if they both are negative then f(x)/g(x)>0 as well oh wait that can't happen. Thanks. I understand now

Since f(x) and g(x) have to be first defined

[midnight haven]

yea

[quiet goblet]

Okay I've reached enlightenment. Will close the channel then 👍

.close

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[tribal compass]

There is a question with a total of 9 points. The Mean score was 4.48 with a standard deviation of 3.19.

I calculated that at least 7.8% of students got 9/9. I also calculated that at least 4.3% of students got at or below -1/9 as a score, and that at least 4.1% of students got 10/9. Even though neither of those makes sense

How can I correctly calculate how many students got 9/9 if the score can only go from 0 to 9?

[plush bramble]

calculate the probability of getting 9 or higher

[stoic dock]

lmao amazing response

[tribal compass]

I wasn’t sure if the Bell Curve going outside of 0 to 9 would make a difference. Which is why I was not sure if I had the right answer since the area from 0 to 9 would not be the same as the area under the bell curve

[stoic dock]

do you have access to a computer for your answer?/

[tribal compass]

Yes, I was just curious since they are last years test statistics for the test I am writing in a few hours

[stoic dock]

ah ok

use desmos and substitute in the mean and standard deviation into gauss' bell curve formula

and you can integrate and find the probability very quickly in desmos

if you want to find 9, integrate the curve from 8.99 to 9.01 or something like that

and you'll get an answer that's correct down to a few decimal places

[tribal compass]

gotcha

[stoic dock]

have you done poisson's stuff yet?

[tribal compass]

no, I’ve done basic integrals in calculus. We have not done much statistics

[stoic dock]

ah ok

however

@tribal compass

if you're studying for an exam, a question like this may require you to use z scores

which is something like $z = \frac{x-μ}{σ}$

[jolly parrot]

Zouni

[stoic dock]

typically they make you remember some value's for z that represent probabilities in the z distribution

[pearl pond]

@tribal compass Has your question been resolved?

[pearl pond]

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[flint drum]

yo

[flint drum]

for a

is it v = 5?

[onyx lynx]

yes

[flint drum]

bet so photomath does work

will be good for my math test then

and also chatgpt

[onyx lynx]

haha gl

[flint drum]

but i fucked for my exam

@onyx lynx for this

is the answer

wait

x = -5/2 and y = -1/2.

?

[pearl pond]

@flint drum Has your question been resolved?

[pearl pond]

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[jolly parrot]

Dark_123

[pearl pond]

Channel closed

Closed due to the original message being deleted

[strong gyro]

wrong one

how would you solve this partial fraction $\frac{3x-1}{(x^3+2)(3-x)}$

[jolly parrot]

Dark_123

[strong gyro]

i understand how you get $\frac{Ax^2+Bx+C}{(x^3+2)} + \frac{D}{(3-x)}$

[jolly parrot]

Dark_123

[autumn fossil]

This channel will automatically close soon

maybe claim another one

[strong gyro]

ok

.close

?

[autumn fossil]

it's closed. You deleted original message

[strong gyro]

its still occupied?

[autumn fossil]

it will move to avaible/hidden soon

[pearl pond]

Available help channel!

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Remember:
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[flat perch]

how would one find the answer to this?

[flat perch]

dude can I get actual help

[pearl pond]

@flat perch Has your question been resolved?

[pearl pond]

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[pearl pond]

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[vast portal]

how do i do this?

[vast portal]

i got to here but idk what to do with the 3

I can't cancel out the bottom h with h^2 because then i would be left with a 4h

not sure what else to try

[warm current]

!show

[pearl pond]

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

[hollow cobalt]

Looks like you haven't expanded correctly, so, yeah, show your work

[pearl pond]

@vast portal Has your question been resolved?

[pearl pond]

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[worthy belfry]

Hello could I please have some help with this question

[sour moat]

As i can understand its asking about how many scores there are that are at least 70

from 70-71 = 6
72-73 = 2
74-75 = 2
6 + 2 + 2 = 10

wrong channel

you have to make your own

[midnight haven]

sorryy

[sour moat]

#❓how-to-get-help

[worthy belfry]

so the answer is 10

[sour moat]

yea

im not good a maths but maybe i might help you make a channel

oh i see

[worthy belfry]

ok next question

[sour moat]

you see those numbers which are out of where the circles intersect

[worthy belfry]

yeah 23,46 and 41 right

[sour moat]

what could you do with them to get total number of students that like only one type of food

[worthy belfry]

multiply

[sour moat]

and what would you get ?

[worthy belfry]

23x46x41=43,378

[sour moat]

quite a big number isn't it?

[worthy belfry]

yeah

[sour moat]

so do you think its needing multiplication?

[worthy belfry]

no

[sour moat]

exactly

[worthy belfry]

so maybe division

[sour moat]

so you need to get an exact number from the column that you have

[sour moat]

your trying to guess

the numbers that are shown can be obtained by adding

sorry

i meant adding

[worthy belfry]

so its 110

[sour moat]

there you go

just be attentive to small details

[worthy belfry]

also completely unrelated question, what anime type do you watch?

[sour moat]

if your asking about my banner

its Tokyo Revengers

[worthy belfry]

ok do you watch it

[sour moat]

mostly science anime and fighting like tokyo revengers or black clover

[sour moat]

i did even read manga

[worthy belfry]

well if you like science anime think the black bullet is one

it's kind of like a sci-fi monster type of anime

ok bye

.close

[pearl pond]

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[fossil harbor]

How do I find these solutions from this set of equations?

[pearl pond]

@fossil harbor Has your question been resolved?

[pearl pond]

@fossil harbor Has your question been resolved?

[pearl pond]

@fossil harbor Has your question been resolved?

[plush bramble]

you have 3 equations and 3 unknowns

try solving for lambda in the first eqn, plug into the second eqn and solve for y or x and plug into the third eqn

[pearl pond]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[midnight haven]

I don't really understand what's going on here?

[midnight haven]

what exactly happened in the middle two lines of that 😅

[lost flax]

Do you mean where the 2^12 appears?

[midnight haven]

yes

although a clarification of the entire question would also be very helpful

as I am confused with that too seemingly

[lost flax]

When you sum all the $\binom{n}{k}$ with $n$ fixed for $k$ between 0 and $n$, the result is $2^n$

[jolly parrot]

d឵

[lost flax]

That can be proved by induction, or using that 2^n is the number of subsets of a set with n elements, and \binom{n}{k} is the number of subsets with k elements

[midnight haven]

oh wouldn't that be similar to permuting it then?

[lost flax]

It is different

When permuting elements, the way they are ordered matters

[midnight haven]

oh i think i get it hmm

also may i ask what does this really entail? @lost flax

both sets share no elements and are not the same?

if i read it correctly?

[lost flax]

It means that they share no elements

They might however be the same

If both are empty

[midnight haven]

how can two sets share no elements while also possibly being the same

oh

hmm

[midnight haven]

but wouldn't the $i \ne j$ portion indicate that they are not

[lost flax]

Actually no
i \neq j only implies that the indices are different
Think of it as a sequence of numbers. You can have x_1, x_2 and x_3. If x_1 = -1, x_2 = 4 and x_3 = 4, then for i=2 and j=3, i \neq j but x_i = x_j

[midnight haven]

ohh how did i not see that

my gosh

[lost flax]

In this particular example, of course, all the A_i are different, but you only know that after studying them

[midnight haven]

yep! thank u!

.close

[pearl pond]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[tribal locust]

This is wrong I need help

<@&286206848099549185>

[hollow condor]

test?

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[tribal locust]

No?

Why

@hollow condor

It’s a practice test and I only got like 30 minutes left of this period

I don’t understand

My teacher said it was wrong

@hollow condor help

[hollow condor]

you seem oddly impatient

what was your logic for that answer

[tribal locust]

Cause it’s a practice test that’s due this period

And I have like 3 more after

[hollow condor]

.

[tribal locust]

?

[tribal locust]

Ok

So

11.31/8.2 = about 1.37

Then

1.37 x 7.8

=

10.76

But it ain’t right

@hollow condor

Ah shit

Safety patrols are leaving

I got 20 minutes

[hollow condor]

stop tagging me

[tribal locust]

K

[tribal locust]

Do you have an explanation?

11.89 maybeee

Wait no

Ahhh

<@&286206848099549185>

Help pls

[pearl pond]

@tribal locust Has your question been resolved?