help-10

[fallen zenith]

not ((1+m)^(m))^-1

[lyric ember]

$a^{ax} = (a^{x})^a$

[warm shale]

tobi

[fallen zenith]

yes

but -1/m = 1/m * -1

[lyric ember]

if a is an integer

[fallen zenith]

not m * -1

i have been left

dam

🦫

[lyric ember]

nah u are correct im trying to find my mistake

[fallen zenith]

okay

sorry

lemme just rewrite to check

[lyric ember]

ah

[fallen zenith]

$\lim_{n \to \infty} (1- \frac{1}{3n})^{3n}$

[lyric ember]

it still works mb

just confused myself

use \infty

[fallen zenith]

ah okay

[fallen zenith]

oh

its just latex

dam

ty

[warm shale]

McMake

[lyric ember]

so now substitude 3n = u

[fallen zenith]

3n = u

ye

then

$\lim_{u \to \infty} (1- \frac{1}{u})^u$

[warm shale]

McMake

[lyric ember]

-1/u = v

[fallen zenith]

$\lim_{v \to -0} (1 + v)^{-\frac{1}{v}}$

[warm shale]

McMake

[lyric ember]

$\lim_{v \to 0} (1 + v)^{-\frac{1}{v}} = \lim_{v \to 0} ((1 + v)^{\frac{1}{v}})^{-1}$

[warm shale]

tobi

[fallen zenith]

isnt the -0 important?

for when you go towards infinity you know which one?

(i just started math university. idk)

oh

oh dam

i see tho

[lyric ember]

thats actually an intresting question, this "proof" relies on that, there is actually a more rigorous way to show what this limit is, I am not sure if its important but I have never looked at that

[fallen zenith]

hm

[lyric ember]

I dont think so, the reason we wrote lim v->0 is because lim u->inf -1/u = 0

[fallen zenith]

but its true that when its approaching 0 it is also approaching +0

[lyric ember]

limit definition, if you have an epsilon, thats very close to 0- its also very close to 0+ so im not sure

[fallen zenith]

||tbf dividing by infinity also only approaches 0, since 1/0 would take on every number at once||

[fallen zenith]

yeah thats true

but

lemme just

[lyric ember]

actually no

[fallen zenith]

task said is correct

[lyric ember]

it doesnt mean 0-

[fallen zenith]

it doesnt?

[lyric ember]

I remember that my prof talked about this in analysis 1, let me look it up rq

[fallen zenith]

ok

[lyric ember]

yeah, the +- here in this specific case is irrelevant, because "-1/inf = 1/inf = 0" (im not 100% sure tho)

here is a correct proof for this

[lyric ember]

this is only true for integers!

very important

idk if you have to specify 0-, I never had to in 3 semester

maybe not in this case because e^x is continuous

[fallen zenith]

idk why, but ill take your word

oh because roots

yeah fair

ty

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[unreal cloud]

When a white and a red dice are thrown twice in a row, how many times will the total values of the dice be the same?

[unreal cloud]

the answer is 2 (1^2 + 2^2 + 3^2 + 4^2 + 5^2 ) + 6^2

but how to get it

<@&286206848099549185>

[lone sky]

wat

[lone sky]

wat do u mean by white and red dice?

is the color immaterial?

is this a probablity question

i would say six times

[unreal cloud]

for differentiate

them

[lone sky]

by total values do u mean sum?

[unreal cloud]

but actually they are regular dice

[lone sky]

sum of values of 1 and 2?

poorly worded question

[unreal cloud]

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[unreal cloud]

How many four-term sequences are there, consisting of numbers, where the sum of the first two terms is equal to the sum of the last two terms?

[royal basin]

numbers?

what kind of numbers

if you meant real numbers the answer is infinitely many

@unreal cloud

so i think you meant something else but forgot to tell us

[unreal cloud]

numbers are from 0 to 9 @royal basin

[royal basin]

so only integers between 0 and 9?

i.e. digits

[unreal cloud]

4325 is example for this question

4 + 3 = 2 + 5

[royal basin]

ok and do you require them to be different from each other?

for example is 1111 legal?

[unreal cloud]

no

yes

[royal basin]

so repeats are allowed

[unreal cloud]

yes

[royal basin]

is the first digit allowed to be 0?

[unreal cloud]

no

[royal basin]

so you're looking for 4-digit numbers where the sum of the first two digits equals the sum of the last two

[unreal cloud]

and there is answer but i do not know how to get it

[royal basin]

hm. might bee a little bit annoying

you

you may have to split into cases based on this digit sum

so how many numbers have each pair of digits sum to 1, now many have each pair sum to 2, etc.

up to 18

[obtuse pebble]

@unreal cloud Has your question been resolved?

[unreal cloud]

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[timid silo]

whats the difference between lower bound and upper bound vs infinimum and supremum? and also we are always checking the <= inequality, if the set X is partially ordered right

[sage geode]

An upper bound is just an element that's greater than or equal to all of the elements inside a given set

And the supremum is the lowest upper bound, i.e., an upper bound of the given set that's lower than every other upper bound

Similar definitions for lower bounds and infimums

[timid silo]

thank you!

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[outer moat]

Im a potato when it comes to matrices but I am tasked to find c and d. Any ideas how to do it?

[outer moat]

the best I can do here on my own is calculate the det of A🥰

Its 6, I can do it either by using calculating 1*4 - (-2) = 6 or I can add 2 times the middle row to the last row and get a triangular (or whatever you call it) matrix and the det will be 1 times 1 times 6 = 6

dang. Someone was typing and then stopped

[ivory estuary]

there might be some shenanigans you can pull along the lines of $6I = A^3 + cA^2 + dA$

[warm shale]

Quixz

[ivory estuary]

not sure

[outer moat]

that looks smart (dunno how it helps)

Does it matter from which side I multiply Inverse of A by A tho?

Like from left or right, do I still get identity?

[ivory estuary]

Yes

[outer moat]

hmm

I mean basically I could calculate the inverse of A, then A^2 put that all in the equation and find c and d

but im lazy as hell and I know that there must be a shorter way

[ivory estuary]

surely A^3 is easier than A^-1

[outer moat]

I doubt it

I could use this magic thing I found on youtube and get the inverse of A in a few steps, its a simple matrix

wait

the det of A is 6

and we have 1/6 there

[ivory estuary]

ah

this is now out of my depth

[outer moat]

mine too

I have no idea who that Adj(M) is

[ivory estuary]

https://en.wikipedia.org/wiki/Adjugate_matrix?

[outer moat]

it was a stupid idea ngl

I just wasted 6 minutes to calculate the Inverse

😎

[outer moat]

and I got it wrong too👍

[outer moat]

<@&286206848099549185>

[obtuse pebble]

@outer moat Has your question been resolved?

[outer moat]

<@&286206848099549185>

how am I even supposed to do cA+d

how do I add an integer to a matrix

"Adding a scalar (or number) to a matrix is not defined." (Google)

[ivory estuary]

d * I then

[outer moat]

oh that makes sense

from the answer key I guessed that I need to add d only to the diagonal (or whatever you call it)

thanks

I maually calculated A^2 and A^(-1) and got the answer but imma leave the chat open in case someone has a better way

@ivory estuary thanks for help!

<@&286206848099549185>

[obtuse pebble]

@outer moat Has your question been resolved?

[simple plaza]

After that you'd get one equation from the diagonal term 6 = (something) * c + (something else) * d

And another one from the extra diagonal terms 0 = (something) * c + (something else) * d

Where "something" can be easily computed from A, A^2 and A^3

[outer moat]

Wdym by "diagonal term"

And isnt it 6=A^3+something c + something d?

[simple plaza]

https://en.wikipedia.org/wiki/Main_diagonal

[outer moat]

I know whats a main diagonal, im not sure whats a diagonal term and the link you gave doesnt say anything about it

[simple plaza]

A diagonal term is one of the numbers in the main diagonal

[outer moat]

Oh fair. I thought its something more complex😂

[outer moat]

I still dont understand how we grt an equation from the diagonal term and why you missed A^3 here

[simple plaza]

Wait, c and d are supposed to be 3x3 matrices ?

[outer moat]

No, just numbers

But we concluded that when they add +b in the end they mean b times identity matrix 3x3

Otherwise it wouldnt make sense

[simple plaza]

Yeah, that got me confused as well

[outer moat]

So what about this?

[simple plaza]

You have two options here. Since you only need necessary conditions (if c and d satisfy this equation, then they also satisfy this, then also this, etc.)

You can multiply both sides by A, so that instead of computing A^-1 (not very complex), you'd have to compute A^3 (easy when you already have A^2 since A^3 = A^2 * A)

[outer moat]

Oh I see

Well I did the same thing but I calculated inverse of A and found c and d but I thought there is an easier option

[simple plaza]

And since A is invertible, the equation you get by multiplying both sides by A is actually equivalent to the original

[simple plaza]

Yeah, you only need two independent equations to find c and d

You don't need to compute the full inverse of A, my intuition tells me that evaluating coefficients at row 1, column 1 and row 1, column 2 would be enough. That's probably why someone brought up this formula with the adjugate matrix, since it allows you to compute the inverse matrix coefficient by coefficient and requires inverting 2x2 matrices if my memory is correct

[outer moat]

Well in fact you just need the second rows of all matrices to calculate c and d

Because in the first row we only have 1 0 0 so we only get one equation and two unknowns

[simple plaza]

Yes, this gives 3 equations, 2 of which are probably proportional

[outer moat]

Hmm I guess im just overreacting and trying to find a faster way but in reality this is the best option

Thanks a lot to all of you guys. All the best!

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[flat ginkgo]

Hello, could someone help me with the first one? (it's about winding number).

Without loss of generality, I assumed that the path has domain [0,1]. I came to this:

$n(\overline{\gamma}, \overline{z})=\frac{1}{2\pi i}\int_{\overline{\gamma}} \frac{d\zeta}{\zeta-\overline{z}}=\frac{1}{2\pi i}\int_0^1 \frac{\overline{\gamma'(t)}}{\overline{\gamma(t)}-\overline{z}}dt=\frac{1}{2\pi i}\int_0^1 \overline{\frac{\gamma'(t)}{\gamma(t)-z}}dt$

But I don't know how to continue from here

[warm shale]

robertrs2206

[flat ginkgo]

Ugh, it was hard for me to write that.

[obtuse pebble]

@flat ginkgo Has your question been resolved?

[flat ginkgo]

<@&286206848099549185>

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[timid silo]

Hi

[wary badger]

what’s the question

[timid silo]

My teacher said this graph is prohibited

I don’t get why that can’t happen

[wary badger]

what can’t happen

what about the graph

[timid silo]

Yes he said you can’t make a graph like this

[wary badger]

a parabola?

[ivory estuary]

can you give any more context

maybe which unit

[wary badger]

what was the original question tho

[timid silo]

No, a speed vs time graph that looks like this

[wary badger]

if it’s constant acceleration then he’s right

speed is linear if it’s constant acceleration

displacement is parabolic

[ivory estuary]

mhm

[timid silo]

No question, he was just showing us in general the graphs that you can’t draw because

[timid silo]

No specifications

[wary badger]

what course r u taking

is it a first year physics course

like ap physics 1

[timid silo]

He just said that this can’t happen because this graph says the overall distance travelled decreases as you go on to certain points

[wary badger]

huh

the distance traveling isn’t decreasing

traveled

it’s increasing speed

until

the vertex

maybe that’s y he wrote no

[timid silo]

Idk I feel like he just confused it with something else

[wary badger]

because in the context of the problem it was always increasing speed

[timid silo]

No, as in don’t make it

[wary badger]

if so then yes

because after the vertex the slope is negative

still

and it’s distance is always increasing here

because it’s moving

[ivory estuary]

surely he confused it with something else

having "distance" instead of "speed" makes sense

[wary badger]

this isn’t even velocity

yea

y were u drawing the graph

[timid silo]

Wait he made a another one like this that he approved

[timid silo]

Learning to interpret graphs. Draw out conclusions

[wary badger]

like meaning of slope and area

slope is acceleration for ur graph

area is displacement

[timid silo]

But he had already discussed that before he mentioned this

[wary badger]

yea the second graph is always increasing

but idk

if it’s speed

and it’s still positive

distance is still increasing lol

if ur moving at 10 m/s then go to 5 m/s ur distance is still increasing

[timid silo]

I feel like he just confused it

[timid silo]

Yea

[wary badger]

he did

the y axis should be distance

if he means always increasing

because the second graph is always increasing

the first one isn’t

[timid silo]

Hmm

I agree with you

Thank you so much

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[stone totem]

how can i find $\lim _{x \to \infty }\frac{(1+x^2)}{x*e^{x^4}}$?

[warm shale]

nana

[tardy epoch]

Do you know that exponential grows faster than all polynomials

[stone totem]

yes, but im not allowed to use that 😢

[tardy epoch]

Why not

[fervent cradle]

you could probably use that e^x > x always at least

so e^x^4 > x^4 always

[obtuse pebble]

@stone totem Has your question been resolved?

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[timid silo]

I don't understand this

got to $ab=16$

[warm shale]

mj

[final thunder]

Yeah

now just write 16 as a productsets of products of two integers

[timid silo]

wait is it just any 2 factors that multiply to 16

oh okay

[final thunder]

Yes, but positive factors.

[timid silo]

oh okay thanks

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[final birch]

i dont get the transition from the 1st to 2nd line, there's an extra (x+1) and (x-1) in the denominator now

[finite pollen]

In 2nd step it should be 2x^3 everywhere not just x firstly.

[final birch]

it's done by partial fractions so the coefficients were found

[viscid gull]

You partial fractions (x^2-1)^2 = (x-1)^2(x + 1)^2

Technically it should be of the form

A/(x - 1) + B/(x + 1) + C/(x-1)^2 + D/(x + 1)^2

But I guess A and B work out to be 0

[final birch]

yeah i agree

the language is in dutch but they did this

[marsh geyser]

Aren’t you allowed to sub or this is just the method provided?

[final birch]

there were 2 possible methods it was sub or this yeah

[finite pollen]

Yeah subbing is easier

[viscid gull]

yeah they're equivalent, for instance A/(x-1) = A(x-1)/(x-1)^2 so you'd have (Ax + (B-A))/(x- 1)^2, then just rename B - A to be C or whatever so then you have (Ax + C)/(x - 1)^2

[final birch]

yeah exactly

thank you

[viscid gull]

Np

[final birch]

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[harsh sentinel]

how do I calculate the grade I need to get at least 40% overall grade? the module is split 50% 50% between coursework and final exam

coursework is split into 3 assignments:

assignment1 weighs 40% I got 96.5%
assignments2 weighs 40% I got 78%
assignment3 weighs 20% I got 0%

[shut lagoon]

Let's first calculate your coursework grade :

With the given weights, your coursework grade would be 0.965*0.40 + 0.78*0.40 +0*0.20 = 0.698

So 69.8%.

Now this counts for 50% of your grade, meaning your coursework has so far gathered 34.9% of your total grade. Then the final comes in, and you're missing 5.1% to get to the 40% minimum.

This means that you need 5.1% / 50% on the exam, or 10.2%.

[harsh sentinel]

I don't get the last part so do I need 5.1% or 10.2%? sorry this might seem straightforward to you but it's getting over my head

[shut lagoon]

What I mean is you're missing 5.1% of your TOTAL GRADE, and the final is worth 50%, so realistically, you need 5.1% of those 50%, so really you need 10.2% on the final at the very least to get 40% total.

[harsh sentinel]

that makes sense so if the final is 100 marks I would need to get 10.2 marks to get at least 40%

[shut lagoon]

Yeah that's it

[harsh sentinel]

thanks a lot

[shut lagoon]

No worries.

[harsh sentinel]

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[slim cove]

What have you tried so far?

Also, do you know what R means? It doesn't appear to be labeled on the diagram

[obtuse pebble]

@fallow latch Has your question been resolved?

[obtuse pebble]

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[lethal ridge]

when you factor with two numbers like this x(y) can you factor the number in the equation with either number?

[junior flame]

can you send an example of what you're trying to ask

[lethal ridge]

the red on the left is the factored

or the number being factored

wait no

sorry thats worded wrong

the number they are factoring by

and they are factoring the denominators

[wary badger]

they’re getting a common denominator

x^2-x=x(x-1)

[lethal ridge]

yeah but I was wondering about the brackets in the factoring

x(x-1)

can factor x and x-1?

[junior flame]

no, x and x-1 don't have any common factors

x(x-1) is the LCD of x and x-1

[lethal ridge]

oh

i c

i c

[junior flame]

is that what u were asking

[lethal ridge]

hm

so could I do 3(5) for like 1/3 and 4/5

like in concept

[junior flame]

yes so 3(5) is the lowest common denominator of 3 and 5

[lethal ridge]

ahhh

yay tx

ty

[junior flame]

but not always is the LCD of 1/x and 1/y x(y)

so just be careful

[lethal ridge]

tysm

[junior flame]

np

[lethal ridge]

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[drifting wraith]

you can write it as an equation

suppose P(B) = 0.2, and X is average number of marbles one draw from B is worth
then X = 0.8 + 0.2(1 + 3X)

magically it works

it worked every time so far, no reason for me to question it

[fluid needle]

<@&286206848099549185> any insights?

Humbly calling for help again... <@&286206848099549185>

[obtuse pebble]

@fluid needle Has your question been resolved?

[narrow vault]

i don't see the number of draws you get to start with
and i assume you want to be drawing for as long as possible

[fluid needle]

we start with 3 draws

"Suppose a golden marble has been drawn the last turn, so we have 3 draws."

[viscid gull]

a golden marble just resets to 3 draws left rather than adding +3 riggt?

[fluid needle]

no, it adds +3

[viscid gull]

oh hmm

[drifting wraith]

but we don't get to choose anything

we draw from A exactly once

[fluid needle]

Yeah

[drifting wraith]

doesn;t matter if it's the first draw or second or third

[fluid needle]

and then it's all Urn B

yeah

[drifting wraith]

okay

i have nothing to add

[obtuse pebble]

@fluid needle Has your question been resolved?

[obtuse pebble]

Channel closed

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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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[river gale]

Hello, I want help in understanding the steps.

I have this question:

m = ((r/1200)(1 + r/1200)^N)/((1 + r/1200)^N - 1)*P

And tells me to rearrange the question to give P in terms of m, r and N.

[tall wind]

this looks hard on a first glance

[royal basin]

but in fact isn't

there should be some answer options to follow tho.

[tall wind]

but luckily I'm going to write the big fraction in brackets

[wary badger]

B

just divide by the giant fraction

[royal basin]

!nosols

[obtuse pebble]

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.

[wary badger]

on both sides

my fault ann

[royal basin]

you didn't let op figure it out for himself

[tall wind]

okay, you gave them the answer knief - not good, but I'm going to ask junie why it works

[timid silo]

i thought nosols mean no souls

[wary badger]

lol

[royal basin]

it does not. it's short for no solutions.

[timid silo]

ik i just figured it out

[tall wind]

@river gale

[river gale]

hello

[tall wind]

why would you think B is the right answer?

[river gale]

Oh before that umm I did find out the answer but I'm like watching a maths tutor do it

but the steps he did doesnt quite make sense

[tall wind]

okay, here's something you can do

I'm going to let the big fraction be K

[wary badger]

pretend the giant fraction was any other variable

yea

[tall wind]

so now your equation looks like $m = KP$

[warm shale]

_Kookie

[wary badger]

🧠

[river gale]

im following

[tall wind]

your question says you need to isolate P

what would you do here?

[river gale]

I would divide both sides by K

[tall wind]

very good

$\frac{m}{K} = P$

[warm shale]

_Kookie

[tall wind]

Now let's put the giant fraction back where K is

it's going to look messy but please bear with me

[river gale]

yes sir

[tall wind]

$$P = \frac{m}{\frac{(\frac{r}{1200})(1 + \frac{r}{1200})^N}{(1 + \frac{r}{1200})^N - 1}}$$

oops

[wary badger]

dividing is just multiplying by the reciprocal

[tall wind]

lmao

[wary badger]

just flip it

[tall wind]

I need to correct that

[wary badger]

gotta be hard to write with latex

[warm shale]

_Kookie

[tall wind]

there

that's better

notice that we have m divided by a fraction

[river gale]

yeah

[tall wind]

do you remember how to deal with dividing fractions

[river gale]

multiply the reciprocal

[tall wind]

that's right

$$P = m \times\frac{(1 + \frac{r}{1200})^N - 1}{(\frac{r}{1200})(1 + \frac{r}{1200})^N}$$

[warm shale]

_Kookie

[wary badger]

yo kookie what’s ur rating in chess

chess.com

[tall wind]

that's what happens if you multiply by the reciprocal

[tall wind]

below 1000

[river gale]

riiiight

[wary badger]

oh

[royal basin]

!redir

[obtuse pebble]

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

[tall wind]

junie can you see what's going on here?

[royal basin]

(#chess-go-shogi in this case)

[wary badger]

lol

[river gale]

waikt let me grasp the concept

i se

[tall wind]

let's take a step back

[river gale]

isee

i see whats going on omg

[tall wind]

$$\frac{4}{5} \div \frac{6}{7}$$

[warm shale]

_Kookie

[tall wind]

to compute this

you equivalently write it as $$\frac{4}{5} \times \frac{7}{6}$$

[warm shale]

_Kookie

[tall wind]

which is what we did earier

[river gale]

yeah i see that

its just this part being a complex fraction confused me when its just m divided by

[wary badger]

flip

[tall wind]

yeah, here I can write it as $$m \div \frac{(\frac{r}{1200})(1 + \frac{r}{1200})^N}{(1 + \frac{r}{1200})^N - 1}$$

[warm shale]

_Kookie

[tall wind]

and now you should be able to see where the flipping action comes in

[river gale]

yeah

[tall wind]

right, now, you can agree with me that our work over here looks like answer B doesn't it

[river gale]

yeah

[tall wind]

beautiful

is there anything else you'd like to ask me?

[river gale]

oh jujst one more thing

[tall wind]

sure

[river gale]

this is how the video im watching did it

[wary badger]

same thing

they just cross multiplied

then divided

[river gale]

first i didnt understand how the P got multiplied to the numerator

[wary badger]

kookie method is better

[river gale]

but it actually is just the fraction * P/1 right

[wary badger]

yea

[royal basin]

bad notation on the penultimate line there

[tall wind]

knief, my method is not better, it's only different

[river gale]

sorry whats penultimate haha

[tall wind]

the last step

either way

lemme try and explain this to junie

basically

[wary badger]

well it’s one step vs two

[tall wind]

instead of trying to write the entire giant fraction as K

why not write the numerator as N

and the denominator as D

[royal basin]

the one before the last

[wary badger]

not sure why you would choose a method which makes it longer

[tall wind]

so we have $m = \frac{N}{D}P$

[warm shale]

_Kookie

[royal basin]

they forgor the parentheses around old num and it looks like they multiplied only the 1 by m

[tall wind]

because this method can make more sense in some circumstances

[wary badger]

but it’s slower

making it objectively worse

[tall wind]

now, @river gale, can you isolate P for me again?

[wary badger]

what matters most is what’s most intuitive for them tho

[tall wind]

I think comprehensibility trumps speed in every way

[wary badger]

yes

[wary badger]

^

[river gale]

I divide by D to get N x P
Then I divide N x P by N to isolate P

[wary badger]

me personally i just see it and flip it in my head

KFC

[tall wind]

I think you meant to multiply by D first junie

[river gale]

oh right

[tall wind]

right

now

[river gale]

i multiply d to get N*P

and divide by N

[tall wind]

very good

we now have $P = \frac{D}{N}m$

[warm shale]

_Kookie

[tall wind]

all you literally have to do now is replace D and N with their original expressions

and it will look exactly like answer B

[river gale]

OHH

[wary badger]

is this algebra 2?

[river gale]

(D/N)m is the same thing as (D * m)/N

[river gale]

Student in new zealand studying SAT maths

[wary badger]

facts

[wary badger]

new zealand accents 🔛🔝

[river gale]

would u say sat is easy

[wary badger]

yes

it’s 7th grade level math

basic algebra 1

[river gale]

so thats like middle school

[wary badger]

this is as complicated as it gets

eh well maybe the geometry can be harder

[river gale]

aight bet my mistake was spending my last few years only doing stats haha

[wary badger]

there’s a good amount of stats too

[river gale]

okay thank you everyone my question has been solved

[tall wind]

no worries mate

[wary badger]

always some questions on mean median mode and standard deviation

[tall wind]

have a good one

[river gale]

those are fine io think ive done that a lot

[wary badger]

i think anyone can get an 800 math

[river gale]

you too tysm

yeah i applied for march 9th but theres 2 more after that my deadline is june-ish

and i saw the sat english and it was pretty easy imo

[wary badger]

yea i think the math is easier tho

[river gale]

damn

[wary badger]

i got bored reading the humanities passages

[river gale]

yeah i need to aim for over 1500

[wary badger]

i really don’t care about a mosaic

i fall asleep reading them

[river gale]

hhaaha

i should close this channel tho right

[wary badger]

yea

[river gale]

.close

[obtuse pebble]

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[quartz trench]

We got a solution in the book but it doesn't make any sense

We pretty much have an impossible question unless otherwise

[surreal forge]

i have not tried this problem yet but there appears to be missing information

[quartz trench]

You think so?

[eager niche]

Yeah you only have one number you should need two?

[surreal forge]

yes. with the given information, there exist infinitely many possible areas for PQR. what does your book say?

[fierce lagoon]

Missing information. We need to at least know the length of SQ or BT

[quartz trench]

Alr I'm gonna give the book

This is what it says

[obtuse pebble]

@quartz trench Has your question been resolved?

[obtuse pebble]

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[brisk arrow]

ayy

[fickle turret]

Hi, do you have a question?

[brisk arrow]

.close

[obtuse pebble]

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[obtuse pebble]

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[wanton dune]

I need to find determinant of matrix

[wanton dune]

I tried doing it with pascals expansion but I did something wrong

What?

Lambda is real number

.close

[obtuse pebble]

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[wanton dune]

how did they get last expression from 3th one?

[random burrow]

its a very long process i think, hopefully someone else can explain

[elfin burrow]

,tex .sum2prod

[warm shale]

Tushar

[elfin burrow]

third line

[wanton dune]

ah shit those frickin sum formulas

but i see it now thx

.close

[obtuse pebble]

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[gloomy vector]

im trying to check if my answer is correct and i got 18

[timid silo]

i have a question

[gloomy vector]

if the area of the shaded region is 1, then what is the area of the hexagon

[versed stratus]

!occupied

[obtuse pebble]

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).

[timid silo]

wait how do i make my own room

[versed stratus]

go to one of the help channels and type your question

[timid silo]

help-0?

[versed stratus]

yes

[gloomy vector]

im trying to check my answer, i got 18

[gloomy vector]

#❓how-to-get-help

oh wait nvm, the teacher said my answet was correct

.close

[obtuse pebble]

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[alpine gull]

Is this a correct way to check the divergence of a series?

[alpine gull]

I thought that it made sense since for odd numbers an = 0 so it doesn’t impact the sum

[obtuse pebble]

@alpine gull Has your question been resolved?

[obtuse pebble]

@alpine gull Has your question been resolved?

[obtuse pebble]

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[fringe spear]

I tried to solve $\int_{0}^{1} \frac{1}{1 + x^2} , dx$ using u sub and substituting u as $1 + x^2$ and im getting 2 * ln|2| but when i checked my answer online it said 1/2 ln|2|. how do i solve this?

[warm shale]

OptimusWaffle🧇

[inland matrix]

How did you proceed after taking u as that?

[fringe spear]

i calculated u to be 2xdx

[inland matrix]

du*

[fringe spear]

yeah

ill just write it down and send it

[inland matrix]

Sure

[rocky flicker]

5

[inland matrix]

,rcw

[inland matrix]

dx is du/2x tho

[fringe spear]

so then what do i do?

[surreal forge]

this is not a u-substitution

[inland matrix]

Yeah

[fringe spear]

how come?

[inland matrix]

x=tan theta will be best

[surreal forge]

you simply use the fact that the integral of 1/(1+ x^2) is arctan(x)

[fringe spear]

then you will get pi/4

so thats it?

[surreal forge]

yes

[fringe spear]

ok thx

how do i close this?

[brazen gorge]

type ".close"

[fringe spear]

.close

[obtuse pebble]

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[tame violet]

7m + 10n = 25k
5n - 4m + 13k = 0
m, n, k, are prime numbers

[inland matrix]

You've to find m,n,k is it?

[robust bloom]

write 5m - 4m = -13k

[tame violet]

Sup space

[robust bloom]

Gud hbu?

[tame violet]

Im stuck again

[robust bloom]

after this you can solver the two equation

[tame violet]

Yes sir

[tame violet]

Wait why did it come to that

[inland matrix]

If you multiply the second equation with 2 and subtract it with eqn 1, you get
15m=51k

[tame violet]

Then what

[inland matrix]

m/k = 51/15 = 17/5

We can take m=17x and k=5x for some real number x
But it's given that m and k are prime

[tame violet]

So whats the possibke step from here?

[inland matrix]

Ideally you need 3 equations to solve for 3 variables, but in this case the other info is that m,n,k are primes

[brazen gorge]

if $m, k$ are prime, that means that $m$ and $k$ are co-prime, which means that $\frac{m}{k}$ will be in its reduced form, thus $\frac{m}{k} = \frac{17}{5} \implies m=17, k=5$

[warm shale]

artemetra

[surreal forge]

... which then implies n = 0.6

[tame violet]

Decimals

[surreal forge]

i believe then the problem is flawed 🤔

[tame violet]

Qh

[tame violet]

This is a flawed equation?

[surreal forge]

you posted two different problems

[tame violet]

Wdym

Aight lets change it

.close

[obtuse pebble]

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[timid silo]

I need help with this geometric sequence

I have no idea where to go from here or if this is even correct so far

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[wintry swift]

the area of ABC is AB x LC/2 = AB * s / 2.

So you need the length of AB, which you should be able to express as geometric series.

[timid silo]

thank you do you know any sites where i can look up how geometric series work or how i can express them? I have never worked with them

[wintry swift]

a geometric series is something like 1 + q + q^2 + q^3 +...., which means a sum of terms where where the factor between two such (following) terms is a constant factor (like q in my example).

wiki is a good starting point. https://en.wikipedia.org/wiki/Geometric_series

[timid silo]

Is this the correct form ?

[timid silo]

Thank you :)

[wintry swift]

i would assume its (c^2+s^2)/c = a and s^2/c^2 = q and then a+aq+aq^2+aq^3+... which is nearly your formula, with one change: with my terms you would get a s^6/c^5 (instead of c^6)

[timid silo]

yes thank you i didnt look right

[timid silo]

is it correct to write it like this?

[timid silo]

I think I went in the right direction but made a mistake along the way…..

[wintry swift]

instead of the las step i would cancel one c in the numerator and the denominator.

you are very near to the result then. only c^2+s^2 is too much. hmmm, let my think about it.

[timid silo]

another problem is, i still need to multiply it by s ..

[wintry swift]

no thats ok.

you have a (c^2+s^2)c/(c^2-s^2) mutlipy this with s/2 then you have the result from your example, with only c^2+s^2 is too much.

[timid silo]

okay but how did i mess the c^2+s^2 up :(

[wintry swift]

as i said: let my think about it.

look at the first trianlge. and you will know what c^2+s^2 is.

your result is right 😉

[timid silo]

its 1!

[timid silo]

thank you so much youve been such a big help

.close

[obtuse pebble]

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[wintry swift]

youre welcome

[obtuse pebble]

Available help channel!

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[mild osprey]

Can someone explain this to me

[tender tusk]

which parts?

[mild osprey]

the whole thing i couldnt understand it

[tender tusk]

aight do you know what sin cos and tan are?

[mild osprey]

yes

[tender tusk]

aight

aight so we know that sin = opp/hyp , cos= adj/hyp and tan = opp/adj

u fine with these?

[mild osprey]

yes

[tender tusk]

ok now lets take sin/cos

that is equal to (opp/hyp)/(adj/hyp) right

from there we can cancel both hypotenuses out

and we are left with opp/adj

which is equal to tan

is that understandable?

[mild osprey]

yes

[tender tusk]

now when cos is equal to 0 we would have sin/0, anything divided by 0 is undefined so we say tan is undefined at cos = 0 or cos x = 0 in this case.

[mild osprey]

okay

i am starting to understand

[tender tusk]

now we are looking for when cos x is equal to 0 to find out when tan is undefined

do you know when cos is equal to zero?

[mild osprey]

no

[tender tusk]

hmm, have you seen the sin and cos graphs?

[mild osprey]

yes

[tender tusk]

here we can see that cos is 0 at 90 and 270 right

[mild osprey]

yes

[tender tusk]

we know that cos repeats every 360 degrees right

[mild osprey]

yes

[tender tusk]

that means if we add 360 degrees to 90 and 270, cos will still be zero, that means cos is 0 at 90, 270, 450(90+360), 630(270+360)..... etc.

[mild osprey]

okay

[tender tusk]

if we look at those numbers, the gap between each number is 180, that means that we start at 90 and we keep adding 180 cos will still be 0

[mild osprey]

okay

[tender tusk]

so hence we get that tan x is undefined at x = 90+180n where n is an integer, basically what that means is take the number 90 for every integer (whole number) that we replace n with, we will get a value of x where tan is undefined, lets take 0 for example x = 90+180(0) = 90 which we know is true, now lets take the number 2, x = 90+180(2) = 90+360 = 450, which we know is another value where tan is undefined

is that understandable?

[mild osprey]

yes thank you so much

wallah u explain better than my teachers

[tender tusk]

lol thx, i think ur teachers will do a great job, just ask some questions if u get stuck.

[mild osprey]

trust me i did and they explain like toddlers

[tender tusk]

oh did u need asymptotes explained?

[mild osprey]

yesssssss

[tender tusk]

ok

[mild osprey]

the period the asymptotes

the trasformation

x-intercept

midline

graphing

[tender tusk]

aight

[mild osprey]

Alot of stuff but just explain every one a bit and its okay if u couldnt

[tender tusk]

don't fret about why it looks like this for now

most importantly look at those dotted lines

[mild osprey]

okay

yeah

thats the asymptote

[tender tusk]

yeah, tan is undefined at those points, which is impossible for us to graph, since how can you graph something that has no set value, since its not graphable we draw dotted lines in their place to indicate where tan is undefined, which we named the asymptote

we know that tan is undefined when cos = 0, so we know where the asymptotes are located, we just have to know when cos is 0

[mild osprey]

okay

that i understand

[tender tusk]

the period of anything just means how long until that one thing starts to repeat itself

are we looking at the period of sin, cos and tan?

well lets start at tan first, since its the easiest to see, with tan we can start at any asymptote, take 90 for example, when does the asymptote repeat itself? At 270 there is the next asymptote, we know that the distance between 90 and 270 is 180, so every 180 degrees tan starts to repeat itself, so it is said that it has a period of 180

you can also start at 0, you see that at 0 tan is touching the horizontal line, the next time is touches the line again is at 180, which means it has repeated itself, yet again we can see the distance from 0 to 180 is 180 and therefore we can see again that tan has a period of 180

[tender tusk]

you said you already knew this so hopefully you can see that cos has a period of 360 degrees, since it repeats itself every 360 degrees, same can be said with sin

are you aware of what x intercepts are in general?

[mild osprey]

yes

[mild osprey]

ye

okay now i understand the cos and sin but since tan is a bit different i need help with that

[tender tusk]

aight hopefully you understood that tan period explanation

the x intercepts of tan are when tan is equal to 0, the only way for tan to be equal to 0 is when sin is equal 0 since tan = sin/cos and if sin was 0 then tan = 0/cos and we know that 0 divided by any number is still 0, if you know the x intercepts of sin then its just the same for tan

[mild osprey]

okayy

[tender tusk]

mid lines are like the name suggests the line that runs through the middle of the graph, with sin and cos the mid line will sit halfway between the highest point and lowest point, on sin its highest point is 1 and lowest is -1 half way between both of them is 0, this is the same with cos, the mid line with only change if a transformation has been applied to it

now onto transformations

im assuming u have seen transformations before

so then we'll just look at the basic forms

[mild osprey]

okay

[tender tusk]

y=Asin(x-B)+C, hopefully from previous knowledge of transformations you will know what a lot of these do, A is a vertical dilation and therefore will strech or compress sin

you can see it here

B will shift it left or right, remember that if the sign is negative it moves to the right, and the positive will shift it to the left

C slides everything up or down, relatively simple

now for graphing, unfortunately the best way to graph them is to memorise the original graph

oh i forgot one of the transformations

horizontal dilations, probably the most important one too

they occur sin(Dx), just before the x

these will actually change the period of the graphs

the red line is the original y=sinx and the blue line is y=sin(2x)

you can see that the 2 has made the graph more compressed, to be more precise its twice as short

sin normally repeats itself every 360 degrees or 2π, but we can see with y=sin(2x) since its twice as short it repeats itself every 180 or π instead

to find out what the period is, we take the number to the left of x and divide the original period by it, for example y=sin(2x), the original period of sin is 2π or 360, we can then divide it by 2 to get π or 180. Another example is tan(x/2), this can be rewritten as tan(½x), tan's original period is 180 which we can divide by ½ to find our new period, 180/½ = 360 or 2π

[obtuse pebble]

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[obtuse pebble]

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[winged mica]

<@&286206848099549185> do this

[royal basin]

!noans

[obtuse pebble]

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

@winged mica Has your question been resolved?

[sharp forum]

there are 2 nested clusters

[winged mica]

Can you please elaborate this

[obtuse pebble]

@winged mica Has your question been resolved?

[obtuse pebble]

@winged mica Has your question been resolved?

[meager mural]

@winged mica what have you tried so far? What is the root of the dendrogram?

[obtuse pebble]

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[winged mica]

idk i didn't read clustering this one from a sample qs

[winged mica]

and tomorrow my exam

[meager mural]

Can you review your course notes for what a dendrogram is?

[obtuse pebble]

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[merry wave]

my question is why isnt it adding the 22.5, but subtracting it

if we are trying to find the original cost, wouldnt you add it

[obtuse pebble]

@merry wave Has your question been resolved?

[tardy epoch]

216.68 is the amount with the coupon

coupons give a discount

[obtuse pebble]

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[wanton kraken]

so I have ψ(x1e1+x2e2+x3e3+x4e4)=(6x1+x2)e1-(6x1+x2)e3+21(x1+x4)e4

[wanton kraken]

and I have to find the basis of the kernel

I did (1, 0, 0 ,0), (0, 1, 0, 0), ...

and got matrix
(6 1 0 0)
(0 0 0 0)
(-6 -1 0 0)
(21 0 0 21)

this is the matrix for ψ, right?

to find the basis of ker(ψ) I have to simplify the matrix and get fundamental system of solutions

right?

<@&286206848099549185>

[obtuse pebble]

@wanton kraken Has your question been resolved?

[wary vigil]

$\begin{pmatrix}6 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 \ -6 & -1 & 0 & 0 \ 21 & 0 & 0 & 21 \end{pmatrix}$

[warm shale]

Katharine

[wanton kraken]

yeah?

first I have to RREF the matrix, right?

[wary vigil]

just made it a bit more easy to read :D

[wanton kraken]

oh x)

[wary vigil]

to find the kernel you find the vectors that when psi is applied give 0

i'm not sure if RREF is a way of doing that sorry

[wanton kraken]

yeah I mean you're right

I was just told to RREF it first

[wary vigil]

ok

[wanton kraken]

got this:

the way I understand it

[wary vigil]

that seems right

[wanton kraken]

I make it a homogenous system of equations

I have 2 equations and 4 variables

that's 2 parameters

and substitute, say, x4=q and x3=p

right?

then find the other parameters

[wary vigil]

you have x1 + x4 = 0 and x2 - 6 x4 = 0

if i'm not mistaken

[wanton kraken]

yeah

oh

[wanton kraken]

so two equations, two variables

no parameters

[wary vigil]

3 variables

x1 x2 x4

[wanton kraken]

oh yeah

3 variables

so x4=p

x2=6p

x1=-p

we get

(-p, 6p, p)

then I have to test for (1, 0, 0) (0, 1, 0) (0, 0, 1)

right?

[orchid mason]

Can you guys help me

😭😭

I forgot how to solve this and later is the deadline

I left my notes on school

[wary vigil]

You should open a new one

[orchid mason]

Im struggling about the new topic

It's really hard

[wanton kraken]

you can take a free channel

there's plenty

this thread is already deep into

[orchid mason]

Where

Oke

[wanton kraken]

people probably won't read

it's just me and @wary vigil here

[wanton kraken]

so I have to test for (1, 0, 0) (0, 1, 0) and (0, 0, 1) right?

[wary vigil]

i'm sorry i don't know what you mean

you know now that the kernel of psi

is of the form

$$- p \hat{e}_1 + 6 p \hat{e}_2 + (ANYTHING)\hat{e}_3 + p \hat{e}_4$$

[wanton kraken]

(1, 1)
(1, -6)?

[warm shale]

Katharine

[wary vigil]

there it is

[wanton kraken]

oh

e3 is 0

[wary vigil]

?

the bit in front of e3 becomes 0 when psi is applied

always

that's part of what you show by solving $\psi \vec{x} = \vec{0}$

[warm shale]

Katharine

[wary vigil]

recall

[wary vigil]

this

when you apply that to x, y, z, w

the vector

[wanton kraken]

🤔

[wary vigil]

you will get 0 z

0z in all of the equations

0 * z

and so what that means

like when you calculate the eigenvectors

is that there is a freedom in choosing z

for example if you get the equation

x + 0 * y = 0

it means the eigen vector is

(0, ANYTHING)

the important bit is

the kernel of psi

[wanton kraken]

so basis is (-1, 0, 0, 1) and (0, 1, 0, 0)

[wary vigil]

has to have p in the 4th position

-p in the first

6p in the second

and if that is the case

[wanton kraken]

ohh

yeah

[wary vigil]

then if you apply psi to it

you get 0

aka it is in the kernel

i hope i haven't confused you more

[wanton kraken]

so (-1, 6, 0, 1) is the basis of the kernel

[wary vigil]

no

as (-1, 6, 1, 1) is in the kernel but cannot be found

as a combination of that basis

[wanton kraken]

alright but like

isn't it like

x1+x4=0
x2-6x4=0

x4=p

then x2=6p

x1=-p

(-p, 6p, 0, p)

x3=0

[wary vigil]

remember that the kernel is all vectors that are mapped to 0

[wanton kraken]

yeah

[wary vigil]

that includes more than just x3 = 0

[wanton kraken]

right, but what does it have to do with the thing above?

[wary vigil]

a basis has to be able to represent all elements does it not?

[wanton kraken]

yes

[wary vigil]

these are all part of the kernel of psi

(-1, 6, k, 1)

[wanton kraken]

right

[wary vigil]

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