help-39

[rough forge]

all numbers that are in 0 < y < 1 are numbers greater than 0 and smaller than 1

[short orchid]

ohhhhh

so we use the second case

omg i get it

btw why does the sign flip again

bcs i divide both sides by -8?

[rough forge]

yes

[short orchid]

OHH

OMG YAY

[rough forge]

multiplying/division by negative

flips

[short orchid]

so i can use this method to find range?

[rough forge]

yea

[short orchid]

like the method of making y the subject

omg yayayayy

[rough forge]

yes

[short orchid]

thank you so much

im in tears

[rough forge]

🦕

[short orchid]

🤩

[rough forge]

https://tenor.com/view/cat-cuddle-gif-27222129

[short orchid]

thanku so much again

thank you so much too revita

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[river lodge]

Why are we doing this

To determine the phase shift

Can someone please explain

<@&286206848099549185>

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[short orchid]

hi guys, "find the total number of different arrangements of the 8 letters in the word TOMORROW that have an R at the beginning and an R at the end, and in which the three Os are not all together."

[short orchid]

can anyone pls explain why does "R at beginning & end" - "R at beginning & end and 3Os together" is the answer?

what is complement

[pseudo oxide]

because

opposite, essentially

[short orchid]

ohh yes

[pseudo oxide]

all the arrangements w/o the 3 O's

and all the arrangements w/ the 3 O's

must make up the entire set

[short orchid]

wait 3O's at the end?

like R's at the beginning and end?

[pseudo oxide]

mb that's a typo

and yes just assume i mean it begins and ends with R's as well

[bronze coyote]

like ROOO---R, R-OOO--R, R--OOO-R, R---OOOR

[short orchid]

oooo yes

[short orchid]

wait i still dont get it

i have a hard time imagining numbers

[inland ivy]

Inclusion-Exclusion Principle

The negation of "the three O's are not together" is "the three O's are together

So you're calculating (Total cases) - (Unfavorable cases)

[short orchid]

ohhhhhh

so we consider the "R at beginning and end" as the total case?

[short orchid]

thank you so much

[short orchid]

thank you so much

[short orchid]

thank you so much

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[rough forge]

https://tenor.com/view/cat-cuddle-gif-27222129

[pearl pond]

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[golden jetty]

could someone help me sketch this graph

[golden jetty]

i know i did this so wrong

[compact ridge]

In other words when x < -3

The slope is getting steeper and steeper upwards

Then it gets less steep but the function is still increasing

[steep saddle]

x < -3
slope is increasing and f'(x) > 0 so f''(x) > 0
smiley face curvature
-3 < x < -1
slope is decreasing f''(x) < 0
frowny face curvature
x = -1
horizontal tangent
x > 1
derivative = slope, slope is decreasing f''(x) < 0
frowny face curvature

[golden jetty]

so smiley face curvature is concave up?

i think

[golden jetty]

where is it getting less steep at? x<-3 also?

[compact ridge]

No

Just -3 < x < -1

[golden jetty]

oh

[golden jetty]

so if a function is increasing, it doesnt necessarily mean that its slope is increasing also?

[compact ridge]

Exactly

[golden jetty]

so increasing just means positive? above x axis

or

[compact ridge]

Just compare y = x^2 and y = sqrtx

[compact ridge]

The function is going up

It can start anywhere

[golden jetty]

ohhh

so when it goes up thats concave up

right?

[compact ridge]

Above x-axis means y > 0

[compact ridge]

No

Concave up is like a U shape

That's when y'' > 0

[golden jetty]

oh

[compact ridge]

https://m.youtube.com/watch?v=iOvorLc5NgQ

You might want to watch this

[golden jetty]

thank you

i understand it its just graphing is hard for me since

these questions are too conceptual

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[glass remnant]

Why isnt there a rel min at x=0, and why isnt there a rel max at x=2

[glass remnant]

im very confused

the blue is my highlighting and the red and black is my teachers work

[summer imp]

I think they just don't count absolute extrema as relative extrema

It depends on the definition, although I agree with you that technically, x=0 and x = 2 should be relative extrema as well in the usual sense

[glass remnant]

so if i wrote them down i wouldnt get marked wrong?

[summer imp]

Again it depends on their definition, so you should probably check with them to be sure about marking

[glass remnant]

alright thanks for the clarification

il ask my teacher tommorow

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[short scaffold]

how would i prove the following

[short scaffold]

if all the elements in the nth row of pascals triangle

except the first and last element

are divisible by n

then n is prime

spent alot of time and got no where

[rose robin]

have you tried using the binomial coefficient?

[short scaffold]

sort of

not rlly

u mean using coeffecient of (a+b)^n?

[rose robin]

you can arrange the binomial coefficients to form a Pascal's triangle

idk how to latex this so

the n here is the row
the k is the column, how much to the right, k=0 is the leftmost, k=n is the rightmost value in the triangle

here's a better picture

[pearl pond]

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[sacred hollow]

Help with question 2 please

[sacred hollow]

,rccw

[compact ridge]

oh you know that $\det(AB) = \det(A) \det(B)$ right

[jolly parrot]

south, just south

[sacred hollow]

Ye

So, |A^2| = |A||A| which is -4.-4 ?

So 16?

[compact ridge]

yep!

[sacred hollow]

Thank u!

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[compact ridge]

no worries

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[tepid ice]

At the end of yesterday's soccer game between Team Why and Team Zed, Team Why had scored 3 goals and Team Zed had scored 2 goals. At half-time of the game, Team Why had scored y goals and Team Zed had scored z goals. If y≥0 and z≥0, how many possibilities are there for the ordered pair of integers (y,z)? (In soccer, each team's score is always a non-negative integer that never decreases as the game proceeds.)

[tepid ice]

I calculated 12

can someone confirm

[dusk horizon]

Yes. Is that all? What made you doubt this?

[tepid ice]

I thought it could be 7 as well

I wasn't sure; but thanks for the confirmation

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[dusky imp]

Hello , im currently doing practice paper as the exam are close . So there is this question that i dont know how to do, specifically

How does the curved line and the straight line information can be used to solve the problems,i cant see where it connects

[pearl pond]

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[junior vapor]

i dont get the highlighted part, i thought for the first part i already proved that it's true for when n is a whole number like whats the difference?

[vital estuary]

so youre saying that if n is a whole number, 8n²+14 is divisible by 4?

[full lotus]

cant you just prove that false by just putting n = 1

14+8 is 22

5.5

[junior vapor]

i hv no idea what im trying to say

[junior vapor]

oh that makes sense actually

alright thanks guys

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[lone fox]

can i get help finding the period for this or just the equation, i've gotten amplitude of 2, centerline at y=3 but im stuck

[gilded loom]

Ok so

Where is the first maximum in this figure?

@lone fox

[lone fox]

(pi/8, 5)

[gilded loom]

Mhm

Next max at?

[lone fox]

3pi/8, 5

[gilded loom]

Mhm

Subtract both x coordinates

[lone fox]

to get pi/4

[gilded loom]

Yes

That will be the period

[lone fox]

alright got it

jsut 4 more questions and i can go to sleep its 2 am

[gilded loom]

It's 4pm for me lol

Mhm

[lone fox]

can you help me speed through these i just wanna sleep

[gilded loom]

Mhm

What is amplitude?

[lone fox]

2

[gilded loom]

Yep

[lone fox]

phase shift i think is 6pi

[gilded loom]

Yes

[lone fox]

no verticle shift

[gilded loom]

Mhm

Period?

[lone fox]

not sure

since theres no 2 peaks

[gilded loom]

Ok so they said it's sine right

[lone fox]

is it 24pi?

[gilded loom]

Yepp

[lone fox]

since half is 12pi

[gilded loom]

Yes

Correct

[lone fox]

so b is 2pi/24pi

or pi/12

[gilded loom]

Yes

[lone fox]

so it would be y=2sin(pi/12x)?

[gilded loom]

Yepp

[lone fox]

desmos is showing a differnet graph

[gilded loom]

Wait

Yes

[lone fox]

oh if its whole number pi

nvm

[gilded loom]

It's x/12

Not πx/12

[lone fox]

pi/12x

[gilded loom]

What is this?

1/12 right

[lone fox]

ye

[gilded loom]

So sin(x/12)

Amplitude is 2

So 2sin(x/12)

[lone fox]

oh

[gilded loom]

Yep

[lone fox]

pi cancled out

[gilded loom]

Yes

[lone fox]

its a cos graph tho

idk why my teacher keeps labeling the wrong things

[gilded loom]

Ah yes

Cos(x/12)

Even I didn't check mb

[lone fox]

its fine this is the second question labeled wrong on this paper

[gilded loom]

Hmm I c

[lone fox]

last one

has amplitude of 1

[gilded loom]

Or maybe he wants it in terms of sine?

[lone fox]

maybe but idc

i just wanna finish and sleep

[gilded loom]

Lol ok

[gilded loom]

Yes

[lone fox]

phase shift 3 i think

with a verticle shift of 2

[gilded loom]

Idk this man

Idk what's the point where it's touching y axis

[gilded loom]

Yes

[lone fox]

with period of 6

[gilded loom]

Yes

[gilded loom]

-1?

Move 1 unit to the left

[lone fox]

it can be negative?

[gilded loom]

Idk prolly

[lone fox]

how do imake it more stretched

which one changes for that

[gilded loom]

Wdym?

Just write down how u did it for the previous question

And add the vertical shift thing

[lone fox]

y=cos(pi/3x)+2

but it doesnt work fully

[gilded loom]

What's the issue?

[gilded loom]

Phase shift

U forgot

[lone fox]

how do i use it

[gilded loom]

Put it inside the angle

[lone fox]

how

im confused

[gilded loom]

Wait a sec

[gilded loom]

So u have this

I told phase shift is -1

So put it in the thetha

Instead of πx/3

[lone fox]

how did you find -1

[gilded loom]

Make it πx/3 - 1

[gilded loom]

Look here

This thing will look like cos if u move it 1 unit to the left

Agreed?

[lone fox]

yeah

since it starts at the max

[gilded loom]

Yep

So 1 to the left is -1

[lone fox]

finally done with that

2 word problems

[gilded loom]

Nice

[lone fox]

and i can get like 4:30 hours of sleep hopefully

[gilded loom]

Hmm

[lone fox]

first one was easy

second one has to do with bearings so gonna be a lil tricky

[gilded loom]

I c all the best

[lone fox]

thanks for your help

[gilded loom]

Anytime!

[lone fox]

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[compact ridge]

,w square of matrix {{0, 1, 0}, {1, 0, 0}, {0, 0, 0}}

ah shit

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by sooouth.

[agile ridge]

Uhm

Wdym..?

[pearl pond]

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[hearty gate]

for 3 and 4

Since 3.) is just positive 2

if going from right to left we get values like 2.1, 2.01, 2.001

which is infinitely bigger and bigger

then 3/infinity -2 = +inf

thats how it works?

pls @ me when u reply

[pearl pond]

@hearty gate Has your question been resolved?

[pearl pond]

@hearty gate Has your question been resolved?

[sage escarp]

I don’t really understand what you have done but when you get that if you substitute it and denominator gets 0, then you will need to study it separately: first x->2+ and x->2-

X when approaches to 2 from the positive side and negative side

And if you calculate that you will find out that one is +infinity and the other -inf. therefore limit does not exist

[hearty gate]

I see

That makes sense

Thanks for the clarification

Have a nc one

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[noble knoll]

not sure how to solve this

[autumn fossil]

So you need to find the probability that the business doesnt close, given that it was started by a woman

Try writing everything you were given in the right notation

[pearl pond]

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[dawn arrow]

So i know i should use washer method for this but i have no idea how to set this up this is what i have so far

[pearl pond]

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[gleaming moss]

Sketch the ring

[pearl pond]

@dawn arrow Has your question been resolved?

[pearl pond]

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[lament parrot]

The example is using 7 letter words.
The first one is easy, but i am having trouble in the second, there are too much possibilities idk

[plush bramble]

what's a "boring word"

[lament parrot]

A word whose all letters are the same

Sorry i forgot to say that

[plush bramble]

what else is given about U

[lament parrot]

U is just a word

Any

In this case, with 7 letters

[autumn fossil]

I'd try a contradiction

[plush bramble]

Yea this is easy then?

[lament parrot]

Yeah i was trying but idk

[lament parrot]

Yeah it should

[plush bramble]

U = s(U) = s^2(U) = ... = s^6(U) and compare 1 letter at a time

[lament parrot]

No its not for all r

Just for one

I think

[plush bramble]

nah

[lament parrot]

It would not make sense

The second would just be the same as the first

[plush bramble]

oh i'm wrong

i guess prove it for r=1 first then generalize

[autumn fossil]

are you?

[lament parrot]

Yeah for 1 is the first

[autumn fossil]

once you prove the first one, then the second one literally follows from it

[lament parrot]

How

[plush bramble]

haha yea

[autumn fossil]

the seconjd one includes s(U) = U as one of the assumptions

[lament parrot]

Hmhmhm

[autumn fossil]

so if you prove that s(U) = U -> U is boring, you have also proved the second one

[tropic saddle]

its meant that s^r(U)=U for some r

not necessarily r=1

[lament parrot]

Why

[tropic saddle]

and thats also enough

[autumn fossil]

oh that would make more sense

yeah

[lament parrot]

Yeah

[tropic saddle]

as a hint, consider where the first letter ends up

and then apply s^r again

[autumn fossil]

I'd try the first one first

[lament parrot]

I did

The problem is for r

[ancient robin]

yeah since 7 is prime you can find r^k=r^1, where k is a multiple of 7

[autumn fossil]

how did you do the first one?

[lament parrot]

Hmm but what if the first letter when it ends is the same as the first in the start

[tropic saddle]

?

[autumn fossil]

where k is a multiple of 7?

[ancient robin]

I mean to say r is a multiple of k my bad

[lament parrot]

If there is at least one different letter, when it moves it will go to a different spot, but anyways i think it is not right still

[tropic saddle]

lets say the first letter is a, the other we dont know

I'm gonna write those as x

[ancient robin]

fuck im misspeaking like a dummy. I meant to say that k is a multiple of r

[tropic saddle]

so axxxxxxx

and say r=2

[lament parrot]

Whaat 😭

[tropic saddle]

what is s^2(my word)

[lament parrot]

Ok

xxaxxxx

[tropic saddle]

good and we know thats the same as the old one

[lament parrot]

So x = a and a = x

Thats it?

[tropic saddle]

I put x as placeholder for all of them, they arent necessarily equal

[autumn fossil]

x is a placeholder, not a letter

[tropic saddle]

but the point is, the first letter needs to be a again

so axaxxxx

[lament parrot]

Oh yeah

So the first x = a

[ancient robin]

Since 7 is prime. For any such r you can find t, such that s^(rt)=s^1

[tropic saddle]

what if you shift it again

[lament parrot]

How

[lament parrot]

Hmhm

Idk what answer you want

[ancient robin]

Z7 is a field, so every element has a multiplicative inverse

[lament parrot]

Indeed

Oh you mean congruence?

[ancient robin]

yeah

So rt is congruent to 1 mod 7

[lament parrot]

I thought it was literally equal

[tropic saddle]

we found out that U=axaxxxx

what is s^2(U)

[lament parrot]

Yeah

xxaxaxx?

[ancient robin]

well, the S function applied rt times is literally equal to the function s

[tropic saddle]

no

[lament parrot]

Yeah it happens each 7 times

[lament parrot]

Why not?

[ancient robin]

so if r=1, t=1, is trivial
r=2, t=4, so s^8(U)=S(U), so by the first statement, U is boring

[tropic saddle]

previously the a's where one letter apart

now they are two

[lament parrot]

Breh yeah sorry

[tropic saddle]

ok good

but again we know that s^2(U)=U

so we can again compare axaxxxx and xxaxaxx

[lament parrot]

Yeah

[tropic saddle]

so what now

[lament parrot]

Sorry i dont understand 😭

[lament parrot]

First x is a

[tropic saddle]

good. so U=axaxaxx

and we go again

what is s^2(U)

[lament parrot]

xxaxaxa

Yeah and so on

[ancient robin]

Say r=2.
Then S^2(U)=U. Apply S^2 again to get
S^4(U)=U. Apply S^2 again to get
S^6(U)=U. Apply S^2 one more time to get
S^8(U)=U.

[lament parrot]

But what if an a goes to the first slot again

[ancient robin]

Now you can realize that S^8(U)=S(U), since U has 7 letters

[lament parrot]

Yeah

[ancient robin]

So now you know that S(U)=U, so U is boring

by the first theorem

[lament parrot]

Oh wow

But can you help me to prove the first theorem rigorously? I just barely proved it

[tropic saddle]

do what we did

but for r=1

[lament parrot]

But i need to generalize for the next

I cant do it say p times

And idk if an a is not going to the first slot

Instead of an x

[ancient robin]

To generalize, you will have to use S^r t times, where t is the multiplicative inverse of r mod 7

But you know you can do that for every such r, since 7 is prime so Z7 is a field

[lament parrot]

Yeah but i am talking about the first

[ancient robin]

Oh sorry

[lament parrot]

My proof was bad

Its ok

I just thought about the case where one letter is different

But what if there are more

[lament parrot]

Oh so that is why the number must be prime

[ancient robin]

Exactly

Well. What I did was call U a 7 diimensional vector U=(a1,a2,a3,a4,a5,a6,a7)

and S(U)=(a7,a1,a2,a3,a4,a5,a6)

And once you start comparing indices, transitivity does all the work for you

[lament parrot]

What is a 7 dimensional vector 😭😭😭

[ancient robin]

oh, nvm then. Just think of U as a list of ordered letters, as in how I wrote them above

U=(a1,a2,a3,a5,a6,a7)
S(U)=(a7,a1,a2,a3,a4,a5,a6)

So if they are the same:
a1=a7
a2=a1
a3=a2
a4=a3
a5=a4
a6=a5

By transitivity, all of them must be the same

[lament parrot]

That is pretty smart

But why cannot we do that for s_r?

[ancient robin]

the transitivity breaks, since you cannot make a "transitivity chain" so to speak

[ancient robin]

because the advantage we have for r=1 is that every next line contains an element from the previous line

but if we do for example r=2 we get
U=(a1,a2,a3,a5,a6,a7)
U=(a6,a7,a1,a2,a3,a4,a5)
a1=a6
a2=a7
...
etc.
So we cannot make use of the transitive property

[lament parrot]

But the order does not matter, does it?

[ancient robin]

It does

[lament parrot]

You can just rearrange the equalities no?

[lament parrot]

Why?

[ancient robin]

oh yeah you can do that

I just went in order for simplicity

[lament parrot]

Yeah so transitivity would work no?

[ancient robin]

Umm, not sure, maybe it does

Let me try and build this transitive chain

[lament parrot]

Yeah it works

[ancient robin]

Yep, you can still build the transitivy

[lament parrot]

You have ak = aj = ai and so on

But the multiplicative inverse is more elegant lol

You are so smart 😮

[ancient robin]

Yeah, I'm just not so sure about the background of your course

[lament parrot]

Yeah i learnt that

Thank you so much you all guys!!!

[ancient robin]

np mate!

[lament parrot]

Also no idk if the transitivity would work

Because if the number of letter is not prime that is not true sometimes

Like for 6

[ancient robin]

yeah it should work for prime numbers

Otherwise you would get cycles

[lament parrot]

Yeah

[ancient robin]

disjoint cycles

[lament parrot]

So we probably missed something

[ancient robin]

no, no, it just works for 7-letter words, so the proof still works

[lament parrot]

Yeah for 1 it would work

[ancient robin]

You wouldn't find an r that would create such cycles, since 7 is prime

[lament parrot]

r=1

But idk if for generalization

Since we did not use the fact that the number of letter is prime

[ancient robin]

No, it wouldn't work to generalize. For example if your word is 6 letters long and you know s^3(U)=U, you cannot deduce that U is boring

[lament parrot]

Yeah

[lament parrot]

And we did not use that 7 is prime for this

But for r=1 it of course works

And then we use the multiplicative inverse thing

[ancient robin]

yep

[lament parrot]

Thanks again!!!

[ancient robin]

Yeah the first statement works for any word length. The second statement only works for primes

[lament parrot]

Yeah

:))

.close

[pearl pond]

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[pearl pond]

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[chrome marsh]

I need to solve for this interval

I factorized

Now what do I do

[carmine star]

Hi my name is sandy and i am in pre algreba in 7th grade randall

[chrome marsh]

Just realized it’s sin

But still lost

[spare lark]

2 = 1/(1/2)

And the other one leads to sin(x) = -3 so

Since -1 < sin(x) < 1

No sol for second one

[pearl pond]

@chrome marsh Has your question been resolved?

[pearl pond]

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[covert pilot]

Hartshorne exercise 3.3.1 : Let $X$ be a Noetherian scheme. Suppose $X_{red}$ is affine. How to show that $X$ is affine?

The hint says to use Serre's cohomological criterion. Let $N$ be the nilradical sheaf. For a coherent sheaf $F$ on $X$, consider the sheaf $F \supseteq N F \supseteq N^2 F \supseteq ...$.

I think I need to prove that $H^1(X, N^k F / N^{k+1} F) = 0$. How do I do that?

[jolly parrot]

DavidL1450

[pearl pond]

@covert pilot Has your question been resolved?

[covert pilot]

<@&286206848099549185>

[midnight haven]

What do you need help with

[covert pilot]

How do I prove $H^1(X, N^k F / N^{k+1}F) = 0$

[jolly parrot]

DavidL1450

[pearl pond]

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[buoyant garnet]

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[buoyant garnet]

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[versed mica]

"use analytic methods"

[buoyant garnet]

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[versed mica]

🤔

don’t use a graph

[buoyant garnet]

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[versed mica]

you do though

do you know EVT

[buoyant garnet]

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[versed mica]

my point exactly

[buoyant garnet]

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[versed mica]

check the endpoints and then relative extrema

[buoyant garnet]

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[versed mica]

local max/min

[buoyant garnet]

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[versed mica]

kenzo

[buoyant garnet]

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[versed mica]

first derivative test or second derivatives test

[sterile turtle]

table of values or 2nd derivative

[buoyant garnet]

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[versed mica]

of

[buoyant garnet]

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[versed mica]

find the critical points, find the local max/min values, find the values at the endpoints, pick the absolute minimum and absolute maximum

[buoyant garnet]

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[versed mica]

kenzo

are you ok

genuinely

[buoyant garnet]

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[versed mica]

…

you’ve been doing this for like 2 months

f’ = 0?

[buoyant garnet]

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[versed mica]

or undefined

[buoyant garnet]

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[versed mica]

finding critical points isn’t exclusive to evt

you’ve literally done it this whole time you’ve done derivatives

[buoyant garnet]

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[versed mica]

do what

[buoyant garnet]

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[versed mica]

$-\frac{1}{x^2} + \frac{1}{x} = 0 \iff \frac{-1+x}{x^2} = 0$

[jolly parrot]

knief

[buoyant garnet]

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[versed mica]

did you forget the negative

[buoyant garnet]

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[versed mica]

🤔

[buoyant garnet]

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[versed mica]

your calculator is wrong

[buoyant garnet]

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[versed mica]

brother

do not

and i mean

do not

use calculators

it’s making you shit at math

you realize there’s a no calculator section right

[buoyant garnet]

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[jolly parrot]

Kenzo

[versed mica]

bro copy and pasted

[buoyant garnet]

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[versed mica]

anyways

[buoyant garnet]

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[versed mica]

x = 1 and x = 0 but 0 isn’t in the interval so fuck that dude

[buoyant garnet]

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[versed mica]

let’s solve for 1

[buoyant garnet]

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[versed mica]

yep

[buoyant garnet]

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[versed mica]

huh

[buoyant garnet]

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[versed mica]

because the absolute extrema could occur at the critical points

[buoyant garnet]

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[versed mica]

compare it to the endpoints

[buoyant garnet]

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[versed mica]

yep

[buoyant garnet]

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[versed mica]

by EVT

[buoyant garnet]

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[versed mica]

it isn’t

the absolute minimum/maximum on the interval have to occur at the endpoints or relative max/min

i challenge you to draw a function that doesn’t satisfy that

[buoyant garnet]

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[pearl pond]

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[pearl pond]

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[formal comet]

where do i start for questions like these

[blissful salmon]

Usually this is a common roots of unity problem. So there’s a general form for these type of equations that you should know

[formal comet]

this is my first time seeing this kinda thing

not the 2nd part js the bit where it says find the six roots

[blissful salmon]

Yeah, I recommend you looking at roots of unity for this problem. There is a general form of the solution, and it’s quite nice :))

[formal comet]

alr ty ill check it out

[blissful salmon]

No problem, also have a read of this if you want: https://en.m.wikipedia.org/wiki/Root_of_unity

[formal comet]

is this it

[blissful salmon]

Yes, this looks right

[blissful salmon]

This is the general form. And for your problem, it’s n=6

[formal comet]

what is the k in that screenshot meant to be

[blissful salmon]

All the possible distinct solutions

It’s perfectly shown in a visual picture of the unit circle where the solutions are on this circle equidistant from each other in the complex plane

[formal comet]

oh i think i get it

so for z^6 = 1 i just sub in 0 - 5 for k?

[blissful salmon]

Yes

[formal comet]

oh alright

ty i get it now

[blissful salmon]

no problem!

[pearl pond]

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[flint wyvern]

this is my question

i already did first two parts

i answered part C

did i answer everyhting correctly

the graph shows those points so i'm certain i am

but double checking

<@&286206848099549185> please if u can help fast i'm on a time limit

@vocal isle if u can just double check

[vocal isle]

hello

what is up

[flint wyvern]

sorry about the tag 😭

i just wanted someone to double check

[vocal isle]

x intercepts -3 and -1

from the graph

what do i check

[flint wyvern]

is all this good

[vocal isle]

yeah i just scrolled and saw it

looks great

[flint wyvern]

thank you

[vocal isle]

except that its on notepad

😭

do you want me to help teach u latex

[flint wyvern]

i cant write dude

[flint wyvern]

ong i need to leanr

[vocal isle]

people do latex on overleaf.com

[flint wyvern]

is it hard

[vocal isle]

no

you can make ur own documents there

theres a latex help channel in herer somwehre

[flint wyvern]

oh cool

ill check it out

well thanks again

[vocal isle]

np

[pearl pond]

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[midnight haven]

we have this D.E

[midnight haven]

and the solutions of the D.E are these

what do they mean by "in general" here

[acoustic path]

that's the 'general solution' for arbitrary A, B

[midnight haven]

https://tenor.com/view/head-scratch-confused-gif-14061427

[midnight haven]

could you put it in a different way

[midnight haven]

oh so a D.E can have infinite solutions

[midnight haven]

every solution is represented in this form

is that conclusion right ?

@acoustic path

[midnight haven]

if we give a certain value to c in this case means that we are giving a initial condition

[pearl pond]

@midnight haven Has your question been resolved?

[sharp vigil]

basically it's like the +C from integration (but there are two of them because it's second order)

[midnight haven]

thats matches this conclusion :)

thanks!

.close

[pearl pond]

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[timber tartan]

Literally f'(x) = e^x

f(x) = e^x + C

$ \bb{C} \in \fraktur{R} $

how did u write blackboard and frakturs again...

anyways C belongs to the reals

[midnight haven]

you add math before it

its its just frak

[midnight haven]

i had a doubt from yesterday,which made this quite confusing

$\mathbb{C} \in \mathfrak{R}$

[pearl pond]

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[cyan coral]

In this task, we prove some fundamental properties of divisibility. We assume that a, b, and n are integers and that c is a positive integer. Prove the following statements and, in each case, provide a numerical example.

a) If c∣a, then c∣na.

HOW DO I PROVE THIS

[limber glacier]

If c|a, what can u write relating c and a

[cyan coral]

huh?

[wraith river]

$c \vert a \Rightarrow a = rc \Rightarrow na = nrc \Rightarrow c \vert na$

[jolly parrot]

LooseEthics

[wraith river]

Not sure if it's possible to break down any further.

[cyan coral]

Does the arrow mean implies?

[wraith river]

yes

[cyan coral]

But how does this prove anything more than what we knew in the beginning

i dont get it

[wraith river]

Proofs like this are just about explicitly showing the chain of logic behind a result that's pretty much self-evident.

[cyan coral]

allright, but why we just throwing an n on the other side illegally

[wraith river]

What do you mean?

[cyan coral]

nobody said we could put n on the other side

na = nrc

[wraith river]

a = rc

multiply both side by n

na = nrc

perfectly legal

not like you're dividing or taking roots

[cyan coral]

Allright allright

[wraith river]

then just note that trivially c|nrc

therefore c|na

[cyan coral]

Oh im kinda starting to get it now

Thanks so much

.close

[pearl pond]

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[cyan coral]

wtf is this vertical line slash thing

or is it a bug

[solid pier]

with a slash it means “does not divide”

so a+b is not a multiple of c

for example

[cyan coral]

Ok

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[cyan coral]

So its a statement, not an operator?

[solid pier]

yeah that’s right

[cyan coral]

alright thx

.close

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[shy island]

I know that this is probably very simple, but I can't figure out how to get it done.
Given D(x) = (picture), I have to prove that D(x) grows by calculating D(x+1)-D(x), which should be greater than 0.

[shy island]

(it's messy, I didn't think I would end up stuck)

Trying to rationalize got me nowhere, since I can't say tell concretely whether the result is smaller or greater than 0 due to the square roots that don't match.

I realized that what I sent originally wasn't very explicit. How do I prove that the expressions is greater than 0 when the square roots don't match?

[pearl pond]

@shy island Has your question been resolved?

[lofty aspen]

Hey @shy island ... Are you sure that's the only way to prove that D(x) is growing??

I know another way...

[shy island]

I know i wrote it like a function, but D is a row, like d1, d2, d3 ... . I have to show that it's growing by either calculating dx+1 - dx, and comparing it to 0, or by calculating dx+1/dx, and comparing it to 1

[lofty aspen]

Sorry for the late response and ping btw..

[lofty aspen]

Maybe we should try the second option??

or do you know differentiation??

[shy island]

No, can't use that

it's a trick, probably

[lofty aspen]

Hmm... I'll see what I can do...

hold on..

do we know that rt(x+1) > rt(x)

mars??

@shy island ??

I'll leave the solution here then...

[shy island]

Sorry, I was busy with something

[lofty aspen]

Oh alright

nevermind, i didn't get the solution..

[shy island]

I have to leave, I'll close this, you can send me the solution if you find it.

[lofty aspen]

Alrightt

[shy island]

.close

[pearl pond]

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[pearl pond]

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[unborn wave]

hey i need help graphing a rational function, and finding asymptotes

[unborn wave]

it’s wrong but i don’t exactly know why

[bronze coyote]

r u supposed to graph $\frac{-x^2-5x+3}{x+1}$

[jolly parrot]

pppoopoo

[unborn wave]

yes u are

[bronze coyote]

the graph looks very wrong

first find the vertical asymptote

then find horizontal asymptote

but this one doesn't have one

do u know how to find an oblique asymptote

[pearl pond]

@unborn wave Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[meager trellis]

is there more context than this?

[urban sun]

thats it

[plush bramble]

show the solution

[urban sun]

ok

[surreal relic]

Cool question

[meager trellis]

i'm not convinced i believe this is a question and not just, the letter E

[urban sun]

the second image comes before*

[surreal relic]

Makes sense

[urban sun]

i don't understand it

[surreal relic]

Basically all rows should have atleast 1 alphabet

[urban sun]

can u not leave 2nd and 4th row empty

[surreal relic]

Since 2nd and 4th row

Have only 1 space

It must be filled

[urban sun]

og

oh

[surreal relic]

These are rows

[surreal relic]

See it now?

[urban sun]

yea

[surreal relic]

Cool

So let's assume we placed 2 letters in them

We left with 4

And 3 rows

[urban sun]

2?

[surreal relic]

1 in each

[urban sun]

i think im confusing rows with columns

[surreal relic]

Yeah

[urban sun]

so theres 3 rows and 5 columns?

[surreal relic]

Red rows

Yellow columns

[urban sun]

oh

so where are u placing 2 letters at?

[surreal relic]

You are confusing them

[urban sun]

ye mb

[urban sun]

oh

[surreal relic]

It's alright

[urban sun]

yes

[surreal relic]

Ok we good upto here?

[urban sun]

yea

u placed bcz they never remain empty right

[surreal relic]

Left with 4 letters

[urban sun]

yea

[surreal relic]

Yup

[urban sun]

yeye

[surreal relic]

Now for 1st 3rd and 5th row

We have 4 letters

So 1 row is gonna get 2 letters

The other 2 rows are gonna get 1 letter each

[urban sun]

yea

or

what next

[surreal relic]

Purple is that 1 row out of 1st 3rd and 5th

That gets 2 letters

Yellow is the remaining 2 rows out that gets only 1 letter

Red represent giving 1 letter each to row 2nd and 4th

[urban sun]

i thought it was 3c1 for selecting a row out of 3

[surreal relic]

Orange represents their arrangement

[surreal relic]

Yes

The 1 row

That's gonna get 2 letters

That's why we selecting it

To give it 2 letters

[urban sun]

whats the 3c2 for

[surreal relic]

Make sense?

The 2 rows

That gonna get 1 letter each

[urban sun]

the other rows?

Oh

[surreal relic]

Cool ?

[urban sun]

why won't we consider 4c2

to give 2 letters out of 4

[surreal relic]

To assign to who

??

Selection of letters isn't the issue

Placement is

[urban sun]

3c1 × 4c2

and 3c1 × 4c1 × 2

i thought this

and ofc the letters arrangement

[surreal relic]

The orange part accounts for all this

[urban sun]

oh

i think all cool

thanks for ur help

[surreal relic]

Great

Ur welcome

[urban sun]

have a great day

[surreal relic]

You too

[urban sun]

yea thanks

. close

.close

[pearl pond]

Channel closed

Closed by @urban sun

Use .reopen if this was a mistake.

Channel closed

Closed by @surreal relic

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• 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.

[dusk laurel]

yo sup

[calm wing]

hello, do you have a question to ask?

[surreal relic]

yo

[pearl pond]

@dusk laurel Has your question been resolved?

[peak vine]

it seems like you should use the hint

[plush bramble]

woulda been faster if you wrote that on paper and mailed your response to him.

[peak vine]

.remindme 2 months

[pearl pond]

Created reminder #help-39 message for <t:1737308725:f>