help-27

[devout snow]

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[restive river]

Is there any general method of knowing how many homomorphisms and isomorphisms exists between a group and itself?

[restive river]

Between a group and a different group it is easy, either there exists 0 or the question is equivalent to my question

[unreal thunder]

It's quite not easy in general

but we can use certain properties of homomorphisms like if |a|=k, then |f(a)| divides k, this helps to count homomorphism between cyclic groups

between Zm, and Zk, there are gcd(m,k) homomorphisms

i suggest someone else to comment on for better idea

[restive river]

What if we know the number of elements is prime?

Does it become somewhat easier

[unreal thunder]

if number of elements is prime then it's isomorphic to Zp

[restive river]

Yes

So I have to count the number of homomorphisms between Zp and itself

And isomorphisms

[dapper tiger]

Zp doesn't have a lot of subgroups so the problem is only counting isomorphisms

also, Zp is cylic, so an isomorphism is mapping a generator onto another

[restive river]

Hmm

Are homomorphisms only between a subgroup and a group?

[dapper tiger]

that's the definition of homomorphisms yeah, they preserve the structure, the image is a group

so if the image is in Zp, it's 0 or Zp

[restive river]

The image is a group

Ok yeah that makes sense

[restive river]

And any element of Zp generates Zp

Except 0 I guess

Soo

[dapper tiger]

be careful about orders

[restive river]

p-1 isomorphisms?

And p homomorphisms

[violet wind]

Yes

[devout snow]

@restive river Has your question been resolved?

[restive river]

Thabks

[devout snow]

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[subtle matrix]

help in probability

[subtle matrix]

I didn't find the answer

there is no helpers thx

.close

[devout snow]

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[quiet breach]

can someone tutor me with speed maths and some tricks

[devout snow]

Please don't occupy multiple help channels.

@quiet breach Has your question been resolved?

[supple knot]

don't use math help channels for non math questions

.close

[devout snow]

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[worthy talon]

If both repeated limits exist in a 2-variable function does the double limit exist as well?

[worthy talon]

I can't think of a counterproof but it feels wrong

[lone ravine]

Let f(x,y) = {1 if x or y = 0, and 0 else
lim x->0 f(x,0) = 1 and same for y. Still f is not continous

[lone ravine]

Do you mean something like that?

[worthy talon]

Yea

Guess that works

Do you think theres a function that isnt piecewise defined like that

I was thinking x^y

But im not sure if that works

[lone ravine]

Yes, we had a similar counterexample

[lone ravine]

We had a counterexample like that

[worthy talon]

Thanks

.close

[devout snow]

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[lone ravine]

You're welcome

[lone ravine]

Not sure if that works though

I think it does, yeah

[devout snow]

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[fathom jungle]

I have no clue how to do this 😭

please help

[verbal vector]

did you draw a diagram?

[devout snow]

@fathom jungle Has your question been resolved?

[fathom jungle]

i tried

i took the angles as 10x x and x

12x = 180

x= 15 i did something

i dont understnad

[verbal vector]

i don’t think the other 2 angles are both x

[fathom jungle]

if i take it as another variable how will i solve it then ?

oh like 10x + 2y = 180 ? since its isosceles y = 45 and...

no then it will be in decimal

[verbal vector]

i’m assuming BC is strictly the smallest size

side*

[fathom jungle]

yeah

[verbal vector]

so angle B would equal (180-10x)/2

just based on angle manipulations

[fathom jungle]

hmm

[verbal vector]

understand so far?

[fathom jungle]

yep

[verbal vector]

if angle B was 60, what would angle C and A be

[fathom jungle]

6 ?

[verbal vector]

60 each

[fathom jungle]

oh

[verbal vector]

which would be an equilateral triangle

[fathom jungle]

then it wud become equilateral

yeah

[verbal vector]

yep

then BC wouldn’t be the smallest side

[fathom jungle]

yeah

[verbal vector]

could it be 65?

[fathom jungle]

50 each

[verbal vector]

not 50 each, C would be 65 and A would be 50

[fathom jungle]

ahh

[north plaza]

You don't have to do this by trial and error.

[verbal vector]

pretty simple tho

[north plaza]

It's tedious to test for every value.

[fathom jungle]

how do i approach it then ? ive been stuck on this for 15 mins 😥

[north plaza]

This simplifies to 90-5x, now based on the information given, x is a multiple of 10, so you can see that x=20 will clearly exceed 90 and you would get a negative number so x = 10 is your only option. Now solve for b.

[devout snow]

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[fathom jungle]

.reopen

[devout snow]

✅

[fathom jungle]

it wud be 40 then

[north plaza]

Oops my bad.

Let's start over.

x+2y=180
y=180-x/2
y=90-x/2

[fathom jungle]

then ?

[devout snow]

@fathom jungle Has your question been resolved?

[devout snow]

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[empty geode]

what is the span of the 4x4 matrix which has 3,6,5,6 in the first row and the rest are only zeros?

[vapid nest]

buddy

[empty geode]

yes?

[vapid nest]

span of the rows of the matrix

span of the columns of the matrix

or the span of the matrix?

[empty geode]

the span of the matrix.

i need the vectors to get to the jordan normalform

i calculated core of(A-Eigenvalue * I) and now i need the span of the core

this is what im trying to achieve

except the matrix is 3,6,5,6 and the rest of the rows are 0

[dapper tiger]

Span{(1, 0, 0, 0)}

if only one row is non-zero, it means the dimension of the span of your matrix is 1, so you only need to find a vector that is in the span

and since the span can only have vectors where the first coordinate can be non-zero and the rest 0

well it's done

[devout snow]

@empty geode Has your question been resolved?

[devout snow]

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[devout snow]

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[light fox]

I need help understanding how to make the rational function

[devout snow]

@light fox Has your question been resolved?

[upper schooner]

,rcw

[devout snow]

@light fox Has your question been resolved?

[devout snow]

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[surreal stream]

For this question, I have two critical numbers, x=9 and x=-9

How would this work?

oops the length and width cant be negative

.close

[devout snow]

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[surreal stream]

I need assistance with this question. Let me send what i have so far.

If the critical number is x=0 (when derivative becomes und) how would I use it when domain doesnt include 0

[dapper tiger]

the area of plexiglas is 4xy

you forgot the top

[dense jay]

it is indeed 3xy, there are only 3 sides

[dapper tiger]

see sentence 2

[dense jay]

oh deary me, i recuse myself

should be 3xy+x^2 though in that case

[dapper tiger]

no

the top is xy too

[dense jay]

left, top and right are xy each

the back face is x^2

[dapper tiger]

oops I got backface wrong

[surreal stream]

wait i dont understand how its 4xy

[dapper tiger]

yeah it's 3xy+x² you're right

backface is x²

[surreal stream]

ohhh i see

[dapper tiger]

others are xy

[surreal stream]

makes sense

[dapper tiger]

so you want to minimize x²+3xy

[surreal stream]

im assuming becauyse it says 3 plexiglass sides and a plexiglass top

ok

[surreal stream]

cn is x=9

okay so length = 9 width = 6 height = 9

[dapper tiger]

,w critical points of x²+3*(486/x)

[woven radish]

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[dapper tiger]

yeah x = 9

and then y = 6

[surreal stream]

thanks

.close

[devout snow]

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[gentle wagon]

how do i solve this

[gentle wagon]

24x^2-8x-14x-7

-8x(-3x+1) -7(2x+1)

[dense jay]

24x^2-22x-7=0

what drove you to split it into -8x and -14x

[gentle wagon]

i used grouping

[brisk totem]

just throw it into quadratic formula

[gentle wagon]

its a method of factoring no?

[dense jay]

it is but why -8 and -14

[gentle wagon]

quadratic formula is too much math

because they add up to -22

and are divisible by 24 and 7

[dense jay]

they arent factors of 24*(-7) though

which is a requirement if youre going to use said method

[gentle wagon]

oh i have to do a * c?

[dense jay]

yeah

[gentle wagon]

ohh

and then what

[dense jay]

get 2 factors of ac that sum to b

then you can split b into said factors

then you can do your grouping

[gentle wagon]

so 24x^2 - 22x - 7
24x^2 -22x -168

is there a method for finding out the sum to b with large numbers

[dense jay]

no, youre not changing the equation like that

[dense jay]

just write out the factors of ac

[gentle wagon]

x^2 - 22x - 168?

[dense jay]

its still just 24x^2-22x-7=0

youre just looking for factors of -168 that sum to -22

[gentle wagon]

so...how would i do that

i have 2^3 * 3 * 7

[brisk totem]

by trial and error

[gentle wagon]

ah i figured

so is it really just skill issue

[brisk totem]

it’s doable but that’s why i said just throwing it into quadratic formula if you don’t see it in a few seconds

(also you have a -1 in there)

[gentle wagon]

yea i was going to factor it in later

[dense jay]

hmm

[gentle wagon]

doing 22^2 by hand and then -4(24)(7)/48

i feel like im missing something

like is there supposed to be a more efficient way to do this

[brisk totem]

i mean if you see the right grouping you can factorise it in like half a minute

[gentle wagon]

(not to mention squaring it)

[brisk totem]

which then comes back to this

[gentle wagon]

i tried it with different grouping

like -15 and -7

didnt work either

[brisk totem]

that being said, whenever grouping applies the quadratic formula usually isn’t bad

as in the sqrt(b²-4ac) usually is a whole number

(and it is here)

[brisk totem]

it’s also usually much faster to consider factors of ac instead of what numbers sum to b

[gentle wagon]

these seem like a pain to calculate by hand

[brisk totem]

that’s extremely fast because you get (22+-34)/48

[gentle wagon]

i guess

then i could just plug it in atp

but again you still have to calculate it

[brisk totem]

well ye, and then you have the choice of throwing it into quadratic formula or trying to factor by grouping

[gentle wagon]

okay well how do i spot the grouping for -168 and -22

2, 84
2,42
21,2
7,3,

[brisk totem]

trial and error as previously mentioned

you just haven’t got there yet

[gentle wagon]

guess not

[brisk totem]

usually we would recommend trying factors from smallest to largest first instead of randomly darting around since you might miss factors or accidentally try some twice

[dense jay]

apologies for the ping

[gentle wagon]

can you show how that would look like

[brisk totem]

1 168
2 84
3 56
4 42
..

[gentle wagon]

so 6 and 28

[brisk totem]

ye there it is

[dense jay]

👏

[gentle wagon]

damn im honestly a little slow without a calculator

thanks guys

🙏

.close

[devout snow]

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[devout snow]

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[wicked egret]

I get stuck right up to this point (the circled)

[wicked egret]

When i do (2x-1)(x+5) I get 2x^2+9x-5

They managed to somehow get 2x^2-9x-5

[devout snow]

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

[supple knot]

(-1) * 5 =?

[wicked egret]

?

[supple knot]

Do you know the answer?

[wicked egret]

Yeah

Its right there

[supple knot]

What is it

[supple knot]

To this question

[wicked egret]

-5

[supple knot]

So how did you get +10

[wicked egret]

Sorry my bad

I mean

2x^2+9x-5

3x + 15 + 2x^2+9x-5 = 2x^2+12x+10

but still the 2x^2+9x-5 is a positive 9 not a negative

@supple knot

[supple knot]

This is still wrong

[wicked egret]

How?

(2x-1)(x+5) = 2x^2+9x-5

[supple knot]

Oh no you're right

As in they're wrong

,w expand 3x + 15 + (2x-1)(x+5)

[woven radish]

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[wicked egret]

yea see

thats what i got

and yet they gave a compteley diff answer

[supple knot]

Yea you're right

[wicked egret]

But when I put the original question into Symbolab I get this answer

I get their answer

[supple knot]

Then they messaged up earlier

!noadvert

[devout snow]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, and no one person can be prioritized over other people, so please patiently wait. Anyone who chooses to help you is a volunteer who is doing so out of their own kindness.

[wicked egret]

Okay so just to confirm my answer is right, right?

[wicked egret]

This

[supple knot]

You simplified correctly from the previous line yes

[wicked egret]

What about the overall answer

[supple knot]

No idea

There was a mistake earlier

Find it

[wicked egret]

Soeone earlier said it was this

the (2x+1)s in green

[supple knot]

Do you understand why

[wicked egret]

Why what

I assumed they just cancelled them out

But I also thought its not allowed to cancel

So idk

[smoky nimbus]

Expand the denominator and see

[wicked egret]

Which part

[smoky nimbus]

You are questioning the green part

Expand the denominator that involves the green part

[wicked egret]

2x^-4x-5?

[smoky nimbus]

How did you get that?

Do you know how to expand?

[wicked egret]

No

Wdym by expand

[smoky nimbus]

Expand (2x + 1)(x - 5)

[wicked egret]

Oh

Thats 2x^2-9x-5

Sorry yeah ik what expand means

I just dont call it expand

[smoky nimbus]

And does that match the original problem?

[wicked egret]

Yeah but it doesnt say that here

it says (2x-1)(x+5)

[smoky nimbus]

What was the original problem?

[wicked egret]

Full answer solution

[smoky nimbus]

And you expanded (2x + 1)(x - 5) to get 2x^2-9x-5

Does that match the original problem?

[wicked egret]

It does but where did you get (2x + 1) from

they removed it at the top

then it was just (2x-1)(x+5)

[smoky nimbus]

I don't care about the stuff after, you asked about this part, the green (2x + 1), the first step from the original problem

The first step is factoring

[smoky nimbus]

If you expanded the denominators, you would get the denominators from the original problem

You just expanded (2x + 1)(x - 5) to result in 2x^2-9x-5

Does that match what the original problem gave?

[wicked egret]

Why does the (2x+1) magically disappear in the next line tho

[smoky nimbus]

Stop focusing on it disappearing in the next step. You asked why the green part is the mistake

I am trying to get you to see why

[wicked egret]

I only said that bc someone else said that so idk

[smoky nimbus]

And that's what I'm asking you

If you expanded the denominators, you would get the denominators from the original problem
You just expanded (2x + 1)(x - 5) to result in 2x^2-9x-5
Does that match what the original problem gave?

[wicked egret]

No I dont get what ur saying

[smoky nimbus]

What is the original problem given?

[wicked egret]

Look

[smoky nimbus]

I'm not asking for the entire problem, I'm asking for the original expression, what is it?

[wicked egret]

This?

[smoky nimbus]

Now you asked about the green part because someone said it was wrong, correct?

[wicked egret]

No I didnt say they were wrong what i meant was they highlighted this

[smoky nimbus]

And he was pointing out that the green part is wrong, and you wanted to figure out why, correct?

[wicked egret]

Yeah I wanted to figure out why it didnt appear in the next line

[smoky nimbus]

It doesn't appear in the next line because it got canceled in that work, but that green (2x + 1) is completely wrong in general

That is what that helper was telling you

[wicked egret]

Okay then what about after itgot cancelled

In the next 4 lines

How does (2x-1)(x+5) give me 2x^2-9x-5

[smoky nimbus]

You are not listening

[smoky nimbus]

This part, the (2x + 1) is wrong

The book factored it wrong

[wicked egret]

Ohhh so why didnt you just say the book factored it wrong

[smoky nimbus]

Because I was trying to get you to see it

Because the first step from the original problem is to factor

So I told you to expand that

[wicked egret]

2x^2-9x-5 is supposed to be 2x^2+9x-5

[smoky nimbus]

To compare it with the original

[wicked egret]

I guess the book is setting me up for failure

[wicked egret]

So if I fix the factoring they did wrong

Will I end up with the answer they have

[smoky nimbus]

As it was mentioned here, the denominator should have been (2x - 1)(x + 5)

[wicked egret]

Okay

[smoky nimbus]

Try it yourself

[wicked egret]

Okay 1 sec

[wicked egret]

Oh yeah

Its (2x-1)(x+5)

@smoky nimbus Am i allowed to cancel (2x-1)/(2x-1)

when adding / subtracting

rationals

[smoky nimbus]

The screenshot tells you

[wicked egret]

okay

Can you also help me with this

Can you explain why they got 6t and 3t^2 when they were multiplying them together in the first step

[smoky nimbus]

Because that's what you need to multiply by to get the LCD

[wicked egret]

ohhhh yeah i get it

ight thanks for ur help

Wait last question

How did they get 30t^3 as the LCD @smoky nimbus

why couldnt it be 30t or 30t^2

[smoky nimbus]

Because of the last fraction

It has t^3

[wicked egret]

so whichever fractions exponent has the highest power the lcd has to match?

@smoky nimbus

[devout snow]

@wicked egret Has your question been resolved?

[wicked egret]

<@&286206848099549185>

Someone explain why 30t^3 is the LCD

Please

pls'

[supple knot]

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:rational-functions/x9e81a4f98389efdf:adding-and-subtracting-rational-expressions/v/least-common-multiples-of-polynomials

[wicked egret]

Thx

[wicked egret]

Im still kinda confused even after watching that

I know the LCM of 5, 10, and 15 is 30 but

why t^3?

[wicked egret]

<@&286206848099549185>

[supple knot]

Try setting t=2. What's the LCM then

[wicked egret]

30 * 2 = 60?

[supple knot]

No

[wicked egret]

or 30(2)^3

[supple knot]

What are those fractions for t=2

Actually just denominators

[wicked egret]

5(2)^2 = 20
10(2) = 20
15(2)^3 = 120

[supple knot]

What's the lcm of all those

[wicked egret]

lcm of 20, 20, and 120 is 120

@supple knot

[supple knot]

So is it 60

[wicked egret]

30(2) is 60

30(2)^3 is 240

[supple knot]

And is that the LCM?

[wicked egret]

well the lcm of 20 and 120 is 120

[supple knot]

?

Does 60 = 120?

[wicked egret]

No

But u said the lcm of 20 and 120

im still pretty confused 😭

[supple knot]

Try it again for t=7

[supple knot]

You thought it was 30t

[supple knot]

Then you demonstrated the lcm with a specific t that the lcm = 120 which is clearly greater than 30(2)

[wicked egret]

well i just wanna know why it has to be 30t^3

and not 30t^2 or 30t

[supple knot]

Do examples to convince yourself

[supple knot]

You already proved it can't be 30t

Why do you still think that's an answer

[wicked egret]

Which example proves its 30t^3

[supple knot]

.

And t=11

[wicked egret]

Are those just random numbers

Or do they haev meaning

[supple knot]

Try it

[wicked egret]

5(7)^2 = 245
10(7) = 70
15(7)^3 = 5145

[supple knot]

Well is the LCM 30 * 7?

[supple knot]

What's the lcm here?

[wicked egret]

the lcm is

10290

according to my calculator

[supple knot]

,calc 30 * 7^3

[woven radish]

Result:

10290

[wicked egret]

yes but i still dont understand why that means that its t^3

[supple knot]

What

[supple knot]

You see this right

[supple knot]

And this

[wicked egret]

yes

but like

what if i didnt know that t = 7

[supple knot]

????????

[wicked egret]

You said

try for t = 7

How would I know where 7 came from

[supple knot]

I don't know how else to convince you

[wicked egret]

bro u said try t=7

but how do i know to get 7

how did u get 7

[supple knot]

Try as many t as it takes you to be convinced

Prove yourself as much as you need to

[wicked egret]

so i can do t = 1 all the way through t = 100

[supple knot]

that the LCM is not 30t and not 30t^2

[supple knot]

.

13, 17, 19, 23, 29

Maybe 6 more will be enough

[devout snow]

@wicked egret Has your question been resolved?

[devout snow]

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[restive river]

How do you derive associativity from the Huntington postulates for a Boolean algebra?

[restive river]

Huntington postulates being:

[lusty sapphire]

What the heck is this?

[sonic smelt]

Distributivity of + over *

[lusty sapphire]

ah boolean algebra I see

[restive river]

yeah this is not the field of real numbers

[lusty sapphire]

You need to prove a number of things before you can prove associativity.

e.g.

• x''=x
• If x+y=1 and x*y=0, then y=x'
  and a few others.

[sonic smelt]

By the second statement do you mean uniqueness of the complements?

Ah wait nvm what I was going to say is probably wrong

[devout snow]

@restive river Has your question been resolved?

[restive river]

okay sorry

[restive river]

for the first statement can we not use 5b postulate twice?

let x + x' = 1

hmmm

[lusty sapphire]

Maybe proving the second statement first will help, that inverses are unique

[restive river]

let's assume there exists a y_1 and y_2 that satisfy

x+y_1 = 1
x+y_2 = 1

that implies x+y_1 = x+y_2

which thereby implies y_1 = y_2 proving the uniqueness of the Inverse

[lusty sapphire]

does it?

Which axiom states that if x+y=x+z, then y=z?

[restive river]

okay fair

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[simple fossil]

Hopefully this isnt off topic, I need a big list of limits with their solutions for practice

[devout snow]

@simple fossil Has your question been resolved?

[simple fossil]

eh ill go shid myself I guess

.close

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[opal vale]

Could someone help me solve this modular arithmetic problem

[opal vale]

not sure how to continue when i get to there

the example my prof has showed us is that typically you would end up with 9x congruent to 1 mod 23

but i have a 19

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[tribal moat]

who can write solve with picture pls

Point K does not belong to the plane of the trapezoid ABCD, the bases of which are AD and BC. Prove that the line AD is parallel to the plane (BKC)

[restive river]

In this server, we are not allowed to give out solutions

We are only able to guide and help you reach the answer

[tribal moat]

okie, can you tell me how i can solve this

[eternal dune]

Lines AD and BC are parallel

And when you figure that out, you are almost done

[tribal moat]

hmm its the right decision?

Since AD is parallel to BC (by definition of a trapezoid) and BC intersects plane BKC at point B, AD must be parallel to plane BKС that if a line is parallel to a line intersecting a plane, it is also parallel to the plane

[eternal dune]

BC is inside the plane BKC, thats why AD and BKC are parallel

[tribal moat]

oh

thanks

.close

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[midnight sluice]

i seem to find alot of P's around in probability notation

which P is this?

our good old P?

i see P,p,\Pr,\mathbb{P} probably more i cant recall

can anyone highlight the differences please?

is this P the cumulative probability?

.close

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[south elk]

How do you find the equilibrium points of a first-order non-linear differential equation

[south elk]

For an example

dy / dt = sin(pi*x)

Okay wait, change of question, I understand that part so set equation to 0 and find solutions

How would you go about drawing a phase line with this?

[devout snow]

@south elk Has your question been resolved?

[supple knot]

Do you mean dx/dt

[south elk]

For the example I have it’s dy/dt

Although I suppose it doesn’t matter too much

[supple knot]

Then what's x on the right side

[devout snow]

@south elk Has your question been resolved?

[devout snow]

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[nimble tide]

you dont differentian the entire thing bro

it's a composite function

so you plug in value of x=3 in f(x)

which gives you 3

so now you differentian g(3)

wrong wrong

you''ve never heard of composite functions?

it's when multiple functions are presnet in one whole big function

in simple words^

[prime egret]

f’(3) is not 5 otherwise correct

[devout snow]

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[serene birch]

how can i find bijection from the set of solutions of x0+x1+x2+...+xm = n, where m and n and xi are from N, and the set of subgroups of {0,1,2,...,m+n} with n elements 😄

[devout snow]

@serene birch Has your question been resolved?

[topaz axle]

take x0, skip 1, take x1, skip 1...

so that you leave out m elements

that sounds wrong, then you always leave out the last one

it has to end in m+n-2 or m+n−1 or something

[serene birch]

the set of solutions is like the set of non decreasing sequences of n elements

i think

x0 is the number of times we have 0

x1 is the number of times we have 1

and so on

set of non decreasing sequences and all of the elements are less than m

{(s1,s2,...,sn) | s_i<=s_{i+1} and s_i<=m}

so bijection from this set to the subgroup set 😄

do they even have the same size

m+n+1 choose n

[topaz axle]

no

[serene birch]

damn D:

[topaz axle]

m = 2, n=3
{0,1,2,3,4}
there would be 10 solutions and 5c3 subsets

so it should end in m+n−1

[serene birch]

ohh

[devout snow]

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[serene birch]

ok then we remove 0

[serene birch]

or ends in m+n-1

yeah

so if x0, x1, ... , xm is a solution we have x0 zeros, x1 ones, .... , xm ms. Number of 0s + number of 1s + ... + number of ms = n, so we get a subset of {0,..,m+n-1}

0+1+2 = 3

0 zeros, 1 ones and 2 twos

(0,1,2) subset of {0,1,2,3,4}

😄

and if there are repetitions

we add +1 somewhere

idk

1+1+1 = 3

this will map to {2,3,4}

somehow

(a1,a2,a3)->{a1,a2,a3} if they are all different

(a1, a1, a2) -> (a1, a1+m, a2)

😄

or ->(a1, n+1, a2)

or

if we have (a1, a2, ...., ai, ai, ai, ..., ai, ...., an) -> {a1, a2, .... , ai, n+1, n+2, n+3, ...., n+k, ....,an}

if they keep repeating we keep adding 1 😄

idk

if they all repeat

(a,a,a,a....,a) -> {a, n+1, n+2, ..., n+m-1}

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[devout snow]

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[swift ledge]

need help with this inequality question

I tried using this modulus function on desmos but honestly i have no idea where to start

@amber wyvern

sorry this might seem a bit early but i posted my question wrong at 7:31

[rough owl]

yo

nahh wdf is this question

[swift ledge]

bro this is a normal question in my country

at least i think

[rough owl]

what country u in

[swift ledge]

im in bottom set maths😭

[swift ledge]

uk

[rough owl]

bro and das a normal question for u

im top set

wait wtf

im in the UK asw

what df is going on

[swift ledge]

idk man im just built dum ig

[rough owl]

nah idk how to do this either

this some weird question

broo where the helpers at

when u need them

[swift ledge]

honestly😓

@amber wyvern

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[devout snow]

@swift ledge Has your question been resolved?

[devout snow]

@swift ledge Has your question been resolved?

[devout snow]

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[marble quail]

,rotate

[marble quail]

Where did I go wrong

I thought i could use this formula to solve it

[dense jay]

whats going on here

the integral of ab isnt int(a) * int(b)

simplify to x/2

then integrate

[marble quail]

But isn't f(x) = x, g(x) = ln(4x)?

[dense jay]

yes? what does that have to do with this integral here

[dense jay]

what you did here is nonsense

[marble quail]

[dense jay]

its like saying the integral of x * x^2 is 1/2 x^2 * 1/3 x^3 = 1/6 x^5

[marble quail]

What i am doing there is this part of the formula

[dense jay]

i know im just saying you did it wrong

[marble quail]

What specifically was wrong? Because afaik F(x) = x^2/2 and g'(x) = 1/x

[dense jay]

im not saying your set up is wrong

the actual integrating was wrong

[marble quail]

oh

[dense jay]

x^2/2 * 1/x = x/2

integrating that gives you x^2/4

oops

there we go

[marble quail]

$\int \frac{x^2}2 * \frac1x dx = [\frac {x^3}{6} * ln(x)]$

[woven radish]

Merineth

[dense jay]

doesnt

you can prove that yourself by differentiating the right side
product rule:
x^2/2 ln(x) + x^2/6

thats not what you started with, is it

[marble quail]

No, you''re right

Then i have no idea how to find F(x) of that

[dense jay]

youre over thinking it

$\int{\frac{x^2}{2} \cdot \frac{1}{x}}dx=\int{\frac{x}{2}dx}$

[woven radish]

AℤØ

[marble quail]

How did you get come up with that?

[dense jay]

x^2/x=x

[marble quail]

x is the deriviate of x^2/x

oh

[dense jay]

it isnt

[marble quail]

oooh i see what you did

Didn't realize i could combine them

to x/2

lol

[desert arch]

Bro did allat and couldnt cancel 😭

[marble quail]

so that integral comes out to 3/4

If i did it correctly

Not getting the correct answer however..

$2ln(8)-\frac{ln(4)}{2} - \frac34$

[woven radish]

Merineth

[marble quail]

i got that

but the right answer is

$5ln(2)-\frac34$

[woven radish]

Merineth

[desert arch]

Cant u like put (2 * 6ln(2) - 2ln(2))/2

And its 10ln(2)/2

5ln(2)?

@marble quail

[marble quail]

i just realized

they are the same

ln(4)/2 = ln(2)

.close

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[pearl valley]

I have been trying to solve $sec^2(x) - 5 = 3tan(x)$ for x for like 2 hours now, I seem to being going in circles

[woven radish]

james

[dense jay]

what is it you have tried

[pearl valley]

well, a lot. I have tried everything I can think of which mostly involoves splitting the tan into sin/cos and then seeing how I can shuffle things to cancel others out

i ended up with sec x - 5 cos x = 3sin x

[dense jay]

that could work eventually sure ig

did you ever consider just using that:

1+tan^2=sec^2

[pearl valley]

Oh! I forgot this was an identity.

I will give this a shot tysm

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[vast canyon]

do we say the derivative exists at end points because we consider one-sided derivatives in it's definition?

[devout snow]

Please don't occupy multiple help channels.

[lusty sapphire]

yes

[vast canyon]

and do we consider undefined points for the first and second derivative for critical and inflection points?

[lusty sapphire]

?

!xy

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[vast canyon]

because they are used for graph changes, and undefined points can have changes in graphs

like this or something

the inflection point is undefined, because of a vertical tangent

but we would still say its an inflection point and the concavity changes from increasing to decreasing slopes

right??

[acoustic leaf]

we usually only define inflection points for functions with a continuous second derivative

[vast canyon]

I thought they were defined as points where concavity changes

[wraith flare]

Hi can i have help:3

[acoustic leaf]

you can change concavity without having an inflection point

e.g., the graph of 1/x changes concavity at a vertical asymptote

[vast canyon]

oh right, thanks

"point" meaning it has to be defined

got it

this function has no relative min/max, right?

[acoustic leaf]

in the case of a cube root function, which looks like this, you can consider it an inflection point, since the point itself is defined

[vast canyon]

well if you graph the second derivative its an asymptote isnt it

infinite y values

[acoustic leaf]

the second derivative graph has an asymptote, yes

[vast canyon]

so f''(0) doesnt exist

[acoustic leaf]

the inflection point itself is on the graph of the original function

the main condition for an inflection point is that f''(x) changes sign, which must mean it's either 0 at that point or discontinuous at that point

[vast canyon]

but it has to be defined there

[acoustic leaf]

the function must be defined there for it to be a point, not necessarily (but usually) the second derivative

[vast canyon]

this is conflicting

definitions differ I guess

my teacher says that it needs to be twice differentiable too

kind of bizarre because she will say a critical point can be undefined in the first derivative, but if continuous on f, then it's a valid point

anyway, @acoustic leaf do you mind answering my other question?

[devout snow]

@vast canyon Has your question been resolved?

[devout snow]

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[glass birch]

How do I solve this trigonometry question?

It says I have to find the height

[lusty sapphire]

what is <BAC?

[glass birch]

what do u mean

[lusty sapphire]

what is the measure of angle BAC?

[glass birch]

It just gives the angle

Which is 27 degrees

Doesn’t give anything else

[lusty sapphire]

Not angle ABC, I'm asking for angle BAC

[glass birch]

Idk

[lusty sapphire]

what about angle ACB?

[glass birch]

27

[lusty sapphire]

that's angle ABC again, I'm asking for angle ACB

[glass birch]

What’s the difference

😭😭

[vast canyon]

the middle letter is the angle you are looking for, its just the convention they use

[glass birch]

There is no middle letter

We did this in class and my teacher said it’s similar to this question but idk how I did this

[lusty sapphire]

The middle letter of angle ABC means we are looking at the angle at point B created by the two lines AB and BC

<ABC is green
<BAC is blue
<ACB is red
<ACD is orange
<ADB is your right angle

[devout snow]

@glass birch Has your question been resolved?

[glass birch]

Yeah I am super lost

[devout snow]

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[jaunty abyss]

If I’m doing like bases and exponents what do I do if the exponent is a negative? Does it change anything?

[jaunty abyss]

Oh wait does that mean I put a 1 over the number I get

[solar goblet]

$x^{-a} = \frac1{x^a}$

[woven radish]

FungusDesu

[jaunty abyss]

Oh ok ty

What do I do with that

.close

[devout snow]

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[solar goblet]

given m is a positive integer, there exists how many possible values of m for $\log_3{(x-3)} - \log_9{x^2} = \log_3{(m-9)}$ to have a root

[woven radish]

FungusDesu

[desert fulcrum]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[solar goblet]

i simplified everything to $\frac{x-3}x = m-9$

[woven radish]

FungusDesu

[solar goblet]

but from here im stumped

[devout snow]

@solar goblet Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
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[weak cove]

For a question like this, would the expected value of |v+w| be the same as, finding the expected value of one root, say, |z| and then just doubling it

I wanted to think that at first, but I'm not certain

linearity of expectation

or something

[desert fulcrum]

Simplify z, v and w as cosx + isinx (different theta for all)

So you basically have √(2+√3)<=√((cosa+cosb)²+(isina+isinb)²)

Simplify it down so you'll get something in terms of (a-b) I think

[weak cove]

I'll try this

[desert fulcrum]

cos a-b

Alright let me know where it takes you

[weak cove]

Not really seeing anything simpler after this point @desert fulcrum

[desert fulcrum]

False

[weak cove]

hmm

[desert fulcrum]

The expansion seems wrong to me

[weak cove]

at which step?

[desert fulcrum]

Real part squared + imaginary part squared

So it's √(cos²x + cos²y + 2cosx*cosy + sin²x + sin²y + 2sinx*siny)

cos²x and sin²x get simplified, so do cos²y and sin²y

Then you get an equation in terms of cos(a-b) (cosx*cosy + sinx*siny simplified)

Square both sides

[weak cove]

shouldn't it be

[desert fulcrum]

And then you know that the angle between each of your roots would be like 360/degree of the polynomial (here, 2023), so you have an angle, you have relations between the angles you need, then just divide the probable cases/total cases

[weak cove]

$$\sqrt{(cos{(\theta_1 )}+i\sin{(\theta_1)})^{2}+(\cos{(\theta_2)}+i\sin{(\theta_2)})^{2}}$$

[woven radish]

Austin

[desert fulcrum]

not really

When you're taking the modulus of a complex number, it's the root of real part squared + imaginary part squated

[weak cove]

Oh

[desert fulcrum]

And my bad, you won't put the i while taking the modulus

If your complex numbers are x+iy and a+bi, then your modulus of the sum of those is √((x+a)² + (y+b)²)

Think of it as a vector if you ever get confused, real part is your x component and imaginary is the y

[weak cove]

So then the norm is $$\sqrt{2}\sqrt{1+(\cos{(\theta_1)}\cos{(\theta_2)}+\sin{(\theta_1)}\sin{(\theta_2)})}$$

[woven radish]

Austin

[weak cove]

and cosacosb+sinasinb probably has a trig identity

[desert fulcrum]

yes this is right

[weak cove]

yes and there is a nice trig identity

it comes out to

$\sqrt{2}\sqrt{1+\cos{(\theta_1-\theta_2)}}$

[woven radish]

Austin

[desert fulcrum]

correct

Square both sides

[weak cove]

I might aswell bring the sqrt2 back in

why square both sides

[desert fulcrum]

so that you can eliminate the squares and get everything outside the root

[weak cove]

I want it in the root no?

it looks similar to our desired form

[desert fulcrum]

why though

[weak cove]

if I bring sqrt(2) in

we have

[desert fulcrum]

It's an inequality and you need the probability

[weak cove]

oh nvm

yes

this problem reduces very nicely

[desert fulcrum]

Great

[weak cove]

$$\frac{\sqrt{3}}{2} \leq \cos{(\theta_1-\theta_2)}$$

[woven radish]

Austin

[weak cove]

to this

[desert fulcrum]

I didn't check the calculation but I'll presume it's correct ✅

[weak cove]

and theta1 is the angle for the root v and theta2 for u

[desert fulcrum]

Consider theta1 - theta2 as another variable and find the angles at which this will be true

[weak cove]

well this is true when we are inside -pi/6 <= ..... <= pi/6 and +/- 2pik to both pieces

[desert fulcrum]

remember that this is greater than equal to

So your angle is 0 to π/6

Or

π to 7π/6

no wait

[weak cove]

no

[desert fulcrum]

11π/6 to 2π

[weak cove]

,w plot x=-pi/6, x=pi/6, y=cos(x), y=sqrt(3)/2

[desert fulcrum]

Yes this

[woven radish]

Results provided by WolframAlpha

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Requested by austinu

[desert fulcrum]

Now if you have 5 roots of unity (here), then angle between each root is 2π/5

So 2023 roots means 2π/2023 or 360/2023

[weak cove]

sorry what?

[desert fulcrum]

If you have 5 roots then the angle between each of them (if you plot it on the graph) is 2π/5

[weak cove]

yes

that I understand

we need roots to be chosen

such that they have a distance of pi/3 (or less) between them?

and also these roots need to satisfy that their difference is in some -pi/6+2pik <= their difference <= pi/6+2pik

or am I wrong

[desert fulcrum]

not really

Think of it like

If you have n roots

Then -30<=n*(360/2023)<=30

[weak cove]

but as n gets large that will clearly never be the case?

[desert fulcrum]

It's an inequality

so like

[weak cove]

<=30 is the top of it

[desert fulcrum]

look, if you simplify this you get -168<=n<=168

[weak cove]

I'm not sure what "this" is

[desert fulcrum]

okay look

Exclude the fourth quadrant temporarily

What do you have

0 and 30 as your extremes

Now if you have 10 roots, then the angle between each of them is 360/10 or 36°

[weak cove]

Why does the angle between each of them matter

[desert fulcrum]

Because that shows how many roots you have between the angles 0 and 30°

[weak cove]

0

[desert fulcrum]

Okay make it simpler

If you didn't get it

Let the angle between each root be 5°

Then how many roots would you have between 0 and 10? Both included

[weak cove]

2

but

also there are

[desert fulcrum]

3

[weak cove]

more intervals we care about

[desert fulcrum]

Not 2

[weak cove]

oh yes

I mean 3

dumb mistake

[desert fulcrum]

yes, so we'll double it

[weak cove]

but why double it?

[desert fulcrum]

Because there's another quadrant

Which we're not including in the original inequality

[weak cove]

no but there's even more intervals than that

-30 +2pik <= ... <= 30 + 2pik

[desert fulcrum]

why would you consider second and third quadrant? Cos is negative in that

[weak cove]

not what I mean

not different quadrants

but larger angles

[desert fulcrum]

okay see

If you had an angle of 5 degrees between each root

And you had the same inequality

[weak cove]

sqrt(3)/2 <= cos(P)

[desert fulcrum]

You'd have 0<=n*5<=30

[weak cove]

I actually don't see where this is even going

because even if we find the number of roots included in such an interval

[desert fulcrum]

Ok take it from this step

Hear me out

Simplify this

[weak cove]

their difference could not be in thtat interval.

[desert fulcrum]

Your question now is roots of x^72-1 = 0

[weak cove]

like what if we had 0 and 10

both in 0<= .. <=30

[desert fulcrum]

yes yes hear me out for once

[weak cove]

then 0-10 is no

Okay sorry

[desert fulcrum]

you have x⁷²-1=0

[weak cove]

Sure

[desert fulcrum]

So you have 72 solutions

[weak cove]

yes

[desert fulcrum]

Angle between each of your solution is 360/72 = 5°

[weak cove]

yes

[desert fulcrum]

And all your solutions are uniformly distributed

[weak cove]

yes

[desert fulcrum]

Throughout 1 circle of 360°

[weak cove]

ohhh

all of the solutions are in 0<= 2pi

[desert fulcrum]

now, if any of your solution lies in the second quadrant, would it satisfy the √3/2? It wouldn't

[desert fulcrum]

Correct

[weak cove]

that is what I was forgetting

[desert fulcrum]

So your angle is 360/2023

[weak cove]

we want to know

how many of these are in

0 to 30

and then double it

[desert fulcrum]

Yes

Correct

[weak cove]

and that tells us what?

that tells us all the possible P

for sqrt(3)/2 <= cos(P)

[desert fulcrum]

That gives the number of solutions between -30 to 30