help-10

[glacial obsidian]

It is a definition

[honest forum]

Just fancy notation

[lucid plover]

what does the colon mean

and arrow

[glacial obsidian]

What do the parentheses mean when you write f(x) = x^2 -3x?

[lucid plover]

ah okay

i see

.close

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[next widget]

How do u do b and c

For c I got 36km/h which was wrong

And idk how to do b

[fresh trail]

For part b, the instantaneous velocity is the slope of the line at t=3. So you just need to calculate the slope of that part of the line and that's the answer in km/min, you'll just need to convert it to m/s after.

For part c, they don't specify the units they want, but the process is you find the slope between the end point and the start point. If you do rise/run, you get (6-0)/(10-0) = 3/5 km/min

[next widget]

Alr thx

[next widget]

Yea I had 1ms^-1 but it was wrong apparently

Maybe had something to do with “after 3 minutes” or smthn

[fresh trail]

The slope is definitely 1, but the units are km/min and you have to convert it to ms^-1. It's a bit of a trick question

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[toxic vigil]

why did chain rule not work

[toxic vigil]

i tried it

[gilded needle]

show what you did

[fierce vale]

!show

[obtuse pebble]

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

[toxic vigil]

ok wait

[obtuse pebble]

@toxic vigil Has your question been resolved?

[toxic vigil]

@fierce vale

[gilded needle]

i mean it looks fine up until the point with the -->
i didn't check after that

do you really need to multiply it all out

[toxic vigil]

Idk

But the answer was wrong

[gilded needle]

said who?

[toxic vigil]

The teacher

[gilded needle]

what does the teacher claim is the right answer?

[toxic vigil]

His answer was
3x^2 + 3 -3/x^2 -3/x^4

[gilded needle]

wolfram agrees with your result right before the -->

[fierce vale]

It's the same

[toxic vigil]

weird

[gilded needle]

but that's exactly the same as yours

[toxic vigil]

but the 3 doesnt have a demonerator?

[gilded needle]

just carry out the division in yours, you get the teacher's answer

exactly the same

[toxic vigil]

ohh

ok ty i didnt realize that

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[north anchor]

How to determine if a matrix has one solution, infinite or none?

[upper shuttle]

you put it in rref

if the matrix has a row of zeros = to a non zero, it has no solutions. if it has a row of zeros = 0 you have infinite. if you have the identity matrix you have one solution

[north anchor]

i dont u8nderstand

[upper shuttle]

do you know what is rref?

[north anchor]

no

[obtuse pebble]

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[granite laurel]

No

[obtuse pebble]

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[civic bone]

guys I'm finishing up this proof

https://i.imgur.com/t5ErpGp.png

[civic bone]

when you need to show something is less than epsilon

but you end up showing something is less than C * epsilon

I've seen proofs that go back and select a convenient initial epsilon like epsilon/3 in this case

but I've also seen proofs say what I did at the end

which is since epsilon is arbitrary we've shown what we need to show

is there an actual difference between the two

and cases where you can do one but not the other

[latent walrus]

its fine either way realistically, it just looks nicer

[civic bone]

ok cool cuz I was stressing

and then when I said fix epsilon and take N

is the usage of fix correct there

and also is the usage of fix necessary there

[obtuse pebble]

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[sudden pasture]

Can someone please help me understand why my teacher marked this wrong

[sudden pasture]

@ me if you can please

I need this done before morning and it’s already 10pm

[untold badge]

ok

[untold badge]

the measure of the central angle is the same as the measure of its arc and vice versa

so that would mean he measure of arc ac is 125

the measure of major arc cba is 55+180 which is 235

using central angles

[sudden pasture]

Wahhhtt wait I thought it was twice the amount

Ohh that makes so much sense

Tysm

OHHH

Ok

Ty

@untold badge could you also help with this one

[obtuse pebble]

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[dusk fox]

Help me out please

[dusk fox]

So I know that the distance from the origin to the lomg part of the triangle is 1

Is this right so far

<@&286206848099549185>

Yo <@&286206848099549185> is it too hard for u people

[exotic latch]

shagger help

u don’t need help

#help-48

no

[dusk fox]

<@&286206848099549185> im ready whenever u peopl are

[exotic latch]

wish i could help but i forgot all abt that unit stuff

[teal ledge]

What are you trying to do here?

[dusk fox]

.close

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[oak cliff]

Any subset of S that contains 1 has the form {1} U (some subset of {2, ..., n})

[obtuse raven]

so any set with 1 is a subset of s

?

[oak cliff]

Any set with 1 and other elements of S is a subset of S

[obtuse raven]

so how am i suppsoed to find how many

[bleak eagle]

Do you mean a subset of S has to have 1 as well as additional elements? Cuz that’s not true, {1} is a subset of S

[oak cliff]

The empty set is a subset of {2,..., n}, so with what I said you can have {1} = {1} U {}

[bleak eagle]

Ooops I goofed

My b. Yes that’s 100% correct

[oak cliff]

If every subset containing 1 has the form that I described, then the problem is equivalent to finding how many subsets of {2,..., n} there are

[obtuse raven]

wha

[oak cliff]

Because for every subset of {2,...,n} you can construct a set containing 1, and this will give you all possible sets containing 1

[obtuse raven]

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[rough marten]

any proper website for me to draw loci points?

[rough marten]

desmos doesnt seem to have

for example plotting this

or how to finnd the arg z of a circle

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[median tangle]

how do i find alpha

[obtuse pebble]

@median tangle Has your question been resolved?

[median tangle]

<@&286206848099549185>

pls help i dont even know how to start

[azure karma]

2+3i= alpha

is what i think since...

imaginary numbers can be roots in pairs only

[median tangle]

but i think 2+3i is the 4th root

no?

or can a cubic have only 3 roots?

hello?

[azure karma]

yeah..

wait

[median tangle]

ok

[azure karma]

okay see,
(2+3i)(2-3i)(alpha)(alpha^4)= whatever that eqn is

then you'll get alpha

[median tangle]

ok ill try that thanks

but there is an unkown in the eqn

[azure karma]

just put z=alpha in there then, it should equate to 0 since it's a root

[median tangle]

but c is unknown and alpha is unkown so how i find them?

[azure karma]

no waiiit

see we can do this...

[z-(2+3i)][z-(2-3i)][z-(alpha)][z-(alpha^4)]= whatever that eqn is

do this oncee

and then you'll get cz^2= (something)z^2
then get c and solve for alpha

c=22, for me...

Check it... If you have the answer

[median tangle]

by z you mean the eqn including c yes?

ah ok

let me try

[azure karma]

This is what I solved...
At the end, you'll compare the two equations to get k=1, A=1, B=-2

That's... The whole solution

[median tangle]

thanks a lot for the help

[azure karma]

did you get it though?

[median tangle]

yes

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[delicate relic]

could i get some help on (a)

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[robust charm]

hi yall can someone answer this question

[robust charm]

If someone explains question 4, i can do the 5 they’re similar anyways

[obtuse pebble]

@robust charm Has your question been resolved?

[timid silo]

Put the system in its matrix form and calculate its determinant, for the system to have a single solution its determinant must be different from zero

[obtuse pebble]

@robust charm Has your question been resolved?

[robust charm]

but then i got like millions of solution?

oh okay now i got it its k != -3

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[robust charm]

.reopen

[obtuse pebble]

✅

[robust charm]

can u also answer one more question?

[drifting roost]

just ask away

[drifting roost]

yeah k = -3 gives a contradiction

aka no solutions

u can't have infinite solutions for this system so just k != -3 is sufficient

[robust charm]

Can someone help me with the pink

I did this but i think i dont need to use b^2 -4ac here

The answer is 5/2 or 2 and when i used the b^2-4ac formula i didnt get something like this

[timid silo]

,rotate

[timid silo]

the equation you reached is correct

plug in 2 and it works

also 5/2

[robust charm]

oh i guess i did my calculations wrong then

anyways thank you so much

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[cyan tree]

what does the double bracket mean at x

[rocky shale]

nothing's mentioned

ill assume it to be GIF ig

[cyan tree]

what is GIF

[sacred goblet]

https://www.cuemath.com/algebra/greatest-integer-function/

[whole dock]

GIF is hard to understand

Floor function is much better

[cyan tree]

I know floor function

[whole dock]

Yeah it's the same

Function

[sacred goblet]

yeah ive actually never seen it called this LOL

or that notation, for that matter

[cyan tree]

but how did u knew it was floor function tho?

isn't the notation a bit different

[whole dock]

Yeah

But nothing has been mentioned so I guess that's the only thing you can consider

[high lily]

context, experience

usually when this notation is use, they'd state what it represents

[cyan tree]

but this is floor notation

[high lily]

that's the most asthetically pleasing, and people generally understand that

but there are other alternate representations like the one in the question

[cyan tree]

but how did u knew that notation was floor and not ceil?

[high lily]

context, experience

only ever seen it used for floor

[cyan tree]

and part b is also floor right?

[high lily]

yes, same notation

[cyan tree]

okay I will now give it a try to solve it, thanks

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[fluid snow]

#1195800467880558672 message

[fathom pecan]

@everyone

[obtuse pebble]

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[north hemlock]

cosx=lnx+1 , how do i find x?

[versed stratus]

Graph it

[north hemlock]

i didn't think about that, ty

[fervent cradle]

there shouldn't be a nice solution

[north hemlock]

how do I know the actual number though?

[versed stratus]

by graphing it, and observing that cos(x) is bounded between -1 and 1 , whereas ln(x) isn't

[fervent cradle]

you should only be able to find an approximate solution

there shouldn't be an exact one

[north hemlock]

it is there according to GeoGebra

[fervent cradle]

right

[north hemlock]

you're right regarding the actual question, the answer had to be an interval

i missed that too

it wasn't as hard as i thought, i just made it harder myself haha

[versed stratus]

that's just an approximate soln though

[north hemlock]

thank you both

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[sterile vale]

How does the 2 infront of the theta affect how I answer this question?

[timid silo]

you can solve the equation for 2theta as normal

and then divide your expression by 2 to get the values of theta

[sterile vale]

I got 2pi/3 and 4pi/3

If I divide them by 2, how do I maintain that 3 in the denominator?

(Answers)

[timid silo]

if you get $2\theta = 2\pi/3$ then $\theta = \pi/3$

[warm shale]

ampl

[timid silo]

similarly for $2\theta = 4\pi/3$ youd get $\theta = 2\pi/3$

[warm shale]

ampl

[obtuse pebble]

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[weary furnace]

Sketch the graph x+(8/x^2)

[weary furnace]

Can someone help me here

[obtuse pebble]

@weary furnace Has your question been resolved?

[obtuse pebble]

@weary furnace Has your question been resolved?

[obtuse pebble]

@weary furnace Has your question been resolved?

[fervent cradle]

just let x = 0, 1, 2, 3, 4, etc.

and -1, -2, -3, etc.

and plot the points

and then draw a curve through them

there's more but that'll do to start

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[brisk grove]

Hint: ||what's sin(1) + sin(181)?||

[spice chasm]

well sin181 is sin(180+1)

[brisk grove]

Yeah, ok

Try to draw it on the unit circle

[spice chasm]

which is sin180cos1+sin1cos180

[brisk grove]

You are overcomplicating it

[spice chasm]

wait u just get sin1

so 0

[brisk grove]

The answer is 0, yes, but I don't see how that helps

sin(181) = -sin(1)

Look at the unit circle

,w sin(181)

Ah no picture

[spice chasm]

press more

[brisk grove]

here

[warm shale]

Results provided by WolframAlpha

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

Requested by redstoneplayz09

[spice chasm]

oh vertical angles

so ye theyre the same

[brisk grove]

With a negative

sin(x) = -sin(x + 180) basically

Try using that

[spice chasm]

so doesnt everything cancel

[brisk grove]

Everything indeed does

[spice chasm]

ok ty

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[sand chasm]

can someone just tell me the answer for c real quick

[sand chasm]

need to verify if i am right or not

[gilded needle]

what was your answer?

[sand chasm]

1/12

@gilded needle

[gilded needle]

i agree with that answer, which method did you use

[sand chasm]

i didnt use a method i used gauthmath

which method should i use though?

[gilded needle]

you could use l'hopital's if allowed

[sand chasm]

dk what that is

im in grade 12

[gilded needle]

oh

what methods do you know?

[sand chasm]

the f(a+h) -f(a) /h

[gilded needle]

i see

well can you see what f(x) would be in this case?

[sand chasm]

no

[sand chasm]

wouldnt it just be the equation in the question?

(x+4)^1/3 -2....

[gilded needle]

oops lemme rewrite that

can you find a function f such that:
f(a+h) = (x+4)^(1/3)
f(a) = 2
a = 4

(you'll need to relate h to x for the first equation)

[sand chasm]

you mean (x+4)^(1/3)/x+4 right?

[gilded needle]

why are you dividing by x+4

[sand chasm]

my friend telling be do u=(x+4)^1/3

[gilded needle]

well if you want to use the (f(a+h) - f(a))/h thing, you just need to do pattern matching

no need for letting u = anything

your denominator is x-4

you need denominator = h

so let h = x-4

therefore x = h + 4

etc

[obtuse pebble]

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[dapper dock]

How do I do question 8

Idk what’s going on

[keen crest]

do you know the limit definition of continuous

@dapper dock

[dapper dock]

No

I mean

I do but

I don’t

@keen crest

Jim;-;

.close

[obtuse pebble]

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[keen crest]

@dapper dock

im back

do you still need help

[obtuse pebble]

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[obtuse raven]

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[restive relic]

whats 1+1?

[restive relic]

ok

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[obtuse pebble]

@winged estuary Has your question been resolved?

[winged estuary]

<@&286206848099549185>

[obtuse pebble]

@winged estuary Has your question been resolved?

[winged estuary]

<@&286206848099549185>

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@winged estuary Has your question been resolved?

[royal mulch]

@winged estuary working on your question rn

So when simplifying we can remove the first option since there is no negative sign. From here we need to note two things: (1) $\sqrt{a^4} = a^2$ since $a^2$ is positive, and (2) $\sqrt{a^2} = |a|$ since we don't know if a is positive or not.

[warm shale]

twitch.tv/pencenter

[royal mulch]

From here you can rule out choices 3, 4 and 5, and settle on choice 2 and the answer.

[obtuse pebble]

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[onyx forum]

I do not understand why on 9 and 11 I set to 2 & 1 respectively, but on 10, 12, and 13 I set to zero. Do I set to zero when I have an x or y value that is squared?

[onyx forum]

This is in reference to my work photo ^

Thanks in advance for the help!

[stable hill]

i guess its a convention, but it shouldnt really matter

if it was put in a form that was easiest to understand i would opt for slope-intercept form or general form for circles

[onyx forum]

Oh okay

That makes sense

Thanks!

[stable hill]

youre welcome!

[onyx forum]

😄

[obtuse pebble]

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[spare stump]

whats the difference between:

${n+k-1}\choose{k}$ and ${n+k-1}\choose{k-1}$

[warm shale]

JWIVY

[bronze mica]

in the first you're choosing k, and in the second you're choosing k-1, in both cases out of the same amount

(the formulas are different too to represent this difference, but I'm too lazy to write them out)

[spare stump]

for thios question i use the k-1 one

and another one id use just k

like this id use the k one

so how can i tell when to use

[bronze mica]

in general, I'm not sure I can be more helpful other than 'depending on whether you need to choose k or k-1' which I realise is basically useless

out of curiosity, what did you do for 1 though

[obtuse pebble]

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[rapid fog]

Every integer is either prime or a composition of primes right? Is the true the same for polynomials?

[timid silo]

every integer greater than 1

[rapid fog]

Let $\large F$ be a field and let $\large p(x), a_{1}(x),a_{2}(x), \dots,a_{k}(x)\in F[x]$, where $\large p(x)$ is irreducible over $\large F$. If $\large p(x)\mid a_{1}(x)a_{2}(x)\dots a_{k}(x)$, show that $\large p(x)$ divides some $\large a_{i}(x)$.

[warm shale]

M. P. HOBSON

[rapid fog]

I know that $F[x]/<p(x)>$ is an integral domain and there fore p(x) is prime

[warm shale]

M. P. HOBSON

[rapid fog]

Where can i go from there? If p is a prime polynomial that means that one of the factors being divided must be prime aswell?

[obtuse pebble]

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[rapid fog]

No

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[left pawn]

see those last 2 steps?

where does the first -x go?

i understand this is just distribution, so I understand why you'd have -1, but to me its like the x just disappeared

[warm canopy]

They divide the top and bottom of the fraction by x

[twilit meteor]

they divided each terrm iin the numeratorr and denominator by. x

[left pawn]

ohhhhh I didn't even see the numerator, that wouldve given it away

ok thanks

[obtuse pebble]

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[tough flax]

Can someone help me with this question pls

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[obtuse pebble]

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[dapper dock]

Idk what I’m doing wrong for 20

Oh

Nvm

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[potent bramble]

i have this question and i think the answer is U Ai = {2n | n "in" Z} but i am not sure ?

[fathom flicker]

how do you know if two sets are equal

[potent bramble]

the set of 2n where n is in Z {.....,-3,-2,-1,0,1,2,3....}

[fathom flicker]

?

[potent bramble]

do you know how to solve this ?

[fathom flicker]

Yes

I am asking you a guiding question

[potent bramble]

how please ??

[fathom flicker]

how do you know if 2 sets are equal?

[potent bramble]

same members ?

[fathom flicker]

let's be more specific

suppose A and B are sets

what do we need to know to say A=B

[potent bramble]

every element in A is also in B

[fathom flicker]

and

[potent bramble]

idk ??

[fathom flicker]

every element in B is also in A

[potent bramble]

yes

[fathom flicker]

so for you to check if your answer works

you'd have to show those 2 things

[potent bramble]

and here i think that A_i = 2n and n is an element of Z

i is an element of N so there are no negative numbers

[fathom flicker]

well hold on

A_i is the set {-2i, 0, 2i}

[potent bramble]

thus the set is = {..., -4,-2,0,2,4,...} = {2n | n is an element of Z}

[fathom flicker]

the set {2n | n in Z} is correct but you're saying it incorrectly

[fathom flicker]

so for example if i=1, {-2,0,2}

there's definitely negatives

[potent bramble]

yes but what i meant is that i>0 or i in N

[fathom flicker]

sure

[potent bramble]

so how can i write it correctly ?

[fathom flicker]

this is fine, I just thought you were saying

Ai={2n | n in Z}

but I see that's not what you meant

[potent bramble]

okay cool ! is there a better way to write it ?

is this right ?

[dull linden]

Yes

Well the union is

[potent bramble]

yes i just need it for the union

[dull linden]

$\bigcup_{i \in \mathbb{N}}A_i = {2n : n \in \mathbb{Z}}$

[warm shale]

aldo booze

[dull linden]

You were right from the beginning I'm pretty sure

[potent bramble]

and intersection is {0} right ?

[dull linden]

Yeah

[potent bramble]

Thanks guys !

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[graceful marten]

So
I am combining the
16 x S3 = a/ 1-r

With s3 =
a(1-r^3) / 1-r

Why is my
[red] wrong

Are you not supposed to multiply
16 x S3.
Both separately by (1-r)

But instead like this
(16 x S3) (1-r)

Bc

(16 x S3/1-r) x 1-r

I guess does not give you 16 (1-r) x S3

hello frien

[crude coral]

S infinity = 16 * S 3

why are u saying 16 * S 3 = a/1-r?

[graceful marten]

Cause s infinity is a/1-r

[crude coral]

no?

[graceful marten]

Its for a convergent seires

[crude coral]

oh ok

[graceful marten]

Srry

16 * S3/(1-r)= a/1-r

[crude coral]

i mean then what i sent would be correct

[graceful marten]

I dont see it?

[crude coral]

i deleted it

[graceful marten]

I guess if you move over the 1-r from the right side to the left

Would it be

( 16 * S3/(1-r) ) (1-r) =a

Or
16(1-r) * S3 = a

[crude coral]

this and 1-r cancels

[graceful marten]

Oh then it would make

16x s3 = a

Withouth the (1/r) of s3

[crude coral]

how u get S3/1-r?

[graceful marten]

Oh no S3 = like a(1-r^3)/ (1-r)

I just gave S3 a value

just to make it a bit clear

On what im getting lost at

(aka division/ canceling part)

[crude coral]

u know what S3 is and u know S infinity = 16 * S3

[graceful marten]

Oh yeah u can just do 16 x a(ect)

lol

That works?

Oh yeah i guess so

welp

Thank you

[crude coral]

it should

[graceful marten]

Ill try both ways

Lets see

:0

I dont think so

I dunno

Ill try it

I need water tho first

thanks

:0

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[obtuse pebble]

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[wary dagger]

i have done 8 and a half hours of schoolwork today my brain is mashed potatos please explain how to turn 229.5/70.5 into a percentage

[ember frost]

express the fraction with decimal

multiply by 100, add %

[wary dagger]

thanks

[ember frost]

anytime

[wary dagger]

.close

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[runic arrow]

How do you prove c

[timid silo]

,rccw

[timber pine]

are these complex numbers

[timid silo]

yes

[runic arrow]

yh

[timid silo]

they are complex numbers

[civic sapphire]

You could just write both as a+bi and see the values match

[timber pine]

what if z1 = a+bi z2 = c+di

since that would be a bit confusing

ok

[ember frost]

<@&268886789983436800>

[timid silo]

Why, what is wrong co-worshipper?

What did I even do

[civic sapphire]

Yes, something like that Cosmo

[timber pine]

you do it then

[ember frost]

ignore him

go on as normal

[timid silo]

is z your mom's body count?

🤡

[timber pine]

z1.z2 = a(c+di)+b(c+di) = ac+adi+bc+bdi

wait

what does the line mean

above the z

[timid silo]

congugate

[timber pine]

conjugate or

oh

[civic sapphire]

conjugate I think

[runic arrow]

yea

[timid silo]

its congugate

[runic arrow]

so i used a - but then idk what the left side means

[timber pine]

z1.z2 = a(c+di)+b(c+di) = ac+adi+bc+bdi = ac+bc + i(ad+bd), conj = ac+bc-i(ad+bd)

[timid silo]

$ z=$a+ib then congugate $z=$a-ib

[warm shale]

EinPest

[crude coral]

first simplify the lhs and then simplify rhs and check if they are equal
let z1 = a+bi and z2 = c+di

[civic sapphire]

c+di

[timber pine]

hold on

[timid silo]

u can check multiplying

[timber pine]

this is wrong

[civic sapphire]

you're off by an i

b(c+di)i

[timber pine]

z1.z2 = a(c+di)+bi(c+di) = ac+adi+bci-bd = ac-bd+i(ad+bc), conj = ac-bd-i(ad+bc)

[timid silo]

multiplying the congugates, you will get $ac-bd-i(ad+bc)$

[warm shale]

EinPest

[timid silo]

same thing cosmos got

[timber pine]

a(c-di)-bi(c-di) = ac-adi-bci-bd = ac-bd-i(ad+bc)

z1.z2 = a(c+di)+bi(c+di) = ac+adi+bci-bd = ac-bd+i(ad+bc), conj = ac-bd-i(ad+bc)
conj z1.conj z2 = a(c-di)-bi(c-di) = ac-adi-bci-bd = ac-bd-i(ad+bc)
LHS=RHS

[runic arrow]

ok ty

.close

[obtuse pebble]

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[boreal cloak]

i dont know how to solve this differential equation

i try to separate the variables but i dont get the same result

[versed stratus]

what's sen(y)?

[boreal cloak]

??

[versed stratus]

you have sen^2(y) in you're DE

is that sin?

[boreal cloak]

yes yes

is in spanish

sorry

i did this

[versed stratus]

oh, sen is sec?

OK

[boreal cloak]

nop

sorry

sin is sen

i integrate both sides

but i dont get the same result

[versed stratus]

$\left(\tan x\right)\sin^2\left(y\right)dx=-\cos^2\left(x\right)\cot\left(y\right)dy$

[warm shale]

Why am. I here

[versed stratus]

can you do it from here

you'll get $\tan\left(x\right)\sec^2\left(x\right)dx=-\cot\left(y\right)\operatorname{cosec}^2\left(y\right)dy$

[warm shale]

Why am. I here

[boreal cloak]

yes

[versed stratus]

which can be integrated using a u-sub on each side

[boreal cloak]

yes

[versed stratus]

so what do you get?

[boreal cloak]

let me do it again

but if is u sub

this is what I get

[versed stratus]

you should get $(tan(x))^2=(cot(x)^2+C$

[warm shale]

Why am. I here

[boreal cloak]

i did this

i will try to write in teXit

so first i divide and get this
$[ \frac{\sin x}{\cos x} \cdot \frac{1}{\cos^2 x} dx + \frac{\cos y}{\sin y} \cdot \frac{1}{\sin^2 y} dy = 0 ]$

[warm shale]

Reist

[boreal cloak]

$[ \int \left( \frac{\sin x}{\cos^3 x} \right) , dx - \int \left( \frac{\cos x}{\sin^3 y} \right) , dy = 0 ]$

[warm shale]

Reist

[boreal cloak]

and then i use u sub, for u=cos (x) and for the right side sin(y) $[ -\frac{\sec^2x}{-2} + C_1 = -\frac{\csc^2y}{-2} + C_2 ]$

[warm shale]

Reist

[boreal cloak]

is the same thing

[obtuse pebble]

@boreal cloak Has your question been resolved?

[obtuse pebble]

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[graceful marten]

Quick Question

Ik you can do it via this

But shouldnt there be another method available?

Something like

2 + 2n + ??? .... 2n

and then

2n .... ??? + 2n + 2

Something like this

if you dont understand what im going on about
Basically isnt there another method to prove this

Ty :3

[dusk widget]

looks like you should use induction for this

[graceful marten]

How would I go about it?

:3

[supple stump]

helper is here

[timid silo]

I think you can use the geometric series

[supple stump]

answer is....

23.....

jk bro

[wise lava]

Induction have two parts

1. Prove that the base case(n=1) is true
2. Assuming that some case is true, prove that the next case is also true

[drifting roost]

why do you need another way btw?

induction takes more effort than what you posted tbh

[supple stump]

Sn= n+n2

[obtuse pebble]

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[open mountain]

What does it mean that the set of real numbers is bounded from above?

[upper shuttle]

it isn't

[open mountain]

If it were a hypothesis what would it mean?

[upper shuttle]

there exists a number that is bigger than all numbers

[open mountain]

ok

.close

[obtuse pebble]

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[open mountain]

.reopen

[obtuse pebble]

✅

[upper shuttle]

major? N?

[open mountain]

What is the minor major of N?

[wise lava]

what is minor major

[open mountain]

majoring

the smallest

understand?

[wise lava]

but what is major

[open mountain]

il piu piccolo maggiorante

in italian

[wise lava]

oh you meant supremum? least upper bound?

[open mountain]

yep

[wise lava]

Assuming the N here is the set of naturals
the supremum does not exist since N is not bounded above

[open mountain]

yes but n <= M

[wise lava]

and?

[open mountain]

so M is the least upper bound

?

[open mountain]

why

[wise lava]

n is a natural number

[open mountain]

like 1 2 3 4 5 6 7

?

+infinity is not a number

[wise lava]

did you mean a subset of N

[open mountain]

.close

[obtuse pebble]

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[wise lava]

If its is a finite subset of N then yes

[open mountain]

i understand

thx

[obtuse pebble]

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[vivid ivy]

I need to show that 5^2n+3 -4^n+1 is divisible by 11 by recurence

[obtuse pebble]

Please don't occupy multiple help channels.

[cunning breach]

You should probably emphasize your terms. Is it $5^{2n+3} - 4^{n+1}$ ?

[warm shale]

Good

[obtuse pebble]

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[open mountain]

\textbf{PROPOSITION.} --- $a_n$ converge to zero if and only if $|a_n|$ converge to zero.

\textit{Proof:} Given $b_n = |a_n|$, $b_n$ converge to zero if and only if $\forall \epsilon > 0$, $\exists v: |b_n| < \epsilon$, $\forall n > v$. Since $|b_n| = ||a_n|| = |a_n|$, $\forall n \in \mathbb{N}$, which is equivalent to the convergence to zero of the sequence $a_n$.

[warm shale]

alee

[open mountain]

i dont understand this

all

[obtuse pebble]

@open mountain Has your question been resolved?

[obtuse pebble]

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[open mountain]

\textbf{PROPOSITION.} --- $a_n$ converge to zero if and only if $|a_n|$ converge to zero.

\textit{Proof:} Given $b_n = |a_n|$, $b_n$ converge to zero if and only if $\forall \epsilon > 0$, $\exists v: |b_n| < \epsilon$, $\forall n > v$. Since $|b_n| = ||||a_n|||| = |a_n|$, $\forall n \in \mathbb{N}$, which is equivalent to the convergence to zero of the sequence $a_n$.

[warm shale]

alee

[open mountain]

i dont understand all

<@&286206848099549185>

[tardy epoch]

do you know what $|a_n|$ means?

[warm shale]

riemann

[open mountain]

yes absolute value an

[glossy estuary]

Wtf is this shit

[tardy epoch]

!redir

[obtuse pebble]

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

[glossy estuary]

Oh mb

Typed in wrong channel

[tardy epoch]

okay so what part of the proposition don't you get?

[open mountain]

i have a problem

\textbf{PROPOSITION.} --- $a_n$ converge to zero if and only if $|a_n|$ converge to zero.

\textit{Proof:} Given $b_n = |a_n|$, $b_n$ converge to zero if and only if $\forall \epsilon > 0$, $\exists v: |b_n| < \epsilon$, $\forall n > v$. Since $|b_n| = ||a_n|| = |a_n|$, $\forall n \in \mathbb{N}$, which is equivalent to the convergence to zero of the sequence $a_n$.

[warm shale]

alee

[open mountain]

ok fixed

[tardy epoch]

you really don't need to type the same thing

[open mountain]

it was ||a_n||

[tardy epoch]

just edit the old one

[open mountain]

it was || an ||

[tardy epoch]

what about it

[open mountain]

on discord i cant type double ||

[tardy epoch]

you said you know absolute value

use `

[open mountain]

it was double absolute value

[tardy epoch]

||a_n||

[tardy epoch]

why is that a problem

[open mountain]

i dont understand the part

where is the ||

a_n

[tardy epoch]

you said you understood absolute value

find the absolute value of the absolute value

[open mountain]

i dont understand the proof

[tardy epoch]

you also don't need to keep repeating yourself

you need to be more specific

[tardy epoch]

.

[open mountain]

WAIT

[tardy epoch]

i told you the double absolute value bars here

[open mountain]

i dont understand

why it is

[tardy epoch]

.

[open mountain]

which is equivalent to the convergence to zero of the sequence $a_n$.

[warm shale]

alee

[open mountain]

this

[tardy epoch]

which words in here don't you get

equivalent?

[open mountain]

yes

[tardy epoch]

,w define equivalent

[warm shale]

Results provided by WolframAlpha

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

Requested by rie.mann

[open mountain]

but why

[tardy epoch]

why what

[tardy epoch]

.

[open mountain]

why its equivalent

[tardy epoch]

they tell you in the proof

you need to understand the proof sentence by sentence

you can't just skip to the end and expect to understand

[obtuse pebble]

@open mountain Has your question been resolved?

[obtuse pebble]

@open mountain Has your question been resolved?

[open mountain]

If $x \geq 0$ then $|x| = x$, and if $x$ is arbitrary, then $|x|\geq 0$
so if you take absolute value twice, you just get absolute value

so $||a_n|-0| = |a_n| = |a_n-0|$
if you plug in what it means for $|a_n|\to 0$ and what it means to $a_n\to 0$, then you'll see the two statements are precisely the same

[warm shale]

alee

[open mountain]

@tardy epoch

That's the definition. $a_n\to 0$ if and only if $|a_n|\to 0$.

[warm shale]

alee

[open mountain]

is correct?

.close

[obtuse pebble]

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[obtuse pebble]

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[hollow frost]

How to solve this

-1 exponent

[delicate kestrel]

bring the number in the denominator

[hollow frost]

Wait

[delicate kestrel]

but if zero is in denomiator then the value is not defined

[hollow frost]

Wrong pic

X=0

[delicate kestrel]

here the -1 exponent cancel out

[hollow frost]

why

[delicate kestrel]

but you could bring the -1 exponent on the numerator where it would become +1

[delicate kestrel]

because its on numerator and denominator both

and the terms arebeing multplied

[hollow frost]

Ohh

So

0/0

🏃🏼‍♂️

[delicate kestrel]

what is 0/0?

[hollow frost]

Indeterminate

[delicate kestrel]

yeah

[hollow frost]

Thankz

[delicate kestrel]

you cant just substitude 0 here

[hollow frost]

oh?

[delicate kestrel]

is this is a limits question?

[hollow frost]

Yes

Lim
X=>0

[delicate kestrel]

if you are getting an indeterminate form then you should try to simplify the equation

[hollow frost]

Oh

So it's not indeter.?

[delicate kestrel]

yeah

[hollow frost]

What is it then

But

It's all 0

[delicate kestrel]

it is if you just substitute

[hollow frost]

Ah

[delicate kestrel]

you need to simplify the equation first

[hollow frost]

Ahh

[delicate kestrel]

now when you substitute 0

you get 2/4

= 1/2

[hollow frost]

Waut

Wait

How did u erase 0

2/0

[delicate kestrel]

i multplied the numerator and denominator both by x

[hollow frost]

When are we going to do that

[delicate kestrel]

wdym

[delicate kestrel]

in this step

[hollow frost]

Ah

Tyy

[delicate kestrel]

least i can do :)

[obtuse pebble]

@hollow frost Has your question been resolved?

[obtuse pebble]

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[cedar kettle]

i dont understand how he got y

[timid silo]

by evaluating the function on x

[cedar kettle]

Yes.

but i put it through g

and i get 39

[timid silo]

for which x

[cedar kettle]

-2

[timid silo]

,calc (-2 - 4)^2 + 3

[warm shale]

Result:

39

[timid silo]

oh well then

ig he made an error

[cedar kettle]

oh what

what error do you think he could have made

[timid silo]

adding 4 instead of subtracting it i think

[cedar kettle]

how am i supposed to graph 39?

what the heck

[timid silo]

don’t

because it’s outside the graph you don’t have to graph it

instead you find values of x that produce values of y inside the graph and graph those

[obtuse pebble]

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[vast torrent]

isnt it just 5 x 2 and 6 x 3

so 10 v18?

[arctic spade]

You can simplify √18

[arctic spade]

But yeah it's correct

[vast torrent]

ahhh thats why the correction said i was wrong

so id just break it up

so it becomes 10 v2x3x3
then 30v2

[arctic spade]

Exactly

[vast torrent]

ty

.close

[obtuse pebble]

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[obtuse pebble]

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[winged timber]

Can someone help me learn how to solve this please

[devout solar]

What

[obtuse pebble]

@winged timber Has your question been resolved?

[winged timber]

Yep, sorry for late response

[obtuse pebble]

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[unkempt ridge]

can someone help with this

ohh

because it implies

that its the inverse

and thats not possible

<@&286206848099549185>

[obtuse pebble]

@unkempt ridge Has your question been resolved?

[cinder basalt]

Well, check if A^-1BA^T times the given inverse is I

[obtuse pebble]

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[high quiver]

Bruh

[polar fossil]

@winged copper stop

[obtuse pebble]

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[dull edge]

Please help with the DFA chart.

[dull edge]

I keep redoing the graph but losing myself after starting the DO and D1 digits for the graph

!status

[obtuse pebble]

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

[dull edge]

2

!status 2

[obtuse pebble]

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

@dull edge Has your question been resolved?

[dull edge]

<@&286206848099549185> Apologies for the ping!

[timid silo]

That’s pretty cool

Is this for discrete math 2 or something?

[dull edge]

its for lexical analysis

[obtuse pebble]

@dull edge Has your question been resolved?

[dull edge]

.close

[obtuse pebble]

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[obtuse pebble]

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