help-13

[fervent pike]

theres no examples either in the textbook

so I just googled an example

and 'followed' it

but it makes sense I think... cause ep is a small number close to 0

[cerulean sail]

But what did you do to get from the first line to the second there?

[fervent pike]

rationale..🥹

[cerulean sail]

How'd you mean?

[fervent pike]

I just assumed lol

yea

I mean, I could do n <1/ep^2

but I need to get it so that its n >

[cerulean sail]

it would already be a > though...

e.g. multiply both sides by n (a positive integer), then divide by eps^2 (which is also positive)

[fervent pike]

oh

omg

ahhhhhahhahahahahhahaaahahhhhhhhhhhhhhhhhh

[fervent pike]

inequality

got it

[cerulean sail]

hehe

[cerulean sail]

What about the other way round?

[fervent pike]

ehhh

i'll worry abt that later, i'll write down sol for the first way first

ok

hm

show by counterex

|an| = (-1)^n converges

thats -1,1,-1..

then an is convergent, but its not

so its a counterexample

[cerulean sail]

[cerulean sail]

(could be phrased slightly differently, I'm sure you have, but of course, you could say that if a(n) = (-1)^n, that |a(n)| = 1 converges whereas the original doesn't)

[fervent pike]

I think I need some guidance for these questions

I know how to do limit proof when it converges(minus my inequality mess up🤨 its ok shhh) so for converges, you have: given eps>0, there is an n_0 s.t n>n_0 for |an-L|<eps

[cerulean sail]

[you may want to explain definition 4.7 that they're referring to ]

[cerulean sail]

Cool

Was thinking it'd be that

[cerulean sail]

(for these, it's more algebra and stuff I would think)

[fervent pike]

wtf how am I supposed to isolate n in n-sqrt(n) > M

M, P same thing

[cerulean sail]

hint: how does n relate to sqrt{n}

[fervent pike]

(sqrt{n})^2= n

hmmmm

if I square both sides, I get a quadratic for the n side

thats not going to help isolate for one n

hmmm wait

ima try smth

im stupid i cant do algebra

[fervent pike]

quadratic indeed works

lovely😃

holy this took an hour bruhh

[cerulean sail]

oh wow

[cerulean sail]

Yep, my hint would've been you could have completed the square for it

(if I go quiet please do ping me if you need me )

[fervent pike]

if they say a sequence is bounded, it means both upper and lower right?

so convergent

@cerulean sail

[cerulean sail]

Yep, there's an upper and lower bound, but...

[fervent pike]

oh doesnt have to converge

cause you can have sinx

[cerulean sail]

Yep, also (-1)^n is bounded too

[fervent pike]

yes ok

[cerulean sail]

However, all bounded sequences have a convergent subsequence, that might be what you're thinking of?

[fervent pike]

oooh yes we did cover that in class

I dont remember anything tho😅

this is the question

I'll review notes for subsequences

[cerulean sail]

Awww any ideas for it, knowing the definition of bounded?

[fervent pike]

knowing the defnition of bounded? I dont know the definition of bounded except it has a sup, inf, min, max

[cerulean sail]

sup and inf will do

[fervent pike]

I should restate that, bounded as in bounded sequences. cause bounded intervals doesnt have to have a max or min if its open interval

[cerulean sail]

Neither do bounded sequences

(you could e.g. say a sequence is bounded if its terms live in a bounded interval)

For example, the sequence 1/n has no minimum, but is convergent to zero (hence bounded), without attaining that limit anywhere

[fervent pike]

I dont know what counts as suffice

for a proof

this is more like an explanation

[cerulean sail]

Try and prove it from the definition as per here

(note that the fact there's an upper bound is kinda irrelevant, we don't really care if it does, and being bounded below alone would have been sufficient)

You don't even strictly need the greatest lower bound, though you can take it if you wanted(!)

[fervent pike]

not sure how to word it or show it

but because bn is bounded, then you can just add/multiply/whatever operation on n_0 with a finite number

[cerulean sail]

Do you require M to be positive?

If so then it's not a massive deal, but anyways

[fervent pike]

M is positive because its going to infinity right?

negative if its negative infinity?

[cerulean sail]

Remember that b(n) has some lower bound, say, d

[cerulean sail]

I meant more as in the definition: it makes not a massive difference anyway, if P can be anything, it can be anything [strictly] positive, and if M must be positive, if you were given something negative, say P, then of course anything positive is greater than P

[note the last part that I said!]

[fervent pike]

P can be anything, what do you mean strictly positive

so has to be p >0?

By M I mean P

I just like using M instead of P

[cerulean sail]

As in, if P can be arbitary, then you can also choose P > 0 and fit the definition you wrote

[fervent pike]

anythinig positive is greater than P?

[cerulean sail]

As in, if your definition is that a(n) goes to +infinity, then for any M > 0, it's like you said here, then if I tried to give you a negative M, you could instead replace it with something like 1, and say "the sequence is bigger than 1, which is bigger than that negative M you picked, so ha!"

[fervent pike]

right.. that makes sense

i get that

[cerulean sail]

anyways, for now, let's just assume the definition is that a sequence $c_n$ tends to $\infty$ if, for any given $M > 0$, there is an $n_0 \in \bN$ such that $c_n > M$ whenever $n \geq n_0$

[wraith dagger]

@cerulean sail

@cerulean sail

@cerulean sail

[cerulean sail]

any ideas? remember that b(n) >= d

[fervent pike]

because b(n) >=d, I guess we can choose any n > d. let n_0 be d

idkkk

i cant do it without a value???

how am I supposed to choose a n_0, if I cant solve for n

[cerulean sail]

You don't need an explicit n0 really here

[cerulean sail]

(and that wasn't the idea I was thinking of )

if we have $b_n \geq d$, then what can we say about $a_n + b_n$?

[wraith dagger]

@cerulean sail

[fervent pike]

an+bn >= an+d

oh so show that an+d tends to infinity

then since an+bn is bounded above it must tend to infinity as well

[cerulean sail]

a(n) + b(n) isn't bounded above (even if b(n) were bounded above by itself), think about it like the a(n) "overpowering" any boundedness

[fervent pike]

ok

[cerulean sail]

Anyways, let's think about this here this is a good idea, if we can show that this must be bigger than the M we picked, then so must the a(n) + b(n)

[fervent pike]

because d is an arbitrary real (not negative inf), cant you just say the limit of d is itself?

cause its not a sequence persay, it was a value we picked from the sequence of bn

[cerulean sail]

it is, sure, I guess

Though we kinda wanna show this directly, so we want to show that a(n) + d must be greater than M (no matter what we picked M as)

That is, we want a(n) + d > M...

[fervent pike]

well then an > M-d

[cerulean sail]

Uh-huh, and since a(n) tends to infinity, what could we say now(?)

[fervent pike]

ohhhhh

Can I just set P = M-d and since P can be any given value, where P>0, I say that an > P and it tends to infinity. Then an+d > an >P = M-d

umm idk if thats what you had in mind

so we have found a P>0 s.t an+d > P for all n > n_0

[cerulean sail]

That is kind of what I had in mind butttt...

[cerulean sail]

...remember that the lower bound d could be anything, as could M, what should we do if we happened to have that M - d wasn't strictly positive?

[fervent pike]

ahhhh

hmmmm

[cerulean sail]

Got a game for ya

Think of a strictly positive number, any positive number you want

[fervent pike]

ok

yea

got it

[cerulean sail]

it as in a number, or how to go about the question

[fervent pike]

huh?

how to go about what question

you mean set the number to d? or a diff thing

[cerulean sail]

I'll take that as you just thought of a number then what number did you think of?

[fervent pike]

5

[cerulean sail]

Alright, cool, now, we have that a(n) tends to infinity, doesn't it?

[fervent pike]

yeaaa?

[cerulean sail]

Uh-huh, and 5 is a positive number, isn't it?

[fervent pike]

yeaa?

[cerulean sail]

And the definition of tending to infinity was?

[fervent pike]

for M>0 there exist n_0 s.t an>M for all n>n_0

tending to infinity means its not bounded in the positive

diverges

[cerulean sail]

Yep, any M we want

[cerulean sail]

As long as it's positive

[fervent pike]

right

well if it was negatuve infinity, it'd just be M<0 ....... an<M....

but in this question it cant be negative infinity

because an "overpowers" d

so even if d is negative, its still going to be M>0

[cerulean sail]

But remember the fact that I can pick any positive M

Say that the lower bound d was -1 and that I give you M = -1/2

Anyways, you might be saying what I want you to, if phrased a bit ambiguously

Let's go back to the fact that 5 is positive

[fervent pike]

ok

[cerulean sail]

If we can do this for any M > 0, and 5 > 0, then...?

[fervent pike]

we can say an>5>0?

but even if d is negative you have an>0>d, an+(-d) > d+(-d) =0

[cerulean sail]

and, in turn, is greater than M - d

[fervent pike]

so theres always going to be a value inbetween (an and 0) >M

d is in (0, inf)

an > 5...d >0 > M-d

if its negative then thats a diff case

but an -d >0

so even if an > 0>..d..>M-d

it still shows that an +d >M

[cerulean sail]

Baaassically, sure

[cerulean sail]

The idea is that if M is strictly greater than d, then this is good enough...

...but if it isn't, and M is not greater than d, then pick a positive number of your choosing and use that in the place of M - d, then you can find an n0 such that all terms a(n) from then on are above that positive number, and consequently above M - d

in either case, we can always find that there's an n0 such that for any n that's at least n0, we have that a(n) > M - d, so that a(n) + d > M, so that...

[fervent pike]

means a(n)+d>M goes to inf.

and if that goes to inf

because an+bn >= an+d

an+bn also goes to inf

[cerulean sail]

There you go

[fervent pike]

🙂 thank you!!!

ok im gonna sleep now

big day tmr!!!

thank youuuuuu

I really appreciate your help :))

@cerulean sail

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[spark shore]

The first one of 10.29

Does anyone know easy way of solving it

<@&286206848099549185>

[digital cliff]

factorise everything first

!15m

[cedar kiln]

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[spark shore]

Wdym

[digital cliff]

factorise all the denominators

[spark shore]

Im not good at English maybe show me

@digital cliff

[digital cliff]

eg i could write x^2+3x+2 as (x+2)(x+1)

that would be factorisation

[spark shore]

Oh

Idk how to factorise

Bruh I'll just go to chat gpt this discord never explains stuff fully

2/10

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[muted bear]

thats 1/5 because you can split the denominator of 10 into 2*5, since 2 is the same as 1*2, the fraction can be converted into 1*2/5*2, the factors of 2 cancel to get 1/5

thats explained fully :D

[cedar kiln]

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[real field]

need help Figuring out B and C.I struggle setting up the equation to solve the problem

[real field]

I have it set up as x*2-2x^2 but this is giving me a negative and my answer should be positive.Im not seeing why the 2x^2 would go first

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[clear halo]

Correct way to do it:

[clear halo]

the way i did it:

why is it wrong the way i did it?

[livid hound]

squaring each individual term isn't a valid step in manipulation

[clear halo]

so i cant always do the ^2

[livid hound]

not to each term like that

[clear halo]

when is it allowed then

i know its allowed when theres like x^2=4

i mean \sqrt x=4

**

[livid hound]

you can square individual terms if you have a single term on each side
(but even then you should note any domain and range restrictions)

[clear halo]

so its only when there's one on each side

[livid hound]

well more generally you're allowed to square the whole of each side

a + b = c
squaring both sides would result in
(a+b)^2 = c^2

[clear halo]

why isnt it a^2+b^2 = c^2

[livid hound]

by definition of squaring, distributive property and/or binomial theorem

apply the definition of squaring to
(a+b)^2
then expand

[clear halo]

so the a+b on the left side is thought of as one entity, thats why they are grouped as (a+b)^2

[livid hound]

yes

the whole side is a+b
(a+b)^2 represents the square of that

[clear halo]

ok

thanks

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[verbal swift]

I ask you for some help on understanding how to work this out
James Cooke, an undercover agent, has received a tip from an anonymous entity. It’s come in the form of a code, and with enemies around every corner, Agent Cook must decipher the hidden number before it’s too late. Unfortunately, Agent Koek is, uhm… Not the tallest tree in the forest, so he’s going to need your help!

[To: Agent Cake]

{ 3 X 2ABCDE = ABCDE2 }

[cedar kiln]

@verbal swift Has your question been resolved?

[verbal swift]

<@&286206848099549185>

[cedar kiln]

@verbal swift Has your question been resolved?

[verbal swift]

<@&286206848099549185>

[sleek locust]

How to even solve this , seems like there is no way!

Imo

[verbal swift]

dam

[sleek locust]

Multiplication is Commutative.isn't it?

[verbal swift]

it is

[sleek locust]

"X" 👈 this is multiplication symbol?

[verbal swift]

yes

[sleek locust]

It's" 3 x Abcde2"

Hidden number is 3

[verbal swift]

[Each letter represents a digit, making a complete six-digit number]

[verbal swift]

?

[sleek locust]

My mind is fecked up , sry bro...

[verbal swift]

it's okag

.close

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[snow schooner]

Yall How tf do I do this question, I can kinda see what direction I'm supposed to go with it but every time I do I Can't Get the right answer, can anyone help?

[sonic robin]

ah

whats the question

whats this

[snow schooner]

Addition sign

[sonic robin]

oh

root 12 can be written as 2root3

then simplify accordingly

[snow schooner]

What

Can you walk me through that?

[burnt vapor]

12 is 4 times 3, right?

[snow schooner]

Yeah

[burnt vapor]

So $\sqrt{12}=\sqrt{3\cdot 4}$

[wraith dagger]

d

[burnt vapor]

Now, remember that $\sqrt{ab}=\sqrt{a}\sqrt{b}$

[wraith dagger]

d

[snow schooner]

What about the 3 outside of the root symbol

[burnt vapor]

Do you know that property?

[burnt vapor]

We'll simplify it later

[snow schooner]

I think so? It's firmiliar

[snow schooner]

Alright

[burnt vapor]

If you have a square root and a product inside, then you can do the square root of each number and then multiply

For example

$\sqrt{4\cdot 3} = \sqrt{3}\sqrt{4}$

[wraith dagger]

d

[burnt vapor]

And this is very useful because the square root of 4 is a simple number

[snow schooner]

2

[burnt vapor]

So $\sqrt{12} = 2\sqrt{3}$

[wraith dagger]

d

[snow schooner]

Then would we multiply the 3 and the 2?

Or add?

One of those

[burnt vapor]

You mean the 3 outside the square root of 12, right?

[snow schooner]

Yeah

[burnt vapor]

You multiply

It is 3 times sqrt(12)

So you obtain $3\cdot(2\cdot\sqrt{3})$

[wraith dagger]

d

[snow schooner]

and that's 6sqrt3 right?

[burnt vapor]

Yes

[snow schooner]

So then after I add that to the 4sqrt3 I'd get 10sqrt3?

[burnt vapor]

Correct

[snow schooner]

Do I simplify anymore after that?

[burnt vapor]

I believe that is the most simple form

[snow schooner]

Alright. Thank you so much!!!

[burnt vapor]

Maybe it's not, it depends on who you ask

Some people consider that the 10 must go inside the square root

Like $2\sqrt{3}=\sqrt{12}$ before

[wraith dagger]

d

[burnt vapor]

$10\sqrt{3}=\sqrt{300}$

[wraith dagger]

d

[snow schooner]

Mhm, 10sqrt3 is on the answer bank so I think that's what my teacher wants

[burnt vapor]

Ok

[snow schooner]

Thank you again

[cedar kiln]

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[unborn bobcat]

The book is wrong..or am I?

[muted bear]

can you specify what

[dire geode]

Did you even say anything?

[cedar kiln]

@unborn bobcat Has your question been resolved?

[unborn bobcat]

How is Table 1.6 a summary of table 1.7 if the numbers don’t add up correctly

[unborn bobcat]

How is Table 1.6 a summary of table 1.7 if the numbers don’t add up correctly

[muted bear]

they seem to add up, which data point doesnt seem to add up

?

[muted bear]

...ok

i can see that

[unborn bobcat]

Can you please explain how the numbers in table 1.6 are accounting for table 1.7?

[unborn bobcat]

Where are the 2 & 3 coming from for the male gender?

[muted bear]

the ones that are male are marked with 1

there are 5 1s in the male column

two of them dont always wear their seatbelt, hemce the two 1s in the other column

[unborn bobcat]

But there’s also five 1s in the other column

[muted bear]

the other three 1s have a 0 in the male column

this corresponds to the three females who dont always wear their seatbelt

[unborn bobcat]

What do you mean the “other 3 ones”?

[cedar kiln]

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[frigid sinew]

i got (-12x^(5/4))/5+24x^(7/4)/7

[frigid sinew]

for

is it right?

nvm

i frogot C im actually losing brain cells

.close

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@merry nacelle Has your question been resolved?

[cedar kiln]

@merry nacelle Has your question been resolved?

[merry nacelle]

Nopeeee

[merry nacelle]

<@&286206848099549185>

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[upbeat socket]

<@&286206848099549185>

[dire geode]

Don't open help channels without a question

[upbeat socket]

Oh my bad

[dire geode]

Also don't ping until 15m after you post a question

[dire geode]

Read this

[upbeat socket]

Ok

I was having trouble with this problem from my homework and I think AC is the diameter so I’ve been putting in 26 which hasn’t been working

And tried the radius as well

[void flax]

24

Circunference is 26π, and 2πr = 26π, is r=13

And 13 is h, 5 is c, and the other c IS the incognite

Apply pitagora's law and 12 is the botton slide, 12+12, 24

I think

I don get It, 23 excersise

[upbeat socket]

It worked

Thank you very much

[void flax]

Your welcome

[upbeat socket]

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[void flax]

<@&286206848099549185>

[crimson sedge]

i think you have to make a new channel in the open help ones

[void flax]

how?

[crimson sedge]

just put your question in #help-7｜zen1thxyz , #help-17 or any other of the open ones in the MATH HELP (AVAILABLE) category

again i am new here

[void flax]

ohh

thanks

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[fiery ermine]

Do I find n by dividing 30 by 3 to get 10 and then using the formula?

[digital cliff]

sure, you never said what formula though

[fiery ermine]

the sum formula for geometric things

[digital cliff]

sounds good

[fiery ermine]

I did 8*((1-6^(10)/(1-6)) and got 96745880

is it really that big?

[digital cliff]

yeah, dont underestimate how fast things can spread lol

[fiery ermine]

mmmm

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[molten kernel]

how to calculate this

[drifting marlin]

${}_nC_r = \frac{n!}{(n-r)!r!}$

[wraith dagger]

Steakanator

[molten kernel]

that makes sense thank you

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[nocturne nexus]

in this sample problem why is 8pi being used instead of 2pi?

[half iron]

It's not that 8pi is being used...

2pi is being used but they took 4 common... From 4-4y^2

[nocturne nexus]

oh im braindead 💀

i didnt not see that 😭

tysm

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[thorn monolith]

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[thorn monolith]

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I need help with a)

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[thorn monolith]

i need help with question a)

[grave robin]

Well the way I learned the first one is to just memorize it

Deriving it would take forever

[mighty shuttle]

try $u=asec(y)$

[wraith dagger]

Why am. I here

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[true sphinx]

did i do 13 correctly?

[raw gulch]

your limit should be equal to 1/2

not zero

[true sphinx]

why would it be 1/2b

[raw gulch]

1/2

take 2^n out in numerator and denominator

[true sphinx]

yeah but what about the +1 in teh exponent

[raw gulch]

do not forget

to divide

1

by

2^n

ok ?

and 2^1

wil be still there

i write you , moment

[true sphinx]

so then it’d be 1/3

[raw gulch]

read it

[raw gulch]

is it clear for you now?

[true sphinx]

not really lol

made it more confusing

[raw gulch]

lol

division , cant you see? division by 2^n

[true sphinx]

yeah i see that now

[raw gulch]

numerator and denominator, both, have been divided by 2^n

🙂

[true sphinx]

why do you beed to divide

[raw gulch]

it i snecessary

[true sphinx]

cant you just set the limit without dividing

[raw gulch]

you can but, you must tell result 1/2

if you can see it in yoru mind

[true sphinx]

but if you set the limit without dividing you’d get 0 right?

[raw gulch]

o wud be wrong

0 is wrong there

[true sphinx]

because 2^(n+1) is bigger than 2^n

[raw gulch]

it is bigger but that is same family, not too big

[true sphinx]

i don’t understand why you have to divide

[raw gulch]

if you had:

[true sphinx]

why cant you set the limit without dividing

[raw gulch]

2^n + 12, divided by

3^n + 35

then your nsweer zero

wud be accepted by me

as a teacher

but

in yoru acse

zero is wrong

your answer must be 1/2

[true sphinx]

is it because the base is the same and that +1 in the exponent is negligible

[raw gulch]

the initial problem of yoru limit is:

then both numerator and

denomiantor reaches plus infinity

so to eliminate this undefined, form

you ned to cancel , remove

or how oyu can say, those factors 2^n

and only divison

r3move them

do you understand me ?

[true sphinx]

no but lets say we have 2^n/(2^(n+1))

[true sphinx]

and plug in any random number

lets say 50

[raw gulch]

look at it, i made for you

50 is too small

50. 00000000000000 is nothing in compare with infinity

[true sphinx]

isnt it growing 1 step ahead in the denominator at all times

[raw gulch]

no

too small

[true sphinx]

yes i do know it’ll be inf/inf

so its just negligible

that extra step

because ultimately it’ll be inf anyways in the exponent

[raw gulch]

tomake yoru teacher in yoru school accepts yoru work, you must divide

zero wud be wrong

[true sphinx]

but is (n+1) not greater than n

do it’d be growing faster on the bottom

regardless of the base

you’d always have a larger number

but it’ll always be 2..

OHHHH

i get it now

thank you!

[raw gulch]

yvw)

[true sphinx]

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[pliant willow]

I'm stuck in the x^2 + y^2 +14y = -40

[surreal cave]

the standard form of the equation of a circle is $(x-k)^2+(y-h)^2=r^2$, yes?

[wraith dagger]

PajamaMamaLlama

[pliant willow]

Yes

[nova mountain]

ig make perfect squares?

with y

[surreal cave]

^

[nova mountain]

do you know how to make a perfect sq?

[surreal cave]

do you know the expansion of (y+b)^2

[pliant willow]

i dont..

[nova mountain]

ig then learnt it

from yt

[surreal cave]

lmao pretty much (y+b)^2=y^2+2by+b^2

notice how in your equation you have y^2+14y?

[pliant willow]

oh

I subtract both side by 40

[surreal cave]

what does this imply your b value is?

[surreal cave]

nope we actually need it on the LHS

[nova mountain]

not exactly but yeah you can do that

[pliant willow]

im like grade 7 rn im confuse so much T-T

[nova mountain]

come in a vc

if there is

[pliant willow]

dont think so..

[surreal cave]

a perfect square is (y+b)^2=y^2+2by+b^2

in your equation you have y^2+14y

Can you figure out what your b-value is?

[pliant willow]

0 i guess?

[surreal cave]

why would it be 0?

[pliant willow]

wait its y

[surreal cave]

how so?

[pliant willow]

sorry sorry

I thought its x+b

[surreal cave]

lmao no worries

so in your equation you have 14y correct?

[pliant willow]

yup

but i think there must be no y on the 14 so idk how to fix that

[surreal cave]

nope that 14 on the y is fine

in the perfect square expansion we have 2by

now because each of these temrs have a y they must be equal

so we have 14y=2by

[pliant willow]

okay

[surreal cave]

now can you find the b-value?

[pliant willow]

wait can I cut it first

[surreal cave]

wdym?

[pliant willow]

I mean like

cut what you are talking atm

if thats ok

I'm just thinking maybe if I show you what I learned it can help you like to explain in the way I studied?

[surreal cave]

man completing the sqaure I really don't like teaching it that way but it's seems to be popular

ye no worries

[pliant willow]

ty ty

[surreal cave]

so we have y^2+14y

[pliant willow]

yup

[surreal cave]

we have to add a value so that way we can get it to (y+a)^2

this value is (14/2)^2

[pliant willow]

yup

[surreal cave]

notice that's the same 14 from the 14y

so y^2+14y+(14/2)^2=(y+14)^2

now because you added it to one side in order to keep the equation balanced you must also do you want to the other side?

[pliant willow]

yup

[surreal cave]

lmao so what do you have to do to the other side?

[pliant willow]

+14/2?

[surreal cave]

🤏

close

[pliant willow]

+14/2^2

[surreal cave]

[pliant willow]

o

[surreal cave]

so now we have x^2+(y+7)^2+40=(14/2)^2

now just do some simplifying and Bob's ur uncle :)

I do have to go but I hoped this helped al ittle

[pliant willow]

good thing i learned some british words ;-;

i would be liike what is bob' your uncle..

[surreal cave]

lmao just learned it too

[pliant willow]

sry if i get confuse I also asked chatgpt but prob im just confuse like why 14y+(14/2)^2 is not equal to 14y+49 which ie likss 63y

oops

do you think you can explain that to me?

[pliant willow]

I changed ssome stuff and the bot told me this but I just dont get how did the 6y gone

why its not like 15y.. im sorry if this sound very stupid

i know what like the step of how to fix this but I just want to understand it too

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[pliant willow]

;-;

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[crimson sedge]

Hey, how do I find all vectors $v$ such that $<1, 2, 1> \times v = <3, 1, -5>$?

[wraith dagger]

00

[upper abyss]

I assume you're using the cross product?

[crimson sedge]

Yes

Where am I supposed to take things from here?

I made a mistake for the second component

Should be x - z, not z - x

can someone who is good maths friend me 😭

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@crimson sedge Has your question been resolved?

[upper abyss]

So far so good! That top equation only has x as an unknown, and can be solved for x.

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[wintry arrow]

Ive worked out distance from G to F is 16 miles

but im stuck on the rest

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[sage goblet]

Hi I have another problem with this one

[sage goblet]

Emily has 20 kinds of fruits (strawberries, peaches, lemons, kiwi, ...) available. She wants to prepare two fruit salads using for the first one four different fruits but without the lemon and for the second one five different fruits, avoiding those used in the first fruit salad and without kiwi. How many pairs of fruit salads can she prepare?

I thought that C(19,4) times C(15,5) would be right but it looks like it’s not….

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[sage goblet]

<@&286206848099549185>

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@sage goblet Has your question been resolved?

[crimson sedge]

the reason your answer dosen’t work is because what if the first salad has kiwi in it? then, it is C(16,5), but if the first salad dosen’t have kiwi, it is C(15,5)

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[hardy venture]

can someone explain why these are the answers

[drifting marlin]

when you're looking at limits at infinity, one thing you can usually do is disregard everything other than the highest power in the numerator and denominator

consider d. For very large x, 2sqrt(x)+1 has negligible difference from 2sqrt(x), so we can really just ignore the +1 term

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[west kindle]

This question is just to make sure I did the operations correctly (I did get the correct answer to the function but I have a feeling I may have wrote something that shouldn't be written)

I need to find the domain, (can't find the translation for this word in my language) monotonicity interval and local extremes of the following function (look top left of image)
I've managed to get the correct answers but I have a gut feeling the way I wrote the first derivation seems...wrong idk.
Everywhere else I go and look for the answer I get different results for the simplified variant of the derivation, but I still think I've done something wrong even though I got the right results.
Any help is much appreciated, thanks!

[azure horizon]

the derivative looks almost right except it should be 2e^2x (-x^2 + x + 2)

[west kindle]

oooooh right right right forgot about that

[azure horizon]

it doesn't change the answer because it's just a multiple of the polynomial so it still has the same roots

[west kindle]

but other than that it's ok?

[azure horizon]

i didn't check the below bit

it looks about right

yeah

[west kindle]

awesome, thank you very much!

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[errant loom]

Hello,we are finding surface area and I believe I did the problem right but are the 2 answers in the square equivalent to each other?

[raven shard]

yes

they’re also both correct values for the surface area if that means anything

[errant loom]

Okay awesome. The bottom one is the teachers answer but honestly I have no idea how to simplify the answer I got to that one

[raven shard]

oh okay

yeah in the top answer in the square you essentially combine the fraction subtraction since there’s two fractions with denominator 3

[errant loom]

I see, where would that 145 sqrt 145 and 27 on the denominator come from after?

[raven shard]

so do you know how fractional exponents work

and also, to get the 27 you factor out the 2/3

2/3 times 1/18 = 1/27

[errant loom]

Ah I see!

No I don’t remember how fractional exponents work except for 1/2 haha

[raven shard]

oh okay

so when you have x^(p/q), this is equal to (x^p)^(1/q) or equivalently (x^(1/q))^p

in the case of 145^(3/2), this is equal to (145^3)^(1/2) which is the square root of (145 cubed), and this is 145sqrt145

[raven shard]

the thing I’m saying in the first sentence is true because x^(ab) = (x^a)^b = (x^b)^a

[errant loom]

Ah I get what you’re saying!

Thank you for the easy explanation

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[restive light]

what else can you write in place of Tan x

[ruby maple]

so I believe tanx is sinx/cosx

[restive light]

yes so write tanx as sinx/cosx in the left hand side

[ruby maple]

okay so sinx/cosx + 1/tanx

[restive light]

write the other tanx as sinx/cosx too

[ruby maple]

so LHS is - sinx/cosx + cosx/sinx?

[restive light]

yes

now what can you do with the LHS

think about it a bit

[ruby maple]

so my first thought is to make the common denominators the same

[restive light]

so u would take LCM

[ruby maple]

sorry what does LCM stand for

[restive light]

lowest common multiple

u make the denominators same thats all

[ruby maple]

brain is a bit mashed form a couple of hours of maths problems I couldnt think lol

[restive light]

happens to the best of us

so what did u get after making the denominators same

[ruby maple]

okay so to make them the same am I right in thinking that because they are both different factors I have to combine them

so it would be sinx/cosxsinx + cosx/cosxsinx

[restive light]

yep

but u got the numerators wrong

[ruby maple]

does the numerator become ^2 because I have multiplied them

[restive light]

yes

[ruby maple]

so sin^2x/sinxcosx + cos^2x/sinxcosx

[restive light]

yep

[ruby maple]

and then add them together

[restive light]

yes

the problem is almost done

do you know the 3 trigonometric identies

[ruby maple]

I believ so if I am correct sin^2a + cos^2b = 1

[restive light]

it would be sin²a + cos²a =1

the angles must be same

[ruby maple]

so the LHS once complete is 1/sinxcosx which proves tanx+1/tanx = 1/sinxcosx

[restive light]

yep

[ruby maple]

thank you so much been struggling with this one for a while

[restive light]

nah the beginning is tough you will get better with practice

[ruby maple]

100% I am glad I found this place, I have used chatgpt before but personally I find it much easier having someone go over it with me

not so much go over but verbalise and confirm what I am thinking

[restive light]

that's true a human explaining it will always be better than an ai

[ruby maple]

right on to the next one thank you again

[restive light]

your welcome

[ruby maple]

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[ionic wolf]

NGL I was NOT paying attention

Like at all

In class

So I'm gonna wing it fr

[outer lily]

which problem do you need help on

[ionic wolf]

Lim x->inf f(x)->0

[ionic wolf]

umm all of them

prob

[ionic wolf]

is this right for the first one fr

do I do negative infinity too

[outer lily]

The first set of problems is just essentially asking where it is continuous, so you need to find numbers for x where there is no solution

For problem 1, what values of x make the denominator equal to 0 (and thus make the function undefined)

[ionic wolf]

3

@outer lily

fr

[outer lily]

ok

so for the domain then

you know every number works other than 3

do you know domain notation

[ionic wolf]

(neg inf, 3)U(3, inf)

[outer lily]

yes

[ionic wolf]

[outer lily]

now we need to figure out the behavior as x approaches 3

[ionic wolf]

O

[outer lily]

that is where you use limits

[ionic wolf]

O

[outer lily]

figure out the limit as x approaches 3. consider what happens when you approach it from the left (meaning starting from some point left and going right) or from the right (visa versa)

then follow a similar approach for problem 2

[ionic wolf]

idk how do that

[outer lily]

which part, the limits?

[ionic wolf]

yes

uhh it gets really big

[outer lily]

yes exactly

[ionic wolf]

how write that

[outer lily]

the limit as x approaches 3 is infinity. but thats not the entire answer, it approaches two numbers

lim(x->3) = inf

[ionic wolf]

do I put like that + or - thing in the superscript area

To say the direction

idk

[outer lily]

yeah

so it would be x->3^+ or x->3^-

[ionic wolf]

so cool

[outer lily]

ok then for 3, 4. just remember to factor the expression and figure out for what values of x does the function become undefined. from there you can determine the domain. i gtg but i hope that helps

[ionic wolf]

positive direction is like

coming from

0

right

[outer lily]

that is backwards

think of it this way

[ionic wolf]

wat

[outer lily]

sorry i'm typing

think of these values:

denominator only:
f(3.1) = -2 ÷ (3.1 - 3) = -2 ÷ 0.1
f(3.01) = -2 ÷ (3.01 - 3) = -2 ÷ 0.01
f(3) = undefined
f(2.99) = -2 ÷ (2.99 - 3) = -2 ÷ -0.01 = 2 ÷ 0.01
f(2.9) = -2 ÷ (2.9 - 3) = -2 ÷ -0.1 = 2 ÷ 0.1

basically just dont forget that the numerator is negative

because the numerator is negative if you take an x value just barely bigger than 3 and subtract it, you will get a negative devided by a positive. which as that sum becomes infinitely small it becomes negative infinity

[ionic wolf]

wat does the plus mean at the top

ngl idek what I wrote

does it mean like going right

idk

wat is the plus direction

[outer lily]

it's a division symbol ÷, sorry if it doesnt translate across devices well

[ionic wolf]

no wat I wrote

a

[outer lily]

no thats right

[ionic wolf]

wat does it mean

I forgor

[outer lily]

it means you're approaching from the right

[ionic wolf]

o

[outer lily]

think of it this way

[ionic wolf]

so like 3.0001

[outer lily]

yes

[ionic wolf]

so cool

[outer lily]

imagine putting your pencil on the right side of your graph and then following the graph to the x value we're testing

[ionic wolf]

OK

I fixed it

I thought it was left😭

[outer lily]

no worries

just remember that likewise x values are positive on the right side of the graph

right is always positive

anyways i hope that helps

[ionic wolf]

do I make 4 limit thingies for question 2

cuz like 2 And -2

[ionic wolf]

<@&286206848099549185>

[cedar kiln]

@ionic wolf Has your question been resolved?

[ionic wolf]

<@&286206848099549185>

[cedar kiln]

@ionic wolf Has your question been resolved?

[cedar kiln]

@ionic wolf Has your question been resolved?

[ionic wolf]

<@&286206848099549185>

[ionic wolf]

<@&286206848099549185>

[gentle pecan]

Hey @ionic wolf here if you have any questions

[ionic wolf]

OMG HI

[gentle pecan]

Hey

Heard you were doing limits

[ionic wolf]

fr

[ionic wolf]

idk if I interpreted it right

.close

[cedar kiln]

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[cedar kiln]

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[lusty marlin]

Hi, I have been doing basic combinatorics and I am stuck at what to do after this step.

got the answer, nvm

.close

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[rigid gorge]

no idea

someone please help

[cedar kiln]

@rigid gorge Has your question been resolved?

[dry latch]

is this the question you have a doubt in

the amplitude will be 8-2/2

[cedar kiln]

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[warm crescent]

is there any easier way (maybe only using algebra) to derive why sinx >=a with |a| <=1 implies arcsin(a) + 2kpi <= x <= arcsin(a) + 2 (pi/2 - arcsin(a)) + 2kpi

geometrically it is easy to see

but i would prefer if you can derive that only using algebra

[cedar kiln]

@warm crescent Has your question been resolved?

[warm crescent]

Ok somehow I figured it out

but how do I make sure those are all the x's that satisfy that

how do I know I'm not missing any x

[cedar kiln]

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[runic fox]

How is 2/5 times 100/1 equal to 40?

I know the answer but I was told this way of solving fractions to percentages is easier

[runic fox]

Isnt 2/5 times 100 just 200/500?

[clear umbra]

the 100 only gets multiplied to the top

not top and bottom at the same time

[runic fox]

ohh i get that

but 200/5?

wait

[clear umbra]

200/5 is 40

[runic fox]

ohhh

no wonder

that was an easy question

I just got mixed up with the 100/1 part

[clear umbra]

if you multiply 100 to top and bottom you're actually only multiplying by 1

[runic fox]

so say something like 32/64

how would i solve that

oh wait thats easy

uh

57/97

how would I do that with that technique?

because 5700/97 how am i supposed to simplfy that

nvm this is stupid

uh

mb

[clear umbra]

maybe long division

[runic fox]

yea i realised

lemme find something that suits this situation better

nvm

i got it now

.close

[cedar kiln]

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[runic fox]

.delete

do i

delete it

[clear umbra]

nah it'll delete this channel off the occupied list on its own

[runic fox]

oh ok

thanks tho

[cedar kiln]

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[bold mauve]

How do I show $W = {P(0) = P(1) = 0 : P \in P_3(\mathbb R)}$ is a subspace of $V = P_3(\mathbb R)$

[wraith dagger]

o.O

[bold mauve]

I cannot figure out how to factor it into the form I would want

Suppose $x, y \in W$ and $\lambda \in \mathbb R$, then it follows $x$ and $y$ are in the forms:
$$x = a(x - 0)(x - 1)(x - c)$$
$$y = b(x - 0)(x - 1)(x - d)$$

for some $a, b, c, d \in \mathbb R$

[wraith dagger]

o.O

[bold mauve]

I wanted to show it was a subspace by showing $x + \lambda y \in W$

[wraith dagger]

o.O

[bold mauve]

yet I cannot figure out how to factor this thing into the desired form (...)(x)(x - 1)(x - ...)

[royal loom]

it's a subset

Don't you need to show

[bold mauve]

to show its a subsspace I need to show a linear combination is in the subspace also and there is a zero vector in the subspace

[royal loom]

Show what you tried here

[bold mauve]

if you expand everything out you will get $ax^{3}-acx^{2}-ax^{2}+acx+\lambda bx^{3}-\lambda bdx^{2}-\lambda bx^{2}+\lambda bdx$

[royal loom]

don't expand everything out though

a(x-0)(x-1)(x-c)+blambda(x-0)(x-1)(x-d)

[wraith dagger]

o.O

[royal loom]

=(a+blambda)(x-0)(x-1)(x-c)

it suffices to show that a+blambda are in R

[bold mauve]

wait let me use L for lambda for conveniencee

$ax\left(x-1\right)\left(x-c\right)+Lbx\left(x-1\right)\left(x-d\right)$

[wraith dagger]

o.O

[bold mauve]

so we started with

[royal loom]

yes

oh

apologies, I forgot that the c/d are not the same

let me take another look

[bold mauve]

maybe it will be easier to just find the third root then reconstruct the equation

[royal loom]

no it will be fine

same thing but slightly more complicated

=(x(x-1))[a(x-c)+Lb(x-d)]

it factors as this instead

[sacred grail]

[royal loom]

so then it suffices to show that the a(x-c)+Lb(x-d) is a linear factor i.e has a root

which as he sum of two linear polynomials , it will

[bold mauve]

ok I think I see

I have $x\left(x-1\right)\left(a\left(x-c\right)+Lb\left(x-d\right)\right)$ and showed $$a(x-c)+Lb(x-d) = 0 \iff x=\frac{Lbd+ac}{a+Lb}$$

[wraith dagger]

o.O

[royal loom]

:o

I would just

[bold mauve]

then we can reconstruct it

[royal loom]

(a+Lb)x-ac-Lbd

group like terms

and then since it's linear you're done

[royal loom]

do you maybe want to show easier way

[sacred grail]

you can do it

[royal loom]

@bold mauve Would you like to see another way to do it or are you happy with this

[bold mauve]

I went with the solution

$\left(a+Lb\right)x\left(x-1\right)\left(x-\frac{Lbd+ac}{a+Lb}\right)$

[wraith dagger]

o.O

[bold mauve]

[royal loom]

it works :p

[bold mauve]

.close

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[proven oar]

This might not be the most Math related question, but I just can't seem to wrap my head around the Shrodingers Cat. Simply explained, if i have a cat hidden in a box with poison, there is the possibility the cat could be either alive or dead. but because i have not opened the box, and can not confirm whether the cat is alive or not, Shrodinger implies it is both dead and alive at the same time. It stays this way until the point of measurement (opening the box).

I dont understand how he came up with the idea that it is both dead and alive, just because he could not confirm it for himself. Isn't the cat just one or the other? Whether you open the box or not? Just because I can't confirm his state, doesn't mean his state is everything all at once no? I understand if it is a philosophical concept, but it seems that this is a genuine thing people in quantum physics do calculations with.

[fair geyser]

the idea that something is already something before you check is called "local hidden variable"

they just disproved it, they did experiments that don't make sense if it's already something

i don't know what experimetns, it's all 2 hard

[cedar kiln]

@proven oar Has your question been resolved?

[fair geyser]

basically things behave like they affect other things at a distance, as if it sends some faster than light signal or they conspire ahead of time, so if you try not to make too strong statements but only report what you see, you see that things aren't already something before the checking happens

[proven oar]

this is all very vague hahhaha

"basically things behave like they affect other things at a distance" you mean like entanglement?

[fair geyser]

i just mean entanglement i think

[tardy hull]

I think some important context is that Schrodinger's cat was originally proposed as an intentionally absurd consequence of a certain interpretation of quantum mechanics

[fair geyser]

by schrodinger

[tardy hull]

Schrodinger was arguing against an interpretation where a Schrodinger's cat kinda thing would be possible

[fair geyser]

yes

so we think he was wrong

[proven oar]

oh! I didnt know that

[fair geyser]

einstein also said it was stupid

[proven oar]

But I am not sure if i have ever read the original paper though. I might have but I have forgotten most of it

[fair geyser]

no one came up with the non stupid explanation that they would accept

[proven oar]

hmm. So the shrodinger cat idea is not really applicable in real life then?

[tardy hull]

I mean quantum superposition is certainly applicable

[fair geyser]

define schrodinger cat idea

[cedar kiln]

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[fair geyser]

you have to think that it's alive and dead, because any other current explanation is more complicated

schrodinger was like, you can't accept this bullshit