help-19

[cursive field]

...

THERE

[solid niche]

Oh , thank you!

[cursive field]

no problem

i'm leaving the latex in case you need it!

:)

[solid niche]

there is a minus in the third equality though

[cursive field]

so?

[clever fjord]

ren

[cursive field]

there

better?

[solid niche]

both minuses dissapeared XD

[cursive field]

...?

yk u can edit it yourself

lol

just copy the message

[solid niche]

$E(x) = \arcsin ( \sin (2x)) ; \forall x \in \R =>\newline \sin (E(x))= \sin (2x) ; \forall x \in \R <=>\newline -\sin (E(x))= -\sin (2x) ; \forall x \in \R <=>\newline\sin (E(x)- \pi)= \sin (2x- \pi ) ; \forall x \in \R <=>\newline E(x) - \pi = \arcsin(\sin(2x- \pi)) + 2k \pi ; \forall x \in \R ,; k \in \Z ; <=>\newline E(x) = \arcsin(\sin(2x- \pi)) + \pi (2k + 1) ; \forall x \in \R ,; k \in \Z; <=>\newline\arcsin(\sin(2x))=\arcsin(\sin(2x- \pi)) + \pi(2k+1) ; \forall x \in \R , ; k \in \Z \newline (FALSE)$

[cursive field]

its fine

[clever fjord]

Peel3
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cursive field]

ill do it for y- oh

oka

[solid niche]

there

THANk you again!

[cursive field]

np

[solid niche]

X is in R , k is in Z!

it seems they dissapeared

.close

[odd edge]

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[odd edge]

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[elfin warren]

"Let g(u, v, w) be differentiable on R^3
and let f(x, y, z) = .... Use the chain rule to
show that for all (x, y, z): xyz ̸= 0 holds:..."

[elfin warren]

could someone help me solve this?

[odd edge]

@elfin warren Has your question been resolved?

[elfin warren]

<@&286206848099549185>

[odd edge]

@elfin warren Has your question been resolved?

[wanton bison]

Is r a vector or what exactly

[elfin warren]

I am not entirely sure

[wanton bison]

ok can you try to get f_x, f_y and f_z?

[elfin warren]

Yes

Do I use the chain rule to get them?

[wanton bison]

Yes, for example:

$\f_x = g_x \cdot \left ( \frac{1}{y} -\frac{z}{x^2} \right )$

[clever fjord]

𝔸dωn𝓲²s

[elfin warren]

F_x = g_x * y + g_z * z?

[wanton bison]

How

with respect to y?

[elfin warren]

Well I don’t know

Maybe I am way off

[wanton bison]

$\frac{\partial f\left (x,y,z) \right )}{\partial x} = \frac{\partial g \left ( \frac{x}{y},\frac{y}{z},\frac{z}{x} \right )}{\partial x} \cdot \ \frac{\partial}{\partial x} \frac{x}{y} + \frac{\partial g \left ( \frac{x}{y},\frac{y}{z},\frac{z}{x} \right )}{\partial x} \cdot \ \frac{\partial}{\partial x} \frac{y}{z} + \frac{\partial g \left (\frac{x}{y},\frac{y}{z},\frac{z}{x} \right )}{\partial x} \cdot \ \frac{\partial}{\partial x} \frac{z}{x}$

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

HAHA

[elfin warren]

That looks complicated

[wanton bison]

I have to think if it's like that

If f(g(x) we get f'(g) • g'

If f(s,t) we get f'(s,t) • (s' + t') or

[elfin warren]

Hmm okay

[wanton bison]

Seems to be this

[wanton bison]

So this is it if you do it very detailed

[elfin warren]

Ohh okay!

I think I will have to continue tomorrow unfortunately but maybe I will see you then!

Thank you for the help so far!

[wanton bison]

np

[elfin warren]

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[azure spruce]

id appreciate any help this ones been hurting my head for a min

[royal herald]

are u familiar with lagrange mutlipliers?

@azure spruce

u may not need it, but i always find it to be the easiest method for these types of problems

[azure spruce]

yeah i am

[royal herald]

alr, why dont u try applying that here?

[azure spruce]

im up to 8y-6(8x-4)-6=2Lambda

[royal herald]

alr hold on lemme check that rq

i dont think im getting that exactly

can u show ur work

[azure spruce]

yh sure gimme a sec

[azure spruce]

been struggling with this topic for a while 😭

[royal herald]

check ur partial derivative wrt x again

after that, ill show u how to do this

[azure spruce]

ill try see where i went wrong

[royal herald]

$<8y-6x - 3, 8x-4> = L<2,3> \newline 2x+3y = 4$

[clever fjord]

Stephen

[royal herald]

this is what i got

$8y-6x - 3 = 2L \newline 8x-4 = 3L \newline 2x+3y = 4$

[clever fjord]

Stephen

[azure spruce]

i see so i messed up by writing -6

now that i look at it

idk why i done that

[azure spruce]

although i am a bit confused with the 2x+3y =4

wait no

im not thats the constraint

[royal herald]

yea

so now id take the first two equations

solve for L

then set them equal

upon doing that we get

$\frac {8y - 6x - 3}{2} = \frac {8x-4}{3} \newline \newline 2x+3y = 4$

now we have a system of equations that we can solve

[azure spruce]

i see

would you mind helping me understand how to get from the first two equations to when we set them up to be equal

[royal herald]

take the first equation and isolate L

what do u get

[azure spruce]

im not exactly sure how

[royal herald]

$8y-6x - 3 = 2L$

[clever fjord]

Stephen

[royal herald]

just get L by itself

how would u do that

how would u get rid of the 2

[azure spruce]

/2?

[royal herald]

yea

/2 both sides

[azure spruce]

ohh fair enough

/3 the other side

the other equation i mean

[royal herald]

yea /3 for both sides in the second equation

and once they are both equal to L, we can set them equal to one another

[azure spruce]

and theyd both be equal to L ?

[clever fjord]

Stephen

[royal herald]

and now we have this

so just solve

[azure spruce]

trying to rn

[azure spruce]

i got

wait solve for L right ?

[royal herald]

no, why?

[azure spruce]

sorry im confused

[royal herald]

we have 2 equations in terms of x and y

why solve for lambda

we dont need it

[azure spruce]

yeah your right

[royal herald]

just solve for x and y

[azure spruce]

i got x = 24y- 19 / 16

[royal herald]

just sub to other eqn

[azure spruce]

into the constraint ?

is y - 17/16 ?

thats what i got im not too sure tho

[azure spruce]

sorry i meant y= 17/16

@royal herald is it correct ?

<@&286206848099549185>

[ember geyser]

question?

[azure spruce]

.

this one

[azure spruce]

got up to here

[azure spruce]

id really appreciate if you can help me understand

[plain python]

24y-18x-9 = 16x-8

you have 2 equations

[azure spruce]

oh i see

cross multiplication

<@&286206848099549185> ngl im really stuck any help if possible ?

[odd edge]

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[thorn imp]

Hi

[thorn imp]

Where there’s u*v

Isn’t it supposed to be -35 and -12

[honest turtle]

Do you got the original question?

[thorn imp]

Calculate the area of the triangle

[honest turtle]

Which part?

[thorn imp]

The one which I took photo of

The one in the middle

[honest turtle]

-8-(-35)

-14-(-12)

Doublecheck your steps.

[thorn imp]

What?

[honest turtle]

The solution you showed is correct.

So you might’ve done some mistakes.

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[twilit pecan]

Hello! I dont understand how the original 2 fractions in the numerator were turned into the 1 fraction in the first step. Can someone please explain? Thanks

[ember geyser]

uh

they got a common denominator

[twilit pecan]

How? I know this is basic and I normally do this flawlessly but for some reason in this scenario it doesnt make sense to me

[ember geyser]

so the left fraction gets multiplied by 2/2

making the numerator 2

the right fraction is multiplied by x+2/x+2

making the numerator x+2

however since there's a minus

the numerator is 2 - (x+2)

making it 2 - x - 2

[shy sundial]

d'y know how to add fractions

[twilit pecan]

But why 2/2

[twilit pecan]

thanks! I really do not why this stumped me

[shy sundial]

prob the second way to do it is easier

[twilit pecan]

ive had a lot of work to do this week so my brain is tired maybe

[odd edge]

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[faint tusk]

I've been getting into card games lately and I want to use this as an opportunity to learn more about probability.

The scenario I'm on is the following (assume a shuffled standard 52 card deck):

1. After drawing 5 cards, what is the probaility of drawing an ace?
2. Remove 5 cards from the deck
3. Draw an additional 5 cards. From the 10 cards drawn so far, what is the probability of drawing an ace?

I understand how to do part 1. The idea is to find the probability of drawing NOT drawing an ace and subtracting that from 1.
So you get the following:
P(at least 1 ace) = 1 - 48C5/52C5

I struggle with understanding how the question changes after I remove 5 from the deck, then draw 5 more cards.

Any help is appreciated.

[odd edge]

@faint tusk Has your question been resolved?

[shell dawn]

i assume that you haven't found an ace after number 2?

[shell dawn]

also, which 5 cards are removed from the deck?

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[faint tusk]

.reopen

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✅

[faint tusk]

No. The question is asking for the probability of finding at least 1 ace. Whether you found an ace or not, it makes do difference for the probability.

[sleek moth]

So you draw 5 cards without putting them back and you want to calculate the probability of getting at least one ace?

[faint tusk]

yes

[sleek moth]

yup your first calculation seems correct

Part 3 would be very easy if you knew the probability distribution for the amount of aces in the deck prior to drawing the last 5 cards

but figuring out the distribution is a little challenging

distribution of aces left
4 aces => none got chosen in step 1 and none got chosen in step 2
3 aces => one got chosen in step 1 and none got chosen in step 2
OR none got chosen in step 1 and one got chosen in step 2
2 aces => one got chosen in step 1 and one got chosen in step 2
OR none got chosen in step 1 and two got chosen in step 2
OR two got chosen in step 1 and none got chosen in step 2
1 aces => 0 and 3, or 1 and 2, or 2 and 1, or 3 and 0
0 aces => 0 and 4, or 1 and 3, or 2 and 2, or 3 and 1, or 4 and 0

It's a bit of work but it's doable

[shell dawn]

^^^^ yeah

[odd edge]

@faint tusk Has your question been resolved?

[faint tusk]

thank you 🙏

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[analog tundra]

what do I do, right now I think having the chart only gives me one estimate

[cold swift]

this is riemann sums right?

[analog tundra]

ye

[cold swift]

so we have two options here, LRAM and RRAM

since it says that the runner’s speed is steadily increasing

which one is going to be the overestimate and which is the underestimate?

[analog tundra]

I think right is upper and left is lower

[cold swift]

yep exactly

so we just need to perform both RRAM and LRAM using the table of values it gives us

and that’ll be it

[analog tundra]

ok ty

i havent done rieman problem yet so ill watch a vid then try and see if i get it right

[cold swift]

alright

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@analog tundra Has your question been resolved?

[analog tundra]

@cold swift for the lower estimate I did (0.5 x 0) + (0.5 x 5.7) + (0.5 x 11.2) + (0.5 x 14.1) + (0.5 x 18.8) + (0.5 * 19.8) = 93.04 what was I supposed to do?

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[round shore]

Can someone tell me if this is right

[round shore]

Asap

i just do the first one then on the second equation i turn it into standard form?

Or a other form idk the name

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@round shore Has your question been resolved?

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[granite root]

Can someone please help me solve question g

[granite root]

.close

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[lyric dust]

This might be a rather general question but this is largely due to my lack of understanding in linear algebra.

I'm self studying from the gilbert strang textbook and don't really have access to a teacher because Australian highschool doesnt teach linear algebra at that level.

However, for the cross-product, it seems like finding the determinant of a 3x3 matrix with i j k going across as its columns or rows (whatever is considered conventional) does the trick.

My question is: is this purely notational/nmnemonical or is there an actual explanation for putting unit vectors into the matrix

[lyric dust]

Im interested in learning it because vectors are part of the highschool assessment course and it seems matrices simplifies a lot of these questions significantly but it is required for me to be able to formally define and explain the processes if i intend on using it in my exams

i have some limited understanding of duality from the textbook and some other readings but i dont think i understand it fully either

[mystic forge]

cross product being evaluated by the determinant is just a coincidence as far as i know

determinant is a much more general concept in R^n

cross product is only defined in particular dimensions

[lyric dust]

im only aware of it being defined in 3d

or maybe other odd number dimensions ig

[mystic forge]

there seems to be more to it but i dont understand it either since i havent formally done lin alg

https://physics.stackexchange.com/questions/133288/why-can-the-cross-product-of-two-vectors-be-calculated-as-the-determinant-of-a-m

[lyric dust]

i would've thought that if you do a 3x3 matrix with ijk and ur 2 vectors, it geometrically generates a parallelpipen

[mystic forge]

yes that is true

parallelepiped

the area would be a parallelogram

|a x b|

[lyric dust]

hmm ok

im just not sure how to correlate that with the fact that the determinant computes the vector perpendicular to both tho

because afaik the determinant generates an area

[mystic forge]

oh the thing is

we dont actually compute the determinant when we use that method

we just perform cofactor expansion and leave it in that form

its purely for technical purposes

[lyric dust]

oh so like i(minor) + j(minor) + k(minor) ?

[mystic forge]

yes

we wouldnt compute whats inside the ijk but in normal determinants we would and ijk would be numbers rather than basis vectors

[lyric dust]

ok that seems fair enough, from the stackexchange post you linked there seems to be some actual geometric significance but i dont think i will understand at this stage, for now i will probably just treat it as a notational trick for computing it and write it in the expanded form

thx for ur time!

.close

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[vapid valve]

can sm1 help me

[empty epoch]

yea post your question

[vapid valve]

with these

[empty epoch]

whats your understanding of probability?

[vapid valve]

im not rly good

at it

i forgot

it

[empty epoch]

probability basically refers to the "chance" of something happening

[vapid valve]

how do i work it out?

[empty epoch]

alright

a probability is expressed in a fraction right?

[vapid valve]

yea

[empty epoch]

and what does the problem ask

we do this first btw

[vapid valve]

the chance of the sweets beingred

[empty epoch]

yea, and how many sweets are red

[vapid valve]

do i minus15 by 4?

ohh

[empty epoch]

nah

[vapid valve]

4 are red]

[empty epoch]

all you have to know is to how many sweets are red

compared to the total sweets

[cursive field]

the question is

what is the chance of you getting a red sweet

if you pick a random sweet

[vapid valve]

the chance of me getting a red sweet it 4 / 15 ?

[cursive field]

from this specific jar ofc

[empty epoch]

yea thats right

[cursive field]

mhm

very good

now the next one

if you have green and blue counters

and you pick a counter RANDOMLY

the chance of the counter being green is 0.3

now what u gotta realize is the fact that you can only get green or blue counters

so the chances of both added together (chance of green) + (chance of blue) = 1, because you cannot have any other outcomes!

[vapid valve]

oh now i get it

is it kinda similar to ratio?

[cursive field]

yes

sort of

[vapid valve]

but how do i know the other color

ohh

0.7?

for question 14 how do i do it

6x 4 - 5?

24-5

19

ohh

ty

i gtg

too

bye ill come back tmrq

[cursive field]

y'all should i .close this

...

should i tho

it's only been 20 mins

...

i feel bad

what if they come back

.close

[odd edge]

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[cursive field]

fine

[odd edge]

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[winter tapir]

Hello. That is an Olympiad task for 11th graders from 2001. For real numbers x,y,z which are greater than 1 each, prove this inequality

[winter tapir]

.rotate

,rotate

[cursive field]

translate

[winter tapir]

I tried squaring both sides which didn't give me anything

I already trasnlated

Also 1/x + 1/y + 1/z =2

Also I think it comes to everyone's mind that x+y+z > 3

[gritty oar]

if you're looking for a solution, the olympiads provide solutions

[winter tapir]

really? But where I find it😭

[gritty oar]

is this russian

[winter tapir]

yeah

I FOUND IT

For some reason it is located in the middle of the book

No paper economy)

Gonna send you all a solution here

Also. How these types of statements are called and where to find those? For example here an already proven inequality is used

I want to see those, is it number theory or iam wrong?

.close

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[primal crane]

Question 11 please help

[odd edge]

@primal crane Has your question been resolved?

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[slow patrol]

i need help with this, i have a test next week abt it and i might be screwed 😭

[nimble blaze]

look up
inside angles theorem circle

[slow patrol]

i just now did and it’s still confusing me, i can’t watch videos right now

[nimble blaze]

what's confsing about what you read

[slow patrol]

it’s too much to process

😭

[nimble blaze]

which part of that is confusing

theorem is just saying that
green = (red + blue)/2

[slow patrol]

okay that makes more sense

is that all?

[nimble blaze]

yes

[slow patrol]

it was obtuse so it couldn’t be that so i subtracted it from 180 (102)

and it was correct

im confused on this too

[nimble blaze]

outside angles theorem circle

[slow patrol]

okay its 2 secants

but im having trouble solving them

[nimble blaze]

what are you seeing when you searched it up

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[pallid valve]

How do I show associativity for this?

I feel like there's a better way than just showing all combinations of reflections and translations

[low locust]

associativity of composition is well known

[pallid valve]

rrr
ttt

trr
rtr
rrt

rtt
trt
ttr

Would be all the combinations but it seems like quite a bit

[pallid valve]

Do you think it would be sufficient to prove that function composition is associative and just state that all combinations of reflections and translations would be composition of functions?

[low locust]

well reflections and translations are functions

if the composition of all types of functions is associative, then certainly the composition of only a few types of functions is associative aswell

[pallid valve]

Oh okay, I see what you mean
Thank you!!

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[wintry viper]

how would i know the line of Centre isnt horizontal or perpendicular?

[forest sky]

one of them is moving diagonally, and they have to collide somehow, so if they were directly vertical/horizontal to each other you wouldn't expect them to hit each other (given their velocities)

[wintry viper]

oh ok

thank you

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[rain tundra]

This is the question, I attached the wrong one 😭

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[silver fossil]

how do i do iv)

[silver fossil]

im so confused

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[ashen drum]

[
\begin{align*}
f(x) &= -2x^2 + 7x - 4 \
f(a + 2) &= -2(a + 2)^2 + 7(a + 2) - 4 \
&= -2(a^2 + 4a + 4) + 7a + 14 - 4 \
&= -2a^2 - 8a - 8 + 7a + 10 \
&= -2a^2 - a + 2
\end{align*}
]

did i do it right?

[clever fjord]

mcor42
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[nova ginkgo]

correct

whats the problem

[ashen drum]

just wanted to make sure. i keep making small mistakes

ty

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[desert marlin]

Hi, I was hoping someone could check my proof for this problem

[odd edge]

@desert marlin Has your question been resolved?

[supple cloak]

Looks fine to me. The only comment I have is that you seem to have split this up into three cases even though that was unnecessary because there are really only two (|a| is finite and |a| is infinite). The third case you showed is really just the contrapositive of the first statement.

actually it's not strictly the contrapositive but it is implied by it

[desert marlin]

Ah okay. Thank you

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[mystic saffron]

what is this??

[mystic saffron]

what kind of black magic is the prof doing here

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

<@&286206848099549185> sorry, any of you guys know gaussian quadrature?

😭 im so screwed

[haughty stream]

https://www.youtube.com/watch?v=w2xjlPwYock&pp=ygUUZ2F1c3NpYW4gcXVhZHJhdHVyZSA%3D

This might help

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@mystic saffron Has your question been resolved?

[mystic saffron]

oh my god

i stared at this question

for 3 hours

and it finally clicked

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[frozen atlas]

Hello

[frozen atlas]

Merp

[steady tide]

whats your question?

[steady tide]

!xy

[odd edge]

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.

[frozen atlas]

How do to add a log that have diefferent in basid

Basis

Addition of logarithms with different bases, how?

[steady tide]

you can use change of base, but that would be ugly

[frozen atlas]

If its the only way

Then how?

Ive searched google, not satisfy

Ping me please

[odd edge]

@frozen atlas Has your question been resolved?

[frozen atlas]

<@&286206848099549185>

[open knot]

What exactly is written in the last line

1/2 * (a*log(b) + b*log(a)) ?

[open knot]

what are the basis and what are the log values?

when you write "alogb" what do you mean

log_b(a)?

log_a(b)?

a * log10(b)?

[frozen atlas]

I found these

[open knot]

what the hell is that notation

[frozen atlas]

"Its known that" bla bla "where" bla bla "and" bla bla, "so the value of" bla bla

[open knot]

ok I presume log_b(a²) and log_a(b²)

[nimble blaze]

other way around

[open knot]

log_a²(b) & log_b²(a)?

[nimble blaze]

$^{a^2}\log(b) = \log_{a^2}(b)$

[clever fjord]

ℝαμΩℕωⅤ

[open knot]

who invented that

[frozen atlas]

Nope

Wait

Whut

[open knot]

nywys the only property you need is:
log_a(X) = log_b(X) / log_b(a)

which you can use to change to any basis

[frozen atlas]

How do we know the value of a and b?

[odd edge]

@frozen atlas Has your question been resolved?

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[frozen atlas]

.close

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[blissful depot]

If X1, X2,..., 1991 are strictly positive numbers and x1 + x2 + .... + X 1991 = 1, show that 2(√x1(x2 + x3 + ... + X1991) + √ x2(x1+x3 + ... + x1991) + ... + √x1991 (x1 + x2 + ... + X1990)) <1991

[blissful depot]

I have no idea how to begin

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@blissful depot Has your question been resolved?

[shell merlin]

I got up untill 2(√x1(1-x1) + √x2(1-x2)...+√x1991(1-x1991))<1991

🥲

[blissful depot]

<@&286206848099549185>

[odd edge]

@blissful depot Has your question been resolved?

[blissful depot]

<@&286206848099549185>

[wooden pasture]

the 2 up here is really messing with me

otherwise it seems easy enough

[blissful depot]

Right

Dividing it all by 2 makes it worse

[blissful depot]

<@&286206848099549185>

[odd edge]

@blissful depot Has your question been resolved?

[manic jungle]

Hint: $2\sqrt{a(S-a)}\le S$

[clever fjord]

Caroline

[supple mural]

are you sure this is true? i dont think equality ever holds

but anyways just use jensens

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[craggy canyon]

in this question, i have to solve for n, and i got n=6, which is a valid solution
however, by guess and checking, you also find that n=3. i seem to have missed this solution, probably when i cancelled (n-3) from both sides in the middle
how would i mathematically get n=3, and how do i avoid missing solutions in the future for other questions?

[craggy canyon]

2nd and 3rd line is pascals identity btw

[runic swift]

move it over and factor it

[craggy canyon]

wdym by move it over?

[runic swift]

(n-1)(n-2)(n-3) = 20(n-3)

move the 20(n-3) over

(n-1)(n-2)(n-3) - 20(n-3) = 0

[craggy canyon]

oh subtracting 20(n-3)
i tried to divide 😭

[runic swift]

well if n=3 then you're essentially dividing by 0

[craggy canyon]

after a bit of factoring and everything, i end up at $(n-3)(n-1)(n-2)=0$, but n=1 and n=2 aren't valid because n>=0 for combinations

[clever fjord]

ـDraedon

[craggy canyon]

*because n>=r

so >= 3

so how do i get n=6 from this?
$(n-1)(n-2)(n-3)-20(n-3)=0\newline$
$(n-3)(n-1)(n-2)(-20)=0\newline$
$(n-1)(n-2)(n-3)=0$

[clever fjord]

ـDraedon

[runic swift]

you factored it wrong

it should be (n-3)((n-2)(n-1) - 20) = 0

[craggy canyon]

oh i see

[runic swift]

so you automatically get n = 3 as an answer

and the rest is (n-2)(n-1) - 20 = 0

which is what you did

that you got your n = 6

[craggy canyon]

ok thanks

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[modern hinge]

why does n have to be even

all values of n would give the same value ?

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[elder vault]

Given a and b are not zero vectors.

Will the equation always hold

[spiral basalt]

That's called linearity of the dot product yes

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[winged sinew]

Given that:
A = { x : 5 < x < 12, x ∈ R }
B = { x : – 3 < x < 8, x ∈ R }
C = { 2x : –1 < x < 5, x ∈ Z }
(a) Express in similar set-builder notation the following sets:
(iv) A’ n B’
(b) Find
(ii) n ( B U C )

[winged sinew]

for part (a), i got x smaller or equal to 3 or x is bigger than 12

how do i show it in a similar notation?

[honest quest]

set builder notation is just {some_expression: condition_to_be_satisfied}

[winged sinew]

i have 2 two different conditions?

or should i express it as it can't be bigger than -3 and smaller than 12

[honest quest]

You can use logical operators i.e. (or, and): e.g., condition 1 or condition 2.

[winged sinew]

ohh ok

[winged sinew]

thx

@honest quest btw for part b, is it possible to find the number of elements in a set B?

since it's real numbers

[honest quest]

You can put down infinity as the answer

[tepid pelican]

x is a set of real numbers 💀

[winged sinew]

oh

[winged sinew]

is the ans 7?

[tepid pelican]

there should be something like "x is a member of the set of real numbers"

or "x is a real number"

but not x is a set of real numbers

[winged sinew]

oh it was like x E R

[tepid pelican]

yes, that means x is a member of R

[winged sinew]

but when i pasted it, it was a box or smtg

oh

[tepid pelican]

∈

[winged sinew]

yeye that symbol

[tepid pelican]

Also suspicious

[winged sinew]

n ( B U C ) = 7?

for part b

[honest quest]

no

[tepid pelican]

(ii) n ( B U C )

what does this mean?

the n

number of elements?

[winged sinew]

number or elements

[tepid pelican]

are you sure its B union C

[winged sinew]

yeah

im sure

[tepid pelican]

then there are infinitely many numbers

[winged sinew]

oh

[tepid pelican]

because B itself is infinite

and B union C must be larger

[honest quest]

I don't think it be larger necessarily because you would need to consider bijectivity but it does not matter

[tepid pelican]

(larger or equal is what i meant)

not strictly larger ofc

that's also because the elements in C might already be contained inside B

although checking the sets, its not the case

[winged sinew]

anyways, thank you @tepid pelican and @honest quest for your insights

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[elfin warren]

"Let g(u, v, w) be differentiable on R^3 and let f(x, y, z) = g(x/y, y/z, z/x). Use the chain rule to show that for all (x, y, z): xyz ̸= 0 holds:"

[elfin warren]

can anybody help me solve this?

[wanton bison]

Chain rule!

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

So you need $\begin{pmatrix} x \ y \ z \end{pmatrix} \cdot \begin{pmatrix} f_x \ f_y \ f_z \end{pmatrix} =
\begin{pmatrix} x \ y \ z \end{pmatrix} \cdot \begin{pmatrix} g_u \cdot u_x + g_v \cdot v_x + g_w \cdot w_x \ g_u \cdot u_y + g_v \cdot v_y + g_w \cdot w_y \ g_u \cdot u_z + g_v \cdot v_z + g_w \cdot w_z \end{pmatrix}$

[clever fjord]

𝔸dωn𝓲²s

[elfin warren]

is that the same thing as this?

[wanton bison]

yes

I didn'T want to use the partial notation

[elfin warren]

yeah okay

[wanton bison]

so let's do f_x

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

What is u_x v_x and w_x

[elfin warren]

is u_x not just y?

[wanton bison]

not qute

u = x/y

now diff. wtr to x

the constant is 1/y not y

so u_x = 1/y

can you follow?

[elfin warren]

okay i think so

[wanton bison]

imagine x/3

what is that differentiated

[elfin warren]

1/3

[wanton bison]

yes

so here it's of that 1/y

[elfin warren]

okay

[wanton bison]

v_x?

[elfin warren]

0?

[wanton bison]

yes

w_x?

[elfin warren]

1/z

or z?

[wanton bison]

careful

what is 2/x differentiated

x in denomiantor now

[elfin warren]

hmm -2/x^2?

[wanton bison]

yes

so z/x?

[elfin warren]

-z/x^2?

[wanton bison]

,,f_x = g_u \cdot \frac{1}{y} + g_w \cdot \left ( -\frac{z}{x^2} \right )

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

Now let's do f_y

,, f_y = g_u \cdot u_y + g_v \cdot v_y + g_w \cdot w_y

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

same game

[elfin warren]

uy = -x/y^2?

vy = 1/z?

wy= 0?

[wanton bison]

yes!!

,, f_y = g_u \cdot \left (-\frac{x}{y^2} \right ) + g_v \cdot \frac{1}{z}

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

,, f_z = g_u \cdot u_z + g_v \cdot v_z + g_w \cdot w_z

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

same game

[elfin warren]

uz = 0, vz = -y/z^2 , wz = 1/x?

[wanton bison]

yes

,, f_z = g_v \cdot \left (- \frac{y}{z^2} \right ) + g_w \cdot \frac{1}{x}

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

$\begin{pmatrix} x \ y \ z \end{pmatrix} \cdot \begin{pmatrix} f_x \ f_y \ f_z \end{pmatrix} =
\begin{pmatrix} x \ y \ z \end{pmatrix} \cdot \begin{pmatrix} g_u \cdot \frac{1}{y} + g_w \cdot \left ( -\frac{z}{x^2} \right ) \ g_u \cdot \left (-\frac{x}{y^2} \right ) + g_v \cdot \frac{1}{z} \ g_v \cdot \left (- \frac{y}{z^2} \right ) + g_w \cdot \frac{1}{x} \end{pmatrix}$

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

Now we need to do the dot product and simplify the terms

at best factor g_u, g_v and g_w

[elfin warren]

is the simplification not just going to look like this?

[wanton bison]

yes

haha

and you see it cancels to 0

[elfin warren]

Yes

[wanton bison]

Yes

[elfin warren]

So we have proven it now?

[wanton bison]

If you did the dot product and simplified it to what you just sent, yes

See math is so easy

jk

[elfin warren]

Haha

Okay well thank you

[wanton bison]

thank you too

[elfin warren]

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[alpine swift]

Since Reddit hasn't gotten the exact explanation on how, I'll just post these here until someone can (unless no one sees this)

This is quite hard tho so idk be prepped for this one

[gritty oar]

you're taking 5 marbles, 4 of which are colored, and the remaning if colorless

you can have 4r,0b,1c; 3r,1b,1c

[gritty oar]

check these probabilities again

[odd edge]

@alpine swift Has your question been resolved?

[alpine swift]

0.0162+0.0004+0.0027 < 0.03645

[alpine swift]

P(4r, 0b, 1c)+ P(3r, 1b, 1c)+ P(3r+ 0b+ 2c)< 0.03645

i have a bit in mind

if the majority in the sample are red, then wouldn't the fact that only 1 or 0 blues exist?

P(0r, 0b, 5c)+ P(1r, 0b, 4c)+ P(2r, 0b, 3c)+ P(3r, 0b, 2c)+ P(4r, 0b, 1c)+ P(0r, 1b, 4c)+ P(1r, 1b, 3c)+ P(2r, 1b, 2c)+ P(3r, 1b, 1c)

0.00781+ 0.02256= 0.03037

this is close woh

hmmm

wouldnt it be also called majority if 2 red and blue balls happen

P(0r, 2b, 3c)+ P(1r, 2b, 2c)+P(2r, 2b, 2c)

0.03816

[alpine swift]

<@&286206848099549185> still stuck btw

[odd edge]

@alpine swift Has your question been resolved?

[alpine swift]

<@&286206848099549185>

[stiff sonnet]

so

I will see

[clever fjord]

Flux Fissionist

[stiff sonnet]

@alpine swift should I continue ?

[alpine swift]

do it yeah

ah wait wait

[alpine swift]

only 14(C)

ive done the rest

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[orchid thunder]

Hi, can someone help me with this one?

[unique maple]

that should be f(u(x))*u'(x)

I believe

use Fundamental theorem of calculus

[clever fjord]

Δyssrupt

[orchid thunder]

do I understand it correctly that it asks me to find a derivative of integral?

[unique maple]

ye

[orchid thunder]

how did u come up with this?

[unique maple]

uh

I believe u can prove FTC 2 using chain rule

like u know the integral of f(t)dt is a function right?

so let F(x) denote the corresponding functino

to f(t)

integral of f(t)dt

then substitute F(t) with F(u(x))

[orchid thunder]

why u(x) ?

[orchid thunder]

why u'(x) appears there?

[tall veldt]

this

[unique maple]

cuz chain rule

[orchid thunder]

okayy

thank you!

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[dreamy wasp]

Find all nilpotent elements of Z/18Z.

[foggy cloak]

in the ring Z/18Z an element is nilpotent if it can be raised to some power and yield the additive identitywhich is 0
Since ( (a + 18\mathbb{Z})^n = a^n + 18\mathbb{Z} ), we need to find all integers ( a ) such that ( a^n \equiv 0 \pmod{18} ) for some positive integer ( n ).

alr so the properties of modular arithmetic, we know that ( 18 = 2 \times 3^2 ). and any element ( a ) such that ( a ) is divisible by either 2 or ( 3^2 ) is nilpotent.

the nilpotent elements of ( \mathbb{Z}/18\mathbb{Z} ) are those elements whose residue class modulo 18 is divisible by 2 or ( 3^2 )

[clever fjord]

ويسالوني بعشقء مين؟🫨

[dreamy wasp]

What is the answer?

[foggy cloak]

I cant provide the answer

We are here to help not to answer

If i tell u i would get banned

[dreamy wasp]

What does it mean whose residue class modulo 18 is divisible by 2 or 3^2

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[oblique rapids]

??

I’m getting 8/25

But is wrong

[cold swift]

thats correct for f(x_0)

but we stil need f'(x_0) and all the other bits

i assume it wants it as an equation

[oblique rapids]

Deriv of 8/25 is 0

?

[cold swift]

derivative of f(x)

evaluated at 5

[oblique rapids]

So I deriv it first and then plug in 5

?

[cold swift]

yes

[oblique rapids]

I got 21/125x -13/25

But it’s still wrong

[cold swift]

so f(x_0) is 8/25

f'(x_0) is -16/125

if we plug in all of that into our formula

[clever fjord]

y0shi

[oblique rapids]

-16/125x+24/25?

[cold swift]

yeah

if that doesnt work, then it might want it in the form i had above

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[hoary sparrow]

how x can be equal to tan(θ)

[lofty spade]

where?

[lofty spade]

It says “x = tan(θ)” not “x = tan(x)”

[hoary sparrow]

oh

:))

but still does not makes sense

<@&286206848099549185>

[lofty spade]

Which part

[hoary sparrow]

I only see two equalitys for x

x=cos(θ)

or

x=cot(θ)

like how x can be equal to tan(θ)

<@&286206848099549185>

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[hoary sparrow]

.reopen

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✅

[hoary sparrow]

<@&286206848099549185>

[stiff sonnet]

@hoary sparrow

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[stoic citrus]

𝐴 = {𝑥 / 𝑥 is a professional soccer team and belongs to Bogotá} 𝐵 = {-4, -2, 0, 2, 4...}

Based on the selected items, the following responses are provided:

According to the selected item, determine by Extension the set given by Comprehension and determine by Comprehension the set given by Extension.

By Extension:
Set given by Comprehension: {𝑥 | 𝑥 is a professional soccer team and belongs to Bogotá}
By Comprehension:
Set given by Extension: It is not possible to define the set by Extension without knowing the specific elements that satisfy the condition given by Comprehension.
Find the cardinality of each of the sets:

Set 𝐴 is undefined in size (a specific list of professional soccer teams belonging to Bogotá is not provided), so its cardinality is unknown.
Set 𝐵 is an infinite set, as the given sequence extends infinitely in both directions. Therefore, its cardinality is infinite.
Identify the types of sets:

Set 𝐴: Infinite (no specific limit on the number of professional soccer teams belonging to Bogotá is specified).
Set 𝐵: Infinite (it is an infinite sequence of integers).

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[icy kindle]

.close

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[near wave]

I've got 2 questions, how would I find a1 for a geometric series that's described? The other one is how would I write and equation for the nth term of a geometric sequence

[near wave]

One example for the 1st question would be Sn = 1550 , n = 3 , r = 5

An example for the second would be 1, 4, 16

[clever fjord]

xd_senBugha

[near wave]

Which one would this be

[junior ether]

well

in geometric sequence

one term divided by the previous term is the common ratio

[junior ether]

wait

[near wave]

I'm just trying to figure out how to solve it

[junior ether]

you need to find a_1 first right

[near wave]

Yes

[junior ether]

plug in everything u know here

[near wave]

Then for the second one need to find the nth term of a geometric sequence

[junior ether]

geometric sequences have the form $a_n=a_1(r)^{n-1}$

[clever fjord]

Jash

[near wave]

Hmm ok

[junior ether]

you know a_1=1

and a_2=4

[near wave]

Yessir

Then it times x4

Everytime

[junior ether]

ye so r=4

and the first term is 1

[stuck ember]

hello need help :c

[near wave]

And n?

[junior ether]

!help

[odd edge]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[junior ether]

you only need the general term right

[stuck ember]

-9√6+11√7+35√7-16√6

:c

[near wave]

Sorry for some reason this is confusing, I'm just trying to make a problem for it

[stuck ember]

!help

[odd edge]

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[near wave]

Setup the equation

[near wave]

So if I just fill this in that would be good for making an equation

For the nth term?

[junior ether]

yes

nth term should be in terms of n

[near wave]

What would I write for n in this case

Just n?

So like

[junior ether]

yes

[near wave]

1(4)^n-1

[junior ether]

nth term just means ur making a formula to get any term

[near wave]

Ok that makes a lot of sense

Back to the 1st rq I have one more question

[near wave]

after you fill this in

Which in my case it would be

1550= a_1(5^3-1) / 5-1

I'm sure you know the command and how to make it an image

[junior ether]

solve for a_1

[near wave]

Yes

So I would go like

A_1(124)/4

Correct?

[junior ether]

i think

[near wave]

Then if that's right where do I go from there

That's my only problem

[junior ether]

1550=31a_1

[near wave]

Ok so I divide the 124/4

[junior ether]

ye

[near wave]

Alright

That makes so much more send

[junior ether]

and divide by 31 on both sides

[near wave]

Sense

Yep yep

[junior ether]

50=a_1

[near wave]

Tysm bro

[junior ether]

np

[near wave]

Have a nice night

[near wave]

Oh crud

I just stumbled upon another thing

On my worksheet

It's the same thing

Find a_1

But instead if n = ?

It's a_n=?

What do I do here?

I have 2 of these problems with a_n=?

[junior ether]

wait can u show the whole problem

[near wave]

Ye

S_n = 3478.2 , r = 2 , a_n = 3481.6

Find a_1 for each geometric series

[junior ether]

$3478.2=\frac{a_1(1-2)^n}{1-2}$

[clever fjord]

Jash

[junior ether]

$a_1(2)^{n-1}=3481.6$

[clever fjord]

Jash

[near wave]

I'm so confused

Where the hell did you get 3481.6

😭

[junior ether]

so this is using the sum

[junior ether]

they give a_n=3481.6

[near wave]

Ohh

It's a_n

Ok ok

But how do I find n

[junior ether]

you have a system of equations

[near wave]

So I can do this

[junior ether]

here a_1=3481.6/2^(n-1)

[near wave]

I'm so confused 😭

[junior ether]

so you have this right

[near wave]

mhm

[junior ether]

you can divide both sides by 2^(n-1)

[near wave]

Which gets this

[junior ether]

and you can substitute that for a_1 in this

[near wave]

And that will solve n

?

[junior ether]

yea u will have an equation in terms of n only which u can solve

[near wave]

3478.2=3481.6/2^(n-1)(1-2)^n / 1-2

[junior ether]

ye

[near wave]

3481.6/2^(n-2)(-1)^n / 1-2

/ -1*

So would I just put everything in parenthesis and make it negative

Than continue to solve

[junior ether]

where did you get n-2

[near wave]

Wait one second

I mightve mistyped

It's n-1*

3478.2 = -(3481.6/2^(n-1)(-1)^n)

Cuz since the fraction is over -1 I can just put it in parenthesis and make it negative

But that doesn't do it cuz 3478.2 is positive

I'm backtracking wtf

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[ember shard]

If they didnt give me the information in red that it was centered at 0
How would I determine c?
This is finding for a Lagrange Error Bound

[forest sky]

the polynomial shown is the first few terms of the maclaurin series for cos(x), which you are expected to know

note that maclaurin series are taylor series centered at x=0

in any case, for a power series centered somewhere else, you would expect to see (x-c)^n in the way they represent the polynomial

[ember shard]

So Taylor Polynomials if not states to be centered at a different value, its always centered at zero?

[forest sky]

,w taylor series for cos x centered at x = pi/6

[clever fjord]

Results provided by WolframAlpha

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Requested by cloudedhorizons

[forest sky]

note how the taylor polynomial for any other center would have (x - c)^n whereas if it's centered at 0, it will just have x^n

[ember shard]

Ah I see, missed that some how 0_0

ty for clarifying i got it

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[rapid ice]

Question 8 a, b, and c

[rapid ice]

Once I figure those I should be able to do the rest

[pastel steeple]

16 is 10 in hex
1 * 16^1 + 0 * 16^0

25 is 16 + 9

if you know what 16 is in hex then you could figure out what 25 is

same with 44

just break it up into units of 16