help-0

[gritty pond]

2

[vapid shuttle]

!show

[lone heart]

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

[vapid shuttle]

!show

[lone heart]

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

[gritty pond]

iso need function, bijective and homo

[vapid shuttle]

ah sorry it lagged

[gritty pond]

idk how to function

but i need function to start

[gritty pond]

how do u invent the function for this one?

[vapid shuttle]

Try listing an arbitrary element from both spaces

[gritty pond]

Ok

Example for 1 is (1,2,3)

exam,ple for 2 is (24,4)

[vapid shuttle]

no an arbitrary element

those are specific elements

[gritty pond]

(a,b,c)

(d,e)

(a,b,c) -> (d,e)

Domain=(a,b,c)
Codomain=(d,e)

[vapid shuttle]

how many possible elements could a,b, or c take on?

likewise how many possible elements could d, e take on?

[gritty pond]

a,b,c,e: 5 elements
d: 25 elements

[vapid shuttle]

Okay so how many elements are there in Z5 X Z5 X Z5

and how many are in Z25 X Z5

[gritty pond]

125 elements in Z5 x Z5 x Z5

125 elements in Z25 X Z25

WAIT did we just prove bijection?

[vapid shuttle]

If you can think of a 1-1 map

it's finite so injective => surjective => bijective

right

[gritty pond]

Ya

Oh so we have to invent a function first to say its bijective?

How the heck we do that

Looks too complex w the crosses

[lone heart]

@gritty pond Has your question been resolved?

[vapid shuttle]

well there's only 125 elements in each

figure out some way that you want to list them

[gritty pond]

wait

[vapid shuttle]

and then just pair them off

[gritty pond]

i found a much easier way i think

cant i just simplify z25 to z5xz5 and identical groups r iso

[vapid shuttle]

I don't know, but this way is also easy

what I mean for example is like maybe an obvious way of listing the elements of Z5 X Z5 X Z5 would be

(0,0,0)

(1,0,0)

(2,0,0)

...

[gritty pond]

yeah that makes sense

Oh wait legit just do it manually?

Lol

[vapid shuttle]

and then make a coorespding list for Z25 X Z5

you don't need to make the entire list

[gritty pond]

I completely forgot I can do it manually LOL

[vapid shuttle]

it will be clear what your bijection is

[gritty pond]

I always thought that I had to find a formula o some shit

[vapid shuttle]

well you do

but that doesn't mean it has to be hard to find

you can make it n your own

[gritty pond]

i cant just list them manually?

isnt that technically func if it is small enough

[vapid shuttle]

Look I mean like

when I listed those above

you obviously knew what the rest of my list was going to be right?

and I only listed 3 of the 125

now if I made a similar list of Z25 X Z5

like

(0,0)

(1,0)

(2,0)

...

(24,0)

...

and I said

now map (0,0,0) to (0,0)

map (1,0,0) to (1,0)

...

then it's clearly injective right

and so on and so forth

[gritty pond]

yeah that makes sense

oh so u agree i dont have to make an actual formula

i just gotta mlist all the values and mappings

[lone heart]

@gritty pond Has your question been resolved?

[lone heart]

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[misty shell]

this is the graph

[misty shell]

this is the equation

I mostly got it right, but how is there a phase shift of two to the left?

[naive valley]

notice that the graph has a peak at x=2

[misty shell]

sine graphs begin at the mid point, and there's a midpoint on x=o right?

[naive valley]

so you know at x=2 the argument to cosine needs to be 0, or a multiple of 2pi

[misty shell]

oh shit it's a cos graph......

that was careless of me

[naive valley]

haha yea

[misty shell]

thanks Bungo

[naive valley]

well they ask for cosine

you could of course express it as a sine as well

one is just a shifted version of the other

[misty shell]

yeah sine looks prettier for this so I just went for it, shouldve read the question though

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[latent lantern]

Someone help me solve for y please

C_1 is a constant equal to e^C

I'm trying to solve a separable differential equation and got super stuck

[tiny fable]

that's the neat part, you don't

[latent lantern]

i might've messed up somewhere along the way 😭

[minor bone]

What was the original equation

[tiny fable]

well if you did your steps right, this just means only an implicit solution exists

doesn't mean it's wrong necessarily

[tiny fable]

my god

[latent lantern]

there was a help video but it only got to here

and i need to solve for y

my professor is torturing me

[tiny fable]

you literally cannot solve for y if that's correct

that's a valid solution right there

were you given an initial condition?

[latent lantern]

i'll try and if it doesn't work i'm just gonna go to office hours the day its due :(

[latent lantern]

nope

[tiny fable]

then you're done lol

[latent lantern]

this is it

[tiny fable]

you can try solving it, but any time exponentials and non-exponentials are together, it won't work

strange that you were given that, that sucks

[latent lantern]

i'm just gonna. move on. and curse web assign.

.close

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[peak briar]

i got 90.8 and the answer is 87.7

[tiny fable]

what steps did you take to get there?

[peak briar]

I first did this

[tiny fable]

okay sure, that looks fine

[peak briar]

And then i used the formula

To find the area

Pretty sure i messed up on this step

[tiny fable]

that looks fine to me, it's possible you just entered something in the calculator wrong

[peak briar]

Lemme check

[tiny fable]

yeah you did lol

[peak briar]

Now i got my answer as 100💀

[tiny fable]

lmao, when i did this in my calculator i got t = 22.6089

so no clue what you entered but it was just a calculator mistake

[peak briar]

1 sec

Omg

My calculator was is grad mode

Instead of deg

[tiny fable]

that would do it lol

it's shocking you got that close anyway

what a coincidence

[peak briar]

Gonna check and see if it calculates to the answer given rq

yeah it works now

idk how my calculator switches its mode out of nowhere

thanks for the help tho

[tiny fable]

lol no prob

[peak briar]

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[digital sierra]

The 20-person student leadership committee recently voted for a retro theme for a school dance. Committee members could vote for one or more themes. Two students were absent from the meeting and didn’t vote. After much negotiation, two ideas made it onto the ballot: Enchanted Evening (E) and Under the Sea (S). Enchanted Evening received 13 votes, and 6 people voted for both Under the Sea and Enchanted Evening.

14. Find the probability that someone voted for both themes, given that they voted for the theme Enchanted Evening.
    Using B for both
    i did P(B|E) = (6/13) x (7/20)/(7/20) which equals 6/13 is this correct?

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[digital sierra]

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[balmy grail]

If f(x)=x^5 then the derivative f’(x) = 5x^4

How would you find f(3) ?

Does this make more sense? f(x=3) = .

The same way you would use f(x) to find the output for x, you can choose a value of x, such as 3, to find its output. E.g. f(x)=x^5 -> f(3) = (3)^5 = 243

So using that logic, and what you found for f’(x) can you plug in those values?

Yeah, it works everywhere that the derivative exists

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@alpine sable Has your question been resolved?

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[light hinge]

I don't know how to solve this one :c

I thought it was a express as x problem but I don't know where to start

[hearty quartz]

So this is an optimisation problem. What you want to do is formulate the objective functions and any constraints. From there you can try and solve the problem.

Basically you need to know what the problem is before you solve it

[light hinge]

Can you work with me here since I'm kind of fried?

I'm working with A=lw here and probably perimeter

[hearty quartz]

There is no point in me doing this?

You want maximise the area so what is the expression for area

[hearty quartz]

This is correct but we need it in terms of the decision variables the values that we are deciding

[light hinge]

Alright So let's say for the pen it will be 40 + x by y

The problem I come across is the barn obstructs the perimeter

Will it just be 40 + x multiplied by x?

If we remove 40 from the perimeter of one of the sides?

[hearty quartz]

No, (40+x)*y is the area which is the objective function

[light hinge]

Ok

[hearty quartz]

Now you need to come up with constraint involving the perimeter

[light hinge]

P = 2y + 40 + 2x

[hearty quartz]

That is the perimeter but what is the cnstraint?

[light hinge]

The 40 from the barn no?

I don't know how to express that since we haven't done it in class yet

Or is the constraint 200 feet of fencing material

200 = 2y + 40 + 2x

Let's say we want to find the area expressed as y because it's easier to work with?

[hearty quartz]

y or x should not matter either way

[light hinge]

80 - x = y?

[hearty quartz]

So now we substitute

[light hinge]

I got 80x - x2

[hearty quartz]

$(40-x)y = (40-x)(80-x)$

[ocean seal]

team132

[hearty quartz]

??

[light hinge]

errrr

Why do you have 40 - x

[hearty quartz]

Ah sorry 40+x

[light hinge]

Ohhhh

-x2 + 40x + 3200

So since I have the restriction

80 > x

I can plug in that for x in this new equation right?

[hearty quartz]

-(x+40)(x-80) = -(x^2 -40x -3200) = -x^2+40x +3200

There is actually no need to expand. You can take the average of the roots of the parabola and that will give you x-value such that the y-value of the parabola is maximised

[light hinge]

Average is that the axis of symmetry?

Ohhhh

It is

[hearty quartz]

yep axis of symmetry

You can use differential calculus but it is more complicated

[light hinge]

This is in a pre calculus course

I am not there yet

Soon though

[hearty quartz]

I see

[light hinge]

will 3600 be our y max?

Since axis of symmetry is 20

[hearty quartz]

So x= 20

We now need to find y

[light hinge]

Which is simply plugging it back in correct

[hearty quartz]

Correct

[light hinge]

So you get the point that correspodns to 20

Alright so now can I plug these values back into a = lw?

20 x 3600

[hearty quartz]

yep

[hearty quartz]

area is 3600 feet ^2

What they want is the dimensionsof the pen

[light hinge]

Why is 3600 feet squared?

[hearty quartz]

area of a rectangle

[light hinge]

lw?

Ohhh

the units

ft2

72000ft2

[hearty quartz]

no

[light hinge]

Hmm?

[hearty quartz]

I don't know why you are multiplying by 20?

[light hinge]

Because that's the area of a rectangle no?

x was our 20

Or do we just work with the max now

[hearty quartz]

x=20, y =60, the objective function is $(x+40)y$

[ocean seal]

team132

[light hinge]

Alright

I see

60 ft

Well we found the dimensions is that it?

[hearty quartz]

Yep it is 60 by 60 which will give an area of 3600

feet^2

[light hinge]

I will write out my units for these problems

So our dimesnions are 20 by 60?

Or do we express this with the picture of the fence in mind

Er pen

60 by x

[hearty quartz]

no dimensions are the side lengths of the rectangle

[light hinge]

Right

So it is 60 x 60

[hearty quartz]

yep

[light hinge]

Thank you I will not forget how to solve these problems now

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[ashen ginkgo]

what is 12 multiply by 21

[ashen ginkgo]

i will give you 1 star\

[old bronze]

252

[hushed locust]

https://www.desmos.com/scientific

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[fossil rune]

i

[steep pendant]

how far have you gotten in solving the problem ?

[fossil rune]

i solved 4

but 5 wording is confusing

[steep pendant]

i agree actually im not so sure what it means in question 5

i think it means that you choose a 6th number independent of the numbers you chose in the first problem and if that number is also correct then you win the lottery

[fossil rune]

so it would be 60 choose 5 * 60 choose 1?

[steep pendant]

yeah

at least that's how I interpreted the problem

[fossil rune]

hmmmmmm

[zinc haven]

you got the answer from problem 4 correct?

[fossil rune]

yeah 60 choose 5

i mean thats what i got

[zinc haven]

the problem 5 essentially states that you choose another number from 1-60, and ask the new number of possible tickets

each ticket now has 60 more possibilities

the rest, you should be able to calculate

[fossil rune]

so 60 choose 5 * 60?

[steep pendant]

mhm thats the same thing that you got before

[zinc haven]

now i have never played lottery before, so i wanna ask something here

can you choose the same number?

is 1 1 1 1 1 a valid combination?

[fossil rune]

if there is that combination in 60 numbers

[steep pendant]

my interpretation of the question is that the first five numbers you have to choose must all be different numbers (so 60 choose 5 would be correct) but if they could be the same number then it would be 60^5. the way the problem is worded makes it sound more like the first one though

[fossil rune]

yeah cause it says choose 5 so

[steep pendant]

well

the way the problem is worded doesnt completely exclude the possibility that it might want you to be able to pick multiple of the same number, but it does feel like thats what it wants you to do

[zinc haven]

i believe it must be different

since its powerball

[zinc haven]

therefore, it should be 60c5 * 55

since there is only 55 ways to pick the sixth number

unless you can choose the repeating number as the powerball, then it would be * 60

otherwise, 55

[steep pendant]

i can see how you can think of that

[fossil rune]

oh i see that

and im p sure thats what the question is intended for

[steep pendant]

and it wants you to think of getting the answer by choosing the powerball first versus choosing it last and if you check:
60 * (59c5) is the same as (60c5) * 55

[fossil rune]

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[safe pecan]

anyone know how to do this question?

i'm trying to use integration by parts

[grave pier]

I can tell you that integration by parts will not work

[safe pecan]

yeahh i got to the end of it and it started looping 💀

what approach should i use?

[grave pier]

And I'll also say that the bounds of this integral were chosen carefully

[safe pecan]

hmmm

[grave pier]

What I mean by that is that you would not be able to solve this if this was an indefinite integral (I don't think)

[safe pecan]

ahh i see

do you know how i can approach this question?

where do i start?

[grave pier]

Ok I'll give a big hint think about even and odd functions

[safe pecan]

hmmmm

i'm still stuck 💀

i haven't touched calc in a long time

[grave pier]

No worries this kind of question is a little strange since you're not able to like do the integral

But consider whether the inner functions (e^-x^2, sin^3(x)) are even or odd, and what that would say about the integral

Alternatively plotting the function might help

[safe pecan]

ooh okok tyty

i gtg now but super helpful stuff

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[cinder breach]

How do i simplify this?

[cinder breach]

Is there any property?

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@cinder breach Has your question been resolved?

[left isle]

besides just plugging in each value of k and summing up the values?

[cinder breach]

Yes, like suppose we didn't have 5 but we had 200

[pallid scarab]

Then we would have used that nC2 = n(n-1)/2

And we use sums of integers and squares of integers

[cinder breach]

Uh sorry what is this formula?

[pallid scarab]

$\binom n 2 = \frac{n(n-1)}{2}$

[ocean seal]

rafilou2003

[cinder breach]

Oh nice thanks

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[static wagon]

Can someone tell me where I went wrong here? I’m not getting the answer that I need

[north stag]

replace small with large

[static wagon]

wait why

[jagged cobalt]

'the quantity of metal used in the larger size is 15.625 times... in the small size'

you said the smaller size was 15.625 times the larger one

[limpid igloo]

ax>x
as long as a>1

[static wagon]

ohh alr

and how would i do question 2 part c

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[static wagon]

<@&286206848099549185>

[dawn tundra]

From Part b you know that Triangle PQT and RSQ are similar.

PT= 6cm and QR= 4cm

so the sides of RSQ are 2/3 of PQT

A=8.9 for PQT

Since $Area=\frac{ab}{2}$

[ocean seal]

GamingCrossroads

[dawn tundra]

where a and b are edge lengths, the area for RSQ is 8.9(2/3)(2/3)

[static wagon]

what would be the reasoning behind this

corresponding sides of similar triangles in same ratio?

[dawn tundra]

Similiar triangles are just scales of each other. So when comparing any two edges on 1 triangle it will have the same ratio as the same two respective edges on a similiar triangle.

[static wagon]

ooo

alrlr

but then what about to the 2

ab/2

[dawn tundra]

the 2 is just that the area of the triangle is half of the area of a rectangle with the same side lengths.

since all triangle area have divide by 2, the divide by 2 can be ignored.

[static wagon]

oh fair enough

alr thank you

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[fair coyote]

How did they calculate it to be sqrt 3

cuz like if u do 2x5 = 10

So then u will get 10pi/6

Which you can divide to 5pi/3

[gray isle]

do you know the sine of that?

[fair coyote]

No

Omg

I’m so dumb

The unit circle

Lol

Sorry

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[fair coyote]

.reopen

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✅

[fair coyote]

nvm I still don’t get it 😭😭😭😭

[gray isle]

are you looking at the unit circle now?

can you identify the value of sin(5pi/3) from that?

[fair coyote]

-sqrt3/2

But then shouldn’t the answer be -sqrt3

[gray isle]

and what happens when you multiply that by -2

[fair coyote]

Got it !

Thank you !

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[hoary bolt]

I need help finding the amount of combinations for passcodes

[hoary bolt]

I need the formula to find the amount of combinations possible for all the 5 different types of passcodes

A fingerprint has been placed on the digits that have been used

So for an aabc passcode there are 3 finger smudges on the phone so you know which digits are used but you don’t know which digits are being repeated

I know the answer is 36 and I used the permutation formula but the formula only gives 12 as you have to times it by the amount of digits known

Just don’t know the correct formula to use for this

Can anyone pls help

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@hoary bolt Has your question been resolved?

[hoary bolt]

Anyone helping?

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@hoary bolt Has your question been resolved?

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[blissful rapids]

I've noticed that in a lot of proofs of trigonometric identities, we sometimes cancel out a trigonometric function from both sides, but 0 is a possible value for them, so why do we still cancel them out from either side?

[rancid jay]

we cancel them out only after considering them as solutions

otherwise u have an incomplete set of solution

unless the question itself restricts the solution set

[blissful rapids]

Wdym?

[rancid jay]

we do consider them as solutions

[blissful rapids]

So we take both cases, of the cancelled part being equal to 0 and of them not being equal to 0?

[livid tundra]

I dont think we usually do that If im not wrong

[high shuttle]

Well, that depends on the nature of the solution, some 0/0 limits do have a non singular vale when x approaches the singularity

for example, the case sin(x)/tan(x) with x -> 0

so sin(x) = tan(x) * cos(x) becomes a valid equality when x -> 0 since sin(x)/tan(x) is non singular in the limit x-> 0

[blissful rapids]

Idk calculus yet

[high shuttle]

Oh, I see.

[blissful rapids]

Ill revisit this later

.close

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[undone trellis]

If you connect the four points A, B, C, and D on the graph of quadratic functions y=x2, y=-1/2x2 as shown in the figure on the right, you can find the area of this square when it becomes a square.

Since I'm Korean, The question is translated into Korean. I'm sorry,,,

[livid tundra]

What do u exactly gotta do btw.?

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[subtle cloud]

A 0.83kg billiard ball initially at rest is given a speed of 15m/s during a time interval of 5.0ms.
what average force acted on the ball during this time?

[subtle cloud]

physics question. I don't even know what topic this is, i just want to help my sister with her homework. I'm taking a physics class right now so I can pretty much understand you if you explain it to me and guide me through the solving

$F=0.83kg\frac{15\frac{m}{s} - 0\frac{m}{s}}{5ms}$

[ocean seal]

sly5372

[subtle cloud]

is this correct huhu

<@&286206848099549185>

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[autumn quarry]

Yo, I just wanted to get some stuff clear, why is it that if this function is even, there is a two in front ?

[autumn quarry]

Also, I know that this fct is periodic so we only have to study it on one bound (interval idk) only, but why is it necessary to make this change in the problem ? Would it be a problem to just do it on the whole bound ?

[shy dove]

It's especially clear when you look at the graph of even functions. Think y = x² for example. if you were to integrate from -1 to 1, the area that's to the left ([-1, 0]) is exactly the same as the area to the right ([0, 1]). So, as to avoid writing negative signs in the bounds, people prefer to write it as twice the integral from 0 to whatever. It doesn't affect anything really.

[autumn quarry]

Ok, so, the 2 comes from the other bound [0;-pi/2] because it's the same, right ?

[shy dove]

yeah! More generally, if e(x) is an even function, then $\\int_{-a}^{a}e(x) = 2\int_{0}^{a}e(x)$

[ocean seal]

ℑμΤ𝛄𝛗θ

[autumn quarry]

Ok, thanks !

.close

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[low moat]

for question a, root1024/16 (radical = 3) is the same as root1024/16 (exponent 1/3), why is my calculator giving me 2?

[low moat]

I understand this may be hard to understand but I rlly cant be bothered to fix my gramamr its like almsot 2 AM 😭

[echo socket]

,w (1024/16)^(1/3)

[ocean seal]

Results provided by WolframAlpha

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

Requested by alonelybean

[low moat]

my calculator gotta be trippin

wait

[echo socket]

Are you sure you are inputting it correctly?

[low moat]

I put (root1024/26)^1/3

[echo socket]

Pretty sure that root is extra

[low moat]

huh

[echo socket]

(1024/16)^(1/3)

Input this

[low moat]

damn

I didnt need the root?

yeah it gave me 4

why do I not need the root

is that like a rule or smth

[echo socket]

Why would you need a root there?

Aren't you exponentiating to the power of 1/3 in order to take the cube root already

[low moat]

oh

oh yeah

wait are u japanese

[echo socket]

No

[low moat]

cuz ur username suggests it

oh.

rahh I thought I was talking with somebody in my country

anyways uh ty bro

.close

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[alpine sable]

.help

[lone heart]

Commands:

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• factoids: .tag
• help: .help
• version: .version

Type .help <command name> for more info on a command.

[alpine sable]

.help math

[lone heart]

No command called "math" found.

[alpine sable]

it's special right triangles im a freshman and i am terrible at geometry i need help on how to do it

[limpid turret]

In the future, you simply need to ask your question to open a channel. No bot commands required

[alpine sable]

oh alr

[limpid turret]

the 45-45-90 right triangle is easier to explain, so let's do that first

[alpine sable]

alright

[limpid turret]

You know the two legs are equal

[alpine sable]

yes

[limpid turret]

So use that, and pythag, to solve for their lengths

[alpine sable]

but i only have hypotenuese

and sometimes it asks like 5 sqrt2

[limpid turret]

you only have hypotenuse, yes, but the two legs are equal, so there is only one unknown

[alpine sable]

oh ok

[limpid turret]

both angle are 45 degrees, so the triangle is isosceles, which means the sides are equal, right?

[alpine sable]

yes

[limpid turret]

well there you go

[alpine sable]

hold on

[alpine sable]

4.24?

thx so much

how would i do the 30 60?

[limpid turret]

one sec

[alpine sable]

alr, take your time

[tropic nest]

hello

i could also help with your question

i did geometry last year

[alpine sable]

alright i solved the one on the right i just need the one on the left

i have the test tmr lol

[tropic nest]

what topic in geometry are you doing?

[alpine sable]

trig and special right triangles

i got trig down

[tropic nest]

alr okok

[limpid turret]

@alpine sable a 30-60-90 triangle is just half of an equilateral triangle

[alpine sable]

ok

omg that helps so much

there was equalateralk triangle questions i didnt understand

[limpid turret]

So if ΔABC is an equilateral triangle, and AC=10, can you find AD?

[alpine sable]

5?

[tropic nest]

do you know how to use sin, tan, and cos yet?

[alpine sable]

yes

SOH CAH TOA

[tropic nest]

yeah

[limpid turret]

Cool. You know AC and you know AD, so can you find CD?

[alpine sable]

15?

[limpid turret]

how'd you get 15?

[alpine sable]

idk

[limpid turret]

then don't guess.

[alpine sable]

ok

[limpid turret]

Try to actually think about it

[alpine sable]

ok

7.5

wait

nvm

[tropic nest]

whats the questio?

question?

[alpine sable]

wait

Pythagorem theorom works on 30-60 or no

[alpine sable]

this one on the left

i have the short side = 5

long side

[limpid turret]

As you can see, triangle ACD is a right triangle, so yes, pythag works

[alpine sable]

ok

8.66?

or 5 sqrt 3

so the long side will always be the short side with times sqrt 3?

[limpid turret]

yup. One of the fundamental facts of trigonometry

[alpine sable]

you teach better than my teacher

[limpid turret]

teacher gotta teacher 30 students

and probably stop them from sexting each other and sharing amung us twerking videos, or whatever your generation does these days

[alpine sable]

lmao

i don't do that thankfully

there probably is people that do that in my class unfortanetly 😭

thank you so much, you made this so much easier have a good day

[limpid turret]

no worries

have a good one

[alpine sable]

.close

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[zealous drift]

im trying to calculate the domain and range of this function f(x)=nroot(x,5) it seems like the domain and range are all real numbers am i correct
g(x)=nroot(x,5)+6 if 1 is correct is this also all real numbers or (-inf, inf), [0, inf)?

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@zealous drift Has your question been resolved?

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@zealous drift Has your question been resolved?

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[ocean seal]

IV_Intel

IV_Intel

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[surreal meteor]

By induction and I have done it but I keep getting the wrong answer

It should be n^2-n-2 not 4…

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@surreal meteor Has your question been resolved?

[alpine sable]

it appears that on the 3rd equals sign you miltiplied the left side by 2 but didn't do the same for the right side

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@surreal meteor Has your question been resolved?

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[fierce ruin]

For context this is a question aiming to find the derivative of the function however in the provided solution they do this and I don't seem to understand why it works or even what it's really doing

[alpine sable]

double subtraction is addition with the last sin(x) and the minus sign in the middle

[surreal meteor]

i multiplied by 2/2

[fierce ruin]

so does the rule in this context not mean "negative A * negative B "

[alpine sable]

it can

so x - y is actually x + (-1)y

so then the first one is ((2 + cos(x))sin(x) + (-1)(11-cos(x))(-sin(x)))/(2 + cos(x))^2

then applying that rule cancels out both minus signs on the right upper half of the equation

[alpine sable]

make a new channel i don’t see your original image

[surreal meteor]

i found the issue, it had to be (k-1) and not (k-2)

[fierce ruin]

still in chat just trying to rationalize this in my head

[alpine sable]

ok

oh

[fierce ruin]

im confused why it's only the sin that seemingly changes sign in this particular case cause wouldnt -1(11-cos(x)) be (-11+cos(x))

[alpine sable]

also the minus sign in the middle changes

[fierce ruin]

with it being - (-11) i'd get it but the cos would flip to be positive no? so like - (-11 - cos(x)) isnt the same as - (-11 - cos(x)

i feel like im probably overcomplicating this but im really unfamiliar with algebra involving trig functions

[alpine sable]

that part isn’t being manipulated at all

the trig functions don’t matter replace them with letters if it helps you understand

imagine it like this: A = (2+cosx)sinx, B = (11-cosx)

so then its A - B*(-sinx)

which is just A + B*sin(x)

[fierce ruin]

wait hm i originally was treating it as A = (11 - cos x ), B = ( sin x)

so like reversing it to be (-11 + cosx ) * (-sinx)

i was mostly doing this cause i wasnt seeing the first part edited at all

ok been fiddling with it on paper and i think i largely get it now after some clarification but one thing im still stumped on at least conceptually is why the sin is treated as "extra" and not part of B

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[fierce ruin]

.close

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[strange jolt]

How would I go on about such problem. Suppose that G is a group and let a, b ∈ G. Prove that if |a| = m and |n| = n with gcd(m, n) = 1, then <a> ∩ <b> = {e}.

[placid zinc]

<a> ∩ <b> is a subgroup of both <a> and <b>.

How might their orders relate?

@strange jolt

[strange jolt]

Idk

[placid zinc]

Know Lagrange's theorem?

[strange jolt]

Were barely covering it

.close

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[urban island]

ive sent both the question and the answer and i kinda need help understanding the solution

especially the part i bracketed at the bottom

i understand the expansion but the simplification not so much

[jagged cobalt]

theres a 2 missing, and then later a ^2 missing

for it to make sense

[urban island]

Careless mistake mb

[jagged cobalt]

oops missed one

[urban island]

Ohhh that makes more sense now

Alright

Then the last line?? How do you equate it to 8Re(z)Re(w)

[jagged cobalt]

re(z)=x, re(w)=a
8re(x)re(w)=8ax

the left side is just cancellation

[urban island]

I’m trying to understand how re(z)= x and re(w)=a

[jagged cobalt]

thats just how you defined z and w

if z=x+jy then Im(z)=y and Re(z)=x

the imaginary and real components respectively

[urban island]

OHHH

[jagged cobalt]

this is on the assumption that x,y,a,b are real numbers

[urban island]

okok i get it now

thank you so much man

[jagged cobalt]

no worries

[urban island]

.close

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[tropic onyx]

The answer is (a), there is no explanation, so I need help.

I don't even know where to start.

[grave cedar]

yo

you know the property where take numerator to denominator by switching a and b

@tropic onyx

you online?/

[tropic onyx]

Yes I do

[grave cedar]

then apply it

what you get?

[tropic onyx]

wait a second

[grave cedar]

ok

quick

typing

.......

[tropic onyx]

Sorry. I think im stuck

[grave cedar]

send a pic where you stuck

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[cyan moss]

How to find the antiderivative of this

[lethal belfry]

try taking the $x^2$ out of the square root, and then sub 1/x=u

[ocean seal]

Why am. I here

[cyan moss]

I am sorry but how are you supposed to take the x^2 out? The only thing i can think of is sqrt[1+9/x^2 ]

Can I solve this if i change the differential?

Like dx to d(x^2+9)

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[cyan moss]

.reopen

[lone heart]

✅

[nimble fern]

if you don't know how to take x^2 out, you can try with u-sub
u=x²+9

[cyan moss]

Oh this this was my first method and it worked out well

Although

I am mostly interested in changing the dx

Is it possible ?

[nimble fern]

I'm not sure if i understand what you mean

[nimble fern]

you mean this?

oh, it's basically u-sub, without the u involved

[cyan moss]

Ohhhhh

[nimble fern]

one of the pros in such substitution is that for definite integral, it will be easier

[cyan moss]

Can you write it out ? I am stuck here I can do it with the u-sub but not the other way around

[nimble fern]

ah

lemme try

[cyan moss]

Yup yup i got all the way there and then what ?

[cyan moss]

Oh damnnnn

Nice, appreciate it bro

[nimble fern]

i guess that's it

[cyan moss]

I get it

Again thank you

[nimble fern]

Cheers!

[cyan moss]

@nimble fern uhh 1 small question just to be sure. You got d[sqrt(x^2 +9)] because of this right?

[nimble fern]

reading now

thinking

technically yes, but my thinking process is like (I'll write it down now):

for me, what i was taught by my teacher back then is that
the feeling of this method is to integrate the terms you want into the d( )

let's take a simpler example:

$\int \frac{x}{x^2+1} \dd x$

[ocean seal]

Biscuity

[nimble fern]

what i would think of:
integrate x into the d( ) w.r.t. x so that we can get something like x^2

[cyan moss]

:0

[nimble fern]

since we know $\int x\dd x=\frac{x^2}{2}(+C)$

[ocean seal]

Biscuity

[nimble fern]

I'll just do $\int \frac1{x^2+1}\boxed{x\dd x}$

[ocean seal]

Biscuity

[nimble fern]

into $\int \frac1{x^2+1}\boxed{\frac{\dd (x^2+1)}2}$

[ocean seal]

Biscuity

[nimble fern]

since we can add any C, what i did is add C=1/2

very similar to u-sub, but reverse the thinking process from du/dx to finding anti-derivative

[nimble fern]

[cyan moss]

Wow that is super easy to understand

You are even better than my teacher

[nimble fern]

don't compare XD
teachers at school are teaching the whole class, while I'm just helping you

[cyan moss]

Good point

Well it is done, have a nice day

[nimble fern]

you too!

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[zinc summit]

hep

help

pls

<@&286206848099549185>

[lyric notch]

My solution is a little convoluted I think

But

[tacit arch]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lyric notch]

To show A or B, you can show that if not A, then B, and if not B, then A

[zinc summit]

oh sorry

[zinc summit]

pls tell me how to show it

[lyric notch]

First suppose the second statement is false. If there are not 3 of them who know each other, then any 3 you choose will not all know each other

[zinc summit]

wait so you just need to write it down

i thought you consider it with algebra

[lyric notch]

I think technically this is best done with graph theory

I'm not sure what algebra you would do

Drawing a picture would probably help though

[zinc summit]

ohh kk

i dont know graph theory

[lyric notch]

Me neither

[zinc summit]

havent learned it

might look into it

welp still need the solution tho

[lyric notch]

So continuing from here, choose 3 at random. Either none of them know each other, in which case you are done, or there is some pair that does know each other

I'm not really sure how to explain this without a picture

I think if you draw out a picture and try to draw the possibilities it will help

Trying to break things basically

[lyric notch]

Like from here, in the latter case there must be a pair that does not know each other, because they cannot all know each other. If you take that pair and choose someone new from outside the group, you will get a new set of three and you can repeat, but eventually you will run out of people

[zinc summit]

ok imma close this because i dont think anyone can explain this

thanks anyway

.close

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[alpine sable]

g'(x) is the derivative function

question is find the set of x values where the function is increasing

shouldn't it be x>1 U x<-3

or am I high

idk where they got 1.098

and -4.098 from

[split mantle]

Should've changed the order to <inf, -3)U(1, inf>. But it's right that g(x) increases in those sections.

[alpine sable]

oh yeh

oh okay thanks

.close

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[royal meadow]

strange

[split mantle]

I'm not sure if the x>1.098 or x<-4.098 is the answer or not, but that is in the area.

[alpine sable]

?

[split mantle]

Same here. I assume that they just took two values that are valid.

[alpine sable]

but it's wrong right

the 1.098

and -4.098

[split mantle]

1.098>1, so that's valid.

So does -4.098, -4.098 < -3

[alpine sable]

but is 1 and -3 the more valid answer

to put here

[split mantle]

Yes.

[alpine sable]

ok good

ty

[split mantle]

Welcome.

[lone heart]

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[tight nest]

I think i got a corret

But im sutck with b..

[zealous lichen]

you want the values of k such that the series converges

so |r|<1

[tight nest]

uhm...

Can you givie me an equation?

of how to do?

[brittle ember]

May I ask what are u_1 and u_2 for?

[lone heart]

@tight nest Has your question been resolved?

[tight nest]

u1 means its the first term of the sequence

u2 means second term

[brittle ember]

so r is ratio of geonetric series

geometric series converges if |r|<1

or would it be more comfortable in korean?

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[lone heart]

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[alpine sable]

Any hint?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[livid sage]

consider how you would solve this problem if it was a rectangle instead of a parallelogram

well. a rectangle is a parallelogram, but still

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[thick agate]

bonjour est-ce que quelqu’un peut m’aider svp

[alpine sable]

what is the notation $p_A(B)$? $P(A | B)$?

[ocean seal]

45

[alpine sable]

ah yes

it says in french conditional probability

[thick agate]

oui

[alpine sable]

est-ce que vous parlais englais?

my french is shit

XD

[thick agate]

a little bite

[alpine sable]

ok

[alpine sable]

c'est vrai?

[thick agate]

yes

[alpine sable]

ok

you know that $P(A | B) = 0.5$ and $P(B | A) = 0.3$, and $P(A) = P(B) + 0.3$

[ocean seal]

45

[thick agate]

yes

[alpine sable]

are you familiar with Bayes' theorem?

[thick agate]

no

what is

[alpine sable]

$P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)}$

[ocean seal]

45

[thick agate]

ah yes

[alpine sable]

utiliser ca

[alpine sable]

avec ca

0.5 = 0.3 * P(A)/P(B)
0.5 = 0.3 * (P(B) + 0.3)/P(B)

[thick agate]

i don’t have P(A)

[alpine sable]

P(A) = P(B) + 0.3

c'est une equation

[thick agate]

no P(A)=P(B)-0,3

[alpine sable]

oops

pardon

oui c'est P(A) = P(B) - 0.3

donc

utiliser ca

[alpine sable]

0.5 = 0.3 * (P(B) - 0.3)/P(B)

[thick agate]

ok 👍

[alpine sable]

,w solve 0.5 = 0.3 * (x - 0.3)/x

[ocean seal]

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[alpine sable]

....

wtf?

[alpine sable]

c'est $P_A(B) = P(A | B)$ ou $P(B | A)$?

[ocean seal]

45

[alpine sable]

@thick agate

[thick agate]

is not possible

[alpine sable]

yeah

probability cannot be negative

[alpine sable]

?

[thick agate]

i don’t now

c’est probabilité de B sachant A

[alpine sable]

je pense que c'est P(B | A) pour P_A(B)

oui

[thick agate]

ok

[alpine sable]

P(A|B) = 0.5
P(B|A) = 0.3
P(B) = P(A) + 0.3

0.5 = 0.3 * (P(A) + 0.3)/P(A)

,w solve 0.5 = 0.3 * (x + 0.3)/x

[ocean seal]

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[alpine sable]

9/20 c'est P(A)

donc P(B) = P(A) + 0.3

0.3 = 6/20

je pense que c'est 0.75

[thick agate]

d’accord merci 👍

[alpine sable]

@thick agate pardon c'est $P(B) = P(B | A) P(A)/P(A | B)$, donc P(B) = 0.5 * (P(B) - 0.3)/0.3

[ocean seal]

45

[alpine sable]

,w solve x = 0.5 * (x - 0.3)/0.3

[ocean seal]

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[alpine sable]

oh

non c'est vrai

[lone heart]

@thick agate Has your question been resolved?

[lone heart]

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[violet shore]

I straight up just can't figure out what i'm supposed to do with this

[alpine sable]

what have you tried

[violet shore]

to me it looks like i'm supposed to solve arcsin(1/3) and that doesn't seem correct for my level

[alpine sable]

yeah but that's not all solutions

[violet shore]

right so i'm lost because i don't know what i'm being asked for at all

[alpine sable]

let's try something simpler like sin(x) = 1/2

what are all the solutions to that equation

[violet shore]

π/6 and 5π/6

[alpine sable]

yh

those are 2 solutions though

keep in mind that sin is 2pi-periodic

[violet shore]

-+ any integer k * 2π

[alpine sable]

yeah

you dont need to say -+ since integers can take on negative values, but yes

so what are two solutions to sin(x) = 1/3?

[violet shore]

i don't know

[alpine sable]

what's arcsin(1/3)

[violet shore]

we're not supposed to use a calculator and i've never run into anything with a sine of 1/3

[alpine sable]

you have to use a calculator for that question

arcsin(1/3) is not a table value

[violet shore]

okay well this entire course doesn't use a calculator so like wtf

[alpine sable]

ok then

what we can try is

instead of finding the value of arcsin(1/3) we can just leave it as arcsin(1/3)

so one of the two sets of solutions is $\arcsin(1/3) + k2\pi, k \in \mbb{Z}$

[ocean seal]

45

[violet shore]

that feels wrong too

[alpine sable]

well there's really no alternative

if you can't use a calculator

[violet shore]

wack

doesn't arcsin have a limit? are we allowed to ignore that if we're writing it as a set?

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[strange fractal]

hi

[strange fractal]

how i draw dis

oh i see

.close

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[valid smelt]

incredible

[rancid jay]

llol

[strange fractal]

.reopen

wait

[lone heart]

✅

[strange fractal]

now how do i do the question..

[valid smelt]

you drew it wrong, that picture doesn't have the circle cutting through C

[strange fractal]

oh

[strange fractal]

waddaheck

[valid smelt]

keyword extended if neccesary

[strange fractal]

wat dat mena

mean?

[valid smelt]

CD here is the infinite line that passes through points C and D

like you extend the side basically

[strange fractal]

o