help-0

[half epoch]

In other words this proposition only holds under the assumption that neither of your vectors are the 0 vector

Yes, I assume the question just wants you to pretend both vectors have length > 0 though

Ah, if we are just doing an exercise then you have to add that both vectors have non-zero length to our initial assumptions, otherwise the proposition is wrong

Mh, I wouldn’t, wlog is usually when you want to split out a particular case but say that the proof in that case would be sufficiently similar

Here it’s more so that the thing we want to prove only applies to non-zero vectors.

Yes, you only proved that the dot product for perpendicular vectors is 0

Now you need to prove that if the dot product is 0 the vectors must be perpendicular

I don’t see how you’d go about the other direction without the cos definition, but I’m sure someone more crafty than me with proofs can figure something else out too 😄

Yes, just to be consistent in notation I usually keep the vec arrow inside the norms

[lone heart]

@thick lynx Has your question been resolved?

[half epoch]

Don’t bother expanding the norms. Notice our added assumption makes sure both |v| and |l| > 0. So our dot product is 0 iff cos(\theta) = 0

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[languid bolt]

do you know the equation for a line

[grave island]

do we know what it means for a line to be parallel

[languid bolt]

why face palm...

[grave island]

idk

yes

if a line is parallel to another

[languid bolt]

yes

[grave island]

it means it has an eqaul gradient

do we know what the gradient of that line is

oki

im not bothered to explain

yes we do

[languid bolt]

it's already in y = mx + b form (nvm ima let mothy explain gbye)

[grave island]

we use point gradient form

point-slope form

i mean

[grave island]

lmao

y − y1 = m(x − x1)

that is point gradient form

use this

so gradient look online

how to use it

​y-intercept 3 -> means it is (0,3)

ur x value is 0 ur y value is 3

so

y-3=m(x)

ehh

not rlly

where m is ur gradient

and

know that parallel lines have equal gradients so

yea different terminology prolly learnt it different words

y = 5x + 6 is ur original line

so the parallel line would have a slope of 5 as well

[grave island]

then we sub into here

y-3=5(x)

y=5x-3

yea

point slope form equation

https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut28_parpen.htm#:~:text=In other words%2C the slopes,they have different y-intercepts.&text=In other words%2C perpendicular slopes are negative reciprocals of each other.

u do not gain x and y values

just sub in any value for x

and u get a y value

say u sub in 1 for x

y=(5)1+6

so ur point would be (1, 11)

there is an infinite amount of values for x and y in a line

y = 5x + 6​ is ur equation right?

so what did u do here

that is not how u do it.

yes it is.

no

like u can say y2-y1

but usually its y-y1=m(x-x1)

where y is an unknown value

and x is also the unkown

so sub ur thingies into there

y-3=5(x-0)

understand?

thats ur equation of the line

notice if u sub in value of (0, 3)

it works.

idk what u dont understand

but im going to bed.

form satisfying the given conditions.
Parallel to y = 5x + 6​; ​y-intercept 3

u dont need to find if its parallel

it says in the question it is parallel

do we not understand how we got our slope?

so i dont understand why u hvae that question?

using the y intercept of 0,3 the equation is that

it the slops are the same

if m1 = m2

where m is the slope of the lines

so lets say u have y=4x+3 and y=4x+71

theses lines are parallel bc the slopes are equal

it has a slope of 4 for both

understand?

any other questions

or are we all g

oki gn lol

[lone heart]

@potent rain Has your question been resolved?

[lone heart]

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[remote agate]

So if i have -x^2 and i know that x=5/2 Does that mean that -x^2= -(5/2)^2 = -25/4 OR -x^2 = (-5/2) ^2 = + 25/4 ?

[ornate condor]

the first

[remote agate]

Thank you!

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[ornate condor]

np

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[abstract badge]

is the following statement always true ? $<Ax,y>=<A^Tx,y>$
where A is a matrix and x, y are vectors

[ocean seal]

dark

[abstract badge]

if yes how do i prove it

[keen plinth]

what do you think

does it seem plausible

[ornate condor]

if no uve to find a counter example

dark knight~

[keen plinth]

you should probably also specify the vector space you are working in

[abstract badge]

hmm

weird thing is it came in a question and we were asked to prove it

then use it to solve something

the way the question was set up makes it look like it should always be true

[abstract badge]

supposedly for all n in R^n

wait i think i moved incorrectly to begin with

imma try something

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[chrome plank]

$\int\frac{1-x^2}{1+x^2}dx\
=\int\frac{1 + x^2 - 2x^2}{1+x^2}dx\
=\int\left(\frac{1+x^2}{1+x^2}+\frac{-2x^2}{1+x^2}\right)dx\
=\int dx-2\int\frac {x^2}{1+x^2}dx\\
=x + C_1-2\arctan(x)+C_2\\
=x-2\arctan(x)+C$

[ocean seal]

Nonna

[chrome plank]

I can't spot my mistake, the answer should be -x + 2arctan(x) + C

On the second line i did +x² - x² in the numerator so it simplifies when the fraction is split

[keen pasture]

Doesn't the sign flip after 3rd last step?

[true pulsar]

This isn't the derivative of arctan

[keen pasture]

Because you do +1-1 in numerator

[chrome plank]

Oh you are right

[chrome plank]

What do you mean?

[keen pasture]

He's right

To fix it you do +1-1 in numerator and split fractions

[chrome plank]

Yeah ahah, that's nice

[keen pasture]

That changes the sign

[chrome plank]

Yeah that's why I put that -2 before that second integral

Is that wrong?

I'll think about it

Thanks a lot to everyone!

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[ornate condor]

whats ur answer

also isnt this physics

[alpine sable]

isn't 37 and how you did calculate 13.37 and i don't think this is related with mathematics. it's just measuring.

[hardy badger]

its 37, how did you get 13.27

[frosty sinew]

yeah, it is TT

[frosty sinew]

its 37

[frosty sinew]

why its not 37? 😦 I just followed the red line

[alpine sable]

you typed 13.37

13 not 37

[frosty sinew]

uh wait I'm confuse

[hardy badger]

the answer is 37

you typed 13.37

[frosty sinew]

ohhh okayokay thank u

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[worn hinge]

How do I simplify 2 fractions?
$$y = \frac{(x+5)}{(x-6)}$$

1. $$y + ∆y = \frac{(x + ∆x +5)}{(x + ∆x -6)}$$
2. $$∆y = \frac{(x + ∆x +5)}{(x + ∆x -6)} - \frac{(x+5)}{(x-6)}$$

[ocean seal]

LrnzDc3696

[worn hinge]

I should be able to remove x, +5 right? Because when you evaluate they are all the same?

$$y = \frac{x+5}{x-6}$$

1. $$y + ∆y = \frac{x + ∆x +5}{x + ∆x -6}$$
2. $$∆y = \frac{x + ∆x +5}{x + ∆x -6} - \frac{x+5}{x-6}$$
3. $$∆y = \frac{∆x}{∆x} = 1$$
4. $$ \frac{∆y}{∆x} = \frac{1}{∆x}$$
5. $$lim (∆x -> x) = \frac{1}{∆x}$$

[ocean seal]

LrnzDc3696

[worn hinge]

Here

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@worn hinge Has your question been resolved?

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[royal cloak]

why -√3 rejected?

[vale wigeon]

this is something real raised to the 2nd power

how do you think it could possibly be negative?

[royal cloak]

hm

but (-√3)² is +ve

[gray isle]

did you mean why -sqrt(3) was rejected for the stuff within the parentheses?

[royal cloak]

hmm yes

[gray isle]

your question should've been more clear

as it was misinterpreted as why -sqrt(3) wasn't the answer to the actual question

[royal cloak]

mm

[gray isle]

anyway

note that sqrt(3)-sqrt(2) is positive

hence the denominator will also positive and the whole inside will be positive

[royal cloak]

hmm

uh why not -√3 then

[gray isle]

wdym

hence the denominator will also positive and the whole inside will be positive

-sqrt(3) is not positive

the whole inside can't be that

[royal cloak]

mmm oho

yeah

always +ve

ok

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.

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[rain sundial]

Hey, why is time squared in acceleration?

[rain sundial]

is it like a certain speed add every second?

like 5 meters per second added every second (or 5 m/s /s)

[vocal hawk]

exactly

it's 5 meter per second, added every second

[rain sundial]

okay thanks

simpler than i thought it would be

[vocal hawk]

other units won't be this simple tho xD

[rain sundial]

thats scary to know

but i was actually also wondering

how does it get to time squared

right now i have m/s/s

thats the opposite of squared

[vocal hawk]

(a/b)/c = a/(bc)

[rain sundial]

so double division somehow becomes multiplication?

[vocal hawk]

a justification for this is when considering the multiplicative inverse

a / b = a * 1/b

that's one of the defining properties of division(sometimes what literally defines division)

so (a/b)/c = (a/b) * 1/c = (a/bc)

[rain sundial]

i think i understand it

cause when you divide 2 division (a/b) / (c/1) you flip the second one and multiply

(a x 1) / (b x c)

[vocal hawk]

and that's because dividing by something is equivalent to multiplying by its inverse

[rain sundial]

okay thanks

.end

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[tacit arch]

s is a scalar, b1 and b2 are the components of the vector b

The diagram label is misleading because it shows a dot product between s and b

Solve for s

Yes that's a number too

Do you know what the hat means on b?

Look at it in the picture

Right

Can you write out what the components of one simple b perp is?

You just need to know b perp dotted with b equals 0

Yea super confusing diagram

[wheat roost]

yea the addition formula (for subtraction)

[tacit arch]

Poorly drawn diagram or learn trig identities, you can't avoid the pain

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[alpine sable]

how did this happen?

[tacit arch]

Where exactly

[alpine sable]

what is D arrow

and what is delta v

[tacit arch]

Change in v

[alpine sable]

so Darrow v

[tacit arch]

What arrow exactly

[alpine sable]

is this a vector?

[tacit arch]

Wait I'm not sure

Find it in your book/notes

It is a vector though

[alpine sable]

never encountered arrows in this course

except limit

so confusing

maybe D-> = delta

[tacit arch]

Email your teacher.

[alpine sable]

dont have a teacher

[tacit arch]

You have a course without a teacher?

[alpine sable]

its open course ware

sheesh ima skip this for now ig

if anyone knows ping me

[limpid spade]

i guess some taylor approximation last step is hell

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@alpine sable Has your question been resolved?

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[supple laurel]

Hey

[supple laurel]

How can I get a and b by knowing all values underlines in yellow

[remote heron]

trig ratios, probably

[supple laurel]

hmm

lemme search that

[remote heron]

i mean

[supple laurel]

wth

[remote heron]

lol excessive toby

but cool

law of sines 😌

[ocean seal]

Toby

[supple laurel]

wait

that reminds me from math from this year

a/A + b/B = c/C

is smt like this?

[supple laurel]

oki ty

but wait

I don't know a and b

[lilac nest]

You know C and c though

[supple laurel]

oh

now I get it

lemme try by myself

[lone heart]

@supple laurel Has your question been resolved?

[supple laurel]

I am still thinking

[wintry sage]

Any of those three equations equate to any other

So you can just drop off one

So that you only have one unknown you are solving for

[supple laurel]

i am trying to think how to implement it to my game

ik I need that but im thinking how to

I have a question

If A is always 90 and B is for example 30

I can establish the side c as any number I want?

[wintry sage]

No because C must be 60 to make 180 degrees

[supple laurel]

So I im working with coordinates and I want to increment x and y by n values

so I want side c to be 5

[wintry sage]

Well yes as long as you solve for the other sides

[supple laurel]

how can I get a and b only by side c

and knowing A is 90

is it possible?

[wintry sage]

Is B still 30?

[supple laurel]

B is a random number from 0 to 90

[wintry sage]

Well for the law of sines you need to know at least one corresponding side length and angle

[supple laurel]

well

no

let me explain my goal

[analog bane]

can somebody help me

[supple laurel]

so c = 5 A = any number from 0 to 90 and C = 90

[wintry sage]

Do you know length a

[supple laurel]

I know everything except a and b

[wintry sage]

Then yes you just plug it into the sine rule

[supple laurel]

ok let me try

[supple laurel]

claim an unclaimed help channel

[wintry sage]

This channel is occupied

[analog bane]

how do i

claim

a channel

[supple laurel]

#help-9

you send any message

[analog bane]

it wont let me

[supple laurel]

sines rule is sin(A)/a = sin(C)/c right?

or sin(b)/b...

[wintry sage]

You have to find b first if you don’t know the angle A

[supple laurel]

Yeah I know

if C is 90 and I A is any number from 0 to 90 (I know this number I just generate it randomly so rn I don't know it)

[wintry sage]

Ohhhh I get what you mean

Yeah then you have enough to solve for it you just need to plug it all in

Then solve for a

Then you equate sin(B)/b = sin(C)/c and do the same thing

[supple laurel]

is the same if I do

sin(B)/b = sin(C)/c b/sin(B) = c/sin(C)

<@&286206848099549185>

[wintry sage]

Is that the product of both?

That’s not the same

[supple laurel]

why tho?

wait

WAIT

IT WORKS

!close

.close

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[warm ibex]

Hello I’m having some troubles with a early level Calc problem

[warm ibex]

The key seems to think that the limit on the bottom left results in 0, with the final answer being 0*1=0

But wouldn’t that limit be 0/0 cause 1-1/0 is undefined? I feel like I need to mess w the problem more but idk where to go from here

Haven’t learned differentials or any other fancy stuff yet just in 2nd week of calc 1 at the moment

[serene junco]

That limit on the bottom left is indeed equal to 0, you can prove it with the squeeze theorem

Have you talked about that in your class?

[warm ibex]

Oh wait is that one of the special limits?

[serene junco]

Well, yeah generally that limit is one you'd probably prove in class and then just memorize

[warm ibex]

I didn’t think to check that, thanks for the help

[serene junco]

Also

Your result in the last line is okay

but your work before is not right

You seem to be treating "cos" and "x" as two separate variables

cos(x) is a function of x, you can't separate the x out of it

[warm ibex]

yeah I’ll fix that I just didn’t think the notation mattered thanks

[serene junco]

so for example x*cos(x) is not cos(x^2)

[warm ibex]

.close

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[abstract yarrow]

Question: For example we have function F with inputs a,b and it should be easy to calculate F(a,b) = x, but if you know x and b for example (and the function it self of course) you should not be able to easily calculate a. How would the function F look like?

[tacit arch]

there's a cs server in #old-network

[abstract yarrow]

I don’t need help with the programming just the maths

[alpine sable]

let F be a (normal) one way function and a,b inputs. Then F(F(a)F(b)) works, right?

[abstract yarrow]

I'm not the one to judge idk

I came across this: https://mathworld.wolfram.com/One-WayFunction.html and the definition is exactly what i need but the functions provided as an example don’t exactly fit the requirements

[tacit arch]

what's your full math question then

"exposed" is not a math term

[alpine sable]

kinda is a crypto term

[abstract yarrow]

Neither programming term by that i meant that for example we have function F with inputs a,b and it should be easy to calculate F(a,b) = x but if you know x and b for example (and the function it self of course) you should not be able to easily calculate a

[mortal trellis]

well which requirements do you have

if those don'tfit

[abstract yarrow]

No those are the requirements the functions in the link don't fit it

[alpine sable]

what exactly are you planning to do with this function?

[analog bane]

Can somebody help

Me

With

[abstract yarrow]

wrong channel bro

please

[analog bane]

Oh

[abstract yarrow]

claim your own channel

[analog bane]

sorry

[abstract yarrow]

np

[analog bane]

i didnt know

[mortal trellis]

well I mean it depends on what you set as a and b

if you set b=e in the case RSA or something then still a=(p, q) is hard to compute

[abstract yarrow]

Yeh i forgot to say that i don't really want to use RSA that's why i came for help

[alpine sable]

^

[abstract yarrow]

Implement it in my app that in process takes two strings one constant (for each user constant) and second each thats each time different and generates one unique string from those two so that the constant string cannot be calculated. We can think of strings as numbers (I can do some conversion stuff)

[tacit arch]

type your requirements exactly with examples

[abstract yarrow]

ok gimme minute

[alpine sable]

(crypto stackexchange might be better suited to help)

[abstract yarrow]

So im writing a function generate() (programming) which takes two strings as input str1 and str2 and it generates a third string uniqueString from them. Therefore uniqueString = generate(str1, str2). Requirements for this function is that you should not be able to easily calculate str1 if you know the function and str2. Inside this function generate() i convert the strings to numbers doesn't matter how and then put those numbers to math function that generates a third number that gets converted to string and is uniqueString now. So in conclusion:
Requirements:

1. Given function has to be one-way (If you know the function and one of two input values (can be a specific one) you should not be able to calculate the other one)
2. Given function should accept all numbers or just primes
3. Given function should not be RSA

[alpine sable]

whats wrong with rsa lol

[abstract yarrow]

It is used a lot

[alpine sable]

uwu https://en.wikipedia.org/wiki/Elliptic-curve_cryptography

[mortal trellis]

just use some hash function? hash(str1+str2) with like xor of the strings or whatever?

[abstract yarrow]

I feel stupid for not thinking about that

[mortal trellis]

happens

[lone heart]

@abstract yarrow Has your question been resolved?

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[lone heart]

Please don't occupy multiple help channels.

[native silo]

i need to find the perimeter of this wacky object

[proud pike]

you need to find the length of each individual segment of the perimeter and add them up

where are you having trouble?

[native silo]

very

i never was taught how to find a quarter circles perimeter and then they throw this at me

[proud pike]

do you know how to find the perimeter of a whole circle?

[native silo]

i assume it would just be 0.25 x pie x r

hence "quarter"

[proud pike]

what's the perimeter of a whole circle?

[native silo]

2 x pie x r

[proud pike]

ok, so what's a quarter of that?

[native silo]

0.25

[proud pike]

no, that's a quarter of 1

[round sable]

It's better to use * for multiplication, because x is often a variable

[native silo]

okp

[proud pike]

what's the quarter of 2πr?

2πr/4 = ?

[native silo]

ok so to find a normal semicircle the formual i ahve been using is 0.5pier

0.5 x pie x r

this has worked

so i assume that it this quarter circle qwwould be half of the this formula

[proud pike]

are you getting mixed up between the radius and diameter?

because 0.5 π d is correct

[native silo]

yes

[proud pike]

the diameter is just twice the radius, so 0.5 π d = π r

[native silo]

so the radius of the large quarte circle id it contiuned to a semicircle is 628.32

so i assume i would divide that by 2 and that is perimeter of the quarter circle

which is 314.16

[proud pike]

it's the length of the outer bit, yeah (the entire perimeter would also include the straight line parts)

[native silo]

ok so would i add the 200 as well as the 100

ahhhhhhh

what i got was incorrect

I GOT IT

LETSSS GOOO

all i had to do was add the outer perimeters and the straight lines

thank you for clarifying

[lone heart]

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[quaint karma]

help

[quaint karma]

so far i really have nothing, and i don't even know where to start

i tried putting z into rectangular form and going from that

but now im stuck

nvm

found answer

.close

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[dawn maple]

how to solve 3/2 = 9/10k

[dawn maple]

what is k

<@&286206848099549185>

[fathom kite]

divide both sides by 9/10

[dawn maple]

Oh lets see

i got 4/26 for 3/2

and

k

so would that be

4/26k

?

and simplified

2/13

[fathom kite]

3/2 divided by 9/10

is 3/2 * 10/9

[dawn maple]

Oh

My bad

30/18

=

[fathom kite]

yeah simplify that

[dawn maple]

which is 1 1/3

i mean 1 1/3

i mean

1 2/3

ok

[fathom kite]

yeah

[dawn maple]

and which would be 4

[fathom kite]

what

its 1 + 2/3

[ancient harbor]

wait what is the question?

[dawn maple]

oh yeah

so 1 2/3

Got it

Thanks!

[ancient harbor]

Oh it's 9/10k not 9
----
10k

[dawn maple]

k is variable

You can make this avalible now

im donew\

[fathom kite]

u have to close it

with .clseo

.close

[dawn maple]

ok

and thanks btw

[fathom kite]

np

[dawn maple]

.close

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[dawn maple]

byye

[lone heart]

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[rugged lintel]

17 pls

[wary stream]

,rotate

[rugged lintel]

?

<@&286206848099549185>

[lone heart]

@rugged lintel Has your question been resolved?

[rugged lintel]

nope

.close

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[opal wyvern]

I am working on this problem and I am a bit confused by other people's responses to the first question. I figure A and B are units of mass, and t = time, however people are saying that A has units mass/time^2, and B has units mass/time^3.

Where is the division come from? Like I understand saying A has units mass * time^2, but mass divided by time^2? I'm lost

Maybe if I try to isolate A and B I might understand further

That works if I isolate A & B in the equations m = At^2 and m = Bt^3, but I still don't understand, because that is not the equation given

Also, people are saying t has units A/B. But when I isolate t from the original equation I get t is equal to the cube root of (At^2-m)/B

Or t is equal to the square root of m/(A-Bt). But that doesn't help

[urban tree]

This response is correct

[urban tree]

Normally, for constants, trying to understand the units is quite meaningless, it's not like speed, where m/s makes intuitive sense. The dimensions must be equal so that the left hand side and right hand side have the same dimensions.

the left hand side has units in kg

so the rhs must also have units kg

this means at^2 and bt^3 individually have units kg, the subtraction doesn't affect it

we know the dimensions of t^2, it's second^2

we don't know the dimensions of A, but we know that whatever it is, let's say k, ks^2 = kg, so k = kg/(s^2)

[urban tree]

The expression involving dimensions, and the actual equation isn't the same

[opal wyvern]

@urban tree What do you mean by dimensions? And is it fine to say m = At^2 and m = Bt^3 when trying to isolate A and B?

[urban tree]

this will preserve the dimensions of the quantity, so for a, thats fine

[robust geode]

Be careful to equate either dimensions (meters, grams,...) with each other, or quantities with each other, or the writing becomes horrible 👀

So t is time

So in an equation of dimension, you would replace it with what dimension?

What unit?

[opal wyvern]

Seconds

[robust geode]

Yup

[opal wyvern]

I haven't taken a math class in two years, its very hard getting back into, I understand why people in high school and middle school always claimed to hate math now

Its fun when you can speed through it. Its hard when you get stuck on every step of the way

[robust geode]

So t^2 is in what unit?

[opal wyvern]

Seconds

Well

Seconds squared

[robust geode]

You replace t with seconds, so yeah you get s²

Same thing goes with the t³

And remember, you can't add tomatoes with potatoes

So As² and Bs³ has to have the same unit, which iiiis?

[opal wyvern]

I assume mass but I don't know why its not mass * seconds to a power

[robust geode]

It's a mass that you get by adding them together, because your function m is a function that measures what?

[opal wyvern]

I just don't understand how a unit of mass can include time, a completely unrelated measurement

Its a function that measures mass over time

You input a point in time into it and recieve mass at that point in time

[robust geode]

You are measuring a mass of sand

Yes it's through time

For sure

But it's not like the amount of mass per second, or the amount of mass added every second

[opal wyvern]

Are you saying A & B are units of mass because t is essentially just a multiplier of whatever mass it is?

[robust geode]

It's a punctual mass

Masses are in gram

As simple as that

[robust geode]

No

I'm just saying what our function is supposed to measure is a mass

The mass of the sand

[robust geode]

Btw I'll help you on that a bit

What's velocity?

[opal wyvern]

Speed and direction

[robust geode]

Hmmm

[opal wyvern]

A vector

[robust geode]

It can be yeah

Doesn't have to though

[opal wyvern]

Velocity is like "7 meters per second West"

I think

[robust geode]

Yup

The west is here to bring a direction so let's not matter about that

Because it's not really what I'm interested to show you here

Okay

You said it

"meters per second"

"m/s"

That's speed

Speed is a well known unit measuring something quite usual

[opal wyvern]

Correct, but I feel like Speed is a measurement of how two units correlate in a way

I'm bad at explaining myself

[robust geode]

No worries, if you want we can vc lmao

[opal wyvern]

I'm down for that

[lone heart]

@opal wyvern Has your question been resolved?

[lone heart]

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[obsidian plover]

how do i solve this

[alpine sable]

Take xsquare common out from the denominatore

[obsidian plover]

what do you mean? i dont understand

[alpine sable]

Take xsquare common from denominator it would like this x(1+4/x^2)^1/2

Then xsquare approaches inf so 4/xsquare approaches 0

[obsidian plover]

how can u pull it out if its inside a root tho

[alpine sable]

Like this( xsquare(1+4/xsquare))^(1/2)

[abstract fractal]

They're saying to factor out an x² from x² + 4

[obsidian plover]

alr

now i have this

then thats infinity times 1+ 0

hmm

[abstract fractal]

Remember (ab)^c = a^c * b^c

[obsidian plover]

i got 3 but the key says -3

[abstract fractal]

Technically (x²)^(1/2) = |x|

That's important

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@obsidian plover Has your question been resolved?

[alpine sable]

Oh I see

[lone heart]

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[alpine sable]

how come

{a} is not a subset of {{a}}

cuz wouldnt the powerset of that be

{0, {a}}

same thing here

why is 1 not a subset of 2^s?

[rose sigil]

because a is not an element of {{a}}

[alpine sable]

so how come in my second screenshot

isnt {1} an element of {1, 2, 3}

[rose sigil]

because 1 is not an element of 2^S lol

the elements are {}, {1}, {2}, etc.

[alpine sable]

so when it says like {1} is a subset of 2^S

its saying that 1 is just an element

or does it mean that {1} is a set containing the element 1

[rose sigil]

{1} is a set containing the element 1 but that's not what the thing you said before that means

{1} is a subset of 2^S means: every element of {1} is also in 2^S

but that's not the case

[alpine sable]

but {1} is in 2^S no?

the thing i circled

oh wait

no ur right

[rose sigil]

{1} is in 2^S but that doesn't say the same thing as {1} is a subset of 2^S

[alpine sable]

i forgot there's outer brackets

so why does {{1}} work and not {1}

sorry for the noob questions bro

[rose sigil]

can you ask that more precisely?

[alpine sable]

so why is {{1}} a subset of 2^S but {1} isnt

Because {{1}} is just a set containing a set that contains the element 1

erm i guess

so like here for example

why is {a} a subset of 2^S

shouldnt it be {{a}}?

[rose sigil]

I don't see it written that {a} is a subset of 2^S

[alpine sable]

oh that means its an element of

right

[rose sigil]

I see {a} is an element of 2^S, which is right

[alpine sable]

and thats simply because it appears in the power set

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[hardy skiff]

can someone show me the derivation of this?

[hardy skiff]

Let y = cos^-1 (z)
cos y = z

e^yi + e^-yi/2 = z

e^yi + e^-yi = 2z

e^yi + e^-yi -2z = 0

then multiplying both sides by e^yi

I'd get (e^yi)^2 - 2z(e^yi) + 1 = 0

then I don't know what to do anymore

[worn fox]

Looks like a quadratic equation to me...

[hardy skiff]

yes but whenever I perform the quadratic equation it doesn't look like similar

[worn fox]

Show ya work

[alpine sable]

Is this correct

Wiat

[worn fox]

Delete

Dont post solutions

[alpine sable]

Ok

I found my mistake nvm

[worn fox]

Ty

[hardy skiff]

so using quadratic equation

I'd get

[alpine sable]

I don't understand why people here just don't post solution directly

Lol

[hardy skiff]

-b +- sqrt of (b^2 - 4ac)/2a

Let a = 1
b = - 2z
c = 1

[worn fox]

Because if you had bothered to read the guidelines of the server that's not the point of the server

[alpine sable]

Ohh lol

[hardy skiff]

-(-2z) +- sqrt of ((-2z)^2 - 4(1)(1))/2(1)

2z +- sqrt of (4z^2 - 4)/2

[ocean seal]

ΣAC

[hardy skiff]

yep

we could make it as z +- sqrt of (z^2 -1)

[worn fox]

Good

[hardy skiff]

can you texit that

[worn fox]

$z\pm\sqrt{z^2-1}$

[ocean seal]

ΣAC

[hardy skiff]

thanks

and that is equal to e^yi

right?

[worn fox]

It is

We have z²-1 in the sqrt and we want 1-z²

Any ideas

[hardy skiff]

that's what I am wondering hahaha

multiply by -1? hahaha

[worn fox]

Nearly

[hardy skiff]

hmm

[worn fox]

factor out a -1

[hardy skiff]

oops

what would it look like?

[worn fox]

$\sqrt{-(1-z^2)}$

[ocean seal]

ΣAC

[hardy skiff]

but

how can we remove the -1 inside the sqrt?

[worn fox]

$\sqrt{ab}=\sqrt{a}\sqrt{b}$

[ocean seal]

ΣAC

[worn fox]

(As long as a and b aren't both negative)

[hardy skiff]

so you mean we should leave it there?

[worn fox]

Hm?

Split the sqrt into two

[hardy skiff]

ohh got it

sqrt of (-1) then sqt of (1-z^2)
the sqrt of (-1) is equal to i

[alpine sable]

Yep

[worn fox]

Yes thanks for repeating me

[hardy skiff]

so we will have

z+- i sqrt of (1-z^2)

[worn fox]

Looking good

[hardy skiff]

can you texit this

e^yi = z+- i sqrt of (1-z^2)

[ocean seal]

ΣAC

[hardy skiff]

then natural log of both sides?

[worn fox]

Yep!

[hardy skiff]

ln e^yi = ln z+- i sqrt of (1-z^2)

is that right?

[hardy skiff]

we don't have -i beside the ln

[worn fox]

Keep going

[hardy skiff]

I don't know the next step hahaha

i have no idea

[alpine sable]

1/iota = what

[worn fox]

$yi = \ln (z\pm i \sqrt{1-z^2})$

[ocean seal]

ΣAC

[hardy skiff]

what's the next step?

[alpine sable]

Dividing by iota

[hardy skiff]

iota is the i?

[alpine sable]

Yes

[hardy skiff]

so ill get y = ln (z+-isqrt of (1-z^2)/i

[alpine sable]

Yes

And 1/ iota = WHat

[hardy skiff]

-i

bingo

[alpine sable]

Yep cool

[hardy skiff]

thank you guys

[worn fox]

.close

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[alpine sable]

I dont understand what im doing wrong.

[alpine sable]

AB is equal to B-A which is (-5, -1, 17/4)

And the dot product of AB and n equals to -25 -9 +17 = -17

[rigid smelt]

d asks for the absolute value of the dot product

so | | does not technically imply the module of a vector

even tho yes, you can still understand it as a module of a number

and therefore, -17 definitely is not the answer to d

[alpine sable]

Ah im stupid lol thank youuu

.close

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[next thorn]

I am trying to solve this number theory puzzle but I feel completely clueless...could anybody help to offer some hints? I think I need to try to factor the number into primes but the problem is I cannot tell who got the sum and who got the product...here is the problem:

Given are two integers both greater than one, for example, 2 and 3. There are two people, A, and B. One of them knows the sum of the integers (e.g., 5), and the other knows the product (e.g., 6). Both A and B` know whether they know the sum or the product, and the rules of the game. A says to B: You cannot infer the value of the two numbers B to A: But, now I know the two numbers. A to B: But, now I also know the numbers!!!<@&286206848099549185>

[lone heart]

@next thorn Has your question been resolved?

[gray isle]

looks like golbachs conjecture

[next thorn]

Jesus...sounds even more congusing

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@next thorn Has your question been resolved?

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[manic harness]

Two bodies of masses 4kg and 6kg are moving in the same direction with speed 10m/s and 5m/s respectively. After collision they stick together and move as a single body, what is its final velocity?

[manic harness]

I’m having difficulties imagining this

If they move in the same direction, how do they even collide?

[coral moss]

One is going faster than the other

The 4kg mass will eventually catch up to the 6kg mass

[manic harness]

Oh, ya makes sense.

[manic harness]

Wouldn’t the 4kg one be ahead?

[coral moss]

Well it's an assumption that the 4kg mass starts behind the 6kg mass and they both move in the exact same direction otherwise they wouldnt collide and the question tells you they collide

[manic harness]

True

Thx

.close

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[keen pasture]

What have you tried?

[fair ingot]

well I did this

prob wrong

[keen pasture]

,rotate

[keen pasture]

1. What happens if 2y+1 = +6

2. You solved the linear equation incorrectly

[fair ingot]

Oh

nic

thanks

.close

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[olive timber]

hey

ca you guys help me

*can

I've a test right now

which includes mathematics

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[olive timber]

in how many ways can we change the sign * with + or - so that 123456789*10=42

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@olive timber Has your question been resolved?

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[primal parcel]

Hey guys, just going over a test I did recently, but there's one questions I dont completely understand. This is what my teacher gave as the answer, but I dont understand how it is done

[lone heart]

@primal parcel Has your question been resolved?

[languid bolt]

hmm

(hint: do not do this from memory)

and yet the teacher uses a formula that she memorizes smh

ok so

linear rule

first of all, ima just do it your way

ima denote F as x and C as y

the 2 points you have is

(212, 100) and (32, 0)

find the gradient

[primal parcel]

Is it 5/9

That's where they got it from

Oh

[languid bolt]

yes, but show working

[primal parcel]

Okay cool, then just substitution

[languid bolt]

?

show working on how you got 5/9

[primal parcel]

Like when she used the point (32,0)

Rise/run

100/180

[languid bolt]

ok

[primal parcel]

Thanks a lot mate

[hollow egret]

ok hi

[languid bolt]

wait..

i did something wrong i think nvm

[hollow egret]

oh sorry

[languid bolt]

oh no

nvm

go on

[hollow egret]

may I help too?

[languid bolt]

yeah

[primal parcel]

Ye

[hollow egret]

ok u go on

I didn't do this question

I will try it for sure

[languid bolt]

have you found the y intercept? @primal parcel

cuz the form for a linear is y = mx + c

where c is the y intercept

or y = mx + b

same thing tho

[primal parcel]

Well so far I have y=5/9x+c

[languid bolt]

yes

[primal parcel]

But I understand what she did to get c

I think

[languid bolt]

ok

[primal parcel]

Is that all there is to it?

[languid bolt]

yes

[primal parcel]

Awesome thanks for your help

[languid bolt]

but does using equations count as doing from memory tho

cuz

(hint: do not do this from memory)

[primal parcel]

I think what it means is that don't use the formula for Celsius to Fahrenheit you remember

[languid bolt]

oh ok

[hollow egret]

what I have learned is c/5 = F-32/9

but we cant use memory here

[languid bolt]

multiply both sides by 5 and you have the equation

c = 5(F-32)/9

[primal parcel]

#help-0

Done

.close

[lone heart]

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[knotty mountain]

Hey, I'm not sure what to do in this circle theorem question to find k. So far I've found that angle DCA is 44, but i have no idea what to do then.

[hollow egret]

hi sir

is there any congruent triangle rule in circle theorem?

@knotty mountain

[knotty mountain]

I'm not sure exactly

[bitter torrent]

i need help with some math question but ill quickly do urs

[hollow egret]

Sir

lets just say

[bitter torrent]

the point where CA and DB intersects we will call it X

[hollow egret]

The intersection of line Db and Ac is O

[hollow egret]

Ye that's what I was sayin lol

[bitter torrent]

DXC = 117

[languid bolt]

dont just do their homework btw, try to explain it

or guide

[bitter torrent]

because the total degree in triangle is 180

u do 180 - 19 - 44

[hollow egret]

AOB = 117 degree

tes triangle sum property

[knotty mountain]

yeah

[bitter torrent]

CXB = 180 - 117 cos it is a straight line

[knotty mountain]

oh

[bitter torrent]

angle DBT and angle ACT = 180 - 44 = 136 cos it is also a straight line

[south mantle]

How do u know that DB is a straight line ;-;

[bitter torrent]

then u do the last part

[knotty mountain]

im not sure how i would get k from that though

[hollow egret]

I think

[knotty mountain]

because it isn't a proper triangle

[hollow egret]

triagnle AOB should be congruent to triagnle DOC

ye now I got it

Why they should be congruent

[knotty mountain]

oh wait since it's a quad lateral i would subtract them?

[hollow egret]

because they have a common intersecting point

O

So we can say that angle B = angle C

gotcha

[bitter torrent]

u do 360 - the rest of the angles that are found since the formula (n-2) x 180, n being the number of sides the quadrilateral has

[hollow egret]

and DT is a straight line

just do 180-44

U'll get angle C of triangle ACT

[knotty mountain]

so k would be 25?

[hollow egret]

ye I think so

check answer

[knotty mountain]

ok yeah it is

thanks a lot!

.close

[hollow egret]

my pleasure sir

[lone heart]

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[fleet mist]

Hello!
Let be integers 1 =< p =< n. Using Pascal's triangle formula, calculate the sum

[fleet mist]

https://cdn.discordapp.com/attachments/727221905085694052/1014146491088125972/unknown.png

I have no idea as why Pascal's triangle formula can help me

i know that the first and last term of the sum are equal to 1 but idk about the others

[vale wigeon]

does this say $\sum_{k=p}^n \binom{k}{p}$?

[ocean seal]

Ann

[fleet mist]

yes it does

[vale wigeon]

right

i had trouble reading your handwriting

but ok

[fleet mist]

oh sorry

[vale wigeon]

i suggest drawing out a pascal triangle and circling the terms in the sum where they appar in the triangle

you may notice something

[fleet mist]

idk how this should help me

it is something like (2^n)*p maybe?

p is not a variable

is it 2^(n-p)

<@&286206848099549185>

[lone heart]

@fleet mist Has your question been resolved?

[lone heart]

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

<@&286206848099549185>

[wheat roost]

Don't ping helpers, just ask a question

[robust copper]

ok sry

i am stuck

[wide raven]

just take 4 sinA cosA common and see the magic

[robust copper]

ok

[fast marlin]

hey

could you help here

to

in help 15

[worn fox]

Wait and be patient, don't invade other people's channels

[ornate condor]

invasion ~

[robust copper]

how can i take 4 sin a and cos. a there is sin(2a) and cos(2a). i cant take common.

[wide raven]

wait what?

are you sure?

[robust copper]

its not square

[wide raven]

it looks like square

[robust copper]

its not

[wide raven]

lhs

it is

it should be

not square, cube, but yeah

[fast marlin]

aaudaina hola

[robust copper]

i dont get it

[wide raven]

you should

you can take 4 sin(a) cos(a) common from lhs

try

[robust copper]

ok