help-13

[smoky whale]

Do you mean like this

[viscid mulch]

Yes exactly

Now applying the same logic of earlier

This logic

What do you get?

[smoky whale]

I have (e) and (x) do i multiply them

I mean

Divide

[quaint eagle]

mhm

[smoky whale]

Sorry

[viscid mulch]

Divide yup

You divide them

[smoky whale]

So x/e

[viscid mulch]

Nope

[quaint eagle]

Achilles, I like the way you explain 😛

[viscid mulch]

Close

[smoky whale]

Other way around

[smoky whale]

He is great

[viscid mulch]

Ty 😆🍵

[viscid mulch]

Yess sir

[quaint eagle]

so stallion, want me to send it written on a paper ?

or Mr Achilles made you understand with his perfect skills ? 😛

[viscid mulch]

[smoky whale]

, rotate

[quaint eagle]

HELL YEAH DUDE!

[smoky whale]

Like this?

[viscid mulch]

Yea

Now what's the next step?

[quaint eagle]

Yup, as Achilles said, we can ignore the logs now

[viscid mulch]

Yes

[quaint eagle]

it was a thrill to watch! good luck to you both!

[viscid mulch]

[smoky whale]

Thanks a lot for the support @quaint eagle

Let me finish solving it and I will check with you @viscid mulch

[viscid mulch]

Yes

If there's any problem, I guess you can ping me 🤷

[smoky whale]

Hi again @viscid mulch I tried again but i couldn’t find x

[smoky whale]

Could you please do that yes

I thought I was capable of solving it halfway through

[quaint eagle]

can you send the problem again

oh okay

achilles and stallion, I got you!

[smoky whale]

Yep

You are fire 🔥 @quaint eagle

This is the problem

[smoky whale]

I just did not understand how you got this

Could you please elaborate a little bit

[soft owl]

it's just an approximation of e/7

[smoky whale]

What

I don’t understand

[quaint eagle]

e is euler's number

[smoky whale]

Ohhhhhhh

I got it

2.718 is e

Right right

Thanks so much for your help

@quaint eagle I really appreciate your effort with me

[quaint eagle]

yeah!

[smoky whale]

How should I start this?

[dire geode]

Put everything on the right side into one log

[smoky whale]

I’ll try

Is this what you meant @dire geode

[dire geode]

You removed the log on the right but not the left

[smoky whale]

Is taht better

What is the next step?

Oh that’s it

Just need to remove logs

Thanks

[cedar kiln]

@smoky whale Has your question been resolved?

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[dim hearth]

This is a mathematical explanation from a physics question, specially on magnitude of vectors. How did they get from step 3 to step 4? Seems like they took the square-root of both the squared V_x and squared V_y separately and added them together to make a single V variable. Is this a valid operation?

[dire geode]

What's the definition of V

[dim hearth]

V_new is the magnitude of a vector, whose components are V_x and V_y

[dire geode]

V

Not V_new

[dim hearth]

I'm not entirely sure actually ...

[cosmic steppe]

Looks like V is the magnitude of a vector

[dim hearth]

I'm assuming they combined V_x and V_y into one variable called V?

[cosmic steppe]

Well Vx looks like the x-component of V, Vy is the y-component of V

Such that V: <Vx, Vy>

[dim hearth]

hmmm...

[sacred grail]

sussy equations

[cosmic steppe]

Among us

[dim hearth]

is the issue math related or is it my understanding of the concept of vectors?

[sacred grail]

well the image shows a bunch of symbols which could be interpreted as some equations calculating the length of a vector once it's been doubled

but the notation is unclear

and the variables have not been defined

so we are missing some key pieces of information

[dim hearth]

"If the x- and y-components of a vector are doubled, what happens to the magnitude of the resulting vector?"

this is the question

[long swan]

@dim hearth is this a 2d vector

[dim hearth]

It isn't specified, but the equation they used is for 2d vectors

[long swan]

what's happening here is the definition of the magnitude of a vector

sqrt(Vx^2 +Vy^2) = |V|

[dim hearth]

and by doubling Vx and Vy, you'd get sqrt{4Vx^2+4Vy^2} = V...

[long swan]

nope

= 2V

in general for some positive real scalar c we can say that:
|cV| = c|V|

Consider:

$$V = \langle V_x , V_y \rangle$$
$$||V|| = \sqrt{V_x^2 + V_y^2}$$
then:
$$cV = \langle cV_x, cV_y \rangle$$
$$||cV|| = \sqrt{(cV_x)^2 + (cV_y)^2}$$
$$||cV|| = \sqrt{c^2 (V_x^2+V_y^2)}$$
$$||cV|| = |c| \sqrt{V_x^2+V_y^2}}$$
$$||cV|| = |c| \cdot ||V||$$

[wraith dagger]

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

[dim hearth]

why would you double the resultant vector in the third step?

[long swan]

exponents distribute over multiplication

[cosmic steppe]

2 must come from somehweee

[long swan]

(ab)^1/2 = a^1/2 b^1/2

[dim hearth]

this is the distributive property of exponents?

[long swan]

yeah

[dim hearth]

I guess I am confused about what is the question is asking. Usually when given an algebraic equation, and asked what would happen to side A if side B of the equation is modified in some way, we don't apply the same modification to side A.

[long swan]

just read this

really it just comes down to applying definitions

what is the magnitude of a vector?

[dire geode]

Show the question

[dire geode]

This was rhetorical. V is just the sum of squares of its vector's components by definition

[dim hearth]

So, using the Pythagorean theorem as an example
$$C=\sqrt{A^2+B^2}$$
$$C=\sqrt{2(A^2)+2(B^2)}$$
$$C=\sqrt{4(A^2+B^2)}$$
$$C=2\sqrt{A^2+B^2}$$
since we can't combine the variables A and B, we would have to substitute it with the definition of $$\sqrt{A^2+B^2}$$, which is C; therefore making it ...
$$C=2C$$

[wraith dagger]

Pizza

[dim hearth]

is this a comparable approach?

[honest kayak]

2 needs to be in the squares

[long swan]

but that's not C

C cannot be both sqrt(A^2+B^2) and sqrt(2A^2 + 2B^2)

at the same time

you're contradicting yourself

[dim hearth]

C is not equal to 2C, but if the question is asking what would happen to C if one side of the equation is doubled, C would "become" 2C

[long swan]

nope

that's not what the question is asking

it's asking what happens if you double each component

you've got the right idea but you're notating it wrong

[long swan]

maybe take another look at this

[dire geode]

You still haven't done this

[dim hearth]

#help-13 message

[dim hearth]

so the solution in the screenshot should technically show 2V_new=2V?

since doubling one side of the equation requires doubling the other side

[long swan]

which step does the 'doubling' occur in

[dim hearth]

first step

[long swan]

this is a definition, not an algebraic operation. just define V_new such that V_new = <2Vx, 2Vy>, then its magnitude must be sqrt((2Vx)^2+(2Vy)^2)

[crystal raptor]

I give you a random amount of pennies, you work out you have V currency. I then give you the same amount MORE pennies, the new amount of money you have, V_new, is double the old amount you had, V_new = 2V

thats whats going on here but im doubling your vector instead

[dim hearth]

Ok, I think I am confusing the definition of vector magnitude:
$$V_{resultant}=\sqrt{V_x^2+V_y^2}$$

with the "new" defined equation from the question:
$$V_{new}==\sqrt{(2V_x^2)+(2V_y^2)}$$

[wraith dagger]

Pizza

[crystal raptor]

its the literal same definition, just with different x and y components of the vector

one vector has components <Vx,Vy>, the other has components <2Vx,2Vy>

[long swan]

honestly i don't like this notation because you're missing the magnitude bars. i'd perhaps write:
V = <Vx, Vy>, so |V| = sqrt(Vx^2 + Vy^2)
then:
2V = <2Vx, 2Vy>, so |2V| = sqrt((2Vx)^2 + (2Vy)^2)

[crystal raptor]

Maybe do it with some numbers, let V = <3,4>, find the magnitude of that and compare it with the magnitude of 2V = <6,8>

[dim hearth]

I think it's starting to click now. I think I was confused about V_new. It's just a label. The question is asking what would happen to the resultant vector when we double its components. The resultant vector is 2V and it isn't "equal" to V_new. V_new is just the name associated with 2V.

[crystal raptor]

the question asks what happens to the magnitude of the new vector

you need to differentiate between your notation for a vector, which sensibly should be V and the notation for the magnitude of a vector, which you really should be writing |V|

[dim hearth]

so the answer should technically be $$V_{new}=2|V|$$?

[wraith dagger]

Pizza

[cedar kiln]

@dim hearth Has your question been resolved?

[long swan]

|V_new| = 2|V|

[dim hearth]

Thank you so much for your help, Mr. Gamer.

and thanks to everyone who contributed to my understanding

Have a good night!

.close

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[obtuse moth]

this might not be so closely related to math, but can someone with both english and vietnamese proficiency tell me what the english terminology are for these words:
đường vuông góc
đường xiên
hình chiếu
note that all of these words are to be put in context of a triangle(-ish.)

[obtuse moth]

rough translation of the text:

from point A not on line d, create a line perpendicular to d at H. On line d, take point B not overlapping H. Then:

• segment AH is called đường vuông góc (what is this in english?) point H is called the "foot" (???, literal translation of "chân") of the
  đường vuông góc or the hình chiếu of point A on line d.

• segment AB is called a đường xiên from A to d.

hình chiếu might have something related to projection, but that's all i know.

[muted bear]

Perpendicular

Foot is the correct translation

[dense hornet]

đường vuông góc means perpendicular line, đường xiên just means slant line basically, and hình chiếu would translate directly to projection

here, projection would be in the right context. "H is the projection of A onto d"

[obtuse moth]

ah, i see. thank you so much! @dense hornet @muted bear

(not me unsubtly trying to hunt for vietnamese people on discord)

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[manic acorn]

first part would be 18c2-5c2+1?

and second part 18c3-5c3?

[cedar kiln]

@manic acorn Has your question been resolved?

[silent finch]

This looks right

[manic acorn]

also pretty confused abt this one

[silent finch]

A regular polygon is just a polygon with equal angles and sidelengths

If I pick 2 vertices, I can pick the 2 opposite vertices to create a rectangle

But this is over counting

I can also pick the 2 vertices creating the width of the rectangle so I divide by 2

[manic acorn]

Ive got it

I think i understand

Thanks

.close

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[wary hamlet]

how would u differentiate this

[lusty grotto]

product rule

and chain rule

[tropic oxide]

$A = \frac{1}{2} x \sqrt{25 - 10x}$?

[wraith dagger]

Ann

[lusty grotto]

$\sqrt{25-10x}=\left(25-10x\right)^\frac{1}{2}$

[wraith dagger]

SilverSoldier

[lusty grotto]

if that helps

[wary hamlet]

wait so which do i produt rule first

1/2 and x

or x and sqrt 25-10x

[lusty grotto]

differentiate the first function, then multiply its derivative by the second function

differentiate the second function, then multiply its derivative by the first function

and add them together

thats the product rule

pick any function u like as the first function and the other as the second

[tropic oxide]

1/2 is a constant

if you wish to apply the product rule in the case that one of the factors is a constant then be our guest

[wary hamlet]

so i can just leave the 1/2?

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[manic acorn]

5c4.12c7?

[tropic oxide]

seems correct

4 bowlers and 7 non bowlers

[manic acorn]

Thanks, the solution says 5c1.12c7💀

.close

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[tropic oxide]

5c1 = 5c4

[manic acorn]

I will

End it all here

.

Why would they

[cedar kiln]

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[manic acorn]

no.3 is 50 ?

[tropic oxide]

yes

handwriting though

[manic acorn]

This is not mine, also it looks so much better than mine

[tropic oxide]

the one in the screenshot is still bad

[manic acorn]

🗿

Bueraeucrat

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[normal pawn]

what ?

[cedar kiln]

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[manic acorn]

So x and y have no restrictions and z belongs from 3 to 50

[crimson sedge]

There are loads of solutions to this. Did they maybe mean to ask for the number of integral solutions?

[pine igloo]

these questions, where we gotta find solutions
i hate them

[manic acorn]

So i thought id calculate no. Of solutions such that z max limit is 50 and subtract no. Of solutions such that z min is 2

[manic acorn]

Yea

No. Of solns such that z is min 2 is 50c2

OH

Oh

No

Nothing

[pine igloo]

which chapter is it?

[manic acorn]

Combinations

Hm for z max 50 if i try to calculate

I get weird stuff ill try a diff approach to calc z max 50

[pine igloo]

permutations and combinations?

[manic acorn]

Yea

[normal pawn]

z is constant tho

[manic acorn]

I know

Ill show an example

This is how i usually calculate the maximum restriction

But in this case if i try to do it

This is what i get 🗿

Oh

Im dumb

[manic acorn]

Yea this should be z<=50 but its still

Not working

I think ive got my mistake

[crimson sedge]

If z = 3, we need x + y = 47
So the possible values for (x, y) are (0, 47), (1, 46), ..., (47, 0)
So 48 possible pairs.

If z = 4, we need x + y = 46
So the possible values for (x, y) are (0, 46), (1, 45), ..., (46, 0)
So 47 possible pairs.

At the end, when z = 50, we need x + y = 0
Which has only 1 pair - (0, 0)

So the answer is 48 + 47 + 46 + ... + 0 = 48*49/2 = 1152

[manic acorn]

,w c(49,2)

[wraith dagger]

Results provided by WolframAlpha

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Requested by idontsleepanymore#8460

[manic acorn]

I get this

It is 49c2 💪

I found out z max 2

Then i found out z max 50

Then i did z max 50-z max 2

Gives solutions to z belong to [3,50]

[crimson sedge]

Oh, I had a calculation mistake at the end.
48*49/2 = 1176.

[manic acorn]

Thanks, ill close

.close

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[crimson sedge]

is 1/0 infinity in the extended complex plane?

[cedar kiln]

@crimson sedge Has your question been resolved?

[latent bloom]

https://www.geogebra.org/m/VWSN8Rp7

Here is a graph you can play with

[cedar kiln]

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[feral juniper]

Proof: "Nested Intervalls" Define unduktiv series with these Intervalls

Why is (2) defined like it is

Its the proof that if f(a) < 0 and f(b) > 0 that at f(p) = 0 exists if the function is continues at the intervall a,b

[feral juniper]

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[crimson sedge]

wheres the difference?

[cedar kiln]

@crimson sedge Has your question been resolved?

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[sand meadow]

this is part of solution.. but i dont understand what happened here

[crimson sedge]

since phi is 1 to 1, |Y|=|im(phi)|≤|S|

[sand meadow]

still don't get it

how is phi 1-1

[crimson sedge]

this shows phi is 1-1 using the definition of 1-1

[sand meadow]

o know definition of 1-1...

" Q is dence in R " so f(x)=g(x) for all x=R

is this a result?

[crimson sedge]

yes, its a result from topology

[sand meadow]

can you tell me

what is whole result

[crimson sedge]

if two functions agree on dense subsets of R, then they agree on R

[sand meadow]

umm okie

[crimson sedge]

I think we only need Hausdorff spaces, but im not completely sure

https://en.wikipedia.org/wiki/Hausdorff_space#Examples_of_Hausdorff_and_non-Hausdorff_spaces

[sand meadow]

i am not aware of Hausdorff spaces 🥲

[crimson sedge]

R is Hausdorff, so it works on the reals

[sand meadow]

okie

i think this topic will come in continuity

.close

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[plucky holly]

if there are 8 people, is it 4x3x2x1?

If the first person is called A, he can only shake hands with 7 other people, making 4 connections?

[cold briar]

ok so

Ima try to explain a case and you try to see a pattern

@plucky holly

Ok in the case of 9 people

The first person shakes hand with 8 other people

The second person shakes hand with 7 other people (7, because it excludes himself and the first person)

The third person shakes hand with 6 other people (6, because it excludes himself and the first and second person)

The fourth person shakes hand with 5 other people

So far, there has been 8 + 7 + 6 + 5 handshakes

and it continues

are you here?

[inland ocean]

Just wondering can we solve this by using permutations?

[cold briar]

Not really, I think we just add.

[cold briar]

+permutation will count the order they shake hands too

which isn't asked for in the question

[inland ocean]

I see

[cedar kiln]

@plucky holly Has your question been resolved?

[plucky holly]

so for 8 people, it should be 7+6+5+4+3+2+1

which equals to 28

so that is wrong?

for 12 people, it would be 11+10+9+8+7+6+5+4+3+2+1

= 66

14 people would be 13+12+11+10+9+8+7+6+5+4+3+2+1

= 91

did i do smthing wrong?

<@&286206848099549185>

[cold briar]

wait.. was my logic wrong???

[plucky holly]

when i tried for everyone one of them, it didn't match any of the answers

choices

[cold briar]

ok i think i might be in the wrong here

wait ima recheck

[plucky holly]

if it was 8, lets say there are people A,B,C,D,E,F,G,H

A can shake hands with B,C,D,E,F,G,H

creating 7 handshakes

isn't that like ur logic?

[cold briar]

yes but it's wrong

wait...

[plucky holly]

when it reaches H

there is no more handshakes

[cold briar]

because everyone already shook hands with him

but i think im wrong

[plucky holly]

but it's still 7+6+5+4+3+2+1

but it equals 28

[cold briar]

yeah that should be correct

28 people

[plucky holly]

for 10 people, 9+8+7+6+5+4+3+2+1 = 45

so both of those aren't correct

[cold briar]

ok idk tbh

[plucky holly]

for 14 people, it is 13+12+11+10+9+8+7+6+5+4+3+2+1 = 91

off by 1

12 people is 11+10+9+8+7+6+5+4+3+2+1=66

then none of them are correct

[normal pawn]

[cold briar]

i feel like my logic is correct

but idk

hmm

we could probably use a formula nC2, to pick how many handshakes combinations are there.

,w 10C2

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[cold briar]

wait it's still 45

nvm

am i wrong or is the question wrong...

12C2

,w 12C2

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[cold briar]

,w 14C2

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[cold briar]

,w 8 choose 2

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[cold briar]

huh???

are you sure the question is correct?

[plucky holly]

yea...

that's all the option

[cold briar]

well... i guess just wait for someone else

[cold briar]

i meant the options itself

is it correct...

[plucky holly]

yes

[cold briar]

welp

we might need someone smart here

[plucky holly]

i didn't edit it or anything

[cold briar]

confirm with the professor ig

or wait for someone else

[plucky holly]

k

ty for trying to help me

[crimson sedge]

nC2 is the maximum number of handshakes

its asking whether the given numbers are possible or not

ie is it smaller than nC2

[spice cedar]

normally I would be doing guess and check because Im a stupid lazy ass

guess that aint helpful

[plucky holly]

so how owuld i do this?

[crimson sedge]

is 92 smaller than 14C2?

[cold briar]

no

[crimson sedge]

then rinse and repeat

and only one of them is smaller

?

[cold briar]

wait yes

[wraith dagger]

The following error occured while calculating:
Error: Undefined function b

[cold briar]

,w 12 choose 2

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[plucky holly]

yea

but the option says 12 people and 60 handshakes

[cold briar]

i thought they all had to shake hands...

[crimson sedge]

60≤66

so its possible

[cold briar]

the question simply stated that they can't shake hands more than once, it doesn't say that they "have" to shake hands.

[crimson sedge]

I dont think thats what the question asks

[cold briar]

yeah exactly

[crimson sedge]

yeah

[cold briar]

i misunderstood it

[crimson sedge]

[plucky holly]

ohh, ok, i get it, thank you everyone

.close

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[crimson sedge]

How can i solve this ? It's a geometric series, i have to proof if its convergent or not

have you tried any tests? like ratio test?

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[tulip stream]

prob do limit version of comparasion test by taking bn as (10/9)^n , didnt test yet i have a high feeling it works

[tulip stream]

/abondon

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[tulip stream]

@crimson sedge

[crimson sedge]

I can't use any tests because we didn't learn it yet

even comparison test? like finding a sequence b_n, such that |10^n/...| ≤ b_n?

nop

this one should be pretty intuitively true

[tulip stream]

just realized i typed the test wrong, it was limit version of comparasion test

oh well limit of this is DNE

and since that means limit isnt 0

[crimson sedge]

so you only know about the geometric series yet right?

[crimson sedge]

yes

I could learn tests and solve it but I would be cheating I guess

[tulip stream]

we can say by the nth term test, our series diverges

[crimson sedge]

i think it does not converge

[tulip stream]

your teacher must have showed you nth term test at least

that is taught before teaching you geometric and telescopic

at least thats how my professor did it

[crimson sedge]

can you at least use that if the sequence 10^n/... itself is not a null sequence, it does not converge?

I can use the nth term test

[tulip stream]

yeah

by the nth term test limit DNE

thus limit is NOT equal to 0, hence our series diverges

because limit of (-1)^n

[cedar kiln]

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[wispy spoke]

how do i solve?

[wispy spoke]

i dont understand A and C

[clear berry]

For A, you are given that p = sina

[wispy spoke]

yeah

[clear berry]

Now you want to find out other values only in terms of p ie sina

[cerulean sail]

(hint: recall the picture I think I sent you last time)

[wispy spoke]

solve the first one

pls

[wispy spoke]

which one

[clear berry]

just use the formula for sin(A+B)

[wispy spoke]

whats that'

[wispy spoke]

never heard of thast

[cerulean sail]

OP won't know that yet

[clear berry]

oh, okay

[wispy spoke]

im doing sum else

[clear berry]

This is basically it

[wispy spoke]

whatttt

im lost

[cerulean sail]

Remember what we said about the graph

,w plot y = sin(x)

[wispy spoke]

2p-180?

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[wispy spoke]

Q A)i)

[cerulean sail]

The symmetry of the sin graph

The purple on $x$ axis is $\alpha$, and the blue on $x$-axis is $180 - \alpha$

[wraith dagger]

chartbit

[wispy spoke]

okay i get that

[cerulean sail]

Both will have the same output $p$ (and remember that $\alpha$ was between 0 and 90)

[wraith dagger]

chartbit

[cerulean sail]

Which is effectively why this statement is true

[wispy spoke]

you lost me 5 different times

wait

[cerulean sail]

• you should have this

[wispy spoke]

can you try solving A)i)

and show me your working out

[wispy spoke]

i know this

[cerulean sail]

Yeah, so can you write something about $2\sin(180 - \alpha)$ using that?

[wraith dagger]

chartbit

[wispy spoke]

i dont know

[cerulean sail]

Well, you know that $\sin(180 - \theta) = \sin(\theta)$ for any $\theta$...

[wraith dagger]

chartbit

[wispy spoke]

it does?

so sin180-theta

brooooooo

i hate thissss

[cerulean sail]

Literally, all you gotta do is replace $\theta$ with $\alpha$, and put that in...

[wraith dagger]

chartbit

[cerulean sail]

That's what all the pictures before have stated

[wispy spoke]

im dying😭

[cerulean sail]

So if $\sin(180 - \alpha) = \sin(\alpha)$, then $2 \sin(180 - \alpha)$ is equal to what?

[wraith dagger]

chartbit

[wispy spoke]

so sin(180-a) = p

[wispy spoke]

2p

[cerulean sail]

Yes, that's it! Pretty much that!

[wispy spoke]

okay but look

how does sin(180-a) = p

shouldnt sin(a) = p

what do we do with the 180

[cerulean sail]

$\sin(180 - a) = \sin(a)$, and $\sin(a) = p$, therefore you get that!

[wraith dagger]

chartbit

[wispy spoke]

how does sin(180-a) = sin(a)

WHERE DOES THE 180 GOOOOOO

[cerulean sail]

See that last video about ambiguous angles - they both give you the same output

[cerulean sail]

Or this diagram

[wispy spoke]

oh wait

so sin(180-a) is the missing angle?

thats why theyre the same

right?

[cerulean sail]

Hmm, you're familiar with solving trig equations in a given interval, right?

[wispy spoke]

show me a question

idk by names

[cerulean sail]

Something like this

Or more simpler, this

[wispy spoke]

like a

sin-1= 5/7

right?

[cerulean sail]

That's one solution, but there's another one

[wispy spoke]

oh yeah

cast

one in t and one in a

right?

[cerulean sail]

One in s and one in a

[wispy spoke]

180 + sin-1= 5/7

[wispy spoke]

but its tan?

[cerulean sail]

Where [s]in is positive and [a]ll are positive

[wispy spoke]

why s

[cerulean sail]

(a) or (e)?

(a) is $7\sin(\theta) = 5$

[wraith dagger]

chartbit

[wispy spoke]

yes that

no whta

its tan

oh shit wtf

bd

i read wrong

[wispy spoke]

yes tah t

[cerulean sail]

Yep that's it!

[wispy spoke]

ik how to do that

[cerulean sail]

Basically it's kind of the same idea here

You know the angle $\alpha$ in in an area where $\sin$ is positive

[wraith dagger]

chartbit

[wispy spoke]

yesss

[cerulean sail]

And as it's between 0 and 90, you know $180 - \alpha$ is in between 90 and 180

[wraith dagger]

chartbit

[cerulean sail]

Which is where sin is still positive

[wispy spoke]

wait

do you wanna go on call😭

[cerulean sail]

Sorry, I don't do calls unfortunately

[wispy spoke]

😢

its okay

[cerulean sail]

But does this at least make a bit more sense?

[wispy spoke]

not really

[cerulean sail]

What may help is to trace it on a CAST diagram

If you give me a second

[wispy spoke]

ok

[cerulean sail]

So initially this is the mental picture we have, right?

And for 180 - a, it's like this

[wispy spoke]

yes

[wispy spoke]

okay

what about this?

[cerulean sail]

Then using the same ideas as you would as if you were solving for something you knew, you can figure out here that sin(180 - a) = sin(a)

For the next question, the mental picture is like (give me one moment)

This

[wispy spoke]

yes

[cerulean sail]

And that's in a region where only tan is positive (so sin is negative)

So that should give you -p, if that makes it any clearer?

(as there, sin(a - 180) = -sin(a) )

[wispy spoke]

omg

i understanddddddddd

thank you

[cerulean sail]

Nice nice, hopefully that makes it clearer! Now try out iii maybe?

[wispy spoke]

okay sir

[cerulean sail]

Probably easy to work on working out that angle first then add 3 to whatever you get

[wispy spoke]

nvm

fuck this

i hate ths

3-p

imma go kms rq

[cerulean sail]

That's what you get?

[wispy spoke]

i think

[cerulean sail]

But that's it, no?

[wispy spoke]

it is but i said anything tbh '

[cerulean sail]

Basically almost the same picture as last time

Or if you're feeling lazy and just want to take something as is, here

[wispy spoke]

how did it become -p

[cerulean sail]

Remember, you're in a region where sin is negative

[wispy spoke]

ohjhhhhhhh

holy shit i get iy

thwnk you thank yoyu

[cerulean sail]

Yep, that's pretty much the idea for these questions really!

[wispy spoke]

okay i understand now

[cerulean sail]

Wherever you can, try and just finesse these if you can

[wispy spoke]

i will i will

thanks

[cerulean sail]

No probs, my pleasure

[cedar kiln]

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[tawny citrus]

how do you solve equations like x^x= 27

[crimson delta]

Well here the answer is easy to guess

In general, computer

[crimson sedge]

lambert W function

hmm one sec

[tawny citrus]

alr

[crimson sedge]

x^x = 27
x = 27^(1/x)
1 = 1/x * 27^(1/x)
1 = 1/x * (e^ln(27))^(1/x)
1 = 1/x * e^(ln(27)/x)
ln(27) = ln(27)/x * e^(ln(27)/x)
W(ln(27)) = ln(27)/x
x = ln(27) / W(ln(27))
x = 3

thats at least the only real solution

[tawny citrus]

ooo okay thank you

.close

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[tranquil folio]

how to solwe this one?

[thick cipher]

Write it as the imaginary part of something

[cedar kiln]

@tranquil folio Has your question been resolved?

[thick cipher]

$\sin(x) = \mathcal{Im}(e^{ix})$

[wraith dagger]

black_couscous

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[cedar kiln]

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[native wedge]

Evaluating the limit

[native wedge]

Solved up until

-x-4 / x+2 / x+4

Dont understand how its simplified to
-1/x+2

[red pumice]

brackets are important

but what i think you mean is

$\frac{\frac{-x-4}{x+2}}{x+4} = \frac{-x-4}{x+2} \cdot \frac{1}{x+4} = -1 \cdot \frac{x+4}{x+2} \cdot \frac{1}{x+4} = \frac{-1}{x+2}$

[wraith dagger]

Gijs

[native wedge]

im dusty on my alg, isnt it supposed to be (-x-4)/(x+2) 1/(x+4)

[red pumice]

well thats what i wrote

[native wedge]

oh lol

[native wedge]

yep 😦 gimme a sec

what happened here

[red pumice]

x+4 cancels with x+4

[native wedge]

FUCK

ok thank you bye lol

.close

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[narrow fiber]

not sure how to go ahead from here

or should I have taken a different path?

[dusty hazel]

All good.

Let's restart from

$3^{\frac{n-6}{3} - 1} = 2^{2-\frac{2n-12}{3}}$

[wraith dagger]

What the hell am I doing here?

[dusty hazel]

So you have this, yes?

[narrow fiber]

yes

[dusty hazel]

Try to isolate the terms with "n" and not 3 and 2.

For example.

[wraith dagger]

What the hell am I doing here?

[narrow fiber]

got it

thank you

[dusty hazel]

You're welcome.

[narrow fiber]

.close

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[dull elbow]

root 3 sec^2 x = csc x

[dull elbow]

how do i solve this?

[latent bloom]

You can use an iterative method

[opal basin]

maybe rewrite sec^2 as 1/(1 - sin^2) and solve the quadratic

in terms of sin

[upbeat lotus]

Do you know what's 1-sin²x?

[crimson sedge]

cos²x

[coral imp]

+2 social credit

[brisk dirge]

[upbeat lotus]

Yeah so how can you write sec²x as 1/(1-sin²x)?

[upbeat lotus]

69*

[opal basin]

@dull elbow

[somber jackal]

Tf

Bruh

[coral imp]

[somber jackal]

Write properly da

[upbeat lotus]

Oh god bas ek baar dila de

💀

[coral imp]

[brisk dirge]

Tharki chhokro aayo re

[somber jackal]

Lmaoo

[somber jackal]

Tum kaha se ho?

[brisk dirge]

Lucknow

[somber jackal]

Matlab, in India, where are you from?

[somber jackal]

Oh acha

[brisk dirge]

Do baar kaahe bol raha

[crimson sedge]

Mods ?

[upbeat lotus]

@somber jackal @brisk dirge @coral imp #chill

[brisk dirge]

No

[lusty grotto]

hey vanilla

[coral imp]

Na ttn 💀

[lusty grotto]

so i still cant make graphs

[crimson sedge]

#latex-help we should talk their, this is help channel

[lusty grotto]

maybe we should chat in #latex-help or #latex-testing ?

yeah okay

[dull elbow]

I can undetstand where ur coming from @opal basin

[somber jackal]

Nah bruh

[dull elbow]

didnt even think of that way

[cedar kiln]

@dull elbow Has your question been resolved?

[cedar kiln]

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[dry swallow]

Calculus

I'm looking to make a graph that behaves like so:
0 <= x < 3 -> f(x) = 1
3 <= x < 6 -> f(x) = -1
6 <= x < 9 -> f(x) = 1
... And so on. Basically every 3 "x units" the function should switch being positive and negative.

How would I go about creating something like this? What would be the condition?

[crimson sedge]

By making a graph do you mean find a function which behaves like this?

As in something that isn’t just a piece wise definition

[violet night]

Play around with trig functions and different ways to floor and upper something

Should be a good starting point

[hushed maple]

One way is to make a fraction function with

[hushed maple]

absolute value of sin(x/(3pi))

over sin(x/(3pi))

[crimson sedge]

Fourier Series also

[hushed maple]

wait bo its xpi/3

[violet night]

yeah theres a bunch of ways

[hushed maple]

think this is one of the more basic ways

[cedar kiln]

@dry swallow Has your question been resolved?

[crimson sedge]

Does it need to be continuous for all rational numbers ? Like what about 6.3 or 1.34 ?
Is it just natural numbers only ?

[dry swallow]

Thank god you pinged me lol

[dry swallow]

For any real number, yeah
Of course the function doesn't have to be a single expression, it can contain conditions if necessary

[dry swallow]

Huh, that's really clever

Would've never thought of it myself tbf

I'll go stare at it a little while longer until it makes sense
Cheers!

.close

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[untold sierra]

Guys think fast
What is the proper the way to set a boundary size for a surface?
I mean to give it a min_size and a max_size and a tendency, which means I want the surface to grow as much as it can or shrink as much as it can
Here is what I achieved so far 🤓

[untold sierra]

Think of the surface as a Vector2

for example we have A=(100,300)

and we have min_size=(50,50)

and we have max_size=(300,800)

If I wanted my Vector2 to grow|shrink as much as it can, what should I do?

So if I wanted A to shrink it would be (50,150) and if I wanted it to grow it would be (266.66,800) Do you get the Idea?

<@&286206848099549185>

[cedar kiln]

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.

[untold sierra]

😭 😭

<@&286206848099549185>

[cedar kiln]

@untold sierra Has your question been resolved?

[cedar kiln]

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[untold sierra]

No screw you

[untold sierra]

.close

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[split briar]

I need to show the running time of this algorithm which is apparently log2log2 n

[split briar]

I was thinking that it could be solved by finding the amount of times that i * i * i^2 * i ^4 * i^8............... = n

[crimson delta]

yes

it might help to simplify the LHS and see how the exponent looks like

[cedar kiln]

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[cedar kiln]

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[zealous torrent]

Prove by contradiction that for xER,
$x>0\implies(x+(1÷x)\le{2})$

[wraith dagger]

b0a

[crimson delta]

Multiply both sides by x, get everything on one side, realize that it looks familiar (hopefully)

[cerulean sail]

,w plot x + (1/x)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by chartbit#3847

[cerulean sail]

Am I misreading or missing something?

[crystal raptor]

should be >=

[cedar kiln]

@zealous torrent Has your question been resolved?

[cedar kiln]

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[smoky whale]

Hi guys, how do I sketch this

[zealous torrent]

Ty

[manic acorn]

do you know graph transformations

[smoky whale]

Yeah

This has a dilation and a transformation

Translation ******sory

[manic acorn]

so what did u do

[smoky whale]

I tried getting y=0 to get x intercept

But I never done this type of question only simple ones

[manic acorn]

what did u get

[smoky whale]

So I wanna know how it should be done

[manic acorn]

this is fairly simple yes

[manic acorn]

line 1 is correct

line 2 is also correct

but what does line 2 have to do with 1

[smoky whale]

Hold on I’ll write it clearer

[dire geode]

x intercept

[manic acorn]

i meant e^0=1 as line 2

[dire geode]

Ah I see

[manic acorn]

that doesnt look correct, at the surface at least🥸

[manic acorn]

take ln term to the left hand side and divide both sides by 2 then take e to the power both sides

[dire geode]

That's wrong. You need to use log/exponent rules

[smoky whale]

Which ln term

[manic acorn]

log base e is also referred to as ln

[smoky whale]

Oh

Ok

Like taht?

[manic acorn]

This picture

ln term goes to the left hand side

Then divide by 2

Then e to the power both sides

If you dont understand this, then review log/exp rules

[smoky whale]

Like this?

[smoky whale]

Do you mean divide ln too

[manic acorn]

Looks wrong

[manic acorn]

What

[smoky whale]

Like this?

[manic acorn]

You had 0=1-2ln(x-1)

ln term goes to the lhs

2ln(x-1)=1

Divide by 2

ln(x-1)=1/2

Now raise both sides to the power of e

[smoky whale]

e^1/2

[manic acorn]

Thats the rhs yes

[manic acorn]

Right

So whats x?

[smoky whale]

X=e^1/2-1

[manic acorn]

Are you sure

[smoky whale]

No

Can I use logs

[manic acorn]

this is not about using logs

[smoky whale]

Finding x

Sketching then

[manic acorn]

This is wrong

[smoky whale]

Yeah

[manic acorn]

Do you get why?

[smoky whale]

Is x -2.64?

[manic acorn]

I have to go sorry hopefully someone else can help

[manic acorn]

Upto here is correct, work from here

[smoky whale]

Can you complete it so I can understand

Please

[manic acorn]

Im not sure im allowed to do that

[smoky whale]

It’s ok, so I can get better picture

.close

[cedar kiln]

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[narrow swallow]

I am using the integration of parts but it keeps looping and I can't get to an answer

[upper garnet]

Show work

Pretty sure after 2nd IBP your right term will look exactly like the initial integral.

[crystal raptor]

yeah if its looping then set call your original integral I and then you'll have an equation in I that you can solve

[cedar kiln]

@narrow swallow Has your question been resolved?

[narrow swallow]

Wdym?

[crystal raptor]

what exactly doesnt make sense

[dire geode]

Your original integral appears again with some constants. Factor out the constants. Solve for it now using the suggestion above

[narrow swallow]

How do i set call the original integer I

[dire geode]

It's just a variable name

[dire geode]

I = the stuff in this image

[narrow swallow]

So i dont use the integration of parts?

[dire geode]

...

[narrow swallow]

How do i break the loop?

[dire geode]

You don't

You solve for I using algebra

I = stuff + c * I

[narrow swallow]

Isnt that the original equation?

[dire geode]

No

It's a new equation

If x = 2 + 3x. Can you solve for x?

[narrow swallow]

Ye

[dire geode]

Your integral is in this form after two integration by parts. Figure out stuff and c

Clean up your equations

[narrow swallow]

Is the I in the first loop?

[dire geode]

.

[dire geode]

.

[narrow swallow]

So the box represents C?

Because its the 3rd time?

[dire geode]

You only need 2

[dire geode]

After IBP twice, your integral comes back and your equation looks like this once you clean up your algebra

[dire geode]

It's still too messy for you to see it here

[cedar kiln]

@narrow swallow Has your question been resolved?

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[cedar kiln]

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[crimson sedge]

hi

[crimson sedge]

can i ask a question here

[patent cape]

go ahead

[crimson sedge]

how do i solve 4 . 3x = 12

i cant seem to get rid of that 4

[tropic oxide]

do you mean 4 * 3x = 12?

or do you mean 4.3 as a decimal?

[patent cape]

like $4\cdot3x=12$?

[wraith dagger]

Kel.plush

[crimson sedge]

yup

multiply

[tropic oxide]

well this is just 12x = 12, no?

[crimson sedge]

i know but i wanna know how to get rid of that 4

[tropic oxide]

the 4 is multiplying the 3x

how do you undo multiplication?

[crimson sedge]

yup

oh wait

i think i get it

thanks

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[dim dune]

How can I find the area of a shape (triangle) defined by some points in a coordinate system where y axis was rotated about the origin for some angle alpha?

[neat meteor]

multiply the shape of the original by the determinant of your linear transformation matrix

[dim dune]

The problem is that I don't have the original

The thing is, I am given three distinct, non-collinear points (x1, y1), (x2, y2) and (x3, y3) in that coordinate system, where x the same and y rotated for some acute angle. This is the only info I have, can I somehow determine the area in this coordinate system?

[neat meteor]

I think so yes, the length of the base would be the norm of the vector connecting two points

then I would calculate the height by projecting another vector onto the normal of the base vector

Then by A = 1/2 bh you would get the area and could calculate the final area after transformation

[dim dune]

hm

I am know thinking of how to connect it with coordinates

given

[neat meteor]

(ΔABC) = (1/2) |x1(y2− y3) + x2(y3− y1) + x3(y1 - y2)|

pulled this of a random website

[dim dune]

This is right for orthogonal coordinates innit?

But for some transformed?

[neat meteor]

well what would the transformation be that rotates the y axis but not the x axis

[dim dune]

So the area of a triangle is defined as a determinant of the matrix above

What changes in the third row if j^ rotates for acute angle alpha?

[cedar kiln]

@dim dune Has your question been resolved?

[dim dune]

Ok, anyways, thanks Door Van Beveren for help!

Much appreciated!

.close

[cedar kiln]

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[novel cloak]

Hi, i want to count how much password with 8 numbers that start with "3" it well if i do it like that ?

[lusty grotto]

how many passwords of 8 numbers there are, that start with 3?

[novel cloak]

Yea

[lusty grotto]

i believe there are as many passwords as there are numbers between 30000000 and 39999999

[novel cloak]

https://cdn.discordapp.com/emojis/1037770587235434567.webp?size=48&quality=lossless&size=40

[lusty grotto]

can u find that

[novel cloak]

Maybe by doing 9! / 9!

[lusty grotto]

how many numbers between 1 and 10

for example

[novel cloak]

9

Nah wait

10 if u count 1 and 10

[lusty grotto]

11?

can u list the 11 one by one

[novel cloak]

10*

[lusty grotto]

right

so how many numbers between 2 and 10

[novel cloak]

9

[lusty grotto]

yes

how many between 2 and 200