help-38

[runic estuary]

another help channel?

[pliant tree]

I figured it out, the answer we get after adding and subtracting 1/(x-1) to the original sum is :-
1/(x-1) - (2^{n+1}) / (x^{2n+1} - 1)

[urban copper]

its because helpers keep helping people when they post here even when they know its not the original question, now the progress of your question is disturbed and lost in between messages

this is why this help channel is taking so damn long

[runic estuary]

seems like it, ill have to ping mods if things derail too much

[zinc ginkgo]

.close

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[mossy summit]

nice graph btw

[mystic pilot]

Ty

In this graph, as the slope in negative and decreasing, velocity is negative and decrease. Is the acceleration -ve or +ve and increasing or decreasjng

acceleration should be -ve and decreasing

[mossy summit]

accelartion should be -ve

since velocity is decreasing

@mystic pilot do u have the curve equation

[mystic pilot]

No

[mossy summit]

isnt finally dy/dx becomes 0

value beomes less and less negitive

[glass kite]

slope negative so velocity negative, but ROI +ve so acceleration positive

[mossy summit]

so velocity is increasing

yeh

accelartion is positive

[glass kite]

velcoity is increasing in -ve x direction

[mossy summit]

velocity is just in opp direction

[mystic pilot]

It's not mathematically decreasing with sign

What is ROI

Wth

I can't read the particular graph for acceleration

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[tulip sapphire]

In this graph the slope of each tangent is negative, therefore the velocity is negative, we also see that the velocity is decreasing, therefore we can say that: "The object is decreasing its speed in the negative direction", which occurs when the acceleration is positive

[tulip sapphire]

@mystic pilot

.close

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[tulip sapphire]

.reopen

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✅

[mystic pilot]

Reading

[tulip sapphire]

continue your question here if you have further inquiries

[mystic pilot]

Got it

Yes I have further

[tulip sapphire]

sure! what else do you need help with?

[mystic pilot]

We have to convert this into x-t graph

[tulip sapphire]

what have you tried so far?

[mystic pilot]

In first interval 0-10sec the velocity is positive and constant and after that velocity is 0

So

if velocity is positive and constant

[tulip sapphire]

in the first interval the velocity isnt constant but linear

as you can see, its constantly increasing

[mystic pilot]

Can we not say v = tan(alpha) = constant

[tulip sapphire]

not really, this graph is VELOCITY with respect to TIME, not POSITION with respect to TIME, v = tan(alpha) is true only when we have POSITION with respect to TIME

This graph shows how the velocity of the object changes over time, and as you can see, its changing

in this case, the acceleration would be tan(alpha)

[mystic pilot]

sorry mb

[tulip sapphire]

all good

Lets look at the second interval first as its easier

What would be the velocity in the second interval?

[mystic pilot]

In 2nd interval velocity is const and a = 0

[tulip sapphire]

Correct, and what is the velocity?

You can read that directly from the graph

[mystic pilot]

Velocity is positive

[tulip sapphire]

Yes, but what is its value exactly

[mystic pilot]

20m/s

[tulip sapphire]

Correct

So if the velocity is 20m/s

then when looking at the graph of **position **with respect to time, we know that between 10 and 15 seconds, tan(alpha) = 20m/s

by position with respect to time, i'm talking about the x-t graph

[mystic pilot]

Yes I'm getting it

I have to draw x-t graph when the velocity is 0 there

And +ve

For the second interval

[tulip sapphire]

sorry i didnt understand

[mystic pilot]

I'll follow you

you can continue

[tulip sapphire]

Okay :)

Lets start from the beggining now, so when t = 0s

For t = 0s, we know that the velocity is 0m/s, in other words in the x-t graph tan(alpha) = 0m/s => alpha = 0.
This means that in the x-t graph, at the point t=0s, the function is flat.

[mystic pilot]

agree

[tulip sapphire]

Then, till t=10s, velocity increases linearily

How would that look?

[tulip sapphire]

Yes!

thats right!, now try to continue it, adding the part from 10s to 15s, knowing that velocity is constantly 20m/s, therefore tan(alpha) = 20

[mystic pilot]

The flat one

Which has +ve vel

[tulip sapphire]

Correct, the velocity is positive

and its constant

[mystic pilot]

Yes

[tulip sapphire]

So how would the graph look?

[mystic pilot]

I sent

where is it

[tulip sapphire]

This is only from 0 to 10s, not the rest

[tulip sapphire]

Yes!

Thats correct! Good job :)

[mystic pilot]

Thank you friend

Ill brb in 10 min and I'll ask one more

so that I can test myself

[tulip sapphire]

Go ahead :) I might not be active though cause i have to leave soon

[mystic pilot]

How did you study physics

Like from book or from lecture

[tulip sapphire]

I'm mostly self taught

[mystic pilot]

That's a proud thing to say

[tulip sapphire]

I havn't done any of these stuff yet in school

[tulip sapphire]

thanks :)

[mystic pilot]

You can self taught the same thing that i asked, while I didn't understand in the class

[tulip sapphire]

Its ok hahah, different people are good in different things, while I'm good in math and physics, I risk failing my class due to my grades in history, german and italian

[mystic pilot]

Oh

but from which book did you study

[tulip sapphire]

Not really any books, mainly youtube, wikipedia, and the internet in general

[mystic pilot]

Ooh

Interesting

[tulip sapphire]

yea nowadays online resources are so wide spread that learning anything has become way easier

[mystic pilot]

Even the cheapest way

[tulip sapphire]

exactly haha

[mystic pilot]

Till how much you have studied physics

[tulip sapphire]

Physics I havn't really studied too much, this is pretty much math disguised as physics and for mathematics I've studied quite alot

[mystic pilot]

mathematics till how much

[tulip sapphire]

For mathematics I would say till the level of a bachelor although I need a more formal understand of measure theory, right now my understanding is very hand-wavy

But of course it varies a lot, right now I'm writing a paper which requires some knowledge in analytical continuations etc...

[mystic pilot]

PAPER

Wow

[tulip sapphire]

Yea hahaha, its just a side project for now

[mystic pilot]

Are you doing this as your own choice and not assignment?

[tulip sapphire]

It's by myself, but I'm collaborating with another member of this server

yea as my own choice

[mystic pilot]

You have mathematician mind

[tulip sapphire]

Hahaha, thanks :)

[mystic pilot]

nice to talk to yu

[tulip sapphire]

It was a pleasure for me aswell :)

Do you mind if I close this channel now?

[mystic pilot]

Sure

[tulip sapphire]

I think it's best you open your own channel

.close

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[plain tusk]

how is htis alternate segment theorem?

[plain tusk]

i dont see it

[burnt mulch]

$\angle CAD$ is standing on arc $CD$; $\angle CDE$ is the angle formed by tangent $DE$ and the chord $CD$ that forms arc $CD$.

[solid kiln]

Civil Service Pigeon

[plain tusk]

i understand the angle cde is formed by tangent de and the chord cd fdorms that arc

but i dont understand the first bit

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[main meteor]

Does y=((1-x^2)^1/2)/x have any vertical tangent lines or cusps? I took the derivative and I get -1/(x^2(1-x^2)^1/2) so that means plus and minus one are defined in the original but not in derivative so I would think to check the left and right side of them. But I can’t because the limits don’t exist of both sides. So does that mean there is or isn’t a vertical tangent line or cusp?

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@main meteor Has your question been resolved?

[main meteor]

No

[light delta]

Has neither

Because x=0 is undefined

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@main meteor Has your question been resolved?

[main meteor]

Yes ik. It’s obvious that there is none at x=0

[bright oracle]

Okay so you basically want the derivative to tend to +inf or -inf

[main meteor]

Ye ik

[tardy hemlock]

If on one side, it doesn't exist, then you should be done

[main meteor]

Ok thank you

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[tardy hemlock]

(maybe wait for Greenie to finish though)

[bright oracle]

It's okay. You can close

I was going to look into it more closely but I'm super tired lol

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[vernal drum]

Hi can someone please explain part c of this question

[vernal drum]

its not in my textbook

[tawdry bloom]

Just calculate both areas and compare them in a ratio

Like if for example area triangle YBX is 15 and triangle OAX is 60 then the ratio is 1:4

[vernal drum]

how do I calculate the area tho?

like cuz there are not lengths

[exotic pine]

did you get b?

if i were to give yoh tvis and asked you to find the ratio of [ABD] to [ACD], do you know how to?

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@vernal drum Has your question been resolved?

[vernal drum]

yep

[exotic pine]

what did you get?

[vernal drum]

i got 4:6

2:3

[exotic pine]

ok

[exotic pine]

.

[vernal drum]

no I don't know

i unfortunately have a skill issue

[exotic pine]

its fine

alright, say i give you the height of ABC is 1 with the base being BC, do you know how to find it then?

[vernal drum]

it would just be 4:3

[exotic pine]

good

what if the height is 2? what about 3?

[vernal drum]

it would still be the same because they share a side?

[exotic pine]

specifically, they share a height

[vernal drum]

like you could factor out the 2 or 3 or whatever and then the ratio would be the same

[exotic pine]

yes exactly

so like, [ABD]:[ACD]=1/2×h×4:1/2×h×3=4:3

alright, try doing something simmilar for c

[vernal drum]

they share BX and OB

[exotic pine]

eh not quite

try to think about triangle ABX

[vernal drum]

ABX to YBX is 4 : 1 right?

[exotic pine]

correct

[vernal drum]

and then ABX to OAX is 2:4?

[exotic pine]

almost

[vernal drum]

what is this topic called?

[exotic pine]

dunno

[vernal drum]

I dont understand like what to do after that

[exotic pine]

for ABX:OAX, the base is 2 and 6 respectively, so its 1:3

so, we have YBX:ABX=1:4 and ABX:OAX=1:3

and we want to find YBX:OAX

any ideas from here?

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[vernal drum]

1 : 12

[exotic pine]

yeah thats it

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[rotund owl]

,, \int \sqrt{x^2+a} \dd{x}

[solid kiln]

taebek

[rotund owl]

I used tanu = x

And you get sec^2(u) under root and some other stuff

I solved by cancelling the root but realised it needs to be in absolute value idk how to remove it

And apparently its not about splitting into cases

[forest needle]

You can either directly use integration by parts or put x=(sqrta)tanu

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@rotund owl Has your question been resolved?

[forest needle]

Just assume that the value is positive don't use the absolute value

[rotund owl]

How?

Cant assume some without proof

[rotund owl]

You cant use integration by parts on this

Needs trig sub or some weird hyperbolic sine sub

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@rotund owl Has your question been resolved?

[nimble niche]

Why do you say that though i really think it should be sec x is periodic

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[rotund owl]

Cause i searched in symbolab and its like that

.reopen

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✅

[rotund owl]

Its one function

Not 2 sections of it

[nimble niche]

secx>0 for x belongs to [0,pi/2]
Secx<0 for x belongs to (pi/2, pi]
And the period of sec x is 2pi.

I do not know why you cant just split the function

[rotund owl]

I can but result on symbolab isnt split

Thats why im thinking there is a mistake

[nimble niche]

!nogpt

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[rotund owl]

Its not gpt

Its liturately with steps

Reliable source imo

[honest magnet]

yes you can solve this without using by parts if you substitute x = √a sinhu i think

[rotund owl]

But everywhere i searched its not 2 sections on the function

[rotund owl]

Whats sinhu?

Oh

Hyperbolic sine

with e^x e^ (-x)?

Thats what i also found but im just trying to see if tan has an issue

[honest magnet]

yes

[rotund owl]

Cause i found different result i believe with tan

[honest magnet]

tan doesn't have an issue, you just need integration by parts

[rotund owl]

Oh you mean afterwards

Like on sec^3x

[honest magnet]

yes

[rotund owl]

For that yea ik

The issue is with removing absolute value

Can i remove it or i need to take cases

[honest magnet]

you don't need to remove the |something|, that's what you want to know right?

[rotund owl]

Yea

[honest magnet]

yes you don't need to

[rotund owl]

Cant really integrate with the ||

Oh

[honest magnet]

no it's only coming at the end

[rotund owl]

But the integral is outside of it

[honest magnet]

you are not needing this while integrating

[rotund owl]

How will i integrate by parts if there is | |

Cant derivate without removing it

[honest magnet]

split cases

[rotund owl]

Fire

Alr

Thanks

Ig thats the only way

Maybe i did mistake and its sums up to 1 type

@nimble niche thanks aswell

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[nimble niche]

I'm still so confused why you cant split but ok yw ig lol

Ah ok nm u decided u can

All g

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[rotund owl]

It probably at some point sums up to 1 section cause they are the same maybe

[rotund owl]

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[fierce tiger]

hello

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[modest rain]

does anyone know what this question is asking ?

[modest rain]

i dont think its determinant but when i tried to find inverse it doesnt match any of the answers either

ans isnt d either its b

[main sigil]

it looks like determinant

[supple copper]

it's determinants

[main sigil]

they are asking you which of these matrices has equal determinant

[supple copper]

is there a topic

[modest rain]

but wouldn't determinant mean a single number answer

[tidal kestrel]

Column ops on cols 1 and 2

[honest magnet]

they're probably asking you if you can perform row and column operations to match one of the options there

[main sigil]

Yeah, they are asking which of the matrices in options have the same determinant as the given one

[modest rain]

wait so is it column operations or equal determinants

[fierce lake]

Either is fine

[honest magnet]

same thing

[main sigil]

column operations preserve determinant

[tidal kestrel]

Column ops don’t change dets

[fierce lake]

You can convert it into question det or evaluate individual dets in the options

[main sigil]

they are asking about determinants, but using column ops seems to be the way to solve it

[modest rain]

umm i dont get what u guys mean by column operations but like u said i just calculated the det of the question is -13 and the det of b is also -13 is that why its the answer ??

[main sigil]

Yes

thats one way to do it

[tidal kestrel]

Add col 1 and 2

[main sigil]

notice that you can get (b) by adding the 2nd column of the given matrix to the 1st column

and by doing this, the determinant isnt changed

so you can change the given matrix into (b), without changing the determinant

hence they must have the same determinant

and you can avoid lengthy computation

[modest rain]

ohh okay i get it now i didnt know u could do that

[modest rain]

got it thank you !

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[marsh forum]

I don't get why the highlighted part is important

[marble wharf]

wdym why its important

the theorem wants that some polynomial exists

and that highlighted part gets you there

by showing that some related polynomial exists

[marsh forum]

I see

okay

Thanks

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[worldly wing]

what's 2.19?

[marsh forum]

.reopen

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✅

[worldly wing]

I don't see how that gives us that u, Tu, ..., T^(m-1)u is lin. ind.

[marsh forum]

it's LI because it's a spanning list the size if dim(V)

[stark oasis]

Hi guys

Someone here

?

😩

[main sigil]

If it was linearly dependent, then there would exist some k <= m-1 such that T^k(u) is in span of {u, ..., T^(k-1)}

[urban copper]

!occupied

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[worldly wing]

no it isn't, the list has m elements which is less than dim V

[main sigil]

contrary to the fact that m was chosen to be the smallest such integer

[worldly wing]

ahh right

okay

m is the smallest degree of a polynomial of T which vanishes at u

[main sigil]

Well, m was initially chosen like this

which literally means that its the smallest integer such that T^m(u) is in the span of {u, ..., T^(m-1) u}

[worldly wing]

fair enough

[main sigil]

and the contradiction with this is then clear

[marsh forum]

I'll close this now?

.close

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[forest needle]

$\int\sqrt{x^2+a};dx=x\sqrt{x^2+a}-\int\frac{x^2+a-a}{\sqrt{x^2+a}};dx=x\sqrt{x^2+a}-\int\sqrt{x^2+a};dx+a,arcsinh(\frac{x}{\sqrt{a}})\rightarrow\int\sqrt{x^2+a};dx=\frac{x\sqrt{x^2+a}}{2}+\frac{a;arcsinh(\frac{x}{\sqrt{a}})}{2}$

[solid kiln]

Aman

[forest needle]

$\int\frac{1}{\sqrt{x^2+a}};dx=arcsinh(\frac{x}{\sqrt{a}}) is a standard result$

[solid kiln]

Aman

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[full dock]

i dont know how to start

[trim lichen]

under what conditions would x = su + tv describe a line and not a plane

[full dock]

if u,v are scalar multiples

[trim lichen]

ok

so can you write down the sentence "Au is a scalar multiple of u"

as an equation

[full dock]

Au = lambda*u for real lambda

[trim lichen]

bingo

[full dock]

so this is just the definition of the eigenvector

[trim lichen]

indeed

[full dock]

so to formalize everything we can say x = su + t(Au) describes a line through the origin if Au = lambda*u for real lambda. Because u is a nonzero vector u is therefore an eigenvector of A?

[trim lichen]

u is an eigenvector of A.

[full dock]

ohh

yeah

.solved

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[full dock]

what the

sir ts not a math question

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[lime tusk]

How can a function be differentiable everywhere on R³ but its partial derivative are not continuous ?

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[wanton nebula]

How do i find the area of this

[floral warren]

Either integrate or understand basic geometry.

[stoic garden]

Try splitting the shape into a rectangle and a triangle.

[floral warren]

Or think of it as one half of a giant rectangle

And solve for the area of the rectangle and divide by 2

[wanton nebula]

So 10.5

[stoic garden]

✅

[wanton nebula]

Alr thx that helped

@stoic garden i got it wrong

🗿

[stoic garden]

Oh, I realized that the y-axis increments differently.

[wanton nebula]

It said it was 21

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[chilly blaze]

how do i tell right away the number of solutions

[zinc ginkgo]

define "right away"

[chilly blaze]

like ik from the direction vectors they're scalar muiltiples so theyre either coincidental or parallel but like how do i know the amount of solutions?

[chilly blaze]

idk it just says determine the number of solutions without solving

but hwo??

[simple panther]

Aren't they equations of lines?

[chilly blaze]

yes they are

direction vectors im referring to s[-2,1,5] and t[-6,3,15]

[simple panther]

So don't they intersect at either 0 or 1 or infinite points?

[zinc ginkgo]

they're parallel lines so they only intersect if they're the same lines

[chilly blaze]

well theyre parallel so that eliminates the second ans

[chilly blaze]

wait

hu

huh

how do yk theyre parallel tho

[fossil folio]

because one of the direction vectors is a scalar multiple of the other

[zinc ginkgo]

do you know the slope of a line in parametric form

[chilly blaze]

dont you just know that they're scalar multiples of each other

[zinc ginkgo]

slope is the wrong word

[chilly blaze]

thats it? but doesnt that mean they can either be parallel or coincidental?

[chilly blaze]

yeah but doenst that tell us that theres either 0 or infinite solutoons

[fossil folio]

they either touch everywhere or nowhere

[chilly blaze]

the direction vector

[zinc ginkgo]

you can tell by looking that they're the not same line

[chilly blaze]

yes but how can i determine that

after noticing that their vectors are scalar multioles

[chilly blaze]

uhh

wait

so if theyre coincidental

nvm idk

[chilly blaze]

yea but

if you look at this, theyre not the same line and their "slopes" are scalar multoples

but in this case theyre coincidental if we solve it

but without solving how would you know?

if you just compare this pic to the one i first sent, both "slopes" are scalar multiples but the first one is parallel and the second pic i sent is coincidental

just how do you know which is parallel/coincidental without solving

[fossil folio]

how are they coincidental

[chilly blaze]

no this one isnt coincidental

[fossil folio]

you just solve first equation = [1, -4, 4]

[chilly blaze]

the first pic is parallel

[fossil folio]

Yes

[chilly blaze]

this is coincidental

[fossil folio]

You have to solve it 💀

[chilly blaze]

but

huh..

😭

[fossil folio]

It’s not even that much work

Don’t be lazy

[chilly blaze]

are you sure there's no way to tell right away?

im not being lazy, it's what the question is asking 😭

[fossil folio]

Maybe if you have good intuition you can see it

[chilly blaze]

intuition is crazy

[fossil folio]

But you’ll have to justify it someday

[chilly blaze]

ok wait

so

in order to see if they're cxoincidental

[chilly blaze]

can i just see if there is some t value that will get me to the point [-2,3,12]

is that how to check

[fossil folio]

yes

from the other equation

[chilly blaze]

ok

[chilly blaze]

so if it was in 2d then

same thing right

[fossil folio]

yes

[chilly blaze]

ty

[wraith hinge]

One quick and small calculation you can do is that you can calculate the difference between [-2,3,12] and [1,-4,4] and check if its direction is parallel with [-2,1,5] or not (if yes they must be parallel and coincidental)

[chilly blaze]

wow whatt

why does that hold true tho?

like whats the prooof behind that

never learned to d that so its interesting

[wraith hinge]

Geometrically, [-2,3,12] and [1,-4,4] are points (more exactly the position vectors of those points) that lie on those lines respectively (just put s or t = 0)

So if the parallel lines coincide, geometrically the displacement vector of those PVs must be parallel/antiparallel with either direction vector

Draw it out to see

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[marsh forum]

I was just wondering, is there anyway around writing down a system of linear equations here

[marsh forum]

wait, I'm getting

\lambda =0?

[zinc ginkgo]

surely you can find the matrix for n=2 or 3

[marsh forum]

yea, it's a matrix with all 1s

[zinc ginkgo]

and did you find the char poly

[marsh forum]

Trying to

[zinc ginkgo]

for n=2 or 3?

[marsh forum]

2 for now

[marsh forum]

I'd say x^2 is the minimal polynomial

for T x^n

[zinc ginkgo]

did you learn how to find eigenvalues by finding the roots of the characteristic polynomial

[marsh forum]

Trying to

[zinc ginkgo]

what do you have so far

for n=2

[marsh forum]

yes

[marsh forum]

this doesn't work?

[zinc ginkgo]

this seems easier to me.

[marsh forum]

wait

[honest magnet]

perhaps a little another way of seeing the eigenvalues. you can see that the image space is generated by (1,1,..,1), so the rank of T is 1. what that means is 0 is an eigenvalue of am = gm = n-1. the remaining eigenvectors must lie within span(1,1,..,1). let x = (1,1,..1), one can easily see Tx = nx, so n is an eigenvalue.

[marsh forum]

I haven't done char polys yet

[zinc ginkgo]

[marsh forum]

just realised.

Axler introduces them in chapter 8

atleast theat;s the first reference I can find

[zinc ginkgo]

[marsh forum]

I mixed up minmal polys with characteristic ploys

[marsh forum]

hmm

makes sense

thanks

I think I have to revise the poly part

oops

.close

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[marsh forum]

Thanks again

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[vernal drum]

Can someone help me spot a mistake here

[vernal drum]

I don't understand what i have done wrong

nvm i got it

.close

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[torpid trellis]

would b be 18.6km?

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[torpid trellis]

help

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[drifting glacier]

i don’t understand how to continue after calculating sigma

[round imp]

it would help if you provided a translation

[drifting glacier]

A caterer is providing lunch at an event with 200 attendees. The lunch consists of soup, a snack, and sandwiches. Experience shows that 15% of the attendees do not want soup. There is enough soup for 175 people.

10. Calculate the probability that there is not enough soup for all the attendees who want soup.

The organization would find it embarrassing if there were not enough soup for everyone who wants it. Therefore, it has been agreed with the caterer that there must be enough soup so that the probability that all people who want soup can actually have it is at least 99%.

11. For how many people should the caterer provide soup?

this is what i have so far

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[drifting glacier]

<@&286206848099549185> i’m crashing out

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[hidden vortex]

the expected amount of ppl who want soup is 170 and they prepared 175 soup

the standard deviation of proportion is root ( p (1-p) / n )

so the z score is (175-170) / standard. dev.

themn use a z table

voila.

[thorn bay]

better to use 175.5

[hidden vortex]

meh justt the same ( i never learned that in my stats course tbh)

[drifting glacier]

what would the answer be?

[hidden vortex]

calculate it yourselfs

hhelpers cant tell u the answers duh

[drifting glacier]

i don’t get it. i’m supposed to use a graphic calculator

can you not show me the calculations?

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[hidden vortex]

Huh

Why graphic calculator

We are supposed to use z table in statistics

[drifting glacier]

i have no idea

something with binominals

[thorn bay]

z table is precomputed values, you can use a scientific calc instead of course

[thorn bay]

its obvious though, for binomial {X>=176} and{X > 175} are the same event, for Gaussian they arent so you pick the middle value

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[chilly blaze]

which is correct im so so so confused

[chilly blaze]

these 2 formulas are different righr?

which is right 😭

[limber minnow]

Whats the value of n in the first formula?

[viral merlin]

right one is absolutely correct , the first img shows similarity with shortest distance between a point and a line , but yeah whats n ?

[limber minnow]

Is it a perpendicular vector to the plane? If so then the inclusion of it in |n•PQ|/|n| should basically be accounting for the difference of Q being "any" point in the left, and Q being a point on the perpendicular line in the right

Look into the relationship between the dot product and cosine

[ionic pendant]

those are the same formula?

[viral merlin]

i dont get it , if its normal it would be right no?

[ionic pendant]

n is presumably any normal vector to the plane

[nimble niche]

Both are

One is in vector form and one is in cartesian

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@chilly blaze Has your question been resolved?

[chilly blaze]

how tho

[chilly blaze]

yes

[chilly blaze]

but like

doesnt the dot product give you ax+by+cz?

the other formula has a +d?

[ionic pendant]

they prove the second formula starting from the first formula, in the second image

[nimble niche]

The second one is from the first

So its essentially just the first

[chilly blaze]

oh so i can jus remember the first one

also like

[viral merlin]

one is solved the algebraic way the other vector , both are interconvertable

and handy

[nimble niche]

Yeah but if you need to give answers in cartesian form second is useful

[chilly blaze]

if vector projection is the distance between a line and a plane, if we're asked to find the distance between a line and a point, aren't we technically using the same formula?

[ionic pendant]

since $(x_0, y_0, z_0)$ is on the plane $Ax + By + Cz + D = 0$, it satisfies that plane equation. therefore we have the equation [ Ax_0 + By_0 + Cz_0 + D = 0 ]

[chilly blaze]

uhh why tho cuz dont they both give magnitude

[solid kiln]

cloud

[nimble niche]

.

This doesnt give only mag

[chilly blaze]

what

what else does it give lol

[nimble niche]

Also gives you the normal vec (A,B,C)

And the d, which is the only thing that changes for parallel planes

[chilly blaze]

oh

ok that kinda makes sense

[nimble niche]

Also lets you know if a certain point lies on the plane by plugging it in exactly like with a line

Lots of uses

[chilly blaze]

ohh

ok

also like

this is the formula for distance between 2 lines

am i tripping or isnt that the same formula as distance between a plane and a line

or distance between a point and a line

are they the same formulas 😭

[nimble niche]

It is

Since only the perpendicular dist mayyers5

[chilly blaze]

so there's kjust 1 formula for distance with lines and planes?

[nimble niche]

Matters*

[ionic pendant]

if two lines don't intersect then they lie in parallel planes

so the distance between the two lines is also the distance from any point on one line to the plane of the other

[nimble niche]

i forgot if its the same for skew lines sorry

Lemme google

Ah yeah ofc it isnt lol

Their direction vectors are different

But yeah for the other stuff you can use that formula

For skew lines its different

[chilly blaze]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhh

[chilly blaze]

WHAT

uhh what

[nimble niche]

Dyk what skew lines are

[chilly blaze]

ya

o wait

isnt the closest distance between skew lines just 0

cuz they intersect

😹

[nimble niche]

no

[chilly blaze]

nevm

nvm,

[nimble niche]

Thats intersecting lines

[chilly blaze]

ioops.

[nimble niche]

Not skew lines

[chilly blaze]

oooooops

ok

so hat do we o then

[ionic pendant]

no, skew lines don't intersect, they are precisely the "parallel plane case" i mentioned but excluding parallel lines

[chilly blaze]

how is it different tho

why do we ened a different method of solving??

[ionic pendant]

so this formula works provided you take QP as the distance from any point on one line to any point on the other line, and n as the cross product of direction vectors

[nimble niche]

The vector form of SD betn skew lines is very handy btw keep that in mind

[ionic pendant]

due to the parallel plane property i mentioned

[chilly blaze]

oh

i think thts what ive always been doing

[chilly blaze]

what

what is sd

[nimble niche]

Shortest distance

[chilly blaze]

also if i use the vector projection formula for 2 lines that intersect it should = 0 right

[nimble niche]

Vector projection of one on the other?

No thats not zero

[chilly blaze]

yea

bro what

doesnt vector projection = distance??

if they intersect..

doesnt distance = 0...

[nimble niche]

Use the formula

[chilly blaze]

im so confused

[nimble niche]

No it doesnt

[chilly blaze]

i thought vector porjection formula gives you closest dostance

.

whar the hll

whzttt

[nimble niche]

Unless im getting confused

[chilly blaze]

im dying

[nimble niche]

Wait lemme reread

[ionic pendant]

if you use this formula for the distance between intersecting lines, then QP and n will be perpendicular and you will get 0

[chilly blaze]

oh

oml

okay

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[chilly blaze]

and just to clarify then that formila (vector projection) gives you the shortest distance?basically distance formula for lines and planes?

ops

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[pure fulcrum]

how do i graph the stuff like sohcahtoa

[pure fulcrum]

im stuck on where to find a period and how to label the graph

or what a period even is

also what is the amplitude and how do i find that

[north girder]

lets see

are you graphing only the left side?

[pure fulcrum]

2 periods

i think from 0

[north girder]

okay

well i like to start with the point on the y axis

[pure fulcrum]

yes

[north girder]

yk where it starts?

[pure fulcrum]

cos(0)=1 so 1

[north girder]

yes yes

so youre familiar with the unit circle right?

[pure fulcrum]

yup

just did a few problems on it

im doing a whole semester review

and i am so lost with a few units

[north girder]

Okay my dms are always open

but ok

[pure fulcrum]

alr ty

[north girder]

So the period is defined

using the formula [(2pi)/B]

Use

[pure fulcrum]

would b be (2theta) 2 being b

[north girder]

yes

[pure fulcrum]

alr

[north girder]

Acos(Bx) + C

[pure fulcrum]

period is pi

[north girder]

Correct

which means it gets squeezed

so instead of graphing the point (pi, -1)

[pure fulcrum]

so if the number is below 1 it gets stretched if its above it gets shrunk

[north girder]

you would graph ((pi/2), -1)

Correct

thats really the ides with periods

[pure fulcrum]

when would the graph be on the y=0

pi/4 in this case?

[north girder]

yes

you got that

[pure fulcrum]

alr i got a big W

cause i went to 2 pi

2 periods

[north girder]

Ya

[pure fulcrum]

but im not completely understanding how

so it starts at x=1

y=1*

then goes down

if i were to use a unit circle how would i use it to help

cos = x on the unit circle

[north girder]

I mean youve been using it

[pure fulcrum]

yes

why would the line go through y=0 at pi/4

[north girder]

Because of the (2x)

[pure fulcrum]

ohhh

[north girder]

If you plug in pi/4

you get 2(pi/4)

which simplifies to cos(pi/2)

and whats cos(pi/2)?

[pure fulcrum]

which would be 0

[north girder]

right

[pure fulcrum]

so thats why pi/2 is -1

vice versa

[north girder]

right

[pure fulcrum]

ahh that makes so much more sense

okay

now what is the amplitude and what does it do

[north girder]

thats not part of the question is it lol

uhm

[pure fulcrum]

ive got this if it helps

[north girder]

oh oj

ok

It changes the range

So, whatever answer you get from sin(4x)

you multiply by 3

which changes the range

[pure fulcrum]

so its just 3 up or 3 down more?

wouldnt that be range?

cause domain is x axis

<-----------> domain

[north girder]

sorry it chances the ragne

meant to say range

[pure fulcrum]

alr just making sure

now since its sin

well be starting at 0

cause sin of 0 is 0

wait how is cos of 0 1?

[north girder]

Unit circle

And yes you are right sin(0) = 0

[pure fulcrum]

cos is x on a unit circle

so x= 0 it would be -1 or 1 right

also i think i finished, its like 2 s's sideways and period is pi/2 amplitude is 3

reflection of the x axis

[north girder]

yep

[north girder]

mhm

[pure fulcrum]

alr im figuring it out

like the way it works from a parent graph

last one

for now

i have analytical trigonometry next and idk what im doing there

[north girder]

loll

This is a horizontal translation

[pure fulcrum]

crud

wait

lemme guess

or think i know what im doing

so first itll be stretched by 2

second needs to be moved right 2 cubes

[north girder]

left

[pure fulcrum]

left alr

[north girder]

(x-b) to the right (x+b) to the left

[pure fulcrum]

alr alr

now it looks like a biger w

[north girder]

nice

[pure fulcrum]

period is 2pi

amplitude 2

[north girder]

yes

[pure fulcrum]

left 1/4 of a period

[north girder]

so the range is [-2, 2]

[pure fulcrum]

ah i get it now

dont get these

[full dock]

Establigh

[pure fulcrum]

its my teacher 😮‍💨

[north girder]

uu

[pure fulcrum]

i think i know b

i dont know 100 % but isnt cot -1 = tan

[north girder]

so

ok i got the first one

[pure fulcrum]

what do i do first

[north girder]

uhm

i dont think so

There's (1 + (cotx)^2) = (cscx)^2

So

I used to do these all the time

The trick is to change everything into sin and cos

So csctheta = 1/sintheta

cotx = cosx/sinx

and go from there

[pure fulcrum]

alr

yo i got a whole cheat in my calc of all idenitys

nvm

i dont

shucks

[north girder]

How are ya doin

[pure fulcrum]

2(1/sin)-2(cos/sin)(cos)=2/csc

[north girder]

chance that other csc too

[pure fulcrum]

i was told by my teach we can only change one side at a time

[north girder]

uh

Why is that?

Cuz mathematically not sure where that comesf rom

[pure fulcrum]

idk

would it just equal sin?

[north girder]

2sinx yes

[pure fulcrum]

/2

[north girder]

(2/1/sinx) = 2sinx

[pure fulcrum]

oh yea

[north girder]

so

actually can you close this chat

were gonna get banned

[pure fulcrum]

why?

.close

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[vague summit]

Hi I need help with this question

[dark sundial]

have you studied the quadratic formula?

what have you tried?

[vague summit]

I tried plugging into desmos

I’m trying smth else right now

Finding if d> 0

[dark sundial]

you have to use the formula x=(-b+sqrt(b^2-4ac))/2a

[round imp]

Do you know the condition for d such that there is only one real root?

[dark sundial]

to find the roots

[vague summit]

Greater than 0

[dark sundial]

the points where the curve touches the x axis

[round imp]

Thats for 2 real roots

[vague summit]

Oh

Sorry

Equal to 0

I have 900 - 4ac = 0

[dark sundial]

a is -9

just isolate the c

[round imp]

Yes

Nice

[vague summit]

900 - 4 (-9)(c) =0

[dark sundial]

isolate the c in order to get the answer

[round imp]

now can you solve for c?

[vague summit]

Subtract 900 both side

[dark sundial]

no

[round imp]

mhm

[vague summit]

Oh

?

[round imp]

It works

[dark sundial]

I saw it now

[round imp]

36 c = -900

[vague summit]

Not -900?

[round imp]

O shit

yes indeed

[vague summit]

Ooo

I got -25 = c

Do I have to memorize the d= 0 stuff for the sat😞

[dark sundial]

I'm not familiar with SAT

but as far as I know

[round imp]

d = 0 if theres one real root only

[dark sundial]

it's a university entrance exam

[round imp]

I advise you to remember

[vague summit]

👍

[dark sundial]

so in this case, yes

[vague summit]

Thank you

.close

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Channel closed

Closed by @vague summit

Use .reopen if this was a mistake.

[round imp]

np

[vague summit]

😭struggling again

I’ve written 16ax + 4bx / 64x^2

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[vague summit]

🧐

[vague summit]

I’ve written 16ax + 4bx / 64x^2

Ohh wait a = 64

Do I still distribute the x to b?

[mellow tendon]

No need for x to be present when you are calculating -B/A

[mellow tendon]

How did you get to that?

[vague summit]

In the equation

It says 64x^2

[mellow tendon]

Oh ok I understand, it is the a coefficient of the quadratic equation not the value of the variable a
You need to clarify that : )

So as you just said coefficient doesn't include x in them so just remove the xs
Then complete your work

[vague summit]

Alr I got 16a + 4b / 64

[limber island]

X^2- 4/64(4a +b)x + ab/64=0

[mellow tendon]

Yeah

So this would be equal to what?

[vague summit]

K(4a +b)

[limber island]

Just think

[mellow tendon]

Ok right can you get k now?

[mellow tendon]

She doesn't want to factor it, she wants to solve for k in the question.
This won't help very much except when it comes to factoring

[limber island]

You can get that

[vague summit]

I got k = 1/4

😞

The key says it’s wrong

How to get that

[mellow tendon]

Yeah but as a helper don't give the solutions right away

[limber island]

Yup

[vague summit]

??

[mellow tendon]

(16a + 4b)/64 = k(4a+b)

[vague summit]

Yes

I have that

[mellow tendon]

What did you do after that

[limber island]

Take 4 common out then

[vague summit]

Ohh

[mellow tendon]

4(4a+b)/64 = k(4a+b)

[vague summit]

Oh

I thought u meant to simply it

[mellow tendon]

So I think you can see it now

[vague summit]

I got 4a+b/16 = k(4a+b

[mellow tendon]

Divide both side over (4a+b)

I understand what you wanted to say, but it is better to write in proper notation to prevent confusion
So put (4a+b) in parentheses or brackets

What have you got?

[vague summit]

I got

[limber island]

But this could be solved by equating -B/A=k(4a+b)

[vague summit]

4a+ b / 64 (4a+b)

= k

So now I cancel out

Them

[mellow tendon]

Yeah but it was 16 not 64

[vague summit]

Whoops yes

Sorry I did right that on spper

[mellow tendon]

Yeah which is what she did

[vague summit]

Paper

[mellow tendon]

So what have you reached to

@vague summit ?

[vague summit]

I got

1/16 =k

[mellow tendon]

Yeah

[vague summit]

Ayee ty

.close

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Closed by @vague summit

Use .reopen if this was a mistake.

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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wooden hamlet]

How are any of these correct

[wooden hamlet]

Because I know this is arthmetic so it has to either be 3 or 4 but then the equestrian is not completed for either of those so they r also wrong

Because they are multiplying and your not supposed to multiply and it only includes one number not two

[waxen drift]

Are we focusing on 1c?

[wooden hamlet]

Yea

Like these equations are not set up correctly

[verbal gulch]

,rccw

[waxen drift]

Yeah, the instructions says to select all expressions that work
So some of them wont

[wooden hamlet]

I know this is arthmetic. A recursive equation for that is f(n)=f(n-1)+GF for when n is equal to or greater to the second value

I know 1 ,2 ,5,6 don’t work because it is saying for when n is equal to or greater to 1 and it is supposed to be 2 not 1

Like I think 3 works but besides that I don’t know what other one could work since 4 is the same as 3 but with the wrong rate of change

[waxen drift]

Aside from actually looking like the general formula, have you also plugged in numbers to confirm your thought process?

[wooden hamlet]

No idk how I would know what number to plug in

[verbal gulch]

you wrote up a table of n and D(n), yeah?

[wooden hamlet]

And if another one is right then I’m screwed because I thought for when n is somthing it is the second step but four of these are equally the first step so idk what I’m doing wrong for all my other work

[verbal gulch]

if you try all of the suggested D(n)s on your table, they should match

the ones that don't match are definitely not correct

[waxen drift]

So you basically replace the variable with a number, preferably one thats on your table
And if it matches and its a good explression

[verbal gulch]

that's not to say that the ones that match are definitely correct, but that's at least a start

[wooden hamlet]

I’m seeing six works but why does it work if it has the wrong n equals to

Because now I’ve been doing this whole unit wrong

And I’m panicking my final is on Wednesday