help-19

[supple shuttle]

But dont know What to do after would appreciate help

[shut flower]

You can do something similar with the denominator

Like you did with the numerator

[supple shuttle]

U want me to factorize the numerator?

Hmm

[shut flower]

Yes

The denominator

[supple shuttle]

Do we make times 1/10^15 with 10

so it becomes 10/10^16

?

10/10^16 - 1/10^16

[shut flower]

Yes, that's a good idea

[supple shuttle]

What did u think of?

[shut flower]

$10^{-15}-10^{-16} = 10^{-16}(10-1)$
It's essentially the same

[clever fjord]

d

[supple shuttle]

Ojhhh so doing it into fraction was unessecery

[shut flower]

Yes

[supple shuttle]

999 x 10^15 / 9/10^16

How do i solve that

[shut flower]

The /10^16 in the denominator becomes *10^16 in the numerator

And you can divide 999 by 9

[supple shuttle]

so keep change flip

get it

thank you

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[graceful parrot]

Am I Doing number one correctly

[mystic saffron]

https://tenor.com/view/tank-gif-26631637

[quasi sparrow]

,w plot sin(x) + cos(x) for 0<x<2pi

[clever fjord]

Results provided by WolframAlpha

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

Requested by rie.mann

[quasi sparrow]

Don't round your answer

You should know the exact values

[graceful parrot]

Oh I just used a calc

[west hull]

Use the unit circle

Or, if you already recognise the decimals, use …

[graceful parrot]

So just do square root 2

Then

?

[west hull]

Yes

[graceful parrot]

.close

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[graceful parrot]

Thank

Tank

Tank

Idk what the emoji is

.close

.reopen

[odd edge]

✅

[graceful parrot]

Am I dojng 2 correct? It’s decreasing from -3,3 and 3, inf and increasing on -inf, -3

[odd edge]

@graceful parrot Has your question been resolved?

[graceful parrot]

Positive negative positive

Sorry

[odd edge]

@graceful parrot Has your question been resolved?

[graceful parrot]

.close

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[graceful parrot]

.reopen

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✅

[graceful parrot]

Problem 4, how would I find my inflection points on the y axis? I know they are when x is equal to 0,2, and 4

I think I would plug it into the original equation but it doesn’t seem right, all I get are positive numbers

I get 0, 100 and 2, 356 and 4, two thousand something

[wanton bison]

hi

[late dust]

(0, 100) and (2, 356) are correct

[graceful parrot]

Really???

Wait

[wanton bison]

x=4 is also correct

f(4) is just messed

[graceful parrot]

Is it 4, 612

[wanton bison]

yes

[graceful parrot]

Oh they all seemed big so it seemed weird to me

[late dust]

3*4^5 is big

[graceful parrot]

Thank guys

Thank

Tank

Tank

. Lose

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[wanton bison]

serial number

[graceful parrot]

8263728648383

.reopen

[odd edge]

✅

[graceful parrot]

Is this my critical number for problem 5?

[wanton bison]

it's asking for y'=0 btw

that's wrong the prelast factorization

You should calculate y'=0

and then use the 2nd derivative to see if these sol. are local max/min

you are such a serial number

[graceful parrot]

Yea I forgot my ba

[wanton bison]

ba

[graceful parrot]

But X would still be 0 right

Or no

[wanton bison]

wdym

[graceful parrot]

If I’m using second derivative don’t I find critical for second

[wanton bison]

huh no

[clever fjord]

bacc the sigma😔🤞

[wanton bison]

the critical numbers of a function f is where the function has slope 0

so basically the solutions of f' = 0

and now additionally you should verify with the 2nd derivative if these solutions are local max/min

[graceful parrot]

I see

Thank

[wanton bison]

No P

[graceful parrot]

Are my critical points 1 and negative 1

Oops

Uhh I kept one x in

[wanton bison]

what about x = 0

factor

[graceful parrot]

From where?

[wanton bison]

-x²+2x = x(2-x) = 0

[wanton bison]

this math aint mathing

-x²/-2x = x/2

[graceful parrot]

X is 0 and 3

2

Not 3

[wanton bison]

yes serial number

[graceful parrot]

Okok sorry what do I do once I have my critical numbers

And thank u

[wanton bison]

you wanna see if x=0 or x=2 is local max/min so you use the 2nd derivative to verify that

And you're welcome

[graceful parrot]

I see thank

[wanton bison]

no pro

[graceful parrot]

In order to find the y values for that would I plug the critical numbers from first derivative to original?

I plugged it into second derivative

[wanton bison]

yes

[wanton bison]

yes to see if these are max/min

the original function tells you what value the max/min has

[graceful parrot]

And got y’’(0) = 2 and y”(2) = e^-2(-2)

[wanton bison]

yea seems about right

[odd edge]

@graceful parrot Has your question been resolved?

[odd edge]

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[wintry bison]

this isnt really math well kinda more of a finance question that idk what the math to do with

[wintry bison]

bond you follow has face value 100, has a coupon rate of 5% (paid once a year) and matures in 5 years
current and the expected yield curve
Calculate the price of the bond based on the current yield curve? (use 2 decimal digits)

Hello?

<@&286206848099549185>

[nimble pumice]

?

oh uh sorry

[odd edge]

@wintry bison Has your question been resolved?

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[long barn]

help

[long barn]

help!

i need to learn it aswell

[urban salmon]

Well for y and x intercepts
There's a y axis and an x axis

Do you know what "intercept" means?

@long barn Actually first Is there specific ones you need help with?

[long barn]

all of them

do you think you could hop in a call with me also? i dont think ill be able to learn it just off of my sight rn

[urban salmon]

Okay well it's vocab so if i ask if you know what something means i am not trying to be rude or sarcastic

Also no I can't do voice chats you'd have to wait for someone else

[long barn]

alrighty

[urban salmon]

Do you want to wait then?

[long barn]

i guess so

but

before then

could you teach m,e how to find]

domain

and range

[urban salmon]

Well both of those are just definition based

Do you know the definition of either?

If not do you know what your inputs and outputs are?

[odd edge]

@long barn Has your question been resolved?

[long barn]

yea ik

both

but no
clue how to find

[urban salmon]

Is there a specific question you're trying to find them?

[urban salmon]

Okay there is a formula for finding domain* but i don't remember it

However it doesn't seem like you need it

Where on the graphed equation does x stop on the negative side?

[odd edge]

@long barn Has your question been resolved?

[jaunty sundial]

@long barn Ping me if u need help I can explain it

[long barn]

yes

[jaunty sundial]

What do you need to know about domain

[long barn]

how do i find it

on stuff like that

[jaunty sundial]

Like this right?

Ok you kinda wanna ignore the y position for domain

You are looking at where there is an x value along the graph. So in this scenario, we see that the minimum or least x value is -5

Do you know about the open or closed ending to signify whether it includes that number or not?

[odd edge]

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[hollow thunder]

Assume that f'(c) = 3, Find f'(-c) if (a) f is an odd function and (b) f is an even function.

how would I do this?

[sand night]

do you know even and off function idnetites

[hollow thunder]

these?

[sand night]

ye those

[hollow thunder]

well then would that just mean f(-c) is -3

[sand night]

so for even

[sand night]

not necessairly

thats true for an odd function

but not for an even

[hollow thunder]

how do we know which they are talking about

because (a) is odd, (b) is even

[sand night]

wait is

f is an odd function and (b) f is an even function. or does it say f' in problem

bc we gotta do something else first

[hollow thunder]

pretty sure this one is b

as in its labeled that

i just didnt know if it correlated with this problem

A looks like this

im confused

[nimble blaze]

start with those even and odd function definitions
differentiate those

[sand night]

what is the deriative of an odd function
what is the deriative of an even function

then you can apply the rules sorry

[hollow thunder]

they're swapped

[sand night]

wdym by that

[hollow thunder]

like derivative of an odd function is even?

or am i wrong

[sand night]

ye

thats right

so

[hollow thunder]

yeah so then like

[sand night]

if f is odd

then its derative is even

and you know f'(c) = 3

[hollow thunder]

then they're both 3

[sand night]

no

remember

deriative of even function is then odd

[hollow thunder]

yes but f'(3) as an odd is still = 3

right?

[sand night]

ye

if f is odd

then

f'(-c)=3

so now if f is an even function

what is the derative of even function

[hollow thunder]

odd

[sand night]

ok so then the rules of odd function are

f'(-c)=-f(c)

right

[hollow thunder]

yes

[sand night]

so what is

-f(c)

in this case

if f(c)=3

[hollow thunder]

f'(c)=3 ?

[sand night]

ye

oops

if f'(c)=3 what is -f'(c)=3

[hollow thunder]

wait i still dont get if its going to be odd or even

[sand night]

deriative of even function is odd

remember

[hollow thunder]

ok but an odd function is even

so which do i use

[sand night]

we already found if f is odd

so we are doing even now

[hollow thunder]

so were looking for two f'(c)

one even and one odd

[sand night]

.

you already solved when its odd

[hollow thunder]

so i needed to find both even and odd to come to f'(-c)=3

[sand night]

you already said if f is odd

then f'(-c)=3

because deriative off odd function is even

then therefore
f'(-c)=f'(c)=3

now just apply the same logic to when f is even

bc
f'(-c)=-f'(c)=?

[hollow thunder]

-3

[sand night]

ye

so if f is odd, then f'(-c)=3
then
if f is even, then f'(-c)=-3

does that make sense?

[hollow thunder]

so i had to solve for each value when its odd or even

okay yeah

thanks

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[tired ginkgo]

is this a fair proof?

[tired ginkgo]

oops made a mistake one sec...

heres the fixed version

does this make sense"

?

[latent scaffold]

For the backwards direction, what you've shown is that A <= AnB.
Even though it is more or less trivial, I would also add that AnB <= A, to then conclude that A = AnB.

It's also weirdly written. You pick x in A and then say that all x is in B.
You can just say that the x you picked in A is also in B.

[tired ginkgo]

Does x epsilon A means for all x in A or just a singular x?

I'm not the greatest at proofs so bare with me

[latent scaffold]

When you say "Let x in A" it means "Pick some x in A"

"epsilon" just means in

[tired ginkgo]

Hmm should I write it as "for all x" then?

[latent scaffold]

No, because the goal is that if you show that for some arbitrary x in A, x is in AnB, then this works for any x.

[tired ginkgo]

ahh I see that makes a lot of sense

thank you for your help brother

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[mellow current]

ok wait i got still 110 - n = n^2

and didnt know where to go next ;-;

[fast halo]

set one side to zero and solve the quadratic

[mellow current]

wait do i need to get rid of the roots-?

[fast halo]

yes

[mellow current]

was this section ok?

so now i would have -n^2 - n + 110 right?

[fast halo]

yes

[mellow current]

alright! do i leave things negative or make em positive?

[boreal crag]

= 0?

[mellow current]

oh yes!

[brittle beacon]

("making them positive" might make your life a bit easier)

[mellow current]

alr alr

ok! so i got (n + 11)(n - 10)

meaning that n is either -11 or positive 10?

wait but how is the answer only 10 guys

ty cats ❤️

[brittle beacon]

Be careful, was gonna mention as well-

[mellow current]

oh i messed up didnt i

[brittle beacon]

Well, you didn't, but-

[brittle beacon]

You need to check in the original equation you had, so put the n = -11 and n = 10 into here

And check if it holds, squaring can introduce extra solutions sometimes (in general you should always check the solutions you find!)

[mellow current]

ohhh that makes sense now

so sometimes theres only one solution, but you get two answers and you HAVE to check over

okok ty!

[brittle beacon]

Yep, it's also a good habit, it makes sure you haven't messed up along the way (and also forces you to check the original question you were dealing with, sometimes it's easy to forget that really what you're doing was part of a larger problem!)

[mellow current]

oh smart- ok ill remember this if i ever have a paper 1!!

ty so much chartbit!

love from me and all of the students in maths ❤️

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[glossy cairn]

why is this A?

[orchid torrent]

No speed in the horizontal direction before -> no speed in the horizontal direction after

[glossy cairn]

ic

thx

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[latent ether]

Could someone help explain how my professor simplified (X+8/2)^2 ? Just wondering how the dominator turns into a 4 while X+8 stays squared, thanks ! (:

[true lodge]

Because

U apply the POWER 2

to both numerator and denominator

( (x+8)/2 ) ^2

is Basically just this

(x+8)^2 / 2^2

If u simplifiy the denominator u get 4

hope that helps

[latent ether]

OHH okay, just visualized it. This helps, thank you !!

.close

[true lodge]

Welcome!

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[mental epoch]

hello,

i need help in forming an equation from a graph, specifically finding the tan and cot equation and the features of them. For the questions, I am provided a graph and have to identify the equation (tan, cot, sec or csc)

[mental epoch]

i want to be able to understand how to find the functions features, which includes the a, k, c & d values

i also want to know how to determine the trig function that should be used.

for example, how would I know that i would have to use csc and not sec, or vice versa?

and the same for tan and cot

in addition, i would appreciate to be helped in the tan/cot functions than the sec/csc!

thanks!

<@&286206848099549185>

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[opal path]

im completely blanking for some reason

i was going to sub in the sin/cos related t functions

but how the hell does the dx and bounds work

[lyric dust]

let t= tanx/2

if x = pi/2, t = tan(pi/4) = 1

if x = pi/3, t = tan(pi/6) = 1/sqrt3

for your differential, dx = 2dt/(1+t^2)

and just sub everyhting into ur integral and crunch algebra

[opal path]

o

oohh

ah makes sense

wait

[opal path]

i get (1+t^2)/(2-2t) dx though

its flipped

somehow

[lyric dust]

huh

[opal path]

the t^2 part has to be on the bottom for me to make it dt

[lyric dust]

$t=tan(\frac{x}{2})$

[clever fjord]

Dootud

[lyric dust]

rearrange if u want, to form
$2 arctan(t) = x$

[clever fjord]

Dootud

[lyric dust]

diff both sides respectively

and u get 2dt/1+t^2 = dx

[opal path]

wait i think im high

yeah im high

it becomes 1/1-t dt right

[lyric dust]

what

whats $\frac{d}{dx}arctan(x)$

[clever fjord]

Dootud

[opal path]

no like in the integral

[lyric dust]

oh

[lyric dust]

do u mean the entire integral?

no

[opal path]

wait no i fked it up why is this indefinite

[lyric dust]

denominator should be t^2 + 2t

[opal path]

ok i think i just forgot how to do this

bounds become 1 and root3/3

the 1/1+sin-cos

becomes

(1+t^2)/(2-2t)

and dx becomes 2/(1+t^2) dt

yes/no?

oh

wait

i fucked up a sign

[lyric dust]

what

[opal path]

yeah got it i had a - instead of a + oopsies

[lyric dust]

thats still not right?

ur integral should become

[opal path]

1/t(1+t)?

[wind coral]

Can anyone pls help me

[lyric dust]

yeah thats correc

[opal path]

ye i fked up a sign in my work lol

[opal path]

did you even bother to read #❓how-to-get-help

[wind coral]

Uh

[opal path]

.cose

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[loud rune]

I'm back 😅 this time I have a question about long division with polynomials.

[loud rune]

I tried doing this with long division, but I think I may be missing a step. Do I need to create a 0 for the 1st degree of x in the dividend to make this work?

oh wait, this may just be an issue with my computation

nvm, still not sure what i'm doing wrong

[vapid lantern]

u just need make 4x = 256x^2 right

so that u can cancel the 2nd degree term

the only way u can do it is by multiplying 64x

[loud rune]

so, i got 64x+48 and then there's the remainder of 96

which would be 96/4x-3

so i think the answer is 64x+48+(96/4x-3)

i was adding the 96 back without that when i came to 64x+144

but that answer with the remainder isn't being accepted either 😐

AH! it's the quotient that's the slant asmpytote

i see, don't need the remainder - that was the issue

figured it out! thanks 🙂

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[outer wadi]

what exactly is part a asking?

[outer wadi]

transforms [0,1,1]^t to what?

are they just asking me to find A?

im not sure why the [0,1,1]^t would be there then

[lofty spade]

that’s what they’re asking you

[lofty spade]

they’re asking you what the image of [0,1,1]^t is, under A

[outer wadi]

oh ok

so find A

then find the image

[lofty spade]

you don’t even need to find A, you can do it just by linearity

[outer wadi]

whats that 😭

ill try looking for it in my book

[lofty spade]

maybe you know it as distribution of matrix multiplication over vectors?

[outer wadi]

hmm I think I see

I dont see 😭

this?

or like

represent Ax as

its A1x1 + A2x2

where A1 is column 1 of A

[tacit wasp]

No no

$$A(\lambda_1 \vec{v_1} + \lambda_2 \vec{v_2}) = \ldots$$

[clever fjord]

Alberto Z.

[outer wadi]

hmmm

[outer wadi]

Im a bit confused are you able to explain a bit more

@tacit wasp

A is our 2x3 matrix right

but whats lamda and v

[tacit wasp]

Lambdas are some numbers you have to find out

v1 and v2 are two vectors

[tacit wasp]

But if you don't know what there should be in place of ... you can't do the exercise

[tacit wasp]

So first you should check if you're 100% sure you haven't been taught this linearity property of matrix product

[outer wadi]

I dont believe so I went through the book and hasnt mentioned it but we may have gone over it in lecture conceptually

[tacit wasp]

A(u + v) = Au + Av

[tacit wasp]

Have you seen this at least?

[outer wadi]

yeah

right distrbutivity

[tacit wasp]

Alright, that's it

[tacit wasp]

Yeah exactly

You can also pull out constants

[outer wadi]

whats the . . .

[tacit wasp]

$$A(\lambda_1 \vec{v_1} + \lambda_2 \vec{v_2}) = \lambda_1 A \vec{v_1} + \lambda_2 A \vec{v_2}$$

[tacit wasp]

Well you should figure it out on your own 😅

[clever fjord]

Alberto Z.

[outer wadi]

I think I'm just a bit confused on the idea of what we are doing 😭

Were distrubuting A over some linear combination of v1 and v2 right

ill try just looking up on linearity

[tacit wasp]

I have to go now, I let you with other helpers

If you want you can tag in like 5 mins if no one helps you now

[odd edge]

@outer wadi Has your question been resolved?

[quiet kettle]

Hello

Do you know linearity

[outer wadi]

I just found the part of the book that talks about it

[quiet kettle]

Oh good

[outer wadi]

example 3.3 goes over it so im gonna try and use that to solve it

though I dont really understand why it works

but I'll probably look up a visual explanation after

[quiet kettle]

Ok the solution is similar to example 3.3

So you’re thinking right

You can do it in similar way

After you solve it, I’ll tell you what linearity is more specifically

[odd edge]

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[craggy anchor]

im haley on my pc

but can someone explainwhat im doing wrong

and how my answer is different

[lofty spade]

can you send your working?

[craggy anchor]

its a calculator problem

so prehaps im typing something wrong

[rustic night]

I don't think it's a calculator problem. I verified by calculating it myself and arrived at the same number. What is the given answer?

[onyx cloak]

this ig

[craggy anchor]

yea

that would be the answer

[rustic night]

,w D[x*sin(3x^2 - 2), x] at x = 7

[clever fjord]

Results provided by WolframAlpha

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

Requested by eththorn_98569

[rustic night]

there might be a rounding error in the actual answer

[craggy anchor]

i see

thank you for the help

[proven crystal]

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[vivid kelp]

Find the locus of point of intersection of two lines xtan(theta) -3ysec(theta) - 5 = 0 and xsec(theta) + 4ytan(theta) + 2 ; where theta is parameter

[signal flame]

ur cooked it 4 am

[vivid kelp]

not where i leave

[vivid kelp]

can you help ?

[signal flame]

im cooked...

[vivid kelp]

ooh

[signal flame]

how are you

[vivid kelp]

fine you are going to help or not ?

[signal flame]

why cant u just graph it

[vivid kelp]

its a subjective solution they want the soln

[rustic night]

Show working, where are you confused?

[warped glacier]

try doing $(x \tan t - 3y \sec t)^2 - (x \sec t + 4y \tan t)^2 = 5^2 - (-2)^2$

[clever fjord]

higher's secret twin brother

[warped glacier]

using the identity $\sec^2 t - \tan^2 t = 1$

[clever fjord]

higher's secret twin brother

[odd edge]

@vivid kelp Has your question been resolved?

[vivid kelp]

does not work

i get stuck there t is not getting eliminated

[warped glacier]

show your working then

[vivid kelp]

my working is literally not here i serached the eqn on wolfram and got this

[vivid kelp]

pls help

[warped glacier]

idk man

[warped glacier]

note that $x^2 \tan^2 t - \sec^2 t = x^2 (-1) = -x^2$

[clever fjord]

higher's secret twin brother

[warped glacier]

similarly, you can get $-16 y^2 \tan^2 t + 9y^2 \sec^2 t$ in terms of $y^2 \sec^2 t$ and a constant

[clever fjord]

higher's secret twin brother

[warped glacier]

then it's a quadratic equation

in the variables (x tan t) and (y sec t)

[odd edge]

@vivid kelp Has your question been resolved?

[vivid kelp]

oh thnx

[odd edge]

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[heady plover]

what thms does this question might use? It looks feyman but I'm not sure how to use that

[true lodge]

is that calc 2?

[odd edge]

@heady plover Has your question been resolved?

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@heady plover Has your question been resolved?

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[lost rain]

hello @here
im havin trouble with this quest.
its like the answer for x when inputted in the 4cos^2(x), it should satisfy the range 0< and <1

[odd edge]

@lost rain Has your question been resolved?

[sterile blaze]

what have u tried

[odd edge]

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[lost rain]

yea

[lost rain]

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[mystic saffron]

Can someone tell me why the instantaneous rate of change is a ratio of dy and dx
By expressing it into a ratio are we essentially 'comparing' the both changes?

[dense sonnet]

yes its a small (infinitesimal) change in the y component divided by a small (infinitesimal) change in the x component

[mystic saffron]

Alright, thank you 👍

.closw

.closw

. close

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[robust wyvern]

Let (C) and (C'l be two circles with the respective diameters [TA] and [TB]. The circles are tangent interiorly at point T. Qe design by M and N two distinctuve points of (C) \ {T}.
(TM) and (TN) cut the circle (C') respectively at P and Q.

[robust wyvern]

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[ancient rampart]

Arranging Objects with Repetitions:

If I have ( n ) objects, with ( p ) of them alike (one kind) and ( q ) of them alike (another kind), the number of distinct arrangements is:

[
\frac{n!}{p!q!}
]

Distributing Distinct Objects into Groups:

If I have ( m+n+p ) distinct objects and I want to divide them into 3 groups of size ( m ), ( n ), and ( p ), the number of ways to do it is:

[
\frac{(m+n+p)!}{m!n!p!}
]

why does this work?

[clever fjord]

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

[ancient rampart]

i kinda get the first one second one not sure

[weary pelican]

anytime you count "N!" ways to rearrange a total of N objects, it's like you treat each object as distinct from the others

[ancient rampart]

the second one

[weary pelican]

yeah I'm explaining for any instance of that

here N can be equal to m+n+p for example

[ancient rampart]

but they are distinct tho

right

[weary pelican]

ok for the second one

if you want to separate into 3 groups

it's just as if you would rearrange them in a line from left to right

the first m go into the first group

the next n go into the second group

and the last p go into the third group

but

again, you're overcounting

by the amount of ways the first m objects can be rearranged between each other

and by the amount of ways the next n objects can be rearranged between each other

etc...

for any rearrangement from left to right

there are exactly m!*n!*p! configurations

that lead you to the formation of the exact same groups

[ancient rampart]

um makes sense i guess

[weary pelican]

so let's recap

for each group distribution, there are exactly m!*n!*p! rearrangements that can form this distribution

there are (m+n+p)! rearrangements in total

[ancient rampart]

yeah and m!n!p! are internal arrangements which are counted in (m+n+p)!

so we disregard the m!n!p! internal arrangements from the grouping arrangements right

[weary pelican]

yeah if you count a total "A" of rearrangements

and "B" possible rearrangements for a single group distribution

you'll count "A/B" different group distributions

[ancient rampart]

both are similar concepts we divide by the repetetion to disregard overcounting

makes me think why does that even work mathematically i get the logic

shouldnt we substract the repetetions from total?

i know we dont but just wondering

[weary pelican]

if you really wanna see it as subtraction

for each group distribution

[ancient rampart]

we are removing the overcounting right

[weary pelican]

there are m!*n!*p! possible ways to reach it

so we are overcounting by m!*n!*p! - 1

for THIS distribution

if you call X the number of different group distributions

you're then ovecounting by X* (m!*n!*p! - 1) in total

so, (m+n+p)! = X + the amount we're overcounting by

(m+n+p)! = X + X*(m!*n!*p! - 1)

doesn't change much from "division"

[ancient rampart]

okay

thanks

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[cursive hedge]

i need help with 6b
my reasoning for it is that you and your friend's calls are independent but the total amount of calls sum to 4, so it can be that

you call 0 times, they call 4 times
you call 1 times, they call 3 times
....
you call k times, they call 4-k times etc
then it should be the sum of all possible ways to call

[gray bronze]

yeah that looks right

[cursive hedge]

alright, thanks

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[wheat plank]

I think I know what I'm doing, just need to check my answers and clarify.
For 1a. I got -.5
For 1b. I got -e/2 -1
For 1c. The wording is really stupid but I think it means I take what I got in 1a and 1b and plug them into that formula?

[quasi sparrow]

show your work

[odd edge]

@wheat plank Has your question been resolved?

[wheat plank]

Nevermind ig

.close

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[trim leaf]

what am I missing here???

[boreal crag]

Magnitude cannot be negative

[trim leaf]

thank you!

[boreal crag]

np

[trim leaf]

do you happen to know how I can find the phase angle

[boreal crag]

Idk the terminology enough

[trim leaf]

I figured it out, the phase angle is pi since it is -3 in the equation

[boreal crag]

So 180?

[trim leaf]

yeah

[boreal crag]

Nice

[trim leaf]

it made sense after you told me that mag cannot be negative

how do I close the chat?

close

-done

[boreal crag]

.close

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[odd edge]

@sour geyser Has your question been resolved?

[sour geyser]

<@&286206848099549185>

[proper tapir]

Prove that the function is surjective

Using the definition

Please help me man this problem got me tripping balls

[sour geyser]

?

<@&286206848099549185>

[quasi sparrow]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[odd edge]

@sour geyser Has your question been resolved?

[sour geyser]

<@&286206848099549185>

[amber pendant]

i think there might be a typo or smth, somewhere

because $A^{+} A = I \quad \implies \quad \vec{x} , ^{\ast}=\vec{0}$

[clever fjord]

Emily

[amber pendant]

which satisfies $A \vec{x}=\vec{0}$

[clever fjord]

Emily

[amber pendant]

but does not satisfy the minimum..

[odd edge]

@sour geyser Has your question been resolved?

[odd edge]

@sour geyser Has your question been resolved?

[amber pendant]

comment on what i said

do you guys solve gradient=0 for the min?

[sour geyser]

for the stationary point yeh

[amber pendant]

this problem doesnt make sense to me, because of what i told you earlier about A* A = I

i did calculate the gradient also

and x=0 does not even make the gradient=0

[sour geyser]

yeh idk i think theres another definition for the Moore-Penrose inverse

that makes A* A not equal to I

might be like some condition

this question is got me annoyed

[amber pendant]

there are 2, yeah, one is called the left-pseudoinverse

which is used when having equations of the form AX=0

and theres one called the right-pseudoinverse

which is used when having eqs of the form XA=0

but were dealing with the first case here

,, A^+ = (A^T A)^{-1} A^T

[clever fjord]

Emily

[amber pendant]

which indeed makes x=0

doesnt make sense lol

[sour geyser]

does that stationary point make this function like strongly convex or something?

[amber pendant]

dunno, but stationary points are what you gotta determine always when asked to find min/max

for instance, could you calculate the gradient of that expression?

[sour geyser]

of x* or f?

[amber pendant]

the top

lets call it f(x1,x2,x3 ... , xd)

[sour geyser]

the min ?

[amber pendant]

the expression

1/2 |x|² + <b,x>

[sour geyser]

i mean this is the work i've done so far

[sour geyser]

yes

the derivative is here

[amber pendant]

yeah, this doesnt make sense to me

[sour geyser]

which part or just the question?

[amber pendant]

the problem, the given x*

[sour geyser]

oh ok thats fine i dont rly either tbh

[odd edge]

@sour geyser Has your question been resolved?

[odd edge]

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[copper quarry]

no idea how to start

[copper quarry]

part i)

[narrow crypt]

All right

Do u agree that the costheta and sintheta should be treated like constants

[copper quarry]

hm if theta is a fixed angle then yea

[narrow crypt]

All right

Do u agree that the line tangent to the circle and the line OT are perpendicular

[copper quarry]

yes

[narrow crypt]

What is the slope of OT

[copper quarry]

uh im not sure

[narrow crypt]

sintheta/costheta, right?

Recall from y/x

[outer coyote]

water beam, do you know your circle geometry

[copper quarry]

oh is it like this

[narrow crypt]

yes

[copper quarry]

okay

[narrow crypt]

So the slope is sin/cos

[copper quarry]

so tan

[narrow crypt]

Then what is the slope of the perpendicular line

[outer coyote]

you need to use the fact that PT is normal to OT

[narrow crypt]

.

[outer coyote]

do you know how to find the gradient of a perpendicular line?

wait not meant to respond to u k lmao

[copper quarry]

yeah passing thru a given point

[narrow crypt]

Like

What is the slope of OT

[copper quarry]

ohh

its cot right

-cot

[outer coyote]

you might wanna use $m_{\text{perpendicular}} = - \frac{1}{m_\text{initial}}$

[clever fjord]

poudel

[narrow crypt]

Yes

Express in sinx and cosx for simplicity

[copper quarry]

$m_{PT}=-\frac{\cos\theta}{\sin\theta}$

[clever fjord]

water beam

[narrow crypt]

Yes

[copper quarry]

$m_{OT}=\frac{\sin\theta}{\cos\theta}$

[clever fjord]

water beam

[copper quarry]

so whats the next step

[narrow crypt]

What

Yeah

[copper quarry]

yeah?

[outer coyote]

ill give you a hint

find what point T is

and then use point-gradient form to express OT

[copper quarry]

ohh okay

[outer coyote]

that should give you your answer

[mellow geyser]

can someone help please

[copper quarry]

yooo i did it

lets go

[outer coyote]

colddd

gj broski

[copper quarry]

thanks bro

so part ii)

[outer coyote]

💀

[copper quarry]

any ideas

🤣

@outer coyote

this is the solution they provided

but why does that give me length BQ?

i dont get it how thats BQ

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[small sail]

Having trouble with a physics equilibrium problem

[sterile blaze]

could u send the problem as well

[small sail]

I put the X component in term of -T1 and put that into the Y component equation so I could solve for T2

[sterile blaze]

this is.. clearly graded

[small sail]

I just realized what might be my mistake

Is the Y components T1 and T2 just mg?

Yes?

It’s class work

[sterile blaze]

so u should not be asking here

[small sail]

I’m not asking for an answer I’m asking for help catching a problem

[odd edge]

@small sail Has your question been resolved?

[small sail]

So am I allowed to ask for help on this or?…..

Like I’m probably missing some easier way cause I’m trying to solve for t2 first but it’s asking for t1 but I can’t find it

[mystic saffron]

Ofc u can get homework help

[small sail]

Above the guy said no 😬

[mystic saffron]

as long as it’s not an exam or we’re blatantly doing the question for you it’s allowed

[small sail]

It’s homework

[mystic saffron]

the help channels are for teaching

[mystic saffron]

it’s totally fine then if someone explains how to go about the questions

[small sail]

I have like 999 tries per question

[mystic saffron]

yh idk what they were talking ab u are allowed

[small sail]

Ok

[mystic saffron]

Just because it has a due date Orr

[small sail]

Yeah that’s a week long deadline

[mystic saffron]

all homework has a due date I don’t get your point 😭

[sterile blaze]

cool then

[small sail]

Brooooooooo

[mystic saffron]

alr good luck aaron

[sterile blaze]

what happened to u helping lol

[small sail]

I was hoping for help too ngl

[mystic saffron]

Girl help idk how to solve that 😭

Ping helpers

[small sail]

Idk what I’m even doing wrong anymore lol

[sterile blaze]

they sum to mg

[small sail]

Yes

But

It’s not working

[sterile blaze]

i believe in this question u draw a triangle for the forces

[small sail]

Idk if I broke some sort of algebra rule or what

I got the components of X and Y

Then put X in terms of T1

Plugged t1 into the Y component

And tried to solve for T2

But the question asks for T1 first

[small sail]

That’s what I drew tho

The thing on the left

[sterile blaze]

does it not work :/

[small sail]

No like the diagram and components are correct (I think)

I asked my TA to take a look at this before and he just said that the X component on T1 was negative

[sterile blaze]

yea

[small sail]

So I made it negative

But something’s still wrong

Maybe I could put it in terms of t2

[sterile blaze]

what answer did u get

or did u not manage to get an answer

gravity is 9.81 or 10?

[small sail]

9.8 or 9.81 works on this software

I’m using 9.8

[sterile blaze]

your answer was

[small sail]

301.79 for t1

Says it’s off by a factor of two

So I’m messing up the algebra somewhere

Lemme rewrite my work and show

[sterile blaze]

whats your working

anyway, its a lot easier if u used sin rule

[small sail]

I have no idea what that is but here’s my work for T2

[sterile blaze]

rather than breaking it up into x and y components and finding one then using that to find T1

more steps and thus more prone to error

[small sail]

,rccw

,rccw

[small sail]

I find the components and put the X comp into terms of T2

[small sail]

I plugged that into the Y comp and solved for T1

[sterile blaze]

who rotated it to get a copy :/

[small sail]

But like idk where I’m messing up

[sterile blaze]

whats alpha and gamma

[sterile blaze]

just use this :/

u really should learn the sin and cos rule

this is sine rule and

[small sail]

But like idk how it affects the rest of the formula

[sterile blaze]

yea these 2

[sterile blaze]

wdym affects the rest of the formula

[small sail]

Physics :(

I just follow the steps I see I’m not good enough at it to be adding stuff yet

[sterile blaze]

theres multiple ways to solve the same problem

[small sail]

Yes

But I was shown the way I’m doing rng

[sterile blaze]

theres nothing wrong with breaking it up into components

but ive no idea what u mean by alpha and gamma component

[small sail]

Shoot I can’t plug them in

The y components are all negative

[sterile blaze]

just use the sine rule

u drew the triangle

[small sail]

I’m not even at that point yet

[sterile blaze]

u know the values

[sterile blaze]

that is literally what u wrote

[small sail]

But idk any of the values

Just the angles

[sterile blaze]

yes

u know all 3 angles right

[small sail]

I don’t know how that rule works

[sterile blaze]

just copy paste it then

[small sail]

I still don’t know what’s going on with it

[sterile blaze]

anyway this.. u mean x and y component right?

[small sail]

Yes

[sterile blaze]

it should both total to 0

as the block isnt moving

[small sail]

For X it’s 0 and for Y it’s mg

[sterile blaze]

your working is

really messy

[small sail]

Come on I’m trying 🥲

[sterile blaze]

no really

why do u even have an alpha and gamma component in the first place

rather than x and y component

and then i see an x component as well

[small sail]

Bro it’s just how we were taught how to do it for now

[sterile blaze]

just use this

its one step

could u give an example of a question being done with that method

[small sail]

I just plug t2 into the Y comp and that’s kinda it

[sterile blaze]

well good luck to u

[small sail]

It’s physics 1?

[small sail]

.close

[odd edge]

Channel closed

Closed by @small sail

Use .reopen if this was a mistake.

[odd edge]

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[unique meadow]

Can we work through this using mesh analysis?

[odd edge]

Please don't occupy multiple help channels.

[unique meadow]

I don't have a lot of understanding of mesh analysis or kirchoffs law

[odd edge]

@unique meadow Has your question been resolved?

[meager juniper]

Mesh analysis generally involves summing around small loops in the network. But this network has diodes, so I would probably approach it only using Kirchoff's rules.

[unique meadow]

Im pr sure those arent diodes, they are just indicating the direction of current

[meager juniper]

Then you can use mesh analysis

How helpful

[unique meadow]

nice so can we work through that?

[meager juniper]

Probably

[unique meadow]

probably?

[meager juniper]

I wonder why they bothered with the arrows though

Because those are dependent on the values on the resistors

[unique meadow]

I assume just to help with understanding whether or not variables in KVL are positive or negative

[meager juniper]

Well, you can just assume a direction

And if you guess wrong the current will be negative

Nbd

[unique meadow]

ig so but I assume that the arrows are there as guides

[meager juniper]

There are cases, probably not this one, where the circuit is not stable and not statically analyzable. That's why I said probably. They don't really mention it in most circuits classes at the ugrad level though.

This one is simple enough that it's certainly solvable

[unique meadow]

I see

[meager juniper]

I just haven't worked through the steps yet to prove it

[unique meadow]

nice, can we work through the steps together?

[meager juniper]

Alas, I cannot right now

I was only online for a few minutes between errands

[unique meadow]

oh

[meager juniper]

Just keep in mind that the voltage change around any loop is zero

[unique meadow]

right

[meager juniper]

If you set up the problem, and try to solve it, and show your work, next time I get a few minutes I can check it, assuming some other helper doesn't first

[odd edge]

@unique meadow Has your question been resolved?

[odd edge]

@unique meadow Has your question been resolved?

[unique meadow]

I dont know how to set it up

[odd edge]

@unique meadow Has your question been resolved?

[odd edge]

@unique meadow Has your question been resolved?

[verbal lynx]

What is asked and given?

[unique meadow]

find all the missing values

you assign the resistors

with resistor values

[verbal lynx]

First instinct is to Y-Δ transform if that can lead us anywhere

[unique meadow]

how do i do that

[verbal lynx]

https://en.m.wikipedia.org/wiki/Y-Δ_transform

Hard to explain by myself

[unique meadow]

i was hoping we could work through the steps

[verbal lynx]

So, like, the R1, R2 and R4 resistors form a triangle

[unique meadow]

yep

[unique meadow]

still there?

[verbal lynx]

Idk, man

The values seem kinda bad for an exact answer