help-42

[plain moss]

error in addition of fractions

[remote mural]

🤦‍♀️

so its +1/2

y=-1/2x+1/2

[plain moss]

yep

so using y=mx+b why wasnt that working?

[remote mural]

lets gooo

[remote mural]

hmmm

[plain moss]

cuz y=mx+b is giving us -1/2x+1

fr thank god its just matter of practicing and this server is clutch

[remote mural]

ill try to fgure out why y=mx+b dont work

[plain moss]

bet ty

[remote mural]

fr lol

a different question?

[plain moss]

uh yes

even questions no answer in textbook

[remote mural]

br

bro

y=mx+b does work

[plain moss]

fr?

how so

[remote mural]

yes

so (-1,1)

y=1

x=-1

m=-1/2

1=(-1/2)(-1)+b

1=1/2 + b

no

what

??

[plain moss]

lol

[remote mural]

bro

[plain moss]

oh

yeye

[remote mural]

^

y=1

m=-1/2 (slope)

x=-1

cux the point is (-1,1)

solve it ut, 1/2=b

y=mx+b

plug in b and m

[plain moss]

oh in this u solve for b gotcha

[remote mural]

yes

u have (x,y)

[plain moss]

i see yea

[remote mural]

and u have m(slope)

[plain moss]

makes sense aii bet

[remote mural]

yeppp

so it does work

[remote mural]

now this

[plain moss]

easy to view

[remote mural]

so same concept as previus questions?

or what

[plain moss]

derivative

thats it

[remote mural]

okok

so same as that 1st

question

[plain moss]

uh yeah

[remote mural]

lets see

so u can use sum rule to seperate it out

constant rule, pull ut the constants

[plain moss]

yep

[remote mural]

multiple the 1st by 7/7 to get common denominator

and combine them

[plain moss]

yep

[remote mural]

seems right

[plain moss]

bet

🫡

[remote mural]

ok

so constant rule

pull out 1/2

product rule im guessing?

[plain moss]

yea

[remote mural]

and then put the constant back innn

[plain moss]

i think its good

[remote mural]

seems correct

[plain moss]

yea

bet

appreciate it

[remote mural]

ofc ofc

[calm coral]

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[mossy orbit]

if i have g(f(x^2)+1) how would I write out the chain rule? I put f(x^2)+2x

[calm coral]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[mossy orbit]

I'm going to do quotient rule and im trying to find derivative for numerator right now

thats what I'm doing but I pretty sure I did it wrong
I got: f(x^2)*2x

I got rid of the g cause the outside is g^1 so the derivative gets rid of it

This is what I wrote

But wouldn’t g’ = 0

[calm coral]

@mossy orbit Has your question been resolved?

[mossy orbit]

oh I get it ty

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[paper wharf]

for this question how do i enter sec^2 in my calculator

[weary kayak]

sec = 1/cos

[paper wharf]

i got the second line and everything in my working out but i just don tknow how to simplify further

[leaden thunder]

!show

[calm coral]

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

[weary kayak]

(1/cos(x))^2

[paper wharf]

yup this worked for me

thank you

[weary kayak]

awesome

[tall moon]

wait im just curious is the tangent line at some point x just the derivative after plugging in x in the graph with the point x?

[plain moss]

if i can understand what u are saying then i think yes

[tall moon]

cool

[plain moss]

u find the derivative of function sub in the value of x in the derivative

u get a slope of the tangent then from there u can find the equation of the tangent line

for that u need x1 and y1

[calm coral]

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[latent zinc]

not sure how to begin with this

i have only found orthonormal basis for vectors

[halcyon rock]

Orthonormal based on what inner product

[latent zinc]

wdym?

Thats all it says

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@latent zinc Has your question been resolved?

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[urban yoke]

I considered E[X_1 + X_2 + … + X_50] where X_i is the number of packs until i comes out

This isn’t consistent with the hint though

And the math seems crazy (E[N] = 2500 this way)

[royal basalt]

Do you know the final answer ?

[calm coral]

@urban yoke Has your question been resolved?

[spare beacon]

Your answer is wrong. You need to think about it in terms of the hint because the first pack you open always yields a card to add to your collection, the second pack you open yields a card to add to your collection with probability 49/50 etc.

You want to consider the time, measured in opening packs of trading cards, to the next time you open a card you haven't seen before, not a specific card.

[calm coral]

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[mental scarab]

just a quick but how do they get 3/2?

[half raven]

1/2(1)(3) = 3/2

[crystal forge]

you understood everything else in that problem but not the final step on getting 3/2? 😅

[mental scarab]

how do they get a number when the bound for y is 0 to 3x?

[solar spruce]

because x goes from 0 to 1

[mental scarab]

https://tenor.com/view/redneck-country-no-nope-gif-4614286

[solar spruce]

if you draw the area you should see how they got a number

[mental scarab]

ok but my think is using this equation

[solar spruce]

that is for a rectangle

[mental scarab]

ok then how do i get it for a triangle? estoy muy stupido

[crystal forge]

this area is just this, x goes from 0 to 1 and for each x, y goes from 0 to 3x think about how that gives a triangle

[solar spruce]

i would try drawing it out. in general [A(D) = \iint_D,dA]

[potent lotus]

maximo

[mental scarab]

i get it when i see the graph

but without

[solar spruce]

so in your case, [A(D) = \int_0^1\int_0^{3x},\dd y\dd x]

[potent lotus]

maximo

[crystal forge]

yup you can just get the area by computing that integral too if you dont see right away its a triangle

[solar spruce]

thought this is overkill since the triangle is just b = 1, and h = 3

[crystal forge]

you get y evaluated from 0 to 3x = 3x now integrate that and its 3/2x^2, plug in 1 = 3/2

[mental scarab]

okay i think i get it 😭

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[paper wharf]

what happened to the root on the RHS? why did it disappear on the second line

[crystal forge]

looks like the whole equation was multiplied by $\sqrt{2}$ to get from the first line to the 2nd

[potent lotus]

Soosh

[paper wharf]

but shouldnt you divide? since in the first line root 2 is being multiplied and to move it to the LHS you have to divide right

[solar spruce]

you don't have to do one or the other. they chose to multiply

[crystal forge]

then you end up with root 2 in the denominator and well to rationalize later you would then need another step

normally "nicer simplified form" is considered to only have roots in the numerator so you would avoid roots in the denominator

namely you would end up with 6y/root2

[paper wharf]

i get what you mean now

they just skipped the rationalise step

[crystal forge]

multiplying by root 2 gets rid of 2 square roots and only intoroduces 1 so you net getting rid of one root while not creating any fractions

[paper wharf]

is it fine to just expand the brackets instead?

so like in this one

[crystal forge]

i don't know the full context of this problem, what you are trying to accomplish with this calculation, the form it expects the answer in etc.

[paper wharf]

its a differentiation of a normal for a curve

[crystal forge]

and this is the equation of the line?

[paper wharf]

yup

[crystal forge]

well technically you can just leave it in any form then 🤷‍♂️

it's not asking for some particular form like point slope or whatever or standard form

all its asking for is "an equation"

or wait, i guess "Exact form" ? is that some form of a line equation? 😛

never heard of it

or it just means don't use approximations i guess?

[paper wharf]

exact form is referring to like the roots

so if you have an answer in radians

you use the triangles with the degrees (forgot the name)

to convert it to like root 3 on 2 and so on

[crystal forge]

so either of those 2 lines would be an acceptable answer

[paper wharf]

my handwriting is pretty bad 💀 but at the red dot is what i was trying to say

can i still expand the brackets and is it still correct (no marks deducted?)

[crystal forge]

yes, i mean its still an equation of a line

should be acceptable since the question doesnt ask for some specific form

as long as you dont make an algebra mistake

[paper wharf]

aight sweet thank you very much for your time

very appreciated

.close

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[vocal elbow]

how do u do this?

[vocal elbow]

my previous thread got deleted or something bc i didnt respondlol

and also this..

[sterile wharf]

is calculator allowed?

[sterile wharf]

y=4

[old falcon]

you dont need a calculator, just use the definition

[vocal elbow]

no

[sterile wharf]

ok y=4 for blue line

[vocal elbow]

so to find the gradient its mx - c?

oops

wrong photo

[vocal elbow]

so the y intercept is the equation?

[sterile wharf]

yes

[vocal elbow]

okay thanks

[vocal elbow]

then how do u do this?

bc i dont really understand the photo u sent for the question

[sterile wharf]

so

look

the y-intercept is -2

you get that right?

if you remember sloper = rise/run means = 6/1

go up 6 and right 1

and put a dot

and create a line

[calm coral]

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[heady jungle]

could anyone explain the product and quotient rules as like simple concepts relating to rate of change and the definition? not sure how to phrase it so I can provide an example:

since the derivative is the rate of change, then with the chain rule, in f(g(x)), one needs to consider how g(x) changes causing f(g(x)) to change, as well as how x changes causing g(x) to change, which is why the chain rule is as it is

this isn't really necessary but I prefer to know the topic thoroughly so if anyone can explain that'd be great

[civic dirge]

Are you sure that description of the chain rule does it for you? Of course you'd need to know how f and g changes, lol

Product rule:
If you know the rate of change of f and g, then you know the rate of change of f*g

[heady jungle]

I mean yeah but to me its better than those explanations that just state dy/dx = dy/du * du/dx and thats it

[civic dirge]

There's no more to it than those explanations, though

[remote mural]

i mean its explained like that for a reason

it doesnt need to be any more complex

[civic dirge]

d/dx(f*g) = (df/dx)g + (dg/dx)f

[heady jungle]

well like I'm trying to understand why it works conceptually with like rate of change

[exotic falcon]

the best way to understand it conceptually is to derive it using the difference quotient

the difference quotient makes sense conceptually

if you derive these rules using it, they must also make sense

[heady jungle]

alright thanks guys

👍

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[arctic grove]

help me out plz

[calm coral]

@arctic grove Has your question been resolved?

[arctic grove]

.close

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[paper wharf]

when i put sin-1 3/5 in my calculator it gives a completely different answer

[harsh siren]

degrees/radians check

[paper wharf]

do i convert the fraction into radians?

the 3/5

[harsh siren]

no

your calculator

check if its in degs or rads

[paper wharf]

its normal i think

wait no degrees

sorry

degrees atm

[harsh siren]

there you go

[paper wharf]

do i swap to radians mode?

[harsh siren]

yes

[paper wharf]

bro ur a lifesaver thank u

i had a few problems earlier about this

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[paper wharf]

.repopen

.reopen

[calm coral]

✅

[paper wharf]

sorry one last question

how do i know when to swap from radians mode to deg mode

is it just the domain they give you?

like if its a radians domain should i swap to rad mode

[harsh siren]

your domain is in pi

so its radians

if it gives you 180 or 360 its degrees

[paper wharf]

alright thank u so much again

have a good day

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[remote mural]

just want clarification on something

[remote mural]

the wording here doesnt quite make sense?

dividing by 0.8 * N gets you a number bigger than the amount of instructions you already have?

similar to here with the 1 core case

am i misinterperting something

[pallid halo]

probably doesn't apply if N=1

[remote mural]

fair enough. I guess calculate the raw values in that case

[harsh siren]

i mean it does say multicore

[pallid halo]

yea
i guess the 0.8 is intended to capture the overhead in managing the multiple cores

[harsh siren]

multi means more than 1

[pallid halo]

"when this program is parallelized and running on multiple cores..."
implying this doesn't apply when N=1

[remote mural]

So for the execution time i assume i just calculate the time that a single core takes to execute the instructions because i am supposing the cores will be working and finishing concurrently

[remote mural]

Yeah right

[pallid halo]

yea, that's probably what they want you to assume

[remote mural]

okay great

one last question

so we know the clock frequency (for which we can find the clock period of) and we know the amount of instructions there are as well

the only thing left to find is the global CPI (cycles per instruction)

Also im operating under this formula for context's sake

but how do i go about finding the global CPI of this?

[pallid halo]

well i guess you assume that, for the types of instructions that get divided among the cores, you just have (global cpi) = (cpi for one core) / (# cores)

and for the types of instructions that don't get divided among the cores, it's just global cpi = cpi for one core

[remote mural]

yeah that i understand haha but i guess my question was finding that "cpi for one core" rather

i am not sure if i just sum up the individual CPI values? like can i just say one core's CPI = 1 + 12 + 5

[pallid halo]

i think you'd want a weighted average of 1, 12, 5, where the weights are the percentage of instructions of each type

[remote mural]

OHHHH

YEAH!

thank you so much bungo

[pallid halo]

sure

[remote mural]

ok i think i can go from here

have a nice one

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[austere moth]

Does someone know of a good way to remember matrix multiplication ?

[winter elbow]

The former 1 means first row of the first matrix

The latter 1 means first row of the second matrix

[austere moth]

my eyes..

[winter elbow]

Bruh

Me no have pen paper on me, please no say me draw bad bad 🥲

It’s only for idea conveyance lol

[mossy gulch]

https://youtu.be/otXuqjoCs2M?si=DEEQYANSCsnW_hXy

https://youtu.be/2spTnAiQg4M?si=63WRcDN1_KeWkUbz

[austere moth]

Ok so there is no easy way to remember?

[pallid halo]

what would you consider an easy way

the element in row i, column j of AB is obtained by taking the dot product of row i of A with column j of B

[austere moth]

A way that not requires me to watch a 20 minute+ videos

[mossy gulch]

Laziness is your problem, then.

[austere moth]

Ok?

Thanks for you input

[austere moth]

What is i,j and AB?

[pallid halo]

AB is the product of A and B

i and j are row and column numbers respectively

[austere moth]

.close

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[blazing coyote]

I suggest you look at 3b1b's linear algebra videos

try this https://youtu.be/XkY2DOUCWMU?si=v3bo5AP_NEBlrWX-

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[fallow tusk]

How to find a if y=2x-2 and y=-x+a
The distance from the point of intersection of the lines to the origin of the coordinates is equal to five

[wheat pine]

You want to solve the equation?

[fallow tusk]

I want to find a

Damn wait

[exotic falcon]

What is "a"?

[fallow tusk]

There's supposed to be a not q

[wheat pine]

Bruh

[exotic falcon]

If y=2x-2
and
y=-x+a
then 2x-2=-x+a

solve for the point of intersection

[fallow tusk]

Yes

How

[wheat pine]

???

[exotic falcon]

solve for the x-value (in terms of a) where the lines intersect

substitute this into the equation of either line, to find the y-value

[fallow tusk]

How to use the five

[exotic falcon]

Then you can compute the distance from the origin to that point

set it equal to 5

and solve

[fallow tusk]

Oh okay

Thank you

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[vernal jackal]

hi, please could i get help with this surface integral question? im new to this topic.

[vernal jackal]

workings so far

[atomic palm]

it looks good so far

so you want your surface integrand to look like $\iint_S G \cdot N dS$

[potent lotus]

AlphaNull

[atomic palm]

you calculated N and ds already, and you are given G

make sure N is normal first

[vernal jackal]

im unsure how to do this do i know i should divide by sqrt(2) right, but which values am i doing that to? am i doing that to (u, 0, 0)?

[atomic palm]

yes, to (u,0,0) which becomes (1,0,0)

[vernal jackal]

yeah okay since my r_u x r_v = u and my | r_u x r_v | = u, i get u/u = 1?

[atomic palm]

yep

[vernal jackal]

so i was wrong about the sqrt(2) part?

[atomic palm]

yeah you dont need to mess with sqrt 2

[vernal jackal]

cool

[atomic palm]

u <= sqrt 2

thats all you should see from it

[vernal jackal]

so now my task is to just compute the LHS? since N = 1?

[vernal jackal]

or to compute this

[atomic palm]

so they are both the same thing actually

its a bit of weird notation

notice in your circle, S is bolded

that bolded S represents both the normal vector and the differential surface factor

[vernal jackal]

oh okay

so i just compute the double integral you said now

[atomic palm]

$\iint_S \bf{G} \cdot d\bf{S} = \iint_S \bf{G} \cdot \bf{N} dS$

[potent lotus]

AlphaNull

[vernal jackal]

cool cool

[atomic palm]

wow it bolded everything

[calm coral]

@vernal jackal Has your question been resolved?

[vernal jackal]

im having trouble with the actual calculationj

[atomic palm]

should be

$$\iint_S e^{{u\cos(v)}^2} \cdot u du dv$$

[potent lotus]

AlphaNull

[vernal jackal]

so i see you plugged in y = ucos(v) into the first term of G, what do we do with the other two terms?

[atomic palm]

we don't need it because the dot product with N eliminates the other terms

[vernal jackal]

OOHHHH YES

i was forgetting it was a dot product damn

[atomic palm]

$G \cdot N = G \cdot \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} = G_x$

👍

[vernal jackal]

let me try to continue now

thanks sm btw

[potent lotus]

AlphaNull

[calm coral]

@vernal jackal Has your question been resolved?

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[vapid stirrup]

We have a convex pentagon in a plane. Two points are located in it. Prove that there is a quadrilateral containing the two points and its vertices are vertices of the pentagon.

[grim blade]

What have you tried?

[vapid stirrup]

not much,i just said A,B (the points in the penatgon) are on a different side,which is not acceptable

cause i think it needs to be proven for anywhere in the pentagon

[grim blade]

a different side?

[grim blade]

Yes, it does

[vapid stirrup]

so can u help

[grim blade]

here you can pick any two vertices and they will form a quadrilateral

[vapid stirrup]

yeah,but u need to prove it

[ivory summit]

Wouldn't that follow from the convexity?

[vapid stirrup]

maybe since it is mentioned

[ivory summit]

Do you remember the definition of convexity?

[vapid stirrup]

no,i remmember something about curviture

[grim blade]

Because no points here are collinear

so they form a quadrilateral

[vapid stirrup]

can u write the proof?

like let A,B be the points in the pentagon since the quadilateral needs 4 verticiews..

[ivory summit]

But what do they mean by "vertices are vertices of the Pentagon"?

[vapid stirrup]

OK,so u have points in the entagon A,B,PROVE that THe verities of the quadrilateral are 4 verticies of the pentagon while A,B the points are in the quadrilateral also

[grim blade]

If the vertices are A,B,C,D,E, then some quadrilateral made from A,B,C,D,E contains the two points

[ivory summit]

Dude you have to put some effort aswell

[grim blade]

like A,B,C,D

or B,C,D,E

[vapid stirrup]

I DO

[grim blade]

I misinterpreted the problem here

[vapid stirrup]

I) We know that the pentagon is convex. Its internal angles are less than or equal to 180. This means that all points in the pentagon lie on the same side of the line connecting its adjacent vertices.
II) Let's mark the two points of the pentagon as A and B.
III) sega zemame bilo koj rab na petagolnikot,pr.AB. we make a line that connects points A and B. This line will cut the pentagon in two other points, which we will mark as C and D.

IV) the quadrilateral with vertices A, B, C and De created by the work AB and the line connecting the points A and B. This quadrilateral contains the two points A and B.

i did this

but i made a mistake A and B needed to be IN the quadrilateral

Draw a line segment connecting points A and B within the pentagon. This line segment might intersect some or none of the pentagon's sides.

No intersection: If the line segment AB doesn't intersect any side of the pentagon, it forms a diagonal entirely inside the pentagon. This diagonal along with any two consecutive vertices of the pentagon (e.g., vertices 1 and 2) forms a desired quadrilateral (A, 1, 2, B).

One or two intersections: If the line segment AB intersects one or two sides of the pentagon, consider the following cases:

One intersection: The intersection point divides the line segment AB into two segments. Choose one of these segments (say, segment AC) and consider the two vertices (e.g., D and E) of the pentagon that are adjacent to the endpoint (C) of the chosen segment (AC) on the pentagon's perimeter. The quadrilateral formed by A, C, D, and E satisfies the conditions.
Two intersections: The line segment AB is divided into three segments by the two intersection points. Choose one of the middle segments (say, segment CD) and consider the two vertices (e.g., E and F) of the pentagon adjacent to the endpoints (C and D) of the chosen segment (CD) on the pentagon's perimeter. The quadrilateral formed by A, C, D, and B satisfies the conditions.

maybe like this IDK

[grim blade]

wow that was fast

[vapid stirrup]

i solved it multyple times none of which i think ae correct

<@&286206848099549185>

[grim blade]

Why not?

[vapid stirrup]

i seem to be making a lot of assumptions

[untold drum]

Can you make a sketch? I do not understand how ABCD should be a quadrilateral and not just a (single) line.

[vapid stirrup]

it was supposed to be A

so the quadrilateral in this case is ACDE with the points A1,B1

[untold drum]

sorry, i still do not understand. you wrote
II) Let's mark the two points of the pentagon as A and B.
What should A and B be? vertices of the pentagon? any points on the lines of the pentagon? any points inside the pentagon?

[vapid stirrup]

well the verticies are the verticies of the pentagon,then we put A1,B1 lets say for example IN the pentagon NOW we need to prove that there exist a quadrilateral whose verticies are 4 of the verticies of the pentagon and IT also contains A1,B1 inside of it

[untold drum]

can you answer my questions?

you wrote
II) Let's mark the two points of the pentagon as A and B.
What should A and B be? vertices of the pentagon? any points on the lines of the pentagon? any points inside the pentagon?

[vapid stirrup]

inside

[untold drum]

i try to understand this:

can you make a sketch which matches this description?

[vapid stirrup]

dude ignore it,I TRIED to solve it

[untold drum]

you asked for help. but ok, i can easily ignore it.

[vapid stirrup]

lets start from scrath

[ivory summit]

For real that's the reason you aren't getting any responses

[vapid stirrup]

FIIIIIINE

idk who asked how i solved it

this is it

[ivory summit]

God how old are you 12?

Bye

[vapid stirrup]

no

keep guessing

[untold drum]

you asked for help.
you said you think your solution is not correct.
i was willing to help you. therefore i tried to understand what you did.
therefore i asked you something about your solution.
you said i should ignore it.
well, i can do this.

dont ask for helpers again.

[vapid stirrup]

.close

[calm coral]

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[buoyant jay]

how to factorise this equation

[buoyant jay]

4x^3 - 4x^2 - 15x + 18 = 0

[calm coral]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[neon sage]

assuming you don't have the cubic formula memorized (like most people), you need to guess a root first and then factorize into (x-a)(quadratic) and solve the quadratic

[rustic osprey]

(Rational root theorem narrows down the list of roots to check)

[untold drum]

and if you assume that one root is an integer you could argue that -15x+18 is divisible by 3, so 4x^3-4x^2=4 x^2(x-1) must be divisible by 3.

[buoyant jay]

I've tried 1,2,3,6

[untold drum]

try negative ones

[buoyant jay]

doesn't work

[rustic osprey]

,w 4x^3-4x^2-15x+18=0

[potent lotus]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[buoyant jay]

oh

so the final answer is (4x^2 - 12x + 9)(x+2)

[rustic osprey]

It says to find when it’s equal to 0

So you actually need to find the solutions

Also ||you can factor 4x^2 - 12x + 9|| as a ||perfect square||

[buoyant jay]

(2x-3)^2?

3/2

Thank you for helping!

Could I have done long division instead?

[rustic osprey]

Yeah

[buoyant jay]

ok thank you!

.close

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[blazing coyote]

$\sum_{n=1}^{10}\frac{n}{1+n^2+n^{4\ }}$

[potent lotus]

Why am. I here

[blazing coyote]

should I perform a substitution first? nvm, this isn't an integral

because I don;t think this can be broken down into partial fractions

[rustic osprey]

Hint: Consider $n^4+2n^2+1$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

The ||denominator can be factored by difference of perfect squares||

[blazing coyote]

ah

oh yeah

got it

thanks!

.close

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[worn anvil]

hey can someone help me on this

i firstly made 4^3x-2 into 2^6x-4

then 1/2rt2 into 2rt2^-1

but idk what to do from here

[blazing coyote]

just to confirm, you wrote $2^{6x-4}=2^{\frac{-3}{2}}$?

[potent lotus]

Why am. I here

[worn anvil]

oh yeah you can turn surds into powers

i completely forgot

thanks

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[proven carbon]

Hello

[proven carbon]

Is there an order to the vectors?

[brave zephyr]

yes

[proven carbon]

Here are the answers

Help me with (a) plsss

Don't understand why it's like that

Mann vector is such a complicated topic

I DONT SEE THE LOGIC FROM THE ANSWERS

[karmic sonnet]

the answers aren't clear

do you know about the loop law of vectors?

[proven carbon]

No

💀

For b the vectors are organised in the order they were initially from the question

[karmic sonnet]

if the vectors form a closed loop , their sum is 0

[proven carbon]

For a idk why it got reordered

[karmic sonnet]

for your question

[proven carbon]

Hmm okay

[karmic sonnet]

AB+BC+CD+DA=0

all are vectors

can't put arrows lol

[proven carbon]

Mhnmm

Is there a vc option here

[karmic sonnet]

now can you see that if you join AC , in triangle ABC , loop law can be applied again

[proven carbon]

I see

[karmic sonnet]

try to reduce the vectors which are sum of other 2

using the equations you get using loop law

idk if its the best method

[proven carbon]

But there's no value given in the vectors

[karmic sonnet]

also , AB = -BA if you don't know

[untold drum]

Think of vectors as a walk you are taking.BA+AC+CD, you start at B, then you go to A, then to C and at the end you go to D, so finally where did you start (B) and where did you end (D)? -> result is BD.

[proven carbon]

But why it begin with B suddenly

In the question it starts from C

[proven carbon]

Knew that thanks 👍🏿🙏🏿

[untold drum]

you ordered the vectors in a way that the endpoint of one vector ist the starting point of the next.

[proven carbon]

Hmm

Okay this

Why reordered

[untold drum]

for the interpretation as "walk".

[proven carbon]

I understand

But why must you start from B

When in question it begins from C

[karmic sonnet]

to walk

[proven carbon]

What

😭

Then why walk from B

Whole thing begins from A right

[untold drum]

do you see anouther way to reorder the vectors such that starting point of one ist the endpoint of the vector before?

[proven carbon]

OUUHJH

OHHH I SEE

Gotta match the ends and starts is it???

DAMN

THANKS YOU

[untold drum]

youre welcome

[proven carbon]

How to end

[untold drum]

type .close

[proven carbon]

.ok buh bye

.close

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[spiral dagger]

This is a problem set question and the solution. Unfortunately, I do not understand the solution. I need help!
The chapter is one the pigeon hole principle

[spiral dagger]

?

[leaden thunder]

Just a spamming troll. Have pinged mods elsewhere

[spiral dagger]

ok

[carmine sorrel]

what part confuses you?

[spiral dagger]

I am confused about whether they did use the pigeonhole principle

[carmine sorrel]

at the "indeed there are too many of them" part. That's by pidgeon hole. They showed there's too many stones for the number of stacks.

[spiral dagger]

in this context, what are the pigeons and what are the pigeonholes?

[carmine sorrel]

stones = pidgeons, heaps = holes

[spiral dagger]

There are a1 + a2 + a3 + a4 stones, how many stacks are there?

[carmine sorrel]

4

[spiral dagger]

, by the pigeonhole principle this implies that atleast one stack contains two or more stones. But the problem asks to prove that atleast two stones are placed into a smaller stack

[carmine sorrel]

that's what the rest of the proof is for.

[spiral dagger]

thanks

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[remote mural]

0.25 were real chocolates.. so, out of 100 chocolates, 25 of em were real. ok, now, its, 25/100, so, 25 chocs got divided by 100 chocs? i dont get the bold text

[remote mural]

like

how did they

get divided?

i cant

imagine

that

yes

25/100 what

ping em

what what

wdym what

25/100

um

like

real chocalates

``25 were real

and the rest were fake

so

yes

well yeah

so whats your question

25 chocs got divided by 100 chocs can u help me imagine this

25 got split by 100

uh

[remote mural]

this is confusing

wait

nvm

.close

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[torpid canopy]

Hello I was wondering if someone could explain to me what this menas

[torpid canopy]

My understanding is $\mathbb{F}_3$ is the field (0,1,2), but I'm not sure what the set of multiples of pi is...

[potent lotus]

Ari

[marsh summit]

...i have no idea

is there any more context than that?

[mortal orbit]

A^B is the set of functions from B to A

[torpid canopy]

So the set of functions that map ${\pi, 2\pi, 3\pi } \rightarrow \mathbb{F}_3$?

[mortal orbit]

yeah I would say so

[potent lotus]

Ari

[torpid canopy]

alr thanks

[calm coral]

@torpid canopy Has your question been resolved?

[torpid canopy]

I'm back

This is the entire question. My professor never really did an example on these types of questions, so I was wondering if someone could help me get started.

<@&286206848099549185>

if anyone does help, tysm and pls ping me so I know you replied ❤️

[calm coral]

@torpid canopy Has your question been resolved?

[calm coral]

@torpid canopy Has your question been resolved?

[calm coral]

@torpid canopy Has your question been resolved?

[torpid canopy]

🥺

[calm coral]

@torpid canopy Has your question been resolved?

[thin peak]

linear transformations?

Did your professor talk about onto functions/ the concepts of injectivity/surjectivity?

[thin peak]

The basis is the linearly independent set of vectors that span the subspace

[torpid canopy]

Yes! But I've only seen pretty basic examples

[torpid canopy]

Right

I'm mainly not sure how we can map polynomials to non polynomial vector space over a field

[calm coral]

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[remote mural]

Hello, i was having a little trouble making sense of the general solution of sin(x)=sin(a) is x= n*pi + (-1)^n a

[remote mural]

so i was hoping someone could help me understand it a little

[winter loom]

what don't you understand?

[remote mural]

i understand the general solution for cosine and tangent

[remote mural]

idk it just doesnt make a lot of sense about where it comes from. i have tried taking the cases where n is even and odd but i couldnt be conclusive about this making sense

[winter loom]

sin(x) = sin(a)
x = 2kpi + a
or
x = (2k-1)pi - a

The first one because sine function has a period of 2pi.
The latter because sin(x) = sin(pi-x)

[remote mural]

the even one makes sense

oh wait the second one does too

oh yes

[winter loom]

And your solution is a way to represent the two together.

[remote mural]

understood

thank you so much

i appreciate it

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[remote mural]

how can i find the angle between A_vec = 3i+2j+k and B_vec = i-j

[remote mural]

do you know the dot product

yea i know

can you try computing it

hold up

1?

yes

so now

,, \vj a \vd \vj b = \norm{\vj a}\norm{\smash{\vj b}}\6\cos\theta

[remote mural]

solve for cos(theta)

oh lmao

im so dumb

gimme a min

oh so like

cos inverse 1/sqrt(28)

just this right?

sounds right

ok yea

thankyou

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[lunar pier]

hi, i was just wondering why for a lot of equations in question 3, the n isn't followed by a -1?

[dull wagon]

wdym

that's just how they chose to define their rule

[lunar pier]

so it doesn't especially matter either way?

[dull wagon]

no

[lunar pier]

aight thanks

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[spiral dagger]

In this proof, i am confused about how they obtained the relation $a_1 +...+a_k > b_1 +...+ b_{k-1} + 1$

[potent lotus]

schrödinger

[calm coral]

@spiral dagger Has your question been resolved?

[spiral dagger]

<@&286206848099549185>

[stray vortex]

What is the relation between the heaps

[spiral dagger]

None of the heaps are empty

[stray vortex]

bruh

but there shud be some inter relation

[spiral dagger]

idk, you can read the proof

[stray vortex]

sry mann cant help

[spiral dagger]

ok

[reef compass]

we already know a1 + … + ak > b1 + … + bk, and as you said none of the heaps are empty

thus bk contains at the minimum one stone

it’s useful to play around with a toy example and set k to something like 3

@spiral dagger

[spiral dagger]

I am still trying it

@reef compass i got nothing

[reef compass]

hmm ok i’ll think through the last part of the proof for a bit

i don’t really understand that last part

if you’re adamant about understanding this specific proof there are probably forums online

[spiral dagger]

i got it

i got the proof

[reef compass]

hmm i got lost after the sentence with k largest old heaps

you could also prove this using a ratio argument from the four piles to the 5 piles

[spiral dagger]

we have that $a_1+...+a_k \geq b_1 +...+b_k + 1$ also $b_1 + ... + b_{k} \geq b_1 +...+b_{k-1}$ then it follows $a_1 + ...+ a_k \geq b_1 +...+ b_{k-1} + 2 $ and from this the relation $a_1 +...+a_k > b_1 +...+ b_{k-1} + 1$ follows

[potent lotus]

schrödinger

[spiral dagger]

K largest old heaps are a1, .., ak

thanks a lot

[reef compass]

i don’t think your statement after “then it follows” is necessarily true

[spiral dagger]

In one of terence tao books, he proved that if $a \geq b+1 $ then $a > b$

[potent lotus]

schrödinger

[spiral dagger]

I used this fact

[reef compass]

oh ok ic, it’s late and i didn’t see the k-1 subscript on the rhs

[spiral dagger]

oh

[reef compass]

yeah that’s valid then

i wonder what they’re doing with the k largest heaps argument

[spiral dagger]

how do you mean

[reef compass]

stones from the k largest old heaps could not all go to the k largest new heaps

[spiral dagger]

theu want to show that some stones went from larger heaps to smaller heap. If you read the proof, the k largest old heaps are all larger than $b_k \geq b_{k-1} \geq ...$. Since $a_1 + ... + a_k \geq b_1 + ... + b_{k-1} + 1$ atleast two stones from the k largest old heap have to go to a heap that is bk or less

[potent lotus]

schrödinger

[spiral dagger]

.close

[calm coral]

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[modest tapir]

can somebody help? I feel like the explanation is lacking here

[calm coral]

@modest tapir Has your question been resolved?

[calm coral]

@modest tapir Has your question been resolved?

[royal basalt]

What explaination ? Everything is written there step by step.

[calm coral]

@modest tapir Has your question been resolved?

[remote mural]

In an implicit derivation you derive both sides of the equation and then isolate y'

[modest tapir]

ok but where did (14x^2y+3) come from? I don't see it anywhere

or is that supposed to be the derivation of the LHS?

[royal cave]

take y' common from the last two terms 🙏

[calm coral]

@modest tapir Has your question been resolved?

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[humble steppe]

why did they ignore the exponents when intergrating this?

[royal basalt]

Because 15/5 = 3

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[humble steppe]

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[frozen dome]

i don't want the other answer if there is one but for the equation cos(x) = cos(1/x) is x = ||aπ+-sqrt(a^2π^2+1) for any integer number a|| the only solution

[frozen dome]

i believe it is the only answer but am wondering if there could be more

[lyric ravine]

,w cos(x)=cos(1/x)

[lyric ravine]

Yes, the negatives of these solutions since cos(-t)=cos(t)

I guess you got these solutions using the difference to product identities

[frozen dome]

oh yeah i forgot about negatives somehow lmao these are all the positive solutions for x and just - for all the negatives

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[frozen dome]

.reopen

[calm coral]

✅

[frozen dome]

the negatives are just flipped so they dont give new answers

.close

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[worn anvil]

im confused on how to start with this

i made a = x^2 and b into Y^2

but idk if i should do that

[gloomy badge]

Well, what happens if you combine denominators on the left?

[worn anvil]

ab

[remote mural]

use am - gm

[worn anvil]

idk what that is

[blazing coyote]

AM>GM

arithmatic mean > Geometric mean

[remote mural]

hint : use am and gm somehow

oh he said it alr

[worn anvil]

i havent learnt that yet

[remote mural]

sequence and series ? arithmetic progressions ?

[worn anvil]

yeah i think i learn that next year

[remote mural]

i see

[blazing coyote]

$(2a+b)^2=4a^2+b^2+4ab)$, can you use this in any way?

[potent lotus]

Why am. I here

[royal basalt]

$\frac{\frac{4a}{b}+\frac{b}{a}}{2} \geq \sqrt{(\frac{4a}{b})(\frac{b}{a})}$

[potent lotus]

Solomaniac

[royal basalt]

@worn anvil

This is the AM > GM inequality.

[worn anvil]

yeah in the mark scheme it shows that but idk what it means

[blazing coyote]

basically $4a^2+b^2=(a+2b)^2-4ab$

[potent lotus]

Why am. I here

[blazing coyote]

so can you tell me which is greater?

[royal basalt]

Yeah

[worn anvil]

ohh i slightly remember learning this

[neon sage]

you mean (2a + b)^2

[worn anvil]

i expanded it and i got a^2+4b^2

[blazing coyote]

yeah, my bad

[worn anvil]

theryre equal are they not then

neither is greater then the other

[blazing coyote]

$\frac{\left(4a^2+b^2\right)}{ab}=\frac{\left(2a+b\right)^2}{ab}-\frac{4ab}{ab}$

[potent lotus]

Why am. I here

[blazing coyote]

do you agree with this?

what can you tell me based off this?

[worn anvil]

yeah theyre equal

[blazing coyote]

not just that

$\frac{\left(4a^2+b^2\right)}{ab}=\frac{\left(2a+b\right)^2}{ab}-4$

[potent lotus]

Why am. I here

[blazing coyote]

right?

and is the LHS always >0?

[worn anvil]

not sure

i might just ask my teacher tommorow

[blazing coyote]

what does the problem statement sat

*say'

a,b>0

right?

[worn anvil]

yeah

[blazing coyote]

oops, not too sure what to do from here

sorry

I was thinking (2a+b)^2/ab >4

but that doesn't help much

[worn anvil]

heres the mark scheme

[blazing coyote]

ah, I get what they;ve done

do you understand it?

[worn anvil]

nah

[blazing coyote]

ok, so

[worn anvil]

i dont get the ... notation

[blazing coyote]

(2a-b)^2 >0

right?

[worn anvil]

yeah

[blazing coyote]

so 4a^2+b^2>4ab

do you see what to do from here

[worn anvil]

no

[blazing coyote]

divide both sides by ab

[worn anvil]

ok

oh i see

yeah i see what i should have dobne

[blazing coyote]

quite a fun problem, ngl

[worn anvil]

thanks btw

[blazing coyote]

np

[worn anvil]

.close

[calm coral]

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[calm coral]

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[blazing coyote]

so here, why is 1 true

intutively it makes sense

[spark stratus]

I remember doing this, you can consider IVT on f(x) - x with y=0

[blazing coyote]

I don't know IVT

is there any other way

[narrow spear]

Yes you use that

Maybe they are other ways but it depends on what you know, it's very basic to use IVT here

[blazing coyote]

hmm, OK. Thanks

.close

[calm coral]

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[marsh agate]

They're all simple corollaries of the IVT

[blazing coyote]

I see, thanks!

[calm coral]

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[orchid palm]

Does this not come out to

It is marked as wrong but I thought that I just needed to apply this identity

[modern condor]

you forgot to take dx=8cos theta dtheta

[orchid palm]

oh I see

lemme try that

that makes the problem cringe all of a sudden

dammit

I get this now

how does the cos theta dtheta cancel out

does it just go away since its a derivative under an integral

?

[calm coral]

@orchid palm Has your question been resolved?

[calm coral]

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[remote mural]

tips

[remote mural]

idk what to do

[spark relic]

Put LHS as one log

log = log

To cancel it out

And be free of logs

[remote mural]

so u mean like

[spark relic]

That's not log = log

That's x - log = log

[remote mural]

u cant make log = log

out of that

oh wait

but no its not

i was thinking

(1-lg5)^{x}

but no

[spark relic]

Nop

Put x with a proper log

[remote mural]

ohh

so i can just

wait

i got it

look

but that doesnt give anything good

[spark relic]

No

lg(x) ≠ x

[remote mural]

how then

i know that lol

i just dont see it

right now

the solution

[spark relic]

$log(10^x) = x$

[potent lotus]

Rub05

[remote mural]

i know

[spark relic]

So instead of x put log(10^x)

[remote mural]

right?

and that would be a 0 there

right

at the end

lg 1

is 0

[spark relic]

No

You must do log = log

[remote mural]

what did i do wrong

ok

so