help-33

[still temple]

240/h=x²

270/h=y²

Factorise 240 and 270

And choose comfortable h to make it whole square

[zenith sable]

Wait heres an example of what my teacher teached us bcs they said if its not like this, then its wrong

[still temple]

Yo ur teacher did nothing they took h as 4 ig

[zenith sable]

Oh

[still temple]

Lemme analyse first

But what exactly are u trying to figure out

[zenith sable]

We have to make a quadratic equation and we have to solve it

[still temple]

Ok

So did they assume bottom is a square

Or not

?

if not then 240/h=xy

Because hyx is volume

For first block

[zenith sable]

Idk

[still temple]

take h as some constant say 4

[zenith sable]

😭

[still temple]

240/4=xy

60=xy

270/4=ab

[marsh citrus]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[still temple]

Ok

U have to know 3 things

First

Volume is hxy for a cuboid

xy is bottom surface area, h is height

@zenith sable .

[zenith sable]

Yes

[still temple]

Now follow the steps they have asked

Read the question properly

Now u got it

Dude ru still stuck?

[zenith sable]

Yes

😭

My classmate said 36 is the constant

Our formula is like

ax²+bx+c (constant) = 0

AND IDK HOW TO GET THE FIRST 2 TERMS

[still temple]

Ok so take 36 constant as height H

[marsh citrus]

@zenith sable Has your question been resolved?

[still temple]

Bro you can lookup google

For exact answer

Because i do not know full context

[marsh citrus]

@zenith sable Has your question been resolved?

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[slender steppe]

Hi

[slender steppe]

Sorry to bother

I’m probably being an idiot

However

However is -infinity,-2

Not included

Or 2,infinity

[slate yarrow]

the function is $f(x)=\sqrt{4-x^2}$

[elfin berry]

convergence

[slender steppe]

I can see that

But I’m still confused

as to why the function wouldn’t just be continuous over - infinity, infinity

[slate yarrow]

now for this function to be defined in $\bR ,f(x)\geq 0$

[elfin berry]

convergence

[slender steppe]

Oh yeah

So I’m an idiot

Sorry

[slate yarrow]

its alright dw

[slender steppe]

.close

[marsh citrus]

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[marsh citrus]

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[hybrid charm]

How would I establish the following identity? I am working through this book for self-study

[south smelt]

Probably just write 1 as sin²x + cos²x

[hybrid charm]

How would that establish the identity?

[south smelt]

Simplify the expression further

[hybrid charm]

My answer key got this, but I am not sure how. Let me get it

[nova sierra]

I did it that method

[hybrid charm]

What does this method show, sorry I am relearning trig identities

[south smelt]

Ah, they messed with the right side

[nova sierra]

So tan(x) = sin(x)/cos(x)

and cot(x) is the reciprocal of tan(x)

[hybrid charm]

Yeah, how do those two come together here

[nova sierra]

well, you are left with sin(x)/cos(x) - cos(x)/sin(x)

[south smelt]

put them on the same denominator

[hybrid charm]

Ah

[nova sierra]

get a common denominator, by multiplying first fraction by sin(x)/sin(x), and the second by cos(x)/cos(x)

[hybrid charm]

What would be the common denominator here?

once you did that

[nova sierra]

sin(x)cos(x)

[hybrid charm]

okay, that makes sense. Thanks for the help everyone

.close

[marsh citrus]

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[marsh citrus]

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[carmine sandal]

How would i go on to show this

[carmine sandal]

I am a bit stuck on what my initial step should be

so i just need a nudge

[quaint elm]

try it for mod 5 first or something maybe

translate (a,n) = 1 into an equation

in fact, try picking numbers for a, b, and n

[carmine sandal]

I did work with a few numbers

in fact those are the previous questions

so i do know its true

however I did notice in the examples preceding this question

that (a,b) = 1

but that is not a specified case in question 16

so im currently seeing if i can work with that

[quaint elm]

actually try it for n = 2

that's a nice and simple case

[carmine sandal]

I jus tried one with n = 7 😭

This makes me think

Im not sure if this is possible to do

with every scenario

[quaint elm]

start with 2 i think, then 3. there's something you might learn

[carmine sandal]

sure ill give that a shot

[quaint elm]

but like fully prove it for n = 2 and arbitrary a and b

(that are coprime to 2)

[carmine sandal]

Sure Ill give that a shot

thanks

.close

[marsh citrus]

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[marsh citrus]

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[hasty hemlock]

can someone help me with this

[marsh citrus]

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

[hasty hemlock]

2

i have begun but got stuck midway

[marsh citrus]

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

[vital oracle]

(k) use the root test
(l) use the direct comparison test with 1/(2^n - 2^(n-1))

[hasty hemlock]

oww oke thanks, for k i used also root test but did a mistake so i got a wrong answer

7. I have completed the problem and don' need help anymore. Thank you

[vital oracle]

np

when youre done with the channel, you (or me) can type .close or .solved to say youre done

[hasty hemlock]

ow oke

yeah im done

.close

[marsh citrus]

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[hasty hemlock]

.reopen

[marsh citrus]

✅

[hasty hemlock]

just had a question about l) so i basically did d.c.test but i have now to prove that b(n) is diverges

[vital oracle]

it converges, not diverges

[hasty hemlock]

yeah converges sorry

but which test do i have to apply then ?

[vital oracle]

for that, you should use that 2^n = 2 * 2^(n-1)

or equally 2^n = 2^(n-1) + 2^(n-1)

[hasty hemlock]

wait but a(n) and b(n) are equal

do i have to use d.ctest or limit c.test

cause for direct comparaison test a(n) should be greater or smaller than b(n)

.close

[marsh citrus]

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[wary bluff]

I have to find the potential and then use FTC

I am getting the correct answer but I just wanna make sure there arent any issues with how Im using the notations

So from the question we know

f_x = yz
f_y = xz
f_z = xy+2z

from f_x I get f(x, y) = xyz +g(y, z) . Now ill take the derivate wrt y to get f_y(x, y) = xz + g_y(y, z) and compare it with f_y from which we see that g_y(y, z) = 0 which implies g(y) = c

Now Ill take the derivative wrt z to get f_z(x, y) = xy + g_z(y, z) and comparing it with f_z we see that g_z(y, z) = 2z which implies g(z) = z^2

So finally,
f(x, y) = xyz + c + z^2. Take g(y) = 0 so f(x, y, z) = xyz + z^2

And now I just plug in f(4, 6, 3) - f(1, 0, -2)

[wary bluff]

especially the part where I say
"g_y(y, z) = 0 which implies g(y) = c"
" g_z(y, z) = 2z which implies g(z) = z^2"

is that a good way to write it

[marsh citrus]

@wary bluff Has your question been resolved?

[wary bluff]

fuckit Ill just write it like that

.close

[marsh citrus]

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[lucid turret]

,w express

[elfin berry]

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Requested by anshulyolo121

[marsh citrus]

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[hybrid charm]

How is polynomial long division possible? How can you divide two terms at the same time? I am relearning this for self study so please use simple words

[fervent rampart]

in regular long division, you subtract multiples of the denominator until you get a remainder. this is the same in polynomial long division

[dapper aspen]

(x+3)(5x+4)+1=5x2+19x+13

[thorny bison]

It's not exactly division rather it's more like factoring the polynomial into its linear Or qudratic Or other factors

[wary kite]

or division

[devout mauve]

remember how you do long division for numbers. you more or less work digit by digit. you subtract something in such a way that the first (or more) digits cancel and then work out what the rest is. and then you repeat

for polynomials is the same idea, just that instead of digits you go by the degrees

[dapper aspen]

its kinda like (2)(3)+1=7

[hybrid charm]

What's the difference

Sorry for being late

[hybrid charm]

but how do you do 2 terms at once with degrees like x+3

[hybrid charm]

how so?

[dapper aspen]

multiply x+3 with 5x+4

what do u get?

[hybrid charm]

5x^2+4x+15x+12?

[thorny bison]

No difference it's just like when you divide a no to find it's factors we divide a polyn to find it's factors

[hybrid charm]

so
5x^2+19x+12

[dapper aspen]

yes

then add 1

u get 5x^2+19x+13

[hybrid charm]

So you would take (y+m)(y+n) for polynomials when factoring?

[hybrid charm]

where did the +1 come from

[thorny bison]

Yes

[dapper aspen]

thats the remainder

[hybrid charm]

That makes sense, how do they get the second line of 5x^2+15x though?

we got the first line of 5x^2+19+13

[hybrid charm]

Thanks

[dapper aspen]

can u vc?

[hybrid charm]

Yeah

[dapper aspen]

is it good if i call u rn?

[hybrid charm]

yeah sure, dm me

[thorny bison]

Wlc

[hybrid charm]

@dapper aspenthanks

[dapper aspen]

np

[hybrid charm]

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

(don't cyberbully me for bad math, also I'm trying my best to be as clear and concise as possible)

so I'm trying to calculate how much a character can lift at a certain power level.

example. Character has 60,000,000 of power and can lift 400,000 kilos. Therefore, 30,000,000 of power means they can lift 200,000 kilos.

Now, what I'm trying to calculate is how much the character would lift at 397 of power, but I can't think of a formula to get their lifting strength in result. (after = symbol if that makes sense)

[carmine sandal]

thats fine ur mathematical ability doesnt matter too much

is ur example

accurate

u can first calculate

[wary kite]

so if you have 60,000,000 you might want to figure out how much 1 power would give

[carmine sandal]

how much weight can be lifted for 1 power

yeah

[wary kite]

then multiply it be 397

that should help you with the algebra

then just use letters to represent whatever you’re looking for

[still temple]

would 400,000÷60,000,000 work?

nvm it does

tysm

.close

[marsh citrus]

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[dull slate]

Okay, simple question, advanced answer, for an axis aligned rectangle like with defined x and y start and x and y stop that may be a line or point and a non-self intersecting not all collinear polygon how do I determine what is inside that rectangle that is part of that polygon from where intersects of rectangle and polygon are? A polygon can be any shape and is 3 or more straight lines and all numbers involved are real, finite, and defined. This is a problem from a 3d algorithm and like my only remaining problem. X E.

[dull slate]

I know that intersection or point in rectangle means part of shape is in, I know that in 2d for intersects, line intersects with t in parametric equations like (end1-start)/(end-start) where end1 is a component x or y and others are in line to intersect possibly, same dimension. I can get what points are in or intersect but I am unsure what to do for things like there is part above intersecting part that is filled like above with no intersect to say there is a part there. X E.

I can find exactly where intersects and inside points are but how do I determine when rest of rectangle in a given direction is filled with polygon?

I also have a working minimum or maximum of x or y intercepts algorithm. That helps I think. X E.

<@&286206848099549185>, sorry but 30 minutes about is seeming long, has anyone seen this? Should I just close because this is complex?

[elder crag]

sorry i think i might be confused as to what the question is exactly

you have a rectangle

and another shape

and you want to know whether this shape is inside the rectangle

and if so what are its intersection points?

[dull slate]

I want to know where that polygon is inside that rectangle, and that polygon is non-self-intersecting. X E.

Sort of yes. X E.

Like imagine above, how to define star part in rectangle?

Sorry, rather advanced apparently. X E.

[elder crag]

lets say the rectangle is the defined by the the following points $(x_1, y_1); (x_2, y_2); (x_3, y_3); (x_4;y_4)$ (these are its vertices)

[elfin berry]

woomy

[dull slate]

We can make it simpler and say xmin, ymin, xmax, ymax and define our rectangle that way. X E.

Thanks for talking, this is like my first time doing this. X E.

[elder crag]

Lets say the shape is parameterised in the following manner: A(x,y). To check where A is inside the rectangle you just need to check when (x_min, y_max) < A(x,y) < (x_max, y_max)

im just not so sure how you would parameterise any shape

[dull slate]

I know how to do intersect and point inside rectangle. Next maybe? Sorry. X E.

[elder crag]

what seems to be missing is a way to parameterise any shape so you can describe it easily only using x and y coordinates

but im not sure i know how to do that

sorry!

goodluck with finding an answer though

i reccomend asking in #algebraic-geometry

[dull slate]

I can make 2d coordinates from 3d, I just need a resultant 2d shape's intersect with said rectangle. X E.

[elder crag]

i'm not too familiar with this topic/geometry in general so i dont think i can help :-(

[dull slate]

I at least know that there is inside by direction of going and outside and anything inside may go outside to form a shape. X E.

I also have how to find centroids so do not need that. X E.

I can project to and from 2d, just need 2d intersect definitions. X E.

Well, 15 minutes again, is it okay to continue with <@&286206848099549185>? Am I too annoying? Should I just close this?

[elder crag]

Theres no problem with pinging helpers! thats what we are here for :) dont think you're annoying! many people dont respond cause they think they cant really help as they dont know.

[dull slate]

I almost hate calling for help. X E.

[elder crag]

If you dont get a response try asking in #algebraic-geometry, if you dont see this channel try #get-advanced-access

[dull slate]

I see that channel but am here which I think is better for now, should not leave this overnight right?

[elder crag]

I understand you a lot, sometimes it just feels like by asking for help you are wasting others time, but one of the main purposes of this server is to help others, so dont feel bad :)

[elder crag]

I'd wait to see if you get a response and if not try the other channel

[dull slate]

7:36PM for me, might leave at 8PM, guess I might close then. X E.

[elder crag]

Sounds good :) I'll go now, have a good day

[dull slate]

You too. X E.

Someone tell me if this needs more explaining. X E.

[deep anvil]

Let me see if I understand the question correctly, you want to find the area of intersection between a rectangle and a polygon?

[coral path]

algebraic geometry???

😮

[dull slate]

I want to define points of where a polygon intersects a rectangle. It is subtle but there is a difference. X E.

[dull slate]

I never get an answer in channels like that. X E.

[deep anvil]

You mean in the contour?

[deep anvil]

Because this is not Algebraic Geometry.

[dull slate]

I am trying to find outlines of where polygon is in rectangle. To that, I guess. X E.

Thank you for talking, any tips?

Rectangle is generated by a common rectangle of two polygons, like my goal is to find centroid z and distance by this and that rectangle method seems way faster. Anyway, rectangle and polygon for now. X E.

Please tell me if anything is confusing. X E.

Also, unsure what to even call this, research turns up about nothing. X E.

[deep anvil]

Are the polygon and the rectangle "filled in"?

[dull slate]

Yes. X E.

I should note that rectangle is possible to be a line or point as I am describing. X E.

At least got any resources?

[deep anvil]

Then I don't see the diffrence. The intersection is the set of points where the polygon intersects the rectangle.

[deep anvil]

One can just do it case by case.

[dull slate]

Note that technically part of this one is in but not with a vertex intersect defining in sort of, unsure how to explain. How to find when to include bottom left corner as in this? X E.

[deep anvil]

I can tell you this is probably a computational problem. You won't find help in any of the "advanced" channels because the problems in those channels are of a different kind. Maybe you'll find help in a #computing-software or a programming server.

[dull slate]

Yeah, I consider this math though and I guess I can approximate. X E.

That channel was not on like my channels list, what do I gotta do to get that category?

[deep anvil]

I think you should have it since you have Undergrad role. But at least it should be visible in the channel list in Channels & Roles at the top of the channel list.

[dull slate]

Actually based on channels list that one was all I should use anyway. I can't keep this all night so .close . X E.

.close

[marsh citrus]

Channel closed

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Use .reopen if this was a mistake.

[dull slate]

X E.

[marsh citrus]

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[idle hinge]

can someone help me with #7 i don’t understand if i did it correctly

[marsh citrus]

@idle hinge Has your question been resolved?

[idle hinge]

@sharp wing

<@&286206848099549185> Help me Guy please im so sleepy

[still temple]

You mean the question in the center of the image?

Just to make sure

[idle hinge]

yes

[still temple]

ok

Firstly, let's define what an exisiting limit is, like what makes a limit exist.

1. The limit from the left equals the limit from the right of that value.

In this case, that value is 2.

So, to check that the limit exists, we can create a table.

To be honest, I'm not entirely sure how to do that here, but I'll try to screenshot a diagram if i can.

[idle hinge]

for me i got a dne

[still temple]

Did you use a table?

[idle hinge]

no

[still temple]

Ok

So, let me get a diagram up real quick, one minute.

[idle hinge]

ok sounds good to me

[still temple]

So here is the rough diagram.

[idle hinge]

so it is 2?

[still temple]

I'm establishing 2 as the center, to check for a trend.

You see, by having values that are very close to 2, we can establish whether or not the right-hand limit and the left-hand limit meet at the same value.

If they do, the limit exists, and you can find it.

Otherwise, it does not exist.

So compute the y-values for those x's based on the piecewise function provided in the question.

After you do so, we can check to see if the y-values meet at the same value.

[idle hinge]

i see

i am back

sorry i took so long

[still temple]

No worries.

Did you fill the table/diagram out on your side?

[marsh citrus]

@idle hinge Has your question been resolved?

[proud kiln]

How is the answer ever 7?

2² + 1 = 5
2 + 4 = 6
it won't approach 7 from either side

[marsh citrus]

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[marsh citrus]

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[steel snow]

Hi I don’t understand the difference between P(t) and r(t), and how the set of all P(t) with x, y, t, (3D) can correspond to a 2D function. Thanks in advance 🙂

[cunning fiber]

t isn’t a dimension

[steel snow]

Yeah but when you graph it isn’t it going to be a 3D graph, or a 2D one which moves as time progresses?

[cunning fiber]

x and y change based on t

Think of it as a parametric function

But written in vector form

[steel snow]

Oh so sub in x = At + B and y = t into y = 2x + 3 and find A and B based on the coefficients of t and the constant

[cunning fiber]

Yeah

[steel snow]

And r(t) is the position vector of P while P(t) is how y changes relative to x

[cunning fiber]

More like r(t) gives P at a specific point t and P(t) models all the possible locations of P

[steel snow]

Oh now I get it, tysm!

.close

[marsh citrus]

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[pure current]

Here is the question

But I don’t get what the answer means

[cunning fiber]

It’s considering said operation in terms of a determinant, then performing elementary row operations to manipulate it to a determinant that equals zero

Then from that point, the conclusion follows by adding (a^2+b^2+c^2) to both sides

[cunning fiber]

Namely $xC_1+yC_2+C_3 \to C_3$

[elfin berry]

Civil Service Pigeon

[pure current]

wait sorry sir can we go one by one my processing speed is bit slow and bad

[pure current]

I don’t get why the first matrix have the “-1” each row

[cunning fiber]

The best I can say is “see what happens when those are 1, notice its off by a negative for the second part, so make it -1”

It’s an intuition/play around til it works kinda game

[pure current]

Ohh

Thank you let me try process now

[cunning fiber]

The determinant is equal to the given thing

Although it’s expressed a bit … messily.

[cunning fiber]

[pure current]

My brain is so confused 😅sorry

why did they make -1 become 0

Is it to see what happens?

[pure current]

but

So are they just trying to show 0 = 0?

[cunning fiber]

??

ax+by=1 -> ax+by-1=0

Same logic for the other two

[pure current]

yes

[cunning fiber]

Which gives you a determinant that is zero

Hence a^2+b^2+c^2-ab-ac-bc=0

[cunning fiber]

Hence why I said this

[pure current]

Ohhh

I think…

I get it now..

wait let me process again

Thank you I get it now

But why is it still kinda buggy

Manipulating zero?

[cunning fiber]

What exactly does this mean?

[pure current]

Wait

[pure current]

sorry I am new to matrix but

Nvm maybe

[pure current]

thank you once again👍

Have a good day or night

[cunning fiber]

[pure current]

.close

[marsh citrus]

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[marsh citrus]

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[lethal bridge]

A circle $\left(x-1\right)^{2}+\left(y-3\right)^{2}=9$ is passed by the line $y=x+d$. Prove that for there to be 2 points of intersection $2-3\sqrt{2}<d<2+3\sqrt{2}$

[elfin berry]

water beam

[lethal bridge]

not sure how to start

[cobalt widget]

Ok

So when u sub the like in the equation

U must get two answers right?

As they r intersecting at two different points

@lethal bridge

[lethal bridge]

i put in x+d into y?

[cobalt widget]

Ye

[lethal bridge]

$\left(x-1\right)^{2}+\left(x+d-3\right)^{2}-9=0$

[elfin berry]

water beam

[lethal bridge]

and then i solve for x?

[cobalt widget]

No

[limber hearth]

d

[cobalt widget]

Expand it first

[cobalt widget]

Niooo

Don't interrupt

We do b2-4ac>0

[lethal bridge]

oh ok

[cobalt widget]

Cuz two different roots

For x

At two different points

[cobalt widget]

So did u expand it?

[lethal bridge]

im doing it

[cobalt widget]

Kk

[lethal bridge]

ok this is what i get

[cobalt widget]

Ok

Now write it in descending order of x degrees

[lethal bridge]

ok

[cobalt widget]

Wait imma send a pic

Wait did u move the 9?

To the left side

Or no?

[lethal bridge]

yes

i move

[cobalt widget]

Kk

,rotate

[cobalt widget]

Ok now I see

This is like a quadratic equation right?

In terms of x

ax^2 + bx +c

[lethal bridge]

where 6d go?

[cobalt widget]

Oh mb

, rotate

[cobalt widget]

There

[lethal bridge]

yeah thats what i have

[cobalt widget]

Get it?

[lethal bridge]

yes

[cobalt widget]

Now do u remember d in quadratic equations

If d>0 the roots are real

[lethal bridge]

yes

[cobalt widget]

And unique

Ok

So here the roots are real nd unique right?

So d must be greater than zero

[lethal bridge]

i think

[cobalt widget]

Cuz there are two distinct points

[lethal bridge]

so i solve for d >0?

[cobalt widget]

D=b^2-4ac

Here yk all of that

Just do that and u will get ur answer

[lethal bridge]

,w simplify (2d-8)^2 - 8 -8d^2 + 24d

[elfin berry]

Results provided by WolframAlpha

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

Requested by waterrbeam

[lethal bridge]

thats what i get if i use b^2 - 4ac

[dapper aspen]

now check if that value is greater than zero between the given limits

[dapper aspen]

.

[lethal bridge]

so i plug it back in?

[dapper aspen]

ok wait

[dapper aspen]

this is wrong

[dapper aspen]

u forgot 2

a=2

u just did 4c

[dapper aspen]

in getting sqrt 13

[marsh citrus]

@lethal bridge Has your question been resolved?

[marsh citrus]

Channel closed

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[marsh citrus]

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[golden zinc]

I’m working on a physics simulation and I’m trying to implement a timescale effect to speed up or slow down the simulation. I’m not sure how to adjust the variables to achieve the desired effect. idk if should reach out to a physicist or a mathematician for this lol

Here are the details of my simulation:

Initial position (pY) = 0
Initial velocity (vY) = 0.5
Gravity (g) = 0.07
Friction (f) = 0.95
Timescale (timescale) = 1 (this is the variable I want to use to control the speed of the simulation)
The update happens every 0.03 seconds with the following logic:

If abs(vY) < 0.003, then vY is set to 0 (this step is optional).
pY is updated by adding vY. (pY += vY)
vY is decreased by g. (vY -= g)
vY is multiplied by f. (vY *= g)
I want to understand how to incorporate the timescale variable to make the simulation run faster or slower. How should I modify the calculations to account for the timescale? do i multiply all the variables by timescale? do some of the variables need to ^timescale? i dont know

[nimble prairie]

First off think of how this must work

Atm how much fps are you getting?

[golden zinc]

this simulation runs at a constant 0.03 seconds

[nimble prairie]

Okay so each step is 0.03 seconds?

[golden zinc]

yes

[nimble prairie]

So this is timescale = 1. I’m assuming you want a slower Timescale if it’s say 0.5?

[golden zinc]

more context: setting the timescale variable to 2 should make the simulation feels like 2x faster (without changing the step rate)

[golden zinc]

yes, and 2 should make it feels 2x faster and so on

[nimble prairie]

Cool. So all you should think about is how something slower would need more steps. I.e a timescale of 0.5 would just half the time it needs for a step.

So if timescale = 0.5.

The each step should take 0.03/2

Then a step should need to take half of

And you should think about it similarly if it’s faster, just that you need less steps instead so double the step time

[golden zinc]

is it not possible to only change the variables and not change the rate?

for example g*timescale

[golden zinc]

or am i wrong?

[nimble prairie]

I mean I guess yeah sure but that feels intuitive compared to actually doing as how a camera would work

[golden zinc]

but about the friction

i can't just do f*timescale right?

[nimble prairie]

Yeah it’s harder that way imo, you have to carefully think about how it must be affected and so on

I wouldn’t recommend that approach

[nimble prairie]

Just multiply by t on 0.003 instead tbh

That should do the trick

[golden zinc]

yes that work i've tried that

but that is not what i wanted

[golden zinc]

can i use math?

or do i have to manually guess?

[nimble prairie]

?

Why would this not be what you’d want?

[golden zinc]

making the step slower will require more processing power

[nimble prairie]

Yeah duh?

[golden zinc]

and collision is hella expensive to run (2 times slower means 2 times collision process)

[nimble prairie]

You can’t make something slower if you don’t actually have the information for it

[golden zinc]

so i want to just change the variables

[golden zinc]

what do you mean by that

[nimble prairie]

If a camera shoots a 1 second clip in different “timescales” what it just does is shoot for more steps when you want something slower essentially

Or else you don’t have the required information for a proper slow motion

You can i guess fake it, but then you’d need to do some sort of interposition between the steps which can look ugly

[golden zinc]

i already use linear interpolation

[golden zinc]

whatt

i thought i could just do
g*timescale
f^timescale

the friction is the main problem

[nimble prairie]

That seems like a hard way of adding more steps to your simulation, if you’re tweaking it incorrectly you’re just changing how the simulation will work as well for the cost of something “appearing” slow, when it’s not at all the same simulation if speed up back again

[golden zinc]

ohhh

so its really not possible?

[nimble prairie]

Yeah unless you fake it with linear interpolation like you mentioned

I’m not exactly sure how youre replaying your simulation, but if you’re doing linear interpolation in between the steps, just separate them so that the playback is slower

[golden zinc]

i did

[golden zinc]

could you correct me on this

i tried this and it doesn't feel accurate

or should i do f^(1/timescale)?

[nimble prairie]

I mean again even if you manage to do it properly, then all you’ve done is increased the steps of
your simulation

[golden zinc]

which is the same as 0.03*timescale right

[nimble prairie]

Yeah

It doesn’t magically work faster I would naively assume, unless there’s some extra cost to precision when doing it with the latter way

But I’m just thinking about it without a computer in mind since I don’t really know how your implementation works in code

[nimble prairie]

I would personally do this if the cost of more steps is too much

[golden zinc]

i will do that

back to the problem

[golden zinc]

can you correct me on this?

btw thanks for helping

[nimble prairie]

My physics knowledge is very limited, so I guess how did you arrive at it?

[golden zinc]

like how do you figure that out?

i just thought that x2 faster mean x2 gravity

[nimble prairie]

Hm what units is ur gravity in?

[golden zinc]

theres no meters or anything

[nimble prairie]

Gravity in physics is usually equipped with an implicit unit of some kind like with respect to netwons or something

[limber lodge]

this probably won't work because your equation is $y_{n+1} = (y_n - g) * f$ right

[elfin berry]

Blona

[limber lodge]

for speed

[golden zinc]

yes that should be the equation

[golden zinc]

the logic i provided is just a simple part of my simulation, it changes the pY upwards while the vY is still on its initial velocity, while decreasing the vY with gravity and friction and slowly pY will decrease too (when the vY is negative), and vY will become 0 sometime when it reach the threshold

so theres no newtons, units, meters, or anything

[nimble prairie]

I’m still sorta confused how tweaking the gravity and friction is more ideal, since if u happen to do it correctly it should be the same as just changing the step amount?!

[golden zinc]

i don't know i just think it's interesting to solve rather than changing the step

[golden zinc]

do you know the right operation should be?

[nimble prairie]

Like essentially you would just be reverse engineering so that your equations behave as how it would if you changed the steps.. which seems so much more complicated

[golden zinc]

yeah

but i feel like its somewhat solveable

with math

[nimble prairie]

Yeah and there’s one elegant way of going about it, by changing the steps directly. But since you’re persistent on doing it the more complicated way (maybe there’s some kind of benefit for that for you) then I won’t stop you! Hopefully someone else might be of more help

[limber lodge]

ok nvm this should work

though idk what you're modelling but it doesn't seem to be the correct equation for fall with air resistance

[golden zinc]

noooo, alright then thank you for helping me @nimble prairie

[golden zinc]

so whats the correct equation?

[golden zinc]

it doesn't

should i f^(1/timescale) instead?

NEVERMIND

ITS BECAUSE THE INITIAL VELOCITY IS NOT CHANGED

.close

[marsh citrus]

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[golden zinc]

.close

.reopen

wait no

[marsh citrus]

✅

[golden zinc]

aight thx guys

@limber lodge @nimble prairie

.close

[marsh citrus]

Channel closed

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[limber lodge]

normally you would make an ODE based on the physics equations and then numerically solve with e.g. euler's method or sth

you can message me later as i haven't woken up yet fully

[golden zinc]

.reopen

[marsh citrus]

✅

[golden zinc]

oh yess i will

i waved at your dms

[marsh citrus]

@golden zinc Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[exotic wolf]

hi

[exotic wolf]

how did they get the number to be 0.2 ?

should it not be 10/5000 = 0.002?

oh nevermind

i got it now

.close

[marsh citrus]

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[keen wren]

Give a,b,c > 0 ; a+ b + c = 8 ; find Min value of the expression F = 2a + 2b + 3c + 4/a + 4/b + 32/c

[keen wren]

My english is bad , im so sorry

[knotty trellis]

Can you use calculus?

[stray fern]

am-gm might work

[knotty trellis]

Didn't really work for me, but it's possible I just didn't try hard enough

[marsh citrus]

@keen wren Has your question been resolved?

[brave spire]

F=16+ c + 4(1/a + 1/b + 8/c)

[marsh citrus]

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[brave spire]

Am suspecting some quadratic thing

.reopen

[marsh citrus]

✅

[brave spire]

Sheesh I can do that

[marsh citrus]

@keen wren Has your question been resolved?

[marsh citrus]

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[lucid turret]

I haven't studied trignometry so i am not able to find it

[lucid turret]

how we got the last ineuqality?

how this is true

[lucid zenith]

Jensen's inequality

[lucid turret]

what does the triple bar mean?

hre

[lucid zenith]

Equivalency

[lucid turret]

wdym

Also like can we extend the last inequality to infinite terms?

inifnity angles

[lucid zenith]

It's just an equals sign

[lucid zenith]

Well no

Look up jensen's inequality

[lucid turret]

ok

[lucid zenith]

It depends on the concavity of the given function (i.e sinx)

[lucid turret]

oh. got it.

.close

[marsh citrus]

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[lucid turret]

. graphs are pretty useful

[agile crescent]

Question no.1

[agile crescent]

<@&286206848099549185>

[calm harbor]

,rotate

[livid swan]

0

[agile crescent]

Why

[calm harbor]

!nosols

[marsh citrus]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[obsidian kayak]

y/x = x is just y = x^2 which is x = sqrt y

[livid swan]

X is a dominator and shouldn't be equal to zero

[obsidian kayak]

now think which value y cant be

[calm harbor]

it's asking for y

[obsidian kayak]

you asked to find y not x

[obsidian kayak]

@agile crescent

[lucid zenith]

No, Ali is right

y can't be zero as well

[livid swan]

Yes y shouldn't be zero as if does then x will be equal to zero

[agile crescent]

But

[livid swan]

Which is wrong
X is denominator shouldn't equal zero

[agile crescent]

0/0 is defined

?

[calm harbor]

what's the problem with x being zero'

[obsidian kayak]

yeah

[lucid zenith]

It's in the denominator

[obsidian kayak]

now it makes sense

[livid swan]

Y/x
X!=0

[calm harbor]

mb

[lucid zenith]

y=x^2 and y/x = x are not identical equations

[lucid zenith]

What?

[obsidian kayak]

hmmmm

[livid swan]

Did you understand?

[agile crescent]

Yes

Yes

[obsidian kayak]

I think y cant be -ve and 0

[calm harbor]

ig it's a multi choice que

[obsidian kayak]

yes

[agile crescent]

Also -1 will be the ans

[agile crescent]

Yeah

Multiple correct

[obsidian kayak]

0 should also be the answer

[agile crescent]

Yeah

[obsidian kayak]

the answer would be 0 and -1 then

[agile crescent]

Yeah

[obsidian kayak]

it was a good question

[livid swan]

Why

[obsidian kayak]

bcoz - 1 doesnt satisfy y = x^2

[calm harbor]

in case y is -1, x would be i

[agile crescent]

If y=-1 then x=√-1

[calm harbor]

which is not a real number

[livid swan]

I see

[agile crescent]

The x is not real number

*then

[livid swan]

Yes

[agile crescent]

.close

[marsh citrus]

Channel closed

Closed by @agile crescent

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[agile crescent]

Help with questions no. 2

<@&286206848099549185>

[lucid turret]

,w expand (x^2 + ax+ b)(x^2 + cx+d)

[elfin berry]

Results provided by WolframAlpha

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

Requested by anshulyolo121

[agile crescent]

Then what

[marsh citrus]

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[timid crater]

whats the domain and range?

[timid crater]

<@&286206848099549185>

[devout peak]

Do you know what a domain is?

?

[timid crater]

no i need help bc im in math 1 and our teacher hasnt tought us yet

[devout peak]

You don't huh?

[timid crater]

yea

[devout peak]

Ok lets see

[timid crater]

its the begining of the year

[devout peak]

I literally just learnt this recently

[eternal schooner]

Set sqrt(4-x^2) = 0 and solve

[still temple]

if your teacher hasnt taught you why are you doing it

[timid crater]

BC SHE ASSIGNED TO US

[devout peak]

So basically domain is the set of values for which the function is defined

[still temple]

wild lmao

[timid crater]

ik

[devout peak]

And what is range

[still temple]

do you know how to solve inequalities

[timid crater]

teachers dont teach

yea

[devout peak]

It should be easy

What is range do you know?

[devout peak]

Are you gonna help him?

Ok help

[still temple]

if ur going to help then go haead

[devout peak]

I'll take my leave

[still temple]

oh

[devout peak]

Whatever you say

[still temple]

u hepl

[devout peak]

You sure?

[still temple]

i have to study now anyway i didnt realise how late it is

yeah

[devout peak]

Ok

@timid crater

You there?

[timid crater]

yea

sry my mom called me

[devout peak]

So let's start from the basics

It's alr

[timid crater]

ty im too stupid to know this

[devout peak]

What is a domain?

Do you know?

[timid crater]

its all possible inputs for a function

ik that

[devout peak]

Yes

[devout peak]

So what are the possible inputs here?

[idle ridge]

Please don't ping Helpers before 15 minutes have passed

[timid crater]

ok

[devout peak]

?

[timid crater]

someone asked not to ping hepers before 15 min thats all

sry

[devout peak]

I saw

So basically domain is teh set of all possible values for which a function is defined

[devout peak]

If we take this question here

What do you think should be the value of x for the function to be defined?

[timid crater]

one thing for sure is that x cannot be 2

bc 4-2^2 =0

[devout peak]

Yes

[timid crater]

so x<1 or x>2

[devout peak]

And why shouldn't it it be 0?

[devout peak]

No l listen

[timid crater]

ok

[devout peak]

Basically it shouldn't be 0

Because denominator can't be 0

And also whatever we have under the sqrt should be a non negative no.

And why is that

Because we are dealing with real valued function

[timid crater]

oh right bc if a square is a negative number we have to use i

[devout peak]

Yes

Imaginary

[timid crater]

as in imaginary number

[devout peak]

Yes

And because we're dealing with real valued functions we need to real numbers

So let's make an equation/inequation

What are our conditions?

[timid crater]

we know that x=1 or x>2

[devout peak]

Why?

[timid crater]

so if x=1 does that mean 1/2 is the minimum value

of f(x)

[devout peak]

Youre thinking about whole numbers

Not rational numbers

[timid crater]

right

[devout peak]

We can have a number like .0000000001

Could you resend the question

It's way too up now

[timid crater]

k

[devout peak]

Ok so

The denominator should be greater than 0 that's our only condition for it

[timid crater]

yea

and not equal to 2

[devout peak]

No

Don't be hasty

[timid crater]

ok

[devout peak]

We ll take it slow

Lets make an equation now

Sqrt(4-x^2) >0

We can square both sides to get rid of the sqrt

What would that result in?

[timid crater]

(4-x^2)^2>0 so (4^2+x^2)-8x>0

[devout peak]

What?

Squaring both sides would just remove the sqrt

[timid crater]

im tryna simplify it more

[devout peak]

So 4-x^2 > 0 only

[timid crater]

no

[devout peak]

Youre complicating it lol

[timid crater]

oh yea

i forgot

[devout peak]

Try writing it on a notebook

[timid crater]

the square root

ok lol

[devout peak]

Yes

And then we can easily find for x

Can you?

[timid crater]

ya its a quadratic equation

[devout peak]

Yeah

[timid crater]

gimme a sec

[devout peak]

And how would you find for x?

Sure

Write it on a n9tebook

And then send it

[timid crater]

by using the formula for solving it

[devout peak]

It's an overkill but ok

Wait bruh

Lemme show it you

Gimme a sec

[timid crater]

lol or we can also find the roots

my brain isnt braining today so yea

[devout peak]

I can see that lol

[timid crater]

-x^2=-4

[devout peak]

Yes

Not the equal sign

[timid crater]

x^2=4

[devout peak]

Greater than sign

Bruh

[timid crater]

x=2 or x=-2

[devout peak]

Bruh

[timid crater]

oh right

yea

just use the greater sign

[devout peak]

Lemme show

Wait

I got my book out

[timid crater]

x>2 or x<-2

[devout peak]

You sure about this?

[timid crater]

i can solve again if u want

[devout peak]

No

[timid crater]

wait

x∈(−2,2)

thats it

[devout peak]

Do you understand this much?

[devout peak]

Yes

[timid crater]

ya

[devout peak]

Ok we have to solve it further

This is the wrong way

This is generally how you do it

We still have a condition at last

[timid crater]

ok

[devout peak]

So basically the product has to be less than 0

That is only if 1 of the two factors is negative

[timid crater]

tysm

[devout peak]

Whole the other is negative

Try solving that

[timid crater]

oh

ok

[devout peak]

Separately

[timid crater]

lemme try

[devout peak]

Yeah go on

[timid crater]

my brain aint braining

i keep getting -2 or 2 or some answer like that

hello?

this is on hold rn

(IM COMING BACK AFTER LIKE 10 MINUTES)

[sacred idol]

ok im here

hello

this is interesting

[marsh citrus]

@timid crater Has your question been resolved?

[timid crater]

<@&286206848099549185>

<@&286206848099549185>

[still temple]

what seems to be the issue

ah domain and range

@timid crater are you here?

[timid crater]

yea sry

.close

[marsh citrus]

Channel closed

Closed by @timid crater

Use .reopen if this was a mistake.

[willow axle]

292919#8382892 ans

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
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[wintry warren]

Here is a question about trig with the correct workings out and answer. Could somebody help me in understanding part of the question please ?

[wintry warren]

I don't understand this point here. The person doing this says that they crossed out 7 and 8 because they were "to big" why is this ?

[cunning fiber]

$-1 \leq \cos \theta \leq 1 \implies \frac{1}{4} \leq 4^{\cos \theta} \leq 4$

[wintry warren]

I see

[elfin berry]

Civil Service Pigeon

[wintry warren]

Thank you that makes a lot more sense and I understand why 7 and 8 were crossed out now. Would you mind explaining how you get this ? The power of cos(vita) is confusing me

[cunning fiber]

$-1 \leq \cos \theta \leq 1 \implies 4^{-1}\leq 4^{\cos \theta} \leq 4^1$

[elfin berry]

Civil Service Pigeon

[cunning fiber]

Also, side note: it’s “theta”, not “vita”

[wintry warren]

Ah my mistake lol. Thanks I get it now, since you gave cos a base of 4 it meant that you had to do it for the lower and upper bound also

I didn't realise that

[cunning fiber]

[marsh citrus]

@wintry warren Has your question been resolved?

Channel closed

Closed by @wintry warren

Use .reopen if this was a mistake.

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

[lean falcon]

am i right with C?

[eager tinsel]

yeah you are

[calm harbor]

correct

[lean falcon]

ty

[marsh citrus]

@lean falcon Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[knotty trellis]

test

[knotty trellis]

.close

[marsh citrus]

Channel closed

Closed by @knotty trellis

Use .reopen if this was a mistake.

[knotty trellis]

.reopen

[marsh citrus]

✅

[knotty trellis]

.close

[marsh citrus]

Channel closed

Closed by @knotty trellis

Use .reopen if this was a mistake.

[lucid zenith]

what was this for lol

[knotty trellis]

#help-38 message

I thought that reopen is broken after the new update

if it already happened

[lucid zenith]

oh

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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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.

[plucky trail]

So I'm aware of the answers 800 I just don't know what the steps are to achieving that result can you help me so I can take notes and retry it

[carmine sandal]

So you're looking to minimise average cost

so C(x) needs to be

at its lowest right?

[plucky trail]

Yes

[carmine sandal]

How would you

go about finding this particular minimum

[plucky trail]

Inputing 100?

[carmine sandal]

not quite

[plucky trail]

Sorry I'm a little slow

[carmine sandal]

have u learnt

[carmine sandal]

no its fine

here to learn afterall

[carmine sandal]

differentiation?

[plucky trail]

Hm it's been about a month so I honestly forgot how to

[carmine sandal]

cuz there's two ways you can go about doing this

so i can tell u which one

depending on whether u have learnt differentiation or not

[plucky trail]

I don't think I did but I could be misremembering

[carmine sandal]

i will just tell you the other way

are you aware of

completing the square?

[plucky trail]

Nope ^^ hah..

[carmine sandal]

Right

so i think u shld revise and take notes on

completing the square

[carmine sandal]

its a way of solving quadratic equations, but in the process it also gives "extra" information about the equation

wont spoil it

but yh

[plucky trail]

Would you be willing to show me the way to do in differentiation? If not no worries

I think we learned it I just exited out of my mind not sure though

[carmine sandal]

oh that is quite simple

so

basically

ok ill just say it

if u dont get it yk u have to revise it

do yk the gradient

of a quadratic graph

at its minimum

[plucky trail]

The formula?

[carmine sandal]

No

so at a minimum

or any turning point for that matter

[plucky trail]

Nopes

[carmine sandal]

(and point of inflections but ignore without confusing urself)

the gradient is 0

so its basically flat

now differentiation

can help u calculate

gradient

at a given x

what u can do is

differentiate g(x)

equate it to 0

[plucky trail]

g(0)?

[carmine sandal]

now differentiation is

[carmine sandal]

no = 0

d/dx of g(x)

so differentiate the function g(x)

commonly written as g'(x)

and make that equal to 0

so

g'(x) = 0

[plucky trail]

Ah

[carmine sandal]

differentiation is simple

u do

ax^b = ab(x^(b-1))

[plucky trail]

a=0.1, b=-1.6 x=0?