help-28

[dim compass]

,rotate

[dim compass]

,rotate

[sick canyon]

For both the acceleration is -g right?

[dim compass]

When ball is going against gravity, it's -g, but if gravity is helping it accelerate it's +g

First ball, +g, second -g

[sick canyon]

Shit yeah

[dim compass]

You have to find the time where the ball is at equal height

As then only collision occurs

[sick canyon]

Yeah

[dim compass]

You can let that distance from either base or top be x

[sick canyon]

And 6 ms^-1 is the initial velocity for B right?

[dim compass]

That's the initial velocity for second ball yes

[sick canyon]

Cool

And im right to assume that g=9.81

Or do i need to assume its 9.8

Ive created two simultaneous equations using the v^2 suvat equation but that doesnt seem to get me the right answer

[full forum]

@sick canyon Has your question been resolved?

[sick canyon]

Help please

[shell marten]

first ball: $2-h=1/2gt^2$

[glossy valve]

zeeeee

[shell marten]

2nd ball: $h= ut-1/2gt^2$

[glossy valve]

zeeeee

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@sick canyon Has your question been resolved?

[full forum]

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[open matrix]

I am not sure how to start

[open matrix]

,rotate

[wild sleet]

same

[open matrix]

Apparently its something to do with integration but im not sure

[wild sleet]

omg i guessed it

[full forum]

@open matrix Has your question been resolved?

[wild sleet]

i think i guessed right in the sense that there's no easier method

https://math.stackexchange.com/a/1281949

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[dim compass]

R={(x,y)|x,y are real numbers and x=wy for some rational number w}. I need to check for reflexive, symmetric, transitive

[dim compass]

I have checked that's it's reflexive but checking symmetric is problematic for me

Because it says x and y both are real numbers. If I put x as irrational and y as rational that means w has to be irrational , but the "for some rational number w" restricts that because because rational * rational ≠ irrational

[spice orchid]

so that means that an irrational cannot be related to a rational

thats fine

[dim compass]

Yes

Oh

[spice orchid]

im not seeing why that would prevent you from showing symmetry

[dim compass]

Because I confused that with x, y belonging to real numbers, my bad

[spice orchid]

yeah all you have to do is show that if x = wy for w rational then y = zx for w rational

[dim compass]

Is 1/w rational if w is rational?

What if w is 0

i can divide w both sides to prove y=zx but that doesn't work for w=0

(x,y) can not be (y,x) if w=0 then

Which prove that symmetric doesn't exist

[dim compass]

Is that right?

I might be overthinking it

[spice orchid]

yes w being 0 seems to obstruct symmetry

[dim compass]

Alright

For transitive , a=bw is first equation, b=cw is second equation, I need to prove a=cw

I proved it transitive

is that correct?

By first finding b=a/w from first equation and then substituting it to second, a=cw^2...

[spice orchid]

w could be zero

[dim compass]

Ohh

[dim compass]

so again no transitive

Thanks

[spice orchid]

not necessarily

[dim compass]

Oh

[spice orchid]

just means your particular argument doesnt hold

[dim compass]

Ohh

[spice orchid]

if you have a = bw and b = cz show that a = (something rational)*c

without dividing

[dim compass]

Okay

w=a/b, and z=b/c, then I've to prove m=a/c.. where m is any rational number. then From first equation I get a=bw-(3), and second equation I get b=cz which I substituted in 3rd equation, and I get a=czw(cz=m) and upon dividing c both sides I get a/c=m which prove the relation is transitive.

[spice orchid]

again, dont divide

you dont know if anything is zero or not

[dim compass]

I didn't divide w, z or m

[spice orchid]

but you did b and c

[dim compass]

Yeah

Okay let me try again

[spice orchid]

nothing stopping them being 0

[dim compass]

Okk

lol I did it this time I think.

a=bw, b=cz and I simply substituted b=cz into first equation, which makes it a=czw

Didn't divide by anything

[dim compass]

Is this correct procedure?

[spice orchid]

yep!

[dim compass]

Wow, thanks!

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[sick canyon]

How do i do a)

Ive tried setting it up in v=u+at but i get an imaginary number when solving the quadratic

Since the velocity would equal 0 when it changes direction i figured doing that and setting it to 0 would give me the answrr

But idk how to do it

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@sick canyon Has your question been resolved?

[sick canyon]

<@&286206848099549185>

[fair oracle]

@sick canyon

[sick canyon]

hello

[fair oracle]

Say you have a acceleration time graph, what possibly could their turning points mean?

[sick canyon]

it means that the gradient is 0 therefore acceleration is 0

[fair oracle]

When does a particle changes direction?

[sick canyon]

when it has a stronger force applied that changes it?

like gravity or something

[fair oracle]

In terms of a particle, if the velocity is negative the particle moving in opposite direction right?

[sick canyon]

yeah

[fair oracle]

right, so we have the acceleration function

Integrate that to get the velocity, add the constant as initial velocity given

[sick canyon]

shit

i get it now

[fair oracle]

What we are trying to achieve is, we will find the stationary points in the velocity function where velocity is coming to zero, check values left and right of the points to find if its changing direction

[sick canyon]

so then use the quadratic to find the values of t?

so 1 or 1.5

[fair oracle]

yes

[sick canyon]

lovely

thanks alot

[fair oracle]

I’m not actually completely sure, whether turning points in a(t) graph changes direction of particle

[sick canyon]

just checked the answer and it is right so i assume so

[fair oracle]

My understanding is the turning points of acceleration function is where the velocity changes from increasing to decrease or vise versa

Right, the reasoning was correct

@sick canyon Changing direction means the displacement has to switch

[sick canyon]

lovely

[fair oracle]

which can be taken from turning points of velocity function

[sick canyon]

just getting last minute revision in for my as exam so the theory isnt the biggest problem

[fair oracle]

as velocity is the derivative of displacement - time

[sick canyon]

since as long as i have a method to get to the answer it should be fine

[fair oracle]

Nahh

Put little effort, you will find it good

Are you able to get the all 3 answers?

[sick canyon]

ive got the other 3, it was more just be getting confused about how to find the change in direction

the other ones are a piece of cake

[fair oracle]

Good luck

[sick canyon]

thanks alot

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[calm light]

What does it mean by oblique point of inflection?

[fair oracle]

Hopefully an image will help

[calm light]

So it’s just a regular poi

[fair oracle]

The above picture i gave has point of tangency as inflection point

because the slope decreases then increases at that point

oblique simply means it’s not horizontal or vertical

[calm light]

Ok

I’m also kinda confused what the hpoi would become as f(x)

[fair oracle]

hpoi?

[stoic kernel]

hpoi?

[calm light]

Horizontal point of inflection

[stoic kernel]

oh

[calm light]

On the f’(x) graph

[fair oracle]

Point of inflection is what the concavity changes

[stoic kernel]

where is differentiability is 0

[calm light]

So if was to draw f(x)

The hpoi would become a local max

[stoic kernel]

i am starting 11th lol i did not learn much sorry

I may not be able to answer that

[calm light]

That’s ok

[stoic kernel]

Sorry

[calm light]

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[dim sky]

can someone help

[mental dagger]

Interesting

Hold on

I will give it a shot

[dim sky]

ok

thanks

[mental dagger]

I don't think there is a way to solve this traditionally

Maybe hit and trial?

[dim sky]

waste of time

4x4x4x4 solutions

there has to be a way to solve it

[mental dagger]

Maybe, but they gave you the percentages themselves

I am unsure of this

[dim sky]

ok

[hot herald]

consider which percentages will lead to integer number of bananas

[dim sky]

not sure what thats ment to mean

[low panther]

He means you should try some combinations and see what doesn't work because it leaves you with a fraction of a banana

[dim sky]

idk

im clueless

[low panther]

Have you tried any combinations at all?
What happens if you put 10 in the first day and 75 in the second day, for example?

[dim sky]

then if u kept on goinjg

the number of bannanas would be less than 1

which isnt valid

[normal tree]

hmm what

[low panther]

Putting 10 in the first day leaves you with 9 bananas for the second day, and you can't get a whole number from 75% of 9, 50% of 9, or 59% of 9, so you know 10 is not a valid choice for day 1

[dim sky]

ye

[low panther]

Keep plugging in answers and ruling out options until you find your solution, there are very few combinations that give you usable numbers so you won't be going through all 4! combinations

[dim sky]

oo ok

[hot herald]

work backwards

[dim sky]

i treid this

it doesnt work

im not sure

bcz

he eat banana every day

1

first day he has 99 banna

[hot herald]

a banana is eaten at the end of the day

[dim sky]

oh

[normal tree]

how many bananas does the monkey have at the end of day 1 if you put 59% first?

see if you can calculate that

if not, try rereading the question carefully

[dim sky]

60

[hot herald]

how are you getting 60

[dim sky]

bruh

59 + 1 = 60

100 - 1

= 59

59 + 1 = 60

[hot herald]

why are you adding 1

[dim sky]

he ate 1

banana

[hot herald]

are you implying that if you had 1 banana, ate that one and only banana you have that
you'll now have two bananas?

[dim sky]

yep

if he had 1 banan at the start

and ate 1

then he will have the one from the start

and the one that he ae

ate

1 + 1 = 2

[hot herald]

no

[dim sky]

why not

[hot herald]

where's he getting this banana you're eating from?

out of thin air?

stealing it?

[dim sky]

no he already had it

[normal tree]

suppose I give you $59

you eat one dollar of it

suddenly you have $60?

[dim sky]

60 dollras

yep

[normal tree]

jeez I might start eating money

so why don't people eat money to get rich then?

[hot herald]

infinite banana glitch

[dim sky]

unhealthy

[hot herald]

only works with bananas it seems

[dim sky]

linen is unhealthy

[normal tree]

honestly can't tell if this guy is trolling at this point lol

[dim sky]

wdym bruh

i just want help

[normal tree]

suppose someone serves you dinner at a restaurant

after eating dinner

do you have two dinners sitting there?

[dim sky]

mayeb

if u order another one

[normal tree]

no

suppose you have a banana in front of you

and you eat the banana

how many bananas do you have in front of you

[dim sky]

2

wait

no

3

[normal tree]

yeah okay why don't you take a banana

put it in front of you

then eat it

and then count the number of bananas in front of you

[dim sky]

ok

i get one

[normal tree]

lmk what you count

[dim sky]

no bananas

i eat it

now it gone

whyy

[normal tree]

glad you figured this bit of knowledge out today

now suppose you have 3 bananas in front of you

and you eat one of them

how many bananas are left in front of you

[dim sky]

2

[normal tree]

suppose a monkey keeps 59 bananas

he eats one banana

how many does he have left

[dim sky]

58

[normal tree]

then why didn't you just give me a straight answer to the question I asked

[dim sky]

idkkkkkkkk

its hard

[normal tree]

no it's not

[dim sky]

ok but anyways

how do i solve that question

[hot herald]

identify which of
75%, 50% or 10%
of 58 is an integer

[dim sky]

50

58 / 2 = 29

[hot herald]

and how many do you have left after eating one

[dim sky]

28

[hot herald]

continue with the same idea with the remaining percentages

[dim sky]

75% of 28 = 21

10% of 21 = 2.1

[hot herald]

don't forget to eat a banana

[dim sky]

oh

20

10% of 20 = 2

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@dim sky Has your question been resolved?

[full forum]

@dim sky Has your question been resolved?

[dim sky]

nope

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[ruby blade]

If I'm given a straight line like : y=x+4, and I was told this system of reference was obtained after translating it originally from (0,0), is it possible to know only by the translated system of coordinates by how much it was translated on its x and y axis to re turn to (0,0)? Or is it not possible and there's an infinity of possibilities unless there's more info given?

[merry swallow]

if we assume no scaling/compression/stretching was done and only translation was done, then the original graph must be y=x, with the same gradient but passing through the origin so thats the only answer

[ruby blade]

yes, I know that, but what I'm asking for is how do we know by how much , vis-a-vis the origin, it was originally displced by?

For example, someone could have displaced it by -3 on the xs, and by 1 on the ys, or by -6 on the xs, and by 100 on the ys.

It seems like there's no solution to this

[merry swallow]

the translation does is the vertical displacement

in this case, it is a translation of 4 units in the positive y-direction

or equivalently a translation of 4 units in the negative x-direction

[ruby blade]

But I did actually (-3,1) when starting from (0,0) though

I translated my whole graph by -3 on the xs, and +1 on the ys

[merry swallow]

ah thats what you meant

yes you can do that

[ruby blade]

yeah

[merry swallow]

but its still equivalent to just translation 4 units in the positive y-direction

translating*

[ruby blade]

yes, how can I know starting from some random line, by how much it was moved vis-a-vis the origin

is it possible without extra info?

[merry swallow]

i guess when doing such questions, by assumption, they question is just asking for a single translation

but no in general if you can ahve as many translations as you awnt

[ruby blade]

no it's actually a question i came up with

[merry swallow]

then there are infinite

translations

[ruby blade]

yes ok, so there's an infinite amount of solutions/possibilities here then

[merry swallow]

you could even just start bouncing back and forth, does this line make sense?

so tranlsate positive y-direction by 4, translate negative y-direction by 4, translate positive y-direction by 4,.......

[ruby blade]

so to restart : if I have y=x+4, unless I'm given extra info, it's not possible for me to guess from how much it was originally moved when starting from (0,0)

because there's many ways it could have been done

[merry swallow]

it really depends on what you mean by originally moved, if you mean "distance", there are infinite

if you mean "displacement", theres only one

[ruby blade]

yes, in this case, -3, +1

is what I orignally came up with

but if I did now know that, it could have been by many other possibilities

[merry swallow]

yep

[ruby blade]

ok, thanks then!

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[zinc cosmos]

How is my answer for the tangent line not correct here?

[merry swallow]

your trying to ask how is your inputted answer different from the answer given?

[zinc cosmos]

I'm asking why my last 3 answers are not right

it's asking for a parametric form of the tangent line at the point where z=20

I plotted my answer as well and it is the correct line

so I'm just wondering why it's not the right answer in the context of the question

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@zinc cosmos Has your question been resolved?

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@zinc cosmos Has your question been resolved?

[full forum]

@zinc cosmos Has your question been resolved?

[full forum]

@zinc cosmos Has your question been resolved?

[umbral dome]

the issue is that you need the tangent line to pass through that point specifically when t=10

currently your line passes through that point when t=0

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[brave sigil]

hey do i still have to apply product rule in this equation?

[brave sigil]

θ(L) = T/GJ [L^2/(L − 2)^2 + 0.5]

t g and j are constant

[eternal schooner]

Huh?

[brave sigil]

sorry mb this is to differentiate to the respect of l

[eternal schooner]

?

[brave sigil]

yes

like would the answer be (uv)' = uv' +vu'?

[eternal schooner]

You don't need the product rule for this

[brave sigil]

or do i ignore the constant?

[eternal schooner]

Ignore constants

[brave sigil]

so the answer should js be the function on the left?

using the quotient rule

[gilded quarry]

Use quotient rule

Are T,J and L constants

[brave sigil]

nope js t g and j

[eternal schooner]

L is the only variable

[brave sigil]

its to diffrientiate to the respect of L

[eternal schooner]

Everything else is constant

so yeah, just use quotient rule

[brave sigil]

then tht shud js be the answer? without the t g and j?

[eternal schooner]

Differentiate

Then put constant back in place

easy peasy

[brave sigil]

sorry so the answer gon be t/gj and the differentiate?

[eternal schooner]

yus

[brave sigil]

ahhh okok tqtqt

[eternal schooner]

kinda like this

differentiate x^3 ignoring the constant 4 multiplication.
then like add the 4 back in later 🙂

idk

[brave sigil]

ahhh okoko got u thank you

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[lunar shadow]

anyone with a clue

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[sleek brook]

in

[sleek brook]

in byparts

how do i know what to take as u and what to take as dv

[severe basin]

Lookup ILATE rule

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[hardy folio]

calc 3

[wild birch]

Use the law of cosine

Formula

I think it was a dot b / magnitude of a/ magnitude b

But I’m not sure

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@hardy folio Has your question been resolved?

[hardy folio]

I'm kind of confused because it wasn't put in vector terms but rather points

I tried to find the distance between them and just solve it like a normal triangle

nevermind I got it, I plugged in the numbers incorrectly

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[supple willow]

@nova basin hey I understand X is a bernouli trial and S is the summation of n bernouli trials therefore it'sbinomial distributed by n,p but what kind of theorem is there used to do this

[supple willow]

or is this just reasoning because I feel like there is some rule used here that I prob forgot

[carmine minnow]

A Bernoulli trial is also BIN(1, p)?

[supple willow]

yes

[supple willow]

but how do I show that S = BIN(N,p)

[carmine minnow]

what sort of definition do you use for BIN(N, p)?

[supple willow]

what u mean?

the pdf?

[ember shadow]

the sum of n independent bernoulli trials is by definition binomial right

[supple willow]

oh is that a definition?

[supple willow]

would it then mean that the sum of r independent geometric trials become by definition a negative binomial distribution?

[supple willow]

wait can't I just solve this by one of the methods like cdf/pdf/mgf method

prob mgf method would be the fastest

[ember shadow]

actually yea i don't know if it's a definition but yeah mgf method should work

[supple willow]

after doing mgf method I should get the mgf of an binomial distribution and therefore it's binomial distributed with n,p right?

[ember shadow]

yes I believe so

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[misty oriole]

So I just learned in my physics class that pressure, given formula force/area, is affected more by the change of area than the change of force. Is there a mathematical proof to this statement?

[misty oriole]

I just learnt differentiation, maybe it can be used to explain this statement? Can anyone give me an insight?

[errant abyss]

not sure if this is right

no this is wrong

applying 2f is equivalent to lowering are to a/2

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[gaunt wing]

Need help with angles ;

How do I convert:

(30.0/0) facing south towards positive Z

To

(??/??) facing east towards positive X

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[final salmon]

How would I go about sketching this graph? I found the stationary points so do I just make a table and substitute values in?

[wise wyvern]

a. The function is an odd function.

b. You can calculate the second derivative for more accuracy. (concavity)

y' = (1-x^2)/(1+x^2)
Use its sign to find the function's nature.

[final salmon]

thanks

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[dense edge]

i know that addition is allowed (left example) with matrices

subtraction is allowed too?

[green lily]

yes

[dense edge]

or is it better practice to change - into +

for matrices

and change signs + to -, - to + within the
e f g h
matrix

[onyx glen]

you can rewrite A - B to A + (-B) whenever you want

it is not enough of a better practice to warrant agonizing over doing it every single time

[dense edge]

OK

what about division? i'm assuming multiplication is the way to go

100% of the time

using reciprocal of the second term

[spice orchid]

There is no notion of division of matrices

[dense edge]

why is that?

we have +, we have -, we have *

but we don't have /

isn't division just multiplication of the second term's recpirocal?

similar to how subtraction is just addition of the negative

[spice orchid]

How do you suppose you take "the reciprocal" of a matrix?

The point is not all matrices have a multiplicative inverse

[dense edge]

the reciprocal of the entries within the second matrix?

[spice orchid]

That is not going to do the thing you want it to do

Because multiplication is not entry wise

And the identity matrix is not a matrix of all ones

[dense edge]

oh right

because to multiply by the second matrix (reciprocal) it's like saying the original is
[1 2 3 | 4]
x + 2y + 3z = 4

but the reciprocal [1/1 1/2 1/3 | 1/4]
is like saying 1/x + 1/(2y) + 1/(3z) = 1/4

[topaz valley]

you can still divide things when multiplicative inverses don't always exist

but matrix multiplication isnt commutative

so when you write something like $\dfrac AB$ it isnt clear whether it means $AB^{-1}$ or $B^{-1}A$

[dense edge]

B is a function tho

i don't think you would want to write a reciprocal of B as 1/B

you would want to write it as B^(-1)

or maybe i'm confusing this with trig functions?
how 1/sin(x) is not the same thing as sin(x)^(-1) = arcsin(x)

[topaz valley]

when multiplication is commutative and multiplicative inverses exist then you can and do write 1/B

but this is not the case for matrices

[dense edge]

so this is the main reason why we avoid division of matrices?

the notation is just a nightmare

but it can still be done? if we use notation correctly

[topaz valley]

left and right multiplication by B^-1 when it exists does make sense

[dense edge]

i guess you are saying inverse of a matrix is the same thing as division?

it represents the reciprocal

[topaz valley]

it has many of the same benefits

[dense edge]

but only if it's AB^(-1)

[topaz valley]

?

[dense edge]

wait nvm

i thought reciprocal has to be second term

but maybe that's a misconception i have

to change multiplication into division

but in this case, B is the matrix we are dividing by

but multiplication of matrices is non-commutative

so AB^(-1) and B^(-1)A would give different answers, and I guess that's what you are trying to explain above

A/B is a mess

what is the difference between an inverse and a reciprocal?

they must mean slightly different things, ya? otherwise why have two words

i suppose when dealing with functions "reciprocal" would be a weird word to use

[topaz valley]

the reciprocal of a number x is 1/x

you probably wouldnt use the word reciprocal for matrices

[dense edge]

and all matrices are functions, ya?

vectors too

even 1x1 matrix is considered a function, i believe? [1]

[topaz valley]

matrices can be thought of as their associated linear transformation but they are also just their own thing

[dense edge]

to my understand, the function analogy for matrices, is also why matrices are non-commutative?

f o g

is not the same as g o f

[green lily]

try it out

[topaz valley]

you can think of it that way

[dense edge]

try what out?

[green lily]

think of two matrices f and g, do f * g, and then do g * f

[onyx glen]

i think this view might be too restrictive, even if it is helpful to a point

but it will take some time (spent learning linear algebra) before you get comfortable viewing matrices as objects in their own right, which correspond in a particular way to a certain type of function

[dense edge]

OK

what's the difference between RREF and Identity Matrix?

[spice orchid]

RREF is not always the identity matrix

It depends on the initial matrix

[dense edge]

does RREF always have 0's and 1's only?

or sometimes other numbers too

[dense edge]

oh, then this screenshot is bullshit

[spice orchid]

Well we can't see the context

[dense edge]

it makes it sounds like all other entries must be 0

but thats not always true for RREF

[ember shadow]

from a textbook I used

[spice orchid]

Yes your screenshot is misleading

[topaz valley]

example

,w rref{{34, -23, 12693}, {-19, 23, -6393}}

[glossy valve]

Results provided by WolframAlpha

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

Requested by 夢雪#2250

[spice orchid]

Oh actually your screenshot is just an example

[dense edge]

taken from this site https://www.instructables.com/Transforming-Square-Matrices-Into-Reduced-Row-Eche/

[spice orchid]

For the matrix on the left

[dense edge]

true, but misleading to understanding RREF and Identity Matrices

pure 0 and 1 is I

[spice orchid]

Well you would hope you have learnt what RREF is before seeing it

[dense edge]

this is RREF

any leading 1 with 0's above and below is the pivot column

[ember shadow]

yes

[dense edge]

so for column 3, there is already a pivot column for this row in column 2, we can ignore 3 2 and move on to next pivot column

in RREF some steps on the staircase going down may be wider than others

but for an Identity Matrix it's a perfectly diagonal staircase, with 0's and 1's only

[ember shadow]

I think you more or less got it

[dense edge]

https://youtu.be/EcgaeUUYV1U credit to this video for helping me to understand

when we talk about m*n

m always refers to rows?
n always refers to columns?

notation wise, all textbooks are the same with that

[ember shadow]

unfortunately no

[dense edge]

lol

[ember shadow]

it can be really frustrating tbh

[dense edge]

but we talk about matrices rows * columns

[ember shadow]

I've seen "m x n", "n x m", "n x p", probably "n x k" as well

[dense edge]

always, right?

[ember shadow]

but yes, I've only ever seen it "rows x columns"

[dense edge]

why on God's Green Earth would a textbook use n * m

English Alphabet = lmnop

m before n

wth man

some asshole out there must have put that into print as an evil thing to do

[ember shadow]

idk, I agree on that

[dense edge]

https://tenor.com/view/fan-thumbs-down-dis-like-mets-mlb-gif-9945612

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[dense edge]

.reopen

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✅

[dense edge]

wait so if the textbook says "suppose n = 1"

how do you know if it's talking about row or column?

[ember shadow]

I guess you wouldn't without more context. though I think there is usually consistency within a book, or a single professor

[spice orchid]

because if you're setting up a linear map from an m dimensional space to an n dimensional space, the matrix is (n x m)

things get swapped so there are valids reasons for either order

[dense edge]

but we always talk about matrices in row * column order

always

[spice orchid]

we do

a ( x ) matrix has rows and columns

[dense edge]

what is it called to modify a matrix to be the opposite? m * n => n * m
i haven't seen that matrix operation before

transpose?

[spice orchid]

transpose is specifically switching the rows and columns yes

[dense edge]

so n * m would be a good example of a transposed matrix of m * n?

but normally m * n would be the default matrix notation, 99% of the time in textbooks?

unless we transpose, then it's written as n * m 99% of the time in textbooks?

[ember shadow]

I wouldn't say anything as high as 99% tbh

[dense edge]

ugh... what a pain...

"suppose n = 1"
COLUMN or ROW?

[spice orchid]

you cant just say an nxm matrix is a good example of the tranpose of an mxn matrix, the transpose is a specific matrix that is nxm when the original is mxn

[dense edge]

i'm just gonna assume n is column, unless the matrix has been transposed

[ember shadow]

what's the context?

[topaz valley]

i dont think people will write suppose n = 1 without context

[limber mural]

Yh what are you trying to do are you trying to transpose a matrix

[dense edge]

well, i guess the only way to multiply matrices is with equal middle terms

so [m * n] [n * m]

we do this to match the middle terms, in this case "n"

so for the first matrix, [m * n], m = rows
and for the second matrix, [n * m], n = rows

so maybe in this case they have to be swapped to get the point across?

either way, [m * n] or [n * m] the the first term is always referring to rows
second term is always referring to columns

as far as I know, that part never changes in textbooks. if it does, it's teaching it wrong

[limber mural]

Where is the question

[ember shadow]

yes, I'd say that is the safe assumption

[dense edge]

understanding matrix notation, specifically with m * n when referring to rows and columns

OK ty

https://tenor.com/view/matrix-neo-gif-22464481

may help to dodge a bullet on that one, on the next test i get

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[torn jolt]

Help please

Ok

[hot herald]

ignore them

[torn jolt]

What that was wrong

[hot herald]

yes

[nocturne kettle]

of course

complete bs

34 + 90 + 88 is not 180

[torn jolt]

Oh NVM I got it

[hot herald]

<@&268886789983436800> being intrusive, obvious trolling

[cloud furnace]

a line drawn from the centre to the point og tangent is perpendicular to the tangent line

[torn jolt]

Stop trolling plz

[formal vale]

“Minus 12 then multiply by 4”

[cloud furnace]

try drawing AO and FO

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[main coral]

.reopen

Anyone know the answer to this

[light sonnet]

Complete the square

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[main coral]

I don’t understand how to

Like I’m not sure how to approach it

[gritty rose]

,tex .cts

[glossy valve]

riemann

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[crystal sparrow]

Help me

Pls

<@&286206848099549185>

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[odd turret]

a)
The coin is flipped 30 times, 23 times the coin lands on heads.
What percentage does 23 of 30 hold? In other words, 23 is how many percent of 30?

Do you know how to calculate this?

I see. So you calculate 23/30 and multiply by 100 to get the percentage.
You get 76.66%

and b) the estimate gets better because 'you have more data' to analyse. More flips = more accurate

yea

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[torn jolt]

so the function could be (0.6)^x

right?

because into years it will be 0.36

I don't know

[gritty rose]

falls by 36% means the ending value is (100 - 36)%

what are you trying to calculate?

[torn jolt]

Idk

[torn jolt]

oh ok

thanks

I think it's an english question more than a math one

[white fern]

welcome to math

[torn jolt]

I understood the question as 0.36 its original value

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[loud dagger]

Can you help me guys?

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[supple willow]

I don't understand the definition of a vertex

[supple willow]

I know they are the corner point

but the definition is a bit hard to undertand

[spice orchid]

"you can't draw a line through the point that stays within P"

[fathom saddle]

A point x is called a vertex of a feasible region P:

if x is in P,

and there are no two distinct points x1 and x2 in P, such that:
x = αx1 + (1 - α)x2
For some α

[supple willow]

I guess I can somewhat understand it now

[supple willow]

do u know if the cT is also called the like optimal solution>?

I don't understand optimal solution and objective value difference

oh wait nvm I understand it optimal solution is the optimal values for decision variable and objective values is the optimal value that the objective function takes

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[supple willow]

can u by any chance hlep me with this one?

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[sweet rain]

im just confused where to start or if its any different than a dif prism

is there a simple formula to use for any prism?

[eternal schooner]

If you'll look at the picture

You'll notice that it's just a triangle ( with depth )

[sweet rain]

thats 1/3Bh right?

area of base

[sweet rain]

no?

i mean not no to that

the formula

[quaint prawn]

area of a triangle is 1/2 * base * height

[sweet rain]

volume

[quaint prawn]

a triangle doesn't have volume

but

if your shape is just a triangle with added depth, then you can think of the base as being the triangle itself

and the height being however deep your shape is

think about how if you know the area of the triangle, and say you just stack a certain height, say (5inches) of this triangle on top of itself

then the volume of that shape will be, the area of the triangular base * 5 inches (in height)

[sweet rain]

just area of base * 5?

[quaint prawn]

for the randomly worded example I just made up yes.

but not exactly for your problem, although it follows the same method

[sweet rain]

5 inches tall

70 for area of base

i mean 35

1/2

[sweet rain]

so 35 * 5?

got it

can i get help with this one

theres just so many shapes. is there a dif formula for all of them or is there something to use for all of them

<@&286206848099549185>

[plush egret]

the shape is called a parallelepiped

if that is helpful

[sweet rain]

i solved it i realized that its the same as triangle

area of base x height

[eternal schooner]

Nice

Try to simplify your problems before solving them

Like the triangle one and this one

[plush egret]

and dont ask if youre not gonna keep us updated

unless youre just hoping we solve it before you

thats abuse of help

[eternal schooner]

If you just look at them as 2D shapes with depth instead of 3D shapes it becomes eaiser

[plush egret]

easier to think of them as pieces cut off a box

this parallelpiped could be solved very quickly in that way

as the difference of the volumes of two boxes

theres a large box

its 17*11*14

2 triangular pieces have been cut off, they can be put together to make a smaller box

[eternal schooner]

How is that easier?

[plush egret]

oh

im misreading

oh well u see what i mean

[eternal schooner]

Instead of just
Square with depth

You're like. Small box big box. Small box inside big box

[plush egret]

eh

personal preference

[eternal schooner]

idk

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[rough tundra]

If this correct?\~\

$\displaystyle \int (x-1)^{\mathrm{d}x}\implies\int\frac{(x-1)^{\mathrm{d}x}}{\mathrm{d}x}\mathrm{d}x\implies\lim{h\to0}\int\frac{(x-1)^{h}}{h}\mathrm{d}x$
Use l'Hôpital's Rule
$\displaystyle \lim{x\to0}\int\frac{(x-1)^{h}\ln(x-1)}{1}\mathrm{d}x\implies\int\ln(x-1)\mathrm{d}x\implies$\fbox{$x(\ln(x-1)-1)-\ln(1-x)$}

[glossy valve]

XxMrFancyu2xX

[rough tundra]

this seems very off, and honestly I was just playing around

[spice orchid]

i can only reiterate what i said last time, with knowing how we're supposed to understand and interpret the first thing, how are we supposed to know if its correct

[rough tundra]

oh damn sorry i didn't see your message, my other channel got invaded ;-;

[rough tundra]

I mean I guess you really can't that's the interesting? (for lack of a better word) thing ig, just attempting to interpret what cannot be interpreted ¯_(ツ)_/¯

[spice orchid]

then i fear that "is this correct?" is a bit of a nothing question

you've certainly pushed symbols around in a convincing way

[rough tundra]

My original goal was to do int x^(dx) because I wanted to abuse the notation as much as possible but my limit ended up being undefined so I opted for int (x-1)^(dx)

[spice orchid]

i assume you've seen the stack exchange post

[rough tundra]

ofc def not an original idea just thought about again and well i know this server exists now

So if I ought to rewrite my original inquiry: Is the multiplying by dx/dx then letting h->0 assuming dx=h but only the denominator and exponent—notation wise—proper?

[spice orchid]

well you would have to check that you can pass your limit inside the integral like that

but we don't even know what this integral is, let alone whether we can exchange limits (if this "integral" even involves a limit anymore)

[rough tundra]

I see, the person on the stack exchange did that, so then the proper answer would be that it breaks/abuses notation so bad it breaks down the fundamentals of the notation itself?

[spice orchid]

possibly? hard to say

[rough tundra]

lol funi

[spice orchid]

lol i just think as a general rule you should work with things that make sense

whatever that means

[rough tundra]

Fine, up until I graduate then I'll start breaking things again

well thanks man, big help!

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[karmic meadow]

how do you do this?

[karmic meadow]

never mind

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[dense edge]

how does Row1 change?

1 4 2 0 4 2 -1 suddenly becomes 1 4 0 0 6 -4 -5

calculator error? i'm using https://matrix.reshish.com/gaussSolution.php

something is going on here and I don't understand what

to make the 2 into a 0 for (r1, c3)

[dense edge]

oh I think I see it

[onyx glen]

R1 := R1 - 2*R2 is the operation that was done

[dense edge]

-2R_2 + R1 -> R1

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@dense edge Has your question been resolved?

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[dim compass]

Is the range of x^12 - x^9, [0, infty)?

[dim compass]

.close

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[onyx glen]

why close immediately after opening

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[white spear]

hello can someone help me solve this problem

[white spear]

its vectors for pre calc

[full forum]

@white spear Has your question been resolved?

[willow sedge]

find initial speed in y direction and displacement in y direction

[white spear]

so i sorta drew a shape

and i dont know if the rock is like curving as it drops

or drops in a linear path

[willow sedge]

yes it is

But that doesn't make it harder

[white spear]

really?

[willow sedge]

yes

[white spear]

because i thought if it was a triangle i could use law of cosines and sine

[willow sedge]

Since the angle is 25 degrees, the ball must curve

[white spear]

oh

[willow sedge]

no way that would be possible

[white spear]

yea ur probably right

[willow sedge]

are you aware of kinematic formulas?

[white spear]

uh no

our teacher taught us like vectors and velocity and magnitude

[willow sedge]

v=u+at
s=ut+1/2at^2
v^2=u^2+2as

[white spear]

and randomly this question shows up

[willow sedge]

Okay, so we should go by vector method

[white spear]

yea

is this what its supposed to look like

[willow sedge]

no

[white spear]

shoot

does it curve up and then down

[willow sedge]

Bad drawing

[white spear]

ohhh 25 is on the inside

i always thought it was outside

im so lost i dont know how to solve this with a curved line

because up until this point we've only learned about using parallelograms and triangles

[willow sedge]

It would have been way easier if you knew kinematic formulas

[white spear]

where i could use law of sin/cosines, alternate interior angles etc

[white spear]

yea...

[willow sedge]

There sure is a method to solve it using vectors but I'm not too aware of it

You don't even know a bit about these formulas?
@white spear

like ut vector, 1/2gt^2 vector

[white spear]

Nope

I havent taken physics

[willow sedge]

Have you missed some classes?

[white spear]

Nope

I dont even know how that question ended up in my practice

[willow sedge]

Higher grade question?

[white spear]

All the other ones were fairly easy bc they were triangles and parallelograms

[white spear]

Honors pre calc

[willow sedge]

What have you learned so far in vectors?

As far as I know you'll need those formulas to solve this

[white spear]

Not much

Our teacher isnt the best at teaching....

[willow sedge]

can you show what previous questions were like?

[white spear]

I just know about velocity and magnitude and forces that can be applied to those problems

Then i can create triangle like shapes to solve

Thats what the worksheet looks like

Im probably just gonna go sleep and ask my teacher abt it tmrw

.close

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[tame bobcat]

Hi everyone , I dont know why the solution to the following question is correct.
Given two random variables X and Y, X is the number we get by throwing a fair dice with eyes 1 through 6, and Y is the number of times we get heads by tossing a fair coin X times.
The exercise wants me to tell the conditional distribution of X given that Y=5.
Now for X=x , x in{1,2,3,4} , thats obviously 0 , alright. But then it says for X=5 we get P(X=5 | Y=5) = 1/4 and then P(X=6 | Y=5) = 3/4
How does that make sense ?
We are supposed to use bayes formula and that would lead me to

P(X=5 | Y=5) = P(X=5 , Y=5) / P(Y=5)
And since X and Y are independent that should leave me with P(X=5) which is 1/6

can someone explain ?

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@tame bobcat Has your question been resolved?

[onyx glen]

X and Y are not independent at all

Y very explicitly depends on X lol

[tame bobcat]

yesterday someone helped me with the exercise that comes right before this one, and we figured they are independent

[onyx glen]

?!

[tame bobcat]

the question with the prior exercise was

one second

to tell the joined distribution of X and Y

[onyx glen]

for these same X and Y?

[tame bobcat]

and then yes

this question I initially posted in this channel is the following question to what I asked yesterday

[onyx glen]

X is the roll of 1d6, and Y ~ Bin(X, 1/2)

[tame bobcat]

yes

[onyx glen]

there is somewhat of a sneaky way of making the distribution table that essentially provides a lot of lubricant (by ridding us of most of the fractions in a smart way)

[tame bobcat]

well what I now dont understand is, when we would for example want to calculate
P(X=3, Y=2) then we'd do
P(X=3, Y=2) = 1/6 * 3 * 1/8 = 1/16
And that to me looks like X and Y are independent and therefore we can simply multiply their probs with onanother. But now you are saying the are not independent
what am I missing ?

[fallen bone]

#help-38 message
^Linked message spreading misinfo.

[onyx glen]

3*1/8 is P(Y=2 given that X=3)

so this is P(A&B) = P(A|B)P(B)

[onyx glen]

ah yes it's fäf

[tame bobcat]

damn thank you xD

[onyx glen]

i have known fäf to be confidently wrong at times

[tame bobcat]

could you elaborate on this one ?

[onyx glen]

why it's true or how it applies?

if both which first

[tame bobcat]

because where do we get P(A|B) from ?

why its true first please

and both, yes

[onyx glen]

P(A|B) is by defn P(A&B)/P(B)

[fallen bone]

Anyway processes are clearly not independent. You only get nice probabilities conditioning on some event.

[onyx glen]

not much else to say on that

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[onyx glen]

3*1/8 is P(Y=2 given that X=3)

.reopen

[full forum]

✅

[tame bobcat]

.reopen

[onyx glen]

lmao

all 3 of us

[tame bobcat]

yes

[onyx glen]

so you know, just turn that around

multiply both sides by P(B)...

[tame bobcat]

ok but where do I get P(A|B) from ?

[onyx glen]

Y's defn

[fallen bone]

From the question

[onyx glen]

the distribution of Y conditioned on X=3 is Bin(3,1/2)

[fallen bone]

The random variable after conditioning is nice

Before conditioning, well, IDK, I never computed it

But you should intuit that Y is more heavily weighted towards 0

[tame bobcat]

in total you mean ?

[fallen bone]

X has a simple distribution

[fallen bone]

Yes, after accounting for every possible X

[onyx glen]

weh

well ok like.

[tame bobcat]

I am still not sure whether I got it right, now lets say for X = 3 and Y=2 , we want to compute P(X=3 & Y=2) = P(Y=2 | X=3) * P(X=3) ?

[onyx glen]

i was gonna show this but it definitely takes explanation

[tame bobcat]

hold on

thats also in my book

its the same table, right ?

[onyx glen]

essentially.

i basically factored out the common denominator outright

[tame bobcat]

oh I just gained an insight

[onyx glen]

also theirs is transposed relative to mine but no biggie

[tame bobcat]

I understand what you meant by
P(Y=y | X = x) follows from Y's definition

its just the binomial distribution of Y with the parameter x and y

so P(Y=2 | X = 3) = 3*1/8

and therefore in total when we do P(Y=2 | X=3) * P(X=3) we get the right result

ok now I see

but I still dont understand why P(X=5 | Y=5) = 1/4
We have
P(X=5 | Y=5 ) = P(X=5 & Y=5) / P(Y=5)
= (1/16) / (1/48) > 1 != 1/4

[onyx glen]

for P(X=5|Y=5) you essentially end up needing Bayes lmao

[tame bobcat]

you mean bayes inverse formula , right ?

[onyx glen]

whatever you call it

the relation between P(A|B) and P(B|A)

[tame bobcat]

but isnt that only needed whenever we dont have P(A) ?

I mean we have P(Y=5) and P(X=5 & Y=5)
why cant we just calculate it

[onyx glen]

recommend not putting yourself into such tight constraints

[onyx glen]

we don't actually have P(Y=5) yet do we

[tame bobcat]

I calculated it in the prior exercise

[onyx glen]

oh so you have

[tame bobcat]

its 1/48

[onyx glen]

,calc 8*48/384

[glossy valve]

Result:

1

[onyx glen]

ok checks out

P(X=5 & Y=5) = 1/192 as per the table

either one

[tame bobcat]

oh I see what I did wrong

for whatever reason I used the value of P(X=3 & Y=2)

instead of the value for P(X=5 & Y=5)

lmao , thats so stupid xD

[tame bobcat]

alright, thank you so so much ann ! And also thanks a lot to you @fallen bone !

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[versed cape]

how do u solve this

?

<@&286206848099549185>

[gilded quarry]

The given limit is wrong

[grave elm]

it should be log(a)

natural log

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@versed cape Has your question been resolved?

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[calm tinsel]

The curve y(x) satisfied t he differentia equation d^2y/dx^2 = 2x+x and has a local minimum at (1,1) find y(x)

[calm tinsel]

I have the 2nd, 1st, and original equations

first being: x^2 + 2x + C

original being: x^3/3 + x^2 + Cx + D

where do I exactly sub in the 1?

[ember shadow]

well how do you find the minimum of a function in general