help-23

[worthy hemlock]

Post that too

[worthy hemlock]

Probably just a typo then

Should be this

[lean otter]

thats what im saying

i was so confused

im like WHAT DID I DO WRONG

.close

[safe radish]

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[slow garden]

Can anyone tell me what it means?

[slow garden]

I haven't seen such an expression

and yes I know it is calculus

[stoic dune]

So one can interpret the derivative as a linear operator

From this point of view, (d/dx)^n means "take the derivative n times"

[slow garden]

ok, I got it

[stoic dune]

d/dx - 1 means "take the derivative, then subtract the function"

And that ^n means "do that n times"

These actually follow nice algebraic laws, like the binomial theorem

[slow garden]

does the expression in here equivalent to this?

[stoic dune]

What are C and D?

[slow garden]

C is the combination

[stoic dune]

Oh I see

[slow garden]

and D is the differential operator

[stoic dune]

This is the binomial theorem I was just talking about haha

[slow garden]

yo wtf

[stoic dune]

Applied to (d/dx - 1)^n

[slow garden]

oh I forgot the 1/n!

[stoic dune]

Me bringing it up at the same time as your book isn't that big of a coincidence - the second you consider derivatives this way, this is a big result you get

Oh yeah the 1/n! doesn't seem to fit

[slow garden]

I was actually studying Laguerre polynomials

but I am a bit confused

like, taking a derivative of this formula but how does it gets the following appearance?

@stoic dune

[safe radish]

@slow garden Has your question been resolved?

[safe radish]

@slow garden Has your question been resolved?

[slow garden]

Can anyone tell me how this is transformed from that to the picture in the latter message?

[safe radish]

@slow garden Has your question been resolved?

[slow garden]

@stoic dune can you tell how it is possible?

[stoic dune]

Sorry haha, we're out of my element

I don't know. Perhaps a textbook on these polynomials might have a proof on why these work

[slow garden]

<@&286206848099549185>

Can someone please fucking help me?

I am here stuck for more than 3 hours

[plucky elk]

Do it by induction

[plucky elk]

Your original question was answered. If you want help with a new question, a better approach is to open a new channel.

[slow garden]

Can you give me an esample of what you mean by "induction"?

[plucky elk]

https://math.stackexchange.com/questions/1060232/use-induction-to-prove-that-legendre-polynomials-solve-the-corresponding-differe

https://youtu.be/dMn5w4_ztSw

[slow garden]

I seriously never used this type of thing

[plucky elk]

So learn it? Idk what you mean

[safe radish]

@slow garden Has your question been resolved?

[slow garden]

.close

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[pine tulip]

A quadrilateral have a diagonal of 1 and the other of 2. What's the surface value?

[fiery fiber]

what is a surface value

surface area ?

[pine tulip]

yes, i meant the surface area

the correct answer of this problem is A<1

but i dont understand what formula its used in this case

I know that the formula to find the area of a quadrilateral using perpendicular diagonals is D_1*D_2/2

but its not the case here

[lilac patio]

Quadrilaterals don't have surface area they are two dimensional

[fiery fiber]

surface area here means area

[lilac patio]

Ah

Use trig

[pine tulip]

how?

[lilac patio]

I think there's a way to find the interior angles with that information

Not 100% sure though

[pine tulip]

I tried looking online but didn't find anything usefull

[lilac patio]

Not sure then if you're formula doesn't work

[fiery fiber]

hey try construction one

or just a rough figure

maybe it will click somehing

[pine tulip]

Well i understand that a single diagonal is able to cut the perpendicular in 2 equal triangles

wait, so i can find the area of the triangles that are generated from these 2 diagonals

Okay, i felt like i was onto something but i don't know what else to do

.close

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[slow fern]

,tex If two numbers when divided by a certain divisor give remainder 35 and 30 respectively and when their sum is divided by the same divisor, the remainder is 20, then the divisor is

[flat frigate]

QuantumBee

[slow fern]

<@&286206848099549185>

[candid storm]

set up some variables and a system of equations

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[gusty wharf]

Hey i am working on a project in which instead of transmitting the signal i want to first convert it into mathematical equations, my question is what is the best way or how can i convert my signal to series of mathematical equation.

[slow fern]

[slow fern]

How?

[candid storm]

let the 2 numbers be x,y and the divisor be d

[slow fern]

So, I need to use Dividend = Divisor*Quotient + Remainder?

[candid storm]

also we should prolly move channels

someone claimed this one already

[slow fern]

umm

is that really necessary?

[candid storm]

i mean if i were you i would because i can

[slow fern]

ok here are the equations
x=dq+35
y=dq'+30
x+y=dq''+20

I'm considering the quotients to be different.

Is that right?

[candid storm]

thats right
now is there another way to produce another equation with x+y

[slow fern]

add the first two equations?

so

[candid storm]

idk see what you get

[slow fern]

x+y=d(q+q')+65

[candid storm]

so we have
x+y=d(q")+20
x+y=d(q+q')+65

[slow fern]

yeah

what's next?

[candid storm]

so what does this mean when we divide x+y by d

[slow fern]

we get remainder as 20

[candid storm]

and from another equation?

[slow fern]

the remainder is 65

[candid storm]

so we have the same number divided by the same divisor

actually you know what heres an easier approach

[candid storm]

so we know that x+y divided by d gives a remainder of 20
which means x+y can be written as d(something)+20

so we should be able to write x+y=d(q+q')+65 in said form

so what happens if we try to write d(q+q')+65 in d(something)+20

[slow fern]

I don't understand

[safe radish]

@gusty wharf Has your question been resolved?

[candid storm]

im sorry; here

so we know that x+y=d(q")+20
and x+y=d(q+q')+65
thus
d(q")+20=d(q+q')+65
can you simplify this equation

[slow fern]

d(q") - d(q+q') = 45

[candid storm]

yes; can you go further

see that both terms on the left-hand side of the equation has d as its factor

[slow fern]

$d = \frac{45}{(q"-q-q')}$

[flat frigate]

QuantumBee

[candid storm]

so d(q"-q-q')=45

what does this tell us about 45 and d

(we know that both d and (q"-q'-q) are integers)

[slow fern]

d is a factor of 45?

[candid storm]

yes

so what are possible values for d

[slow fern]

1,3,5,9

[candid storm]

theres more

[slow fern]

yes

45 has 6 factors

[candid storm]

yeah

so lets look at the original problem

traditionally, can the divisor (d) be less than the remainder (30,35,20)?

[slow fern]

traditionally?

yes I guess

[candid storm]

like normally when you do division can you end up with a remainder larger than the divisor

[slow fern]

no

hmm that is understood

[candid storm]

so what are possible values of d

with that in mind

(d is a factor of 45 and is larger than 30,35,20)

[slow fern]

45 itself?

[candid storm]

yes

thats our answer

[slow fern]

answer is 45?

[candid storm]

here i was gonna say d(q")+20=d(q+q')+45+20 so d must be a factor of 45

[candid storm]

yeah

[slow fern]

Alright, understood.

[candid storm]

for the first thing i was gonna bring up i was gonna say that if the remainders can be 20 and 65 the divisor should be a factor of the difference between these numbers (45) which is a little more complicated to explain

[slow fern]

thanks.

[candid storm]

np

[slow fern]

@gusty wharf sorry for interrupting you!

[slow fern]

.

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@gusty wharf Has your question been resolved?

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@gusty wharf Has your question been resolved?

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@gusty wharf Has your question been resolved?

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@gusty wharf Has your question been resolved?

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[shadow glade]

what's a horizontal limit?

[heavy frost]

I was thinking of something completely different

The vertical asymptotes should be easy bc you take the limit as x approaches the x value of the asymptote

Then you can observe the limit go to inf or -inf. Which proves it was an asymptote

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[lean otter]

could someone teach me how to solve this problem, my teacher gave it to us as a bonus question but never taught us how to solve it, ive tried many times but couldnt get it

[worthy hemlock]

Maybe because it was a bonus question, it means the people who know the concept will be able to do it, but it's not an important concept to know at the moment, so therefore it might not be needed for you to currently know

[lean otter]

but could you teach me how to get the answer for future reference

i really want to know so i could be a bit ahead in class

[worthy hemlock]

I mean, just ask your teacher

It's an extra concept

You're just solving an equation

[lean otter]

Breh a 13 yo ofc knows how to solve equations

[lean otter]

kk

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[prisma wedge]

how do i write csc(theta) in terms of tan(theta)

[frail carbon]

tan only or can you have cos ?

[frail carbon]

ok

so theres the identity cot^2 + 1 = csc^2

[prisma wedge]

right

but how do i go from csc theta

to that

@frail carbon

[safe radish]

@prisma wedge Has your question been resolved?

[prisma wedge]

<@&286206848099549185>

[frail carbon]

sorry i was busy

cot^2 + 1 = csc^2
so csc = +/- sqrt(cot^2 + 1 ) right

but what is cot = ?

[prisma wedge]

im so lost lol

can u draw it out possibly

[prisma wedge]

so is there answer just +- 1tan^2+1

[frail carbon]

theres a sqrt

yes

[prisma wedge]

oh i c

[frail carbon]

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[ancient vessel]

What is the radius of a 1092^2 regular heptagon? i have no idea how to start

[marsh walrus]

wait what does 1092^2 mean

[ancient vessel]

ft

^2

[marsh walrus]

oh, area?

[ancient vessel]

like its area is 1092

yh

[marsh walrus]

ah

@frail carbon

well, can you find the area of a heptagon?

whats the area of a regular polygon in general?

[shadow glade]

divide it into 7 triangles, each of those would have 1/7 of the area

they would be isosceles triangles whose legs are the radius of the heptagon

and you can figure out the angle at the center of the heptagon etc, try going from there

[ancient vessel]

1/2(a)(p)?

[ancient vessel]

156 no?

[ancient vessel]

how do i find the angle at the center

[shadow glade]

there are 7 triangles that are all congruent that you form around that point right? so 360 / 7

[ancient vessel]

o

51.42

43

ohh i see

wat does that help with

[shadow glade]

you have an isosceles triangle with a known area and known angles, the legs of the isosceles are the radius of the heptagon, you can solve for the length of the legs from there

[ancient vessel]

oh

[ancient vessel]

i srsly dont know how to use the area to find the leg

is it trig?

[shadow glade]

ya, just try dividing that green triangle in two down the middle to get a right triangle then work with that

oftentimes when you have isosceles triangles its helpful to just work with half of it since its always a right triangle

[ancient vessel]

ok

ok i got all the angles

of the right triangle

not sure if its gonna help lol

.cloaw

.close

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[feral root]

Ok, I'm playing Backpack Hero and trying to figure out damage maths.

I'll try to explain it in a way that makes sense to someone who hasn't played the game

I've got two items that use the same resource, and I can activate them a total of 33 times. The first item does 10 damage and adds 2 to each activation of the other item. The second does 20 damage, increasing by 1 each activation. What ratio of activations would maximise output?

[shadow glade]

um well since the first improves the second, but the second doesn't improve the first I guess you would just use the first one up then start using the second, if I understand what you're saying

oh wait

i see you can only use 33 total

[feral root]

Yeah

Hence my conundrum

[halcyon copper]

so basically u have items x and y, using x adds 10 damage to total, y adds 20, using x adds 2 to y's damage, using y adds 1 to y's damage

is this the situation?

[feral root]

Yes

[halcyon copper]

ok

[shadow glade]

so using x will add +2 damage to all of y's next activations?

[feral root]

Yes

[halcyon copper]

so first i believe it makes sense to use up the x type items?

since using the same number of x items later will not add to the used up ys which is what we want

for example if the last item u use is x, the +2 effect on y gets wasted

[shadow glade]

yeah you'd definitely do all the x's first but question is how many? : )

[halcyon copper]

i got a simple quadratic

and a reasonable answer

hold on

[shadow glade]

if you ignore the +1 from the y since its kinda insigificant you can try to just maximize (33-x)(20+2x)+10x

thats my suggestion

[feral root]

33 activations between them, so I need to know how many of each to do. 17/18 for example

It's really not insignificant. It's not just +1 per activation, it increases by one each time. So if it were activated 3 times, it'd add 6 damage, 4 times would be 10, so on and so forth

[halcyon copper]

yes it's not

[shadow glade]

i got x=14 with my approximation without considering the +1 for the y

so x = 14 then the other one 19 times

seems reasonable

:p

[halcyon copper]

ok so i got an arithmetic progression sum

so first rename the items as A and B instead of x and y

(just convenience)

[feral root]

Someone in the game server suggested

x = amount of A uses:
(20+2x)(33-x) + 0.5(33-x)(33-x+1)

[halcyon copper]

and let the number of items used be x and y for types A and B respectively

[feral root]

Yep yep

[halcyon copper]

so u have x+y=33

[feral root]

Let me just edit and swap them around

[shadow glade]

ya thats just using triangular numbers to do the part with the +1 but honestly why bother lol

just leave it out and you might be off by 1 : )

maybe its 13 instead of 14, but at least you can do the math with pencil and paper in 10 seconds

[halcyon copper]

now x's contribution = 10x

and y's contribution = y/2(2(20+2x)+(y-1)1)

here 20+2x is the initial 'increased damage' start for y

[shadow glade]

i just approximated it with (20+2x)(33-x) + 10x, derivative is 56-4x=0 so the max is at x=14, simple enough

[halcyon copper]

the exact thing gives 8

so big difference

[shadow glade]

oh : ( haha

that doesnt seem right to me

[halcyon copper]

lol

i used wolfram so

i got 14 too ignoring +1

but it feels intuitively right for more y than x so yeah

[shadow glade]

so whats the total value for when you use 8 vs 14

: )

[halcyon copper]

ye that lemme chack

[feral root]

The person in the other server made a graph

[shadow glade]

im getting a higher value when using 14

😄

[shadow glade]

guess my math is off :p

[halcyon copper]

yeah cuz u didn't incorporate 1

[shadow glade]

i did

[halcyon copper]

lol

[shadow glade]

or thought i did

but im doing lazy math again

[halcyon copper]

xD

@feral root does this make sense?

[feral root]

Uhhhhh, slightly? I'm not very maths-minded, so not a lot of mathematics makes sense to me

[halcyon copper]

we're basically just using arithmetic progression

[shadow glade]

ok so help me understand what im doing wrong with my calculation

with x = 8

[halcyon copper]

u could try out sooshon's method for an approximation

[shadow glade]

so the x's do 8*10 that much is clear

then the first y does 20+16 = 36 right?

[halcyon copper]

yess

then second one 20+16+1

n so on

[shadow glade]

and the last y does 36+24 right (since there will be 24 before it adding 1)

[halcyon copper]

wait lemme check again

[shadow glade]

so on average the y's do 48

in the middle

so the y's should contribute 48*25

so total should be 80+48*25

ah 1280 yeah

ok guess i just put in numbers in calculator wrong first time or something

[halcyon copper]

nono it's correct

u prolly didn't use +1 cuz i got 14 exact without that

[shadow glade]

yeah if it was just for some game purpose, quick approx is easier :p

[halcyon copper]

@feral root ig Sooshon's method (the one now) will be easier to understand

[shadow glade]

that one i could do on paper without a calculator

[halcyon copper]

yeah

[shadow glade]

but i guess the +1's add up more than i thought

[halcyon copper]

than mine

yeah true

cuz it's just half of +2

wouldn't make much diff if it was +10 and +0.1 instead of +2 and +1

[feral root]

Ok so I guess my optimal ratio is 8/25, yeah?

[halcyon copper]

ye

[feral root]

Ok

[halcyon copper]

if u want the final thing yeah

[feral root]

Say I were to add more to that total number of activations. What's the equation, with N replacing 33?

[halcyon copper]

yeah

and 32-x would be N-x-1

$10x+\frac{N-x}{2}\left(2(2x+20)+N-x-1\right)$

[flat frigate]

newbienoob

[halcyon copper]

N is the number of activations

and u differentiate and maximize this to get ur values

[feral root]

Thank you! I guess I could put this into a graph maker to do that?

[halcyon copper]

wolfram alpha

copy that latex

and put it in there

[feral root]

Latex?

[halcyon copper]

so u end up at $x=\frac{N+\frac{19}{2}}{3}$

[flat frigate]

newbienoob

[halcyon copper]

.

latex

[halcyon copper]

and if u don't get an integer, plug in the nearest one

so in the 33 case u get x=7.8

and hence we use 8

[feral root]

Sorry, I'm confused. What does latex mean?

[halcyon copper]

oh

so latex is a kinda language

the fancy stuff u see there

$,\frac etc

it's used for 'writing' maths

[feral root]

Oh, okay

[halcyon copper]

try #latex-help or #latex-testing

if u interested

[feral root]

Wait, (33+19/2)/3=14.16

Not 7.8

[halcyon copper]

wait

mb it's -

$x=\frac{N-\frac{19}{2}}{3}$

[flat frigate]

newbienoob

[halcyon copper]

now see

[feral root]

Thank you! So that's my formula for it, excellent

[halcyon copper]

yeah but remember

for N<10

like 9 or less

this gives negative

can u think of what this means?

unless u too keen on using up many of ur 'items' xD

so if u play in the 100s then this shouldn't bother u

[feral root]

Oh yeah, it's only going up from here. Thank you so much!

Is it .close to vacate the channel?

[halcyon copper]

yes

[feral root]

.close

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[gusty cedar]

Where the angels go

[tall bough]

Heaven

[gusty cedar]

Really I don’t know

[safe radish]

@gusty cedar Has your question been resolved?

[safe radish]

@gusty cedar Has your question been resolved?

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[restive condor]

i got it to
(-6x + 2y) / x

and i subbed in (1,-1.5) and got -9

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[rugged saddle]

is the updated probability 0.35?

[quasi bison]

this is hard to read

[buoyant shadow]

you're not interpreting the question correctly i think

it asks for P(Wed|Tue)

[rugged saddle]

it rained on tuesday

its asking after it rained on tuesday, what is the updated probability of it raining on wed

[buoyant shadow]

yeah

[rugged saddle]

the probability of a occurring and not b is 0.35

[buoyant shadow]

that's just normal conditional

[rugged saddle]

so how do i change it because it rained on tues

[buoyant shadow]

that's also not correct

[rugged saddle]

how so?

[quasi bison]

the text is hard to read because your image has low resolution

[buoyant shadow]

i never get when people say it, i'm the one with the glasses

i guess glasses ftw

[quasi bison]

you think i don't wear glasses?

[buoyant shadow]

i guess glasses are not everything

[rugged saddle]

is this any better?

[buoyant shadow]

yes

[rugged saddle]

sorry

[quasi bison]

P(Tue rain) = 0.5, P(Wed rain) = 0.3, P(Tue rain OR Wed rain) = 0.6, you're asked for P(Wed rain | Tue rain)

[rugged saddle]

how do i calculate P(Wed rain | Tue rain)

0.4?

[buoyant shadow]

you probably want P(Wed rain & Tue rain) first

i had like 6 steps but maybe it can be shorter

[rugged saddle]

P(Wed rain & Tue rain) would be 0.65 correct?

[buoyant shadow]

0.2

[rugged saddle]

how did you get 0.2

[buoyant shadow]

like, P(Tue & Wed) + P(Tue & ~Wed) + P(~Tue & Wed) = 0.6
P(Tue & Wed) + P(Tue & ~Wed) = 0.5
P(Tue & Wed) + P(~Tue & Wed) = 0.3

and that somehow algebras to 0.2

[rugged saddle]

huh?

[quasi bison]

inclusion-exclusion

...or what frownyfrog said, though that's more complicated imo

P(Tue rain) + P(Wed rain) - P(Tue rain OR Wed rain) = P(Tue rain AND Wed rain)

[rugged saddle]

so P(Wed rain | Tue rain) = 0.2?

[buoyant shadow]

0.4

[rugged saddle]

oh nvm its wed and tue rain

so the answer is 0.4

P(Wed rain | Tue rain) =0.4

[buoyant shadow]

yeah

[rugged saddle]

ty .close

.close

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[hybrid copper]

Need to find A such that the function underlined by red is normalised, but am stuck at the integral at the bottom, what should I do differently?

[hybrid copper]

<@&286206848099549185>

[safe radish]

@hybrid copper Has your question been resolved?

[tall bough]

can u latex the integral

[hybrid copper]

whats latex?

[tall bough]

uhhh just take a better screenshot of the integral thats readable

[tall bough]

$Ae^{-\beta x²-iwt}$

is it this

[flat frigate]

$$\frac{talo}{gist}$$

[hybrid copper]

yup

[tall bough]

how did u solve the first integral left term

the one in the form of e^-x²

[hybrid copper]

used this from a table of intergrals

[tall bough]

u should bcome

$\frac{\sqrt{\pi}}{2^{\frac{3}{2}} \sqrt{\beta}}$

right

[flat frigate]

$$\frac{talo}{gist}$$

[tall bough]

the second integral will diverge mhh

[hybrid copper]

hmm yeah, thats the main problem, i think it has something to do with the complex conjugate, |A|^2=(A^*)A

[safe radish]

@hybrid copper Has your question been resolved?

[safe radish]

@hybrid copper Has your question been resolved?

[hybrid copper]

.close

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[viral peak]

any ideas?

[shy temple]

f(x) = (a-x)f'(a-x), and then you can integrate it by parts

[viral peak]

but there is one x that went missing

<@&286206848099549185>

[safe radish]

@viral peak Has your question been resolved?

[halcyon copper]

replace x with a-x and integrate

and replace f(a-x) with what's given

[viral peak]

how can i replace x with a-x

they are not equal

[shy temple]

f(x) = (a-x)f'(a-x), and then you can integrate it by parts

[safe radish]

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[zealous atlas]

Yo

[zealous atlas]

How tf does the answer key say 2,777,777 for part A

I dont get how they get that number

[sharp sage]

you cropped out like 50% of the question 💀

[thorn laurel]

Where do I get help

[haughty jetty]

#❓how-to-get-help

Read

[sharp sage]

ok so to finish the question we need to find (area of palm tree trunk)/(area of tube)

[royal kiln]

Area.

[sharp sage]

area cause it doesn't give u depth

[royal kiln]

Not volume

[zealous atlas]

Which is 0.25^2 * pi for the trunk

[sharp sage]

already caught my mistake

[zealous atlas]

And 0.00015^2 * pi

[royal kiln]

I saw. I'm proud of you

[sharp sage]

Area = πr^2

so for trunk

[zealous atlas]

Ah

I got it

I didnt do the area for the pipe

[sharp sage]

[zealous atlas]

How is this a angle bisector

[safe radish]

@zealous atlas Has your question been resolved?

[woven anchor]

wdym?

[safe radish]

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[abstract comet]

Sorry I left the other channel xd. Yes that is a 5 up there, and no that's not a+b it means a and b are coprimes

[abstract comet]

(The issue is that I can't seem to filter more options from the second condition, which is that pq-5 has to divide 4p and find all possible values for both p and q)

So that would let me find all possible values for a and b, which is what I was trying to find in the first place

<@&286206848099549185>

um

[safe radish]

@abstract comet Has your question been resolved?

[abstract comet]

Can I ping again? Is it because my handwriting is difficult to read?

[abstract comet]

Still no clue on how to solve this

[plucky elk]

For me yes

[safe radish]

@abstract comet Has your question been resolved?

[abstract comet]

Then wait a sec I'll try to rewrite it

Is this better?

[abstract comet]

<@&286206848099549185>

[safe radish]

@abstract comet Has your question been resolved?

[plucky elk]

<@&268886789983436800>

[wild cape]

ty

[plucky elk]

Everything looks right

So you need to plug those back into eqns 1 and 2 to for different values for p, q that make their product in the set in the last line

[safe radish]

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[real zenith]

Quick question, does this look correct?

[worthy hemlock]

Use a calcualator

[real zenith]

I did, I just want a second set of eyes to look over it

what do you think, is it good?

[worthy hemlock]

Two things, if you used a calculator, the answer given, should be correct

[real zenith]

it wasnt it

[worthy hemlock]

Second, the sentence says

Write your answer as a fraction or as a whole or mixed number

[worthy hemlock]

Then you typed it in wrong

[real zenith]

ok

.close

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[pliant orchid]

hi - just wondering what I am missing in relation to conversion rates please. How can there be a 14 cent difference between AUD to USD conversion as opposed to USD to AUD conversion rate?

[pliant orchid]

I would have thought that if 1 AUD is .70 USD that the polar opposite of that would be true I.e., that 1 USD would = 1.30 AUD

can anyone explain what I am missing please?

[worthy hemlock]

Probably taxes

Because exchanging 1 USD to AUD, you would get 1.44 AUD

But 1 AUD to USD, you get 0.7 USD so probably government taxes are involved or something

[pliant orchid]

we have a Government Sales Tax (GST) here in Aus but I wouldn't have thought that applied to a live money market ... like I know the 1st Google result is never 100% accurate but even the live market is pretty close

[worthy hemlock]

Well, ask a broker

[eager crypt]

Nice

[pliant orchid]

I feel like a dummy cause this makes absolutely no sense to me

[haughty jetty]

I guess it's rounding error. I might be wrong
1 USD = 1.44 AUD
(1/1.44) USD = 1 AUD
0.69444... USD = 1 AUD

[pliant orchid]

oh ohhhh ... now I need to wrap my brain round those figures

another quick one then

let's say I am going to make a purchase from a US based website where the price is in USD ... I am in Aus so I pay with a CC ... am I paying 1.30 AUD per USD or 1.44 AUD per USD?

again ... apologies cause I feel like a dope but math and calculating online transaction amounts with conversions is not my strong suit at my age

(or ever for that matter)

[haughty jetty]

1.44

[pliant orchid]

nooooooooooooo!

thanks very much for your time @haughty jetty

[haughty jetty]

Np, mate

[winter pivot]

Hello

[pliant orchid]

ok, so on a 303 USD purchase today it would actually cost me (303 x 1.44) + (3.4 / 100 x 303 x 1.44) with CC conversion rate tacked on after converting 303 USD to AUD with a grand total of 451.15488 AUD charged to my CC at the end of the day (if I got a my brackets right there)

(and hello)

geez the AUD sux!

[safe radish]

@pliant orchid Has your question been resolved?

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[kindred schooner]

Hey, I'm wondering how I would go about simplifying:
((1/x) - (1/y)/(x^2) - (y^2))

[limber patio]

a²-b² = (a+b)(a-b)

Use this and try again

[kindred schooner]

okay, i'm not sure what to do with the top fractions, though

the 1/x - 1/y are throwing me off

I'll give that a go though

[limber patio]

Take the LCM, the method is the same as you'd solve (1/2 - 1/3)

[kindred schooner]

oh that makes sense!

ahh thank you!

[quasi bison]

is this $\frac{1}{x} - \frac{1/y}{x^2} - y^2$?

[flat frigate]

Ann

[kindred schooner]

no

1/x-1/y on numerator

x^2-y^2 on denom

idk how to use that bot

syntax

[simple current]

$\frac{1/x-1/y}{x^2-y^2}$

[flat frigate]

Wishes

[kindred schooner]

yeah that

i think i figured it out tho

maybe

perhapsibly

[limber patio]

What'd you get?

[kindred schooner]

i got -(1/xy(x+y))

but im not sure if it should be x+y or x-y

[limber patio]

Yeah, looks good

[kindred schooner]

i think + since i took the - out

[limber patio]

Yup

[kindred schooner]

awesome, thank ya

while im here, any advice for learning trig/unit circle? It's my worst part of calc/pre calc

actually, i shall sleep now

[limber patio]

It's mainly practice, checkout word problems for trigs

[kindred schooner]

okay. tyvm

!close

/close

$close

[limber patio]

.close

[kindred schooner]

oh

everything but ahahahah

.close

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[restive condor]

does anyone know how to find the unit normal vector ?

[tall bough]

https://www.youtube.com/watch?v=mDUpR1_Qn70

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[young zinc]

this solution is incorrect right?

[young zinc]

what I got is $\frac{-1}{199}(x^{-100}+1)^{199/100} + c$

[flat frigate]

ホタル

[safe radish]

@young zinc Has your question been resolved?

[young zinc]

<@&286206848099549185>

[tall bough]

use an integral solver to verify your answer

[young zinc]

that gives something else

[final halo]

That's equivalent to the first solution you posted

[young zinc]

oh ic

.close

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[kind dove]

is there any neat way how to prove that this is integral domain ?

[final halo]

Depends what you mean by neat

[kind dove]

that i dont have to show all properties of ring for example

[final halo]

View it as a subring of the complex numbers

[kind dove]

then i have to prove only identity of multiplication, commutativity of multiplication and zero divisors ?

[kind dove]

how can i be sure that is subring of complex numbers ?

[final halo]

Show that it is closed under addition and multiplication

It's clearly a subset

Also check that it has 0 and 1

[kind dove]

uhm

ok thanks

.close

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[steel swan]

The systolic blood pressure of a man obeys the normal distribution, with mean 125 and deviation 14. Kimmo's doctor tells that the standardised value of Kimmo's systolic blood pressure is 1,75. What is Kimmo's systolic blood pressure?
Until now I've been finding standardized values but this question came along and I don't understand what to do. Any hints?

[sharp crane]

@steel swan Do you know how to standardize a normal distribution?

[steel swan]

x - mean / deviation?

[sharp crane]

yea

(x - mean) / deviation more precisely

So you originally had a normal variable X with distribution N(125, 14) and transformed it into Z = (X - 125) / 14 with distribution N(0, 1)

They simply ask you to "undo" the transformation, i.e. solve for X

Z = standardized blood pressure
X = actual blood pressure

[steel swan]

Damn still dont get how im supposed to do that

But thanks for laying it out like that, I have a somewhat idea as to what should be going on

[sharp crane]

@steel swan Take the equation
$$Z = (X - 125) / 14$$

[flat frigate]

nvx

[sharp crane]

Now we know that $Z = 1.75$, right?

[flat frigate]

nvx

[steel swan]

We do

[sharp crane]

So can you solve that equation for X?

[steel swan]

125 + 1.75 × 14 = 149.5?

[sharp crane]

Yup

[steel swan]

Thanks, seeing it in english helps. Dno why I can't understand the litterature at all

Much obliged sir

.close

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[dense magnet]

What does it take for a polynomial to be symmetrical about the y-axis, and what are some examples?

[shy temple]

it needs to have only even powers of x. example x^2

[dense magnet]

thank you mate

also are there functions that are symmetrical about the x-axis?

[tall bough]

pretty sure they wouldn't be called functions

[shy temple]

no. they won't be functions then, since a function associates each x with a unique y (i.e for each x in the domain, there is a unique y in the codomain such that y = f(x))

[dense magnet]

so there are no functions that are symmetrical about the x-axis

[lone crest]

do you mean like x^2 turned 90 degrees

if yes then

x = y^2

oh wait function?

[tall bough]

combination of two same but opposite functions can be symmetrical, like y=sqrt(x^2+r^2) and y=-sqrt(x^2+r^2) these two functions fork a circle

[dense magnet]

i also have a text question i need help with

It says Tom lives on a farm and is bothered by flies in the summer, so he buys bug catchers to ease the problem. A model differentiates the number of flies caught in day x in july and is given as: f(x) = x^*2 - 38x + 397, x ∈ [1,31], how do i draw f(x) in in geogebra (digitally)??

please i need help with this now

.close

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[ionic gull]

Hello I recently played around with xⁿ and discovered something about the way it grows.

I'd like to explain further with a picture

[ionic gull]

The bottom row, which I consider to be the first row has the integers starting from 1

The row above that is the row below that, it has just been run through this function:

f(x) = x⁴

[woven anchor]

is what you're interested about the constant difference?

that comes from the 4th derivative of x^4

[ionic gull]

Yes , when I make a comparison between neighboring values and do the same for the comparesions of the comparasions then there is always a constant value

The value is btw n factorial

I would like to know the terminology for that phenomenon

1 is constant at the first layer of comparasion, 2 at 2
3 at 6
4 at 24
5 at 120
...

I looked up to 22 as the exponent, and it holds

[ionic gull]

If there is any

[woven anchor]

actually it doesn't come from that

but you can derive it by binomial expansion probably

[ionic gull]

It even works into the negative using odd exponents

[woven anchor]

the degree decreases with each step

and the 2nd term of the binomial expansion is n choose 1

[ionic gull]

What's non-monic?

[woven anchor]

the 2nd term

[ionic gull]

Oh understand

Thanks dor giving me some terms to look for, that's all I needed

I'll close the help now

.close

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[languid granite]

Hello, I'm substituting for a geometry class but it's been a long time since I've taken it myself. This problem came up and I'm unsure if EH must be parallel to DG and FJ or not.

[languid granite]

I feel like it is not necessarily parallel because you can have two distinct right triangles that have the same hypotenuse length

but i'm not sure if that's completely accurate or the best way to explain it

[safe radish]

@languid granite Has your question been resolved?

[languid granite]

<@&286206848099549185>

[safe radish]

@languid granite Has your question been resolved?

[safe radish]

@languid granite Has your question been resolved?

[languid granite]

<@&286206848099549185> any info would be super appreciated ! don't have anyone i can ask in person at the moment unfortunately

[safe radish]

@languid granite Has your question been resolved?

[jolly kiln]

Depends

If it is symmetrical than yes

But there are no conditions given in this example so... you can made everything up yourself

An example for them to be parallel is when you mirror them in the middle (between EH, DG, FJ) then it has to be identical

[languid granite]

Yeah that makes sense

There’s no other conditions given aside from the picture

If EH is not parallel to FJ or DG would the EHFJ not be a rectangle? Just some other quadrilateral?

[jolly kiln]

Yes

But there could be also a parallelogramm

Where EH und FJ are parallel but it doesnt create a rectangle

[languid granite]

Gotcha

Thanks!!

.close

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[proven field]

Hi all, I have a question about converting parametric equations into an implicit equation involving y and x. One of the methods I was taught involves squaring each parametric equation, followed by elimination of the parameter. My question is, by squaring each parametric equation, aren't we actually representing two parametric equations at the same time? Here is what I mean.

[proven field]

In the second method in the image, squaring the first parametric equation gives x^2 = t ^2 + 2 +1/t^2 . But we are only graphing the positive root of this equation ( x = t + 1/t ), not the negative one ( x = -(t + 1/t) ). So why can we find the implicit equation by squaring each equation, when it introduces an extra parametric equation into the sketch (if that makes sense)?

Going to ask this question in another channel

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[tawdry flicker]

hey I'm in a class of programming and math, and I'm a bit at a loss about how exactly I'm supposed to rotate matrices 90 degrees?

[tawdry flicker]

for illustration this is an example

E = [[-1, 2], [-2, 1], [1, 1], [-2, 0], [-2, -1], [1, -1], [-1, -2]]

A = [[-2. -1.], [-1. -2.], [-1. 1.], [ 0. -2.], [ 1. -2.], [ 1. 1.], [ 2. -1.]]

this is supposed to make a smiley which each [] are the coordinates of a point for rotate 90° anti clockwise

[eager ravine]

E -> A just looks like (x, y) -> (-y, x)

[tawdry flicker]

oh ok

Im given a tip that

"Tips & Warnings What you are looking for is a 2x2 rotation matrix."

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[forest fossil]

how to solve 1st question

[plucky elk]

you're on the right track. you just need to set the equation for mean = 1.5

$\sum_x x f(x) = 1.5$

[flat frigate]

riemann

[forest fossil]

i'm really not familiar with discrete random variable with a table

I know with integrals

Can you help me understand it plz

w8

I'll try smth

[final halo]

With discrete the integral just becomes a sum

[plucky elk]

How did you find f(x) ?

[lone crest]

whattt where did all channels go

[plucky elk]

$f(x) \neq F(x)$

[flat frigate]

riemann

[forest fossil]

ohhh ok wait a sec

I GOT IT

no I don't

[plucky elk]

haha

https://math.stackexchange.com/questions/1518909/finding-the-probability-mass-function-given-the-cumulative-distribution-function

This should be little f(x)

[forest fossil]

is that correct?

[plucky elk]

You can check by using the definition of cumulative distribution function.
$F(x) = \sum_{i \le x} f(i)$

[flat frigate]

riemann

[plucky elk]

In this case, $F(2) = k$, so that means your $f(2)$ is wrong

[flat frigate]

riemann

[forest fossil]

k-0.5 ?

[plucky elk]

.

[forest fossil]

should be correct then right?

[forest fossil]

can u explain more what's this mean?

[final halo]

Youre basically there btw just do this now

[final halo]

Integrals are like the continuous version of sums

[forest fossil]

it seems wrong

[forest fossil]

yep I know this, but when they give us a table should we always convert it from F(x) to f(x) ?

[final halo]

What happened to your x=3 term?

Missing some bits from your table

[forest fossil]

it was F(3) = 1

[final halo]

Ah I see

[forest fossil]

I converted to f(3) by
f(3)= 1-F(2)

[final halo]

But your k should be minus

Your table is fine

The k should be minus in your calculation

[forest fossil]

but the probability can't be less than 0

or negative

f(x) always >=0

[final halo]

When you did 3f(3)

The k in that term should be negative like in your table

[forest fossil]

bro i feel lost haha

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[shell raven]

please explain how did we get T2

[stray socket]

Well

Start by removing the fraction

what would you do to get rid of the 2 in the denominator?

@shell raven

[shell raven]

Multiply?

[stray socket]

by what

[shell raven]

950 ?

Send the simplification part if you have pen and paper please

[stray socket]

I cant give you the answer

But I will provide you the steps

@shell raven here

950\ =\ \frac{2.99T_{2}+T_{2}}{2}

$950\ =\ \frac{2.99T_{2}+T_{2}}{2}$

[flat frigate]

Umbraleviathan

[stray socket]

So you were right, you do multiply, but by what

[shell raven]

950

[stray socket]

No

what number is in the denominator?

[shell raven]

2

[stray socket]

Yeah

so waht do you do

to get rid of that denominator

what would you multiply by

to get rid of the 2

[shell raven]

Don't know

[stray socket]

haveyou taken pre-algebra yet?

[shell raven]

Sorry idk what your saying

[stray socket]

have you taken pre-algebra yet?

stuff like solving intercepts, isolating variables, linear lines, etc?

my main concern is that if youre taking physics, then you should be able to do pre-algebra stuff

[shell raven]

Okay i got the answer anyway , the subject I'm learning now is BME

[stray socket]

whats BME

[shell raven]

Basic mechanical engineering

[stray socket]

Oh i see

Yeah youre gonna need to know how to isolate variables

But you said you got how to isolate T2?

Do you understand the steps?

[shell raven]

Yes taking t2 common

[stray socket]

wdym taking t2 common

i mean its adding like terms

and then multiplying by by 2/(3.99)

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[lyric ingot]

Couldn't this be true for both coses

cases*

[eager ravine]

ye

[lyric ingot]

Ok, I thought maybe I was overlooking something

(Seemed like a really simple question)

[eager ravine]

technically the youngest a student could be is -78 lol

[lyric ingot]

O_O

But not literally

Well, I should try the practice grade 12 MAP

:)

"Grade 12" my ass

6.4 done

[eager ravine]

what is MAP lol

[lyric ingot]

An adaptive test

[eager ravine]

ok

[lyric ingot]

Gets harder the more you get it correct

I did a practice

Pfft whatever, the practice is useless

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[tight creek]

It’s 6.40 you need to account for sig figs

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[unique bison]

I don't think there's enough information to solve this

[thick mica]

This question is from a friend of mine and here's the solution she came up with that I did not quite understand

[unique bison]

Oh yea nvm it's just a differential equation

What step don't you understand?

[thick mica]

Well from the very beginning... how does the equation given in the question relate to the one in the first step?

[unique bison]

Product rule

' means derivative wrt x

[thick mica]

Okay so we take the derivative of x.f(x) but where did the + f'(x) go? sorry if this is a stupid question I might be missing something easy.. if we take the integral of f(x).x + f'(x) for x that should be the square root of x correct? I just couldn't figure out how the solution goes

[unique bison]

Differentiate x.f(x) yourself using the product rule

[thick mica]

Oh one moment

This is how it goes right?

[unique bison]

I mean yea that's pretty much what they wrote

But I'm not sure if you understood why the first line is valid

[unique bison]

I'd say just do this and forget about the question for a moment

Do it without using information from the question

[thick mica]

I found the derivative of x.f(x) with the product rule, 1.f(x)+f'(x).x

Which is also the equation in the question

And then took the integral of either sides I think and put x=1

[unique bison]

Yes

[thick mica]

The integral of d[x.f(x)] is /d[f(x).x]dx which then equals to [f(x).x]dx and thats why I took the derivative of it which equals to the integral of the other side, square root of x

Is this correct?

Pretend / is the integral symbol hahah

[unique bison]

Almost

[thick mica]

Where did it go wrong?

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@thick mica Has your question been resolved?

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@thick mica Has your question been resolved?

[neon island]

pls check this one

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[rich moss]

How do I find n

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@rich moss Has your question been resolved?

[rich moss]

Never mind It seems like no one is around to help

I get the solution now

8^2 = (n+4) x 4

64 = 4n + 16

-4n = 16 - 64

-4n = -48

Therefore n = 12

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[rich moss]

How do I find f

[rich moss]

I found e already

<@&286206848099549185>

[prime rock]

Didn't even wait. just pinged 🤣

[rich moss]

Cause no one responded to my other one

So I had to figure out myself

for the above

[prime rock]

Yea me too. I think it's less busy at this time

[rich moss]

I guess so

because no one is around to help

[dense sierra]

you can use the same way you did finding e to find f

which is PC*PD = e^2

[rich moss]

So I can do 12^2 = (f + 6) x 6

?