math-help

[left ether]

when trying to like "solve" this you'll end up with
y=y

[plain pendant]

Using the replacement method?

my english is bad sorry

[left ether]

[plain pendant]

wtf

[left ether]

n for the second part, idk what ur exactly looking for
but this is how the system looks

[plain pendant]

5 and 2 are below x-y at the same time

Oh, forget it, it will be difficult to explain the issue

but thank u so much for try

[left ether]

like this?

it's still wrong

[left ether]

np

[glass crater]

my brains stuck on it too so i put it in an app to try

for a function x and y would both be 0

idk if that helps tho good luck

[plain pendant]

https://youtu.be/O8zps4mwQT8

[left knoll]

One of them has to be (x+y) so that we can get an answer

[plain pendant]

is this

[plain pendant]

in mine show me different

[left knoll]

Did u copy this from ur text book or anything ? might want to check if u didn't write anything wrong

[glass crater]

^

[plain pendant]

no,my teacher did pass in the class

[left ether]

[drowsy stream]

you have x=(1-5y/2) and you replace it in the first line

[glass crater]

(it shows u the steps on how to solve it in the app)

[left knoll]

how do u project a point?

topic is vectors

[weak lotus]

thats it? nothing else of information they give you?

i would assume you have a vector Esub1 with a certain length

you can use that as an example, but since i do not have enough information on the problem you are working at i would not know

[silent pendant]

vector a projection on vector b is = a dot product b/ magnitude of b

[timber coyote]

calc 2 help if anyone can offer it

currently trying to get the integral of cos(9t)cos(18t)

I know I'm supposed to use product to sum rule

but with my understanding I believe it should be

cos(-9t) + cos (27t) but each time I see the problem solved online it simplifies to cos(9t) + cos(27t).
does anyone know why the first cos is cos(9t) instead of cos(-9t)

[silent pendant]

Cos(-9t)=cos(9t)

[timber coyote]

i feel dumb for being in calc 2 and not knowing this, but can you explain why?

wait nvm i think i understand

i just had to think about it in terms of a circle

ty that saved me a headache ❤️

[silent pendant]

Np : D

[valid carbon]

anyone knows how to show the equality between the two?

[valid carbon]

is it because of the trigo identity fomula (confunction)

[silent pendant]

1. because of definite integral (as shown in the picture)

2)Not a formula, but quadrant change

[valid carbon]

ohh i see thank you!

[silent pendant]

Np : )

[lunar dome]

Hi! could anyone help me with a very simple math question?

[left ether]

sure

[ripe jay]

https://dontasktoask.com/

[languid compass]

Can someone please walk me through how to solve problems like these? Would I find the derivative of the equation and replace the t with x?

[ripe swan]

you need to substitute the expression of dy/dx into that formula for starters :)

and input the integral range of 3 ≤ x ≤ 7

[languid compass]

Ok gotcha thank you! Would it be something like this then?

Oh that wasn't right, I did something wrong 😅

[ripe swan]

right, so you dont do the calculation all the way to the end. you stop at √[(16x^4)-1]

substitute that expression into the formula for the length of the curve. and make sure you replace the integral of 1 to x with your range.

[languid compass]

Ok I got it now, thank you so much!

[ripe swan]

what answer did you get?

[languid compass]

421.35714

[ripe swan]

show the working?

maybe my hand-me-down casio calculator is deciding to retire, i got something close to that

[languid compass]

My bad I just checked and I actually got 421.333 if that's what you got

[ripe swan]

no 😭

[languid compass]

Oh oop 😭 I got marked right on my practice problem system so it has to be somewhere around there

[ripe swan]

it was marked correct?

i think it has that range of leniency then haha

[languid compass]

Probably 😅 what did your calculator give you

[ripe swan]

436.27

[languid compass]

Oh wow yeah I was pretty off

[ripe swan]

meh as long as it's marked correctly

[languid compass]

Yeah haha thank you again I appreciate it

[ripe swan]

no probs, pay it forward :)

[rich pelican]

anyone here can helo me with diferential calculus please

[drowsy stream]

but it is correct

[static pasture]

Hello, math peeps! I have a question.

Initially, I tried doing my revisions according to its chapter sequence beginning with 1. Since I have a limited time, I tried revising the chapters beginning with the one I had the least mastery on (measured by practice tests scores) instead. My scores went by a couple points higher lol.

Next attempt, I tried to factor in its importance in the exam (current mastery/total items in the exam). It made some difference though a bit smaller than last time. Still. Every point counts.

Now, I realize there's also the length of the chapter as in time spent on it. So I kinda came up with this table to try to figure out how I could efficiently go with my revision.

I'm not a math person. This is me being a Type A person and a procrastinator. Is this right? Any input would help! Thanks!

Edit: The last column probably shouldn't be efficiency since I'll be revising the chapter with the least value first...but I also probably did something wrong with my formula.

[left knoll]

So I have online classes, and in the lectures they showed how to solve limits like these:

But they didn't show how to solve an example like this:

what does this problem mean and how do I solve this?

[surreal flint]

that is just a fancy way to write a function

it simply says that whenever x is not 5 the value of function is 4

and when x is 5 the value of function is 1

and as we know x is tending to 5

the answer of the question will be 1

[left knoll]

oh I see, tysm!!

[lucid hollow]

Hey there, im having some trouble finding the correct path to solve this problem and i hope someone can help me out. The figure shows an equilateral triangle ABC. Points D, E and F are the midpoints of the sides of [ABC] and the area of the triangle is equal to 20. Find the area of the triangle [XYZ].

[drowsy stream]

did you draw the triangle ?

[lucid hollow]

Im not able to understand the triangle side size

we have 20 as the area of the triangle

A= ABxEC/2

how am i suppost to get the size

[drowsy stream]

I guess you have to find the height of xyz

The information given for x y z are coordinates imo

[lucid hollow]

But it asks to provide the area of the triangle [xyz]

i am suppost to know beforehand the size of one side of the triangle i think

[supple island]

you do know that

you have the area of an equilateral triangle, so you have the side

[lucid hollow]

Area = ABxEC/2

right?

[supple island]

yep

[lucid hollow]

so how do i get one of the side sizes

[supple island]

you find EC using the triangle EBC

[drowsy stream]

Pythagore maybe

[supple island]

yep

[drowsy stream]

But if you place the points

I think it is easier

[supple island]

EC=sqrt(bc^2-bc/2^2)

then you subsitute in the area

and you find the side

[lucid hollow]

Im still so confused @supple island 🤯

[supple island]

about what

[lucid hollow]

Soo CB ^2 = EC^2 + EB^2

right?=

[supple island]

yes

[lucid hollow]

EC=sqrt(bc^2-bc/2^2) how did you transformed into this?

[supple island]

eb=cb/2

[lucid hollow]

ok that makes sense

[drowsy stream]

X is in the middle of EB, Y is in the middle of FB and z is in the middle of EF i think

[obsidian forge]

Hey guys I'm having trouble understanding how my professor got the answer to this question (I hope I'm putting this in the right chat because this is for my Gen. Biology 1 class):

( I attached a picture of the question) Calculate the mass - volume ratio as % and as mg/ml for the following:

10g of salt in 200ml of water
______%
______mg/ml

The correct answer is 5% and 50mg/ml

However I can't figure out the math in which he did this - I attached a screenshot of how I tried to solve the problem and the answer I got was 2,000 and 0.02 mg/ml

I'm stumped

[tidal kelp]

ik this is simple but im stumped please help. thanks!

y = 1

[mystic thunder]

I got it

wait a sec

@tidal kelp

[tidal kelp]

thanks so much!

[obsidian forge]

Thank you so much!!! Yes I agree my math was just off, I appreciate you catching that mistake and helping out!!

[sly night]

hello

[left knoll]

hi@sly night

[sly night]

hru

[left knoll]

im good

[sly night]

cool gtht

[left knoll]

do u need help?

[sly night]

Do you study the possibilities well

[left knoll]

probability?

[sly night]

ya

[left knoll]

i know a thing or two from my course

[sly night]

I have difficulty with random variable

[left knoll]

Oh show your question

[sly night]

I don't have a problem with the random variable, but with questions related to it

For example, define the probability law for the random variable x, then calculate its mathematical hope in terms of m

Where m is a real number, which takes the value m*m if event a is met and takes the value m if event b is met.

[left knoll]

Ohhh

[sly night]

A=8/16

B=3/16

[left knoll]

i could give u some links to learn from

[sly night]

ya thank you

[left knoll]

the last part is really simple

i guess the finding of m is where ur finding it hard

[sly night]

ya

[left knoll]

https://youtu.be/3v9w79NhsfI

i think this is pretty basic stuff

https://youtu.be/vfqPpai_9jI

this is pretty trustable

[sly night]

good

[left knoll]

the following lectures of this will probably be related so check those out too!

[sly night]

ok thanks bro

[edgy sun]

i need help on this

this is wat i have so far

[left knoll]

did u solve it yet?

[edgy sun]

yea i got it

but im having trouble on the next problem that relies on problem 3

im not sure how to start

[left knoll]

is this formula (3)?

[edgy sun]

wait no that formula was wrong

the upper bound isnt supposed to be x

[left knoll]

Ah

what is the formula 3 tho

oh

it's given sorry

[edgy sun]

yea

[left knoll]

far beyond my reach

[craggy scaffold]

Try plugging it in the Pythagorean theorem

Does anyone know a good system that is time efficient and effective at taking notes and practicing problems for math?

[left knoll]

Don't take notes on the content because they are already in the textbooks, instead focus on doing problems and noting down their solutions.

[craggy scaffold]

I will try this approach

Maybe I will rephrase some of the content

[formal forge]

are non linear inequalities and quadratic inequalities the same

[left knoll]

almost
non linear inequalities may include inequalities with power 3 whereas quadratic inequalities will have maximum of power 2

[formal forge]

TYY

[left knoll]

NP!

[left knoll]

3 or more*

[craggy scaffold]

Pastor are you studying math?

I’ll be taking algebra 3 soon

[left knoll]

I am studying maths, physics and chem rn, indian high school level

final year

[left knoll]

@ebon ravine what level would you consider this? don't mind the line above that's a different question

[spice flare]

integrate with partial fractions

[tidal kelp]

please help, i know its difference of two cubes or something like that but how do i expand then simplify???

[mystic thunder]

isn't it like already in expanded form

or u have to multiple both the brackets

[tidal kelp]

im not sure!!!!

[tidal kelp]

it says the answers a-b\

[mystic thunder]

okay I think

ur supposed to multiply

I'll try once

oh it is (3√a-3√b)^3

so the root will cut

and u get a-b

[tidal kelp]

thx!

[mystic thunder]

there

[tidal kelp]

thank you

[left knoll]

@mystic thunder send here

[mystic thunder]

yeah js a sec im lunch

[left knoll]

yuh

[left knoll]

oh late lunch

[mystic thunder]

yeah cuz

it was rice

me no rice

there

I had a problem ith the b and d part

[left knoll]

ohhh

yeah it'd trick me too

[mystic thunder]

dude

i

[left knoll]

what

[mystic thunder]

this question makes my head go

belususjhveilpushiemlsshks

plus it's so ugly

[left knoll]

did u solve it yet tho

[mystic thunder]

No

im not gonna do it

[left knoll]

oh

lmao

[mystic thunder]

real

[left knoll]

lemme try

i just wanna do smth else than organic

[mystic thunder]

yes

hahah rip

is it really that bad

[left knoll]

lmao i thought of a way but that didn't work

like use the radius to make em all isosceles triangles so the base angles are equal in the triangles and then equate

[mystic thunder]

yeah that's what I did

in the fourth part

snd then I used angle sum of quadi

somehow got the answer but I wonder if it's even correct

[left knoll]

won't bdc be half of 110

oh it's not mentioned

[mystic thunder]

rip

igs okay

[left knoll]

use the

[mystic thunder]

is there circles in 10th also?

the

[left knoll]

angle substended by

semi circle

[mystic thunder]

not in bdc tho

wait what

[left knoll]

like

bcd is 90

[mystic thunder]

bur

don't I need that

angle cbo

[left knoll]

cbo is

40

no 30*

[mystic thunder]

no

is it mentione

s

d

[left knoll]

cbo isn't

mentioned

[mystic thunder]

how did u get ir

[left knoll]

right triangle

[mystic thunder]

WAIT OF MY GOD

[left knoll]

draw a perpendicular from o to ac

[mystic thunder]

I GOT IT

HOMYSGIT

o

[left knoll]

😭

[mystic thunder]

FUCKKCKKXSHAKBD

why didn't I get in exam

whatever

[left knoll]

happens

[mystic thunder]

yeah idc

[worldly spoke]

a,b and c are pozitive int

How many a,b,c are there

a+b+c=12

Answer is 7 , ı can do it by writing every posibility but is there any other solution

Like permutation , comb etc.

[left knoll]

n= 12 r = 3

n+r-1
C
r-1

14C2

14!/2!x12!

14x13/2x1

OH WAIT

i forgot that a is smallest

[left knoll]

= 91

that's total possibilities

I'm not sure about the a<b<c part

[worldly spoke]

I've come this far but thanks anyway

[left knoll]

oops

[worldly spoke]

[kindred prairie]

i personally, do maths while listening songs... like chill songs or sometimes some hype songs... i keeps me focused and dont let me think of other stuffs... so yeah u can give it a try

NOTE: YOU DONT HAVE TO FORCE YOURSELF TO FOLLOW THIS IF YOU CAN'T

and for learning... tbh there are no tips except some topics (depends)... just revise solve and if u are getting stuck somewhere... run back to basics and doing examples.... and practice again...

[tidal kelp]

how do i do question 5

[mystic thunder]

pythagoras theorem

[green swift]

ab+c = 2020 a +bc = 2021 can someone explain why you have to subtract these

to get to the answer

i watched a video on how to solve

they label it first then subtract but i dont understand

[tidal kelp]

thx

[mystic thunder]

is x=4 and x=12

[tidal kelp]

no its just 12 bc in the x-7 side then it would be negative so u cant do that

[mystic thunder]

OH yeah ur right my bad

[tidal kelp]

all g

[worthy skiff]

hey, ik it’s dumb

but somebody could help me with this?

[craggy scaffold]

Man you got some good handwriting

[grave harbor]

May i know what topic is this

[worthy skiff]

transform the decimal numbers into fractions and then resolve the value of X

[mystic thunder]

is there bar on 5 and 9?

[worthy skiff]

yes

[mystic thunder]

okay I think that first u need to convert the decimal numbers with bar into fractional form (p/q form like a rational number) and then cross multiply

btw ur pfp is cool

[worthy skiff]

oh

okey ill try this

[mystic thunder]

wait

[worthy skiff]

thx for the pfp compliment :3

[mystic thunder]

is it

[worthy skiff]

what

[mystic thunder]

10-5 in the denominator

or 70

[worthy skiff]

10-5

[mystic thunder]

ok cool

[worthy skiff]

ok

let me try that way

[mystic thunder]

is x=4/11

oh my god I'm so blind

waitno

[cursive quest]

Help plsss

[dusk knoll]

Hey, does anybody know a good app/program to study maths with or simply just for documenting mathematical topics because it’s such a bother to always have to copy and paste certain symbols?

[left knoll]

if writing the math symbols is the problem learn LaTeX

[civic osprey]

you'll have to pay a mathematician to solve this xd. What year r un in btw?

[cursive quest]

Bro i am a high school student

And thats an entrance exam level question

Lol JEE

[civic osprey]

We also do integrals in high school but i think ur is way to complex

[cursive quest]

Yup

[civic osprey]

this is one of the easiest on the school book

[cursive quest]

Lol

[civic osprey]

what country are u from and u do such hard integrals in high school wtf xd

[cursive quest]

Just search JEE

[civic osprey]

i thought out school system was considered hard in comparison with other EU countries

[cursive quest]

You'll get to know to know about it

[cursive quest]

T-Series lmao

[civic osprey]

that kind of math is usually met in universities xd. India is on some kind of stuff lol

[cursive quest]

Idk why exam body in my country is obsessed with high level mathematics

[civic osprey]

if they teach that to 07 students then what the hell do they teach to 05 who are in their last year of high school

[cursive quest]

Last year of high school is comparatively easy than that of entrance exam since there are much more students appearing there

We can say 70% per easier

[civic osprey]

that makes 0 sense xd

why make it harder for younger students but easy for older ones

[cursive quest]

I mean this is the question for undergrad entrance exam
Jee advanced

High school students with last year give this exam
I.e class/grade 12th

[left knoll]

i swear i hate the entrance culture here 💀 they are like rejection centred not acceptance

[left knoll]

these are the easier bunch

[civic osprey]

yeah but how many students fail them?

[left knoll]

the thing is the advanced is the second part of the exam which you get to appear if you qualify the first part to get in the premier institute and selection is done on basis of percentile so even if you attempt 1 question and rest all of the students attempt 0 you will get in

first one is comparatively much easier but i think still pretty tough compared to other countries

[civic osprey]

that's weird. So, if u dont show up, then u repeat the class?

[left knoll]

you get to give it another shot the next uear

year

2 attempts

[civic osprey]

we have tough entry exams for unis here. Last year of high school is tough af for the students

[left knoll]

in total

[civic osprey]

but, do u pass the year or no?

[left knoll]

yeah this is one of those

[civic osprey]

ah ok

[left knoll]

yes you graduate high school but to enter in an uni u need to give this

[civic osprey]

as long as u have a paper confirming that u finished high school, then u can take as many tries as u want for the uni exams

[left knoll]

high school finals is a different one

[left knoll]

no only 2

[civic osprey]

im talking about us

we have that here

[left knoll]

the year you graduate and the year following that

ohhhh

i see i see

i mean we have a huge population and a stupidly difficult education system

[civic osprey]

yeah but lets say u pass at the uni u want at the first try, then u graduate from the uni and u want to enter another one. At that case, u take the exams again

[left knoll]

ahh i see i see

[civic osprey]

Ours is also difficult here compared to other countries (meaning that our class material is a couple years more advanced than other countries), but after seeing that Integral that guy sent I freaked out xd

[left knoll]

lmaoo yeah I don't get why it needs to be this hard lmao

[civic osprey]

if school maths is so hard then wtf do u even do at ur unis

u design rocket ships? xd

[left knoll]

idk man

why they do this

[proven mason]

Helo is anyone here in precalc

@ me if u are

or dm

[willow basin]

Does anyone know how to solve this? I'm stuck.

Please ping in reply.

[muted beacon]

Am i correct ? ;-;

Coulombs law:free-diagram

If anyone doesn't understand my writing pls ask

This due tomorrow

-4.77x10^9 is my total for Fnet charge

[left knoll]

ya you fine @muted beacon

i mean u substracted both F tho right?

[reef igloo]

make denominator in terms of tan and make it in terms of sec in numberator then substitute

[tidal kelp]

my math test is on monday, wish me luck!

[left knoll]

goodlucj

[tidal kelp]

thx

[tawny kelp]

hi i need help

i honestly don’t get it

i think it would be this my answer is in red but i’m unsure

[civic osprey]

basically u flip y=h(x), mirroring on y'y and then u do the -4

I think the red one on ur picture is fine

[tawny kelp]

thanks 🙏🏻

would this be correct

[civic osprey]

based on the cords yeah

[tawny kelp]

thx

[bright widget]

good day everyone

can someone help me with this
Please calculate a variation of the golden ratio using the following recursive equation:
a(n+2)=2a(n+1)+a_n,n≥1,
by starting with a_1=a_2=1.

[left knoll]

We use Newton's formula for quadratic equations here

but I don't fully understand wdym by variation

[left knoll]

guess internet doesn't complete agree with that name

[left knoll]

So basically you can suppose a_n as p^n + q^n

Then convert the equation into a quadratic and find p and q

[bright widget]

thank you so much

[left knoll]

Can someone give me a little help?

How can I do potentiation really fast mentally?

I have a math Olympiad next month

[left knoll]

An object is moving along the x-axis. At t= 0 it is at x= 0. Its x-component of velocity vx as a function of time is given by vx(t)=αt−βt3, where α=7.6 m/s2 and β=3.2 m/s4. At what nonzero time t is the object again at x= 0?

[silver rover]

are there any websites that gives you like a bunch of questions to train

practice questions

cause im the type of person who learn tremendously from deliberate practice

[supple island]

really depends on what you're preparing for

[cosmic kestrel]

https://en.akinator.com/

or specify what exact topic you are preparing for

[left knoll]

💀

[silver rover]

Bruh

Oh sorry what i meant was math practice questions

Lmao

[left knoll]

standard? country? topics?

[silver rover]

Idk just basic math? I'm trying to learn calculus

I'm prepping for my country's national college admissions test

[woven orchid]

The table below shows the percentage of young people aged 16 to 24 who smoked in a country.

a) use regression to create a linear function expression for the proportion of smokers r(t) t years after the year 2006.

b) in which year will the proportion of smokers in this age group be 1% according to the model? and find the turning point of the graph and give a practical interpretation of the answer

c) in 1968, the proportion aged 16 to 24 who smoked daily was 35% Can the model be used for the years before 2000? Justify the answer with calculations

[cosmic kestrel]

what country? Google at first previous year problems from this test and send here

[silver rover]

Indonesia

[cosmic kestrel]

wtf is basic math

[silver rover]

I still lack in operating numbers

[cedar storm]

What do you need help with?

[woven orchid]

Oh, i need help with question b and c

i found out about question a, i just needed to put the years to 2006 = 0, 2008 = 2, 2010 = 4

etc...

[cedar storm]

Have you determined the linear regression equation?

[woven orchid]

Yes, i got:

[cedar storm]

gimme a minute

[woven orchid]

totally fine!

[cedar storm]

I think your linear regression equation is wrong, cuz if I put 2006 under the x I got -3464

I should get someting around 29

I calculated the linear regression and got

[woven orchid]

yeah, but i use 2006 = 0

2008 = 2

2010 = 4 etc

[cedar storm]

why?

[woven orchid]

because The problem defined t as the number of years after 2006

[cedar storm]

Oh sorry haha

mb

[woven orchid]

no its fine, i got that wrong too

[cedar storm]

for the b) first you need to put 1 under the y

[woven orchid]

ok, so 1 = -1.7x + 29

[cedar storm]

ye, we are calculating how many years after 2006 we got 1% of smokers in this age group

and then 2006 + x we have in which year the percentage of smokers will be 1%

[woven orchid]

could you show me how?

[cedar storm]

our y is a precentage of smokers in this age group
our x is a number of year after 2006
So if we have a linear regression equation, we just need to put 1% under the y, and the just solve the linear equation and we will have how many years after the 2006 the precentage of smokers will be 1%.

1 = -1.74x + 29.71
x = 16,5 (something like this)

So 16,5 years after 2006 the precentage of smokers will be 1%, so 2006 + 17 (rounds up) = 2023

[woven orchid]

oooh ok

thank you!

[woven orchid]

do you also know for the question turning point and what that tells us? and question c?

[cedar storm]

What do they mean by turning point of the linear function graph?

[woven orchid]

When you double derivative/second derivaitive the graph

maybe its called the inflection point?

[cedar storm]

I won't help with next questions i think, sorryyy

[woven orchid]

do you know anything about probability?

[cedar storm]

gimme questions i'll see

[woven orchid]

@cedar storm There are 29 students in a class. One day, 23 of the students had done their homework. The teacher checks the homework of 12 random students. Let X be the number of students who have done the homework among the 12. a) determine the expression for the probability distribution P(X=x).

i got this:

but my question is, if i need to find P(X=8) how do i do that in here?

[cedar storm]

just put 8 under the x

[woven orchid]

but there is a method of solving this probability? for 23 / x its:

but how do i do that when its 12-x?

[cedar storm]

12-8 = 4

so you just calculate (6 4)

and just use binomial theorem to calculate it

[woven orchid]

ooh

💀 damn im dumb

[cedar storm]

you mean that?

[woven orchid]

yeah that was it thank you 😭🙏

[cedar storm]

no problem

[open tide]

a.(bxc) can also be written as (axb).c

[fair belfry]

could someone help me with 3d please 🙏

[left knoll]

Summation of frequency = Summation(xf)/Mean

[dire aurora]

Can someone please explain why my approach this question is logically wrong? The answer is 36^4, which I agree with but I don't understand how its logic differs from my approach.

[graceful vault]

digits 0-9

so

10 more characters

[dire aurora]

I agree

(26+10)^4

[dire aurora]

but why is the logic behind my approach wrong in comparison?

[graceful vault]

ig

the permutation

coz

when you choose 2 digits

for eg 3 and 6

and 2 letters

a and b

they can be

permuted amongst each other

like

36ab

can be

[dire aurora]

but my multiplication brings about all the different ordered selections too

in each case I use the multiplication principle

and then added all the cases together

[graceful vault]

no like

36ab and 63ba in your case

is one

case

[dire aurora]

are u sure?

[graceful vault]

thats what my brain computes 😅

i might be wrong

[dire aurora]

26^2 x 10^2 should include 36ab and ba63

and 63ba etc

[graceful vault]

thats more like selecting 4 characters

[dire aurora]

sadly my approach is wrong and i dont know why; so in a test I'll have to get lucky with which logic i decide to use as I cant differentiate between the two logics

[dire aurora]

dont agree

selecting 4 characters is 26^2 x 10^2 / 4!

[graceful vault]

wait

i also got confused

if thats the case

this should work right

[dire aurora]

which?

[graceful vault]

wait i got whats wrong

holup

[dire aurora]

ok !

[graceful vault]

i'll send what i did on paper

[dire aurora]

take ur time im in no rush

[graceful vault]

there will be 4 different cases

for 1 letter

codes

ig you took account of only one

[dire aurora]

I dont agree from what i understand but will take a look with open mind

ok will try your way

to see if its the same as 36^4

[graceful vault]

yea

ill do the same too

[graceful vault]

there will be 24 if the numbers interchange

[dire aurora]

the answr is 1.927... x 10^12

too big

[graceful vault]

ok yea

[dire aurora]

10^4 + (4 x 10^3 x 26) + (12 x 10^2 x 26^2)+ (24x10x26^3) + 26^4

is too big

[graceful vault]

36^4 should be around 10^7 right

[dire aurora]

um

[graceful vault]

like 7 digits

ig

havent done math in a long time

[dire aurora]

36^4 is miniscule in comparison yeah

u just need to remember multiplication principle gives ordered selections of objects

[graceful vault]

i still gotta figure out why your logic is wrong tho

🫡

[dire aurora]

i think i mispressed a number on the calculator

your answer is 5600416

not bad

still wrong

gosh i think your approach is more correct than mine

questioning everything i know now lol

[graceful vault]

ayy

i still got it wrong tho

😅

trying another way

[dire aurora]

lets do another question

a group of 5 girls and 6 boys sit in a row

[graceful vault]

yea

[dire aurora]

how many ways can 5 girls and 6 boys sit in a row

so here replacement is not allowed

so the answer is 11!

[graceful vault]

yea

[dire aurora]

can you get 11! with ur approach

idk its been a long night for me

[graceful vault]

yes

ill send

[dire aurora]

if u want u can just do 3 girls and 2 boys or smth

less cases

[graceful vault]

there is only one case

no.of boys and girls dont matter

only the total

matters

[dire aurora]

I know that

[graceful vault]

it shd be 2 sat 9 left

[dire aurora]

we are trying to see if ur previous approach can also work

[graceful vault]

whats different

in this and the prev

[dire aurora]

before it was a sum of cases

mutually exclusive cases

[graceful vault]

here there is only one case tho

like

[dire aurora]

10^4 + (4 x 10^3 x 26) + (12 x 10^2 x 26^2)+ (24x10x26^3) + 26^4

[graceful vault]

every person is different

there

2 letters cd have

been the same

or 2 numbers

[dire aurora]

i dont understand

we arent on the same page

[graceful vault]

uhh

its fine

my communication skills are pretty bad

[dire aurora]

my brain is fried rn

it always happens

at a certain threshold i cant think as clearly

[graceful vault]

coffee helps

[dire aurora]

and this is very abstract topic

its past midnight here haha will try coffee next time

[graceful vault]

oh

you should be asleep

when's your test

[dire aurora]

sleep schedule is messed up rn hence im still up

[graceful vault]

same

i feel sleepy

at the wrong times

and awake at like 2 am

[dire aurora]

i feel sleepy; hop into bed and then insomnia

so hopping into bed wasted time

yep

i think i get it

your explanation is correct

I think my method is wrong as you have explained, but modifying my way to make it a correct method is too hard compared to 36^4

i.e smth is wrong with this calculation: 10^4 + (4 x 10^3 x 26) + (12 x 10^2 x 26^2)+ (24x10x26^3) + 26^4

[graceful vault]

yea

i just tried

the basic way

that is

counting

for all unique

and like

some diff and some unique

[dire aurora]

think some time i will be able to do it correctly the hard way

idk man

10^4 + (4 x 10^3 x 26) + (12 x 10^2 x 26^2)+ (4x10x26^3) + 26^4

this should be correct now but it still isnt

idk why

you still addressed a gaping hole in my logic though

which ur first explanation

thats clear to me now

I was putting letters in a specific place

and assumed otherwise

[graceful vault]

wait

ig i figured out why it wouldnt work

like my logic

gets the wrong answer

because

when i choose like 1 letter 3 no.s

i cd have a code B666

this is one case

and there is a repetition

when there is a repetion u divide by (no.of repetitions)! to the total number of cases

so

1 letter and 3 numbers

can have 1 letter 3 unique numbers

1 letter 1 pair of same numbers and a different number

1 letter and 1 triplet of same numbers

so 3 cases

[dire aurora]

Oh

shit

god

lets try again

so we over counted didnt we?

um

[graceful vault]

1st case will have 4!(26)(10)(9*8 + 9/2! + 9/3!)

that should be around

400000

im guessing

put in the calculator

and check

[dire aurora]

wait whats the problem with ur original method?

[graceful vault]

like

imagine

1 letter 3 no.s

now

in one case

[dire aurora]

ur method includes these arrangments: can have 1 letter 3 unique numbers
1 letter 1 pair of same numbers and a different number
1 letter and 1 triplet of same numbers

(4 x 10^3 x 26)

that includes those 3 sub-cases

[graceful vault]

for the second case

[dire aurora]

doesnt it?

[graceful vault]

i will overcount

since

[dire aurora]

oh

um i think so

[graceful vault]

i cd have

twice the number of cases

because the digits repeat

once

like

once when the first number is equal to the other number

and again when the latter is equal to the prev

ig im juust complicating rn

holup

lemme refresh

[dire aurora]

I dont think this over counts: (4 x 10^3 x 26)

[graceful vault]

yea

i fogured it out

[dire aurora]

its (10^3 x 26) + (10^3 x 26) + (10^3 x 26) + (10^3 x 26)

for 4 places the letter could be

there no overcounting across those 4 sub-cases

because the letter is in a different position each time

1st, 2nd, 3rd then 4th

if its a circle then we are over counting

[graceful vault]

how close is this to the answer

[dire aurora]

67600 short

[graceful vault]

this?

[dire aurora]

exact

[graceful vault]

ayy

ayyy

ayyyyyy

[dire aurora]

!!!!!!!!!!!!!

show me !

[dire aurora]

6

i did 12

[graceful vault]

yea

[dire aurora]

10^4 + (4 x 10^3 x 26) + (12 x 10^2 x 26^2)+ (4x10x26^3) + 26^4

either way this prove everything

why did u put 6 instead of 12

[graceful vault]

6 cases

when 2 letters and 2 no.s

[dire aurora]

i did 4P2= 12

buttttt

whats the but

[graceful vault]

it shd be 4c2

[dire aurora]

i think i see

[graceful vault]

NNLL
NLNL
NLLN
LLNN
LNLN
LNNL

these are the 6 cases

N for number
L for letter

[dire aurora]

yes

and LL are identical

thats why its 4p2/2!

[graceful vault]

yea

[dire aurora]

amazing

Thank you very much

rigorously proved that: number of possible selections in multiplication only gives their arrangements where items of a type are in fixed position.

hence to find arrangements where the selections are in every position, those selections must multiply with a permutation

for some reason I forgot that

also forgot that when items are identical, arrangements of such items must not overcount

this is such a hard topic lol

but i guess I now know that i can in fact take the sum of mutually exclusive cases. All i needed to discover was that my construction of the sum was not diligent enouhg

my main concern was not being able to take the sum of mutually exclusive cases and that was boggling to me hence I came here in desperation

but u showed that i can and that got rid of my doubt

I just need to do it properly

[graceful vault]

sorry for wasting your time with my confusion in bw

😅

[dire aurora]

i.e. number of possible selections in multiplication only gives their arrangements where items of a type are in fixed position.
hence to find arrangements where the selections are in every position, those selections must multiply with a permutation. If that permutation is of identical items r, then u must remove overcounting by dividing by r!

idk how long i had this misconception

taht multiplication takes all cases

[graceful vault]

yea

[dire aurora]

it only has the case of each item in its fixed position

[graceful vault]

if you dont mind me asking

what course are u on rn?

[dire aurora]

high school

[graceful vault]

ayy same

[dire aurora]

awesome

its also interesting to me to note that there's a relativity to the "repositioning" of the items of different types

we repositioned the letters

and that automatically repostioned the numbers

because doing it the other way by repositioning the numbers manually would just be symetrical

assuming our layout is kept constant

[graceful vault]

also idk if you have done binomial theorem but this is nothing but the binomial expansion of (10+26)^4

[dire aurora]

OH

[dire aurora]

WOW

oh

my

god

did u see what i just said

i came to that conclusion

without realising

its a trend

[dire aurora]

with ALL binomials

[graceful vault]

yea

[dire aurora]

taking the sum of mutually exclusive cases not only works (which was my biggest relief), it is the expansion of the binomial with which you use to arrange all cases in intersection

im fucking mind blown

ive been doing weeks of counting

but didnt think it would be this closely related to binomial theorem

btw what software do you use

when u worked with me?

like how do u have such nice equations

whereas I typed mine onto the chat here?

so

If a counting question has the conditions: replacement allowed and items of the same type (identical type), then it will most likely be corelated to the binimial theorem

[graceful vault]

desmos.com

[dire aurora]

because replacement means the number can be brought to the power of smth

[graceful vault]

really good website

[dire aurora]

and items of the same type need to be permutated whilst removing overcounts, yielding combination coefficients of the terms in a binomial when using the logical approach of summing mutually exclusive cases.

so beautiful

and specific

correlation

i only ever used desmos for graphing

thanks for your help!

better highschooler than me

[graceful vault]

ayy it's fine

I also never thought about your approach

i would have just done 36^4

[dire aurora]

Are u in ur final year?

[graceful vault]

and got over it

[graceful vault]

Like

It's a different type of

Education system here

[dire aurora]

How so?

[graceful vault]

But like we have 4 years

In high school

I'm in the third

Will step into 4th in a month

[dire aurora]

How old is ur cohort?

I see

You did this stuff last year i presume?

[graceful vault]

yea

[cosmic kestrel]

when people say "high school"

I am imagining bachelor degree

because here it's called high education

and school is just called school

how old I am jesus

[cosmic kestrel]

https://tenor.com/view/omg-schock-cat-gif-20299170

[left knoll]

Help
Let and ellipse with semi major axis "a" and semi minor axis "b" slides between coordinate axes.Find locus of focus of ellipse.

[left knoll]

heyo im lookin for your notion notes to literally any subject of yours im tryna collect a big amount of data to make an overview over any subject that is out there and bring it together to a big pool of useful notes.If u feel like u could help me out please dm! PEACE

[silver coral]

hiiii, would anyone of you like to help me on a mass and energy balance calculation?

[left knoll]

I know I might botch this explanation, but I will try my best. I am going to do it for the first example, and you can try to apply for all.

We can see that T takes in a vector V = (x, y) and 'outputs' a new vector V' = (2x+y, x-y).
If we want to show that T is a Matrix Transformation, T must follow the TWO rules:
(1) T(u+v) = T(u) + T (v) where u, v are vectors
(2) T(ku) = kT(u) where k is a scalar

Let's consider two vectors u = (x1, y1) and v = (x2, y2).

1. What is u + v? (Hint: it's a vector)
2. What is T(u+v) (Hint: it's also a vector)
3. Can you take this new vector T(u+v) and split it into a sum of two vectors? (Hint: you can and try to split it such away that the first vector is T(u) and the second one is T(v) as you already know T(u) and T(v))

T(u) = (2x1 + y1, x1-y1)
T(v) = (2x2 + y2, x2-y2)

Of course it has to satisfies the second condition
Let's use the vector we already defined u = (x1, y1) and a scalar k

1. What is k.u (Hint: it's a vector, as always)
2. What is T(k.u) (Hint: yep it's a vector)
3. Using basic properties of vectors can you take the scalar out? and what does it leave you with? (Hint: it's k.T(u))

Lay those 6 steps out and you should have the complete proof for the first question.

[pure crescent]

can write them in vector form and then use cross product = 0 as they are perpendicular

[static monolith]

anyone know how to do this?

[left knoll]

Let and ellipse with semi major axis "a" and semi minor axis "b" slides between coordinate axes.Find locus of focus of ellipse.

[nimble bone]

hello, have been stuck on this problem for a while pls help

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis.

y = 8sqrt(x), y = 0, x = 1; about x = - 2

Struggle explanation (if it matters): ||When I graph it out, x = 1 is outside, so it would be (-2-1)^2 - (-2-8sqrt(x))^2, but each time I do the math I end up with integrals that ALWAYS end up negative, and I feel like it has to do with my initial set up but I can't figure it out to save my life||

[nimble bone]

^solved, ty

[tiny vessel]

Hello, does anyone know what the idea is here . i believe its adjacency matrix multiplied by itself -- but i could be way off:

[rain solstice]

we have a game
if you win you get 20$
if you lose you pay 13$
you played 18 times and at the end you lost 3 $
how many times you won?

[broken hinge]

create two linear eqn upon given condition

[rain solstice]

what are the they and how many is it 2 or 3

[broken hinge]

2 equations

first declare two variables

x = no. of games won
y = games lost

now what can you say about x and y?

@rain solstice

[rain solstice]

we will do

18-y=x right

18 number of games played

but the problem here is y since we have 2 variable we need two equation

[broken hinge]

right

[broken hinge]

think about how you can form the second one

they have said winning game gives you 20 and losing takes away 13

in the end you are left with a loss of 3

[rain solstice]

y=-13x-3?

[broken hinge]

not quite

[rain solstice]

should't we do another variable of how much money we have

[broken hinge]

no need

in the end you are left with -3

which would have come by adding all the money you gained and subtracting all the money you lost

remember you have won x games and lost y games

[rain solstice]

20x -13y=-3?

[broken hinge]

you got it

[rain solstice]

fr?

[broken hinge]

now you need to solve the two equations and you will get the answer

[rain solstice]

i feel dumb man

[broken hinge]

nah it’s alright

you solved it yourself

i just guided you on some steps

[rain solstice]

ayo thx for making it simpler

[broken hinge]

glad i could help

good luck, the answers 7

[rain solstice]

thx man

[static monolith]

Wow. Thanks man. Very much appreciated.

[copper heart]

Is it possible to do this math under 75 seconds . "a,b,c,d are in GP are in ascending order such a+d=112 and b+c=48 that If the GP is continued with a as the first term, then what is the sum of the first six term ?"

[broken hinge]

not sure about the 75 seconds part but it could be done pretty quick

[copper heart]

I don't think it could be under 5 minutes

[broken hinge]

i think i did it in a minute or two

[copper heart]

Can u send me a picture of it

[broken hinge]

see 4th term of gp = a * r^3

is that right?

[broken hinge]

i did it in mind calculations so don’t have any solving to show

[copper heart]

Like can you elaborate

It would be helpful if you could show me or tell me how did you solve it

[broken hinge]

is the answer 1456? @copper heart

[copper heart]

Yes

[broken hinge]

ok so i am right

[broken hinge]

@copper heart do you agree with this?

[copper heart]

Yes

[broken hinge]

right

so according to question a + ar^3 = 112

is that right? @copper heart

[copper heart]

Yes

[broken hinge]

what can you take common in this

[copper heart]

A(1+r'3)

[broken hinge]

right

[copper heart]

Then again ar(1+r)=48

[broken hinge]

yes

[copper heart]

Then??

[broken hinge]

i did it with this approach:
find factors of 112

[copper heart]

Yes

112=447

[broken hinge]

here what you need to remember is one of the factors must be a cube + 1

[broken hinge]

which is 28 * 4 am i right?

[copper heart]

Yes

[broken hinge]

and 28 = 27 + 1
meaning 1 + 3^3

is that correct?

[copper heart]

Yes

[broken hinge]

now then if you compare a(1 + r^3) with 4(1 + 3^3) can you tell me what will be a and r?

[copper heart]

Bro u really found it ... Great i understand now🔥🔥

I was cross multiplying and etc

[broken hinge]

well you found it yourself too now ggs

[copper heart]

Thanks again mate👍

[broken hinge]

good luck and im glad i could help

[pulsar terrace]

How can I identify these distributions? I still do not have some clue about this part in probability and statistics for computer science.

[silk sandal]

Have you read the chapters

All the distributions come from the world around us, so i would suggest looking into examples for each one

For example, the binomial random variable is just a counting variable. It counts how many times an event happens over a finite amount of trials

Say, you are throwing a 6 sided dice N times and wish to know the probability of get a 4, k times

Look for other examples online too

[cosmic kestrel]

Just how did he randomly put exterior product on these places

its Kirillov-Kostant-Souriau symplectic 2form as I understand

[copper heart]

Can some one solve this

[left knoll]

use sum of ap formula...n/2(2a+(n-1)d) > 1000
a=3
d=3solve n

[copper heart]

Bro try solving... I couldn't able to using this ...

[copper heart]

I will make it easy for you, 3(n^2 + n) >2000

[left knoll]

okay doing

[copper heart]

Plz try, i really need the help... Remember not to use calculator.. use pure reasoning and formula

[left knoll]

n=25.32
hence least is 26

[copper heart]

How can you send ... Me the solution.. answer is right btw

[left knoll]

transpose 2000 to side of n and apply quadraic formula of sridharya acharya...and get n...u will get roots as n-25.32 and other I guess 26..so (n-25.32)(n+26) >0
apply wavy curve and get n=25.32

[solid anvil]

it is c

[rain solstice]

can i ask u how you are good i math i mean i do a lot of problems but i find difficulties in solving a problem i never saw before i can solve anything similar i already did but i noticed some can get it even if they never saw a similar one like you

[copper heart]

I think he is a topG .🙃

[dusk jetty]

what is the gradient of the tangent to the function y = (2x^2 - 5)/x at the point (1,-3)

[copper heart]

Can someone provide me notes (digital: pdf ) for whole probability... More like revision sheet

[broken hinge]

my classmates always ask the same thing and i really don't have an answer to it lol

but it's probably because i have solved similar problems before and also i just like solving math in general i find it fun

i find problems difficult at times too lol i am a human too, it's just that i knew how to solve the problems asked here

[hexed scaffold]

How to calculate a number ^raised to a number in a calculator?

[rain solstice]

so the key is doing lot of problems as possible thx

[left knoll]

Pattern recognition is the key

[rain solstice]

you got a point i noticed it too also there is a lot of methods to solve a problem to more you know the easier it is

[left knoll]

But stick to one...I would suggest

[rain solstice]

nop i suggest the opposite some problems wont be solved unless u use a specific method some might but other methods will be way faster

[left knoll]

Nah I mean like know all the methods but while doing a particular sum solve it in one of the way..not all...while an exact same sum but which can be done easier by another method, then apply that for that one

[rain solstice]

yeah that's true like master a method and if it didn't work use the others

[left knoll]

Yeah yeah

[fervent rain]

hiya does anyone know how i could solve exp(z) = -exp(1)?

[pulsar terrace]

yes...I read it but I'm still confused that is any difference between these distributions?
like binomial, geometric these two are quite similar....
so How can I know when I could use it?
and the Poisson distribution depends on what situation?

[ancient prawn]

help

pls

[mental yew]

e^(x+e^x) is e^x*e^(e^x) dx
u=e^x
du=e^xdx
leads us to e^u du
which is just e^u sub back u

[primal delta]

yo

[sullen thicket]

Hellooo

I need help guys🫡

[trim saddle]

derivative y on those points

[haughty rivet]

helpppppppppp The system of equations whose translation is written below is written

[left knoll]

happy pi day guys

[left knoll]

3.141😀

[fair belfry]

😭 🙏

[past crown]

@fair belfry (2x-3) is common in equation so you can use it as a substitution and than multiply equation by your substitute, now its just solving quadratic equation

[fair belfry]

wait

my attempt 😭

[past crown]

you can also solve it like that but its much more complicated i think

[fair belfry]

yea

idk how to do it efficiently

[past crown]

use substitution t=(2x-3)^2