help-28

[frozen lark]

ye and i'm kinda happy cuz i could rescue some mis-guided student

[torn jolt]

and ill come back after i understand it

well

[frozen lark]

no first go learn binary, oct, hex

[torn jolt]

i didn't get anything done

[frozen lark]

and then get used to the logic gates

[torn jolt]

but being able to talk about this and my bozo math teacher gave me more potential to do better in this class

[frozen lark]

good

[torn jolt]

what do i look up

[frozen lark]

listen for the moment don't listen to that trashy teacher

go learn binary

pls

[marsh surge]

binary tutorial

[torn jolt]

ok thank you

[marsh surge]

or number systems

[frozen lark]

then get used to the logic gates

[torn jolt]

wait

i actually know that

octa is 8

hexa is 16

binary is 2

[marsh surge]

binary is w

2

[frozen lark]

stop right there

[torn jolt]

i used a calculator for that i learned that in chapter 4 i got a 100 on it

[marsh surge]

decimal is 10

[frozen lark]

binary is 2

[torn jolt]

i mean that yeah

my bad

1110110

pattern of 0s and 1's

bit strings yeah i hate the written question assigments

[frozen lark]

sound good

you'll use them in the truth tables

[torn jolt]

how do logic gates work

[marsh surge]

convert 1011001 to decimal

[frozen lark]

ayo guys sorry but i think it's time for me to close up

[marsh surge]

bye bye

[frozen lark]

@marsh surge please take care of him

[marsh surge]

will do

[torn jolt]

bye

[frozen lark]

k thnx hope you get over this

[torn jolt]

thank you for the help

[frozen lark]

no pb

[marsh surge]

np mate

[torn jolt]

now what

im still confused on the first question

[marsh surge]

learn
binary

[torn jolt]

i know how to do it

binary is 2

its simple

2 10 8 16

[marsh surge]

what

[torn jolt]

binary decimal octa and hexa

[marsh surge]

those are the bases

[torn jolt]

base 10

[marsh surge]

do you know how to convert from binary to decimal, hexa to octal, etc

[torn jolt]

yes

i used a calculator

[marsh surge]

learn how to do it without a calculator

[torn jolt]

but i need help with the logic gate 😦

i only have today to get help

otherwise tomorrow i go to the tutor and get help

[marsh surge]

https://youtu.be/JQBRzsPhw2w

great guy

[torn jolt]

the crash course video made more sense

anyway

ill come back after i understand the material

you can close it

[marsh surge]

only you can do that

[torn jolt]

.close

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[marsh surge]

gl on your logic gate endeavors

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[torn jolt]

.close

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[lyric ember]

post problem please

[eager violet]

!15m

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@tough jackal Has your question been resolved?

[torn jolt]

hello can you help me with math?

[spice orchid]

#❓how-to-get-help

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@tough jackal Has your question been resolved?

[gritty rose]

• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.

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[full fossil]

i can't solve this question , i just dont know where to start . the question sounds simple i just done know what equation to use here

[limber flicker]

hmm lets see

so

a_distancebetweenslits = 0.6e-3

D_distance from screen to doubleslit = 11m

x_distance between fringes = 10e-3

do you agree

@full fossil

[full fossil]

yes

[limber flicker]

and do you know your relationships ?

$\lambda = \frac{\alpha \cdot x}{D}$

[glossy valve]

Clarkie

[limber flicker]

,calc (0.6e-3 * 10e-3)/11

[glossy valve]

Result:

5.4545454545455e-7

[limber flicker]

545nm

[full fossil]

ohhh

[full fossil]

i seee

[limber flicker]

do you know how to approach the second part?

[full fossil]

you mean )d ?

[limber flicker]

yes

[full fossil]

well its basically saying how would the wave would look like if it was blue , (shorter wavelength)? if im not mistaken

[limber flicker]

and what affect would that have on the geometric appearence?

[timber oriole]

another physics question 😢

[limber flicker]

MY FAV

[full fossil]

i dont know

[limber flicker]

do you know about proportionality ?

if I say that wavelength is inversely proportional to frequency do you know what I mean

[full fossil]

yes i understand that part

[limber flicker]

ok nice , so look at our equation and pick out the variables that we are interested in first of all

[full fossil]

lambda?

[limber flicker]

yep thats a pretty important one

which ones stays constant

[full fossil]

the distance

[limber flicker]

yep

[full fossil]

so then we have to change a and x?

[limber flicker]

if lambda decreases yes they must increase, although im not sure if that happens in the real world

trying to think

meh actually should be fine

distance between fringes is proportional to the wavelength

im glad its not asking for a percentage :DD

[full fossil]

oh it is , its just that im not taking high level phy , so im fine lol

i dont have to do that part

i understand all of u said just a small question in the equation , is x the slit separation?

[limber flicker]

yes x is the fringe width

distance between fringes

a is the slit separation distance

and D is how far from thingy

[full fossil]

yeeyey okay

okay thank u very much sorry for the phy question lol

[limber flicker]

all g man

keep them coming

I post mine in here too :))))) to the annoyance of everyone

[full fossil]

wait i have another phy question , is it okay if i send in 2 min and not close this room?

[limber flicker]

for sure

[full fossil]

k thanks , 2 min

sorry for the long wait

[limber flicker]

what is the le problem

just gotta draw the line

but dont force it through the origin

[full fossil]

sorry bad pic , that is not the problem

how do i do the (i) part

[limber flicker]

you drawn your line?

$m=\frac{y_2-y_1}{x_2-x_1}$

[glossy valve]

Clarkie

[limber flicker]

just pick two points that are easy to determine on your line

[full fossil]

yes

[limber flicker]

although

it says at 80 degrees

which implies it changes throughout the graph

which isnt true for a linear best fit line.

you can do

y-b=m(x-a)

y = specific heat capacity

a and b are a point

and m is your gradient

[limber flicker]

this

might have to play around with it a bit

[full fossil]

yes , so y= 100 , a and b i just choise and the m i have to calculate correct?

[limber flicker]

I would just start with getting the gradient first from two points on your line tbh

then can you send a pic

[full fossil]

ok i got a bit confused went it said 80 degree , i can choose any point i want no?

[limber flicker]

I would say any two points

thats my interpretation

because its a line it should be constant throughout

this is a solution to something similar for thermal physics and it uses what im speaking about

to find a specific value

but I dont think thats what this requires

[full fossil]

but if there is a line there isn't the gradient increasing as the tempters rises?

[limber flicker]

no the gradient will remain constant on a line

[full fossil]

okay i see

[limber flicker]

gotta get on with some stuff so ill quickly tell you

part ii) is multidimensional analysis, everything is already in SI units, and its vertical axis against horizontal, think v=d/t graph and the units of that and how they compare here

and c , just put everything into percentage uncertainty and you should get your absolute uncertainty in your final result

and ci is just E=cmdeltat

ci I think is referring to similar stuff of what the screenshot i sent in

or just get reading from graph actually

[full fossil]

okay i go one by one , thank u so much 👍

[limber flicker]

gl

[full fossil]

thanks

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[half lark]

What's the name of the non-euclidian distance from point A to point B? The distance that can be calculated like this: D = (Bx - Ax) + (Ay - By).

For example, if :
Point A = (0;0)
Point B = (1;1)
Then D = 2

[full forum]

Please don't occupy multiple help channels.

[gritty rose]

i closed your other channel

[simple totem]

if you're talking about $$d((x_1,y_1),(x_2,y_2))=|x_1 - x_2| + |y_1 - y_2|$$ it's called the $L^1$ norm and sometimes the taxicab metric

[glossy valve]

maximo

[half lark]

ok thanks

.close

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[proven jay]

i'm trying to prove ciel(x) = floor(x) + 1 , where x doesn't belong Z

[simple totem]

let x be in (0,1). show ciel(x) = floor(x) + 1
then let z be in Z, and we can write any real number (not in z) as x + z. show ciel(x + z) = floor(x + z) + 1

[topaz valley]

depending on how you define floor and ceil this should be extremely straightforward

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[quartz monolith]

This is my question , Which method involves the fewest calculation steps? (Gauss Jordan, Gauss elimination, orCramer ?

[torn jolt]

probably cramer's rule

it depends on the dimension though

[quartz monolith]

how about 3x3

[torn jolt]

it really depends on the system

there is no universal fastest method

[quartz monolith]

x+y=6
-3y+z= 7
2x +y +3z= 15

[quartz monolith]

which the fewest calculation method for this case?

[torn jolt]

probably cramer's rule bc you can take the determinant of $\begin{pmatrix} 1 & 1 & 0 \ 0 & -3 & 1 \ 2 & 1 & 3 \end{pmatrix}$ and find $\frac{\Delta x}{\Delta}$ and then just substitute into the first equation

[glossy valve]

45

[quartz monolith]

it is 3x3 matriks? or 3x4?

[torn jolt]

3x3

[quartz monolith]

oh okay thanks

[topaz valley]

cramers rule is seriously inefficient in terms of solving systems of linear equations

calculating determinants is not exactly fast

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[hearty adder]

help

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[eager anchor]

For strictly diagonal nxn matrices

[eager anchor]

is it true if they have the same characteristic polynomial then they are similar?

I know this statement is false for arbitrary nxn matrices

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[pseudo portal]

So I am doing that problem and it throws me incorrect

[pseudo portal]

I have calculated by the following way:
20 is 6C3 and 10 is 5C2

[lyric ember]

okay so you have clearly calculated 20 as 6C3 and 10 as 5C2

[pseudo portal]

Yes

[lyric ember]

you should multiply the two parts rather than add them though

because if you have p possiblities for the first part, and q possiblities for the second part, you have pq possiblities in total

not p+q

[pseudo portal]

Oh ok

Sorry, I just hate math

[lyric ember]

so instead of your answer being 120+20, it will be 120*20=2400

[pseudo portal]

Gotcha

TKS

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[novel thistle]

Is the solution and answer right?

[lyric ember]

@novel thistle could you give a screenshot of the problem?

[novel thistle]

Oh the problem is the part where the problem presents...30 tents

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@novel thistle Has your question been resolved?

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@novel thistle Has your question been resolved?

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[burnt torrent]

why didn't get y get a derivative like x?

[burnt torrent]

do we only take a derivative of one side?

[vast fossil]

Yeah, (xy)' = (x)' * y + x * y' = 1 * y + xy' = y + xy'

[burnt torrent]

Thanks

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[burnt torrent]

so I can also solve it with respect to y

and the answer would be the same right

it doesn't matter

.reopen

[full forum]

✅

[vast fossil]

You need to find the slope

So solve for dy/dx

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[dense edge]

Can someone show me the steps involved to get from top to bottom?

I can get this far and then I am stuck…

[torn jolt]

Yo bro

Whats up

Where did you get?

[torn jolt]

Broooooo

Factor the quadratic bro

Ah okay got it

Lemme show you

[dense edge]

Oh yeah.. you can just factor the quadratic from the second line

Just noticed that now

Lemme see if I can get the same answer as you

[torn jolt]

So i just have to factor the quadratic?

So like you understood how they got there?

[torn jolt]

Im done factoring the quadratic

[dense edge]

What numbers did you get?

Add to -71 and multiply to 1290

[torn jolt]

Hmm

Lemme try again wait a sec

[dense edge]

Wait

My method doesn’t work

I forgot

It has to equal 0

So this is wrong..

I don’t know how to factor power 3 polynomials yet

I can’t just factor the power 2 polynomial and call it a day, that would be too easy..

[dense edge]

I need to find out how this group factor was calculated

3y-2

Because it doesn’t make sense to me.. any <@&286206848099549185> experienced with factoring power 3 polynomials?

[torn jolt]

Hmm lemme try again

[dense edge]

I think this gets tricky as you go higher power.. factoring a power 4 would be like a masters degree level math problem

[torn jolt]

Probably yes

Yo

We can try using rational roots theorem

Lemme try

[full forum]

@dense edge Has your question been resolved?

[dense edge]

Any luck?

[torn jolt]

Nah bro

This is unsolvabele

[dense edge]

It’s solvable .. just don’t know how to do it yet..

[torn jolt]

Shit

I copied the question wrong duddeeeeeee

Solved

I DID IT

[dense edge]

How?

[torn jolt]

Rational roots theorem

Heard of it?

[dense edge]

Can you post answer with steps involved?

[torn jolt]

Just a sec lemme solve the whole

[dense edge]

Ok thanks

[torn jolt]

@dense edge

Bro sorry it took so much time

It took more time to factor the quadratic than the cubic lol

[dense edge]

Oh

But you factored it?

[torn jolt]

Yea

[dense edge]

Yikes

Much more involved than quadratics

I don’t want to imagine a power 4 or power 5 and the work involved

I wonder what the world record is for highest degree polynomial factored by hand? And highest degree polynomial factored by a supercomputer

But I’m gonna guess that the supercomputer is capable of factoring to an exponentially higher degree than the world’s smartest mathematician

[dense edge]

Thanks for this explanation

[torn jolt]

Yeah dude i wanna die

[torn jolt]

Lemme check it out

Nice question

[torn jolt]

Yes true

Or is it?

[torn jolt]

No problem bro

@dense edge type .close

[dense edge]

.close

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[feral sorrel]

how do i find the rate of change

[feral sorrel]

ive found the a when x=4, f, f'(x)= x^3-ax^2-4ax

but im not sure how to find the rate of change when x=-2

im a bit stuck

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[void arch]

I don’t think my solution is valid?

[ivory cairn]

you want 2a+b-5c to equal 0, not 1

and any 3 values that satisfy that equation will work. there isn't a single answer.

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[languid wharf]

Compute the following sum of a geometric sequence: S = 0.9^32 + 0.9^31 + ... + 0.9^2 + 0.9 + 1

[lyric ember]

@languid wharf do you still need help?

[languid wharf]

yes

[lyric ember]

okay so first we will find 1+0.9+0.9^2+... all the way to infiity

do you know the formula for an infinite geometric sequence?

[languid wharf]

no

[lyric ember]

okay

ill just give you the whole formula

[languid wharf]

ok

[lyric ember]

if we start with a term of x, and if we have n terms, with a common ratio of r

the formula is x(1-r^n)/(1-r)

remember this formula

[languid wharf]

ok

[lyric ember]

so what is our first term?

x=1

[languid wharf]

ok x=1

[lyric ember]

what is our common ratio?

r=0.9

how many terms are there? 33 terms (0.9^0 to 0.9^32)

[languid wharf]

ok

[lyric ember]

so just plug these numbers into the formula

[languid wharf]

so n=33

[languid wharf]

1(1-0.9^33)/(1-0.9)?

[lyric ember]

yep

[languid wharf]

is the exponent supposed to be inside the parenthesis?

if so the answer is 9.69096845

[lyric ember]

yep its inside the parenthesis

[lyric ember]

that answer seems correct

[languid wharf]

ok

ty

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[woeful moth]

How would you find a linear equation that goes through the origin (0,0) and has exactly one intersection point with the a parabola? In this case $x^2+2$

[glossy valve]

Ladan

[smoky wing]

assume the line is in the form y=mx then mx=x^2+2 this must have only one solution

can you go from here

[woeful moth]

why can there only be one solution to that?

wouldn't something like y=4x have 2 solutions?

[smoky wing]

wdym

[smoky wing]

they intersect once

[woeful moth]

Oh you are saying I have to figure out how to only get 1 solution

I thought you were saying x^2+2 = mx can only have 1 solution

[smoky wing]

you have to figure out the value of m for which x^2+2=mx has 1 solution

[woeful moth]

I could be missing something simple, but im really unsure on how to start

[smoky wing]

x^2 - mx + 2 = 0 has only 1 solution implies the discriminant is ?

[woeful moth]

ohhhhh -mx

I was trying to figure out how to get them over

I totally missed just subtracting

let me take a look..

m = sqrt(8)

how did I not think about the simple subtraction

thanks you!

[smoky wing]

also -sqrt(8)

[woeful moth]

true actually, symmetry

thanks for helping

🙂

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[full forum]

@ivory moon Has your question been resolved?

[astral sinew]

what happens when y=z=0?

nvm the whole thing blows up

that gives us new inequalities though

0 < x,y,z < 1

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[oblique pier]

How would I find the vertical asymptotes of xln(1+1/x)?

[oblique pier]

how do I get the domain?

[rare dock]

1+1/x >0 is true when x > 0 or when x < -1

look at it as 1/x > -1

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[deft zodiac]

well

a^b = e^(b log a)

then from there use ur normal laws

[woven sand]

e is a constant, an interesting one at that

It's intrinsically the base of the natural log

[deft zodiac]

yes

i mean to convert to base e first

theres no integration or derivative rules involving an arbitrary base a

but there is one involving base e

idk

how to change basis to e maybe

sry

change of base

[deft zodiac]

first

then

$(e^{f(x)})' = f'(x)e^{f(x)}$

[deft zodiac]

where (f'(x)) = (f(x))' which denotes the derivative of f

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[sleek quest]

suppose n is a positive integer such that 2n has 28 postive factors and 3n has 30 positive factors. how many positive factors does 6n have?

[sleek quest]

i need a hint

[forest widget]

relate the prime factorizations of 2n and 3n to the number of positive factors

[sleek quest]

idk if i know enough about n to use it

2n = n but one more power of 2

3n = n but one more power of 3

6n = one more power of 2 and 3

what do i do with thisp

[forest widget]

how does the prime factorization of a number relate to the number of positive factors they have?

then you can do some (tiny amount of) bruteforce to find an answer that matches information given

[wild sleet]

i'm not really getting any bruteforce, it's just generally weird

(x+1)(y)z = 28
(x)(y+1)z = 30
```this is what we actually know

z is just the unchanging rest of it

[sleek quest]

but thats three variables and two equations

[wild sleet]

(you multiply the amounts after increasing them by one to count divisors)

xyz+yz = 28
xyz+xz = 30

xz = yz + 2

and because they are integers it somehow must mean that z is 2 and x is y+1

if i'm not glitching

[sleek quest]

how did you get that

[wild sleet]

i'm guessing the brute force was trying to find z that would divide both 28 and 30

[wild sleet]

i didn;t notice that z=1 also works with that

[sleek quest]

z=1 seems more realistic

[wild sleet]

neither z seems to be making sense...

[sleek quest]

z=1 makes sense

means the number only consists of 2s and 3s

[wild sleet]

oh, got it

z=1

(y+3)y = 28

obviously 4 and 7

[sleek quest]

wouldnt x be y+2 theb?

[wild sleet]

correct

and we have (x+1)y = 28

[wild sleet]

here

[sleek quest]

right

so then 6n has 40 divisors

[wild sleet]

35 i think

x=6

(6+1)(4+1)(1)

is yours cleaner iv?

[forest widget]

yeah i probably should've hinted at factorizing 28/30 first maybe?

if you check that the only way they have 3 prime factors are 2 * 2 * 7 and 2 * 3 * 5 which doesn't satisfy the constraints

then you get n only has 2 and 3 as prime factors and it's immediately a lot simpler

no 35 is right

[wild sleet]

i don't get it

oh i get it

[forest widget]

compare 2n to 3n: 2n has one more factor of 2, 3n has one more factor of 3

so comparing prime factorizations of 28 and 30: these are the products of exponents (+1 but that doesn't matter) of prime factors, so their prime factorizations should be {stuff} {a,b} vs {stuff} {a-1, b+1} where a,b correspond to the exponent of 2 and 3

[full forum]

@sleek quest Has your question been resolved?

[wild sleet]

basically 2 * 2 * 7 / 2 * 3 * 5 are too different, but maybe it was 4×1×7 that became 3×2×5 but even that doesn't work

not any cleaner then lol

[sleek quest]

thanks for ur help!

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[radiant narwhal]

what does F^n mean

[gritty rose]

depends on context

show the original notation

and look at it in the book where you got it from

[gritty rose]

what book/class is this from

[radiant narwhal]

linear algebra

[gritty rose]

F is a field like $\C$ or $\R$

[glossy valve]

riemann

[radiant narwhal]

what does the n represent

[gritty rose]

So $F^n$ is $\C^n$ or $\R^n$

[glossy valve]

riemann

[gritty rose]

n is a positive integer

dimension of the vector space

typically, n=2,3

[radiant narwhal]

so is F2 the same as R2

[gritty rose]

well depends

[radiant narwhal]

on what

[gritty rose]

on the context

the problem where it's used.

otherwise it's a general field

[radiant narwhal]

okay thanks

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[gritty rose]

that

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[astral sinew]

hint: every move must be rightwards

hint #2: think backwards

[mystic frigate]

2!3!2!?

2!3!2! is it

[shell whale]

well you do need to take in count the routes you cannot take

[mystic frigate]

you have 4 steps; however, the fourth step must be either A or B. The first step you have 2 choices of hexagons, so 2!, the second step you have 3 choices, so 3!, and the third step you have 2 choices, so 2!; however there is the middle route so subtract 2

[shell whale]

so it isnt as simple as 2! 3! 2!

since this is done on a small scale, you can prove it like so

[shell whale]

but using your formula here it is incorrect

yes it is 6

you cannot jump from 2 to 3

or 1 to 5

from step 1 u can either go 1 or 2, so 2!

[mystic frigate]

2!3!/2!

[shell whale]

step 2 depending on what you choose

[mystic frigate]

so 3! or 6

[shell whale]

if you choose 3 or 5 then its 1!, but if you choose 4 its 2!

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[rocky sail]

A coffee factory combines 3 different blends to produce a special coffee blend. Mixture A contains 5 grams of coffee, 2 grams of sugar and 3 grams of creamer. Mixture B contains 3 grams of coffee, 1 gram of sugar and 7 grams of creamer. Mixture C only consists of 2 grams of coffee and 3 grams of sugar. How many packs of each of these mixtures must be combined to produce special coffee blend that contains 21 grams of coffee, 10 grams of sugar and 27 grams of creamer?

[rocky sail]

i know the answer

but i dont know the way to get the answer

[atomic blade]

Let A be the number of A-mixture, B be the number of B-mixture, and C be the number of C-mixture packets

Write an equation for the grams of coffee

[rocky sail]

A+B+C =...?

like that

[shell whale]

yea something like that

now write each ingredient with the grams for each mixture

for ex. A = Sugar 2gr, coffee 5gr, ice cream 9gr

[rocky sail]

A = 5c, 2s, 3cr
B = 3c, 1s, 3cr
C = 2c, 3s

[shell whale]

and in your question it asks how many packs of each mixtures combine into 21c, 10s, and 27cr

do you know what to do from here?

[rocky sail]

no

i am bad at math

[shell whale]

ah ok basically just try things out would be the brute force method

[rocky sail]

is there a way, without the manual counting?

[shell whale]

in mixture C we dont have any creamer

yes using the method im showing you right now

[shell whale]

that means no matter how much mixtures of C we add in, there wont be an overflow of creamers

so lets focus on the creamers on mixture A and B then

[shell whale]

and get the creamers to the end result of 27cr

[shell whale]

easy way to see it, either 9 packs of A or 9 packs of B will get us 27 creamers, but 9 packs of A will give us more than 21c

so will 9 packs of B

[rocky sail]

but if we use 9 packs of b

we get more coffe

[shell whale]

oh wait yea we could use 9 packs of B

well no actually 9x3 = 27

[rocky sail]

ohh nvm

so then?

we add the c?

[shell whale]

well since we need to match our cr in A and B

to the end result

we need A and B to add up to 9

so either 8A and 1B
7A and 2B
6A and 3B
etc.

but you also should cross out the ones which arent possible

from 9A to 4A isnt possible

because 4A would be 5B and 4A + 5B= 20c + 15c

[shell whale]

this exceeds our limit

of 21c

that means we use under 4A packs from what we know so far

now lets look at B

wait I myself am confused

[rocky sail]

so b max = 3?

and c max is 3

we didnt use more than 3 packs of c

or b

[shell whale]

3A + 6B = 15c+18c. Over the limit we need

focus back on A

[rocky sail]

ok

[shell whale]

2A + 7B = 10c + 21c

nah something must be wrong here 💀

the A and B mixtures must equal to 9 because the creamers need to be 27 😡

[rocky sail]

wdy

wdym

[shell whale]

whats the answer

A = 5c, 2s, 3cr
B = 3c, 1s, 7cr..

omg

you said ..

[rocky sail]

2,3,1

[shell whale]

u said mixture B had 3 cr 😭

[rocky sail]

oh sry

there must be a typo

[vernal elk]

yeah don't worry my friend is an idiot sometimes

[shell whale]

🤭

thats alright lemme see this again

[rocky sail]

okie

[vernal elk]

btw, if he fails this, he's going to get lectured by someone

[shell whale]

A = 5c, 2s, 3cr
B = 3c, 1s, 7cr
C = 2c, 3s
end mixture : 21c, 10s, and 27cr

since there is no creamer in C we can focus on A and B to get 27cr

[rocky sail]

ok

[shell whale]

we can assume there will not be more than 3B packets because 4b would make 28cr

[rocky sail]

yes

[vernal elk]

He already said the answer

[shell whale]

so if it was 3B it would be 21cr

[vernal elk]

he's just confused how to get there

[shell whale]

how much more A do we need to get to 28

[vernal elk]

2...

2A

[shell whale]

yes

and so 2A + 3B = 27 cr

we dont need to do anything else for A and B

we can either look at c or s now on C

for example look at sugar

we need 10 sugar and we already have 7 sugar

how many packets of C can give us the remaining sugar we need

1 packet

so thats done there

2A+ 3B +1C

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@reef hazel Has your question been resolved?

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[tame crag]

for all elements x of a symmetric group S_n, o(x) <= n right?

[brittle steeple]

(12)(345) has order 6

[tame crag]

i cant get my head around this

but yeah ty

[brittle steeple]

np

[tame crag]

oh wait i see

even though any particular number is gonna get back to itself within 5 compositions

it might take more compositions for every number to get back to itself

[brittle steeple]

yeah

[tame crag]

makes sense now

OK ty

[brittle steeple]

Np

[tame crag]

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[karmic stag]

can someone explain this question in simplify manner ??

[smoky wing]

If we have x1 and x2 where x1 is less than x2 then f(x1) < f(x2) so the function is strictly increasing

[cursive flare]

What have you tried so far

[hasty kelp]

How can I solve this🥲

[karmic stag]

this is assinged to me

rn

[karmic stag]

and how they draw the graph

@smoky wing

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[karmic stag]

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[brittle parcel]

OM/OA=MM1/AA1 via the theorem of Thales

[brittle parcel]

is this correct?

I don't think so

AA1 || MM1

AA1 is height to the plane

but I don't know how OM/OA=MM1/AA1

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[tawny pond]

ABCD is a rectangle with an area of 72 cm2. E and F are the midpoints of their corresponding sides as shown. Find the area of triangle AEF.

[sharp vine]

maybe upload pic

[tawny pond]

ok sure

wait It will take me some time to get it so I'm just gonna close and I'll open a new one later

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[rapid laurel]

still dont get this

how do you write a=1/ (-1)(-2)(-3)...(-n-1) to a = (-1)^n+1 * 1/(n+1)!

and what to do after that

what does Q(n+1) equal that

can someone also explain to me just generally, what it means when polynomial behaves a certain way with a range of numbers and the question expects you to extrapolate the next value

so weird

[torn jolt]

just taken the -1 common from all brackets and factorials well i am assuming you know that

[rapid laurel]

yeh i see that

how about the step afte rit

like Q(n+1)=

does it just plug in the original Q(x)

[torn jolt]

yes what else

Q(n+1)= a(n+1)(n)(n-1)....(1)

Q(n+1)= a. (n+1)!

[rapid laurel]

right

ur right

how about

and thus that

after and thus that

P(n+1)=

[torn jolt]

=((-1)^n+1/(n+1)!) .(n+1)!

[rapid laurel]

yeh i get that step

the P(n+1) step

[torn jolt]

P(n+1) = (Q(n+1) + n+1)/n+2

[rapid laurel]

where did u get that?

oh

original

Q(x) definition

[torn jolt]

Q(x)= (x+1)P(x) - x
(x+1)P(x) = Q(x) + x

P(x) = (Q(x) + x)/x+1

[torn jolt]

..

[rapid laurel]

ehhh yeh thats true

ehh its still so weird

why can u plug n+1 to p now

when original u cant

ik origianly, the range is set to be 1-n

but why when u set up a new polynomial Q(x)

now you can plug p(n+1)

its because

[torn jolt]

that's not range those were the roots

[rapid laurel]

Q(x) has degree

n+1

right?

ah

i see

i see

[torn jolt]

yes

[rapid laurel]

okokok

thank you

[torn jolt]

welcome

[rapid laurel]

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[proper bloom]

how do I find $w(z) = -ilog(z+\sqrt{z^2-1})$

[proper bloom]

branch points and cuts of this

[glossy valve]

Vansh

[proper bloom]

I want to find branch cuts and branch points for this

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[real yacht]

im to find the derivative of U eff

[real yacht]

does this make sense?

can someone check my work

[normal tree]

https://live.sympy.org/ is your friend for checking difficult algebra

or if you want, you can learn Mathematica

[real yacht]

let me try sympy

i hope i figure out how to run it

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@real yacht Has your question been resolved?

[scenic yew]

Is this the derivative with respect to theta right?

@real yacht ?

[scenic yew]

Because if mgl are constant, you can just not include them during your derivative
find mgl (d/d(theta)(-cos(theta)+(w^2)/(4gl) cos^2(theta)))
mgl (d/d(theta)(-cos(theta))+d/d(theta)((w^2)/(4gl) cos^2(theta)))
mgl (d/d(theta)(-cos(theta)) + (w^2)/(4gl) d/d(theta)(cos^2(theta)))
mgl (sin(theta) + (w^2)/(4gl) (-2cos(theta)sin(theta)))
mgl sin(theta) + ((w^2) m cos(theta)sin(theta))/(2)

[real yacht]

yess

[scenic yew]

Sorry about this, don't really know on how to use texit

[real yacht]

do you mean
mgl sin(theta) - ((w^2) m cos(theta)sin(theta))/(2)

a minus in the middle

no?

thanks a ton

[scenic yew]

Because at the second step, you didn't find the derivative of -cos but instead find the derivative of mgl +mgl sin

That's why I asked you if you are really deriving for theta

[real yacht]

right yeah

i was a bit phased out and i notice i had replaced theta with x too

apparently this is what im doing

so the derivative = 0

and then i find an equation in terms of theta?

i guess that is what the part a needs

[scenic yew]

Yeah, and you excluded -cos theta which should be the one you're deriving. and your derivation at the 3rd step left side is that you derived w which is not theta

[real yacht]

i see

yeah gotchu

thankss

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@real yacht Has your question been resolved?

[lament shard]

Can anyone do this? Got an answer repeatedly but it’s incorrect

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[spice kelp]

Hey.

[spice kelp]

Somone help me?

I have to solve equation.

but how

I need to find x and y

13x + 11y = 36
11x + 13y = 12

You are very welcome to give an explanation for each point😄

<@&286206848099549185>

[scenic yew]

Try to equate or equalize the coefficient. You can use any coefficient in both systems of equation

[spice kelp]

I say.

14x * 11 + 11y * 11 = 36 * 11

But my friend say this is not correct

[scenic yew]

Nah, this isn't enough

Yeah

That's not correct

[scenic yew]

I mean, why are you trying so hard to equalize x or y when you can just equalize 36 and 12

What is the LCM of 36 and 12?

@spice kelp , can you find it?

[spice kelp]

What you mean?

[scenic yew]

Do you know lcm first?

LCM, least common multiple

Ok, what number can we multiply to 12 so that we can get 36

[spice kelp]

Can you show me example

[scenic yew]

https://www.mathsisfun.com/least-common-multiple.html

[spice kelp]

the substitution method do you know it

[scenic yew]

Yes

I mean, I already got the answer

[scenic yew]

I think you are wrong here cause you don't know how to do LCM

[scenic yew]

You just randomly multiplied 11 there

[spice kelp]

Correct, im from Denmark

Can't you show how to start with it?

[torn jolt]

he’s basically asking, in the simplest way possible, how do you get the X/Y values to be the same?

in order to do this you need to find the LCM (least common multiple)

[scenic yew]

Multiples are products of a number when multiplied to 1,2,3,4,5,6,....

[spice kelp]

Not the same, we have to solve one of them first and from there we get to the expression.

[torn jolt]

in order to solve it you need one of the x or y values to be the same

well atleast in the substitution method which is what you’re doing now

[spice kelp]

Yesss

[torn jolt]

yes

[spice kelp]

Example

[torn jolt]

technically yes you can solve like that too

i don’t remember that method though

[spice kelp]

Hmm

[scenic yew]

Sorry I went to another channel

[scenic yew]

Ah that's substitution

[scenic yew]

That's why I was trying to teach you earlier how to do LCM but you can use division

[scenic yew]

You can try to find a number that when divided to 36 is equal to 12

@spice kelp still there?

[full forum]

@spice kelp Has your question been resolved?

[torpid stag]

@spice kelp hvilket niveau er du på? Der findes et par forskellige metoder til at løse ligningssystemer, nemlig

[full forum]

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[dark forge]

proof that if a b c are three real numbers such as that a^2 + b^2 = c^2 then c/a + c/b >= sqrt8

[dark forge]

it's ok just delete it

and maybe help if u can :)

[boreal marlin]

Not sure it holds, sqrt(8) comes out of nothing

[dark forge]

no but like b4 that we have a question saying

x is a real positif number bigger than zero proof that x + 1/x >= 2

i managed to proof it

(x-1)² ≥2

but like the second one is hard

[dark forge]

this one is really hard

[viral jasper]

I dunno, square both sides maybe?

[dark forge]

alr did

[torn jolt]

just calculate c from first one and change it in second equation

sqrt(1 + (b/a)^2) + sqrt(1 + (a/b)^2) >= sqrt(8)

[dark forge]

wdym

[spice kelp]

Hej, er du dansk!

[torn jolt]

sqrt(2) + sqrt(2) >= sqrt(8)

sqrt(2) + sqrt(2) >= 2sqrt(2)

[viral jasper]

looks good

[torn jolt]

sqrt(1 + (b/a)^2) + sqrt(1 + (a/b)^2) you should prove that minimal will be

sqrt(2) + sqrt(2)

i dont remember exactly sorry

[dark forge]

i don't get it :)

what did u do here exactly please

[torn jolt]

c = sqrt(a^2 + b^2)

then

c/a + c/b >= sqrt8

sqrt(a^2 + b^2)/a + sqrt(a^2 + b^2)/b >= sqrt8

[dark forge]

but why

[torn jolt]

sqrt(1 + b^2/a^2) + sqrt(a^2/b^2 + 1) >= sqrt8

[dark forge]

the whole point is why

[torn jolt]

u know that sqrt it will be positive and also a and b is not equal to 0 so you should find minimal value

[worthy tree]

i got it

[torn jolt]

u can change (a/b) by k

(b/a) by 1 / k

so k^2 + 1/k^2 minimal result is always 2

as i remember from school

becuase

[worthy tree]

$\sqrt{ 1 + \frac{b²}{a²}} + \sqrt{ 1 + \frac{a²}{b²}} >= \sqrt{2}$

[torn jolt]

its is equal to (k + 1/k)^2 - 2

[glossy valve]

Mehdi_Moulati

[worthy tree]

if a < b

if a > b then $\sqrt{ 1 + \frac{b²}{a²}} > \sqrt{2}$

[dark forge]

mnin jebna a < b

[worthy tree]

fasl l7alat

[dark forge]

a 5ay ana f cinquième chmen fasl l7alat

[glossy valve]

Mehdi_Moulati

[worthy tree]

if a²/b² < 1 then sqrt(1 + a² / b²) >= sqrt(1) and sqrt(1 + b² / a²) >= sqrt(2)

[dark forge]

can you start over

[worthy tree]

so $\sqrt{ 1 + \frac{b²}{a²}} + \sqrt{ 1 + \frac{a²}{b²}} >= \sqrt{ 1 } + \sqrt{ 1 + 2}$

[glossy valve]

Mehdi_Moulati

[dark forge]

step by step

please

because im totally blocked

[glossy valve]

Mehdi_Moulati

$\frac{c}{a} + \frac{c}{b} >= \sqrt{8}$

[dark forge]

yes

[worthy tree]

$c² =a² + b²$ so $c = \sqrt{a² + b²}$

[dark forge]

yes

[glossy valve]

Mehdi_Moulati

[dark forge]

so we replace c in c/a and c/b

got it

then wut ?

[worthy tree]

$\frac{\sqrt{a² + b²}}{a} + \frac{\sqrt{a² + b²}}{b} >= \sqrt{8}$

[glossy valve]

Mehdi_Moulati

[dark forge]

an7aydo racine ?

[glossy valve]

Mehdi_Moulati

[dark forge]

okay

got it

[worthy tree]

$\sqrt{1 + \frac{b²}{a²}} + \sqrt{1 + \frac{a²}{b²}} >= \sqrt{8}$

[glossy valve]

Mehdi_Moulati

[dark forge]

mhm

yes

got it

[worthy tree]

if a = b

$\sqrt{1 + 1} + \sqrt{1 + 1}= \sqrt{2} + \sqrt{2}= \sqrt{8}$

[dark forge]

but where did we get a = b from

[glossy valve]

Mehdi_Moulati

[worthy tree]

i said if

[dark forge]

mhm

[worthy tree]

i will check all the cases

if a = b , a<b and b<a

if a < b :

[glossy valve]

Mehdi_Moulati

[dark forge]

yes

[glossy valve]

Mehdi_Moulati

[worthy tree]

since b²/a²> 1

so $\sqrt{1 + \frac{b²}{a²}} + \sqrt{1 + \frac{a²}{b²}} >1 + \sqrt{2}$

[glossy valve]

Mehdi_Moulati

[dark forge]

@worthy tree ?

[worthy tree]

we will find the same result if a > b

the problem here is sqrt(8) > 1 + sqrt(2)

[worthy tree]

@torn jolt is that correct , Right ?

[full forum]

@dark forge Has your question been resolved?

[torn jolt]

i think you can do that as u did but there should be easier solution

[worthy tree]

if you know share with us

[torn jolt]

proof that if a b c are three real numbers such as that a^2 + b^2 = c^2 then c/a + c/b >= sqrt8

i dont know but i try again

c/a + c/b >= sqrt8

if i find solution i will share

[worthy tree]

thanks

[dark forge]

.close

[full forum]

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[crude salmon]

I need help I don't know what is the next step

[boreal marlin]

Symmetry of triangles?

[crude salmon]

What do you mean

<@&286206848099549185>

[upbeat jungle]

because W and V are the same angle, the triangles are the same

[full forum]

@crude salmon Has your question been resolved?

[crude salmon]

.close

[full forum]

Channel closed

Closed by @crude salmon

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[junior siren]

help

[woven eagle]

,rotate

[junior siren]

help

[woven eagle]

is this math