math-help

[rain onyx]

can anyone help me with this question

[rain onyx]

I found dy/dx

isn't equal that given diff eqn

but answer sheet says yes

where did I go wrong

[sacred mauve]

can someone explain to me what this question is asking me to do and how i can go about the first few vague steps of completing it

[thick haven]

divide each term by e^xy to get the same answer as that of the answer sheet.

[rain onyx]

aahahah thx I couldn't saw

[sand agate]

can someone help me with the question b

i dont know how to get the answer

[atomic dagger]

Guys I have a question. Is this allowed ?

[ember kernel]

I'm more concerned about how you intend to divide a vector by another vector

And how you managed to get to that expression

Oh nvm that's a scalar product in the denominator, so that's okay

No, you can't """simplify vectors"""

But that expression looks very strange to me. In what context did you get to it ? (just out of plain curiosity, I'm not saying it's wrong)

[atomic dagger]

I’m practicing the Gram Schmidt method !

In linear algebra

[ember kernel]

Oh yeah I remember that

No

You can't simplify vectors

Because dividing by vectors makes no sense in the first place (in linear algebra at least. There could be some obscure field of maths where that could exist, idk I'm not knowledgeable enough)

[atomic dagger]

Thank you! I thought so too but I wanted to make sure

[ember kernel]

np

[cobalt mural]

hey; if possible, can u guys take this rly quick survey? it’s for my ap stats project and im in need of 20+ sample size. any help would be great! thanks! https://forms.gle/cUKhwUtEGxMQQREb7

[shy wraith]

I did, hope it helps :)

[cobalt mural]

thank you!!

[fallow sandal]

hey. anyone here german and good at maths?

no clue how to do this one

[left knoll]

Hii, anyone here free to help me out with this addmaths problem? It requires system of linear eqns to solve it

[thin gull]

Hi everyone,
I would like to ask for help
Known function R -> R (real numbers -> real numbers, it's a Real number symbol) defined with a rule as how is it shown on the picture
The question is, what is the **set ** of that function's co-domain?

[left knoll]

This one is a bit tricky you have to do that so that you'll get a answer and then just do that to get another answer then after 2 answers do that again and you'll get your answer hope this helps:)

[night kindle]

hi guys i just want to know how did they make this?

it's hard

oh sorry the number cropped is 12 and 4

[ember kernel]

Looks like they simplified -30/-300 to 1/10 and simplified sqrt(2)/sqrt(2) to 1

Then they calculated 10pi/12 + pi/4 = 13pi/12. Therefore sin(13pi/12)/cos(13pi/2) = tan(13pi/12) = 2-sqrt(3)

Now multiply those three independant simplifications and you get (2-sqrt(3))/10

(Also the absolute value has been implicitly removed since it's always positive)

[slim juniper]

I’m having trouble understanding discrete time dynamicals. How do you come to this conclusion? Why is t an exp for in d (pt = 0.57^t) and for e it multiplied by t (p100= p0 + 0.57t) ?

[ember kernel]

Question (d) :

p0 = 12

p1 = 0.57 * p0

p2 = 0.57 * p1 = 0.57 * (0.57 * p0)
= 0.57² * p0

p3 = 0.57 * p2 = 0.57 * (0.57² * p0)
= 0.57³ * p0

...

From there it is easy to observe that
pt = 0.57^t * p0

Question (e) :
Do the same kind of development, but add 0.57 each time instead of multiplying by 0.57

The observation will be just as easy

[mild imp]

guys i need help with linear algebra in Singular value decomposition if u know this topic well pls DM me

[night kindle]

hmmmmmm understood that's work

[raven halo]

i need help pls

when will this function always decreasing

[outer hamlet]

Are there any French people pls ?

[ember kernel]

Oui

[ember kernel]

Hi ! I don't understand the question. Could you translate it pls ?

[outer hamlet]

on peut parler pv ?

si sa te dérange on peut ici aussi

[gritty agate]

HELLO!!!!! Hope you all doing good, I need help with beamer LATEX , is there anyone who can help ?

[grand gust]

all you have to do is- (excuse the writing). I think this should be correct.

[cinder yacht]

u forgot the square root, so it's [-3/2 ; 3/2]

[copper folio]

Has anyone have any example of funcions(sorry, I don't know how are they called in eng) that would work for a field? I mean something like a ◽b = a + b + 1, and two others so it can be a field. Is it even possible?

Or they must be addition, subtraction, multiplication and division?

I need to show how to check if set of ''numbers'' is a field

[ember kernel]

What's the square operator ?

[copper folio]

It is literally a + b + 1, which creates abelian group inside of real numbers

And for example I have a triangle b = 2ab

Which creates field with previous operation inside real numbers

Now I need more "creative" examples of operators

[hardy kite]

I've been stuck on this question for a while but apparently the answer is a) but I'm not sure how you'd get that answer, do you use one of the log laws and if so which one?

[ember kernel]

You need to use the following logarithm identity :

[ember kernel]

Therefore, you get :

[ember kernel]

Because there's that other log identity :

[ember kernel]

⬆️

[hardy kite]

OHHHH RIGHT

the change of base law!!

[ember kernel]

It's pretty important when dealing with logarithms

Try to remember it

[hardy kite]

so in other words you'd first use the change of base log law to change Log_3 (10) and Log_4 (10) which gives you this and then from there you can tell it's equivalent to a right?

[hardy kite]

definitely tysm for the help! :)

[ember kernel]

Yes, because log_10 (10) = 1. Thus, both numerators simplify to 1

[hardy kite]

ahhh ok ok that makes sm sense

[thin gull]

I appreciate it, thank you!

[hardy kite]

thanks again for the help!

[ember kernel]

np:)

[hardy kite]

ah sorry turns out I'm stuck on another question but for this one, I tried using symbolab for help but the correct answer is 3.2 and I don't know how or why :/

I understand how you'd get 2^log_2 (x multiplied by 5) since you use the product law but I'm just stuck on how to continue solving to get 3.2 if that makes any sense

[cyan night]

2^log base 2(x)+log base 2(5) = 16 => log base 2 (x) + log base 2 (5) = 4 => log base 2 of (5x) = 4 => 5x = 16 => x = 16/5 = 3.2

[raven halo]

They asked when will this function always decreasing

Like with the m you replace in the function

The bigger the x

The smaller the result

[hardy kite]

OHH RIGHT

tysm for the help!! :)

[grand gust]

yes right😅

[ember kernel]

You want a constant negative slope (because you want a decreasing function)
Therefore, you want want a negative derivative

Therefore, differentiate, and using the quadratic formula solve for y' = 0 to get the roots of y'
From there you'll just need to determine the value(s) for m that make(s) y' negative near the roots

[mystic thunder]

I totally just forgot how to solve these type of questions

[left knoll]

it means that if y increase n times, then x will also increase n times and if y decreases n times then x will also decrease n times

y2 decreased 12 times from 60 so x2 will also decrease 12 times

[lapis relic]

with 2 point you can find the line equation, after that you have to substitute

[mystic thunder]

OH right omg thanku guys I appreciate it

[glacial dust]

@ember kernel how did he know that there is a 0.05 value of the nickels (n) and the rest as well

[flat blaze]

Hi could someone help me with this?
the answer is D but I dont know why

[ember kernel]

Because a nickel is worth 5 cents

That's it

[glacial dust]

oh I was thinking to much 🙃 \

aight ty

[ember kernel]

Np

Don't forget the 3rd equation :
d = n+2 (there are 2 more dimes than nickels)

[glacial dust]

ye

I i knew that one

[cinder ibex]

Hi, so f(1)=0 because thats the graphic of f(x), from that graph since the curve is turned down u can conclude f''(x) is negative so f"(1)<f(1). From that graph u can also see that it is "growing" (sorry for my lack of knowledge of certain words of maths in english) and so f'(x) is positive so f"(1)<f(1)<f'(x)

[flat blaze]

oh concave down

ahhhh

im so dumb

thanks a bunch

[cinder ibex]

No u not, it happens forgetting!

[flat blaze]

Many thanks 🙏

[cinder ibex]

U are welcome! : )

[sly sandal]

For 10a this is what I did

Even though the answer is correct, I'm not too sure why 240/12^2 is the acceleration I need to use

Can someone help me understand why it gives me the correct acceleration

[raw iron]

Keşke üni sınavında bunu sorsalar

[left knoll]

is n=10?

[glacial dust]

ye it was thanks for the effort but I got the answer already 😊

[left knoll]

i just did it for practice

[slow cobalt]

no clue whats going on

[cinder ibex]

well with tan u can get the angle, also u need to decompose what u have, in the end u need N and the friction force to get the u

[left knoll]

you need to apply lami's theorem here

[cinder ibex]

i just did this

but im just thinking how u can get aceleration in the x axis

with that u can solve the exercise

[cinder ibex]

never heard about that theorem

<@&717391911132069929> hey sorry to bother but this person multiposted probably a virus or something

[cinder ibex]

oh its in equilibrium so ax is 0

so ur Fa is 22

and u is 0.48

[formal patrol]

hi!! could someone please explain what the tip below the question means? why isn’t the ratio 5:16?

[slow cobalt]

ooooh ok ill try that thnx a lot

[cinder ibex]

np, lmk if u need any further explanation on what i did

[ember kernel]

Because if the ratio was 5 black : 16 green, then the probability of choosing a black olive would be 5 / (5 + 16) = 5/21 (= # of black olives / total # of olives)

But here we know it is 5/16

[formal patrol]

ohh got it! thank you!

[marble crest]

Can someone help me with number 13 please^o^

[left knoll]

Use the quadratic formula

[glacial dust]

Try multiplying coefficients A and C and if there are no factors that either add or subtract to 8 then proceed with quadratic formula

[cobalt mural]

hello!! i just need 7 more responses to reach my required sample size for my stat survey; it's one question, and any help would be much appreciated: https://docs.google.com/forms/d/e/1FAIpQLSeEYt28JNqf6CPf-a1Zt5DQ2SODF7TDHRHerPOCQ-ecs10Xxw/viewform?fbzx=-7323831876303373235

rly close to finishing it,, thank you to all who took it!!

[left knoll]

would this be considered to be an altitude and perpendicular bisector as well? the image got cut off but my answers were:

■angle bisector
□ perpendicular bisector
■median
□altitude

|| ■→selected | □→ blank||

I wanted to select all initially but there was no 90ﾟto indicate a perpendicular segment so I only chose those two options but now I'm second guessing (💢´ºヮº｀)و“

[cyan night]

in an iscoseles triangle, a median bisector would be everything else also

[left knoll]

ohhh okay thank you!

noting that down

[cyan night]

yes its simply provable using congruent triangles

np!

[left knoll]

it was in my physics book, for when an object is in equilibrium, could share a photo if you want

[left knoll]

What is going on for the second part

Cos( a+b)

[glacial summit]

If there’s no solution to this matrix, then what’s the value of a?

• 3
  *2
  *-1
  *7

Can anyone help me solve this problem??

[eager ibex]

Hi what do I do about this if I mustn't use L'Hospital

[ember kernel]

No solution = two lines (or linear combinations of lines) have the same coefficients for x, y and z but different independant values

[cinder ibex]

No need i already searched it up but thank you!

[left knoll]

alr

[left knoll]

Is this right

It’s wrong bruh

How is the denominator 4

Can someone confirm

[left knoll]

whats the question?

okay wait...

[hazy halo]

Ya

There is a typo maybe

[tepid oracle]

mine is 2

[left knoll]

My guy

Styuuuupid donnies

Textbook people

Thank you guys

[tepid oracle]

i'm brazilian , i know a little how text in english and i know a little to calculate , but calculate in english is hard to me , hahaha

[tardy bone]

does anyone know what "s.t." stands for in math equations?

nvm

[ember kernel]

"such that" I guess ?

[left knoll]

two perpendicular lines that aren!t vertical will always be what?

・have equal slope
・have one slope equal to 0
・have slope products of -1
・have slope products of 1

[left knoll]

Ayo

Can u guys help me explain the second but

Cos (a+b)

Bit*

[glacial dust]

do I have to find Voltage drops here

[cinder ibex]

hey, how can i simplify this? im not getting to the right solution

[limpid harness]

all of the second bit?

cos theta = sin (90- theta)

that's how they're linked

and in that case theta is (A+B)

then it just does sin(A+B)

[ember kernel]

log4(sqrt(5))
= log2(sqrt(5)) / log2(4)
= log2(sqrt(5)) / 2
= log2(sqrt(5)^1/2)
= log2(5^(1/4))

[left knoll]

is that just something u have to know?

[limpid harness]

yeah

[left knoll]

i always just use propeties of trig graphs

[limpid harness]

i think you get given it

[left knoll]

bruh

[limpid harness]

in the exam

[left knoll]

u do a levels?

[limpid harness]

yea

[left knoll]

nice

[cinder ibex]

thank you!!

[left knoll]

what year u in

[limpid harness]

12

[left knoll]

ahhhhh damn im year 13 and u know more than me

prepare for year 2 my friend

[ember kernel]

Dont forget to add the log2(3) to that

[limpid harness]

im doing further maths thats why haha

[left knoll]

ooooooooo smart guy

[limpid harness]

we've done year 2 already

[left knoll]

sheeeeeeeeesh

[limpid harness]

😩

[left knoll]

best of luck to u

[limpid harness]

u too

[left knoll]

what else u pick?

[cinder ibex]

yes thank you very much!

[limpid harness]

maths, further maths and economics

what do u do?

[left knoll]

bio chem maths

[limpid harness]

oooh

[left knoll]

i hope we can help each other out with maths

wait lemme add u jic

[limpid harness]

of course

[tropic lava]

How to draw that? This is ok?

[atomic dagger]

We are trying to find the argument and module of z. I don't understand what we're doing starting the third step? and how we found that the argument equals -3 from cos(3)-sin(3)i since 3 isn't on the circle

[ember kernel]

cos(3) - i sin(3)
= cos(-3) + i sin(-3)
= exp(-3i)
(Via trigonometric identities)

And 3 is on the circle. Actually, any real value can be an argument. It's just widespread convention to use multiples of pi when dealing with angles. However, that doesn't make any other real number invalid. (3 is close to pi = 3.1415 afterall, which is """on the circle""" as you say)

[deep kernel]

i'am looking for the min and max of -3sin(x) - exp(x)

why did he put the condition : expression < 1 ?

it says condition fullfilled

ty in advance

[flat blaze]

How do I find B in this situation? Thank you in advance

[willow silo]

Continuous and differentiable at x = 1. Use these conditions to form 2 equations in a and b then solve for them.

[flat blaze]

Ah, thank you

[stable dew]

shut the fuck up lol

[versed badge]

sorry i got hacked trying to fix it rn

[atomic dagger]

thanks!!

[viscid yarrow]

The largest binomial coefficient(s) in the expansion of (1+x)^4

[left knoll]

I must be tripping

Whenever I enter value of sin(3.14/4) I get root 2/2

How r they getting 1/root 2

Can someone please confirm I’m so lost

Just a quick calculation on ur part

[subtle dust]

1/root 2 is root2/2

[tepid oracle]

exactly

this is called of rationalization

[fossil kite]

root 2 and 2 cancel out

root 2 squared is 2

[left knoll]

How does

Become

?

[fossil kite]

is the number 72.28 or 7228?

[left knoll]

72 x 28

[fossil kite]

oh okay

[cloud ocean]

help meeee

heeeeelp

[fossil kite]

also, is the comma a typo?

[left knoll]

The comma is meant to represent a times

72 . 28

[fossil kite]

could u translate, im not that good at spanish

[left knoll]

We write times as a floating .

[fossil kite]

oh okay

thats kinda confusing for me

gimme some time

i'll figure it out

[left knoll]

I think I figured it out

How do I make a root symbol?

[fossil kite]

oh okay

[glacial dust]

Half it to form a right triangle and from there use trig ratios then times it by 2 after getting all three sides use cos rule to find one of the angles then from there just solve for X

the reason why I multiplied it its because it’s only half the lenth just in case if ur wondering why

Remember that the sum of the angles should be 180

[left knoll]

Ohhhh damn thank you everybody for the help

My Brian was not working

[dusk stone]

hellp

[woven orchid]

i need help with physics

how do i solve this question: assume that a gas is enatomic and consists of 6.0 * 10 ^ 23 identical atoms. how big is the change in temperature? and it loses 90j heat energy

it is in a closed container that does not change shape tbh

[ember kernel]

Thermodynamics would fit better in #science-help

[woven orchid]

its thermophysics

[ember kernel]

That's a different name for the same problem, but it's still science

You're more likely to get an answer if you send it to the best-fitting channel

[woven orchid]

💀

aight

[ember kernel]

Thanks have a good day you too :D

(Just trying to help you get an answer btw)

[atomic dagger]

can someone help me find a way to get this answer ? ^^'

[ember kernel]

define x = a², then solve for x using the quadratic formula
Then, knowing the possible values for x, find a using the x = a² relation we defined before

[atomic dagger]

oh that's very clever thank you! i will try that

[cinder yacht]

also later in ur uni years, if u see such a quadratic equation, u can try by testing 1 and -1 as solution before using formulas. Cause if by any chance u find that for example 1 is a solution of x^2 + bx + c, then you know that it's actually equal to (x-1)(x-x2) where x2*1 = c

anyway I guess it's just a detail, but sometimes allows to very quickly solve such equations

[left knoll]

can someone explain whats going on here

first line

isnt the 0.5 meant to be on angle

if its not i still dont get it

[ember kernel]

The formula is :

[left knoll]

wait

[ember kernel]

Now just replace alpha by theta/2 and you get what you have

[left knoll]

i understand the substitution

is this a double angle formula?

ooooooooooo

damn i see it

[ember kernel]

That's one of the fundamental trigonometric identities

[left knoll]

its just rearranged other than that its familiar

right

thank you

[ember kernel]

Yeah

We usually use theta and 2theta, instead of theta/2 and theta

But that's the same, since the value of theta can be anything we want

[left knoll]

that makes a lot of sense now

muchos gracias ese

[ember kernel]

np

[void gazelle]

hello, I had a question regarding statistics: is cross validation done only for unsupervised algorithms (like the k-means algorithm), or it can also be used for super supervised algorithms (like for the k-nearest neighbours algorithm)?

If I'm not mistaken, cross validation is when we separate our dataset into a training dataset and a testing dataset to prevent overfitting so I think it can be used for both supervised and unsupervised algorithms but I'm not sure...

[limpid juniper]

Can anyone help me with this please? im so confused on how to start

[gentle axle]

Does this add up?

i keep getting a total of 69 instead of the 66 it says

i want to know if im the idiot or the question is just wrong

universe is supposed to be 66, am i doing something wrong here?

[ember kernel]

The number of wine merchants that only sell Gevrey Chambertin and Cheap Plonk is not mentioned, but that doesn't mean it is 0 :)

[gentle axle]

legend

[ember kernel]

That looks better, right ? :D

[gentle axle]

yep

love you brother 😳

[ember kernel]

👉👈 😳

[atomic dagger]

i will keep that in mind thank you!

[tropic lava]

"(A × B) ∪ C" How to turn it into a sentence formula ? I mean x∈ A and ....

[ember kernel]

I'm sorry, I don't know what you mean. What kind of "sentence formula" would you want to get knowing only one of its words (the set (A × B) ∪ C) ?
Maybe some context or goal would help us see what you want

[marble crest]

I have a test can someone please explain how to do at least one or two of these

[ember kernel]

Just use the formula :
x²-y²= (x+y)(x-y)

You just need to find appropriate values for x and y

Eg for (a) :

Define c = 8 a^4 and d = 4b² (I took the square roots of each initial term of 64 a^8 - 16 b^4)

Now you have :
64 a^8 - 16 b^4 = (8 a^4)² - (4 b²)² = c²-d²

Using the formula I gave earlier, you can express that as :
c²-d² = (c+d)(c-d) = (8 a^4 + 4b²) * (8 a^4 - 4b²)

BUT you see that there are still even powers in the second factor, so we could go even further

Let's focus on that second factor and forget the first one (which we can't factor anymore)

We see that : 8 a^4 - 4b² = (sqrt(8) a²)² - (2b)²
So let's conveniently define e = sqrt(8) a² and f = 2b

Therefore : (sqrt(8) a²)² - (2b)² = e² - f² = (e+f)(e-f) = (sqrt(8) a² + 2b)(sqrt(8) a² - 2b)
Let's insert this in our previous result, and we finally get :

[ember kernel]

(At the end I just factored out constants, which I should've done in the first place but forgot to tbh)
It seems tedious at first, but it's actually pretty easy once you get the hang of it. With enough training, you won't even need to define intermediate parameters (c,d,e,f)

[marble crest]

Thank you!!

<3

[sterile birch]

can anybody help me understand how me teacher got from that step to the next?

[sharp ferry]

does this help

basically they made the denominator common

you can multiply like dat because basically youre just doing 1/1

[sterile birch]

yo thanks for the reply, and then after doing that they just multiplied the h to the top?

because its like 1/1/h so they then multiplied 1/h i think

Because that would be the reciprocal right

[sharp ferry]

nah nah that gets a confusing at times but think of it like this

now dividing by h is just 1/h

[sharp ferry]

1/1/h can get a bit ambiguous because it can mean (1/1)/h=1/h and it could also mean 1/(1/h)=h depending on where u place your brackets

so to avoid that error the thing that i personally do it while making calculations it to write it using the ÷ or using proper () or just separating stuff out early on so that i dont end up with some calculation error later on

[sterile birch]

oh ok makes sense

yea

i can see how that would become a problem

thank you man : )

[sharp ferry]

welcome 🤗

[tropic lava]

i mean this [photo] we can translate to this [photo2] , but i dont know how to translate this [photo3] to form with x ∈ ... like [photo2]

[ember kernel]

The equality seems kinda wrong to me though

[tropic lava]

but i just need to translate photo3 to something like photo2

this is my try, but I don't think I did it right

[vivid void]

hi I have a really simple question: how is 444...44 (100 digits) written in scientific notation?

[glacial dust]

its because the calc reads the numeircal value as thousands,millions (interms of the digits itself) so its raise to 10

can anyone correct me If its wrong ty

[left knoll]

Is step 3 legal in math? Is that a thing you can do?

[glacial dust]

U cant cross multiply there is an add operation

[left knoll]

i just did a proof, apparently this is legal

[solemn vessel]

@left knoll what are you asking?

is what legal?

[left knoll]

oh i was asking if it was possible to do, apparently it is

it just seems very quirky i guess

[solemn vessel]

What about it confuses you

I could explain it

are you confused about finding the common denominator?

[left knoll]

it just seemed odd to me that you can just multiply and put them in the same fraction with the squares on top. it makes sense now though

[tropic lava]

Is this inductive proof good?

[left knoll]

yeah bro

step 2 to 3 i believe thats fine

u just made them common denominator

and from 3 to 4 u applied trig equation

[ember kernel]

It's totally legal and not 'quirky' as you said. That's one of the many fundamental identities of maths that one must know

[ember kernel]

:|

[left knoll]

me rn

[left knoll]

<@&717391911132069929>

[left knoll]

Somebody please help

It would be really appreciated

[gleaming ridge]

need help with this

[karmic whale]

heeeeeeeeeeeeeellllllllllpppppppppppppppppppppppppp

[vivid void]

what do u have to do?

[cinder yacht]

use second line to deduce x = 3y-3
then replace, in the first line, "x" by "3y-3"
then solve for y in that line
then for each solution y, deduce a value of x with x = 3y-3

[cinder yacht]

replace sin(..) with sqrt(1-cos(..)^2)
that works cause sin(a) = sqrt(1-cos(a)^2) when a idan angle in [0,pi/2]

[fossil kite]

thats not elimination

[karmic whale]

But I'm only required to use the elimination method and not the substitution. I tried the elimination but it gives me a different value.

[fossil kite]

but yea, thats the only way u can solve it as per me

[fossil kite]

wt is ur answer though

it tried smthg related to elimination, not sure bout my ans though

[fossil kite]

yea no

my answer is nowhere close to it

[karmic whale]

I think I got the same answer with a different solution wherein I tried to square both sides of the equation 2.

But still the answer is not the same when u try to use the substitution

[waxen sequoia]

Do u still need help?

[cinder yacht]

ah idk these English terms

[fossil kite]

i did that only

[karmic whale]

same,,, I was trying to manipulate but Idk if the solution is acceptable as long as I've proved the same answer. Do you guys think my solution to this problem is right according the method used?

[weary arrow]

anyone have any good resources for commutators? We use them in physical chemistry but I haven't actually had the math behind them so i struggle to remember the rules and how they work

[limpid juniper]

can anyone help me with 3c. please?

[pearl flare]

Can someone help me with this question 😦
Given 𝑓(𝑥)=2𝑥3−7𝑥2+9𝑥−3, use the Remainder Theorem to find 𝑓(−13). Briefly explain the result and its relation to the given polynomial function and the 𝑓(−13).

[ember kernel]

sec is just 1/cos
Since you calculated sin in 3a, you can calculate cos via the trigonometric identity sin² + cos² = 1
(All angles in the trig functions are x-y btw)

[knotty heart]

I need too calculate low and high point from this function could someone help?

[crisp heron]

hii could someone please help me with a doubt the exercise says to write the equation of the parabola, with axis parallel to the y axis, which has the verx in V and passes through P. I was doing 204 and it didn't come to me, so I checked phoromath to see if I was doing the wrong calculations, but I realized that the vertex coordinate was 0, so I substituted it in the term c, when I did, I put it in the equation of the point through which the parabola passes and it gave me the result , however this does not make sense to me, why shouldn't I put it in the equacion of the axis of the vertex? (I tried it that way and it didn't work)
Why should I replace it in the equation of the passing point with the parabola and not in the vertex?

[ember kernel]

f(x) is local a minimum if : f'(x) = 0 and f''(x) > 0
f(x) is a local maximum if : f'(x) = 0 and f''(x) < 0

The derivative of x^b is b x^(b-1) (with b a certain constant , and it's called the power rule)

The derivative of a sum is the sum of the derivatives : (f+g)' = f' + g'

That's all the technical stuff you need (I'm sure you know how to solve an inequation)

[knotty heart]

Thanks!!

[left knoll]

Find the length of the Normal chord to the parabola y^(2)=4x, which subtends a right angle at the vertex.

(a) 2√3
(b) 7√3
(c) 6√2
(d) 6√3

could someone help?

[limpid juniper]

ty 🙂

[left knoll]

Yes

[solemn vessel]

Anyone here good at multivariable calculus

Ive got three questions I need help on

[sharp ferry]

i dont understand why my teacher stopped here at n+1 while integrating

[delicate fossil]

well since the areas between 0 vertical axis and** (n+1) vertical axis **are counted

||forgive me i dont study math in eng||

[sharp ferry]

OH

tysmmmm!!

[sharp ferry]

nah nah thanks a tonn im dummm mfdsj

[left knoll]

Look at this question

And look at part c here

Here

In the second line why is it -xy

Is this some sort of mistake? What do you guys think?

I’m so baffled ur adding probabilities why is there a -xy?

[ember kernel]

Because it's -P(A inter B)

Because A union B = A + B - A inter B

[left knoll]

Don’t u add A inter B

U add everything in the regions so why would u be subtracting the A inter B

[ember kernel]

You should check a video about that (the inclusion-exclusion principle).
It's considered known in this exercise

And it makes total sense "geometrically". You just need to watch a video to understand it

[tough peak]

Hi

[ember kernel]

Hi

[left knoll]

I can see it appearing in other questions, so is it just something ur meant to know?

[tough peak]

Haw are you?

[ember kernel]

Yup, you're supposed to know about it

[left knoll]

Ello there

[ember kernel]

Inclusion-exclusion principle

[left knoll]

Ahhhh right thank you

[ember kernel]

Np

[left knoll]

It’s been a good year since I’ve revised stats

I’ll be good in some time

[ember kernel]

That explains

It's a very easy-to-understand principle you'll see

[left knoll]

Yeah I knew it but I wasn’t using my eyes

So I was confusing myself

[left knoll]

Textbook is so messed

It covers inclusion exclusion principle couple chapters after they set questions on it

[steady anchor]

hii,anyone know how to do c ?

[weary arrow]

can anyone explain how they went from the 2nd to the 3rd line? Where did the third expression come from?

[left knoll]

wtf is this lmao

[left knoll]

accident

[slim juniper]

Does anyone know why they equate the function to x before finding the equilibrium

[ember kernel]

Needs context

[slim juniper]

The question asks to find inverses of the updating functions involved and discuss the stability of the original and inverse dynamical systems

[ember kernel]

Yeah can't help sorry

[slim juniper]

nw!

[signal nimbus]

Hi! I'm in grade 11 functions and the unit we're studying atm is trig functions. I've come across a problem such as the following. Does anyone know how to go about doing question 3ai?

Here's what I've tried - I've come up with an equation such as -10cos(pi/6 * x) + 15 and tried plugging in 7 for x, however that answer didn't line up with the correct answer which is 20m

[shadow kestrel]

The first step in the sample solution is not quite clear to me, it goes as follows:

Can anybody explain where does this come from, please?

[viral locust]

A quick multiplication of 2+sqrt3 on both sides gives 4-3 which is equal to 1

Just rearranged

[terse niche]

i think you should shift the function to the right by 15

[shadow kestrel]

Yes. But I'd like to know how to rearrange it to get the right side.

[cyan night]

woah

[ember kernel]

⬆️

[ripe bay]

If the message above contains a link to free nitro, it is most likely a phishing link. Please DON'T click on it to protect your privacy. You can notify the peeps by tagging them ( @peeps ) and they will take care of it as soon as they can.

[shadow kestrel]

Wow, thank you so much!! ^^ all is clear now

[magic flame]

Try 10cos(pi/12(x-15))+15

[cinder yacht]

in ur answer, when x=3, it's cos(pi/2), which isn't a minimum

-cos(0) is a minimum tho so we want what's inside the cos to equal 0 when x=3
so u can do :
15 - 10 * cos(pi/6 * (x-3))
aka
15-10 * cos( pi / 6 * x - pi / 2)
aka
15-10 sin ( pi/6 * x)

[hexed zealot]

What concepts do arithmetic sequences lead into?

[left knoll]

How do you do this

I have been trying for an 20 minutes

[delicate fossil]

it's the application of the important algebra formulas

[ripe bay]

If the message above contains a link to free nitro, it is most likely a phishing link. Please DON'T click on it to protect your privacy. You can notify the peeps by tagging them ( @peeps ) and they will take care of it as soon as they can. Please don't ping them if someone else already did in the chat thank you.

[mild valve]

how do you simplify?? im crying rn :'D someone pls explain step by step thank you :D

[ember kernel]

Thanks for trying to help. Unfortunately, that's not how maths work (at all) and the different steps you suggested are invalid and would merely lead to a wrong answer

@mild valve

[fading talon]

Alrighty

[steel berry]

yo does anyone know how valuable a usamo qual is

[harsh oracle]

hii can anyone maybe help me with my math? 😭

[steel berry]

what do u need help with

i may or may not be able to help u

[harsh oracle]

idk if this is right ^

[steel berry]

do u remember the definition of a polynomial?

[harsh oracle]

nope..

[steel berry]

it must have a positive integer exponent

every term

actually let me phrase it again

it must have non-negative integer exponents/powers

it can be 0

wait is this a test

[harsh oracle]

nope

it's an exercise

like a pre thing

[steel berry]

ok then i can help u if it was a test then im kinda fucked

but do u know the answer now

[harsh oracle]

i'm guessing 2?

[steel berry]

yes

its the only option that is a nonnegative integer

just rememeber that the variables and terms need to havae a nonnegative integer as its power in order to make it a polynomial

[harsh oracle]

then this one is the last one right?

[steel berry]

wait

[harsh oracle]

or the second one..

[steel berry]

its the 3rd one

[harsh oracle]

ohhh the one with 0?

[steel berry]

yes

so for the 4th one, u looked at its coefficient

[harsh oracle]

ah so it's NOT a polynomial function?

[steel berry]

yes

it isnt

0 is not a polynomial function

it must be =/= 0

[steel berry]

the coefficient just has to be a real number, and -1/2 is a real number so 4th is valid

[harsh oracle]

wait there's another question here i'm confused with cause it looks a bit messy

[steel berry]

uh ok

im trusting u that its not an exam

[harsh oracle]

i swear it's not 💀 😭

[steel berry]

come back to the definition of the polynomial

[harsh oracle]

we had our exam done and now they're preparing us for the next lesson

[steel berry]

it has to have whole number (nonnegative integer) powers

keep in mind that roots can be rewritten as fractional powers, for example sqrt x = x^1/2 or cbrt x = x^1/3 etc

[harsh oracle]

ohh so the answer the first one in the options?

[steel berry]

no

no

the second one

the first one has a negative power

[harsh oracle]

oh it's not ii?

[steel berry]

it is ii

ii only

[harsh oracle]

yeah i meant in the option thing below haha

wait how do i know the leading coefficient?

[left knoll]

He asked the answer. Its ii

[steel berry]

the leading coefficient is the coefficient of the leading term

where the leading term is the term with the highest power

[steel berry]

yess sorry my bad

[harsh oracle]

ah so it's 4 for here? just to make sure-

[steel berry]

yes

this question was trying to trick u because it didnt arrange the polynomial

[harsh oracle]

and what's a constant term?

[steel berry]

in descending order

[harsh oracle]

yeahh

[steel berry]

the constant term is the one with 0th degree

(without any variables)

wait that didnt sound right

[left knoll]

Yes. Its a basic algebra expression

[steel berry]

a constant term is just the term with no variables

[harsh oracle]

ah so it's for this it's 1 (the constant term)

or 3..

[steel berry]

yes

1

[harsh oracle]

oh okayy

[steel berry]

why 3?

[left knoll]

1

[harsh oracle]

so far i'm getting this lesson

oh wait oh no what's a standard form..

[steel berry]

standard form of a polynomial?

[harsh oracle]

yeah

[steel berry]

it just means that it is arranged in descending orders of power

for example x^2 + 2x^3 is not standard form

because 2x^3 has larger power it should be placed first

[steel berry]

for this u can do it two ways 1. u can expand it 2. if u know the algebraic rule for a^3-b^3 then u can go for that

fyi a^3 - b^3 = (a-b)(a^2+ab+b^2)

[harsh oracle]

hold on i'll try it out

[steel berry]

okok

[left knoll]

It easier with polynomial identities
The answer is 4

[harsh oracle]

ohh i thought it was the second onee

so it's 125x^3-27 ?

[steel berry]

yes

because 125x^3-27 = (5x)^3 - 3^3

= (5x-3)(25x^2+15x+9)

[harsh oracle]

OHHH

[left knoll]

Exponent of negative will receive negative

[steel berry]

this is the list of basic identities that i learnt this year lol

[harsh oracle]

wait so what would be the leading term for this number?

[steel berry]

uhh

just

[harsh oracle]

oh my gosh ill save this thanksss

[steel berry]

the leading term is just the one with the highest power

the leading coefficient is the coefficient of the leading term

[left knoll]

There will be alot in next year called exponent rules

[steel berry]

im in 8th

[harsh oracle]

so the leading term is 125x^3 right?

[steel berry]

correct

[left knoll]

Ik. But let you know to prepare then

[steel berry]

what exponent rules?

[harsh oracle]

and the constant term is -3..?

[steel berry]

check again

[left knoll]

The constant is -27

[steel berry]

^

[harsh oracle]

i was gonna say that too but they didn't but it in the options 💀

ill tell my teacher about it

[steel berry]

Ok

[left knoll]

A set of principles based on which problems based on exponents are solved. You can google it, not hard but very important to understand for calculus

[steel berry]

like just a^m*a^n = a^(m+n) stuff?

[left knoll]

Correct

[steel berry]

or log_b(xy) = log_b(x) + log_b(y)

[harsh oracle]

oh my gosh what in the world is this...

[steel berry]

this was from grade 6 tho

oh geometry

its congruent triangles

[harsh oracle]

how am i suppose to answer these..

[steel berry]

oh, use congruent triangles

have u learnt congruent triangles yet?

what grade u in btw

for 45 i dont know the property of arcs but it is just congruent

[left knoll]

Both. The exponent first then logarithm

[steel berry]

exponents were from 6th grade tho

in 9th grade there arent logs yet unfortunately

[harsh oracle]

nope, they said we have to take this exercise activity to be prepared for the next lesson and this is graded too 💀

[steel berry]

oh

uhh what have u learnt then

[harsh oracle]

is 44 maybe B?

[steel berry]

ya

44 is B

[harsh oracle]

ohh

[steel berry]

wait

B is literally bolded

[harsh oracle]

BHAHAHAHA

[steel berry]

so is 45 A

. . . xD

[left knoll]

I dont learn in 6th, instead i learn how to code pascal based on algebra. I learn exponent rules in 7th and logs in 9th

[steel berry]

okk

[harsh oracle]

i guess since this is just an exercise, they didn't really care whether it was a teacher's copy or not

[steel berry]

@left knoll have u learnt matrix transformations yet im currently struggling hard with it lol

[harsh oracle]

is this one B then too?

[steel berry]

idk

i havent learnt arcs yet

[harsh oracle]

do you maybe know @left knoll?

[steel berry]

give me the shape ill probably figure it out

the answer is C

[left knoll]

I did. But i havent touch for month

[harsh oracle]

there's a shape for this question but not the previous one

[steel berry]

mkay

[left knoll]

I dont study geo since college. Only focus in calculus and statistics rn sorry

[steel berry]

the answer for that q is C

the one above

but i havent studied arcs yet

[harsh oracle]

the one without a shape?

[steel berry]

yh

[harsh oracle]

thankss

[harsh oracle]

no it's fine no worriesss

[left knoll]

But i know document contain the matrix from TheGreatCourse maybe. I can send it if you want

[steel berry]

can u send me it

[harsh oracle]

what is the longest chord in a circle?

is it secant?

or diameter?

i'm kind of confused..

[steel berry]

a secant is essentially just a line that cuts the curves at two distinct points at minimum

the longest chord is the diameter

a circle have infinite secants

in a circle u can only cut 2 distinct points with a secant

[left knoll]

<@&717391911132069929>

[steel berry]

L

no

[atomic dagger]

guys if you have 1/a = 1/b + 1/c how do you isolate a... i know it's easy but for some reason i can't do it...

[steel berry]

1/a = c+b/bc => a(c+b) = bc => a = bc/b+c

[left knoll]

we 1st add 1/b and 1/c and then take the reciprocal of every term on both sides of = sign, giving a = bc/b+c

[glacial dust]

10t2 + 4mt + 2t

does someone know how to simplify this?

cant use quadratic because c is supposed to be constant but itshas a"t"

do I just take their common and am done I think thats the only viable way to factor it out

[steel berry]

u can factor 2t out

10t^2 + 4mt + 2t = 2t(5t + 2m + 1)

[glacial dust]

yah

thats what I did

[steel berry]

yeah

[glacial dust]

Where t is the time taken (in seconds) for machining the plumb bob and m is the cost of the
material.
Simplify the expression in its simplest form.
(d) If the overall cost of manufacturing one plumb bob is 12 and the cost of the material is
2, calculate the amount of time taken to manufacture one plumb bob

12=2t(5t+2m+1)

so if the m is 2 then 2 times 2 plus 1 (combining like terms)=5

so 12=2t(5t+5)

then,10t^2+10t-12 what now?

[atomic dagger]

thank you both! @left knoll

[steel berry]

np

[fossil kite]

shud be C, cuz its doubled (a theorem is there)

[fossil kite]

o = 0
p = 1
q = 3.5 = 35/10 or 7/2
s = -0.5 = -1/2

wts the division between 2 points? is it specified?

[left knoll]

yo

for c

why would u not take into account

doesnt have condition given test is negative

so why not do 0.96*0.98

[ember kernel]

Because we only want the probability of having the condition

[left knoll]

wdym

it just says they were tested for the condition

doesn't say or imply that only people with it are considered

or does it?

[torpid root]

I'm finding a = 1/2, but no matter what I do, I get negative reesults in the second part where I have to draw the function

nevermind I got it, the result is 1 as it should be

[ember kernel]

The probability of having a positive patient and a negative test

is not the same as

The probability of having a positive patient knowing that their test was negative

In the second case, we have an additional information (the negative test) that we knew happened (in the past) for sure. Whereas in the first case we can only speak in terms of probability that this situation (negative test) happens (in the future).

The two situations are different, and thus require different formulas/workings

In the first case, which is the answer you gave but with a wrongvalue, the probability of a positive person being negatively tested is 0.96 * 0.1

In the second case, which is what the question asks you for, the probability is the one in the solution (and it's just a general formula to apply)

[ember kernel]

Actually a = 1/2

Because when you integrate f(y) from -2/3 to 0, you should get 1. That's only the case for a = 1/2

[left knoll]

I see

Thank you

[ember kernel]

Np

[torpid root]

yes I found a without issue but I was having problem with drawing the function

I got it though

[hexed zealot]

What concepts do arithmetic sequences lead into?

[livid crescent]

Does anybody know task 7. Pls

[ember kernel]

Sequences are useless in themselves.
However, they lead to the concept of series, which is used in many fields (economy, complex calculus, numerical calculus, ...)

[cinder yacht]

wot

sequences ain't useless o_o

[ember kernel]

What else have you used them for ?

[cinder yacht]

the evolution of any variable

AI is full of sequences of numbers or matrices for example

numerical resolution of equations

any algorithm really, probably relies on some sequence

numerical resolution of differential equations (which is used in multiple fields) rely on sequences

anyway i dont have a lot of examples in mind but ik sequences are everywhere

especially in computers where nothing is actually continuous and everything is discrete

[ember kernel]

Oh we use series for most of these

[cinder yacht]

series as in the sum of the terms of a sequence?

[ember kernel]

Yup

Obviously we use sequences in the concept of series, but we rarely use the terms of a sequence alone

[cinder yacht]

series are sequences and sequences are series anyway np, but i think more often we don't sum terms rather than summing them, which seems intuitive to me as not every quantity deserves to be summed

I thought u said we care more abt the sum than the sequence itself in general, which I don't think is necessarily true. But for sure exercices on series are more varied and difficult than for sequences when we're studying calculus so we do more of them

like even just the notion of U(n+1) = f(U(n)) and analyzing how U(n) changes with n is sooo common

(U could be a matrix sequence)

[supple island]

ehm, what's the question here?

[left knoll]

I think its germany math but seems like he is on the test so i dont help him

[ember kernel]

Well the test is already graded so ...

[atomic dagger]

in order to prove that m vectors form a base of R^n, is it sufficient for me to say that they are linearly independant?

or do i need to prove that they are generators too?

[supple island]

well if m>n you need n vectors to be linearly indipendant, if m<n they can't form a base

if m=n you just need them to be independant, a system of generators is just a set of independant vectors that can form any vectors of a space

which is also a base

[waxen sequoia]

I'm german and that's definitely not german. 😅

[left knoll]

I mean when he asked i didnt answer because i saw he put the test paper under table so i dont help him

[ember kernel]

Yeah that's a good initiative. But on the picture you could see that the test had already been graded ^^

[atomic dagger]

thank you!

hi! could someone help me understand this, please ? we know that teta = omega*t. and omega(t) = alpha * t

my question is why do we obtain teta(t) by integrating from omega(t) ?

[supple island]

well that's because ω is the angular velocity, and θ is the angle, so ω= dθ/dt, similar to v and x

[atomic dagger]

Thanks !

[atomic dagger]

hi! i remember learning in high school a method to be able to know all the equivalent angles on a trigonometric circle. For example, 2pi/3 = 8pi/3, it's just that we have turned a few more times around the circle. Does anyone know what these equivalent angles are called and how I can find them more easily without doing the whole circle everytime ?

[ember kernel]

Well one turn around the circle = 2pi
And 8pi/3 - 2pi = 2pi/3

You just need to remove/add 2pi until you get to an angle that's contained between -pi and pi (or 0 and 2pi, it depends on your school's convention)

[supple island]

here X is the diagonal matrix with k at the extremes and 0 in between?

[atomic dagger]

thank you!

okay i'm sorry for asking so many questions but i'm very confused at the resolution of this problem:

what I originally did was to transform the right side using Euler formula which made me obtain this:

now according to de Moivre, we should have sqrt(2^8) * e^(i...)

[supple island]

now transform it into the sin, cos and proceed with the resolution of roots for complex numbers

[atomic dagger]

but in order to solve roots for complex numbers, shouldn't we be using the square root of the module?

[supple island]

the mosul remains in that form

[atomic dagger]

why ?

[supple island]

you get 2^8(cos(2pi/3)+isin(2pi/3))

2^8 is the mosule

*module

[atomic dagger]

I understand this, then how do i calculate the roots?

[cyan night]

helo

[supple island]

i just think it's easier this way but with euler formula you can just do z1,2=+-2^4e^(ipi/3)

[atomic dagger]

no i agree it's easier

[cyan night]

z^2 under polar form using de moivre's is z^2(cos(2theta) +isin(2theta))

then maybe u can compare modulus with modulus, argument with argument?

[supple island]

it's just that with the polar form you can represent the solutions on gauss plane

[atomic dagger]

sorry i'm a little confused with what you mean @cyan night

[atomic dagger]

this is what my teacher did, I don't know how she did it though

[cyan night]

u have both the lhs and the rhs under polar form

[atomic dagger]

lhs and rhs?

[cyan night]

left hand and right hand

[supple island]

yes, it's the same thing,

writing e^i(4pi/3) is the same as writing -e^i(pi/3)

[cyan night]

the right hand under polar form is 2^8(cos (2pi/3 +2kpi) + i sin (2pi/3 + 2kpi) i think

nvm

[atomic dagger]

but she didn't write -

[supple island]

well she didn't but she could

[atomic dagger]

and why do we go from 2^8 to 2^4 but the e^(i...) stays the same we just use an equivalent, i'm so confused... i'm so sorry everyone

[cyan night]

dont take my word plz im only 8th grade

listen to physstudent

[atomic dagger]

haha thank you don't worry about it

[supple island]

ok so

first of all

when we are talking about complex numbers we know that if we are looking for the roots of a number we will always find the same amount of solutions as the root we are looking for, so if we are looking for all the z of z^5, we will find 5 solutions, in the real plane we would have only found one

[atomic dagger]

okay!

[supple island]

to represent the complex number we use the gauss plane, to do that we represent the complex number in the polar form

we have 2 informations, the module and the angle, we imagine to draw a circle with radius equal to the module, then we trace a semirect from the origin with an angle in respect to the positive x axis equal to the angle

the intersection is a point on the circle with that angle

that is our complex number

[glacial dust]

Can anyone helop?

[supple island]

a cool thing about this is that if we look for all the Zs of Z^a (where a is a natural number), all the points which represent all the Zs will form a regular polygon, (this is actually a consequence and not where the reasoning should start, but for the purposes of this we will postulate this fact), so all the Zs are distant from each other by an angle of 2pi/a

[atomic dagger]

did you study Ohm's law?

[glacial dust]

I know a

I also know the equation

for b

but the thing is

[atomic dagger]

i didn't know that

[glacial dust]

L = ΦN/I.

[supple island]

it's very cool!

[glacial dust]

this is the formula

I dont have these ΦN

the only given was volt source resistance and freq

[supple island]

ok so first of all

what's ΦN?

it's important you understand it

[glacial dust]

the turns and the magnetic flux

[supple island]

ok, have you studied the Faraday-Neumann-Lenz law?

[glacial dust]

nope

[supple island]

i'm pretty sure you did or you couldn't do this exercise

it connects the elctromotive force and the magnetic fulx

[glacial dust]

the only topic we covered for this was the electromagnetic inductance

[supple island]

i'd say check your notes again or contact the professor, cuz if they gave you this you should be able to solve it

[glacial dust]

thanks for the help

[cyan night]

@atomic dagger have u figured out the complex number yet

essentially you would want to bring both sides under either both euler or both polar form

dont forget + 2pi*k

[raven urchin]

Current is charge over time, both are known, so it's easy.
The second part is also very simple; all you really need is impedance, which is voltage devided by current, but it's also sqrt((internal resistance)^2 + reactance^2). Reactance is 2pif*L. Just solve the equation and that's how you can find your L. :)

[glacial dust]

ohhh thank you been trying to figure this one out for 2 hours

wait which one is my L? for this 2pif*L

and also do I have to convert minutes to seconds for time?

[raven urchin]

L stands for inductance, that's what you need.

[raven urchin]

Of course

[glacial dust]

no I meant for reactance "Reactance is 2pif*L."

how will i get reactance if i dont have L ?

[raven urchin]

But you have U/I = sqrt(Ri^2 + (2pif*L)^2), where Ri = Internal resistance
You have to solve the equation to get your L. Ok?

Everything else is already known

[glacial dust]

oh so i just solve for L in this equation?

U/I = sqrt(Ri^2 + (2pif*L)^2)

[raven urchin]

Yes

[left knoll]

can u guys solve this

i got 12.6 however answer is different by 0.1

[left knoll]

yo lads

u guys r the ceos of maths

it will take 2 seconds

[supple island]

i have absolutely no idea of the context of this and what you are actually asking

[ember kernel]

Same sry

Ceo's can't work without info

[left knoll]

its asking for the median position and the median

so thats to do with linear interpolatio

n

i ccan show working they did hold on

[left knoll]

anything is possible in this life

[ember kernel]

What are Freq and C.F ?

I'm confused

[supple island]

even if freq. is frequency that still doesn't tell us anything

[left knoll]

cumulative frequency

top of line is the cumulative data

below it are the lower and upper bounds

and the quesion is asking for us to find the time at which the median is at

hope that makes sense

[glacial dust]

guys

just one quick question

is it always required when solving a problem to always change it to SI units

[kind sail]

well, mostly questions are based on SI

just make sure, all are in same system. that's it

[left knoll]

yo another question

my methods been fine so far ut idk whats going on

idk how to find x^2 mean

if u guys could help me it'd be very appreciated

[ember kernel]

Denote E[X] as the mean of X

The standard deviation is defined as sqrt(E[X²] - (E[X])² ) where X is the number of days absent

E[X²] = (0² * 12 + 1² * 20 + 2² * 10 + 3² * 7 + 4² * 5) / (12+20+10+7+5) = 203/54

E[X] = (0 * 12 + 1 * 20 + 2 * 10 + 3 * 7 + 4 * 5) / (12+20+10+7+5) = 81/54

Therefore the standard deviation is sqrt(E[X²] - (E[X])²) = sqrt(203/54 - (81/54)²)

If you struggle to understand, the mean of x² is just the mean of this table :

[left knoll]

i got it

calc function

[somber reef]

can anyone help the first one ?

[supple island]

you know that OAC is a right triangle, OC is equal to the radius and since OA=AB and OB=radius, OA=radius/2, in this triangle OC is the hypotenuse

now just use rules for right triangles and you can easily find theta

the proportion to trasform in radians is pi:180=x:theta

[somber reef]

ah okay i get it now,, thankss

[somber reef]

does anyone hve any idea how to do this ?

[ember kernel]

Hint to get started : use a certain trigonometry formula that involves the angle OAB and that could help you calculate the length of |AB|

Full solution (only reveal the titles, then try to do it yourself, then correct your work with the detailed steps)

1. ||Get |OB|||
   ||Using trigonometry :||
   ||sin(30°) = |OB| / |OA| which means |OA| = |OB| / sin(30°) ||
   ||And you know that |OB| = 15cm and sin(30°) = 1/2||

2. ||Get |AP|||
   || |AP| = Radius of the bigger circle = |AO| + smaller circle radius||
   ||Fortunately, you've just calculated |AO| and the smaller circle radius is given to you||

3. ||Get the arc length of PQ||
   ||Arc length PQ = Radius of bigger circle * Angle PAQ in radians||
   || Radius of bigger circle = |AP| and Angle PAQ in radians = 60° = pi/3||

[glacial dust]

whats the question asking for?

is it area of the shaded region?

[ember kernel]

The question is circled on the picture

Correct me if my calculations are wrong pls

[glacial dust]

I got 58.64 for the arc the way I did it was I formulated another bigger right triangle to get the side PB

IDK could be wrong I just did it for the sake of practice

[somber reef]

hurm the correct answer is 47.12

thank youu everyone

[glacial dust]

ye did it in a rush

[limber aspen]

heyo, need some help on this question
(translation: p>=2, q>=2 , p and q are relatively prime numbers. demonstrate that (X^p -1)(X^q-1) divides (X-1)(X^pq -1))

[cinder yacht]

I'll answer in FR :
X^p-1 = produit des (X-xn) où les xn sont les p racines p-emes de l'unité, càd xn = exp(i * 2pi * n/ p) avec n € {0,1,..,p-1}
X^q-1 = produit des (X-yk) où les yk sont les q racines q-emes de l'unité, càd yk = exp(i * 2pi * k / q) avec k€{0,...,q-1}
donc le produit (X^p-1)(X^q-1) donne p+q facteurs (X - truc) où les trucs sont les p racines p-emes et les q racines q-emes

en attendant, X^(pq) - 1 s'écrit comme produit de (X-zj) où les zj sont les pq racines pq-emes de l'unité, càd zj = exp(i * 2pi * j / (pq)) avec j €{0,...,pq-1}
en fait, dans ces pq facteurs, y a les facteurs où j=0q, j=1q, j=2q,... j=(p-1)q=pq-q. Càd où j = nq avec n des entiers dans {0,...,p-1}
donc les racines exp(i * 2pi * (nq) / (pq)) sont dans ce produit, càd les racines xn = exp(i * 2pi * n/p) pour n€{0,..,p-1}
même raisonnement : les facteurs (X-yk) pour k€{0,...,q-1} apparaissent dans le produit

note qu'on a pas encore utilisé "premiers entre eux"
le truc, c'est qu'on a prouvé que dans (X^pq-1) y a le produit des (X-xn) et y a le produit des (X-yk). Mais le problème est qu'on a ptet recompté des facteurs communs (genre par exemple, on aurait compté (X-x4) et (X-y7) séparément alors que c'est ptet le même facteur). donc faut voir si y a des xn qui sont égaux à certains yk. Avec de l'arithmétique,et en utilisant le fait que p et q sont premiers entre eux, tu montre que c'est le cas pour exactement un unique couplé de (xn, yk) : celui de (x0, y0). En effet, ces deux premières racines de l'unité sont égales à 1 (car exp(0)).
Mais là est l'avantage du fait que (X^(pq) - 1) est multipliée par (X-1) dans l'énoncé. Ça permet de compenser le fait qu'on a "recompté" (X-1)

On en déduit alors que dans la factorisation en polynômes de degrés 1 de (X-1)(X^pq-1), on trouve que ça s'écrit [produit des (X-xn)] * [Produit des (X-yk)] * [produit d'autres favteurs dont on se fout]
càd que ça s'écrit [X^p-1][X^q-1][autres facteurs]
ce qui répond à la question

t'as peut être accès à des théorèmes qui permettent d'aller plus vite, mais sans le savoir je propose ça comme solution

btw what do you study? university? prépa?

[late axle]

heyy need a help on this integral : 0 to 1, 1 / (1+x^3) dx

i got sth but the level of possible errors is too high lol

delta<0, canonical form, tan-1 ...

(im fr)

[cinder yacht]

tried writing 1/(1+x^3) = a/(1+x) + (bx+c) / (1-x+x^2)?

and finding a, b, c

[late axle]

ouais j'ai fait ça mais j'ai juste mis b

a / ... et b / ...

pq fallait mettre bx+c ?

[cinder yacht]

car 1-x+x^2 est de degre 2

[late axle]

et donc en décomposant en elmt simples on doit mettre un degré 2 au dessus du degré 2 ?

(je sais pas decomposer en éléments simples mdr)

[cinder yacht]

un degré 1 au dessus du degré 2 irreductible

si ma mémoire est bonne (ça fait 1.5 an j'ai pas fait ça)

et degré 0 au dessus du degré 1

[late axle]

au num on met un degré en dessous de ce qu'on a au denom?

[cinder yacht]

oui

[late axle]

ah je savais pas!!

merci

[cinder yacht]

en faisant en sorte que le denom est irreductible

cad pas factorisable dans R ici

[late axle]

ok ok parce que la dcp j'étais bien parti javais bien decomposé, forme cano, changement de variables et tt j'étais chaud

mais non

en tt cas merci!

[cinder yacht]

bah en fait si la décomposition a marché sans degré 2 elle est ptet correcte (en gros par chance, dans la decomp que j'ai proposé, ptet que b=0)

[late axle]

nn justement javais trouvé a=0 et b=1

[cinder yacht]

mais par intuition j'ai pas l'impression que ce soit le cas, t'es sur de pas t'être trompée ?

ah ouais c faux

[late axle]

ou par chance mon c = 0

ouais je me suis trompé carrément mdr

merci!

[cinder yacht]

jcrois t'en es pas encore la, mais des moyens rapides de vérifier la décomposition pour moi c'est pzr exemple de multiplier partout par x et de faire tendre x vers +infini (dans la decomp que j'ai proposée ça donne que 0 = a + b)

si t'as pas l'habitude oublie cette astuce j'imagine

[limber aspen]

on montre que seulement le couple (x0,y0) se repete en faisant la resolution de l'equation xn=yn, c'est ça?
je crois que je vois la demarche, merci beaucoup

[limber aspen]

prépa intégré, on étudie les polynome en algèbre 1 hehehe

[cinder yacht]

xn=yk

[limber aspen]

ah oui, c'est vrai

[cinder yacht]

cad "est-ce que y a un n/p égal à un k/q, avec n dans {0,...,p-1} et k dans {0,...,q-1}"
cas 1 : n=k=0 : ouais ça marche
cas 2 : au moins l'un de n et k est non nul. selon l'équation ça veut dire que les deux sont non nuls.
Raisonnement par l'absurde : supposons que l'égalité est vraie pour n et k non nuls.
l'equa de taleur donne nq = pk = c (avec c un entier)
nq est un multiple de q et pk est un multiple de p. Ils sont non nuls. donc ils sont supérieurs ou egaux au plus petit commun multiple de p et de q. autrement dit, c>=PPCM(p, q)
or p et q sont premiers entre eux donc ce plus petit commun multiple s'écrit : PPCM(p, q) = pq
donc c>= pq
or c = nq par exemple
donc nq>=pq
donc n>=p
or n est dans {0,..., p-1} : c'est absurde

y a ptet un chemin plus court, mais vla j'ai fait peu d'arithmétique dans ma scolarité xD

[cinder yacht]

ah okk, gl

moi je suis à l'x

[limber aspen]

peut être qu'il faut penser à l'ordre de multiplicité des racines

[limber aspen]

l'ecole polytech? c bien, bonne chance

[cinder yacht]

enfin avec des formules, toutes les phrases que j'ai écrites se font en ~6 lignes je crois

en partant de la factorisation de (X^pq-1) et montrant dans le produit où sont les racines p-emes et les racines q-emes de l'unité

[cinder yacht]

merci toi aussi

[left knoll]

bruh what langauge is this

ig french

[cinder yacht]

yeahh

[left knoll]

YOOOOO LADS

got something i need to check