help-17

[simple urchin]

same thing

same thing i presume?

[river minnow]

Here the triangle is also isosceles

Yes, same approach

[simple urchin]

yeah

94

easy!

now..

this-

we have

5 triangles in one circle

uhh

i don't know this one i can't lie

arc addition postulate

[river minnow]

The angles at the center should add up to 360

[simple urchin]

yeah

[river minnow]

At the same time they are all equal

Equal to x, in fact

So 5x = 360, x = ?

[simple urchin]

360 divided by 5

72

[river minnow]

Yes

[simple urchin]

oh

that's it?

[river minnow]

Yes

[simple urchin]

so in this question

i just

add up all arcs

5 arcs

and then

divide by 360 which is a measure of the whle circle

and that obivously equals to x

am i right?

[river minnow]

Divide 360 by the amount of arcs*

[simple urchin]

yup

not measure amount

alrighty

thank you so much i really appreciate you

have an awesome day man

[river minnow]

You too

[simple urchin]

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[vocal sleet]

@hasty bane Has your question been resolved?

[rugged orchid]

what have you tried

@hasty bane

[hasty bane]

I am not sure how to continue but I know that the probability density equal total 1

[paper depot]

yes so $\int_0^2 f(x) \dd{x} = 1$

[twin meteor]

Ann

[hasty bane]

Yup but idk how to integrate it

[paper depot]

$\int_0^2 f(x) \dd{x} = \int_0^1 f(x) \dd{x} + \int_1^2 f(x) \dd{x}$

[twin meteor]

Ann

[hasty bane]

I got that but not sure how to continue

[paper depot]

what's $\int_0^1 ae^x \dd{x}$?

[twin meteor]

Ann

[hasty bane]

Do we use product rule for…?

[paper depot]

integration by parts is unnecessary.

a is a constant

do you know how to integrate e^x?

[hasty bane]

e^x/2 I think

Still gotta work on integration topic

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[hasty bane]

.reopen

[vocal sleet]

✅

[solid sage]

Have you learnt differentiating yet

[hasty bane]

Yeah but not much with exponential

[solid sage]

e^x is probably the most well known integral/derivative

[hasty bane]

Would it still be the same? e^x=e^x

[solid sage]

The derivative of e^x is e^x and integral is e^x too

However of course integration requires you to sub in the values or add a constant depending on if it’s definite or indefinite

[hasty bane]

That where I got pretty stuck on because the answer solutions shows otherwise

Pls explain the integration part like howww,,,

[vocal sleet]

@hasty bane Has your question been resolved?

[boreal field]

Let's start with the first integral, shall we.

We need to evaluate the indefinite integral first, to do that we can use the property of integrals to take out the constant $a \$
We're left with $a \int e^x dx$

[twin meteor]

Graylen

[boreal field]

As annyeong said the integral of $e^x$ equals itself

[twin meteor]

Graylen

[boreal field]

So we need to evaluate $[a e^x]_0^1$

[twin meteor]

Graylen

[boreal field]

And we're left with $ae - a$

[twin meteor]

Graylen

[hasty bane]

Oh now I understand the -a but how did u get the ae?

Cuz wouldn't get a decimal number ins if sub in 1?

[boreal field]

$[F(x)]_a^b = F(b) - F(a) \$
$\rightarrow ae^1 - ae^0 = ae - a$

[twin meteor]

Graylen

[boreal field]

Every number raised to the power of 1 equals itself so e would remain unchanged

[hasty bane]

Oh it makes so much sense now thanks a lot! That was extremely helpful, I think I can do it now thank you

[boreal field]

Alright!

[hasty bane]

tysm

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[normal loom]

Hey

[normal loom]

how can i solve these type of equations

Cos (4x) +2 Cos (6x) = 0

general solutions

[waxen hawk]

That’s lit

What’s the correct answer

[normal loom]

idk

an online calucator is showing me pretty weird results

[waxen hawk]

Where the question from

[waxen hawk]

I think I got it

[normal loom]

how

[waxen hawk]

What’s your thought at the first sight

[normal loom]

only thing problematic here is the 2

if it was cos 6x + cos 4x then it would be easy

[waxen hawk]

Could you explain why it would be easy

[waxen hawk]

How will you solve that

[normal loom]

we can solve it by transformation formulae

Cos C + Cos D = 2 Cos (C +D)/2 *Cos (C-D)/2

[waxen hawk]

Ohh

[waxen hawk]

That’s the one that I never been able to memorize

However, what we trying to do is making those shitty different trig functions into one variable

[normal loom]

YES

[waxen hawk]

Could you tell me what is the intent to make the different trig functions into one variable

Is there any benefit for doing so

[normal loom]

what exactly are you asking

[waxen hawk]

That’s the key

The formulas in the upper-right is quite important

I’m not sure if I miscalculated sth but that’s the general solution

Transform it into one trig function then treat it as a variable

[normal loom]

i get it now

looks like i need to memorize the formulas again

i did not know Cos 3X formula

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[main ferry]

i get (3*sqrt(3))/(2+2sqrt(2))

but im using hexagon and octagon formulas and theres probably an easier method of doing it

ok so the area of the hexagon should be 6*sqrt 12

help im stuck on the octagon part

[somber portal]

I'd recommend starting by getting SP

[main ferry]

isnt that 8?

[main ferry]

if thats not 8 then i def need more help

yea its not 8 is it

[somber portal]

8cm, not 8. Also, you havent shown your work, so we dont know what you HAVE done

[main ferry]

its 8cm?

[main ferry]

i just thought since the edges are 4 and it has to equal the edges

[somber portal]

Having SP, you should also be able to get the side of the octagon

[main ferry]

using the triangles?

[somber portal]

that's one way

[main ferry]

i cant find the length of the base of the triangle

im not allowed to use a claculator either

[somber portal]

I'm not gonna solve it for you. But let's assume you have SP correctly calculated, and thus you can get SO trivially.
You know it's a regular octagon, so you also know that SO = TO; Since you also know it's a regular octagon, you know the angles OST and OTS

that should be enough information for you to get ST with no calculator

[main ferry]

yea i got that far

both side lengths are 4cm of the triangle

and STO should be 67.5 degrees

same with the other angle

[main ferry]

with no calculator though?

im not sure how to go forward but i have all the angles and 2 side length

[somber portal]

you can divide the triangle in two right triangles

[main ferry]

yes

but i dont have the height of the triangle

i have its angles and one side

[somber portal]

you can also use the triangle SOU as help

or remember the relation between the radius of an octagon and either the apothem or the side

[main ferry]

ok now im lost

i still cant find ST

ok so the radius is 4cm

yupp im completely lost

[somber portal]

SOU is a right triangle (on O), it's isosceles, and you know the two equal sides. With that, you should be able to get the height (a part of OT, let's say it intersects it in X, so OX) with Pithagoras theorem.
Since you know OT, you can get the other part of OT (TX). Since you know SU, you can also get SX. As you have SX and TX, you can get ST again with the Pythagorean theorem

[main ferry]

i dont know the length of SU though

[somber portal]

you can get it applying pythagoras to SOU

[vocal sleet]

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[vital wave]

How do you find the multiplicative inverse of a = 51 over p = 2^27 - 1 using Extended Euclid Algorithm?

[vocal sleet]

@vital wave Has your question been resolved?

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@vital wave Has your question been resolved?

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[fossil dawn]

does this solution make sense

[paper depot]

second paragraph from the bottom, typo: a^2 = 3(3b**^2**)

but otherwise yeah this seems fine

[fossil dawn]

i dont get the part where it says we obtained a contradiction

is a^2 equalling both 3(3b^2) and 3(3k + 1) a contradiction? can you help me understand why it is

[vocal sleet]

@fossil dawn Has your question been resolved?

[vocal sleet]

@fossil dawn Has your question been resolved?

[fossil dawn]

<@&286206848099549185>

[velvet gale]

wsp?

[fossil dawn]

help with discrete math question

[fossil dawn]

.

[severe gazelle]

hello, you still need help?

@fossil dawn

[fossil dawn]

yea @severe gazelle

[severe gazelle]

right

[severe gazelle]

so this is your issue right?

how it is a contradiction?

[fossil dawn]

yep

and the way the solution justified it

[severe gazelle]

when a^2=3(3b^2), then a^2 on division by 3 leaves a remainder of 0, right?

[fossil dawn]

yea

[severe gazelle]

it's also divisible by 9?

right?

[fossil dawn]

yep

[severe gazelle]

now, when a^2=3(3k+1), we can simplify this to a^2=9k+3

this on dividing by 9 leaves a remainder of?

[fossil dawn]

3?

[severe gazelle]

yes

[fossil dawn]

yoooo

[severe gazelle]

that is not possible, right?

[fossil dawn]

yea

[severe gazelle]

that's the contradiction

[fossil dawn]

damn

[severe gazelle]

the book meant to use 9, not 3

they messed up

[fossil dawn]

yea man i got confused by that

[severe gazelle]

😭 i can feel you

[fossil dawn]

cuz i was thinking they both have remainder 0 bruhh when divided by 3

thanks man

[severe gazelle]

sure

np

close the channel if you are done

[fossil dawn]

aight

.close

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[faint fjord]

what is the kernel of the 2x2 matrix?

you will see that it is exactly the image of (1, 1)

the kernel is
A*x = 0

(1 - 1) *(x) = 0
(0 0) (y)

1x - 1y = 0
then x = y

the span is the set* generated by the column vectors
in this case (1, 1)
which is the same as lambda*(1, 1) for all lambda in real numbers

hahaha thanks, it is from one piece

then I think you gotta read the theory

https://en.wikipedia.org/wiki/Linear_span

but geometrically, the span of a single vector, is the line that you get when you stretch that vector

it is the whole line,

and that is the same line that is the kernel of that 2x2 matrix

you are right though

yes

the span of the vector (1,1) are all solutions which A*x = 0

i think so yeah

not really the span of the augmented matrix

but you get the span by solving the matrix

sure

sorry, what is the matrix we're talking about?

you gotta put it in row echlon form

you can simplify even further but we can work from here

then
x - y = 0
-y = 0

the kernel of the matrix is the solution of this linear system

yes R1 = R1 - R2

we have two equations

we have to find x and y such that both of the equations are true

correct

yes

the ker(your matrix) is the zero vector

do you know about the rank nullity yet?

it is a little bit more advanced, but you can skip solutions if you know about this

https://brilliant.org/wiki/rank-nullity-theorem

the dim(image(A)) + dim(ker(A)) = dim(V)

in this case the dim of the image is the dimension generated by the column vectors of A

since we know that there are two linearly independent vectors

we know that the dimension of the space generated by them is 2

and obviously the dimension of R2 is 2

so we have that dim(ker(A)) is 0

we don't even have to solve the equation

you don't need to understand this completely

but this is the way to solve most of the things in linear algebra

by knowing the dimensions

when you get to eigenvalues and eigenvectors

there are geometric and algebraic multiplicity

which you solve by doing the same thing

all about dimensions

dim(Im(a)) + dim(ker(A)) = dim(V)
2 + dim(ker(A)) = 2
dim(Ker(A)) = 0

if you haven't seen this theorem you do not need to understand this

i can explain in another way

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@slate plume Has your question been resolved?

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[unique jolt]

this is my work, could someone check it please?

[vocal sleet]

@unique jolt Has your question been resolved?

[vast shale]

-7, not +7

$\frac{t^2-[(\sqrt{t^2+7})(\sqrt{t^2+7})]}{\sqrt{t^2+7}}$

[twin meteor]

dimpledoink

[vast shale]

$\frac{t^2-(t^2+7)}{\sqrt{t^2+7}}$

[twin meteor]

dimpledoink

[unique jolt]

where did the t^2 in the denominator go?

[vast shale]

that i didnt include i wanted to simplify the numerator first

[unique jolt]

oh ok

then couldnt you just do t^2 - t^2 = 0 and the 7 would be left?

[vast shale]

-7 would be left

open the brackets

it's t^2-t^2-7

[unique jolt]

ohhhh

so like you are supposed to kinda distribute the negative then calculate

[vast shale]

yes

[unique jolt]

ok thank you that makes sense

[vast shale]

the negative sign is on (t^2+7) as a whole, so it's just -t^2-7 in disguise

[unique jolt]

oh ok i thought you could take the parentesis out because its all the same operator\

ok tysm

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[hushed needle]

Hi

[hushed needle]

how i can find the value of X, X1 and Y

<@&286206848099549185>

[outer arrow]

3x +15 is 90 start from there

[hushed needle]

ehh

[gray crescent]

https://en.wikipedia.org/wiki/Cyclic_quadrilateral

[gray crescent]

Wait, where is that written?

[outer arrow]

Its a

[outer arrow]

Idk the English term but AB is diameter no?

[gray crescent]

Ah, right.

That's correct.

[outer arrow]

How do you call that anle

Angle

Thats 2x smaller than Center angle

[gray crescent]

No clue. xD

My geometry sucks.

I think it's this.

[outer arrow]

Peripheral angle

[hushed needle]

damn

[hushed needle]

is that the formule :3?

for search X x1 y

[gray crescent]

Yes.

[hushed needle]

omg

but

there are much formules

what is

[hushed needle]

arguments

how

[gray crescent]

Here.

[hushed needle]

o

[vocal sleet]

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[vocal sleet]

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[hushed needle]

.close

thanks @gray crescent :3

[hushed needle]

.close

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[sinful magnet]

should i let u = arcsin x

?

[split heart]

let x = sin y

[sinful magnet]

huh

[split heart]

Try it

[untold surge]

Yes

Oh same thing with the reply below

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[vast shale]

how can I solve this?

[vast shale]

let's take only "x^2" and log(x) = 3. what should I do with this information?

[glad python]

use log exponent rule

log(x^a)=alog(x)

[vast shale]

so in case "x^2", it will be "2 log(x) = 3"?

[glad python]

Yes

Wait

No

Why is it equal to 3

[vast shale]

should it be 6?

[glad python]

Yeah

[vast shale]

because he says, log(x) = 3

thank you

[maiden iron]

log(x²y³) - log(z) = 2log(x) + 3log(y) - log(z)

[twin meteor]

dldh06

[vast shale]

so now I have "2 log(x) = 3" how can I know what is x ?

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[vast shale]

would it be $6 = 2 (10^x)$

[karmic imp]

No, it's saying that log(x) = 3

[twin meteor]

Chocolate

[karmic imp]

Meaning that everywhere you see log(x), replace it with 3

[vast shale]

thank you

[vast shale]

that's an interesting message, do you think AI typed this or was it one of the mods?

[glad python]

anyone, do !nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[vast shale]

I get -36

$\frac{2(3)*3(2)}{-1}$

[twin meteor]

Chocolate

[vast shale]

plug them in

lol

and use log expansion

(logx^2 + logy^3)/logz

[vast shale]

I appreciate your help but don't type everything you feel, for example this laugh should not be said in the chat, only laugh in your room because it's not necessarily important thing to show, thank you sir

[vast shale]

dang

dude got mad

$3(2)+3(2)-1 = 11$

[twin meteor]

Chocolate

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

thank you guys

[vocal sleet]

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[glass galleon]

this is my question

and this is my answer

am i correct with everything here?

[bright yew]

looks correct

[glass galleon]

is this part easy to understand or should i change it

[bright yew]

you could write 5C1, 5C2, etc if you want

but it's clear enough

[ornate ember]

I mean that's the what it works out to be but if you want to be a bit more clean, maybe start out with the binomial expansion with the combinations expressed

[glass galleon]

theres an example in my lesson i kinda wanted to follow

they basically want it like this

[ornate ember]

that's fine then 🙂

personally, I think it's cleaner if you were to do something like:
$\(x-4)^5 = \binom{5}{0} x^5(-4)^0 + \binom{5}{1} x^4(-4)^1 + \binom{5}{2} x^3(-4)^2 + \binom{5}{3} x^2(-4)^3 + \binom{5}{4} x^1(-4)^4 + \binom{5}{5} x^0(-4)^5$

but to each their own, it's the same thing at the end of the day 🙂

[twin meteor]

MellowDramaLlama

[vocal sleet]

@glass galleon Has your question been resolved?

[glass galleon]

thank you so much for this i like this way better

[ornate ember]

np!

they both end up at the same answer, I just think using combinations makes it look more clean

best of luck!

[glass galleon]

it looks way better now

[ornate ember]

yep I agree 🙂

a lot easier to read imo

nice work!

[glass galleon]

thank u again!

.close

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[weak swan]

In regular falsi method How to know what value must be kept constant and what to change like in [a b] how to know whether f(a) or f(b) must be kept constant

[flat whale]

This?
https://byjus.com/maths/false-position-method/#:~:text=In simple words%2C the method,value according to the result.

[weak swan]

yeah this method there is [a b] value given we need to find f(a) and f(b) right and then we substitue it in the formula and find f(x) right
How do I know if f(a) or f(b) must be kept constant cause in few problems we must keep f(a) constant and other f(b) constant

[chrome raptor]

You have to choose the interval which contains the change of sign

(Bolzano's theorem)

So imagine you have (a, b) and you calculate the regula-falsi's c

If f(a)f(c)<0, then, the root is in (a,c) so you choose (a,c)

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[sudden fox]

so for the domain

can you also write it as

(-2, infinity)?

x>-2 and
(-2, infinity) are both accepted right?

[rugged vortex]

Both accepted

[chrome raptor]

(-2, ∞) is better

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[sudden fox]

gotcha

thanks guys!

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[timber badge]

hi

[boreal remnant]

Bye

[timber badge]

how this even possible?

what it means by a sector

on what basis did they consider sector AOP here

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[plain aurora]

Does anyone have any intuitive explanations of the binomial recurrence formula?

[plain aurora]

$\begin{pmatrix}n \ k \end{pmatrix}=\begin{pmatrix}n-1 \ k-1 \end{pmatrix} + \begin{pmatrix}n-1 \ k \end{pmatrix}$

[twin meteor]

Kalgar

[plain aurora]

A tangible explanation (i.e. picking items from boxes) is ideal, but an intuitive proof (if it exists) is good too (i.e., not induction or grinding out with algebra)

[paper depot]

you have a group of n people, among them 1 person named Bob. you want to find the number of teams of k people that could be chosen from this group.

on the one hand that is just nCk by definition of the binomial coefficient / choose function / whatever else you want to call it.

on the one hand, you have a choice: either include Bob (and then pick k-1 more people from the remaining n-1) or don't (and then pick all k people for your team from the remaining n-1)

[plain aurora]

Thanks

[paper depot]

btw you can write $\binom{n}{k}$

[twin meteor]

Ann

[plain aurora]

oh thx

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[keen ocean]

Can I get help making this idea more rigorous

[keen ocean]

For every sqaure number n^2, there is a difference of sqaures that is a semiprime.

Let k^2 be all sqaure numbers less then n^2
k^2 = (n - 1)^2 to 1^2

Let s be a sqaure free semiprime that is the product of p and q, where p, q > 2

For all n^2 - k^2 I propose that there has to be atleast one semiprime has a result of n^2 - k^2.

I propose all numbers from the sqrt(n^2) to (2*sqrt(n^2)) - 1, are one of the "simple factors" of all n^2 - k^2 for a given n. And the other "simple factors" for n^2 - k^2 are
Between 1 and sqrt(n^2)

To simplify:
let sqrt(n^2) be m

We get
m to 2m - 1

Batrand's postulate states that there has to be a prime p, between n and 2n - 2, where n > 3

Naturally there there has to be a primes p and q, between m/2 < q < m < p < 2m - 1

I propose that if there is always a prime between. m/2 < q < m, and m < q < 2m - 1

There has to be a number that is both
n^2 - k^2 and also the products of two prime numbers

Does this even make sense?

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[keen ocean]

Is this even close to being some sort of proof?

<@&286206848099549185>

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@keen ocean Has your question been resolved?

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@keen ocean Has your question been resolved?

[livid tapir]

What are you tryna prove exactly

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[glossy mango]

can someone please help me with this rollercoaster project bro

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[boreal remnant]

how do i figure this out programmatically

the naive approach is too damn slow

it takes several minutes to even get onto the order of 10**-8

[peak matrix]

cant you use something like lagrange error bound?

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[cunning saddle]

what is c?

[cunning saddle]

my work

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[vast shale]

(I) Train A is 1 km long and traveling at 50 m/s. Train B is 0.5 km long and starts just as the rear of train A passes the front of train B. Train B has an acceleration of
3 m/s2 and a maximum speed of 60 m/s. (a) At what instant does B overtake A, i.e. at what instant does the rear of B overtake the front of A? (b) How far did train A travel during this time?

<@&286206848099549185>

[vast shale]

relative velocity concept

my brain is glitching i'm sorry

what i have understoof

d

is when the back of train b meets the front of train a

train a accelerate of 3m/s

the time it took for train b to meet front of train a by my calculations is 30 sec

then after id ont know to do

by applying eq 1 v=u+at

60=0+3*t

t=20

why

let me solve it for you

train a doesnt automaticly start

train b has to do 1,5km

then a starts

since the velocity 50m/s is constant

1500m/50

30 seconds

[vast shale]

https://www.youtube.com/watch?v=e5xO6rRu47s

what the hell

same exact problem

thank you

[vast shale]

you can solve this question by relative velocity also

and the apply the equation of motion

you'll get the same answer

ur saving me

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[undone cedar]

how do i solve this equation

[glossy maple]

What have you tried?

[undone cedar]

atm nothing, just looked at this exercise and nothing come to my mind

do i solve just the first one and compare?

[glossy maple]

Well if you expand this right now, you will get a very messy equation.

Try first getting the determinant (on the left) to something that's easier to work with)

[undone cedar]

i solved it and got x^3-x^2

[glossy maple]

The left side?

Okay

[undone cedar]

yeah

[glossy maple]

Right, my bad, that works

now simplify the right side

Are you sure that's right though

Let me check

[undone cedar]

i can show you but i think i didnt do any mistake

[glossy maple]

Ok yeah I got that too

[undone cedar]

right side 4x-4 right?

[glossy maple]

Yes

So you have x^3 - x^2 = 4x - 4

Do you see how to continue?

You don't even need the cubic formula for this

[undone cedar]

i think so

i got 3 solutions

x=1, x=2 and x=-2

[glossy maple]

yes correct

good job

[undone cedar]

thanks for the help

[glossy maple]

np

have a good day

[undone cedar]

u2

.close

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[hushed spindle]

What formula should I be using for this problem? Its geometry btw

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[north mist]

i think i figured it out

I believe your answer to be 41

@hushed spindle is that your answer?

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[echo smelt]

someone confirm my answer

2 step equation

[amber vault]

can confirm that that’s the right answer

[echo smelt]

so correct?

[gritty sage]

Yes, it's correct. If you want to confirm it (this will help on tests because if you finish early, you can check your answers), you can

Write the equation:

2x + 3 = 7

Fill in x with what you got:

2(2) + 3 = 7

Then do the math and make sure both sides come to the same number.

[echo smelt]

ah nvm

whyd i make the channel

ty

[gritty sage]

For this one, you get:

2(2) + 3 = 7
4 + 3 = 7
7 = 7

You're welcome.

[echo smelt]

https://tenor.com/view/dog-dog-fling-flung-dog-dog-flung-fling-dog-gif-24868321

.close

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[grizzled musk]

,rotate

[grizzled musk]

i found that $(x^2+y^2)(x^3+y^3)$ can be re-written as $(1-2xy)(x-y)^2+xy$

[twin meteor]

REUBENOOB

[grizzled musk]

$(x^3+y^3)$ = $(x+y)(x^2-xy+y^2)$ = $(1)(x^2-2xy+y^2)+xy$ = $(x-y)^2+xy$

[twin meteor]

REUBENOOB

[grizzled musk]

$(x^2+y^2)$ = $((x+y)^2-2xy)$ = $1-2xy$

[twin meteor]

REUBENOOB

[grizzled musk]

i don't know how to proceed from here

<@&286206848099549185>

[split heart]

Can it be expressed in terms of x and y?

Or does it have to be a number?

[grizzled musk]

number

like the value of x^2+y^2 has to be a number

is there a way to manipulate the equation such that we an find xy?

[gaunt sparrow]

Notice that $$(x^2+y^2)(x^3+y^3) = (x^2+y^2)(x+y)(x^2-xy+y^2) = (x^2 + y^2) (x^2-xy+y^2) = 12$$

[grizzled musk]

why is it 12

@gaunt sparrow

[gaunt sparrow]

It's not wait just fixing something

Ok

Got it

[twin meteor]

Azyrashacorki

[gaunt sparrow]

Oh well

Too long

Anyways, it shouldn't be too hard to see that $$x^2-xy+y^2 =\frac{12}{x^2+y^2}$$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

We good?

[grizzled musk]

i-

[gaunt sparrow]

Ok wait

From second equation

$$(x^2+y^2)(x^3+y^3) = (x^2+y^2)(x+y)(x^2-xy+y^2) = 12$$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

Fair enough?

[grizzled musk]

yeah that's what I got

[gaunt sparrow]

Ok

[grizzled musk]

oh

[gaunt sparrow]

Now x+y = 1

[grizzled musk]

i see it

[gaunt sparrow]

So you get how I got to the previous expression then?

[grizzled musk]

yeah

[gaunt sparrow]

Ok

Now using the first equation again

x+y = 1

So (x+y)^2 = 1

So x^2 + 2xy + y^2 = 1

Now add this $$x^2-xy+y^2 =\frac{12}{x^2+y^2}$$

twice

[twin meteor]

Azyrashacorki

[gaunt sparrow]

to it

So we get

$$3x^2 + 3y^2 = 1 + \frac{24}{x^2+y^2}$$

[twin meteor]

Azyrashacorki

[grizzled musk]

how

[gaunt sparrow]

You just add the equations

We have

x^2 +2xy + y^2 = 1

Now we add twice x^2 -xy + y^2 = 12/(x^2+y^2)

To the first

See how it cancels out the xy terms?

[grizzled musk]

why adding twice

[grizzled musk]

there's only 1 xy term here

[gaunt sparrow]

Right

If you have two equations

You can add them together

without changing the truth value

Especially if they are meant to have the same solutions

Which is the case

this is a system of equations so you can add the equations together

[gaunt sparrow]

That's why you add twice

Right you have x^2 + 2xy + y^2 = 1

Now add x^2 - xy + y^2 = 12/(x^2+y^2)

So we get

x^2 +2xy +y^2 + (x^2 - xy + y^2) = 1 + 12/(x^2+y^2)

And then that becomes 2x^2 + xy + 2y^2 = 1 + 12/(x^2+y^2)

[grizzled musk]

ok

[gaunt sparrow]

Now we do it again

[grizzled musk]

so

[gaunt sparrow]

So it becomes 3x^2 + 3y^2 = 1 + 24/(x^2+y^2)

[grizzled musk]

ok

[gaunt sparrow]

Then you sub u = x^2+y^2 and realize this is just a quadratic equation

[grizzled musk]

what's sub u

[gaunt sparrow]

Replace with

This last equation can effectively be written all in terms of x^2 + y^2

[grizzled musk]

i don't see how

[grizzled musk]

oh u meant substitution

[gaunt sparrow]

Yes

[grizzled musk]

so 3u^2 - u - 24 = 0

[gaunt sparrow]

Yes

[grizzled musk]

im not that familiar with the quadratic formula but i'll try subbing in the values

[gaunt sparrow]

Be sure to keep only positive solutions since x^2 + y^2 >= 0

[grizzled musk]

how do you know >=0

[gaunt sparrow]

Well x^2 is definitely positive

So is y^2

[grizzled musk]

ohh i see

im confused with the quadratic formula

i got

1 +/- sqrt(1- 4(3)(-24)

then divide everything by 6

@gaunt sparrow

[gaunt sparrow]

yes

[grizzled musk]

OHH

i thought the sqrt was going to be negative

but there's -24

[gaunt sparrow]

Yes that's why

[grizzled musk]

answer is 3

[gaunt sparrow]

Yes

[grizzled musk]

so quadratic formula can be used for any equation in the form of ax^2+bx +c = 0?

.close

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[gaunt sparrow]

Yes

[grizzled musk]

alr

tysm

.close

.reopen

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✅

[grizzled musk]

.close

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[whole saffron]

How do I decide on what I have to substitute when doing integrals ? I wanted to integrate the function

tan(squrt (x)) / sqrt(x)

I Split up the tan in sin/cos and I decided the use the sin as u substitution. Didn’t end up with a solution really. Then I tried it with cos as the u substitution and it worked.

How do I decide on what I have to substitute ? I know that I have to substitute something whose Derivative is in the integral. That’s why I chose sin.

[split heart]

To be honest, this is one of those things you learn from experience

You must recognize that if you do a sub with u = sqrt(x) , du = 1/2sqrt(x) dx, which is a part of your integral

so you end up just integrating 2tan(u) du

There are, of course, certain rules that come up commonly, like trig substitutions

[whole saffron]

Okay, I see. I just started learning substitution so I think I might just need more experience. Thank you

[split heart]

Yeah. This is why a good chunk of calc is just doing these "boring" integral tricks. It's there to expose you

Incidentally, you need to do another substitution to integrate 2tan (u) du. You seem to already have a good idea on how to do that

[whole saffron]

Ya. Thanks 🙏🏼:)

[split heart]

messed up my constants there. I edited my answers

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[finite schooner]

need help

[waxen hawk]

What does the “H” stands for

[waxen hawk]

This channel is occupied by someone already.

[waxen hawk]

We know that C is the abbreviation of combination

I wonder what the H is abbreviated to

[karmic imp]

Is there more context? Like where is that from?

[waxen hawk]

I can give you a example, which I think is the origin of H

[karmic imp]

That's not what I was asking for

[waxen hawk]

This is from my textbook

[karmic imp]

I'm asking if it came from a textbook or video to see what the context is used

[waxen hawk]

@karmic imp is that you

Wish I had a smaller cap

[karmic imp]

It does not help if you start off topic conversations. I am asking you to post more context than just that equation

[waxen hawk]

The H is being used in questions like this

How many sets of(x,y,z) there are?

[karmic imp]

I see no H in there

[waxen hawk]

I’m explaining

So here the vertical bar would do sth like “classifying” those small balls

And those small balls stands for “quantity” quantity of x,y,z

For example

That means (x,y,z)=(3,7,0)

By the similar logic to combination we will be able to get the sets of it

That’s the H origin from

[karmic imp]

I'm still not understanding so I'll let someone more competent answer you

[waxen hawk]

But it is more convenient for you to answer my question

[waxen hawk]

As you just standing beside of me

Standing right upon my table

<@&286206848099549185>

[vast shale]

Grammer is English

Ohh

[waxen hawk]

Wdym, grammar is grammar

[digital kraken]

the H stands for repetition, it means that from the "n" set of items, you're taking "m" items allowing them to repeat

[vocal sleet]

@waxen hawk Has your question been resolved?

[digital kraken]

so, to make an example, let's take this, we could think of this as choosing 10 letters, allowing repetitions, so here, you're taking 3 x's and 7 y's, to make it [x, x, x, y, y, y, y, y, y, y], this is equal to saying that x=3 and y=7

or taking 5 x's, 3 y's and 1 z, basically we're asking how many sets can we do that satify the condition allowing repetitions

[dark kiln]

that's not the question

[digital kraken]

the question was what does the "H" stands for, right? I thought giving an example of what it is would help to have it more clear, sorry for that

[dark kiln]

it's a good addition, someone will learn from it, just not what they asked

they just wonder what word starts with H

[vocal sleet]

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[vocal sleet]

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[mossy mantle]

i need help with finding the integral

[mossy mantle]

I understand it up to the power rule

[karmic imp]

You mean this?

[mossy mantle]

Wait so you don’t plug in u

You keep it as u and do the power rule?

[karmic imp]

Yep

[mossy mantle]

Oh

Thank you

.close

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[vast shale]

how

idk how to start

[mild flower]

with these sorts of problems

where they don't tell you details that would seem kind of useful

[mild flower]

it's clearly not important the exact direction the path takes so you can just assume whatever is most convenient

[vast shale]

oh

[vocal sleet]

@vast shale Has your question been resolved?

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[vale wave]

how would this work

[split heart]

Can you factor those numbers so that you get some perfect squares?

[vale wave]

the bottom one into 5

[split heart]

Ok, so far so good

[vale wave]

and the top

listen its 2am

gimme a sec

[split heart]

Not rushing you

[vale wave]

wait im stupid

64

ok

then 8 racical 2

radical

over 5?

[split heart]

$\frac{8\sqrt{2}}{5}$?

[twin meteor]

TooManyCooks

[vale wave]

yeah

[split heart]

Then yes

[vale wave]

ok brb

how about this

@split heart

[split heart]

Go try it first

[vale wave]

where do i start

well that 9

3

uh

what i do with x^20

[split heart]

What the square root really does is divide the exponent by two

[vale wave]

so its 10?

[split heart]

YES!

[vale wave]

im a smart guy

i divided!

uh

this is my channel

anyway

how do i make the bot show a pictue

nvm its fine

but

so the p is 2?

[split heart]

Yes

[vale wave]

3 square root 3x^10p^2?

[split heart]

No, sqrt 9 is just 3. Not 3 sqrt 3

[vale wave]

radical*

[split heart]

Still no

[vale wave]

uh

radicand

[split heart]

Ok let's take a step back

What's your final answer

[vale wave]

thijs but idk the term for the square root symbol

final answer for wha

[split heart]

I just read it as it is "square root of 9"

But as I was saying, $\sqrt 9 \neq 3 \sqrt{3}$

[twin meteor]

TooManyCooks

[vale wave]

wait

why not

[split heart]

Okay. Factor 9 into a bunch of 3

[vale wave]

3 3's

[split heart]

3 * 3 * 3?

[vale wave]

what..

3 + 3 + 3

3 * 3

oh wait

2 3's

sorry

[split heart]

So does that change your answer?

[vale wave]

this is what i have

is it right

[vale wave]

idk does it

[split heart]

Your answer suggests 9 = 3^3 which is not true

Also, after applying the sqrt operator, it vanishes

[vale wave]

my teacher sais that uh]\

[split heart]

$\sqrt 9 = 3$

[twin meteor]

TooManyCooks

[vale wave]

9 goes into 3 and 3 cuz 3 * 3

oh right

[split heart]

Just do the same thing you did earlier

You took the sqrt of 25, and you had 5

[vale wave]

$3 x^10 p^2?$

aww

[split heart]

Might wanna remove the space before 3

[twin meteor]

Kingboo

[vale wave]

LOL

[split heart]

x^{10}

But yes

[vale wave]

you get the point

ok i got it right

ty lol

mr cook

[split heart]

You're good with this already. I know you can do it

[vale wave]

you said they get divided by two

but theyre odd

[split heart]

Aha!

So you need to write it to the nearest even power

What's that number

[vale wave]

10 and 20

[split heart]

hint: -1

[vale wave]

why -1

[split heart]

I meant 19-1

[vale wave]

what abt the 9

[split heart]

You want to rewrite that n^19 to n^(18+1)

[vale wave]

mhmmmm

same with the 9 too?

[split heart]

Yes! Exactly!

[vale wave]

so m^(9+1)

ok

[split heart]

No, no

[vale wave]

so we gotta simply the 45

hm

[split heart]

It has to add up to the original number

19 = 18 + 1

[vale wave]

so 10-1

oh add

8+1

[split heart]

You could do that too, but do you really want to deal with negative exponents?

[vale wave]

nope

[split heart]

Good. Let's stick to addition then

[vale wave]

mhm

so

how do you say sqaure root with the $

sqrt?

[split heart]

"square root" or just "root"

Oh you mean in latex

I thought you meant verbally

[vale wave]

ohhhh

[split heart]

it's \sqrt

[vale wave]

ok

[split heart]

$\sqrt 9$

[twin meteor]

TooManyCooks

[vale wave]

$3\sqrt {5m^8m n^{18}n}$

..

[split heart]

backslash not forward

also, if you want to put something in groups, you need to use curly braces

[vale wave]

oh

[split heart]

so like this $\sqrt{45 m^9 n^{19}}$

[twin meteor]

TooManyCooks

[vale wave]

got it

[split heart]

^ only looks at the first thing it sees, so you need to put {} if you want more than 1 digit

[vale wave]

got it

[split heart]

Not quite

[vale wave]

OHHH

[split heart]

What happened to m^9 a and n^19?

[vale wave]

the extra variables go to the 3

[vale wave]

though we went to 8 and 18 for dividing

[split heart]

Yes, but you can't just arbitrarily remove one factor

[vale wave]

can i not add it back later

?

[split heart]

$\sqrt{m^{9}} = \sqrt{m^{8+1}}$

[twin meteor]

TooManyCooks

[vale wave]

yes

i got that

[split heart]

Your answer up there was missing a factor of m and n

[vale wave]

yeah

now how do we divide the things

[split heart]

$\sqrt{m^{8+1}} = \sqrt{m^8 m}$

[twin meteor]

TooManyCooks

[split heart]

Does that help?

[vale wave]

theres a space

oh kidna

kinda

oops

[split heart]

So your answer earlier looks something like $\sqrt{ 3^2 5 m^8 m n^{18} n}$

[twin meteor]

Kingboo

TooManyCooks

[vale wave]

yes

pretty much the same i think

[split heart]

Anything with an even power gets to go outside

but with their power divided by 2

[vale wave]

m4 and n9

[split heart]

Exactly!

But there are still those that got stuck inside

don't forget about them

[vale wave]

yes