help-49

[raven dome]

the main prob is coming when im applying the limit

[glass dome]

i don't see why you need a limit here

but if you're talking about the limit definition of the integral

then you shouldn't integrate and then do another limit integral

you should just do the first integral with limits

otherwise it's the same as integrating twice

but really it's pretty depressing if they're making you do this completely straightforward integral with limits

[raven dome]

i have to prove it equal to 3- (1/2 ) ln2
and i am trying the same question for over an hpur and 15 mins now and i still dont get the answer right

[raven dome]

after applying the limit it is all messed up and the answer isnt equal to 3- (1/2 ) ln2

[glass dome]

there is no limit needed here

look

$\int_0^3 \frac{x^2 + 3x + 3}{(x+1)(x+3)} dx $\
$= \int_0^3 1 + \frac{1}{2(x+1)} - \frac{3}{2(x+3)} dx $\
$= \left[ x + \frac{1}{2} \ln(x+1) - \frac{3}{2} \ln(x+3) \right]_0^3$ \
$= \left(3 + \frac{1}{2} \ln(4) -\frac{3}{2} \ln(6)\right) - \left(0 + \frac{1}{2} \ln(1) -\frac{3}{2} \ln(3)\right)$ \
$= 3 + \ln(2) - \frac{3}{2} (\ln(3) + \ln(2)) + \frac{3}{2} \ln(3)$\
$= 3 - \frac{1}{2} \ln(2)$

[grand pond]

Kaisheng21

[glass dome]

that's it

[raven dome]

OMGGG THANK U

[glass dome]

the only place a limit could be used is to go from 1/(x+1) to ln(x+1)

[raven dome]

now i got it

[glass dome]

and 1/(x+3) to ln(x+3)

which is just more steps

[raven dome]

thanks bud< 3

u saved my day

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[flat spire]

I’m not sure how to split this up. I tried Ax + B split but I don’t think that works out with the denominator

[fallow scarab]

did you try factoring the denominator first

[flat spire]

well yea that’s what you do when you split it isn’t it. I got x^2(2x-1)

[last slate]

yeah do partial fraction decomposition now

[flat spire]

Ax/(x^2) + B/(2x-1) ?

[last slate]

no

[flat spire]

What should the numerator be?

[last slate]

x^2 is a repeated root

[flat spire]

oh

I didn’t realise that

[last slate]

is that what they call them i forgot

[flat spire]

wait so

Will the numerator just be like

A1 + A2 + A3?

[last slate]

[tribal temple]

hint: the denominator has degree 2

[last slate]

where did u pull that from

[flat spire]

My book

[flat spire]

Hm

I don’t know what to put on the top

[last slate]

yeah

you got that part down

[tribal temple]

If you meant that A1, A2 and A3 were the respective numerators of those, then fine

[last slate]

yeah ^

[flat spire]

so I am right?

[last slate]

saying A_1 + A_2 + A_3 is just vague though so dont put it like that

[flat spire]

so my book didn’t lie to me

[last slate]

i could have assumed you meant [
\4{A_1+A_2+A_3}x
]
for instance

[last slate]

yeah

so uh

do the thing now

[flat spire]

is there a reason we don’t name them A + B + C instead of using subscripts for different A’s

[last slate]

[last slate]

i mean u run out of letters eventually

numbers go forever luckily

[flat spire]

I have the whole alphabet tho

I doubt there will ever be a partial fractions integral that uses all of the alphabet

[last slate]

subscripts are nice if you wanna generalise like a procedure

[flat spire]

I c

ok I’ll do the partial thing now

When I find the denominator for A_3, do I combine it like A3(x)(x^2)

or can I rewrite that as x^3

[tribal temple]

You shouldn't...

[flat spire]

shouldn’t what?

[tribal temple]

Combine it like this

[flat spire]

hm okay

I didn’t know if that was the correct way

[tribal temple]

Think about what the lowest common denominator is here

[flat spire]

There really doesn’t seem to be any good picks for x values

They all result in a 0 if I try to make one of them 0

[tribal temple]

how did you try and get a common denominator?

[flat spire]

Maybe x=1/2 is the only good pick here

yes

[tribal temple]

x = 0 should work if you common denominatored fine

[flat spire]

But it won’t because look

I’m pretty sure I didn’t denominator it wrong

I hope

Anyways if x = 1/2 I get C= 8

[tribal temple]

You did

[flat spire]

where?

[tribal temple]

What did you multiply everything by?

[flat spire]

I just got everything with a common denominator

by multiplying each thing with what it needs

[tribal temple]

no passing go, no $200 for you

[flat spire]

I swear I did that for the last partial fractions question and it worked fine

[tribal temple]

You multiplied some things with an extra x that makes it more "heavy" than it needed to be

Remember that you want the denominator of each to be x^2(2x - 1) from your factorisation from before

[flat spire]

Ohhhhh

Wait

I thought I did that for A

[tribal temple]

You only need one x, not x^2, for it

[flat spire]

Why’s that?

Because I already have an x?

In the denominator

right

[tribal temple]

Yep, try multiplying both top and bottom by x(2x -1) and see the denominator you get!

[flat spire]

oh okay

I c now

I gotta get to class but I’ll see if I can solve it on my own

thank u

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[nimble matrix]

It's a question from physics about 2 block spring system.

If M1 is released, is there a chance that m2 will jump?

M2 is placed on ground

[twilit field]

You'd probably want to find the maximum force in teh spring due to teh compression to determine the same

[nimble matrix]

Yeah

It's at natural length when M1 is released

[twilit field]

hmm, so M1g=kx

[nimble matrix]

Maximum compression should be x=2mg/k

[twilit field]

2mg/k?

[nimble matrix]

Because the energy gained

Further compress

[nimble matrix]

Yes at equilibrium point

That will be the equation

[nimble matrix]

Maximum compression, though I'm not sure about maximum force

[twilit field]

this is SHM , isn't it, maybe you can use that to your advantage

[nimble matrix]

yeah

[twilit field]

as a=-w^2(x) If I remember right

[nimble matrix]

It's original question was about finding time period which I know

[nimble matrix]

Yes

Acceleration would be maximum at maximum compression

[twilit field]

yup

thus so would force

[nimble matrix]

The block m2 might jump

[twilit field]

depends on the values of $m_1$ and $m_2$

[grand pond]

Why am. I here

[nimble matrix]

So do we have to make cases where it may or may not jump

Yeah

My teacher assumed that m2 was fixed but from figure that may not be the case

[twilit field]

Yup, you'll get an inequality

[nimble matrix]

Yeah

I understand

Time period of both blocks M1 and M2 will come out different

but if m2 is fixed then T=2pi*√(M1/k)

[twilit field]

yup

[nimble matrix]

Thank you!

Ill close then

.close

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[keen wadi]

@muted panther Oui, mais en fait je pense que je ne comprends pas tout à fait la question. Il faut trouver a,b,c et d c'est bien ça ?

[muted panther]

Tu dois trouver les coordonnées du vecteur (a,b,c,d) en fonction de a, b, c et d

dans la base donnée

[keen wadi]

d'accord donc en fait je dois trouver les coefficients devant chaque vecteur de la base

en fonction de a,b,c et d

[muted panther]

C'est ça

Tu cherches en gros les constantes $K_1, K_2, K_3, K_4$ tel que $\sum_{i=1}^4 K_i u_i = (a,b,c,d)$

[grand pond]

all matrices are invertible

[muted panther]

Les K_i dépendront de a,b,c,d

[keen wadi]

D'accord, je pensais que ce qu'on appelle "coordonnées d'un vecteur" c'était le a, le b, le c et le d

[muted panther]

Oui, attention, les coordonnées du vecteur (a,b,c,d) dans la case canonique

c'est simplement (a,b,c,d)

Quand on change de base, les coordonnées changent, mais ça reste le même vecteur

[keen wadi]

ah okay

mais dans une autre base a,b,c,d ne sont plus les coordonnées mais juste des nombres comme ça

c'est ça ?

[muted panther]

Si on veut

[keen wadi]

Enfin en tout cas on peut plus les appeler coordonnées d'un vecteur

On les appelle comment alors ?

[muted panther]

On les appelle encore coordonnées d'un vecteur, mais dans la base canonique

[keen wadi]

Okay, merci beaucoup !

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[last slate]

Find the number of ways in which 8 distinct toys can be distributed among 5 children

[last slate]

87654

Is this the answer?

[meager berry]

ah

steps

?

number of cases

[last slate]

8.7.6.5.4?

[meager berry]

?

[lament fox]

This would be correct

however this does not evaluate to 87654

ohh i see you are multpilying

[meager berry]

you have to make cases

[last slate]

Cases?

[meager berry]

yes

there are going to be many cases

is there any restriction?

[last slate]

No this is the full question

[meager berry]

ah

[last slate]

It can be 8.8.8.8.8 also right?

Wait no

[meager berry]

and many more

way too many cases

if im not wrong

[placid spoke]

8.7.6.5.4 would be correct

[meager berry]

uh

what if one child gets all

rest 4 none

[last slate]

is the answer 5^8?

[placid spoke]

oh good point

[meager berry]

there are going to be way too many cases

[meager berry]

.

[placid spoke]

i was assuming each child only gets 1

[meager berry]

that's why I asked if there's any restriction

recheck question

[last slate]

Each toys can be given to 5 .. then will it be 5^8?

[last slate]

No bro this is the full qn

[meager berry]

hold on

[placid spoke]

I looked this up, these are Stirling numbers

have you learned those?

[last slate]

Lemme check

Nope

[meager berry]

there are going to be many cases

[placid spoke]

no wait, that assumes each child gets at least 1. but the problem kinda implies it since it says distributes

wack

[meager berry]

cant assume

he didnt say

0 0 0 0 8
0 0 0 1 7
0 0 1 1 6

and so on

[placid spoke]

wait this is overthinking lmao

[last slate]

@placid spoke @meager berry
Just got an answer from internt

[meager berry]

huh

[last slate]

I don't really understand though

[placid spoke]

the first toy has 5 choices of where it can go, the second has 5, the third has 5, etc

^

lol

[meager berry]

when do we group stuff

then

when do we take cases then

is it for identical?

[placid spoke]

you have 8 toys and 5 kids. the first toy has 5 possible kids it can go to. the second toy has 5 possible kids it can go to, etc

so if you distribute 8 toys, you have 5.5.5.5.5.5.5.5 ways to distribute

[last slate]

I see

[meager berry]

when do we take cases then

[last slate]

Then what is the problem when taking 8^5?

[placid spoke]

that assumes you are assigning kids to toys, not toys to kids

that is saying you have 8.8.8.8.8 ways to do things

[last slate]

Yeah the question says toys to kids huh

Oh

Ah ok thx guyz

[meager berry]

?

[placid spoke]

sorry for the initial confusion

[placid spoke]

you could if you wanted i'm sure but you'd end up getting the same answer either way

[last slate]

Cases ?how?

[meager berry]

in some questions

i think that's grouping

ok mb

got it

thx

[last slate]

👍

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[twilit field]

$\int_0^x [x+1]^2dx$

[twilit field]

where [x] is the floor function

I was hoping someone could guide me through this

[grand pond]

Why am. I here

[twilit field]

so I start by breaking it into multiple intervals

[obtuse basin]

Start by diving the interval into [0, 1], [1, 2], ..., [floor(x)-1, floor(x)], [floor(x), x]

[quiet hinge]

Thats it, you've solved it

You need to know how to evaluate the sum of squares now

[obtuse basin]

I would avoid using the same letter for the integration variable and the bounds though

[twilit field]

hmm, so this method works for any power "n" belonging to Z, right?

[twilit field]

got it?

[twilit field]

*@

*W

[quiet hinge]

Yes, but you need to know what the sum 1+2^n+3^n+...+ floor(x)^n is equal to

In our case we have the sum of squares

Which you might have learnt how to evaluate

[twilit field]

$\int_0^11dt+\int_1^24dt+\int_2^39dt.....$

[grand pond]

Why am. I here

[twilit field]

but what would the last term be

could I have a hint

[quiet hinge]

You know the bounds right?

[quiet hinge]

Its given here

[twilit field]

ah, thanks!

hmm, so I have to evaluvate 1+(8-4)+)(27-18)...

so 1+4+9...

[quiet hinge]

Yep

Be careful with the last term though!

[twilit field]

yeah, that's what I'm trying to figure out

The last term would be x

so $\frac{\left([x]-1\right)\left([x]\right)\left(2[x]-1\right)}{6}$+ x

[grand pond]

Why am. I here

[quiet hinge]

$\int_{[x]}^{x} [t+1]^{2}dt = [x+1]^{2}(x-[x])$

[grand pond]

smidgin

[twilit field]

ah

but is the first part of my resut correct?

[quiet hinge]

Yes

[twilit field]

awesome, thanks!

[quiet hinge]

I think what I've written here is incorrect though, I should work this out properly

Nope nevermind, it is correct

[twilit field]

Can I close the channel now?

[quiet hinge]

Yes, but you should make sure once for yourself this result

I am certain it is correct, but you shouldn't believe me

[twilit field]

.close

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[mellow steeple]

What do they mean by resultant partial fractions?

[mellow steeple]

All combined?

My answers are 1,-2,4

[midnight plank]

@mellow steeple Has your question been resolved?

[mellow steeple]

<@&286206848099549185>

[wary wing]

ghr

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@mellow steeple Has your question been resolved?

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[last slate]

alt exterior angles i thin u do like 5x - 21 = 3x + 17 and thats how to get x

and after u get x plug it into 3x + 17

[sharp laurel]

<@&286206848099549185>

[ocean skiff]

🫣

[sharp laurel]

???

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[last nebula]

how do I find DB

And also does triangle CBA ~ triangle EDA. Work?

[worthy kestrel]

you can find BC via Pythagorean

[last nebula]

yea

[worthy kestrel]

with BC and DE you can work out the ratios between the triangles. with that and AB you can find AD. and subsequently BD with BD=AD-AB

[worthy kestrel]

they are similar triangles if I'm understanding that ~ correctly

[last nebula]

Yea

[last nebula]

wdym ratios tho

how do I find the ratios

like divide?

[worthy kestrel]

yes

[last nebula]

ohh

[worthy kestrel]

given a triangle ABC and another triangle XYZ are similar, we get their lengths are equal to some scaling factor of the other. AB/XY=BC/YZ=CA/ZX=q for some number q

[last nebula]

U got 1.25

*I

[worthy kestrel]

yes

[last nebula]

ok wait so what so I do with that again

[worthy kestrel]

find AD first

[last nebula]

how?

wouldn’t I need DB

[worthy kestrel]

it helps you find DB

[last nebula]

but how do I get AD without DB

[worthy kestrel]

use this property of similar triangles

in this case, AD is AB multiplied by a scaling factor

[last nebula]

so like AE/AC = DA/BA = DE/BC

?

is that what you mean cuz I don’t get this concept rlly

[worthy kestrel]

yes that

[last nebula]

did I do it right

[worthy kestrel]

you didn't do anything. you just asked if you should use that equation and I said yes

[midnight plank]

@last nebula Has your question been resolved?

[last nebula]

oh

[last nebula]

no but i mean like the equation

like do i just fill in the variables

with numbers

@worthy kestrel sorry for the tag

[worthy kestrel]

yes

[worthy kestrel]

note you already got the ratio earlier

[last nebula]

where do i use the ratio tho

soryr but i still dont understand where i need to use it

[worthy kestrel]

fill in all the values you know here

[last nebula]

AE/3.6 = DA/3 = 2.5/1.99

[worthy kestrel]

in this case you want AD (or DA), so which equation would you use from this?

[last nebula]

ohh

cross multply

[worthy kestrel]

yes

did you manage to get AD?

[last nebula]

not yet

its approx 3.8

[worthy kestrel]

it's 3.75

[worthy kestrel]

now you can use this to work out DB (or BD)

[last nebula]

okkkk ty

[last nebula]

wait abt for the similar triangle

its ∆ADE ~ ∆ABC

does that work?

[worthy kestrel]

yes

you can change the starting points or direction you label the triangle, as long as you change it appropriately for how you label the other triangle too

[last nebula]

okk

[cloud jungle]

Do u help with geo

[midnight plank]

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[azure token]

Not a clue

[hearty rune]

look at amplitude, period and center line

also compare to the parent function, obviously

[azure token]

Is there a video you know of in which I can understand those 3 concepts

its an online class and the textbook is gibberish

[cedar coral]

start by the basics. What's the value of cos(0)?

[azure token]

1

[cedar coral]

okay. What's the value of sin(0)?

[azure token]

0

[cedar coral]

and what's the value of tan(0)?

[azure token]

0

[cedar coral]

with that information, obtain the value of each of the cases you're given, at x=0

[azure token]

so like -pi, pi, 3pi, 5pi

[cedar coral]

-.-

let's start by (b), since it's easier to type on discord

you told me that sin(0) = 0

if x=0, how much is 2x?

[azure token]

0

[cedar coral]

so if 2x=0, how much is -3sin(0)?

[azure token]

uh 0

[cedar coral]

now compare it to the graph

at x=0, is the graph on y=0?

[azure token]

y=2.5

[cedar coral]

it's a yes or no question

[azure token]

no

[cedar coral]

(also, the value would be 3, not 2.5)

[azure token]

damn

[cedar coral]

now repeat it for a, c and d

[azure token]

thank you

what was the other guy referring to with amplitude

[cedar coral]

you didnt pay enough attention to the grid. it's not "somewhere between 0 and 5", it's exactly on the grid intersection

[azure token]

I tried to scroll up and down and answer ina timely manner

[cedar coral]

amplitude is how wide, on the y-axis, the curve is

[azure token]

how would i find question 2.?

[cedar coral]

what is secx, in terms of the basic trig functions?

[azure token]

google says 1 cos x

[cedar coral]

1/cosx

okay. So you know that secx= -1 = 1/cosx

knowing that, how much is cosx?

[azure token]

-1

[cedar coral]

and what value(s) of x have cosx= -1?

[azure token]

not sure

[cedar coral]

have you studied the unit circle?

[azure token]

i know 0 about trig

beginning of the semester so I am learning as we go

[cedar coral]

Some values

[azure token]

ah yes I have seen that

[cedar coral]

in this pic, in parenthesis, first value here would be cosine. Second value would be sine. They determine the point coords on the circle

inside would be the angle in both radians and degrees

now, look at the circle for a bit until you can tell me which angle has cosx= -1

[azure token]

180 degree one

[cedar coral]

are you asked to work in degrees, radians, or you can choose?

[azure token]

u can choose

[cedar coral]

okay

so now you have ONE value of x that gives you secx= -1

are there any others in the unit circle?

[azure token]

any other what

[cedar coral]

any other values of x

that give you that cosine (or secant)

[azure token]

doesnt appear so to me but i assume yes bc youre asking

[cedar coral]

i'm asking because i want you to reason the answer

[azure token]

no

[cedar coral]

so you got one on the unit circle, and no more, but you're asked for two. Can you get a value outside the unit circle?

[azure token]

i assume not

[cedar coral]

why

assuming is almost always bad, you need to justify with proper reasoning

[azure token]

i see the arrows so i presume the graph goes on

so yes u can find a value beyond

im gonna research all this later

[cedar coral]

You can. As you can see the trig functions are periodic

specifically, they keep repeating every full rotation of the circle

[azure token]

ah

[cedar coral]

for tan and cotan, it would be every half rotation

that should be enough information for you to get a second value for (b)

[azure token]

so how would the values change if they repeat?

[cedar coral]

look at the graph that you're given

[azure token]

or would the same point be used twice

[cedar coral]

you can see that cos(0) = 1

but so is cos(2pi), cos(4pi), and so on

(you dont ACTUALLY see it in the graph you're given since you're not actually graphing cosx)

[azure token]

ok i should be good

but the first value would be 180 degrees>

?

[cedar coral]

yes

[azure token]

ok

appreciate it

[cedar coral]

i recommend, on the graph, to put the index finger on the first peak, the middle finger on the second peak, and move both fingers simultaneously along the graph

[azure token]

what would i discover

[cedar coral]

you'll notice how your fingers stay in the same position, as the graph is periodical

[azure token]

ah

[cedar coral]

which means that as long as you keep two points that fixed distance (the period) apart, the function has the same values

[azure token]

what does that entail

ah

[cedar coral]

or in other words: f(x) = f(x + kT) with k being a natural number, and T being the period

for sine and cosine, T=2pi

[azure token]

i think i just need to learn what cos tan and sin mean

first

[cedar coral]

wikipedia has several definitions

[azure token]

yea I passed algebra with minimal knowledge and i assumed it would be the same with trig but it seems i might need to contribute time

[cedar coral]

the right triangle one is probably the easiest to understand

you probably should put some effort into algebra, it's the bread and butter of pretty much everything else

[azure token]

honestly i was there until we got to the big E

[cedar coral]

you mean the sum symbol?

$\sum$

[grand pond]

LordFelix

[azure token]

fancy looking E

yeah

[cedar coral]

this one?

that's easy to understand if you know programming

[azure token]

and then the numbers that surround it

im working on my 4th language rn

thats why i need math

[cedar coral]

"For i = whatever it says on the bottom, i++, until whatever it says on top:"

"Add whatever says on the right to the current result"

[azure token]

i dont remember it being that simple

[cedar coral]

it wasnt for me either until i made the connection

[azure token]

there was like a y=6 to the side and then there'd be a 2.654654 on top

and some arbitruary num on the bottom

what does i ++ mean in programming terms

is it liie +=

[cedar coral]

i++ would be c language

[azure token]

dont think python uses that

[cedar coral]

it means increment i in 1 unit

[azure token]

ah yea so i+=

working on java rn

[cedar coral]

example

"from j = 1 until j=n, add j^2"

let's say n=4

[azure token]

'''public class Multadd {
public static double multadd(double a, double b, double c){
return a * b + c;
}
public static double expSum(double x) {
double term1 = multadd(x, Math.exp(-x), 0);
double term2 = multadd(1 - Math.exp(-x), 1, 0);
return multadd(term1, term2, 0);
}
public static void main(String[] args){
double answer = multadd(1.0,2.0,3.0);
System.out.println(answer);
double expression1 = multadd(Math.sin(Math.PI / 4), 1.0 / Math.sqrt(2), Math.cos(Math.PI / 4) / 2);
double expression2 = multadd(Math.log10(10), 1, Math.log10(20));
System.out.println(expression1);
System.out.println(expression2);
}
}
'''

[cedar coral]

we start in j=1: We add 1^2; we increment j in 1 unit.
j=2; We add 2^2; j++
j=3; += 3^2; j++
j=4; += 4^2; We done, since j=n
Result: 1^2 + 2^2 + 3^2 + 4^2

[azure token]

forgot its the weird symbol

    public static double multadd(double a, double b, double c){
        return a * b + c;
    }
    public static double expSum(double x) {
        double term1 = multadd(x, Math.exp(-x), 0);
        double term2 = multadd(1 - Math.exp(-x), 1, 0);
        return multadd(term1, term2, 0);
    }
    public static void main(String[] args){
        double answer = multadd(1.0,2.0,3.0);
        System.out.println(answer);
        double expression1 = multadd(Math.sin(Math.PI / 4), 1.0 / Math.sqrt(2), Math.cos(Math.PI / 4) / 2);
        double expression2 = multadd(Math.log10(10), 1, Math.log10(20));
        System.out.println(expression1);
        System.out.println(expression2);
    }
}

yea i get what ur saying

this was one of our recent projects and i had some struggles with it

especially bc it incorporated trig

mostly just figuring out how to relay an expression into a way the comp understands

[cedar coral]

programming is mostly writing the finnicky way that the computer wants its things

[azure token]

do you think its even worth it to pursue right now

[cedar coral]

pursue what

[azure token]

comp sci

for a career

[cedar coral]

everything has a computer these days

so probably

[azure token]

been massive layoffs as of late

especially with ai

[cedar coral]

meh, someone has to repair that ai

[azure token]

soon even this exchange of complicated math will be automated

[cedar coral]

not anytime soon

[azure token]

in my lifetime at least

[cedar coral]

also, remember that someone will need to understand how to build the hardware that can make the computations

so yeah, not going away anytime soon

[azure token]

machine automation no?

[cedar coral]

there's a reason why that's not allowed in the server

it's very, very bad

[azure token]

like robots not ai

[cedar coral]

anyways, it's bed timee

[azure token]

alr sorry

i appreicate it tho

.close

[midnight plank]

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[last slate]

Hi

[last slate]

can someone explain this step

like i really dont know why they are doing this

this whole proof format is stupid anyways but

yeah

[next rover]

they are saying they proved it in case of ~ϕ

on the line (3)

where it says (3) (→I)

[last slate]

yeah thats fine

but more like

okay yeah so both not phi and phi are true

[next rover]

no

not both

[last slate]

or

[next rover]

just at least one

[last slate]

one is true

yeah

i see what they did

bruh they didnt even need to do

they could have just done the implication

that part is so unecessary

i hate this class and this stupid proof format

[next rover]

what imlpication

[last slate]

like just going to this step

[next rover]

???

we don;t have this implication

that's how you can make one

if assuming left lets you prove right you get yourself an implication

[last slate]

You assumed the LHS not phi -> psi or whatever, and you showed phi or psi so its fine you have the right hting

[next rover]

but there's no unnecessary step anywhere

[midnight plank]

@last slate Has your question been resolved?

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[wild pivot]

The ln is throwing me off

[honest kelp]

calculate the limit of the general term of the series, i.e. check whether the necessary condition for convergence of the series is satisfied

things like this, you just have to notice them

[wild pivot]

Oh ok I'll check it now

[honest kelp]

ok)

[wild pivot]

If It gives me infinity over infinity in the ln can I do hôpital rule only In the parentheses

[honest kelp]

no no, no need hospital at all 🙂

it wud be too much!

look:

5n + 3 / 6n + 1 ---> 5 / 6

you should know it form the shcool

and because logarith is continous function

you can use: lim ln (..) = ln ( lim ... )

so you shoud say now:

the geenral term reaches ln 5/6

do you agree on it ?

[wild pivot]

Yes I see

[honest kelp]

and now

you should say loudly that:

becasue the general term does not go to zero,

that makes your given serives immediately divergent

[wild pivot]

Yes makes perfect sense now thank you

[honest kelp]

yvw 🙂

[wild pivot]

.close

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[twilit field]

$$\int \sqrt{1+x^3}dx$$

I tried two approaches

approach 1:-

$$x\sqrt{1+x^3}-$\int \frac{3x^3}{2\sqrt{1+x^3}}$

[grand pond]

Why am. I here
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

which is

$\int \frac{3x^3}{2\sqrt{1+x^3}}$, $\left(x\sqrt{1+x^3}-\frac{3}{2}\int_{\ } \frac{\left(x^3+1\right)}{\sqrt{1+x^3}}-\frac{1}{\sqrt{1+x^{3\ }}}\right)$

[grand pond]

Why am. I here

[twilit field]

now what?

how do I integrate $\sqrt{1}{\sqrt{1+x^3}}$

[grand pond]

Why am. I here

[twilit field]

alternatively, I re-wrote $(1+x^3)=(1+x)(1-x+x^2)$

[grand pond]

Why am. I here

[twilit field]

but that's even more of a headache

,w integrate \sqrt(1+x^3)

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by physicsrocks

[twilit field]

ok, wth

how was I supposed to know that the integral can't be expressed in terms of elementary functions

Please don't explain differential galois theory or something like that, I just need a rough idea of what to do

,w integrate 1/\sqrt{1+x^3}

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by physicsrocks

[main current]

What's your question?

[twilit field]

how do I determine if a fuction can be expressed in terms of elemntary functions, using just highschool calculus

[main current]

Allow me to explain differential galois theory:

Jk. Really, you'll start to get a sense for "solvable patterns"

Square roots in general are tough to integrate, and VERY few functions involving square roots actually work

Arclengths involve square roots and very few arclengths are actually closed functions

[twilit field]

ah, OK. Thanks so much!

can I close this now?

[main current]

Oh yeah if you want. .close

[twilit field]

.close

[midnight plank]

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[jovial nebula]

sequence and series

[jovial nebula]

i have solved almost all but there is a doubt

so first i did number of 3 digit numbers divisible by 2 i.e 450

then divisible by 3 i.e 300

then divisible by 6 i.e 150

Numbers divisible by 2x7 ---------64 ,,,,,,,,,,,, , 3x7---------43

and then i solve it i dont get the right ans

why so......

<@&286206848099549185>

[midnight plank]

@jovial nebula Has your question been resolved?

[jovial nebula]

<@&286206848099549185>

[lusty elm]

you are probably overcounting here?

[jovial nebula]

pls look into it

[lusty elm]

the lists
14, 28, 42, ...
21, 42, 63, ....
have overlap

[jovial nebula]

ohhhh

yeahh

then i make a 6x7 series

right>??

[lusty elm]

yea that should work

[jovial nebula]

ok so i.e 21 numbers

now what

like do i add or subtract

450+300-150-64-43

-21

or +21

[lusty elm]

let's see...

[jovial nebula]

ok

[lusty elm]

450 + 300 - 150 is the # of nums divisible by 2 or 3

[jovial nebula]

there should be a vc i hate typing this much

[jovial nebula]

yus

[lusty elm]

so yea +21 if that's how many numbers here overlap and if the lists have 64 and 43 numbers

which i didn't check but yea that's the right idea

[jovial nebula]

sorry

why

[lusty elm]

you want to remove the ones that are divisible by 7

[jovial nebula]

uhum

oh

is e

see

i subtacted it twice

yes?

[lusty elm]

so you can count the ones of form 2*k*7 and ones of form 3*k*7

but that counts the ones of form 6*k*7 twice

so subtract those off

[jovial nebula]

yeah

thanks

alot

.close

[midnight plank]

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[twilit field]

$\int \frac{x^2\left(x\sec^2\left(x\right)+\tan\left(x\right)\right)}{\left(x\tan x+1\right)^2}dx$

[grand pond]

Why am. I here

[twilit field]

could I have a hint?

I tried xtan(x)+1=u

but that doesn't work

Should I maybe rewrite everything in terms of sin and cos?

huh, I'm getting $\int \frac{x^2}{\left(x\sin\left(x\right)+\cos\left(x\right)\right)}$

[grand pond]

Why am. I here

[twilit field]

how do I integrate this?

oh, nvm

$\int \frac{x^2\left(x+\sin\left(x\right)\cos\left(x\right)\right)}{\left(x\sin\left(x\right)+\cos\left(x\right)^\right)^2}$

ok, how do I do rthis?

[grand pond]

Why am. I here
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[honest kelp]

.reopen

[midnight plank]

✅

[honest kelp]

read and analyze:

.close

[midnight plank]

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[chrome hill]

Numbers 650 to 900 which are not divisible by 3 or 7

[jovial nebula]

yo

you wanna count the numbers

?

like how many are there'

@chrome hill

?/

[chrome hill]

Ohh it is done

.close

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[twilit field]

$\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac{n^3}{\left(n^2+k^2\right)\left(n^2+3k^2\right)}$

[grand pond]

Why am. I here

[twilit field]

How do I convert this rieman sum into an integral ?

[wary thorn]

rienman

[twilit field]

Could I have a hint please

do I start by dividing by $n^2$?

[grand pond]

Why am. I here

[misty crater]

I think you should start by dividing by n^4

[twilit field]

hmm, yeah, I did think of that, but how would I deal with k^2/n^4

[misty crater]

Think about how you distribute 1/n^4 in the denominator

You are distributing across a product, not a sum

(ab)/n^4 ≠ (a/n^4)(b/n^4)

[twilit field]

$\sum_{k=1}^{\infty}\frac{\left(\frac{1}{n}\right)}{\left(\frac{1}{n^2}+\frac{k^2}{n^4}\right)\left(\frac{1}{n^2}+\frac{2k^2}{n^{\left(4\right)}}\right)}$

[grand pond]

Why am. I here

[twilit field]

so now I factor the 1/n^2 out of each summation?

[misty crater]

^

(a)(b)/n^4 = (a/n^2)(b/n^2)

[twilit field]

$\sum_{k=1}^{\infty}\frac{\left(\frac{1}{n}\right)}{\left(\frac{1}{n^2}+\frac{k^2}{n^4}\right)\left(\frac{1}{n^2}+\frac{2k^2}{n^{\left(4\right)}}\right)}=\sum_{k=1}^{\infty}\frac{\left(\frac{1}{n}\right)}{\left(\frac{1}{n^2}\right)\left(1+\frac{k^2}{n^2}\right)+\frac{1}{n^2}\left(1+\frac{3k^2}{n^2}\right)}$

[grand pond]

Why am. I here

[twilit field]

this?

[misty crater]

No, give me a sec to do the tex

$\lim{n\rightarrow\infty}\sum{k=1}^n\frac{n^3}{\left(n^2+k^2\right)\left(n^2+3k^2\right)} = \lim{n\rightarrow\infty}\sum{k=1}^n\frac{\frac{n^3}{n^4}}{\frac{\left(n^2+k^2\right)\left(n^2+3k^2\right)}{n^4}} = \lim{n\rightarrow\infty}\sum{k=1}^n\frac{\frac{1}{n}}{\frac{\left(n^2+k^2\right)}{n^2}\frac{\left(n^2+3k^2\right)}{n^2}}$

[misty crater]

Okay some of it got messed up but the important stuff is there

$\frac{n^3}{\left(n^2+k^2\right)\left(n^2+3k^2\right)} = \frac{\frac{n^3}{n^4}}{\frac{\left(n^2+k^2\right)\left(n^2+3k^2\right)}{n^4}} = \frac{\frac{1}{n}}{\frac{\left(n^2+k^2\right)}{n^2}\frac{\left(n^2+3k^2\right)}{n^2}}$

[twilit field]

hmm, OK

now what , the integral is $\int_0^1\frac{dx}{\left(1+x^2\right)\left(1+3x^2\right)}$

[grand pond]

Why am. I here

[twilit field]

?

[misty crater]

Yes

[twilit field]

cool

Thanks!

I'll close the channel now, if that's fine with you

[misty crater]

Yeah

[twilit field]

Thanks!

.close

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[pseudo jetty]

can someone help me

[pseudo jetty]

please

pleaseeeewee

[glass dome]

multiply top and bottom to get rid of the square root in the fraction in the top

[pseudo jetty]

how 😭

multiply square root of 9-x^2 with 9-x^2?

.close

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[pseudo jetty]

.close

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[red helm]

D)

Do I turn cos 45 into 2(1/2 root 2)

[torn compass]

no

just use the formula for sum/difference

[midnight plank]

@red helm Has your question been resolved?

[red helm]

This is my working but the answer is 1/2(cos70 + cos 20)

Idk what I did wrong

Or is the answer wrong

[midnight plank]

@red helm Has your question been resolved?

[midnight plank]

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[latent plover]

when you are integrating area between 2 limits

[latent plover]

does it integrate the area clockwise

or anticlockwise

like if i was integarting between pi/3 and 5pi/3 as a and b respectively for my limits

lemme take an ss from desmos of what i mean 1 sec

how do i know

if it integrates to find R

or the area that is unshaded

the purple and green lines are meant to be half lines

buit idk how u do that 😭

[midnight plank]

@latent plover Has your question been resolved?

[midnight plank]

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[twilit field]

I'm not sure I understand how to differentiate a determinant , could someone please explain?

[last slate]

differentiate a determinant?

send the problem

[twilit field]

More specifically why do we have to differentiate one row or column at a time

[visual tiger]

because the determinant is a polynomial in the coefficients of the column vectors (or the row vectors)

f(x) is a sum of (-1)^... *a*b*c, where a is a coefficient on the first column, b a coefficient from the second column, c from the third

and so this is just a product rule

[twilit field]

Thanks!

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[round palm]

Hi, is this the correct "circle graph" that's undirected (edges go in both ways) for my question?

[round palm]

My question: Suppose a "circle graph" has 4 nodes connected (in both directions) by edges around
a circle. What is its adjacency matrix S

Or this?

[shell wigeon]

neither?

Hold on maybe I don't know what a circle graph is

I see you got the image from Wikipedia

If you remove node c you need to remove all three edges with it

[midnight plank]

@round palm Has your question been resolved?

[round palm]

😭 so what's the "circle graph" they're talking about

the adjacency matrix changes for each case

[round palm]

oh yeah sorry i forgot about that but yes yes

if we're talking only about the outer edges then which one is correct?

[shell wigeon]

Either

It doesn't matter what you call each node, they are 4 nodes connected by edges in a loop

[round palm]

Hmm, but the answer is necessarily just one matrix

[shell wigeon]

The question is kind of ambiguous tbh, I don't even think they are talking about circle graphs like that Wikipedia article

[round palm]

This is the answer

[round palm]

This is my configuration^^ so this is my answer

[round palm]

i think the answer uses this

so i'm lost how they determined which one to choose

but in hindsight the answer uses a cyclic order: 1 -> 2 ->3 ->4 while mine doesn't

[shell wigeon]

You can go from one to the other by just swapping two rows and two columns

(in essence renaming two vertices)

[round palm]

right, so would you say my answer is right then?

oh

[shell wigeon]

Yes, it's the same graph

[round palm]

okay fair enough then, tyty-

[shell wigeon]

Just a different vertex ordering

[round palm]

also why is it called a circle graph when there's nothing "circle" about it lol

[shell wigeon]

Idk

[round palm]

it's a square right? any graph with 4 nodes?

[shell wigeon]

You could make a similar graph with any number of vertices (>=3)

[round palm]

alright alright, thanks again

this is a lin alg exercise so i wanted to make sure

i wasn't missing anything from graph theory or whatever

.close

[midnight plank]

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[midnight plank]

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[hot portal]

A site inspection is necessary to determine where and how you will be building your learning hub.

Below are some calculations to determine:
Height of structure (AC)
Length of ladder (BC)
Angle of placing ladder (CBA)
Angle of placing tree (CAE)
Angle of Sun overheard (DBA)
A few things must be worked out before the construction process:
First, determine the tallest height point C of your learning hub/ dream home/ space habitat structure. This will be the Length AC. Point A is a point on the ground. You can set this length yourself.
Next, you are given the distance at which a person is standing from the learning hub which is Length BA = 3 meters. Using the information you have so far, use the Pythagoras theorem to calculate Angle CBA.
Use the Pythagoras theorem again to calculate the Length BC which would be the distance required to determine ladder length.
Next, calculate the Angle CAE to help plan the planting of a tree. To get this angle you must first find Angle EAB. (Hint: use the chart below to help you. Congruent means the angle size is the same as each other)
Lastly, work out Angle DBA, using a protractor to measure. Remember, this is the only angle you must use a protractor to measure. None of the other angles must be measured with a protractor. This angle will help determine the angle of the sun overhead.

Note: For every angle above, you must state what type of angle it is. Is it acute, obtuse, right angle or reflex? If you use angles around parallel lines, then you must use the above chart to point out which type of angles you used. Are they corresponding, vertical, alternate interior or alternate exterior?

Can some one please help me with this!?

[midnight plank]

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[midnight plank]

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[hot portal]

A site inspection is necessary to determine where and how you will be building your learning hub.

Below are some calculations to determine:
Height of structure (AC)
Length of ladder (BC)
Angle of placing ladder (CBA)
Angle of placing tree (CAE)
Angle of Sun overheard (DBA)
A few things must be worked out before the construction process:
First, determine the tallest height point C of your learning hub/ dream home/ space habitat structure. This will be the Length AC. Point A is a point on the ground. You can set this length yourself.
Next, you are given the distance at which a person is standing from the learning hub which is Length BA = 3 meters. Using the information you have so far, use the Pythagoras theorem to calculate Angle CBA.
Use the Pythagoras theorem again to calculate the Length BC which would be the distance required to determine ladder length.
Next, calculate the Angle CAE to help plan the planting of a tree. To get this angle you must first find Angle EAB. (Hint: use the chart below to help you. Congruent means the angle size is the same as each other)
Lastly, work out Angle DBA, using a protractor to measure. Remember, this is the only angle you must use a protractor to measure. None of the other angles must be measured with a protractor. This angle will help determine the angle of the sun overhead.

Note: For every angle above, you must state what type of angle it is. Is it acute, obtuse, right angle or reflex? If you use angles around parallel lines, then you must use the above chart to point out which type of angles you used. Are they corresponding, vertical, alternate interior or alternate exterior?

Can some one please help me with this!?

[midnight plank]

@hot portal Has your question been resolved?

[midnight plank]

@hot portal Has your question been resolved?

[hot portal]

<@&286206848099549185> Can i get osme help

[midnight plank]

@hot portal Has your question been resolved?

[midnight plank]

@hot portal Has your question been resolved?

[hot portal]

;D

can some one please help

[hot portal]

<@&286206848099549185>

please some one im super lost

[onyx plume]

what is the question?

[hot portal]

@onyx plume

this

Height of structure (AC)
Length of ladder (BC)
Angle of placing ladder (CBA)
Angle of placing tree (CAE)
Angle of Sun overheard (DBA)

A few things must be worked out before the construction process:
First, determine the tallest height point C of your learning hub/ dream home/ space habitat structure. This will be the Length AC. Point A is a point on the ground. You can set this length yourself.
Next, you are given the distance at which a person is standing from the learning hub which is Length BA = 3 meters. Using the information you have so far, use the Pythagoras theorem to calculate Angle CBA.
Use the Pythagoras theorem again to calculate the Length BC which would be the distance required to determine ladder length.
Next, calculate the Angle CAE to help plan the planting of a tree. To get this angle you must first find Angle EAB. (Hint: use the chart below to help you. Congruent means the angle size is the same as each other)
Lastly, work out Angle DBA, using a protractor to measure. Remember, this is the only angle you must use a protractor to measure. None of the other angles must be measured with a protractor. This angle will help determine the angle of the sun overhead.

Note: For every angle above, you must state what type of angle it is. Is it acute, obtuse, right angle or reflex? If you use angles around parallel lines, then you must use the above chart to point out which type of angles you used. Are they corresponding, vertical, alternate interior or alternate exterior?

[midnight plank]

@hot portal Has your question been resolved?

[midnight plank]

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[midnight plank]

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[zenith kettle]

.close

[midnight plank]

Channel closed

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[zenith kettle]

<@&268886789983436800>

[pearl idol]

thanks, banned them

[zenith kettle]

.reopen

[midnight plank]

✅

[zenith kettle]

.clsoe

.close

[midnight plank]

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[midnight plank]

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[sage yew]

i feel stupid but plz help

[stable dawn]

Can't

Understand what's written

Is that 8 or 2

Is that a face or an equal sign

[midnight plank]

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[midnight plank]

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[vagrant prairie]

I’m looking at a an example and I’m just confused on how they factored this

[humble nacelle]

do you know long division?

[cerulean glade]

They most likely used synthetic division

[humble nacelle]

thats whatthey did

they didnt show it

[humble nacelle]

whats that?

[cerulean glade]

its division with polynomials

[surreal moon]

Rational root theorem would be the first thing to try.

[humble nacelle]

long division?

[surreal moon]

Then yes, synthetic division would get you the remaining polynomial factors

[vagrant prairie]

Yes that’s what they were doing

but they found that Polynomial through synthetic division

[surreal moon]

I like your pfp

[vagrant prairie]

Do I have to use synthetic division again to factor it?

[surreal moon]

[vagrant prairie]

I LOVE YIURS

ok now answer my question 😁

[surreal moon]

Let's say you had $4x^4-11x^3+45x^2-28x-10$ and let's say you used RRT to get $(x-1)$ as one of the factors of the polynomial. Then yes, you would need to use synthetic division to get $4x^3-7x^2+38x+10$ as the other factor.

[grand pond]

SWR

[vagrant prairie]

so it would be like 1

And then

4 , -7 , 38 and 10 right

[surreal moon]

I'm not really seeing how you're arriving at this

[vagrant prairie]

Is 4X+1 the remainder and the X^2 one the quotient?

[vagrant prairie]

LMFAO

[surreal moon]

There is no remainder when dividing $4x^4-...$ by $x-1$. That's what makes $x-1$ a factor of the polynomial

[grand pond]

SWR

[surreal moon]

$4x+1$ is another root. You'd find it by RRTing the resulting cubic $4x^3-...$

[grand pond]

SWR

[vagrant prairie]

wait what?

Wait do you think you can just do it out ? I’m a little confused

[surreal moon]

Through rational root theorem, you find that $x-1$ is a factor of $4x^4-11x^3+45x^2-28x-10$. Therefore, you can do synthetic division to calculate
$$\frac{4x^4-11x^3+45x^2-28x-10}{x-1}=4x^3-7x^2+38x+10$$
There is no remainder. You can do RRT again to find that $4x+1$ is a factor of $4x^3-7x^2+38x+10$. You do synthetic division again to calculate $$\frac{4x^3-7x^2+38x+10}{4x+1}=x^2-2x+10$$
Again, no remainder.

[grand pond]

SWR

[vagrant prairie]

Ohh I think I get it

Thanks

[midnight plank]

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[wise belfry]

thinking false for A, true for B

[wise belfry]

but whats the different between a vector space and R3, I think it has to be linearly independent to span R3?

[midnight plank]

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[midnight plank]

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[finite jolt]

For lim as n approaches infinity of ((n-5)/n)^n, I understand how to get to e^-5 but i also don't get why it's not 1

[finite jolt]

I can manipulate the equation to get (1-(5/n))^n, but I also don't get why the original function wouldnt simplify to (n^n/(n^n) for large n

[midnight plank]

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[midnight plank]

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[grave zephyr]

hi

can someone help me

with this

can you show me the seperation steps to get it to the P&Q formula

pls

<@&268886789983436800> idk why it doesnt create a channel for me

[last slate]

bots broken malware we think

[grave zephyr]

oh okey

[analog vine]

not "malware" lmao

just an api outage

itll fix itself

[last slate]

ok

ok

[grave zephyr]

can you create a channel for me?

@analog vine

[analog vine]

no

[grave zephyr]

ok

[analog vine]

use the forum

#1021175428326633542

for now