help-13

[crimson delta]

well then they mean (0, m)

[wraith dagger]

kartheek

[restive plover]

But I do not know how to use this 'm' could you please help me

[crimson delta]

while technically correct, not what they want from you here. what is the slope of the line going through (0, m) and (-1, 0)

[restive plover]

1?

I meant $\frac{sin \theta}{cos \theta +1}$

[wraith dagger]

kartheek

[crimson delta]

lets forget that (cos theta, sin theta) also lies on that line

[restive plover]

sorry

its m

[crimson delta]

what is the slope of the line going through (-1, 0) and (0, m)

ok

so now y= ?

[restive plover]

y=m(x+1)

[crimson delta]

ok

next x^2+y^2=1

plug y in and get?

[restive plover]

and should i substitute this into x^2+y^2

yeah yeah

and I will get the result in the second part of the subquestion right?

[crimson delta]

if you do the steps correctly, yes

[restive plover]

god dayumm I love you got the answer hehehe

Thanku sm

[crimson delta]

tho I have to say, not a very well written question

[restive plover]

Exactly I wasted a whole night, It was very confusing

in this part of the question, now i know that (-1,0) and (cos theta, sin theta) is in the same line I meant the points in the line and the circle, So How can I say that -1, and Cos theta are the roots of the equation (1+m^2)x^2+2m^2x+m^2-1=0

[crimson delta]

wdym with "how"

they are solutions cause they are on the intersection of the line and the circle

[restive plover]

I tried up to this, now I have the roots Ig I am supposed to prove this, $cos \theta = \frac{1-m^2}{1+m^2}$

[wraith dagger]

kartheek

[restive plover]

Could you give me some hints?

Should i add the roots or something like that?

[crimson delta]

about what. this is the solution

if you know one root of a quadratic you can easily find the other one with vieta

[restive plover]

vieta?

[crimson delta]

if x^2+px+q=0 and a, b are solutions, then ab=q, a+b=-p

I have to go now

[restive plover]

Thanks for the help @crimson delta

One doubt, should the sum of the roots of the quadratic equation be equal to the coefficient of x , where did that negative sign come from? Sorry if it is dumb

nv,

nvm

got it got it

then again thanks

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[mint raft]

For m I got -1/2 and more c I got -1 is it right?

[mint raft]

Then can I please get help on this one!

[south tundra]

Yes that's right

Well firstly c = -5 here

And if the line y = mx - 5 passes through (3; 1)

It means that 1 = 3m - 5

You can solve for m here

[mint raft]

I got 2

[south tundra]

So m = 2 and c = -5

[mint raft]

Oh okay thanks bro I appreciate it 🙏

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[limber snow]

how would once calculate ln(3pi/2) without calculator

[tropic oxide]

you wouldn't.

[limber snow]

nvm Im tripping. theres a 0 before all that so it should just be 0.

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[floral grail]

.open

3x² + 12x + 23/4

[floral grail]

This is a quadratic function thing

Like the vertex/standard form go to general form

Or either way

Show solutions pls

Thanks

<@&286206848099549185>

The answer should be 3(x + 2)² + 1/2

But it need the solution

Anyone pls

[crimson sedge]

why so desperate for a solution..

[cedar kiln]

@floral grail Has your question been resolved?

[floral grail]

The teacher needs it, also i do need it to understand the topic more

[livid hound]

have you been taught anything about completing the square

[cedar kiln]

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[visual bison]

Stuck on the third one down ,we have to show whether this series converge

I've tried ,comparison test , null sequence test and ratio test

I'm not sure how to simply ot

[earnest socket]

,rotate

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[violet night]

tried to get this done last night but couldnt get my head around it

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[mint raft]

Please help!

[mint raft]

For this one I have to find the length of AB

[sleek condor]

use pythagoras' theorem

[mint raft]

For which one?

[sleek condor]

for both

[mint raft]

so for the first one I’m working with 1 and 5?

[sleek condor]

wdym?

[mint raft]

I’m confused you said I must use phythag…

[sleek condor]

yeah

what does pythagoras say?

[mint raft]

It involving squaring

I know how to do your pythag

But in this case I’m a bit confused

[sleek condor]

can you answer:

"what is the pythagorean theorem?"

[mint raft]

Sure.

Pythagoras is the square on the hypotenuse of a right angled triangle is equal in area to the sum of the squares on the other two sides

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[deep wind]

how to find number of terms (a+b)^2 * (a+b+c)^4

[cedar kiln]

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[deep wind]

<@&286206848099549185>

broke it down to a+b+c <= 6 and a,b <= 2 and c <= 4

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[jaunty raven]

Hello. Can anyone review my work?

This is the question

[dire geode]

,w integral 0 to pi/4 1/(csc(x) -1)

[wraith dagger]

Results provided by WolframAlpha

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Requested by riemann#2529

[jaunty raven]

I got wrong answer, i got 0.414... I need pro version to view steps

[dire geode]

what happened here?

did you skip finding the antiderivative?

[jaunty raven]

i changed csc(x) to 1/sin(x) and cot^2(x) to 1-sin^2(x)/sin^2(x) and replace the x with the limit

[dire geode]

that's not how you do integration

integration means you find a function whose derivative is your integrand

by FTC

[jaunty raven]

mhm ok

[dire geode]

so find an antiderivative of
$\frac{\csc(x)}{\cot^2(x)}$

[wraith dagger]

riemann

[dire geode]

and your other integrand

probably will help to simplify in terms of sines and cosines first

[jaunty raven]

ok I got the answer correct thank you

.close

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[viscid plinth]

I know i need to multiply numbers to these rows and add them or subtract them in order to get them to be 1s and 0s

but where to begin...

[cedar kiln]

@viscid plinth Has your question been resolved?

[viscid plinth]

<@&286206848099549185>

[dawn jacinth]

Are you looking for a regular row echelon form or row-reduced specifically?

[viscid plinth]

row echelon form

[dawn jacinth]

Then all we need to do is to make sure that the first column does not contain two pivot points

[viscid plinth]

what do you mean?

[dawn jacinth]

In other words, we can leave the first row unchanged, and just zero out the -3 on the second row

This way, the second row will have 0x, and isolate y to the constant. This allows you to solve for y individually, then substitute it to the first equation to solve for x

[viscid plinth]

so is the goal to have a diagonal of 1s?

[dawn jacinth]

Not necessarily for non-row reduced row echelon form, but you kind of have the right idea

[viscid plinth]

so whats the end goal?

[dawn jacinth]

The end goal is to isolate variables in each equation, allowing you to solve for one, then substitute that into higher equations

[viscid plinth]

right

so whats the first step?

[dawn jacinth]

Well for this specific matrix, there is only 1 step

You can use the Gaussian Elimination algorithm to solve this equation. This will be kind of difficult to explain if you haven't read up on it, so I suggest taking some time to see it, since it'll help you with every augmented matrix

But I can provide it in this situation

Using the values on the first row, we can zero out -3 by adding (or subtracting) it to some multiple of -8. This is kind of an ugly number, but in this case, R2 = R2 - c*R1, where c is some constant.

In this case, we want 0 = (-3) - c(-8). If we solve, we get c = 3/8

[viscid plinth]

but dont we multiply that entire row by that number?

[dawn jacinth]

Then, for every column in row 2, we do the same operation. For example, -11 - (3/8 * -29) for column 2 of row 2

Then 1 - (3/8 * 2) for column 3, row 2

That's it. You're done for the augmented matrix

Also yeah, you do multiply the entire row. I was just showing it step by step

[viscid plinth]

so i multiply R1 by -3/8

then add it to R2

[dawn jacinth]

Yup

[viscid plinth]

but R1 doesnt change

[dawn jacinth]

ye

[viscid plinth]

so R1 goes back to being -8 -29

[dawn jacinth]

yup, R1 unchanged in this scenario

[viscid plinth]

so i should have

R1 = -8 -29 2

R2 = 0 -1/8 1/4

[cedar kiln]

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[radiant prairie]

Describe as simply as possible the set of all real numbers x, for
which applies (the solution must be specified):

[radiant prairie]

in not sure because i could solve for a -1 < x

or maybe does : -1 turn the < to >

ok i found the answer

next question

on task b

can i make it easier?

cant pull the root of negative numbers

<@&286206848099549185>

[craggy oriole]

Can you write clearly what are the conditions?

[radiant prairie]

Describe as simply as possible the set of all real numbers x, for
which applies (the solution must be specified):

b is the problem

(b) 5 − x^2 < 8

i got -3 <x^2

am i finished or maybe did i do something wrong

[craggy oriole]

You're right, the numbers with square larger than -3 are all real numbers

[radiant prairie]

is the d finished?

[craggy oriole]

Not quite, can you graph that quadratic formula?

[radiant prairie]

didnt understand sry

[craggy oriole]

The exercise asks you where the parabola (x-1)*(x-3) is above 0

[radiant prairie]

parobola?

where?

[craggy oriole]

,w graph (x-1)*(x-3)

[wraith dagger]

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Requested by ralvrz#3477

[craggy oriole]

So, for which vales of x is the graph above the x axis?

[radiant prairie]

0-1 and 3-4

[craggy oriole]

No need to stop at 0 and 4, it goes from minus infinity to infinity

But yes, all numbers smaller than 1 or larger than 3

[radiant prairie]

how sould i write it down?

[craggy oriole]

x<1 or x>3

[radiant prairie]

in the task they want all my steps how i got the solution

i dont know if it is allowed to uses graphs?

[craggy oriole]

If a*b>0 then both a and b have the same sign, either positive or negative

That is (x-1)>0 and (x-3)>0, or (x-1)<0 and (x-3)<0

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[radiant prairie]

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[violet viper]

i am not sure what i need to do past here

[cedar kiln]

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[violet viper]

,open

.reopen

.close

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[winter lion]

these are the questions and what I have tried so far

[winter lion]

I am unsure if my images are correct for the given functions and if I have shown enough work

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[winter lion]

<@&286206848099549185>

[winter lion]

<@&286206848099549185>

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[somber egret]

not sure how to do b, c, or d

i thought i found out how to do b but one of the steps includes finding rref of A and taking the columns that don't have leading 1s

but rref of A leads to all columns having leading 1s

so idk

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[crimson sedge]

struggling on how to find the line of symmetry

[long arrow]

line of symmetry is x = h

in general case x = - b/2a

[crimson sedge]

whats b and a?

[long arrow]

ax^2 + bx + c

coefficients

from your previous question u know h = 0, then x = 0 is line of symmetry

using x = -b/2a

a = 1, b = 0

x = 0

[crimson sedge]

ohhh

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[lethal vale]

.close

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[junior timber]

Ms. Moorey has $220.90 in her piggy back. She keeps only dimes an quarters. If she has a total of 1231 coins, how many coins of these are dimes

[junior timber]

<@&286206848099549185>

[cosmic steppe]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[junior timber]

oops sorry

forgot about that

[cosmic steppe]

Set up a system of equations. Let x = the number of dimes and y = the number of quarters

You can find the equation for the total number of coins

[junior timber]

okk

and then what?

im lost

[cosmic steppe]

What's your equation

[junior timber]

x + y = 1231

???

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[obtuse frigate]

Hi

[obtuse frigate]

How do I do this integral?

[livid hound]

IBP a few times (or tabular)

[obtuse frigate]

What does tabular mean?

[livid hound]

tabular integration aka DI method

[obtuse frigate]

So I let x^3 be u and e^(-2x) be v

[obtuse frigate]

Ohh

What’s the DI method?

I don’t think I’ve learnt that

[haughty wraith]

https://youtu.be/2I-_SV8cwsw

[livid hound]

it's a convenient way to do multiple iterations of ibp

[obtuse frigate]

OH that

[obtuse frigate]

So I need to differentiate the one that will eventually become 0 right

So x^3 for this integral

Then integrate e^(-2x)

[livid hound]

yeh

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[balmy totem]

trying to do this and i am pretty sure you are supposed to subtract the extras but im not sure how to do that

[balmy totem]

What is the probability that a positive integer not exceeding 100 selected uniformly at random is a multiple of 3 or a multiple of 4?

so i just tried writing down all the possible multiples and then i got 26+25

and then since you pick a random integer not exceeding 100 i tried C(100,51)

[cedar kiln]

@balmy totem Has your question been resolved?

[balmy totem]

do i not use choose for this

[buoyant ivy]

can anyone help me

i have a elemtary school triangle question

[balmy totem]

or was i right with 100/3 and 100/4

[craggy oriole]

Do you account for numbers being a multiple of both 3 and 4?

[balmy totem]

do you mean like

12 and 24 and 36 and so on

thats how i got 51 originally i got 58 but then i subtracted 7 because thats the duplicates

ok nvm i got 51% i think you add them

i added 33/100+25/100 and subtracted 7/100

i got 51% that looks right maybe

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[smoky narwhal]

I have an exam tomorrow. There is a really complicated lesson that’s called optimization with linear programming

[smoky narwhal]

It’s a whole chapter talking about Optimization with linear programming

[craggy oriole]

This is a math server, do you need help with something specific?

[smoky narwhal]

Wait what I said is not math?

[crimson sedge]

they're asking what your question is

[smoky narwhal]

My bad. I’ve been trying to understand the linear programming but I don’t seem to get it.

[smoky narwhal]

I assume you thought it was something about computer programming or something but that’s not really it.

[haughty wraith]

do you have a specific problem

[cedar kiln]

@smoky narwhal Has your question been resolved?

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@smoky narwhal Has your question been resolved?

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[crimson sedge]

How do I solve for x?

[earnest socket]

you need the numerator to be 0 and the denominator nonzero

[haughty wraith]

i dont think there any solutions

[earnest socket]

so 2e^(-2/x) = 0 when x != 0

but 2e^(-2/x) = 0 implies that e^(-2/x) = 0

is 0 in the image of the exponential function?

[cedar kiln]

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[earnest socket]

you want to write e_1 as a linear combination of c_1 and c_2

this defines a system of two equations in two variables

[cedar kiln]

@cold patrol Has your question been resolved?

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[ember mesa]

hi,i need someone to check the results of this exercise: Count the elements of order 4 in S_8(Symmetric group).I found a total of 2520 elements.

[ember mesa]

In particular i found: (41^4,420),(4^2,1260),(42^2,420),(421^2,420).Where the first element is a symmetric structure and the second is the number of elements that have that structure

[silver parcel]

shouldn't it be 2940 ? I don't know if there's something I counted twice

[ember mesa]

how did came up with it?

[silver parcel]

for 2520 you counted all composition of two 4-cycles like that: (....)(....) right ?
but there are also the simple 4-cycle

420 of them

so I think we should add them to the 2520

wait there is even more

the 4-2-2 also

errhh

it's definitely more than 2940

[ember mesa]

i thought i counted them,i got 420 4-cycles, 1260 4-4, 420 4-2-2 and 420 4-2-1-1

[silver parcel]

oh yeah for 4-4 it's 1260

[ember mesa]

for 4-cycles i mean 4-1-1-1-1

[silver parcel]

that's where I counted twice

420 4-cycles, 1260 4-4, I agree

but I keep finding more for 4-2-1-1 and idk where I'm making a mistake

4-2-1-1 you're building two cycles
the 4 one and the 2 one
for the 4 you have 8 * 7 * 6 * 5/4 possibilities and for the 2 you have 4 * 3/2

that's way more than 420

you have a problem with the 4-2-2 I think, too

it should be 1260

the total should be the 2940 + the 4-2-1-1

the final total should be 5460 I think

[ember mesa]

i think you are right but i can't find the error in my calculations,mind if i send a photo to check?

[silver parcel]

let's list the cases:

1. 4-1-1-1

2. 4-2-1-1

3. 4-2-2

4. 4-4
   right ?
   so let's do them one by one

5. we choose 4 numbers with order and we divide by 4 because of the 4 shifts for each
   8 * 7 * 6 * 5/4 = 420, done

6. we choose 6 numbers with order and we divide by 4 * 2 by the same reasoning as above, we get 2520

7. we choose 8 numbers with order and we divide by 4 * 2 * 2 as above, and re-divide by 2 because we can swap the 2-cycles between themselves and it's the same element
   so 8!/32 = 1260

8. we choose 8 numbers with order, and we divide by 4*4 because of the shifts, and the order in which we take the cycles don't matter, so we divide by 2, 8!/32 = 1260

we were agreeing on cases 1) and 4), but found different things on 2) and 3)

so here is my reasoning for 2) and 3)

[ember mesa]

for example in 2) dont you have to divide 2520 for 2?Because for example (1,2,3,4)(5,6) = (5,6)(1,2,3,4)

[silver parcel]

no because I'm not counting (5, 6)(1, 2, 3, 4) in the first place

I'm filling (_ _ _ _)(_ _) by choosing 6 numbers

then dividing by 4*2 accounting for the shifts of each

[earnest socket]

4-2-1-1 should be (8 choose 4) * (4 choose 2) * (2 choose 1) * (1 choose 1), right? this was horribly wrong

[silver parcel]

no

order matters in some way

we're talking about arrangements where the shift don't matter

but you're supposing 1 2 3 4 and 1 3 2 4 will be the same

which is not

[earnest socket]

oh right

[ember mesa]

i get it!!!thanks a bunch

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[languid atlas]

can someone please check if my variable are correct

[silver parcel]

it's correct 'til now

[languid atlas]

Alrighty does my process look alright?

[silver parcel]

yes

,w 4*0.8^3 * 0.2

[wraith dagger]

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[languid atlas]

okay thank you so much

.close

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[restive plover]

How can I draw the inverse of sin function in the interval $[\frac{3 \pi}{2} \frac{5 \pi}{2}]$

[wraith dagger]

kartheek

[foggy merlin]

symmetry of sin by the line y = x

[restive plover]

I do not have the idea how it would look

[foggy merlin]

just draw it on a graph

[restive plover]

Can you give me a rough sketch, You flip the axis right

[foggy merlin]

you cant tell me you dont know how to draw symmetry

[restive plover]

lol yup ig

[crimson sedge]

Take the derivative

[restive plover]

Hey I hate chemistry tooo

[foggy merlin]

,w graph arcsin(x), 3pi/2 < x <5pi/2

oops

,w graph arcsin(x), sin(3pi/2) < x < sin(5pi/2)

[wraith dagger]

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[restive plover]

,w graph arcsin(x), sin(-3pi/2) < x < sin(-5pi/2)

[wraith dagger]

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[restive plover]

@foggy merlin what is a maximal interval?

is it the interval in which arcsin is one to one but adding one more point would change it to many one

[foggy merlin]

the domain of arcsin is [-1,1]

[foggy merlin]

wdym by maximal interval

[foggy merlin]

what does "1-1" mean

[dense hornet]

most likely "one-to-one"

[crimson sedge]

Injective

[cedar kiln]

@restive plover Has your question been resolved?

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[shrewd brook]

How should i approach this

[cedar kiln]

@shrewd brook Has your question been resolved?

[cedar kiln]

@shrewd brook Has your question been resolved?

[silver parcel]

notice that h(4/3) = 6.2 and that's the max of h, cause 1 is the max of cos

h(t) <= 3.7 + 2.5 = 6.2

and h(t) can indeed be equal to 6.2 when t = 4/3

for example

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[simple ivy]

useless server

[south tundra]

<@&268886789983436800>

[hushed spoke]

Does this call for a moderator ping?

[calm sierra]

@simple ivy pls dont abuse help channels

.close

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@cunning flume Has your question been resolved?

[cedar kiln]

@cunning flume Has your question been resolved?

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[cedar kiln]

@bronze prairie Has your question been resolved?

[bronze prairie]

No

<@&286206848099549185>

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[warm cedar]

.reopen

hi

[warm cedar]

In ℝ3. Giving Euclidean Vector {𝑢1 = (1, −6,1), 𝑢2 = (2,3,5), 𝑢3 = (3,7,8)} and vector x=(7,-2,m). Find value of m, x is a linear combination of the system {𝑢𝑖
} above (Linear Algebra) I need help

SOMEONE HERE

[crimson sedge]

Uh

I guess just normally row reduce it?

[warm cedar]

really?

[crimson sedge]

I suppose, and set whatever you get for your RREF as equal to that for its row

If you are trying to find if it is a linear combination or not I presume

[warm cedar]

Giving Vector

[warm cedar]

You mean gauss elimination?

[crimson sedge]

Yes

[warm cedar]

after I reduce RREF, I do r different 0

[crimson sedge]

r different 0?

[warm cedar]

I mean det A different 0

[warm cedar]

I mean r(A) = m, and det A different 0

Line Independ, and Line Depend

[crimson sedge]

I mean yeah sure

Whatever way that gets you to the answer

[warm cedar]

how many method to answer this question?

[crimson sedge]

I think you really only need basic gaussian laws

[warm cedar]

ok.

I have another question

[crimson sedge]

Sure

[warm cedar]

In ℝ3. Giving Euclidean Vector {𝑢1 = (1, −6,1), 𝑢2 = (2,3,5), 𝑢3 = (3,7,8)} and vector x=(7,-2,m). Find value of m, x is a linear combination of the system {𝑢𝑖
} above (Linear Algebra) I need help. Do you know what is the name of the solution?

[crimson sedge]

Isn't that what we just talked about?

[warm cedar]

yes.

[warm cedar]

Another exercise. I found out inverse matrix . A^-1/ |A| (det A)

[crimson sedge]

I can't really understand you, sorry

[warm cedar]

Another exercise. I found out inverse matrix . A^-1/ |A| (det A) . I am sorry.

[crimson sedge]

Yeah you saying it again doesn't change the fact that I still cannot comprehend what you are trying to ask sadly

[warm cedar]

This is vietnamese document

You see exercise 2, and 3.

AX=B

X = A^-1 C^t (Inverse matrix)

[sudden garden]

can someone help whit my homework

[warm cedar]

hello

[warm cedar]

what is this math?

[warm cedar]

In
3
for 2 vector systems:
𝐴 = {𝛼1 = (2; 1; 1), 𝛼2 = (2; −1; 1), 𝛼3 = (1; 2; 1)) } ,
𝐵 = {𝛽1 = (3; 1; −5), 𝛽2 = (1; 1; −3), 𝛽3 = (−1; 0; 2) }
a) Show that A, B are the bases of
3
b) Find the base transfer matrix from B to A;
c) Let 𝛼 = (−5; 8; −5). Calculate [𝛼]|𝐴.
d) Calculate [𝛼]|𝐵 in 2 ways: Directly and through the base transfer matrix

[cedar kiln]

@warm cedar Has your question been resolved?

[cedar kiln]

@warm cedar Has your question been resolved?

[warm cedar]

In 𝑅^3 for 2 vector systems:
𝐴 = {𝛼1 = (2; 1; 1), 𝛼2 = (2; −1; 1), 𝛼3 = (1; 2; 1)) } ,
𝐵 = {𝛽1 = (3; 1; −5), 𝛽2 = (1; 1; −3), 𝛽3 = (−1; 0; 2) }
a) Show that A, B are the bases of 𝑅^3
b) Find the base transfer matrix from B to A;
c) Let 𝛼 = (−5; 8; −5). Calculate [𝛼]|𝐴.
d) Calculate [𝛼]|𝐵 in 2 ways: Directly and through the base transfer matrix

[cedar kiln]

@warm cedar Has your question been resolved?

[cedar kiln]

@warm cedar Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[ivory yoke]

How can i find the matrice contains Column space (1,1,0) and (0,1,1), and the null space (1,0,1) and (0,0,1)

[foggy merlin]

??

[copper axle]

If you want a matrix A such that (1, 1, 0) and (0, 1, 1) are in the column space of A, and (1, 0, 1) and (0, 0, 1) are in the null space of A, such a matrix does not exist.

[cedar kiln]

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[brittle parcel]

Hey I’m quite confused with this question

[violet night]

do you know what parallel means?

[brittle parcel]

Yes

The gradients should be same, right?

[violet night]

exactly

gotta be careful here tho

left hand equation is in one form whilst the right hand one is in another

[brittle parcel]

Yeah that’s why I’m a bit confused with it

[violet night]

how would you rewrite the right hand equation as standard y=mx+b form?

[violet night]

exactly, although I recommend to leave things as fractions unless told to do otherwise

and since their gradients HAVE to be the same since we are told they are parallel

what does that mean we can do ?

[brittle parcel]

Oh multiply the gradient of the left hand equation by 2?

[violet night]

not quite

if I say that a must equal b

then I can write a=b

if the gradient of the left must equal the gradient of the rightwhat would I equate ?

[brittle parcel]

I’m not quite understanding what you mean

[violet night]

do you agree that the gradient of the left is 5?

[brittle parcel]

Yes

[violet night]

and the right is $\frac{k}{2}$

[wraith dagger]

Clarkie

[brittle parcel]

Yes

[violet night]

these two things have to be equal because they are parallel right?

[brittle parcel]

Yep

[violet night]

how would you write that out on paper?

[brittle parcel]

5 = k/2 ?

[violet night]

exactly

now its just simply solving for k

[brittle parcel]

Ohh

Okay yeah that’s makes sense

[violet night]

nice

[brittle parcel]

So it practically is the same as 5 x 2 like what I wrote here

[violet night]

uhhhhh actually yes it is,

[brittle parcel]

👍👍 tysm

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[winged token]

$\lim_{x \to 0^+}\frac{(lnx)^2}{ln(sinx)}$

[wraith dagger]

rekraP

[winged token]

At some point I got

$\lim_{x \to 0^+}\frac{2lnx}{cosx}*\frac{sinx}{x}$

[wraith dagger]

rekraP

[long arrow]

aa

sr

it's l'Hospital

[winged token]

Could I take the limit of 2lnx/cosx and sinx/x seperately

[long arrow]

if they're both exist and you won't get indeterminate form then yes

[winged token]

alright thank you

Should I delete the TexIts before I close

[dire geode]

No. Just leave it

[winged token]

.close

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[short blade]

Im having a problem trying to make a input to find x at any given point along two lines, does anyone have any ideas?

[cedar kiln]

@short blade Has your question been resolved?

[short blade]

So I know x=w-(between two lines)-(other side)

My question now would be is how do I find the distance from the perimeter Is the curve, and what kind of curve it is

[short blade]

I think I figured it out

✅

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[dry heart]

meow

[dry heart]

confused

wont there be infinity many A_i near zero

,w x sin(1/x) = x

[wraith dagger]

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[dry heart]

oh hm

oh wait it is not asking for Ai near zero

[cedar kiln]

@dry heart Has your question been resolved?

[dire geode]

yea just reading it off is $x_1 = \frac{2}{\pi (4 + 1)}$ and $x_2 = \frac{2}{\pi (4(2) + 1)}$ ?

[wraith dagger]

riemann

[dire geode]

is it obvious what x_1 should be? could it also be n=0 ?

[dry heart]

why not

why cant x1 be 2/pi

it is >0

also it makes sense because of the way question is designed

,w 2/pi - 2/(5pi)

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[dry heart]

compared to

,w 2/(5pi)-2/(9pi)

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[dire geode]

yea x1 corresponding to n=0 makes more sense. my brain just went straight to x1 meant n=1

[dry heart]

haha i did that to at first glance

anyways

thanks riemann

[dire geode]

fun problem

[dry heart]

.close

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[grizzled zenith]

I want to find the largest size of 4 squares to fit in a circle of diameter 8

[grizzled zenith]

What I was initially thinking is that this single square could be broken into 4 squares with a cross + and that would be the solution

[cold briar]

can the squares overlap?

[grizzled zenith]

No

[cold briar]

OK

[clever lintel]

isn't that just one square in 4 parts then?

[crimson sedge]

Thats a good idea, but that wont be the solution

[cold briar]

i thought of this too

[grizzled zenith]

But I'm thinking it might be possible to get even bigger squares if I put them at angles

Not sure though

[cold briar]

does the squares have to be identical

[grizzled zenith]

Yes

[cold briar]

does it have to be a square? or can it be rectangles

[clever lintel]

Hmm if I use my imagination to rotate those squares they have to get smaller

[grizzled zenith]

Has to be 4 identical squares that don't overlap

[clever lintel]

To account for the empty space in the middle of the circle

[grizzled zenith]

So is initial idea the correct answer?

[clever lintel]

The question is how do you have to prove it lol

[grizzled zenith]

I dunno how to prove it 😦

[clever lintel]

It feels like it has something to do with the longest line you can fit in the circle

[grizzled zenith]

I did the longest line

That's the diameter

[clever lintel]

Which like in the drawing is a hypotenuse of a square, or diameter of the circle

what about squaring the circle? just an idea

[grizzled zenith]

What do you mean?

[clever lintel]

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

Honestly just throwing it out there. Might lead to how to prove that thing exactly

but idnk

[grizzled zenith]

I think that's a whole other thing

I'm trying to figure out maximum square size to pack in a circle

For different number of squares

This one is for 4 squares

https://mathworld.wolfram.com/SquarePacking.html

I found the answer online, but no proof

6 is pretty interesting

[crimson sedge]

Hmm, i think the proof should be that the 4 squares will form another square that will be inscribed in the circle, since that is the largest square possible, then we can divide the side by 2 to get our answer, which would be 1

I think

[grizzled zenith]

The diagonal of a single square is the radius of the circle

[muted bear]

Diameter

Oh nvm

[grizzled zenith]

No

If you take a diagonal of a square that's a 45 45 90 triangle

[muted bear]

Proof involves showing you cant go bigger and proof by contradiction

[grizzled zenith]

Which means that the hypotenuse is s(2^.5)

[crimson sedge]

Yes, i know that, but i was talking about the larger square

[crimson sedge]

Oh true, yes that can be done

[grizzled zenith]

So s(2^.5) = 4

[crimson sedge]

Makes sense

[grizzled zenith]

Therfore s =4/2^.5

That is the size of the sides of the squares

2.8 ish

[crimson sedge]

Yes

That was stupid

Sorry

[grizzled zenith]

Why multiply by 2?

[crimson sedge]

Nvm

[grizzled zenith]

You mean root 2

[crimson sedge]

Lets see

[crimson sedge]

Yes

So what i mean is

When you find the side

Multiply side by 2

Then find the diagonal

If it is equal to 8

We got the answer

[grizzled zenith]

I didn't understand

[crimson sedge]

We got 4/√2 as the side right?

[grizzled zenith]

Yes

Diagonal will be 4

[crimson sedge]

Yes

[grizzled zenith]

2 diagonals together will be 8

[crimson sedge]

And hence 4/√2 is correct

[grizzled zenith]

That's how we calculated the side

You did the thing in reverse

[crimson sedge]

Cross checking

I always make mistakes there

[grizzled zenith]

Oh ok

No worries

[crimson sedge]

Which grade problem was this?

[grizzled zenith]

It's a cooking problem

[crimson sedge]

Lol wut?

[grizzled zenith]

I'd guess a problem like this could be 3rd grade though

[crimson sedge]

Oof

Well maybe

[grizzled zenith]

You want to see the cooking problem?

[crimson sedge]

Yea

[crimson sedge]

Oh well

[grizzled zenith]

I have this thick stainless tube

It has diameter of 8

Bananas are for scale

It weights 16 lbs

[crimson sedge]

Lol

Okay okay

[grizzled zenith]

I want to use it as diy tofu press with some trays

Then I have some rectangular molds

For cooking

[crimson sedge]

Ah i see

[grizzled zenith]

So to increase weight on the tofu pieces I want to fit the steel food molds into the circle and distribute the weight more evenly

[crimson sedge]

I guess good luck cooking the tofu

[grizzled zenith]

But my molds are too big

[crimson sedge]

This became a 3 dimensional problem

[grizzled zenith]

So I need to calculate the correct size of molds to get

Which is now 2.8 ish

[crimson sedge]

Great then

[grizzled zenith]

You know how people tell you the you won't use math in real life?

That's because they don't cook tofu

[crimson sedge]

Slap on their face this is

[crimson sedge]

Hahahahahaha

[grizzled zenith]

It's a new branch of math

Advanced tofu mathematics

[crimson sedge]

Hahahahahaha cool

[grizzled zenith]

@clever lintel did you learn the tofu math from this problem?

[cedar kiln]

@grizzled zenith Has your question been resolved?

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[heavy bramble]

hey

[shrewd sorrel]

Hello.

[toxic moat]

elo

[shrewd sorrel]

Go to #help-10

He needs a helper.

[heavy bramble]

if we have something like
can we linearize this without taking log both sides?

[shrewd sorrel]

I’m almost certain you have to take log.

[heavy bramble]

I was just asking to be sure if there's a way to do it when the exponent lacks "i"

[dire geode]

can you just show the original question

[heavy bramble]

I'm just saying that if the exponent was ib then I can use the property cos b + i sin b.
it's not an exercise, I'm just asking if I have y = a*e^(b)
if it's possible to linearize without taking log

@dire geode

[toxic moat]

...

no alien

alien bad

[dire geode]

define "linearize"

[heavy bramble]

something = something + something

instead of it being a product

[dire geode]

something = ae^b and something = 0

[heavy bramble]

not necessarily = to 0

hang on

if i log both sides I get
log(y) = log (a) + log (e^b)

can I get the same outcome without taking log..

[dire geode]

something ae^b + 1 and something = -1

[heavy bramble]

I don't want a specific case

[dire geode]

ae^b is a specific case

[heavy bramble]

I mean I don't want to set it equal to zero

set it equal to y, any image

[dire geode]

-1 isn't = to 0

[heavy bramble]

ok what can we do then

[dire geode]

to accomplish what?

[heavy bramble]

.

[dire geode]

i already linearized it for you

[toxic moat]

hey

are we looking at complex log

:O

[heavy bramble]

did you get my point?

[dire geode]

no?

y = (ae^b + k) + (-k) is linearized

[heavy bramble]

nevermind

thanks anyway

[cedar kiln]

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[void marsh]

how to do this

[south tundra]

The triangle on the left is isosceles, so 5x - 3 = 2x + 30

[void marsh]

why

[haughty wraith]

same angles

[void marsh]

ohoho

base angles

ok ok

[south tundra]

And in the right triangle the 3rd angle has to be 180 - (3x - 6)

And, since the sum of angles in a triangle is always equal to 180, you should have 40 + 40 + 180 - (3x - 6) = 180

[void marsh]

isn't the exterior angle theroem like 40+40+3x-6 = 180

ya nvm

I got it

wait nvm

[south tundra]

Ye you can use it to write that 3x - 6 = 40 + 40

That's the shortcut

[void marsh]

wouldn't that be a decimal tho

so that would mean 3x-6 = 80

[south tundra]

Yeah the answer isn't an integer

[void marsh]

80.000001 so we can say 80 ig

[south tundra]

86/3 is not that

[void marsh]

no after you subtract -6 from it

if 86/3 is x

we do that multiplied by 3

[haughty wraith]

1/3 * 3 = 1

[south tundra]

Sure but we're supposed to solve for x

[void marsh]

oh ya

mb

[void marsh]

but if we do 86/3 converted to a decimal it is 28.6666666667

oh nvm

[haughty wraith]

well its 28.6 recurring

[void marsh]

ya I messed up something now I got it

thank you guys

.close

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[cedar kiln]

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[solar matrix]

how would i do b?

[cedar kiln]

Please don't occupy multiple help channels.

[sonic thistle]

look at the big triangle

abc

write the relation between it's angles

[cedar kiln]

@solar matrix Has your question been resolved?

[cedar kiln]

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[median raptor]

Find the closure of the subset Z of R

[median raptor]

Is the answer Z? Or no limit points?

[solid juniper]

yes, Z is closed

[median raptor]

The closure is Z?

[solid juniper]

yes

if the space is R with the standard topology

[median raptor]

What if the space is Z?

[solid juniper]

what do you think?

[median raptor]

I thought if the space is R then it has no limit points

And for Z, Z is the closure

[solid juniper]

sure

[median raptor]

Can you explain ?

[solid juniper]

which part?

[median raptor]

Why it has no limit points in R?

[solid juniper]

take an integer n and look at (n - 1/2, n + 1/2)

[median raptor]

Okay

[solid juniper]

need more detail? lol

[median raptor]

No

Got it thanks

.close

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[solid juniper]

np ^-^

[cedar kiln]

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[timber dragon]

Hi, i'm having trouble resolving this limit: -2^n, it's supposed to be - infinite, but, we saw in class that, if q is < 0, then the sequence has no limit

[timber dragon]

However, here the correction says it's -infinite

is it like, -(2^n) so you can do the calculus and then put the minus ? is it allowed to do that ?

[haughty wraith]

-2^n means

-1 * 2^n

[timber dragon]

So it this wrong ? I can do that with all sequences no ? (the sentence means, if q < -1, the sequence has no limit)

[haughty wraith]

(-2)^n is different than -2^n

you have two separate functions you are equating

[timber dragon]

Can I do the red part ?

[haughty wraith]

no

it should be (-1)^m if you are going to factor it out

im sleeping now bye

[timber dragon]

how do you get (-1) ?

[haughty wraith]

ok i cant sleep

(-x)^n

=

(-1)^n * (x)^n

law of exponents

(ab)^m = a^m * b^m

[timber dragon]

and (-1)^n can be either 1 or -1 right ? how can I know which one it is at +infinite, because it would change the final limit ?

[cedar kiln]

@timber dragon Has your question been resolved?

[timber dragon]

I've got to n^2 x (-1)^n x (2)^n

but i'm not sure how to deal with the (-1)^n because it can be -1 and 1 right ?

[cedar kiln]

@timber dragon Has your question been resolved?

[cedar kiln]

@timber dragon Has your question been resolved?

[timber dragon]

.close

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[halcyon patrol]

Hi, how do I solve this?

[cedar kiln]

@halcyon patrol Has your question been resolved?

[cedar kiln]

@halcyon patrol Has your question been resolved?

[halcyon patrol]

.close

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[tender forge]

hey guys

[tender forge]

i need math help

can anyone help

[crystal raptor]

Still not reading?

[south tundra]

Don't ask if anyone could help, just post the question

[tender forge]

hey

i need help

with math equations

o ok

i need to simplify 3rd exercises equations

a) equation got done but i doubt in the answer

lets start off with b) equation right now

[south tundra]

First make the denominators common in the brackets

[tender forge]

okay

okay

what now

[south tundra]

Make the 2 have the same denominator as well

[tender forge]

ight

idk if you understand anything lol

?

[south tundra]

Anyways you should have gotten this in the first brackets $\frac{x(2x^2 + xy - y^2)}{x^3 - y^3} - 2 + \frac{y}{y - x} = \frac{2x^3 + x^2y + xy^2 - 2x^3 + 2y^3 - x^2y - xy^2 - y^3}{x^3 - y^3} = \frac{y^3}{x^3 - y^3}$

[wraith dagger]

A Lonely Bean

[tender forge]

ehh how do you do it bro

[south tundra]

Latex

[tender forge]

huh

[south tundra]

Or you mean how did I get it?

[tender forge]

yes

how can you y - x like that

[south tundra]

Rewrote 2 as 2(x^3 - y^3)/(x^3 - y^3) and y/(y - x) as -y(x^2 + xy + y^2)/(-(y^3 - x^3)

Expanded and collected the terms

[tender forge]

but you cant?

[south tundra]

And in the second brackets you get $\frac{x - y}{x} - \frac{x}{x - y} = \frac{(x - y)^2 - x^2}{x(x - y)} = \frac{x^2 - 2xy + y^2 - x^2}{x(x - y)} = \frac{y^2 - 2xy}{x(x - y)} = \frac{y(y - 2x)}{x(x - y)}$

[wraith dagger]

A Lonely Bean

[south tundra]

Why?

[tender forge]

because how you turn the y - x?

or can you?

[south tundra]

(y - x) = -(x - y)

[tender forge]

yes

and that sign changes the other sign in front of it

is that what you did

[south tundra]

Yes

[tender forge]

could you show me the whole solving process?

[south tundra]

Now that we simplified the fractions we need to divide y^3/(x^3 - y^3) by y(y - 2x)/x(x - y)

Which is the same as multiplying by x(x - y)/y(y - 2x)

[tender forge]

show the full process with TeXit bot

that way i can understand much better

[south tundra]

Will take an hour like that

[tender forge]

no problem

lets do it for hour then

just show first brackers full solving process bro

with TeXit bot

cuz that way i understand much better

whats problem bro

[south tundra]

Let me type it out jesus

[tender forge]

ight bro and then TeXit bot will show it?

is my answer on the picture correct btw?

so lets move on from that instead

i mean its correct, right?

answer lol

what you typin bro?

[south tundra]

$[\frac{x(2x^2 + xy - y^2)}{x^3 - y^3} - 2 + \frac{y}{x}]:(\frac{x - y}{x} - \frac{x}{x - y}) - \frac{x^2 + xy}{x^2 + xy + y^2} = [\frac{2x^3 + x^2y - xy^2 - 2x^3 + 2y^3 - x^2y - xy^2 - y^3}{x^3 - y^3}]:\frac{(x - y)^2 - x^2}{x(x - y)} - \frac{x^2 + xy)}{x^2 + xy + y^2} = \frac{y^3}{(x - y)(x^2 + xy + y^2)}:\frac{x^2 - 2xy + y^2 - x^2}{x(x - y)} - \frac{x^2 + xy)}{x^2 + xy + y^2} = \frac{y^3}{(x - y)(x^2 + xy + y^2)}\cdot{\frac{x(x-y)}{y(y - 2x)}} - \frac{x(x + y)}{x^2 + xy + y^2} = \frac{y^2}{x^2 + xy + y^2}\cdot{\frac{x}{(y - 2x)}} - \frac{x^2 + xy}{x^2 + xy + y^2} = \frac{xy^2}{(y - 2x)(x^2 + xy + y^2)} - \frac{x^2 + xy}{x^2 + xy + y^2} = \frac{xy^2 - (x^2 + xy)(y - 2x)}{(y - 2x)(x^2 + xy + y^2)} = \frac{xy^2 - (x^2y - 2x^3 + xy^2 - 2x^2y)}{(y - 2x)(x^2 + xy + y^2)} = \frac{xy^2 - (-x^2y + 2x^3 - xy^2)}{(y - 2x)(x^2 + xy^ + y^2)} = \frac{xy^2 + x^2y - 2x^3 + xy^2}{(y - 2x)(x^2 + xy + y^2)} = \frac{x(2y^2 + xy - 2x^2)}{(y - 2x)(x^2 + xy + y^2)}$

Probably a ton of mistakes

[wraith dagger]

A Lonely Bean

[tender forge]

first brackets last fraction is y/y-x not y/x

did it mess up the solving process

.close

[cedar kiln]

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[forest steeple]

https://gyazo.com/9f7bffebf671c75fb4563d3c0115c66d

[forest steeple]

can someone explain this

[broken mist]

How familiar are you with slope-intercept form?

y = mx + b

[forest steeple]

its exams so its stuff we did over 6 months ago so could you explain it to me, im noit too familiar with it

[broken mist]

So the y = mx + b form is used to describe many lines that can be graphed, like the one in part b

The y and x usually stay as y and x, but m and b can be replaced with other numbers to describe different lines

The number m (in part b m is -3) describes how steep the line is, and the number b (in part b, b is 0 so it does not show up at all) tells you how high off the x-axis the line is at the origin

[forest steeple]

so in part B the b is 0 like the number at the bottom, and is M always the second number?