help-10

[storm ridge]

and c k-3

[candid yarrow]

yes

[storm ridge]

The discriminant is -4k + 16?

[candid yarrow]

,rotate

[storm ridge]

-4k/-4 = 16/-4

k = -4 (?)

[candid yarrow]

oh

[storm ridge]

Oh positive 4

Alright thanks so much everyone I can finally submit this atrocity 🙏🏽🤍

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[storm ridge]

.close

[obtuse pebble]

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[mild kernel]

am i allowed to ask for help please

[fathom flicker]

No

Just kidding

What are you stuck on

[mild kernel]

well the top of the garage roof is 1.5m above the ceiling and the distance from the top of the roof to each gutter is 3.5m and I'm meant to calculate the width of the garage which I know the answer is 6.3m but I need to write how it got It but that's the thing someone already told me that answer but didn't write how they got it and i have to use the abc formula

[fathom flicker]

Can you find half the width of the ceiling?

[mild kernel]

i am dumb

i dont know man

😭

[fathom flicker]

Do you know SOHCAHTOA

[mild kernel]

yes

[hearty terrace]

do u know pythagoras

[fathom flicker]

Could you use it to find the missing side length of the right side triangle

Oh right, it is just pythagorean theorem, no need for SOHCAHTOA here

[mild kernel]

thats what im mean to do

[hearty terrace]

yeah

this

[mild kernel]

yeahh thats what im doing

but i dont know where to put what?

is it plus or take away

[hearty terrace]

hm?

where to put what

[mild kernel]

because i did √(3.5²+1.5²) and it came up with 3.8m

but the answer im looking for is 6.3m

[hearty terrace]

o

[haughty stump]

alright look at it this way

you have c

you have a

you are not solving for c

you are solving for b

this means b^2 = c^2 - a^2

plug in the values for c and a you get:

b = sqrt(3.5^2 - 1.5^2)

aka b = 3.16227766017

this is half the wdith of the garage

so the width of the garage is 2b

2b = 2(3.16227766017)

2b = 6.32455532034

[hearty terrace]

^

[mild kernel]

oh okay thank youuu!

[haughty stump]

aka 6.3 rounded to the tenths

[mild kernel]

ahh okay

[haughty stump]

add the units

and you get 6.3 meters = width of the garage

[mild kernel]

thank you!!

[haughty stump]

remember to use .close to close

[mild kernel]

close to close?

[haughty stump]

dotclose

replace dot with a period

[mild kernel]

ahh

[obtuse pebble]

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[warped hearth]

Hi please help? Idk how to approach this question, thanks.

[gloomy vector]

can you write it in modular form?

[warped hearth]

I'm supposed to work it out using hcf and lcm

im not so sure what that means.

I'm not too advanced, im almost 14

[frigid prawn]

by the division theorem, a number q can be expressed as
q = ma + b

b is the remainder, 5 in this case

q = ma + 5

[last pilot]

or you can just

calculate the LCM

[warped hearth]

im not so sure what the division theorom is sorry

[last pilot]

multiply until it's 4 digits long

[frigid prawn]

true ... ._. nvm

[last pilot]

and add 5

so

[warped hearth]

I see

so js trial and error?

[last pilot]

modular form isn't really needed lol

[last pilot]

no...?

first calculate the LCM

[warped hearth]

oh wait

LCM of 40 50 and 60

[last pilot]

it's alr lol

[last pilot]

yeah

[warped hearth]

600

[last pilot]

yep

[warped hearth]

Then what do I do, from there?

[last pilot]

find another number

that's divisible by these three

but is four digits long

[warped hearth]

Ok

[last pilot]

aka find a multiple of 600 four digits long

[warped hearth]

I jst hve a question

[last pilot]

yeah

@warped hearth what's ur question dude

[warped hearth]

for the previous question, this.

I did this

[last pilot]

so just 605

yeah

[warped hearth]

Am I supposed to use a method likethis?

my teacher was saying that he cares about the working out more than the answer or sm

so like do i do it with variables do u think?

[last pilot]

no

[warped hearth]

Ohk thankyou

[last pilot]

ur method is perfectly fine

[warped hearth]

alright thanks

so the answer is 1205?

[gloomy vector]

i remember not knowing the fact that something like the top can do that, so i plugged in small numbers and basically hoped it worked lmao

[last pilot]

yep

[warped hearth]

Alright is this k?

Thanks

[warped hearth]

Oh lol ok

[obtuse pebble]

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[opaque dome]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[haughty stump]

Find du

[granite dragon]

2

du = 1/(x-1)^2

[haughty stump]

negative

but yeah

but thats du/dx you need du which is -(dx/((x-1)^2))

[granite dragon]

ye

[haughty stump]

then sub u and du into the integral

and then simplify and yada yada

gtg for like 20mins so ask yakubros or something if you get stuck again

[granite dragon]

but the bottom doesn't have (x-1)^2

<@&286206848099549185>

[neon gazelle]

<@&286206848099549185>

i need help

where can i do anything

[opaque dome]

! Occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[granite dragon]

@opaque dome

do you know how to do it

[obtuse pebble]

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[obtuse pebble]

@granite dragon Has your question been resolved?

[obtuse pebble]

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[devout warren]

Hello

[devout warren]

Please could I have some help understanding differential operators

And how they are shown as a matrix

Is it just a matrix with d/dx down the leading diagonal and zeroes everywhere else

[idle leaf]

Try asking in #linear-algebra or #odes-and-pdes

[devout warren]

Ok

[idle leaf]

You'll likely to get better answers

[devout warren]

Thank you very much

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[ruby nimbus]

i have answer but not sure how to get? i understand that to find the 'chance' we find P(ideal out come)/p(total possibilities)

[fossil crag]

so it's not P(...)/P(...) you're looking for, it's number(...)/number(total)

so suppose you have to sit M1,....,M5 and W1,...,W5 around a table

make M1 sit anywhere you want, it doesn't matter where since you can rotate the table

[ruby nimbus]

oh right sorry

ye i menat that

[fossil crag]

if you want to separate the men

what is the "sequence" of M and W that will ensue?

like is MMWWMMW... possible

[ruby nimbus]

no

because 2 MM

[fossil crag]

ok

so forgetting about the number of each person

like M4 becomes M, etc

what are the possibilities?

(for reference you're starting with M since you start from M1)

[ruby nimbus]

its okay i think i got it now

thanks a lot rafilou2003!

[fossil crag]

alr

[ruby nimbus]

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[wooden narwhal]

Hellow where do i even begin

.close

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[dense tinsel]

I wanted to make sure this was correct

[dense tinsel]

I wanted confirmation that this is right

I thought it should be

Or not

I think i had to divide by 2, not by 4

[rigid skiff]

u were supposed to take the square root of the area

@dense tinsel

or better yet just divide 4.1 by 4

[obtuse pebble]

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[jade leaf]

I've been stuck on this matrices questions for a while, it has to be solved using the gauss-jordan elimination method, could anyone help me out

8 6 4 | 148
6 10 4 | 160
6 10 8 | 180

[unreal musk]

Did you have any ideas that you tried?

[jade leaf]

R1 - R3
R2 - R3
R2 = R1 - 4R2
R1 = R3 - R1

and that didnt work

I got the bottom row to be one, but it didnt match the answer at the back! 😢

[unreal musk]

I see well, I mean, row operations can be done in any order, which might make things easier if you choose a certain choice first

In particular, those last two rows, almost look the same

[jade leaf]

ooh, ill try something about that, thanks!

[unreal musk]

You might find that much easier it'll get rid of things in a different order maybe than you might be used to, but it could be more convenient (imo it is )

[jade leaf]

I solved it, thanks!!

[obtuse pebble]

@jade leaf Has your question been resolved?

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[languid mirage]

I need to make a piecewise function word problem that when is graphed shows a certain shape. I need help we've been trying all day

[timid silo]

What shape?

[languid mirage]

anything, as long as it draws something

and compatible with a word problem that can be used on a piecewise function

[iron edge]

Like when someone solve the riddle or sth, it yields a shape such as star, am I right?

[languid mirage]

not riddle, rather the word problem

but yeah something like that

[iron edge]

World or word?

[languid mirage]

WORD

myb

im so sorry

[iron edge]

It’s ok, you’re good

[languid mirage]

thank you

something like this

but it has to be, a shape when you graph it

we can't seem to make values that matches with the word problem

that can make a shape

[iron edge]

Looks interesting

[hexed gull]

First construct the shape, then the problem.

[weary reef]

grrr

beat me to it

[languid mirage]

wait, we tried to do that earlier

[iron edge]

Well yeah, lunatic has basically given the main approach

[languid mirage]

but i don't know how to make it match to the world problem

wait

here

wait actually

i just found out how

thanks to my team

since its negative, we'll do a debt word problem

ill try hol up

[obtuse pebble]

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[fierce glade]

Is it possible to find the number of Hamiltonian cycles in a tournament graph given that it is strongly connected?

[obtuse pebble]

@fierce glade Has your question been resolved?

[fierce glade]

<@&286206848099549185>

[neon gazelle]

yes

[fierce glade]

.close

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[inland thicket]

confusion

.close

[obtuse pebble]

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[solemn basin]

seriously

[inland thicket]

sorry im stressed

🫨

cos

coffe drip on my t shirt

[nocturne bronze]

cuberoot(x^3) = x^3^(1/3)

[inland thicket]

and i need to wear a very warm one

[nocturne bronze]

so it will be x^1

= x

it should be x^0 for being 1

[inland thicket]

hmm ye

thanks

[obtuse pebble]

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[torpid topaz]

help

i just cant seem to get the equation

<@&286206848099549185>

[lapis wedge]

Let me think about it for a second

[torpid topaz]

yay sure

[lusty sentinel]

a = 3 b = 17

[torpid topaz]

... ik i have the answer, but i need the steps

[lusty sentinel]

a + b = 20 and 3a^2 = b + 10

solve for a and b

[lapis wedge]

Since it is not said to divide 20 into 2 EQUAL parts, we will use 2 different variables x and y.
So x + y = 20.
And now just follow the second sentence:
3×x² = y + 10

[torpid topaz]

oki i got it

thank u so much

[lapis wedge]

You're welcome

[torpid topaz]

.close

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[kindred saddle]

How to solve x^x^3 = 3

[opaque dome]

Log that evil expression

Or let y = x^3

Yeah im more into the second

Cuz would be $x = \sqrt[3]{y}$

[warm shale]

YakuBros

[old lily]

need lambert W function probably

[opaque dome]

I have a way without

[tardy epoch]

Exponentiation is not associative. You need parentheses

[wary badger]

you mean $x^{x^3}$ yes?

[warm shale]

knief

[opaque dome]

so $\sqrt[3]{y}^y = 3$

[warm shale]

YakuBros

[opaque dome]

$(y^{\frac{1}{3}})^y = 3$

[warm shale]

YakuBros

[opaque dome]

Cube both sides

$y^y = 3^3$

[warm shale]

YakuBros

[opaque dome]

Pretty forward now, remembering y = x^3

[wary badger]

ahh that’s nice

good spot

[kindred saddle]

Ty bro

[obtuse pebble]

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[tall owl]

Help I’m lost

[worn yoke]

do you know how to find taylor/maclaurin series?

[tall owl]

no idea

[worn yoke]

well have you been introduced to taylor/maclaurin series yet?

[obtuse pebble]

@tall owl Has your question been resolved?

[tall owl]

Nope

[worn yoke]

ok. well for now you can use the fact that the "best polynomial approximation near x = 0" should have the same value, first derivative, and second derivative as the function at x = 0

[tall owl]

should i. try to teach myself the taylor/maclaurin series or is that something ill learn later in calc BC?

[worn yoke]

it should be covered by your class

[tall owl]

thanks

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[rocky mesa]

I'm having trouble with this problem. I'm not sure how to get the sin out the denominator

[rocky mesa]

I've broken it down to this point

[hardy widget]

This is cos(7x)

[rocky mesa]

right, my bad\

[hardy widget]

But recall that $\lim_{x \to 0} \frac{\sin x}{x}=1$

[warm shale]

Civil Service Pigeon

[hardy widget]

That’s really all you need

[rocky mesa]

Yeah, but how do I manipulate this into that form? Sorry if this if that's a dumb question, I'm struggling here 😭

[hardy widget]

You don’t have any factors of x, so add them in yourself

Where it would be beneficial to do so is for you to figure out

Just make sure that you keep it as an equivalent form

[rocky mesa]

So for my first fraction, would I multiply it by 7x/7x?

[hardy widget]

Yeah you could do that

[rocky mesa]

I guess where I'm confused is what happens to that extra x in the numerator?

[hardy widget]

Hint: You have two fractions with a sin that you need to “force” this into

[rocky mesa]

what do I do with these extra x's?

oops that should only be x in the second fraction

[hardy widget]

sin(7x)

[hardy widget]

sin(3x)

[rocky mesa]

hmmm okay

[hardy widget]

$\frac{7}{3} \cdot \frac{\sin 7x}{7x} \cdot \frac{3x}{\sin 3x} \cdot \frac{1}{\cos 7x}$

[warm shale]

Civil Service Pigeon

[hardy widget]

It should be this

[hardy widget]

I mean that’s what the original question says

[rocky mesa]

So I just end up with 7/3?

[hardy widget]

Yeah

[rocky mesa]

whooof, calculus is confusing. Thanks for the help

[hardy widget]

!done

[obtuse pebble]

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[shell fulcrum]

Hello, basically I'm trying to find the perimeter of this parallelogram but I'm not sure how to find that missing side. The problem gives that triangle for extra info but I'm not sure what to do with it

Thank you

[shell fulcrum]

And for clarification, the line of at extending from (a) to the right side of the parallelogram is supposed to be one single straigbt line

[pliant lily]

Similar triangles adb and cde

With a coefficient of 1/2 => cd = 1/2ad=6

@shell fulcrum

[shell fulcrum]

Thank you

.close

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[obtuse pebble]

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[odd mortar]

dont u have to flip the inequality sign after dividing?

[odd mortar]

or is that something else

[worn yoke]

only if the number you divide by is negative

[odd mortar]

ah

okay

.close

[obtuse pebble]

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[molten spear]

well technically reciprocal also requires flipping

[ember frost]

to be fair its just simplifying both side, you dont do any "dividing both sides by xyz" here

[subtle sinew]

He was referring to flipping the inequality sign

[molten spear]

yes iirc if you get the reciprocal of both sides you have to flip the inequality sign as well, not just negatives

5 > 3

1/5 < 1/3

[odd mortar]

good to know

[obtuse pebble]

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[woven urchin]

can someone explain to me what is wrong with my proof for de moivre's for rational numbers?

[obtuse pebble]

@woven urchin Has your question been resolved?

[obtuse pebble]

@woven urchin Has your question been resolved?

[woven urchin]

<@&286206848099549185>

[obtuse pebble]

@woven urchin Has your question been resolved?

[tardy epoch]

You need to separate what properties you're given and what your work is

[obtuse pebble]

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[brittle holly]

can i ask for help here

[brittle holly]

apparently at least one of these are wrong

im like 99 percent sure the first three are right

[near kraken]

50

5*

6 too

[brittle holly]

it said the fourth one was wrong and that didnt even maje sense to me

[near kraken]

8

[brittle holly]

5, 6 , and 8 are wrong?

[near kraken]

4 seems right

i think yes

5, def not approaching -4

6, same thing

and 8, it does exist (it's asking for the limit not the value)

there could be others wrong idk

[tawdry lark]

Lim x approaches 3 definitely approaches a value

[brittle holly]

i put dne because i thought it was a double sided that are both different values therefore dne

[near kraken]

we're talking about 8 right?

[tawdry lark]

Yeah number 8

[brittle holly]

yes

[near kraken]

both are approaching 0 as x->3

[tawdry lark]

It approaches 0 from both sides like he said. It doesnt need to be defined at the point for the limit to exist

[brittle holly]

do you always start at the closed dots when moving from left to right or vice versa

[near kraken]

u start from wherever

[brittle holly]

i guess i was just tripping out and i was doing it like this

hold on ill send image

[tawdry lark]

Also 5,6 are wrong, im not sure where you got those answers from

[tawdry lark]

As x approaches 1 from the positive side, ie from the right, what does f(x) approach

[brittle holly]

is this question 2

[tawdry lark]

No 6

[brittle holly]

six says positive one approaching from the right

x approaching positive 1 from the right

[tawdry lark]

Yes

So what does f(x) approach

[brittle holly]

is it also 3

[tawdry lark]

Where did you get 3 from

Dawg look at the graph if you need help

You legit circled the point

[brittle holly]

i already put that

it says its wrong

thats why im confused

[tawdry lark]

What did you put

[brittle holly]

-1

[tawdry lark]

Yes

[brittle holly]

oh wait nvm its right i accidentally added a space after and it doesnt accept that

[tawdry lark]

🤦

[brittle holly]

my b

[tawdry lark]

No not ur bad

Whoever made the website’s bad

Ok now use the same logic approaching from the left for number 5

[brittle holly]

so if not 4 then 3

[tawdry lark]

No dont go guessing again😭

[brittle holly]

im not guessing i just seriously dont understand

[tawdry lark]

Ok, same logic, as x approaches 1 from the negative side, ie from the left, what does f(x) approach

So look at the graph as it gets close to 1 on the left end

[brittle holly]

if its not 3 or 4 idk

[tawdry lark]

The y value dawg

[brittle holly]

like im starting from that closed dot then moving from there to the right, and the function ends at 4 from what i see

o

0.5

[tawdry lark]

Huh

Ok chill out a bit

[brittle holly]

the y value of the closed dot?

[tawdry lark]

First thing is that the graph never even goes to y=4

Second, dont think about the endpoints too much

We are talking about just the limits from one side

So look at the graph and scan across it starting from the left and stop at x=1

[brittle holly]

this is how i thought im supposed to look at it

[tawdry lark]

Ok i see why ur confused

Where is x = 1

Circle where it is

[tawdry lark]

Ok

Now lets define “from the left”

[brittle holly]

ok

[tawdry lark]

A point like (0,-2.5) is to the left of the (1, -1)

When i talk about “from the left”, i mean to the left of x=1, not starting from x=1 and moving left

You need to understand that the x approaches 1 doesn’t mean where you start, but rather where you end up

[brittle holly]

but doesnt the - mean from the left then to the right

[tawdry lark]

Yes

Left to right

[brittle holly]

oh

[tawdry lark]

Until you end up at x=1

[brittle holly]

so i can start from anywhere at that function but must end up at x=1

[tawdry lark]

Yep

More specifically anywhere to the left of x=1

[brittle holly]

so i actually dont even worry about that function on the right side and look at the other one instead

[tawdry lark]

Yessir

[brittle holly]

ok ok one sec

so then\

-2

is the answer

[tawdry lark]

Yep👏

[brittle holly]

awesome thx man

[tawdry lark]

Np

[brittle holly]

i was looking at it completely different and didnt realize it bc somehow i was getting like 85 percent of the problems right

[tawdry lark]

It makes much more sense once you understand what its asking for

[brittle holly]

would you say actually sketching the graph is harder than this

this for example

[tawdry lark]

I mean you dont have to sketch it once you understand it, but when you’re starting out it can be helpful just to visualize your work

I would stick to the graph they give you though; it has all the necessary information

[tawdry lark]

Also dont just calculate using desmos😭

You dont need to do allat

[brittle holly]

well i mean in this problem its not me willingly using it its asking me to use desmos and graph it once i solve

[tawdry lark]

Once you get the hang of it the graphing will practice get more tedious. They probably wont ask you to graph for most problems though; they only ask you to do this for your understanding.

[brittle holly]

ok well idek how to do that so im gonna watch org chem tutor on yt

[tawdry lark]

Ight

[brittle holly]

thx for the help it really made more sense after you explained

[tawdry lark]

Ofc, thats what we’re here for

[brittle holly]

do you hagv

[brittle holly]

do you have any video explaining how to do this

i cant find a video on yt

@tawdry lark

[tawdry lark]

Idk i never did it before

[brittle holly]

am i allowed to ping helpers role\

[brittle holly]

<@&286206848099549185>

[grave ingot]

https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-new/ab-1-5b/v/limits-of-piecewise-functions

[obtuse pebble]

@brittle holly Has your question been resolved?

[obtuse pebble]

@brittle holly Has your question been resolved?

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[ember dagger]

Hi

I've been having trouble with this problem on my test

this is my progress so far

I think I did something wrong?

Idk how to keep going 😅

<@&286206848099549185>

.close

[obtuse pebble]

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[nocturne bronze]

!15min

[obtuse pebble]

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.

[ember dagger]

sorry I forgot, i haven't used the server in a while 😅

.close

[obtuse pebble]

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[unique storm]

i am absolutely lost and have made no progress, ive tried all i can but all my answers have been wrong

[tardy epoch]

What did you get for du?

[unique storm]

-2r dr

[tardy epoch]

What about the new limits

In terms of u

[unique storm]

6 to 375, it said to order them so “that the lower limit integration is less than the upper limit of integration”

[tardy epoch]

,calc 631-25^2

[warm shale]

Result:

6

[tardy epoch]

,calc 631-16^2

[warm shale]

Result:

375

[tardy epoch]

Are they swapped or did you already swap them because of the -sign on -2rdr

[unique storm]

i swapped them already

[tardy epoch]

What did you get for the new integrand

And anti derivative

[unique storm]

i got -1/2 (u)^1/2 du

i cant do square roots so im just doing to the 1/2 power

[tardy epoch]

Why is there a -sign

[unique storm]

is there not supposed to be

[tardy epoch]

.

[unique storm]

that was my last attempt

[tardy epoch]

When you swap the limits, you pick up a -sign

[unique storm]

ok maybe i didnt swap like i thought

[unique storm]

i see that was the issue now

thank you😭

[tardy epoch]

Common mistake

Don't make it on your test

[unique storm]

will not make it again, i was stuck on that for too long

thank you again, how do i close this

[tardy epoch]

Type .close

[unique storm]

okok ty

.Close

.close

[obtuse pebble]

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[zenith marten]

If I have a plastic cup that's shaped like this, how would I use optimisation to find the minimum surface area of plastic to hold a specific volume of liquid V? I was trying to use surface of revolution (integration) and then differentiate to find minimum point, but then there are 3 variables - the radius of the base r, the height of the cup h, and how slanted the sides are (difference between the x coordinate of the top corner and the x coordinate of the bottom corner is t, in the image of the graph). I can't figure out how to do this without having to reduce the number of variables down to 2. I am ignoring the thickness of the plastic (assume it is negligible).

[obtuse pebble]

@zenith marten Has your question been resolved?

[zenith marten]

<@&286206848099549185>

[vale saddle]

hi guys

[tardy epoch]

"shaped like this" is too vague to not use two variables

[zenith marten]

should i just set radius as a constant then

[tardy epoch]

It's your problem I dunno

[zenith marten]

no but is there a way to do it without having to set radius as a constant

[wintry swift]

https://math.stackexchange.com/questions/2802516/optimization-over-multiple-variables-with-a-single-objective

[obtuse pebble]

@zenith marten Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[neon gazelle]

can i ask mat questions here

[deft magnet]

go ahead

[junior granite]

If it is related to maths then sure

[neon gazelle]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[neon gazelle]

1

[deft magnet]

do you have any more information? Like a description

[swift stream]

It looks like the differences of the corresponding top and bottom numbers are taken. This works everywhere except the first 5 in 551

[junior granite]

considering the entire number works

915-364

789-543

or |543-789| whatever

[swift stream]

Yeah it just looks like | top nunber - bottom number |

[waxen mirage]

7×8 +4 = 60

9×9 +9 = 90

[waxen mirage]

So 8×x + 6 = 70

Or x = 8

[neon gazelle]

oh

i was seeing horizontal only

thx XOR-11

.close

[obtuse pebble]

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[obtuse pebble]

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[small sedge]

im having trouble with this question
i looked up this solution but heres the issuehttps://cdn.discordapp.com/attachments/1180601068799918220/1279757359828307999/image.png?ex=66d59a6b&is=66d448eb&hm=f7cb6b0f4346d00f39db9e9d32cd49429f8af1b1266756d698d55e3e018a32c1&

they are dividing the quotient successively, not the number
thats my problem

[waxen mirage]

10 works

Consider the number 10

[small sedge]

oki give me a moment

wait

thank you so much wth

this is way easier

i got the ans 0,0,2 as remainders instantly

[obtuse pebble]

@small sedge Has your question been resolved?

[obtuse pebble]

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[ionic oar]

"Let there be a square with sides a + 6, if we cut out an inner square if sides a - 1. if the shape we left with has an area equal to another square with sides k, find the minimum of a + k (a and k are positive integers greater than 1)

[ionic oar]

i got that 14a + 35 = k^2 and the possible (a,k) are (1,7)and (29,21)

but a > 1 so a + k = 50?

im wondering if i missed any a less than 29

[fiery sinew]

can i ask this chapter is based on differentiation or other topic? since finding minimum can have many ways depending on the topic of that chapter

[ionic oar]

this question was included in a standard math education test for 8th grade

[fiery sinew]

ok

[ionic oar]

so i dont think differentiating is used

but id still want to know how to do it with differentiating

[fiery sinew]

cant a be 1?

wait nvm

iam dumb lol

i think u need other ppl to help you, i have no idea on this question

[static furnace]

did you try 14?

its a hunch

[ionic oar]

231 isnt a square

<@&286206848099549185>

[static furnace]

this is probably correct

[shadow copper]

Bengali?

[ionic oar]

its thai

[ionic oar]

is there any faster way to get to the solution than what i did? (testing different a's and so on)

[static furnace]

well

if you rewrite your equation you get a=(k^2)/14-35

and since a and k have to be integers

and 35 is an integer

than k^2/14 must be an integer

[ionic oar]

oh

[shadow copper]

It's a quadratic so you prolly have all of them

[ionic oar]

then k must be a multiple of 7

[static furnace]

right

and since 14 and 7 dont work

then k must be 21

[ionic oar]

thanks

i think this question would be too hard for a standard math test, no?

cause most people arent even taught (a + b)^2

[static furnace]

it is quite methodical

[static furnace]

then this question would be quite difficult, yes

[ionic oar]

anyways thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[surreal pike]

how is (1/k) from 0 to infinity a diverging sequence?

[azure wraith]

Its harmonic

And the partial sum always increases

[opaque dome]

So 1 not 0 becareful

k can't be 0

[tender tusk]

i think a common proof is to take the series and group the terms 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ... and you can see that each of the groups are greater than 1/2 and thus the series is divergent

[surreal pike]

yeah sorry my bad

[surreal pike]

but the series is bounded from below right?

[high lily]

don't conflate sequence with series

[opaque dome]

Series is the sum

Of the terms

[surreal pike]

im sorry, im pretty new to the concept

[opaque dome]

Sequence is the "list" of the terms

[opaque dome]

No prob you're here to learn, don't be sorry

[surreal pike]

okayyyyy

so, the sequence is bounded below right? like it cant go below a certain point. also the limit of the inifite'th term is 0
and the sequence is monotonically decreasing, why then is it not considered converging?

[opaque dome]

Its bounded above too, by 1. And it is converging

The sequence is converging, but the serie isin't

[surreal pike]

ooh so

the sum will never converge to a finite value

but the terms will?

[opaque dome]

Yes

[surreal pike]

okayyy i think i get it

i was just stuck on this part

thank you!

[opaque dome]

And btw

If you want a different case, where both converges (sequence and serie), you have for integers k non zero, 1/k^2

[surreal pike]

how do you prove that?

i can see the series converging

[opaque dome]

To pi^2 /6 indeed

[opaque dome]

That the sequence converges or the serie converges ?

[surreal pike]

series

wait

can we do it with cauchy's principle?

[opaque dome]

It is known as basel problem, if you're interested you can look into https://en.wikipedia.org/wiki/Basel_problem

[surreal pike]

okay thanks

[opaque dome]

There are videos on yt too

3b1b done one

[surreal pike]

really? 💀

i'll look into it, thanks!

[opaque dome]

You're welcome

[surreal pike]

.close

[obtuse pebble]

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[restive coral]

Can someone explain how to get to this answer

[drifting wraith]

| x − a | is distance between x and a

so it's points that are equally far from −321 and 750, so there should be one like that

[restive coral]

but why equally far between -321 and 750

[restive gorge]

,, \abs{x} = \sqrt{x^2}

[warm shale]

bacc

[drifting wraith]

wut

[restive coral]

what

[restive gorge]

Then you can square both sides and solve the quadratic equation

[drifting wraith]

oh okay

[high lily]

there's no need to apply |x| = sqrt(x^2)
you can square directly if you want to go that route

nvm, just thought if it differently

[restive gorge]

How is |x| injective

[tranquil geyser]

What about x=-y

[tranquil geyser]

It isn't

[restive gorge]

Yes

[tranquil geyser]

Let x+321=-(x-750)

x+321=x-750 is not a valid option I think 🤔

[tranquil geyser]

@restive coral

):

[restive coral]

I am thinking but have no idea

[tranquil geyser]

Absolute value gives the same output for -z and z (z is a number)

So I picked the negative value

[ripe shard]

id say the easiest way is to draw a quick diagram to see which branches u need to equate

doesnt have to be to scale

[restive coral]

but what should I do with this

let x+321=-(x+750)

[tranquil geyser]

Solve for x from here

[tranquil geyser]

Yes

[tranquil geyser]

Hmmm what do you mean? 🤔

[ripe shard]

nah like there are multiple solutions right

so like a quick graph of the two graph will show u where they intecept

so u know what branches of each graph u take

[restive coral]

x = -535.5?

x + 321 = -x-750

2x = -1071

x = -535.5

[tranquil geyser]

I re checked the question

It says|x-750|

Not |x+750|

MB

[restive coral]

😭

[tranquil geyser]

#help-10 message

I corrected the error 😞

😞

Now try it again @restive coral

😅

[restive coral]

214.5

lets GO

@tranquil geyser

[tranquil geyser]

Yay ✋

[restive coral]

can you explain the -(x-750) for a very big NOOB

ELI5

why the -(x-750)

[tranquil geyser]

Well |-x| is...

[restive coral]

|x-750| translates TO -(x-750) ??

[tranquil geyser]

No

[restive coral]

Whaaaaaaaaaat

[restive gorge]

If you consider the cases where both are ++ or -- then you got no solution but a wrong statement cause x falls off

[tranquil geyser]

If |z|=|x-750|

Then z=±(x-750)

[restive coral]

.close

[obtuse pebble]

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[azure wraith]

How did go from this to this

[warm shale]

bacc

[azure wraith]

How did u get this inside the log

[restive gorge]

Actually

The more I look at it the more it confuses me

Either it's expressed bad or it's just wrong

[warm shale]

bacc

[azure wraith]

Also

Uhh

Try considering the limit definition of e

[chrome arch]

I'm confused in that as well

I asked for the solution from someone

[restive gorge]

The step is just wrong, judging by the next lines

[chrome arch]

Howcome the answer is correct

@red ice

Could you help me understand this?

[red ice]

pls don't ping me

[red ice]

no thx

[chrome arch]

Alr

[chrome arch]

I'm sorry

[restive gorge]

I would consinder a substitution

,, \lim_{x \to 0} = (1+2x)^{\frac{1}{2x}} = \lim_{u \to \infty} = \left (1+\frac{1}{u} \right )^{u}

[warm shale]

bacc

[restive gorge]

This should let you understand whats going on in the numerator

[chrome arch]

I don't understand

I'll just use the expansion for it

[restive gorge]

Thats literally the limit definition of e

The way I wrote it

You basically have 0/0

[chrome arch]

<@&286206848099549185>

[restive gorge]

Now I understand south

[chrome arch]

Wdym ?

[obtuse pebble]

@chrome arch Has your question been resolved?

[chrome arch]

.close

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[prime imp]

hi so can someone explain what is inverse functions such as arcosine and artan and arsine?
idk really get those

[stone holly]

yes so

we know that

for eg.
sin(π/6) = 1/2
so now understand like
arc sin (1/2) or sin inverse (1/2) = π/6

in short arcsinx or arccosx is an angle for which you get x

so if,
a ==> some constant b/w [-1, 1]
sin x = a

arc sin(a) would be equal to x

(there could be multiple solution, but we take principle values)

[haughty stump]

When trying to solve sin(π) you are trying to find what sin is at the angle π

When trying to solve arcsin(0) you are trying to find what the starting angle is if sin is equal to 0

Since there are multiple places where sin can be 0 you COULD have multiple solutions. However, it depends on the range you are given, but more likely than not for arcsin it will be between -π/2 and π/2. This means the only solution within this range for arcsin(0) would be 0.

TLDR: for sin you are just trying to find what sin is given the starting angle and for arcsin you are trying to find the starting angle given what sin is.

[obtuse pebble]

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[restive coral]

can someone explain how this is even possible

[haughty stump]

Alright so you need to know how radicals work first

rad(3x^2) is the same as rad(3) * rad(x^2)

and rad(x^2) is just x

so they just took out the x

[restive coral]

ahhh so it is sqrt(3) * x?

not sqrt(3x)

[haughty stump]

ye, it matters what the line is above

you can see that it isnt covering the x, only the 3

[marsh geyser]

Do you have more context?

I cant see the domain

Is it positive numbers?

[haughty stump]

i think he was just asking how rad(3x^2) turned into rad(3)*x

but he didnt realize it was rad(3)*x and instead thought it was rad(3x)

hence underlined part

[marsh geyser]

But sqrt(x^2) is not generally x

But |x|

That is why i was asking for more context

[restive coral]

this is the context Write the following expressions, in which x represents a positive real number, as simple as
possible:

(sqrt(3x) - sqrt(x))/(sqrt(3x) + sqrt(x))

[marsh geyser]

Perfect

[alpine bison]

don be to pointy

[marsh geyser]

Positive was the info missing for me

[marsh geyser]

Accurate

[restive coral]

I dont understand what just happened

[alpine bison]

hahah right

[restive coral]

I understand what @haughty stump said but not what you (@samuel) said

[haughty stump]

Oh hes just being technical

Sometimes sqrt(x^2) can only be one thing

if x was given as either a positive or negative integer

[marsh geyser]

When you have only positive numbers, you can indeed say that sqrt(x^2) = x

But nor when you have whole domain

Because for example, sqrt(3^2) = sqrt((-3)^2)

But -3 is not = to 3

[restive coral]

ohhh

yeah

I understand

[marsh geyser]

Instead |3| = |-3|

[restive coral]

yes

Thanks

.close

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[limpid eagle]

I'm currently trying to learn the delta-epsilon proof (which is killing my mind), and in this example with quadratics, I think I loosely understand everything up to the part where I highlighted, which I do not see the logic behind "bounding" |x+3|, what does bounding even mean in this situation? Also if I can get help with the rest of the proof after that, that would be great too

[unreal musk]

You want the |x + 3| term to be at most "something", so that you can "ignore" it

You can then replace the |x + 3| in |x + 3||x - 3| with its upper bound (with the appropriate inequality), and then just deal with what you end up finding, which'll only have the |x - 3| to work with

In their case, they choose to force delta to be at most 1, which means that |x + 3| < 7, so you then can say that |x + 3||x - 3| < 7|x - 3|

[limpid eagle]

how did they go from delta <=1 to -1 < x-3 < 1?

[hexed gull]

|x-3| < delta

and delta <= 1

imply |x-3| < 1, which you can rewrite as -1 < x-3 < 1

[unreal musk]

They did the inequality chains here: the fact that x + 3 is strictly greater than 5 means it's [x + 3 is] positive, so x + 3 = |x + 3|

[limpid eagle]

so they took the |x-3| < delta from the 0 < |x-3| < delta and kinda just ignored the 0 part?

[hexed gull]

they didn't ignore it, they just rewrote the second inequality

[obtuse pebble]

@limpid eagle Has your question been resolved?

[unreal musk]

(as an alternate way, you can use the triangle inequality to say |x + 3| = |(x - 3) + 6| <= |x - 3| + 6 < 1 + 6)

[obtuse pebble]

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[fresh bay]

how do u do this

[fresh bay]

i just dont get the pi thing

[lusty sentinel]

The bounds indicate the quadrant. sin theta indicates the ratio of the opposite side to the hypotenuse of a supposed triangle. Use Pythagoras to find the last side and then solve for tan and cos theta

the values of sin cos and tan change depending on the quadrant they are on

[fresh bay]

so first one is quadrant 2

what is my anwser suppose to look like

[lusty sentinel]

yes

you're supposed to make a triangle in that quadrant of the unit circle

and then use the information to make a right triangle

[lusty sentinel]

solve for cos theta and tan theta

[obtuse pebble]

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[late roost]

Quick question:

[late roost]

If the derivitive of 0 is 0 why is it und at that point?

Like i understand it graphically but idk the mathamatical reason

The 3rd x in the first piece wise is = 0 btw this is just modeling abs(x)

[civic zealot]

because the derivative at 0 isn't 0.
If you look at the limit definition of the derivative at x = 0, then you get different values for the left and right limits. So the limit is undefined, and thus the derivative is undefined.

[late roost]

ah

Ok wait do you have a resource that defines differentiability?

like rules and such

[civic zealot]

This is the classic example for why differentiability says a function must be smooth and continuous to be differentiable.

[late roost]

stuff like this lol

I cant find one

[civic zealot]

any calc book would have it.

https://openstax.org/books/calculus-volume-1/pages/3-1-defining-the-derivative

[late roost]

Alr bet thank you

.close

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[late roost]

ok wait

.reopen

[obtuse pebble]

✅

[late roost]

Wont you get 0 coming from both sides?

Limit as x -> 0 of -x is 0 (left)

and of right is also 0

[civic zealot]

for |x|, yes. Not the derivative limit though.

[late roost]

ahhhhh

So if i wanted to prove this to myself mathamatically

how would I show it

[civic zealot]

you basically get $lim_{h\to0} \frac{|h|}{h}$ which goes to 1 for h>0 and -1 for h<0
So the left and right limits disagree.

[warm shale]

Zybikron

[late roost]

Ok but whats the reason for even taking the limit, for a function to be differenciable does the limit need to exist at that point too?

Or is that just continunity

[civic zealot]

if $f(x)$ is a function, then the derivative of $f(x)$ at $x=a$ is $f'(a) = \lim_{h\to 0}\frac{f(a+h) - f(a)}{h}$.\
here we have a = 0

[warm shale]

Zybikron

[late roost]

h is just deta(x) right

[civic zealot]

yes

[late roost]

ok so breaking this down

what would delta x be in this case

oh

[civic zealot]

the value we're taking to 0 for the limit

[late roost]

it would be |x+h|

so |x+h| - |x|

over h

[civic zealot]

yes

[late roost]

and take that limit to 0

one sec ill work it out

[civic zealot]

since we only care about the x = 0 point, you can just plug in x=0 into that fraction

yes

[late roost]

forget i said that

I plugged it in for x

by mistake lmfao

ok it would be 0/h

or 0/0

[civic zealot]

no. It would be |h|/h

and then the limit as h goes to 0

[late roost]

so just 0 / 0

[civic zealot]

which is indeterminant. That doesn't tell you anything.

[late roost]

you cant h'lopitals it

[civic zealot]

you'd need to do the derivative of |h|, which is the problem to begin with.

[late roost]

ok sorry im more lost

just to break it down

The def of a derivitive is this:

so we are trying to define it a f(0)

[civic zealot]

yes

[late roost]

which leaves us with this :

we cant h'lopitals it because it would just keep being 0/0?

[civic zealot]

you'd need to take the derivative of |h| at 0, which is the problem you're trying to solve.

[late roost]

ok going back what does h represent

is it legit the derivitive

like slope of x

[civic zealot]

yes.

[late roost]

ok that makes much more sense

sorry lol thank you for the patience

[civic zealot]

I'm saying, to do l'hopital. You have to do the derivative of the top and bottom of |h| / h
So you'd need to find the derivative of |h|, which is the problem you started with (finding the derivative of |x|)

[late roost]

Yeah i get it

so just to make sure

h = delta(x) = derivitive of f(x)

[civic zealot]

no

[late roost]

oh

[late roost]

wait so