help-23

[glass gust]

i cancel out on every term?

[stray socket]

_ _

At this point it's just division

[glass gust]

delx+2x-delx

2x

[stray socket]

Not - delx

How does -2delx/delx become -del x

[glass gust]

2/1=2 so -2delx lol

yeah im just so lost rn

the point where 2/1=1

[stray socket]

Not -2delx

$\frac{-2\Delta x}{\Delta x}$

[flat frigate]

Umbraleviathan

[glass gust]

wha

wait

-2

because delx cancels out

[stray socket]

Yeah

So you're left with delx + 2x - 2

The limit is Del x to 0

[glass gust]

so 2x-2

[stray socket]

Yeah

[glass gust]

i feel like it wasnt as hard to come to this conclusion as i made it out to be

[stray socket]

No it isn't, it's just a lot of tedious work

Because like

If you get really long polynomials

f(x) = x^5 - 3x^4 + ....

[glass gust]

yeah i see that now]

thanks a lot for the help

.close

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[onyx pike]

so for the first, i think it's fair to say vector u is <0,1>

for the second, i know i must have sqrt49

for the absolute value of |V| to be 7

[cold aurora]

You don't need to really know exact component of the vectors.

[onyx pike]

so for V, a must be sqrt49/2, and b must also be sqrt49/2

[cold aurora]

You can find them out if you want surely but you don't have to.

[onyx pike]

okay so my linear algebra is rusty

id like to see it your way

[cold aurora]

As long as sqrt is over (49/2), and not just 49.

[onyx pike]

haha yea

but how would you solve it?

[cold aurora]

Well, do you know what the direction will be?

[onyx pike]

northwest

right?

oh sorry

[cold aurora]

No. I meant the direction of cross product.

[onyx pike]

hmm

i don't

so we have these two vectors, one north and one northeast

wouldn't the cross product give a vector orthogonal to both?

[cold aurora]

It would.

Simply by the definition of cross product.

[onyx pike]

so by my limited imagination

i imagine it going into the page or out of it

but thats not right

HmMmM

[cold aurora]

That is correct.

[cold aurora]

Why not?

[onyx pike]

well it's not what they're looking for, at least

\

since these are my options

[cold aurora]

I don't think so.

[onyx pike]

OH

UP OR DOWN

LOL

[cold aurora]

Up and down mean "above out of the page" or "downward into the page" respectively.

[onyx pike]

xD

omg

it's been a long day.

so for the magnitude is there a better way?

[cold aurora]

It's alright.

Well, first tell me. Which direction is it?

Upward or downward?

[onyx pike]

well couldn't it be both?

they're both orthogonal, right?

[cold aurora]

No.

A vector has a unique direction.

[onyx pike]

OH

its upward?

is it the left hand rule

or right hand rule

[cold aurora]

In case you didn't know this - you can use right hand rule to find it.

[onyx pike]

so its downward

[cold aurora]

Align your extended finger towards u and curl them in the direction you'll have to rotate the vector u to move towards v, now, direction of thumb will give direction of cross product.

[onyx pike]

ahhh thank you friend

[cold aurora]

For magnitude, Magnitude of cross product is just product of magnitudes multiplied with sin(theta).

You know that north and northeast are at 45° angle to each other.

[onyx pike]

ah yes!

i always hated how mechanical linear algebra had become to me

i wish i had more geometric intuition

and it still is

clearly

[cold aurora]

Well, you can keep trying to relate. It'll slowly become more intuitive.

[onyx pike]

thank you for your help.

.close

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[onyx pike]

.reopen

[safe radish]

✅

[onyx pike]

OH MY GOD THE MAGNITUDE IS 7sqrt(2)/2

but that's just sqrt(49/2), which is the distance of each component

[cold aurora]

That's what you got.

[onyx pike]

is that just a coincidence?

maybe i did it wrong

[cold aurora]

Yes because other vector is just unit vector.

[cold aurora]

No. Lol.

[onyx pike]

for vector v

[cold aurora]

You got $\sqrt{\frac {49}{2}}$, which is same as $\frac{7}{\sqrt{2}}$, which is same as $\frac{7\sqrt{2}}{2}$.

[flat frigate]

Enemagneto

[onyx pike]

bro im tired

[cold aurora]

Maybe try later. Catch some rest for now.

[onyx pike]

answer was right tho. ill think about it later.

thanks again.

goodnight

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[tardy copper]

Can someone help explain what formula to use in this problem? I dont know how my prof set it up and my set up seams completely incorrect:

[safe radish]

@tardy copper Has your question been resolved?

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[prime nacelle]

I'm not quite sure how to prove the identity and what to do with the RHS, this is all i had

[prime nacelle]

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[ripe plover]

6(3^x-1) = 3^4-3^x. I need to solve for x. Im quite confused on how to make both sides of the exponential equations base the same.

[hazy elbow]

Well, you can do a substitution to transform that into a linear equation

[ripe plover]

could you elaborate?

[hazy elbow]

set 3^x=y

And now try to solve for y

[ripe plover]

oh

ok thank you.

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[past birch]

can someone please tell me what did i do wrong? the answer is 24000*cube root 2

[past birch]

question 15

[hazy elbow]

,rccw

[hazy elbow]

well you minimized the surface area

you need to minimize the cost

[past birch]

oh

wait i think i know what i did wrong

[past birch]

cost = 500 times 2x^2 + 800 times 4xh?

[hazy elbow]

do you mean 500*2*(x^2)+800*4*h*x

[past birch]

yes

howd you do that?

[hazy elbow]

use \*

[past birch]

500*2x^2+800*4xh

oh it works

thanks

[hazy elbow]

welcome

[past birch]

is there any way to simplify 10000*32^(2/3) + 64000*32^(-1/3)?

it is the right answer but i didnt learn law of indices yet

[safe radish]

@past birch Has your question been resolved?

[ripe plover]

Yes, it is possible but by using indices. You could do it like this. 32^2/3 can be simplified into (2^5)^2/3. The powers will multiply, 5 and 2/3. Take the lcm and you will get the answer power. You can do the same thing 64000*32^(-1/3), but in this you will also simplify 64000 into 40^3.

[past birch]

I got to 10000*2^(10/3)+64000*2^(-5/3)

[safe radish]

@past birch Has your question been resolved?

[safe radish]

@past birch Has your question been resolved?

[safe radish]

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[safe radish]

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[fleet current]

hello

can someone help me with part 2

i was doing an approach but i felt it was wrong due to the structure of question

i substituted it above

in right part of equation

and said i (f hat(n)- f hat(-n))=0 since f is real valued

thenn f hat(n)=fhat(-n)

oh wait

nvm

i said that since f is real valued then f hat = conjugate of f hat which is clearly wrong🤣

can someone help

[safe radish]

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[twilit rose]

How do I find the domain and range of this? I need to put it in interval notation

[normal moss]

Do you know what the domain and range of a function tell you?

[twilit rose]

I think it’s what can be used as x

Would the domain of this be [-6, -1/6] U [1/6, 6]?

[normal moss]

You have black dots.

[twilit rose]

Oh right

[normal moss]

Correct.

[twilit rose]

Ok and I think the range is y values so I’ll see if I can figure that out

[-6, -1/6] U [1/6, 6]?

[normal moss]

The lower number should be on the left

[twilit rose]

Oh right I was reading from left to right not bottom to top

[normal moss]

Gotta to the same for the second interval 🙃

[twilit rose]

Oh I thought since it was above 6

[normal moss]

There you go

The lower number in an interval should be on the left 😄

[twilit rose]

Oh ok

[normal moss]

(x1, x2); x1<x2

But yeah you go it

So when you are searching for the domain

You look for x's

And the range are the y's which are represented by the function

[twilit rose]

So I’m this case the domain is the same as the range?

[normal moss]

Correct

[twilit rose]

Huh neat

Thank you for helping 🙂

[normal moss]

No problem 🙂 Have a good day

[twilit rose]

You too

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[onyx pike]

i mean i could find a vector passiong through the points by (x2-x1,y2-y1,z2-z1) right?

but how would i make it parametric?

just add t? that cant be right.

or would the resultant vector, each component multiplied by t and added to either point be the vector im looking for?

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[fickle trail]

wouldn't +C be required for second line too? since it's an indefinite integral

f(x) is the antiderivative of f'(x)?
or is it
f(x) + C is the antiderivative of f'(x)?

which of these is correct?

[final halo]

Honestly at this point I think the best thing for you to do is to take "antiderivative" and "indefinite integral" so be synonymous

This is all very superficial, just call anything that differentiates to your original function the antiderivative AND the indefinite integral and be done with it

And just always write +C for them

[safe radish]

@fickle trail Has your question been resolved?

[fickle trail]

OK

[fickle trail]

Thank you

so it is my understanding...
definite integrals leave you with a number for the result

indefinite integrals leave you with a variable of some sort and +C for the result

[final halo]

Good enough

[fickle trail]

https://tenor.com/view/howareyou-not-perfect-but-good-gif-9262123

also, I don't understand why variable to the power 0 is considered "bad practice" for using power rule with constants

[final halo]

Inefficient use of time to do that every single time

[fickle trail]

meh, i'm not looking to break any speedrun records on my exams

[final halo]

Exams are fundamentally timed events lmao

[final halo]

You also lost the integral symbol and dx after your first equals

[fickle trail]

but not to this extreme: https://youtu.be/zLphAlLQNiE?si=lFmQE96rmCXAJIMF&t=53

[fickle trail]

LOL, this is why I don't use Organic Chemistry Tutor

it's screenshot from his video

[final halo]

You want every second you can get in an exam to think about the actual hard things. Writing x⁰ everywhere is going to rob you of those seconds

This is why videos shouldn't be your primary source for "information" [note I did not say learning]

Texts are more easy to be reviewed and edited

[fickle trail]

if the video is not quite correct, and if they are good teacher, they will annotate the correction at the timestamp

unfortunately for OCT videos, they tend to miss out on that, so I gotta be careful to take as a grain of salt when watching them

[safe radish]

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[onyx pike]

what am i not understanding

[icy lance]

you need to write parametric equations

eg x(t)=0-5t=-5t

[onyx pike]

hmm so i dont want you to just "give" me the ansswer obv but it seems ambiguous to me. if they want the equations when t=0 isn't x(t)=0 and y(t)=3?

etc

im clearly not understanding something fundamental

[icy lance]

it doesnt ask for it at t=0

they say that it will pass through p WHEN t=0

[onyx pike]

okay that makes a lot of sense. because otherwise it would just be a line orthogonal to the plane?

im probably wrong there too

[icy lance]

if youre doing t=0 then you would just be writing the coordinates of p

it wouldnt be a line, just a point

[onyx pike]

ah

thank you again xd

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[open sleet]

Could someone help me understand this question?

[open sleet]

From looking at it I see that it is lower triangular if x = 0

what would my answer look like? I don't think we covered ones with variables

[safe radish]

@open sleet Has your question been resolved?

[open sleet]

<@&286206848099549185>

[open sleet]

.close

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[digital bobcat]

How do i do this surd question

[digital bobcat]

do I divide or multipy

[normal moss]

I would do neither

[digital bobcat]

Is it ratoinalizing

[normal moss]

I would first kick the minuses out and put it infront of the whole fraction

((-a)*(-b))/(-c) = - ((a*b)/c)

[digital bobcat]

you mean making it positive?

[normal moss]

I mean, it's still negative but yeah

[digital bobcat]

Bro we not making it out with passing grades 😭

[normal moss]

You will, believe in yourself!

[digital bobcat]

ok if i kick the minusus what next

[normal moss]

Anyway, once you do that there are some things to be canceled out and simplified

[digital bobcat]

what do i do with the bottom number

[normal moss]

You have $-\frac{12\sqrt{6}*5\sqrt{10}}{3\sqrt{15}}$

[digital bobcat]

is that the final answer

[normal moss]

No

[flat frigate]

USS-Enterprise

[digital bobcat]

ohh so you just have to make it postivie

oh

[normal moss]

Well

There's a minus in front of the fraction

So it's still negative

[digital bobcat]

OHH

[normal moss]

But do you see something that can be canceled out

[digital bobcat]

honeslty no

but arent surds postiive

[gentle oak]

$-\dfrac{12\sqrt{3}\sqrt{2}\cdot5\sqrt{2}\sqrt{5}}{3\sqrt{3}\sqrt{5}}$ and if you write it like this

[flat frigate]

Joseph.P

[digital bobcat]

WHAT IS THAT

[normal moss]

Equivalent to the previous one

Just expanded the root

[digital bobcat]

OHH I GET IT

[gentle oak]

the square root

[normal moss]

Oops

I meant the square root yeah

😂

And you see how everything is just being multiplied

[digital bobcat]

so cancel out the minsus andddd

[normal moss]

This is a prime time to cancel things out

What minus

[digital bobcat]

you said cancel out the minsus

at the start

[normal moss]

No

I said kick out the minus to the front

[digital bobcat]

why though

[normal moss]

Because it looks much neater and is easier to work with

[digital bobcat]

so easier to break down?

[normal moss]

A suggestion, if you ever have just a buttload of minuses see if you can factor them out

Yes

So what would you do now?

[digital bobcat]

multipy

multiply

cause on the paper it says mulitpying and dividing surds

[normal moss]

If you have $\frac{12x}{3}$

[flat frigate]

USS-Enterprise

[normal moss]

Can you cancel something out

[digital bobcat]

the x

[normal moss]

What

[digital bobcat]

huh

[shadow condor]

the 12?

[normal moss]

How about this

[digital bobcat]

guys its joever

[normal moss]

$\frac{43x}{3}$

[flat frigate]

USS-Enterprise

[shadow condor]

yeah that's a good visual

[digital bobcat]

oh so you break it down

[normal moss]

What can be canceled out here

[digital bobcat]

the 4x3 x to just be 12x

cause 4 * 3 is 12

[normal moss]

??

[digital bobcat]

STOP IM TRYING

[flat frigate]

USS-Enterprise

[normal moss]

Oops

[shadow condor]

what are you guys trying to do rn?

[digital bobcat]

to see what to cancel out

[normal moss]

That was a mistake

[digital bobcat]

its okay

[shadow condor]

is this the example?

[normal moss]

Yes

[digital bobcat]

what does that have to do with the original queston?

are we practicing singiling things out

[shadow condor]

looks like an example to learn how to simplify liketerms or smth idk

[normal moss]

You don't see 3 can be canceled out here

[digital bobcat]

yes

[normal moss]

Well then

[shadow condor]

would that need any parenthesis or nah
for the expanded square roots

[normal moss]

And so can sqrt(3) and sqrt(5)

And your whole denominator is gone

[digital bobcat]

OH

[normal moss]

Everything is being multiplied

And you are left with $-(4*\sqrt{2}5\sqrt{2})$

[shadow condor]

yeah nvm i forgor lol

[digital bobcat]

is tht the final answer

[flat frigate]

USS-Enterprise

[normal moss]

Do you not see how things can be multiplied here

[digital bobcat]

yes

[normal moss]

Or are you just trolling

[digital bobcat]

i swear im not im trying i have math test tommorow

we times the radicals right

4 * 5 is 10 2 * 2 is 4

thats what i learnt in school

[normal moss]

4 * 5 ≠ 10

[digital bobcat]

yes i did that

[normal moss]

No

And you have a square root squared

Which is just the number inside

[digital bobcat]

OMG I READ THAT AS 2 * 5

ITS 20

[normal moss]

Yes

[digital bobcat]

sorry my bad

[normal moss]

$-(20*(\sqrt{2})^2)$

[flat frigate]

USS-Enterprise

[normal moss]

And that is ...

[digital bobcat]

40 to the power of 2

its not is it

[normal moss]

What

[digital bobcat]

YOURE ASKING ME THE ANSWER

[normal moss]

It's not 1,600, no

[digital bobcat]

isnt there an invisible 1

oh i did that wrong

20* 2 is 40

[normal moss]

I am having fun too

[digital bobcat]

i bet you arent

[normal moss]

Mhm

[digital bobcat]

i have a rock as a brain

[normal moss]

How come

[digital bobcat]

it does not intake knowledge

everyhting but numbers

[normal moss]

Well

Anyway

[digital bobcat]

ITS ALWAYS MATH

[normal moss]

We got to the answer

-40

[digital bobcat]

YAY

[normal moss]

Took a while

Good luck on your test tomorrow

[digital bobcat]

thank you

[normal moss]

On sunday

[digital bobcat]

i will ask my teacher to revise this

thanks internet man

yes on SUNDAYS

not mondays

weird right

[normal moss]

👍

I don't think your teacher needs to revise it

Just you do

[digital bobcat]

no to help me

we have 10 mintures to revise and she helps us

answe questions and stuff

ok back to grinding

[shadow condor]

oh ok i get this

i havent done radicals stuff in ages

[digital bobcat]

if i saw radicals in real life i would beat him up

[shadow condor]

lol

did you end up understanding the problem

[digital bobcat]

i understood that you need to break it down and stuff

but its a 50/50

[shadow condor]

how about the process

[digital bobcat]

process of what

[shadow condor]

of solving the thing

start to finish

[digital bobcat]

bit confusing but when my rock brain put it together it made some sense

im doing multipying radicalsnow

[shadow condor]

did you have any questions left about it
just to know

[digital bobcat]

i will just ask my teacher

she is smart

[shadow condor]

that's nice

[digital bobcat]

did i do this right

multipyling radials

ricals

i need confirmation

forgot to add the sign for the 3

imagin its there

[shadow condor]

i think it would be 3sqrt(12)
and then you can simplify the sqrt(12) thing idk

also ik this might not matter anymore but this is what i understood of the other thing

[digital bobcat]

OH DO YOU SUBRTACT

whre did you get the 4 frmo

[shadow condor]

12/3

[digital bobcat]

OH

[shadow condor]

the thing the other guy was trying to say with 12x/3

[digital bobcat]

oh i forgot about the diving thing

[shadow condor]

is that you can do the like terms and then apply the other weird stuff to them

12x/3 means
4x + 4x + 4x

[digital bobcat]

so if its the same number under you cancel it out

[shadow condor]

yeah kinda

[digital bobcat]

ya that confused me a wee bit

[shadow condor]

the radicals kinda work like variables

which is why i canceled them separately

[digital bobcat]

OKAY I UNDERSTNAD NOW

[shadow condor]

swag

[digital bobcat]

i will try the other queston and i will chekc

[shadow condor]

and then you check if the sqrt of 12 can be simplified

yeah i rember now lol

sqrt of 12 can be turned into sqrt of 4 and sqrt of 3

sqrt of 4 turns into 2

combine like terms and you multiply 3 by 2
that leaves you with 6 * sqrt(3)

cause that sqrt cant be simplified further

[digital bobcat]

so you simplyify down till you cant?

[shadow condor]

ye

[digital bobcat]

and that is yor answe

[shadow condor]

for radicals

ye

[digital bobcat]

but this one is differnt

[shadow condor]

damn that's some stuff fr

[digital bobcat]

THATS 2 I KNOW THAT

[shadow condor]

first you do the freebies with that huge division

like simplify some stuff ig
the sqrt(3) squared is 3

[digital bobcat]

i forgot to add the other bracket

[shadow condor]

i love ms paint fr

[digital bobcat]

okay what i got frmo that is the 40 has something to do with the 20

and the 5

they are very suspicous

[shadow condor]

did you put the other parenthesis

that it's missing

idk

to make sure everything is cool and ready for the solving

[digital bobcat]

i added he missing thngs

[shadow condor]

oh alr

first we make the three thingy a 3

[digital bobcat]

but why

cause of the little 2?

[shadow condor]

cause it looks nicer i guess

yeah

you know how

[digital bobcat]

oh to make it easier

[shadow condor]

you cancel terms and stuff with opposite operations?

square roots can get canceled by exponents of 2

[digital bobcat]

i remember we cancel out stuff and break it down to make it easier to do the eqatuin

[shadow condor]

the little 2

[digital bobcat]

yes

now do we cancel the 2 threes

[shadow condor]

a little visual to make it understandable

the exponent of 2 cancels a square root

they're opposites

sort of

but it works

[digital bobcat]

OHH

my internet is getting slow

okay okay

[shadow condor]

that's why we have this now so far

[digital bobcat]

now next step is to do the other brackets

[shadow condor]

now we should try and break down the square root of 40 and 20

i think

[digital bobcat]

thats 8

[shadow condor]

we wont touch the sqrt of 2 cause it's cool like that

so imma try that out rn and ill show what i get

ig imma try the sqrt(40) first

[digital bobcat]

8

cause 5 * 40 is 8

[shadow condor]

5 * 40 or
5* 8

[digital bobcat]

oops

yes

[shadow condor]

watch out with that

it happens

[digital bobcat]

its 2 am its oky

[shadow condor]

so we wanna look for the two numbers that give 40

there is 5 and 8

but

[digital bobcat]

THE 20

[shadow condor]

none of their square roots are clean, whole numbers

[digital bobcat]

AT THE BOTTOM

[digital bobcat]

oh

[shadow condor]

yeah we gotta be patient and get the weird numbers into easier cooler looking ones

[digital bobcat]

yes i like that metaphor

[shadow condor]

so far i have this rn

from here ig imma break down the square root of 10 again to match the 5 at the bottom

[digital bobcat]

yes

[shadow condor]

to see if we can do funny cancels

[digital bobcat]

thinknig

the negative saure 2

sqaure

[digital bobcat]

dont we simplyigy th 10

oh. no

NOW WE DIVIE IT

[shadow condor]

i think i see like two ways to solve this

[digital bobcat]

what is the easiest one

[shadow condor]

i wanna make sure it gives a proper answer before telling you

to not confuse you

[digital bobcat]

yes me and my pea brain

i swear im good at other stuff than math

[shadow condor]

happens
skillsets vary

[digital bobcat]

trust me when i get griding it never stops and i will be so good at math ill be helping other people

if i stop responding i fell asleep ok

i will send a letter when i m back

[shadow condor]

ok the first method i tried doesnt work so imma try the other

@digital bobcatalr so

that's much better

then

[safe radish]

@digital bobcat Has your question been resolved?

[shadow condor]

hopefully it helps

[safe radish]

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[frigid ocean]

@final halo
So, essentially what we talked about like 1minute ago, this would mean that even thoug hthose function looked a lot like each other this has both a positive and negative value, which by our theorem means (If the eigenvalues are all nonzero but mixed, then Hf is a saddle point) it's a saddle point?

[final halo]

The second was a saddle yeah

[frigid ocean]

And sorry for tag as well

I usually never do, but I just thought since it was like three seconds ago

,w plot 3Power[Subscript[x,1],2]+4Subscript[x,1]*Subscript[x,2]+Power[Subscript[x,2],2]

[flat frigate]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by simonwin

[frigid ocean]

But, this des not look like a saddle point to me?

,w plot 3 x_1^2 + 4 x_1 x_2 + x_2^2

[flat frigate]

Results provided by WolframAlpha

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

Requested by simonwin

[frigid ocean]

I though't it'd have to look like this

Aaah... By definition..

a point at which a function of two variables has partial derivatives equal to zero but at which the function has neither a maximum nor a minimum value.
Which the function has neither a maximum nor a minimum

[final halo]

,w plot 3x^2 + 4xy + y^2 from -10 to 10

[flat frigate]

Results provided by WolframAlpha

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

Requested by icaird

[final halo]

Yeah I think one direction of a saddle can be flat

[frigid ocean]

Yeah okay, so that functions at the critical point does not have a maximum nor minimum

Which was (0, 0, f(0, 0))

I guess it also looks a bit like it's going down and upwards at the same time so

On the different sides

thanks 💪

.close

[safe radish]

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[thick fog]

explain how to get the answer...

[thick fog]

y=\frac{5}{3}x-\frac{19}{3}

?y=\frac{5}{3}x-\frac{19}{3}

how to use bot

,help

[flat frigate]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[lean thorn]

wrap it in dollar signs

[lean thorn]

$y=\frac{5}{3}x-\frac{19}{3}$

[thick fog]

oh

[flat frigate]

MellowDramaLlama

[thick fog]

ty

[thick fog]

where does this answer come from?

[royal kiln]

It's the perpendicular bisector.

Pick any point on the line (which isn't the midpoint between the towns)

Connect the points to make a triangle

What kind of triangle will that be?

[thick fog]

like literally any point?

oh

so let's say I pick 0,0

[royal kiln]

Any point but the midpo8nt, since that is obviously an equidistant point and would make the triangle degenerate

[royal kiln]

Any point on the line

This will work better on Geogebra than Desmos

[thick fog]

what do you mean? can you give pick a point instead?

umm

hello

:p

[safe radish]

@thick fog Has your question been resolved?

[thick fog]

:(

<@&286206848099549185>

<@&286206848099549185> pls

[royal kiln]

I am sorry

I have AT&T 😭

So anyway

@thick fog you should understand that when you connect the three points in a triangle, you always get an equilateral triangle.

The triangle isn't really important at all, but triangles are neat :3

You can manually measure the segments from the two cities to any point on the line and see that they are equal length

Now, if you are wondering how the line was built,

First you need the slope of the line through cities A(3,10) and B(13,4)

So
m = (4 - 10)/(13-3) = -6/10 = -3/5

Next, get the midpoint between these points (this point will be on the line whose slope we just found)

[thick fog]

oh :0)

[royal kiln]

M = ((A_x + B_x)/2, (A_y + B_y)/2)

= ((3+13)/2, (10+4)/2)

= (8,7)

[thick fog]

ohhhhhh

[royal kiln]

Finally, we need the slope of the line perpendicular to the line through A and B (which is through the midpoint, M)

m_perpendicular = -1/m

= -1/(-3/5)

= 5/3

The equation of that line is through midpoint M and has slope m_perpendicular.

This is called the "perpendicular bisector" of A and B

[thick fog]

I see

[royal kiln]

Use point-slope formula:
y - y_0 = m_perpendicular(x-x_0)

Where (x_0,y_0) is a point on this line. In our case, this is the midpoint, M

[thick fog]

the text book answer is this but like every time I tried I kept getting different answers 😭

[royal kiln]

y - 7 = (5/3)(x - 8)

[thick fog]

y - 7 = (-5/3)(x - 8)

wait

wrong clipboard

[royal kiln]

y = (5/3)x - 40/3 + 21/3

Yeah, somethings gone awry here

[thick fog]

ye...

[royal kiln]

I got sign error

Let me fix

[thick fog]

take ur time issok

[royal kiln]

y = (5/3)x - 19/3

✅

@thick fog

[thick fog]

oh

[thick fog]

@royal kiln thank youuu

.close

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[main rivet]

I need someone who knows how to draw on desmos with equations to help me

[stray socket]

Well we're not gonna do the assignment for you

Just play around on desmos

[main rivet]

Well ive never gotten anything like this where do i even get equations from

[stray socket]

Just model them based on your face

Like

Imagine if every curve you sketched was an equation

You'd start there

Curve of best fit kinda

[main rivet]

.close

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[static hedge]

not sure why this is wrong

[safe radish]

@static hedge Has your question been resolved?

[static hedge]

<@&286206848099549185>

[young nexus]

$g(f(x))=\frac{(f(x))^2+1}{3}$

[flat frigate]

ThM

[static hedge]

ye

[young nexus]

is your question answered?

[static hedge]

no

[young nexus]

why?

[static hedge]

OHH NEVERMIND

f o g = x^2

not f(x)

thanak u !

[young nexus]

exactly

[static hedge]

.close

[safe radish]

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[static hedge]

.reopen

[safe radish]

✅

[static hedge]

@young nexus

it still said it was wrong

[young nexus]

hmm, your g(x) differs from the first picture.

[static hedge]

OH YEAH 😭

oh it was still wrong

[young nexus]

what is (f(x))^2?

[static hedge]

9x^2-6x+1

but after u add the 1 it just becomes 9x^2-6x

ok let me redo the problem

[young nexus]

hmmm, when i add 1 to 9x^2-6x+1 i get 9x^2-6x+2

[static hedge]

OHH

wait omg 😭

AHH

THANK U SO SO MUCH 🙏

.close

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[vital cairn]

hi, how would you do this problem?

[vital cairn]

i found (15, 6) but there are two other possibilities which im not sure how to find

[safe radish]

@vital cairn Has your question been resolved?

[little mesa]

3 different parallelograms they could make

[vital cairn]

yea but how do you actually find the verticies i wants me to use vectors somehow

[little mesa]

if we name the points A(10,22), B(-3.7), C(2,-9) and unknown D, you use different combinations to find different possible D

and in a parallelogram, the parallel sides have equal vectors, right?

[vital cairn]

so you just take the vectors and add them to the known vertices? is there anyway to know how to do this without drawing a picture

[little mesa]

a picture would be the best way, but just picking different combinations is fine

like AB=CD will give you one possibility

[vital cairn]

hmm okay i kinda get it thanks

[safe radish]

@vital cairn Has your question been resolved?

[safe radish]

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[past birch]

how do i do a.?

[safe radish]

@past birch Has your question been resolved?

[past birch]

no

[safe radish]

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[past birch]

how do i do (a)?

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@past birch Has your question been resolved?

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[wind abyss]

Why did the person in this video do distance formula for

That just confused me

I show you how i do a problem like this brb

[plucky elk]

How else do you find the equation of the circle given two points on the diameter?

[wind abyss]

like that is the same type of question

and i didnt use distance formula idk i was used to doing it this way and the video threw me off 😭

and i forgot the + sign inthe answer great.

[plucky elk]

Your work is equivalent

Calculating radius is equivalent to calculating the diameter

Since 2 * radius = diameter

[wind abyss]

O

[plucky elk]

You just calculated diameter/2 for radius

Or radius squared I guess

Both ways give the right answer. So whatever is easier for you

[wind abyss]

Ohk perfect

THnx

.close

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[sour zenith]

The data are real numbers a, b,c Prove that if some two of the numbers
ab + b +1
bc + c + 1
ca + c +1
are equal to 0 so also the third of them is equal to 0

[sour zenith]

I have this problem

And every time i'm transforming equations i end up in form when i need to divide by the third number

[obsidian oracle]

since you want to prove this number is equal to 0, suppose by contradiction that it isn't

this way you can divide by it and maybe reach an absurdity

[safe radish]

@sour zenith Has your question been resolved?

[sour zenith]

I have this equation (a + 1)(b - c) = -1 -c(a + c)

[obsidian oracle]

isn't the 3rd one supposed to be ca+a+1? just to be sure

[sour zenith]

it's -(ca + a +1 )

[sour zenith]

<@&286206848099549185>

[safe radish]

@sour zenith Has your question been resolved?

[sour zenith]

<@&286206848099549185>

<@&286206848099549185>

[young nexus]

if 2 of the three numbers are 0 then the third. so without loss of generality we can assume the first and the second are 0 -> b appears in both. express b from the second and plugin this in in the first. it is just straight forward.

@sour zenith

[sour zenith]

so b= -c -1/c

@young nexus

[young nexus]

yes, and now use this in the first

[sour zenith]

-ac - a/c - c - 1/c = -1

so now i define a?

@young nexus

[young nexus]

wait a moment

[young nexus]

this is not correct, you missed the brackets.

0 = ab+b+1=(a+1)b+1=(a+1)(-c-1)/c+1

so you have 0 = (a+1)(-c-1)/c+1

expand the brackets.

[safe radish]

@sour zenith Has your question been resolved?

[safe radish]

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[split fulcrum]

Let f(x) = 1/x. Complete a Riemann 10 sum to estimate the the area under the graph of f for 1<=x<=3. Use the midpoint of each interval as the sample point

[split fulcrum]

Not sure how to start

[gentle oak]

you need to know $$\int_1^3 \frac 1x$$ ?

[flat frigate]

Joseph.P

[split fulcrum]

Im using Riemann sums

so probably not

so this is 1/x for 1<=x<=3

and i guess im trying to find the area under that curve

[grim plover]

do you know how riemann sums work?

[split fulcrum]

sortaa i know that if you calculate the area of an infinite amount of rectangles you get that area but im confused on how to actually do them

so i assume im trying to find this

[grim plover]

no, get the top side under the curve

[split fulcrum]

then what

huh?

wym

[grim plover]

https://www.youtube.com/watch?app=desktop&v=YTKQswb60Pw

this should explain better

[split fulcrum]

yeah im on that video rn lol

[grim plover]

also ignore my suggestion

Riemann 10 sum implies that n=10

[split fulcrum]

right

so 10 rectangles?

[grim plover]

yes

and if u r watching it (the video)

then can u find delta x?

[split fulcrum]

ill try

@grim plover would it look like this

[grim plover]

ye

[final halo]

The question wants you to use the midpoints

[split fulcrum]

oh

[split fulcrum]

so how do i do that

[final halo]

The height of a rectangle is the value of the function at the middle x coord of the rectangle

[grim plover]

i ignored that 🤦‍♂️

[split fulcrum]

reee lol

[split fulcrum]

wait so how do i do this with only 1,3

my midpoint will be like? 1.5

[final halo]

You have 10 rectangles, you have 10 midpoints

[split fulcrum]

so i can count the rectangles from 1 to 10 normally

i dont have to worry about the interval thing?

@final halo is this what you mean?

@grim plover Yo

.close

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[flint harbor]

I need help, is this correct? I used trigo sub

[little mesa]

you can't just convert $\int 4sin^2\theta d\theta = \int \frac{x^2}4dx$

[flint harbor]

ohh

[flat frigate]

chlamydia

[little mesa]

don't forget $d\theta$ btw

[flat frigate]

chlamydia

[deep matrix]

Could've done it easier

[flint harbor]

Got the answer thank you so much💕

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[harsh estuary]

How do you work out a gradient from a midpoint if one coordinate is a fraction?

[lean otter]

wdym

[lean otter]

,rccw

[harsh estuary]

My lecturer says the question shouldn’t have fraction but if i do it another way with the other coordinates my answer is different from the real answer

[lean otter]

how did you get 1-1

your points are w(-1;9) and m(1; -5/2)

right

$$m = \frac{y_2-y_1}{x_2-x_1}$$

[flat frigate]

sen

[lean otter]

@harsh estuary

[harsh estuary]

Sorry i meant the plus but im still confused about the fraction

[lean otter]

,rccw

[harsh estuary]

Apparently im not supposed to have a fraction but i cant get the correct answer anyway i do it?

[lean otter]

hmm? okay let me verify everything

[lean otter]

what do u get after this

i get a fraction too

[lean otter]

assuming you copied all of your points correctly, i see nothing wrong

[harsh estuary]

Idk cause it’s supposed to be a non calculator question. Others in my class seemed to get it and i asked for help but i never understood it

Ill ask them to send their answers see if i misunderstood a step in the question

[lean otter]

sure

[harsh estuary]

Thank you though! : )

[lean otter]

np

[copper pewter]

What are u finding the gradient of

@harsh estuary

[lean otter]

wm

[harsh estuary]

Gradient of the line of the midpoint y

[copper pewter]

k thanks

[harsh estuary]

Yeah!

[copper pewter]

Wait my brain died

I got -23/4 damn I can't mental rn

[lean otter]

his working is fine, i verified it

[lean otter]

send your working, need to help u too

[copper pewter]

Nah I'm fine I just tried to mental but yeah it's a fraction

There's no way to not get a fraction

The figure is a tad wonky though

[safe radish]

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[rugged surge]

how can i do this?

[rugged surge]

i know the slope will be 2(x+1)

y = 2(x+1) + c

[dull sequoia]

What other constraints are on these tangent lines?

[rugged surge]

0 = 2(0 +1) + c so c = -2

[dull sequoia]

Other than some particular slope

[rugged surge]

they pass through origion

[dull sequoia]

And?

[dull sequoia]

Why is x = 0 here

[rugged surge]

because the tangent line passes through the origion so it should pass through 0,0

[dull sequoia]

What is this saying

Actually

Where did you even get that from

[rugged surge]

the gradient of the tangent line should be 2(x+1) and then comparing with y = mx+c
y = 2(x+1) + c

and because the line passes through 0,0 i substituted 0 for y and 0 for x and solved for c

thats what i did

[dull sequoia]

The gradient of the tangent line is indeed 2(x+1)

This is correct

“And then comparing with y = mx+c
y = 2(x+1) + c”

[rugged surge]

oh wait

[dull sequoia]

Where did this come from

[rugged surge]

y = 2(x+1)*x + c

[dull sequoia]

Now there’s 2 x’s

But they aren’t the same x

The x in (x+1) is about which point is the line tangent at

The x in *x is the variable your function y is dependent on

Subbing in x, y= 0 gives you 0 = 2(x+1)*0 + c

Which tells you c = 0

Very expected

You definitely expect straight lines that go through the origin to have + 0 for their constant

What other constraint do you have for the straight line?

Other than “it’s straight” and “passes through the origin”

[rugged surge]

@dull sequoia do we just do

2(x+1) = (x+1)^2

and solve for x

[dull sequoia]

Why do you want to do this?

[rugged surge]

because 2(x+1) = y
and y = (x+1)^2
2(x+1) is the equation of our tangent and its intersecting the curve so at that intersection y point of the tangent and the curve should be the same so i equated them

[dull sequoia]

Where does it say 2(x+1) = y

2(x+1) is not an equation

y = (x+1)² is true

This was given

[rugged surge]

y = 2(x+1)*x1 + c
is the equation right?

[dull sequoia]

Equation of what

(There’s a reason I’m getting you to specify exactly what you’re saying, a big confusing part of calculus is all the things that look the same but aren’t)

[rugged surge]

for the tangent

[rugged surge]

yeaaa

[dull sequoia]

Tangent of?

Tangent needs a curve and a point

[rugged surge]

tangent to the curve at a point x

[dull sequoia]

To the curve y = (x+1)² at the point…

[dull sequoia]

x is not a point

A point is a pair of values

[rugged surge]

x,(x+1)^2

[dull sequoia]

Right

[dull sequoia]

This is the equation of the tangent

[dull sequoia]

Now where did you get 2(x+1) = y from

(It’s wrong, it should be 2(x+1) = y’)

[rugged surge]

y = 2(x-1) * x1 + c

and we are told that it passes through the point 0,0 so
0 = 2(x-1) * 0 + c
so c becomes 0 so the equation of the tangent becomes:
y=2(x-1)x1

[dull sequoia]

This is correct

[dull sequoia]

.

We’ve gone a bit sidetracked

Something we needed to observe is that what really is happening is this

We have a curve

y = (x+1)^2

If we chose some x

We can use this x to find a particular y

We can get a point on the curve

Yeah?

[rugged surge]

yeaaa

[dull sequoia]

Ok let’s call this x “x₁”

And then we’ll call this particular y as y₁, defined by y₁ = (x₁+1)²

[rugged surge]

okayy

[dull sequoia]

Then the equation of a line tangent to y = (x+1)² (the curve) at point (x₁, y₁) (the point on the curve) will be…?

(Note that we have not yet considered the fact this tangent goes through the origin, this comes in later)