help-10

[worthy dune]

huh

[clear sinew]

idk how universiities around the world are like but at mine they usually provide officuse hours. I call them counselling hours lol xD

[worthy dune]

ok

[clear sinew]

In German we call them Sprechstunde

[worthy dune]

oh are you from germany

[clear sinew]

yeah 😄

[worthy dune]

i live in australia rip

down under

so many wrong/weird stereotypes about australia

[mighty geyser]

yeah it's pretty late there

[worthy dune]

anyway thank you i should really be getting some sleep

[mighty geyser]

definitely sleep

close this channel ig?

[worthy dune]

thank you element and mr brown

yeah

.close

[obtuse pebble]

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[dusky rampart]

jeff get the star block

oops wrong chat

[obtuse pebble]

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[quick sigil]

can someone help me convert this to sin() function

[spiral maple]

sinc(t) := sin(t)/t..

[obtuse pebble]

@quick sigil Has your question been resolved?

[quick sigil]

feeling stupid, but Im not sure how to apply what you wrote to what i have 😫

[high lily]

sinc(this) = sin(this)/this

do a little substitution and simplify

[obtuse pebble]

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[obtuse pebble]

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@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

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[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[woeful cosmos]

I’m confused about this question

How do I find the radius?

[mild ocean]

a circle has a central angle of 2pi radians

so that’s 1/24 of the circle

how would you get the total of something if you already know what 1/24 of it is?

@woeful cosmos

[woeful cosmos]

Do I add something to 2pi?

[mild ocean]

no

there’s no adding going on here

you know what the area of 1/24 of the circle is, which is 3pi cm^2

you want to know the area of the whole circle to continue

what would you do

[woeful cosmos]

Oh so 3pi cm^ is only area of a part of the circle

[mild ocean]

yes

1/24 of the total circle

[woeful cosmos]

Pi r^2

[mild ocean]

so what’s the area of the whole circle

[woeful cosmos]

So 3pi^2 x r^2?

I’m confused

[mild ocean]

we know the area of 1/24 of the circle right

[woeful cosmos]

Yes

3pi cm^2

[mild ocean]

so how would you find the area of the whole circle with that knowledge

[woeful cosmos]

Multiple..?

[mild ocean]

yes

[woeful cosmos]

Something..

[mild ocean]

what times 1/24 is 1

[woeful cosmos]

Ok so it’s 24

no..?

[mild ocean]

yes

what’s 3pi*24

[woeful cosmos]

72pi

[mild ocean]

yes

so the area of the circle is 72pi

keep in mind that the area of the circle is also pi*r^2

so 72pi = pi*r^2

you have to isolate r

[woeful cosmos]

Square root 72?

[mild ocean]

yes

[woeful cosmos]

Oooooooooooooo

It’s correctttt

[woeful cosmos]

But I’m abit confused as to why we did this

[mild ocean]

because we needed to find the whole area of the circle

3pi was the area of 1/24 of the circle

[woeful cosmos]

Here

U did what times what equals to whole circle

Which is 24

[mild ocean]

yeah

[woeful cosmos]

And then I timed that again to the 3pi

Okie

Thank u so much 🥺

[mild ocean]

no problem

if you’re done do .close

[woeful cosmos]

I have one more question

[mild ocean]

ok

[woeful cosmos]

What’s a central angle

Why did the question put pi/12 if we didn’t use it

[mild ocean]

i don’t really know how to say it honestly

probably look it up, because the explanation you’d find would be better than mine

[woeful cosmos]

Okie

How about these

How would I find the obtuse of

36°

I saw the answer key but

I didn’t answer how they got the answer

Understand**

[mild ocean]

sin(x) = sin(pi-x) for radians

sin(x) = sin(180-x) for degrees

[woeful cosmos]

so what’s the question asking us

To convert 36°

To radians?

[mild ocean]

no

[strong vale]

Do you know what an obtuse angle is?

[woeful cosmos]

Yes

Bigger than 180

[timid silo]

Just find an obtuse angle that has the same sine as 36 degrees

[timid silo]

No

[woeful cosmos]

Oh

Oops

[strong vale]

So what's an obtuse angle?

[woeful cosmos]

Greater than 90

😬😬

[strong vale]

So 270 is obtuse?

[woeful cosmos]

Yes

[strong vale]

Obtuse is angle between 90 and 180

[woeful cosmos]

OhHhHhH

Omg

I’m sorry 😭😭

I’m so sleep deprived

ok I’m just stupid

[strong vale]

Lol. Now that you are clear about what the question is asking for

Do you agree that values of sin(x) repeat themselves?

[woeful cosmos]

no

[strong vale]

Have you seen the graph of the sin(x) curve?

[woeful cosmos]

I mean Idk

Yes

Starts from 0 up down

180-36

144

Right

[strong vale]

Yes

Because $\sin(180-x) = \sin(x)$

[warm shale]

azeem321

[woeful cosmos]

What about these

360-40?

[strong vale]

Graph the cosine curve

and see if you notice any patterns

or something

[timid silo]

You don't need to lol

[woeful cosmos]

Oke so I just minus em

Which is 320

Nice nice

[strong vale]

ik, but it's much better to show how things work with graphs instead of just saying cos(-x + 360)

[woeful cosmos]

Graphs r confusing for me tbh

Wait what

-x + 360?

[timid silo]

-x + 360 = 360 - x

[woeful cosmos]

Oh same thing

Almost cried

[strong vale]

The issue is you are relying on memorizing formulae. Try to understand why things work that way lol

[woeful cosmos]

It’s hard to understand over phone

☹️☹️

Question 4, is solving with identities right 😬

or no

[strong vale]

The question wants you to find all values of $\theta$ for which $\sin(\theta) = \frac{1}{2}$

[warm shale]

azeem321

[woeful cosmos]

wut

So no identities

[timid silo]

unit circle

[strong vale]

you think this guy knows what a unit circle is

[timid silo]

idk maybe they knew

[strong vale]

https://www.khanacademy.org/math/trigonometry

watch videos

[woeful cosmos]

I’m a girl

☹️

[strong vale]

my apologies

[woeful cosmos]

And I’m sorry I don’t know

But I care so I’m here trying my best

I just have a hard time answering questions

[strong vale]

i don't mind helping you solve the questions, but my concern is that you don't really understand whats going on

and so when u in an exam

u not gonna do so well

i think you need to find some videos on the topic

to watch

[woeful cosmos]

I have been watching videos

All morning and weekend

But I can’t grasp it I can’t help I’m just stupid

[strong vale]

You're not stupid. You are just a newbie at math. We've all been in that position before.

You should try khan academy really good videos

[woeful cosmos]

I would but some things they teaching in videos is not part of my curriculum or level so I get confused

How would I solve these type of questions

Like 1a

Identities..?

[short spruce]

“to one decimal place” makes me think you’re allowed a calculator

[timid silo]

^^

[woeful cosmos]

Oh nice

I just put that in the calc?

[short spruce]

need to use inverse trig

[woeful cosmos]

I got the first

55.0

But there’s two answers

The other is

306.9

How did they get 306.9

Nvm I got it

How do I solve 3?

[obtuse pebble]

@woeful cosmos Has your question been resolved?

[obtuse pebble]

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[zinc ibex]

Guys do you know how to find the perimeter and area of a inscribed trapezium?

[zinc ibex]

||"An isosceles trapezoid is inscribed in a circle and the major base coincides with the diameter. The radius of the circle is 30 cm and the diagonal of the trapezoid is 48 cm. Calculate the perimeter and area of the trapezoid."||

this one too

||"The isosceles trapezoid ABCD is inscribed in the circle of radius 65 cm, with center O inside the trapezoid. The two bases are 33 cm and 16 cm from the center respectively.
Calculate the area of the trapezoid."||

this is translated from italian so if u dont get something tell me

<@&286206848099549185> can you help me? I need to go to sleep soon

its late here

<@&286206848099549185>

i still cant realize how to do it

i feel so dumb

and useless

.open

i need first the second problem as it comes before the first

[tardy plaza]

Aight...

[zinc ibex]

Do*

This is what i was able to di

[tardy plaza]

I only found the height, sorry!

[zinc ibex]

how did u do that?

that can help me

[tardy plaza]

Bet...
the bases are 33 and 16 centimeters away
That means they are 33 - 16 centimeters away from each other
33 - 16 = 17

[zinc ibex]

ur right

[tardy plaza]

🎉

[zinc ibex]

thanks for that

[tardy plaza]

Np!

Good luck getting the rest 🍀

[zinc ibex]

thanks

@tardy plaza i did it!!

[zinc ibex]

u were wrong on it XD

[tardy plaza]

Ayy 🎉

Nice job

[zinc ibex]

it was 33+16

[tardy plaza]

Sorry about that 😅

[zinc ibex]

np lol

u made me comprehend it

[tardy plaza]

Yay 😅

[zinc ibex]

Here the result

Area is matching with the book

(my handwriting is bad i know lol)

now It’s about drive😈
It’s about power💪
We stay hungry🥵
We devour👅
Put in the work🙏
Put in the hours🕦
And take what’s ours✨
black and Samoan in my veins😤
What’s my motherfucking name🥵
Rock💸

lol

now i gotta do 3 more exercises

ill tell u if i dont get something

another one

||"A rhombus has one diagonal that is
3/4 of the other and their sum is 140 cm.

Calculate the radius of the inscribed circle."||

it counts to me 20

140/(3+4)=20

ops

<@&286206848099549185>

[oblique glacier]

don't ghost ping

and don't ping helpers until after 15 minutes

please follow the rules outlined in #❓how-to-get-help

[zinc ibex]

they arent helping me

neither others

im sorry though

[oblique glacier]

nobody on this server is obligated to help; everyone is a volunteer

[zinc ibex]

i wont ping them

[oblique glacier]

please be patient

thank you

[zinc ibex]

yes i will

for any further updates ping me pls

im going to sleep

its already too late

(0:13 am)

in case u need here are the problems i need to solve

1-
||"A rhombus has one diagonal that is
3/4 of the other and their sum is 140 cm.

Calculate the radius of the inscribed circle."||

2-
||"In a circle whose diameter measures 14 cm,
the inscribed isosceles triangle ABC contains within it the center of the circle. The height of the triangle relative to the unequal side measures 11.2 cm.
Calculate the length of the perimeter of the triangle and its area (round off the result to the nearest hundredth).||

3-
||"An isosceles trapezoid is inscribed in a circle and the major base coincides with the diameter. The radius of the circle is 30 cm and the diagonal of the trapezoid is 48 cm. Calculate the perimeter and area of the trapezoid."||

[obtuse pebble]

@zinc ibex Has your question been resolved?

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[obtuse pebble]

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[timid silo]

I dont know what are the variables

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

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[bronze acorn]

hey

[bronze acorn]

i want to ask what is the derivee?

[strong vale]

derivee?

[bronze acorn]

the dx/dy thing

[junior inlet]

that's the derivative

so

what do you already know about calculus

[bronze acorn]

nothing

[junior inlet]

well, i guess i'll have to do a Calculus 101 thing

@bronze acorn you know about functions?

[bronze acorn]

yes

[junior inlet]

right

let's say we have a function f(x)

and here's its graph

wait a second my desmos isnt responding

[bronze acorn]

it's ok

[junior inlet]

okay, so let's say we have a complicated function like this

this is the graph of f(x)

[bronze acorn]

ok

[junior inlet]

sorry, i mean this

this is a sum of several functions, one of which is $\frac{1}{x^2}$

[warm shale]

Kanga Gang ¬Sam

[junior inlet]

@bronze acorn what is the value of f(0)?

[bronze acorn]

you mean lim?

[junior inlet]

no

the value

[junior inlet]

wait you alredy know limits? that's handy

[bronze acorn]

yes limits

[junior inlet]

so

f(0) is undefined

but what is the limit of f as x approaches 0?

[bronze acorn]

• infint

[junior inlet]

yeah

lemme ask another question

let this be f(x) instead

what's the limit as x approaches 0?

[bronze acorn]

0+ or 0-?

[junior inlet]

oh, you already know one-sided limits, great

but the important thing here is that the limit doesn't exist

only the one-sided limits do

and if the limits as x approaches n+ and n- are different, the limit as x approaches n doesn't exist

now let's move to a seemingly unrelated topic: secant lines @bronze acorn

(not related to secant, as in the trig identity)

@bronze acorn you here?

[bronze acorn]

yes

[junior inlet]

so let's say we have a function f(x)=x^2

[bronze acorn]

ok

[junior inlet]

now i can draw a line that cuts the curve at 2 points

and that's your secant line

but it's not particularly interesting, so we want another kind of line: tangent lines

[bronze acorn]

ok like tangent with which points?

[junior inlet]

for example, this is a tangent line:

(just y=0)

[bronze acorn]

ok

[junior inlet]

y=2x-1 is another tangent line

so is y=4x-6

the derivative is just the slope of the tangent line

@bronze acorn and we calculate it with limits

[bronze acorn]

just?

[junior inlet]

just what?

[bronze acorn]

i mean with what do i need them?

[junior inlet]

oh, a lot of things
like optimization

@bronze acorn also do you want to know how to calculate limitsthe derivative? (sorry)

[mild ocean]

derivatives tell you the rate of change, so they are very useful

[bronze acorn]

oh thank you

i have to go now

bye

[mild ocean]

@bronze acorn do .close then

[bronze acorn]

.close

[obtuse pebble]

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[dense moon]

find the range of values of k for which the expression x^2-8kx+2x+15k^2-2k-7 is never negative for all real values of x

[dense moon]

<@&286206848099549185>

40 mins till exam starts

[mild ocean]

don’t immediately ping helpers

[dense moon]

sorry

i won't from now onwards

[alpine raven]

good luck bro

[dense moon]

thanks

i have it done but i don't get it why the sign is less than or equal to 0?

i did that in class before but forgot teacher's explanation lol

[alpine raven]

because if the discriminant is negative, the sign of the polynomial is the sign of the coefficient of x^2

[dense moon]

find the range of values of k for which the expression x^2-8kx+2x+15k^2-2k-7 is never negative for all real values of x

this is the question

[alpine raven]

and I explained it

read

[dense moon]

ok

[dense moon]

don't get it how it goes with this question

[alpine raven]

they asked you for which values of k the expression is never negative

[dense moon]

shouldn't it be greater than or equal to 0?

never negative

[alpine raven]

what is the main way to study the sign of a polynomial

[dense moon]

its inequalities

not functions

[alpine raven]

?????

what are you saying

[dense moon]

well u can say it is a function

[alpine raven]

inequalities and function arent the same

[atomic bridge]

that polynomial can be less than 0

[dense moon]

its from chapter inequalities

[alpine raven]

so what ?

[dense moon]

pls explain it from beginning

[alpine raven]

Like I'm saying, if you know how to study the sign of a polynomial, you find the discriminant of this polynomial. If the discriminant is positive, you have two solutions that cancels the polynomials so the polynomial can be negative somewhere. Same for when the discriminant is equal to 0.

Now if the discriminant is negative, the polynomial doesn't have any real solutions, so it has the sign of the coefficient of x^2

here the coefficient is 1, so it's always positive, never negative.

you found your discriminant in term of k

now solve the inequality with k you've got

(k-4)(k-2) < 0

find for which values this is negative

and you have the range of k

[dense moon]

don't know why my teacher didn't explain that

[junior inlet]

@dense moon fyi pinging helpers usually won't do anything

[junior inlet]

we can use this?

@dense moon do you know the discriminant of a quadratic?

[dense moon]

yes

b^2 - 4ac

it gives us the nature of the roots of the equation

right?

[alpine raven]

not only the nature but the number of roots

[dense moon]

yes that as well

[alpine raven]

yes

[mild ocean]

@junior inlet i know

[obtuse pebble]

@dense moon Has your question been resolved?

[junior inlet]

what?

[mild ocean]

.

[junior inlet]

uh sorry scott my mother just asked me to do the dishes so i was away for a while...

[obtuse pebble]

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[obtuse pebble]

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[small cloud]

Suppose you have the numbers 1 to 15 (these could be 15 objects or people etc) now, we are going to take 10 of these 15 numbers and then replace them. We will then take another 10 and replace then and then do this a third time. Is there a way of guaranteeing that every single number will be chosen exactly twice? If so, how?

[small cloud]

I know this is a combinatorics question I just haven’t gotten to that section of my textbooks yet and I’d really like to know how to do this without reading an entire section of a textbook

[obtuse pebble]

@small cloud Has your question been resolved?

[small cloud]

<@&286206848099549185>

[small cloud]

.close

[obtuse pebble]

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[quartz citrus]

Hello.

[quartz citrus]

This is a screenshot from my lecture introducing partial fraction decomposition.

I am in particular wondering

How my professor took the integral of

$$
\int \frac{2}{x-1} d x
$$

[warm shale]

Someone Somewhere

[spiral maple]

reverse chain rule.

$2\int\frac{1}{x-1}\dd{x}=2\ln(\abs{x-1})+C$

[warm shale]

Mosh

[quartz citrus]

I am really failing to see how we suddenly get $$
\ln |x-1|
$$

[warm shale]

Someone Somewhere

[spiral maple]

cause... d/dx ln(x-1) = 1/(x-1)

[quartz citrus]

Hmm actually that's interesting.

[spiral maple]

That shouldn't be tbh

[quartz citrus]

If I gave something like..

[spiral maple]

If you're doing integral calc, you should have done differential calc

[quartz citrus]

Well, I have done differential calculus and we learned quite explicitly that an integral is quite specifically a function who's derivative returns the integrand but I haven't been exposed to it in this context, it was immediately obvious what I should do in this instance

But on another note if I had something like
$$
\int \frac{3}{x-4} d x
$$

Would I follow the same rule?

[warm shale]

Someone Somewhere

[spiral maple]

Yes... you should integrate if you have an integral

[quartz citrus]

Would this amount to the following?
$$
\begin{array}{l}
3 \int \frac{1}{x-4} d x \
= \
3 \cdot \ln (x-4)
\end{array}
$$

[warm shale]

Someone Somewhere

[spiral maple]

no

you need abs value for the ln to be well defined on the same domain.

[quartz citrus]

So is this what you're trying to say?
$$
\begin{array}{l}
3 \int \frac{1}{x-4} d x \
= \
3 \cdot \ln (\mid x-4)
\end{array}
$$

[warm shale]

Someone Somewhere

[spiral maple]

+C, but yes.

[quartz citrus]

Excuse my textIt hopefully the intention was obvious

[spiral maple]

\abs{}

[quartz citrus]

I don't think it rendered

[spiral maple]

yeah, cause you didn't use it..

it's not gonna convert something not there.

[quartz citrus]

😔

I tried my best

But thank you..

I appreciate your help, truly.

🙂

.close

[obtuse pebble]

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[weary marsh]

ok i need help with a math question, i am really stumped on it and no it is not a test., this is the image of it here:

[weary marsh]

30 miles per gallon is uh

6 gallons

wow thats little

$20 per fill

126 per 180 miles

this is in a game lol

63 per 90 miles

-$12 per 90 miles, thats 3 gallons

63 - 12 = 51

IT KEEPS CHANGING NOO

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[merry dove]

I have a bit of a question.

[merry dove]

How can one prove that (a^b)^c = a^(bc) for rational number and negative exponents?

[obtuse pebble]

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[merry dove]

.reopen

[obtuse pebble]

✅

[merry dove]

<@&286206848099549185>

It's rather easy to prove this property for integers, but not rational number exponents (e.g. 3^(2/3)) or negative exponents (e.g. 3^-2)

[obtuse pebble]

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[sand osprey]

i'm stuck on q15

i can follow my teacher's work until he gets to the third line

then i don't understand what he's doing

like how can he get the original function without taking the partial derivative of a variable?

[spiral maple]

They just skipped the step.

[sand osprey]

that's what i thought but i don't understand how to skip it

[spiral maple]

what

[sand osprey]

like i get it how to do it with 2 variables but not 3

[spiral maple]

You do the steps if you need to

[sand osprey]

right but i don't understand how to do it

[spiral maple]

same way as w/ 2 var

[sand osprey]

like i don't know what the steps are

[spiral maple]

just.. more variables

partial integrate wrt 1 of the variable, then compare that things partials wrt the other 2 vars

[sand osprey]

oh

ahaha i see ok lemme try that

[spiral maple]

but you can easily note that the vf is almost symmetric, in that xyz will give the xy, xz, and xy parts respecively

then you need something added on that has only z and will differentiate to 2z

so xyz+z^2

then +C

[sand osprey]

okok give me some time to try it out

[mild ocean]

@upbeat plinth

[obtuse pebble]

@sand osprey Has your question been resolved?

[sand osprey]

okkk i think i get it :)

ty mosh!

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[lament vapor]

is someone able to help show me how we show the calculation for standard frm

i've never learnt that

i know how to do it for factored, vertex and y=mx+b form

using another point

but how would we show it for standard form

y = 1/2x^2 -18x+132

^^^^ this example

[obtuse pebble]

@lament vapor Has your question been resolved?

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@lament vapor Has your question been resolved?

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[craggy cobalt]

Hello

[craggy cobalt]

Would you say that the relationship displayed in this scatterplot is non-monotonic?

[spice sparrow]

can anybody help me solve following equation 5/2*x=-5

[obtuse pebble]

@craggy cobalt Has your question been resolved?

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[lean smelt]

Hi im stuck with this promblem

[lean smelt]

I ve tried the pascals triangle but dont know what to do next

My mom tried this up up thingbut i dont get her work and it is still wrong

Practice

For mathcounts

<@&286206848099549185> the answer is 32 but can somone explain this promblem

Yea

Yes

Line 2?

[lean smelt]

Yea

From right to left?

Yes

I get this

Ok and then

Yeayea

Ahhhh wow

Wow thank u alot

And then times them

Ok btw is there other ways that can solve this promblem

Or is it all like slower

Ok ill read over ur explanation again but thanxs

.close

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Channel closed

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[final radish]

.close

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[silent echo]

hello

[silent echo]

.close

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[green moat]

how does one go about calculating/showing if 2 eigenvectors are orthogonal?

[spiral maple]

inner product = 0.

[green moat]

by inner you mean?

[spiral maple]

2 vectors are orthogonal iff $\langle u,v\rangle=0$

[warm shale]

Mosh

[spiral maple]

idk what inner product you're working w/ cause you haven't given a vector space.

if you're in R^n space, then the canonical dot product is likely what you're using.

[green moat]

lets go with
[1,
1]
and
[-1
0]

[spiral maple]

so R^2 space?

[green moat]

both eigenvectors

for checking orthogonality

[green moat]

no other way to check without extra parameters?

[spiral maple]

what?

Can you just answer what I asked?

[green moat]

we just have two linear eigenvectors

[spiral maple]

Are you working w/ R^2 vectors?

[green moat]

idk what R^2 space is

[spiral maple]

the xy plane...

[green moat]

yes

[spiral maple]

Then just use the dot product.

[green moat]

wait then its not xy plane

[spiral maple]

??????

[green moat]

coz dot product yields wrong answers

[spiral maple]

You're speaking nonsense.

[green moat]

now im really confused

[spiral maple]

You take the dot product of the 2 vectors.

If you get 0, they're orthogonal.

No more, no less.

[obtuse pebble]

@green moat Has your question been resolved?

[spiral maple]

🙄

[short spruce]

lol

[green moat]

doesnt this imply that the dot product of eigenvectors can be non zero and still be orthogonal

[spiral maple]

????

dot product itself isn't orthogonal...

it's an operation

[green moat]

fixed

[spiral maple]

no.

the definition of orthogonal is that they inner product together to get 0.

[green moat]

ok ok, and i use the Hermitian Inner Product for complex vector space right?

[spiral maple]

sure.

[green moat]

is that a yes? or a sarcastic "idc do what u want"?

[spiral maple]

Can't recall the canonical IP for C-spaces.

However orthogonality is dictated by whatever IP you use.

[green moat]

bro, i am not majoring in maths, i have not done this stuff for a long time i dont remember what those mean

thank you for the effort doe and for being patient

.close

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[chilly torrent]

hello, i need some help on this question, picture and diagram are there
i alr know the idk what it is called property of tangents that both tangents to s circle are congruent if they share one vertex angle or something

also trying to use radius of incircle for similar triangles sounds kinda stupid but idk where to go after, im at the point where like i figured out part of it only need some hints, thank you

[obtuse pebble]

@chilly torrent Has your question been resolved?

[chilly torrent]

<@&286206848099549185>

[obtuse pebble]

@chilly torrent Has your question been resolved?

[scarlet gale]

@chilly torrent It doesn't state where D and E are placed, so, for example, the tangent point of DE could be on the bottom arbitrarily close to where the circle intersects BC.

It could also be on the top arbitrarily close to where the circle intersects AC.

So, if the tangent point is on the C side of the circle, the perimeter is going to be essentially twice the length from C to where the circle intersects AC.

This is assuming it has a consistent answer regardless of where DE intersects the circle.

If it doesn't, there's not enough information to solve it.

Like this shows two possible DE lines.

So if all possible DE lines give the same CDE perimeter, then it doesn't matter which you pick.

So you might as well pick a DE with an intersection with the circle very, very, very close to where AC intersects the circle.

That's close enough to having DE be collinear with AC, where D is at the intersection of AC and the circle and E is at C.

Then one side of the triangle will be very, very, very close to zero and the other two will be very, very, very, close to the length DC.

So, the perimeter will be 2 DC, but only if it doesn't matter which DE line you pick.

[chilly torrent]

My teacher gave me this challenge problem, he tell me to assume all possible de lines have the same length, aka same cde perimeter

@scarlet gale Thankyou i got the answer

ik what DC is

[obtuse pebble]

@chilly torrent Has your question been resolved?

[obtuse pebble]

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[obsidian isle]

ok so we typically define functions in 3D as z = f(x,y). why can't we do something like (y,z) = f(x)? or does that just not work

[forest sinew]

it works fine

you can define "functions" very loosely in general

they might not even fit the definition any more

but yea i mean just switching indepedent and dependent variables around is fine

[mild ocean]

z = f(x,y) is just what people normally use

[forest sinew]

its a nice convention

since its the same any which way

[mild ocean]

x = f(y,z) and y = f(x,z) work the same

[forest sinew]

you may think it doesnt matter but then like

if you start writing stuff like f(y) = y^2 = x

idk it just looks wrong

if it doesnt matter which choice you make, making the one that conforms more to previous conventions and intuition is probably the better choice

[obsidian isle]

No but can you uniquely determine two values from an expression in a single variable? How would that work

[forest sinew]

? how do you mean

what breaks a function is ambiguous evaluation, if you put in the same input, you need to get the same output

thats the only thing

[forest sinew]

youre putting an additional constraint on it here, i think?

whats important as that we uniquely get one value from 2 input values

only have a single set of inputs for each output value is a more restrictive type of function

(its inverse is also a function over the codomain)

[obsidian isle]

Right that's what I figured you can't do that

[forest sinew]

you can

consider a plane

wait

no dont consider a plane

you can

i think a line works

a 3d line

no

wait

right

still not right

is it

we need another plane

[forest sinew]

you can

its a special kind of function

obviously it gets harder in more dimensions

but it can be done

i mean a trivial example of just like

a function defined over a single point

(1,1,1) being the only thing it hits

this satisfies your condition

a nice straight 3d line thats not parallel to any axis works too

[obsidian isle]

Maybe something like (y,z) = (f(x), g(x)) would work

Which is pretty much just a parametric

[forest sinew]

its fine for a function to not have a unique input for some output

is the main point

but there needs to not be ambiguity about where some input is sent

[obtuse pebble]

@obsidian isle Has your question been resolved?

[obtuse pebble]

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[ruby pendant]

Idk if that's okay to ask another question but i'm having a hard time because i'm trying to get into some discret math for uni entrance exam i'll have in a few months and i have this question on my exercise list of truth table that i never seen, the question is only described as follows:

Find the disjuntives minimal forms (idk if that how it's names in english, if its not please tell me) for the following boolean expressions:

Truth table : a = T(x,y,z;173);

[ruby pendant]

what could the 173 possibly mean ?

[obtuse pebble]

@ruby pendant Has your question been resolved?

[ruby pendant]

<@&286206848099549185>

[obtuse pebble]

@ruby pendant Has your question been resolved?

[nocturne minnow]

Do you have a screenshot of the actual question?

[obtuse pebble]

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[timid silo]

how do you do this

arc length of a function

what is that

[spiral maple]

Find the turning points.
Compute the distance travelled b/w those points
Sum

[timid silo]

oh whoops i misread it

you mean the relative min and max?

[spiral maple]

Yes.

Those are the turning points.

[timid silo]

then do the distance formula between 0 and turning point then between turning point and 7?

sounds like an approximation rather than an exact amount

[spiral maple]

it isn't.

for example let's say (1,7) and (5,3) are the turning points and (0,0) and (7,10) are the endpoints

You'd do |0-7|+|3-7|+|10-3|

and that'll be the total distance

[timid silo]

wait

you dont need to do distance formula?

[spiral maple]

no.

it's (t,x) points

[timid silo]

wait holy shit

[spiral maple]

at some time you're some distance away

and you're told it's confined to 1D

[timid silo]

i got 2090

im clearly doing something wrong

[spiral maple]

,w solve 6t^2-18t-60=0

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Omegabet_#9771

[spiral maple]

so |x(0)-x(5)|+|x(5)-x(7)|

[timid silo]

yeah i got that

i did that too

4 + 271

-271 - 1544

[spiral maple]

??

[timid silo]

totalling 2090

[spiral maple]

where the fuck did 1544 come from

[timid silo]

x(7)

[spiral maple]

that's wrong.

,calc 2(7^3)-9(49)-60(7)+4

[warm shale]

Result:

-171

[timid silo]

huh

[spiral maple]

you messed up computing x(7).

[timid silo]

ok thats still 442 + 275 which is 717 and still too big

which honestly surprises me i thought the issue of this system would be undrestimating

[spiral maple]

yeah... cause |x(5)-x(7)| isnt 442...

[timid silo]

holy fuck

[spiral maple]

it's 100

[timid silo]

is there a name for this topic btw

[spiral maple]

calculus

[timid silo]

i mean

whatever

thx

[obtuse pebble]

@timid silo Has your question been resolved?

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[silk moth]

I have a point P on a circle with radius n. Given a direction, I need to find another point across from P inside of the circle.

[silk moth]

I would love help to figure this out.

[junior inlet]

can you... uh, clarify what already haCve fro his problem?

i mean, what do you already have

like, the equation for the circle and the line?

[silk moth]

I honestly don't really know where to start

[junior inlet]

do you have the equations for the shapes

[silk moth]

oh, so, the point is just a position. And the direction is just a vector.

[junior inlet]

calculste the equations and find the intersections?

[silk moth]

oh ok. thanks.

[silk moth]

I have another question. How can I find the 2 possible positions based on a given distance from P to the point I need to find.

[junior inlet]

intersection of 2 circles?

[silk moth]

will that give me both positions? Or are you saying I need to do it twice?

[junior inlet]

you'll get both...

[silk moth]

ok thanks.

.close

[obtuse pebble]

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[frozen willow]

u = sec(x)+tan(x)

[frozen willow]

What is du?

[junior inlet]

and

taktew athe dreivvative

[frozen willow]

I have an answer, but want other’s too

[junior inlet]

usm rule

sum rule

derivative aof sec is sec times tan

[frozen willow]

I have an answer my dude

[junior inlet]

dervietiva of tan is sec srqQared

[frozen willow]

I’m more here for verification (formatting the answer specifically)

Question: u = sec(x)+tan(x)
What’s du?

[mild ocean]

,w derivative of sec(x)+tan(x)

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by quantum#6320

[junior inlet]

qanfdstum help

[frozen willow]

Hmm

du = {sec^2(x) + tan(x)sec(x)}dx

Is what a professor said

If anything, I’d think it was sex^2(x) dx + tan(x)sec(x) dx

[mild ocean]

same thing

[frozen willow]

Oh bruh

[frozen willow]

To clarify this would be differentiating with respect to x right?

[mild ocean]

yeah

[frozen willow]

Ok

Thanks

.close

[obtuse pebble]

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[hexed inlet]

<@&286206848099549185>

i’m not sure where to start?

could someone just help me find an equation

[obtuse pebble]

@hexed inlet Has your question been resolved?

[hexed inlet]

.close

[obtuse pebble]

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[tame spade]

Anyone able to help me with this?

[tame spade]

.done

.close

[obtuse pebble]

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[obtuse pebble]

@latent nova Has your question been resolved?

[jolly ginkgo]

Gud

[obtuse pebble]

@latent nova Has your question been resolved?

[novel knoll]

$\frac{1}{n}-\sin\left(\frac{1}{n}\right) \sim \frac{1}{6n^3}$

[warm shale]

ScapeProf

[novel knoll]

So converges from comparison

What?

Taylor series

[novel knoll]

,w Taylor series sin(1/n)

[warm shale]

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[obtuse pebble]

@latent nova Has your question been resolved?

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[obtuse pebble]

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[sage geode]

Derivative is defined with limit

df/dx is the same as f'(x)

But if y = f(x), then yeah

dy/dx = f'(x)

[obtuse pebble]

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[sage geode]

Btw it's usually h instead of dx (in the limit definition)

[obtuse pebble]

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[obtuse pebble]

@timid silo Has your question been resolved?

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[dim steeple]

Can somebody help me to understand the signs i have marked in this page please

[dim steeple]

about that ❄️ sticker sign, and that reverse A sign and at last that reverse 3 E sign

[spiral maple]

• is a general operation

$\forall, \in$

upside down A is read "For all"
backwards E is read "There exists"

[timid pelican]

so lets say if you have an element of a set a and a set S, $a\in S$ means that the element a is in the set you read it as "a in S"

[warm shale]

norablackcat

Mosh

[dim steeple]

i see thankyou

[spiral maple]

my bad, yeah the curled E looking one is "in"

[dim steeple]

but the sign you put the reversed E was other sign then the one i am confused about

[spiral maple]

$\exists$?

[warm shale]

Mosh

[dim steeple]

yea thats what i was to ask u bout thankyou

[spiral maple]

Backwards E is "there exists"

[dim steeple]

still thankyou for telling about this sign aswell

[spiral maple]

ex $\exists x\in\mathbb{R}$ st $3x=6$

[warm shale]

Mosh

[timid pelican]

@dim steeple if you go back through the list of things is there anything that is confusing that is not notation?

[dim steeple]

no

i dont know

cause i didn't knew what was going on that page, so how could i understood that, still i think im done thankyou

both of you

.stop

[spiral maple]

.close

[dim steeple]

..close

.close

[obtuse pebble]

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[mighty jackal]

I need some help

[mighty jackal]

I went back to do some revision and I was looking at completing square and this question confused me a little bit

because when I looked at the mark scheme it didn’t follow the completing the square formula

question 4

then in the mark scheme there’s a negative

shouldn’t it be (x+3/2)^2 - 3/2^2 + 6

<@&286206848099549185>

[bitter fractal]

yeah, it should be

the values of a and b given are correct, but the working out is wrong as you say

[mighty jackal]

I got a completely different answer

I had 56/9-4/9

[bitter fractal]

could you post your working?

[mighty jackal]

okay 1 sec

oh nvm

I see where I went wrong

I did 6 x 9 instead of 6 x 4

Thanks so much btw

[bitter fractal]

no worries :)

feel free to do .close if you have no more questions

[mighty jackal]

.close

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[grand hedge]

if i prove that the quadrilateral is a trapezium i am done.
But how how do i prove it

[grand hedge]

this is my diagram

<@&286206848099549185>

[grand hedge]

cud anyone please help me with this question

[spark patrol]

Hello I have a question

Anyone here?

[obtuse pebble]

@grand hedge Has your question been resolved?

[grand hedge]

<@&286206848099549185>

.close

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[rough bough]

How would you know which one is the value

[obtuse pebble]

@rough bough Has your question been resolved?

[rough bough]

<@&286206848099549185>

[strong vale]

Are you aware that $1+\omega+\omega^2 = 0$?

[rugged kite]

(type \omega for the letter)

[strong vale]

thx

[warm shale]

azeem321

[rough bough]

@strong vale yeah

[strong vale]

So... $1+2\omega + 3\omega^2 = -2 - \omega$ and $1+3\omega + 2\omega^2 = -3 +\omega$ Can you figure out now?

[warm shale]

azeem321

[rough bough]

yeah

but how did you make that equation though?

[strong vale]

$1+\omega + \omega^2 = 0$

[warm shale]

azeem321

[rough bough]

yeah okay

[strong vale]

This will always be true when dealing with nth roots of unity

[rugged kite]

(so long as they aren't equal to 1, this is just what happens when you factor out ω^n-1)

[rough bough]

wait that identity 1 + w + w^2 is always true no matter it would be w^8?

[rugged kite]

as long as the root of unity isn't 1, it does work

[rough bough]

okay

[rugged kite]

in this case you have a third root of unity (that isn't 1), so ω⁸ does indeed work

(mostly because it's equal to ω²)

[rough bough]

@rugged kite what about for w^2

[rugged kite]

yes..?

[rough bough]

What is the proof it works for w^2

wouldn't that that only be the case for when w^3 = 1

[strong vale]

I think you are misunderstanding something. We use $\omega$ to denote a complex nth root of unity. $\omega \not= 1$ as $\omega^n = 1$. Therefore let's consider $z^4 = 1$. The roots of which are $1,-1,i,-i$. Notice that the complex roots namely $i,-i$ appear in conjugate pairs. If we let $\omega = i$, then $\omega^2 = -1, \omega^3 = -i$ and $\omega^4 = 1$

[rough bough]

yeah

[strong vale]

Did I answer your question or did i misunderstand your question?

[warm shale]

azeem321

[rough bough]

Can you elaborate on why w cannot equal 1 as w^n = 1

when w = 1

no matter the n, it should always be equal to 1

and we use it for w^4

[strong vale]

Taking powers of one of the complex roots produces the other roots. The why behind this can be seen by graphing the roots on an argand diagram and considering their arguments

[rough bough]

yeah in the argand diagram they would be evenly spaced the arguments

[obtuse pebble]

@rough bough Has your question been resolved?

[rough bough]

@strong vale If you were to take account of the book's solution, i don't get how they are trying to decide which one is either part of the true value solution

[strong vale]

So the roots of the quadratic are $x = \frac{-3 \pm i\sqrt{3}}{2}$

[warm shale]

azeem321

[rough bough]

yeah

[strong vale]

They are defining $\omega = -\frac{-1}{2} + \frac{\sqrt{3}i}{2}$

[warm shale]

azeem321

[strong vale]

$\arg(\omega) = \frac{2\pi}{3}$

[warm shale]

azeem321

[strong vale]

You get that?

[rough bough]

-1/2 + root(3)i/2 ?

shouldn't that be -3/2

[strong vale]

I am talking about the cube roots of unity. The solutions to this $z^3 = 1$

[warm shale]

azeem321

[rough bough]

yeah

[strong vale]

Because of what I told you earlier, $arg(\omega^2) = \cos(\frac{4\pi}{3}) + i\sin(\frac{4\pi}{3})$

[warm shale]

azeem321

[strong vale]

Do you get that?

[rough bough]

why is it representing the arg(w^2) to be equal in mod arg form?

[strong vale]

Remember the cube roots of unity ALL lie on a unit circle

You get that?

[rough bough]

yeah i get that they lie on a unit circle

[strong vale]

So they can all be expressed in this form

[rough bough]

but isn't this considering the argument of w^2

well yoyu can generalise into that since it is a unit circle