help-26

[spark ibex]

the program I'm on that my teach assigned me for summer

I've divided a figure one way but it wants it a specific way

so i get confused

So just do it the way I have it right now?

[fiery badge]

The way you have it now is perfectly legal, yes

[spark ibex]

ok thank you

[fiery badge]

If you're in a tyranical government

then they may ask you to comply with their way

calling yours wrong

do not listen to them, liberate yourself from the shackles of "my way" math

[spark ibex]

that's what my teacher tells us

can you just stay here for a lil longer just in case i get it wrong

[fiery badge]

Sure

[spark ibex]

ok thank you

So according to what I think

I did this right

and it equals 109.5

sq in

[fiery badge]

How'd you get 48?

[spark ibex]

OHHHH SHOOT

i didnt type in the answer yet tho

I just realized my mistake

it's supposed to be 36

[fiery badge]

ye

[spark ibex]

ok I am 100% confident in my answer

it's 97.5 sq in

[fiery badge]

looks good to me now

[spark ibex]

ok

ok I got it right

I need one more and I'm done

but I need to make sure how I split the figure is legal

is this as legal as my birthdate

[fiery badge]

sure looks legal to me

[spark ibex]

ok

i put numbers in

sorry if they bad

but im on a laptop

and using my mouse makes it worse

so it's 104.5 sq cm right?

[fiery badge]

Looks good

[spark ibex]

ok

thanks I'm done

bye

.close

[topaz sinew]

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[blazing mauve]

which question

[cunning kayak]

and also:

[topaz sinew]

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

[slow flower]

b,c,d,e

idk where to start

[blazing mauve]

for b, since the entire expression is equal to 0, you can just consider the numerator, and so x+5=0, and x=-5

[cunning kayak]

for c take x common (dont cancel it out) and then factorize quad eqn

for d, tell me what method you use for solving inequalities?

[slow flower]

for d i could do

[cunning kayak]

and for e take LCM and simplify

[slow flower]

(x+3)(x-3)/x+1 _> 0

[cunning kayak]

correct

then?

[slow flower]

can u help on c a lil more

[cunning kayak]

first we finish d

and then we go on to c

[slow flower]

okay

[cunning kayak]

so yk wavy curve method

[slow flower]

ive heard but not really

[cunning kayak]

aha actually i also don't know wavy curve method, i just solve it with some diff method

and I cannot explain it unfortunately

[slow flower]

dang

[cunning kayak]

I suggest you to learn wavy curve method from yt or just ask your teacher

[slow flower]

ill try from yt cause

this work ahs to be done by tomorrow

haha

its summer work for ap calc

[cunning kayak]

for c,
simplified eqn will be:
x(3x²-14x-5)=0

can you factorize 3x²-14x-5?

[slow flower]

uh

[cunning kayak]

tell me 2 numbers

whose sum is -14

[slow flower]

wait wait

oh

[cunning kayak]

and product is 15

[slow flower]

okay

[cunning kayak]

tell me

[slow flower]

LMFOA

hold up

-15 and 1

wait but the product

isnt 15

er

[cunning kayak]

tell me, before i disturb you at your dreams

yea correct

-15

[slow flower]

oh

thought u were saying pos

[cunning kayak]

product = -15

[slow flower]

i was like

uh

dont think i can do that

[cunning kayak]

i meant my bad

[slow flower]

NP

[cunning kayak]

so we write the quad as:

3x²-15x+x-5

now we take 3 common

3x*

3x(x-5)+1(x-5)
(3x+1)(x-5)

now we keep it in original eqn which is:

x(3x+1)(x-5)=0

can you tell for what values of x, the eqn is satisfied?

[slow flower]

1 .sec

im assuming its -1 and 5

[cunning kayak]

how -1?

[slow flower]

tbh idk

is it wrong

[cunning kayak]

yea

[slow flower]

is 5 right?

[cunning kayak]

yes

just equate them seperately

like

x=0

3x+1=0

x-5=0

[slow flower]

oh

oops

[cunning kayak]

now tell me all values of x

there will be 3

[slow flower]

okay so x=0,-1/3 and 5

[cunning kayak]

correct

and that's your answer

ez

[slow flower]

ez

[cunning kayak]

now I go to sleep, it's 2:30 am here

[slow flower]

ahah

[cunning kayak]

b is really easy

[slow flower]

okay

[cunning kayak]

do it

[slow flower]

i finished b

just neewd to do d and e

[cunning kayak]

aha nice

do e

take lcm

[slow flower]

ok

[cunning kayak]

and tell me what you get

I am waiting for you

[slow flower]

lol

its 2

[cunning kayak]

aha i trust you

[slow flower]

i think it is

[cunning kayak]

would be correct

now go learn wavy curve method

[slow flower]

ye 1-1 = 0

kk

lol

ty

[cunning kayak]

yup np

and ping helpers if you need help

[slow flower]

kk

[slow flower]

<@&286206848099549185> can someone elaborate on d

im confused on the wavy curve method

[pseudo bone]

not sure about the wavy curve method

but what you usually do when you have a situation like this:
f(x)/g(x) >= 0
Is analyze the sign of f(x) and g(x)

In this case you want to see when:
x^2 - 9 is positive, negative, or equal to 0
and when:
x+1 is positive, negative or equal to 0

[slow flower]

hm

ngl im trapped

[pseudo bone]

so what we do is draw a tabel like this:

and ill complete for x^2 -9, and then you complete for x+1

[slow flower]

okay

[pseudo bone]

made a mistake, its -9 lol

oops

ill correct it and show u

[slow flower]

okay

im just confused tbh

and my brain is fried

ive done like 30 problems today

[pseudo bone]

no worries, you should take a break, ill show you how to do it for x^2 - 9 and you can come back whenever you want and try for x+1

[slow flower]

i just need to do it now tbh

i have to have it done by tomorrow so

[pseudo bone]

oh

now you need to do the same for x+1, draw the lines where the function is equal to 0, and then see on the left and right side if its positive or negative

[cunning kayak]

oh yea I use the same method

[pseudo bone]

after doing that, complete the last line, note that the fraction will be positive if both the top and bottom are positive, or they are both negative, and if atleast one is negative than the fraction will be negative. If the denominator is 0, the fraction isnt defined so you will get a strict inequality, if the numerator is 0 the fraction is 0

@slow flower if you're too tired heres the solution:

you just need to see when its positive or equal to 0 since thats when the function is > or = to 0

but this is how you would go about solving these types of inequalities since you can never multiply by x unless you explicitly know its sign

[topaz sinew]

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[earnest glade]

A standard deck of cards contains 52 cards. There are four suits, (diamonds, spades, clubs, hearts), with 13 cards per suit. Given that a card chosen out of a standard deck is a diamond, what is the probability that it is a face card (king, queen, or jack)?

[earnest glade]

this is pretty simple stuff but the wording is throwing me a little

3/52 is right im 90 sure but there are 3 face cards per type of card, right?

idk anymore

[stuck hearth]

Is it not 3/13

[earnest glade]

oh no

[wary tulip]

why would it be 3/52

[earnest glade]

this is gonna be so embarassing let me read that again

[stuck hearth]

You have a condition

[earnest glade]

probability of getting a diamond face card

[stuck hearth]

Yeah

[earnest glade]

3 diamond face cards in a deck

[icy sky]

reread it

it should probability of getting _ given _

[earnest glade]

oh no

[earnest glade]

what do you mean by this

[stuck hearth]

Condition

As in conditional probability

[wary tulip]

i think he meant the condition is a card chosen out of a standard deck is a diamond

[earnest glade]

i'm not helping my own case

anyways-

[rigid schooner]

Let's isolate what we have - we know we pulled a diamond, which means there are 13 cards we could have. Out of these 13, 3 are face cards. Therefore, our answer should be 3/13

[earnest glade]

yep yep

easy

ok this was a little humiliating

thank you all lmao

[rigid schooner]

np, take care

[earnest glade]

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[stuck hearth]

lol no problem

[earnest glade]

nah me too thing is either way they meant it they werent wrong

.close

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[clever citrus]

the difference between these 2 are that small-sample should work for small n and large sample should work for n >= 30? I dont rlly understand the difference

[clever citrus]

because 2 slides after the small-sample CI I see this:

that means the t method will also work for large n?

oh im slow so the first one is for mu, sigma(unknown) and second is for mu, sigma(known)

[clever citrus]

now im only confused about this. it says (small sample) but here it says when sample size is larger than 40, 50 the t method is safe to use

so that means it can work for both large and small samples?

[stuck hearth]

t-distribution

Converges in distribution to the normal

So yes it works with large samples

[clever citrus]

so the sample size doesnt rlly matter? if it works for both small and large

also why is it named small-sample CI then

[stuck hearth]

You can always use t

But it’s jsut more annoying to use

Than the normal

Because in t you have a degree of freedom parameter

And the pdf is uglier

[clever citrus]

oh wait im so dumb. so the z method works for larger samples and t method works for all of them. but for small sample sizes we have to use the t method thats why its called small-sample CI?

[stuck hearth]

Ye

[clever citrus]

I get it

tysm

[stuck hearth]

U could use t for large too

But it’s annoying

[clever citrus]

ye but easier to use z

[stuck hearth]

Yes

[clever citrus]

tysm

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[earnest glade]

There are 12 people that have entered a gaming tournament with prizes for 1st place, 2nd place, and 3rd place. How many combinations of winners can there be?

[earnest glade]

order matters here

so i say 1320

sounds right?

[stuck hearth]

12x11x10

[earnest glade]

nice

thanks for the confirmation!

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[sour egret]

Hello all!
Yesterday I asked this question (photo) and was told to multiply by 4 and 8. Then I asked why specifically 4 and 8 and the kind helper replied with “Well, you have that the lengths of the larger solid are 2x the lengths of the smaller one, which means that the surface areas will be 2^2 larger, and volumes are 2^3 larger”

Okay. Makes sense. But what about this new question where the scale factor is 2:7? How can I know how many times bigger or smaller the objects are?

[craggy ermine]

Surface area scales quadratically with respect to length, and volume scales cubicly with respect length

so you would have $SA = 28 \cdot (\frac{7}{2})^2$

[thorny flame]

shawn_xu

[sour egret]

7 on top?

[craggy ermine]

yes

and $V = 216 \cdot (\frac{7}{2})^3$

[thorny flame]

shawn_xu

[sour egret]

Wait why is 7 on top

[craggy ermine]

because the scaling factor is 2:7

and you are given the surface area and volume of the smaller solid

so the length of the smaller solid to the larger solid would be 2 to 7

and if you solve the ratio the length of the larger solid is $\frac{7}{2}$ the length of the smaller solid

[thorny flame]

shawn_xu

[sour egret]

Ohh okay

[sour egret]

If the scale factor was like 4:8 would I do 8/4 ?

[craggy ermine]

yes

if the problem gives you the volume and surface area of the smaller solid

[sour egret]

But if it gave me the v and sa of the larger solid then I’d put 4/8?

[craggy ermine]

yes

[sour egret]

Oh okay

Thank you

Sorry to bother but I have one more question. For this one I picked 2 sides and put them as a fraction (8/12) and I got 2/3 = 2:3 but idk it’s wrong

[craggy ermine]

well it's similar to the questions above

you have the the length of the big cake being $\frac{3}{2}$ the length of the small cake

[thorny flame]

shawn_xu

[craggy ermine]

so the volume of the big cake is $(\frac{3}{2})^3$ the volume of the small cake

[thorny flame]

shawn_xu

[sour egret]

How did you know to put it 3/2 and not 2/3

Bc they’re asking small to large?

And not large to small?

I’m just confused how you can tell

[fair copper]

Or 8:27

[topaz sinew]

@sour egret Has your question been resolved?

[sour egret]

Why 27

Idk

Where are you guys getting these nunbers

[fair copper]

Since you need ratio of volumes

All I think is to be done is multiply all sides

And then take the ratios

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[blissful python]

can someone explain naive bayes classifier in extremely intuitive terms?

[blissful python]

nvm

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[livid garden]

I can't seem to do this

[livid garden]

btw it is 30 degrees not 630

[pseudo sonnet]

,rotate

[livid garden]

thx

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[livid garden]

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[steep ibex]

I am stuck on this one.

[steep ibex]

Let $\vec{L} = \begin{bmatrix} 1\ -6 \ -2\ \end{bmatrix}$. $proj_L(\vec{v}) = \vec{v}(\frac{\vec{v}\cdot\vec{L}}{\vec{v}\cdot\vec{v}})$

This is what I've established so far.

[thorny flame]

Narutoes

[steep ibex]

I got the dot product fraction to be -6/23

But it's saying it's wrong.

The software is saying the answer is $\begin{bmatrix} -\frac{42}{41}\ \frac{252}{41}\ \frac{84}{41} \end{bmatrix}$ and I have no idea where those numbers came from.

[thorny flame]

Narutoes

[topaz sinew]

@steep ibex Has your question been resolved?

[cinder sequoia]

shouldn't it be v (v dot L) / (L dot L)?

[steep ibex]

How so?

[cinder sequoia]

i'm not positive but iirc that's the formula

[steep ibex]

I thought we were projecting v onto L

[cinder sequoia]

we are

afaik that's how you'd do that

might be wrong

[steep ibex]

Lemme try that and see what happens.

I missed this day of class because I was in the hospital for 5 hours.

So what I have is notes from the guy next to me.

[cinder sequoia]

u spans the space that you're projecting onto, v is the vector you're projecting

so in your case it should be <v, L> / <L, L> v

[steep ibex]

So it would be -42/41? That fits more with what the problem is saying

[cinder sequoia]

i haven't checked the computation

yeah -42/41 times v

[steep ibex]

Ok cool

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[grand cargo]

i need help with homework

[rigid cloak]

Yo

[grand cargo]

<@&286206848099549185>

oh

hi

[sudden temple]

!15m

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

[grand cargo]

i got a math question that needs help

the topic is called plotting linear graphs

and the question is, plot 2 points and use them to draw a linear graph

the equation is y=2x-1

idk how to do this one

i need someone to tell me what i need to do

i dont even know the vaule of x

at the top

it says x vaules -3, -2, -1, 0, 1, 2, 3

ill send the pdf file of what im talking about

[slow cypress]

They ask you to plot some points and link them. You need to find a y-value at a specific x-value

[slow cypress]

For example, what is the value of y when x=0?

[grand cargo]

here is my lesson im doing

you can take a look if you want

my homework is the examples

the first example

[rigid cloak]

Yo

[sudden temple]

!filetype

[rigid cloak]

I'm back

[topaz sinew]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[rigid cloak]

!fishy

I am saying that don't send links which may seem fishy

Send a SS or elaborate

What u want to ask

[grand cargo]

oh oops

sorry

ill show you my question after i ss my math work

[rigid cloak]

Okok

[grand cargo]

here it is

ive already done the first one

its the second one that is confusing

[sudden temple]

they just want you to plot that curve with 2 points

[grand cargo]

how should i do that

or what should it look like

[sudden temple]

It is given that it is a linear graph

[grand cargo]

ok

[sudden temple]

to plot it you just have to consider any points

[grand cargo]

do you have to like make a table with x and y with the equation or not

[sudden temple]

yup

that's what you have to do

[grand cargo]

ok

[sudden temple]

But that fact is that you are just allowed to get two points

[grand cargo]

tho my next problem is finding the vaule of x

with that equation

y=2x-1

would x be 1

or is y equal to 1

[sudden temple]

you just have to take any values that you like

[sudden temple]

for example take x=something and then solve for y
and you'll get a point

[grand cargo]

ok

[sudden temple]

same for other point

[grand cargo]

so how do i solve my equation to find what i need

[sudden temple]

You're not supposed to solve for the equation
you're just given one

[grand cargo]

oh

[sudden temple]

you have to plot it

[grand cargo]

whats my first x and y gonna be

if my y is equal to 1

and my x is any number

[sudden temple]

follow this procedure

[grand cargo]

ok

man linear graphs are confusing

not the 4 quadrants tho

[sudden temple]

or you can do vice versa by taking y=something and then find corresponding x value to get the point

[grand cargo]

so basically

ill make the table first?

ok

[sudden temple]

yup

[grand cargo]

then ill have to make a linear graph which i have already done with 4 equal quadrants and with y and x axis

all i need to do is plot 2 points of the y=2x-1

ok

i think i can do this

[sudden temple]

and do not forget to join those 2 points for the actual curve

[grand cargo]

ok

hmmm

so what is my first dot going to be

where i meant

i dont know the decimals on which one

there negative and positive numbers on equal sides of the quadrants

[sudden temple]

huh?

what are you doing with quadrants?

[grand cargo]

no because a linear graph has 4 quadrants

what is the highest number should be for my linear graph

[sudden temple]

@rigid cloak you here?

[grand cargo]

he is gone

he offline

tho he did chat to me before here

[rigid cloak]

Yo

[sudden temple]

bro you don't have to consider any maxima/minima for this curve

[grand cargo]

what?

first]

[rigid cloak]

Bro

[grand cargo]

idk what you mean by maxima and minima

[rigid cloak]

Plot the straight line

[grand cargo]

straight?

77 told me it was curved

ok i got it ow

now

i got the 2 plots on my cartesian plane

thx for the help

but now i got another homework

its called features of linear graphs

[sudden temple]

bro a curve literally could be any straight line/point or anything else

[grand cargo]

this includes gradients

ill get a ss of my last homework i need help with

heres my first part of my last homework

it says calculate the gradients of the lines

which is confusing for me

my homework is a and b and d

explain what a gradient is pls

hello?

<@&286206848099549185>

ok

so my first gradient should first have to be a calculation of a rise and run?

ok

so all i need to do is find the rise and run

hey i have that formula in my lesson thing

so its the substitution thing

ok

ok

oh

my rise and run for c was 0 being my rise and 5.1 for my run

oh wow

thats cool

i never expected math teachers to use discord if im being totally honest

anyways

my last questio is now b

[rigid cloak]

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[rigid cloak]

.reopen

@grand cargo

See he is helping you

[topaz sinew]

@mild tapir Has your question been resolved?

[neon iron]

<@&268886789983436800>

[topaz sinew]

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[hoary fable]

how do u do part b?

[hoary fable]

do u start by doing this?

[rigid cloak]

Yo

Do u know

Arctan(x)+arctan(y) ?

[brave coral]

I feel like they meant tan^-1 2x

But it's solvable either way

[rigid cloak]

Nah I'd win 🗣️ 🔥

[brave coral]

The hence bit tho is what I'm confused about

[lone nest]

Stop disrupting help channels with nonsense, please.

[brave coral]

Idk what keygen funk is

[rigid cloak]

I'm sry got carried away

[hoary fable]

nope

[rigid cloak]

See

Learn the formula

arctan(x)+arctan(y)=arctan(x+y/1-xy)

Now solve

[brave coral]

You won't be able to use the previous identity if you set alpha = tan^-1 x because then beta = tan^-1 2x which won't work with your equation in b

Which is why I'm half certain there's a typo in that question

Anyway, you can still solve it with the exact same way you derived the equation in part a

[rigid cloak]

U learn the formula you will be able to solve then

[hoary fable]

Wait wdym

[brave coral]

You used compound angle formula to show the equation in part a right?

[hoary fable]

Yes

[brave coral]

Just do the same for part b

Take the tan of both sides

[hoary fable]

So after I do that, I expand LHS and it will get me the answer?

[brave coral]

You'll have to solve the equation you get from doing that

[hoary fable]

I got x=1/3 and the answer is x=1/2

[rigid cloak]

Ok

We see

[reef adder]

But this is for the case when xy<1 tho we have to know about xy to apply arctan x+arctan y formula

[rigid cloak]

Yes sure good enough

For xy=1 it's π/2

[reef adder]

yea

[brave coral]

Okay, so it might be a typo after all

[rigid cloak]

I feel this is the problem

[reef adder]

this is good approach

[hoary fable]

Oh

[brave coral]

Replace the 2 with 2x

[rigid cloak]

Ohok

[hoary fable]

so like this?

[brave coral]

That is what I mean but I solved it and didn't get a 1/2 either

[hoary fable]

Oh

I think I get how do it, just the typo is messing the answer up

Thanks a lot :)

.close

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[brave coral]

Np

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[deft holly]

can I have help with part b

[clear nova]

what have you got so far

[deft holly]

nothing, all i got is the costheta = uv/|u||v|

and idk what to do with it

[wintry kraken]

Isolate u.v

and square it

[deft holly]

:/

wait

can i do this, make u and v into magnitude. by squaring both of them and then cause |u||v| is hyp, its larger than the other one

[wintry kraken]

There is a much simpler way with the calculation

[deft holly]

ah but then ig it cant be equal to

[deft holly]

what do i do with the costheta then

[wintry kraken]

Just write (u.v)² = ...

[deft holly]

(u.v)^2=|u|^2|v|^2 cos^2theta

[wintry kraken]

exactly

what can you say about cos² theta ?

[deft holly]

0 to 1

idk

[wintry kraken]

you got it you know

[wintry kraken]

What does the question ask you?

[deft holly]

?

it asks for bounds

no

sorry

asks for whatever it says there im not bothered to write it

i dont care about the equality part

[wintry kraken]

Do you agree with that? (sorry for the writing)

[deft holly]

fair

ah

i forgot im allowed to do that

thanks for the help

.close

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[wintry kraken]

and btw, (u.v)² = | u | ² * |v| ²

when theta = 0 or 180

but i think you guessed it

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[cyan harbor]

Is {(-3/4), (-1/2), (1/2), (3/4)} function or mere relation?

[cyan harbor]

.close

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[cyan harbor]

{(1/2,5),(-1/2,6),(3/4,7),(-3/4,8)} is it a function or mere relation

Is it okay for x values to repeat if the other one is a negative?

.close

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[shell orbit]

If im not mistaken the only vertical tangent is on the far left

So im not really sure what exactly we do

We need the derivative of something to get the slope of the tangent

And I think we set something equal to 0 cause the slope of the tangent is 0

But I dont know specifically the steps

Im also thinking maybe we ignore y(t) and only do something with x(t)? Because we dont care about the y axis or something? Really not sure

<@&286206848099549185>

[topaz sinew]

@shell orbit Has your question been resolved?

[shell orbit]

<@&286206848099549185>

<@&286206848099549185>

QQ

[silver acorn]

??

[shell orbit]

??

[silver acorn]

its so hard

give me 2h

[craggy jay]

OH SUP

wait

[rigid cloak]

@shell orbit

Vertical tangent line

That like slope is 90

find dy/dx as (dy/dt)/(dx/dt)

,w d/dt 2t^4-t^2

[thorny flame]

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[rigid cloak]

,w d/dt t^3-3t

[thorny flame]

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Upgrade to WolframAlpha Pro!

Requested by oppailol.

[rigid cloak]

So dy/dx=

3(t^2-1)/(8t^3-2t)

Now we need dy/dx=tan90

That is not defined

Only possible when 8t^3-2t becomes 0

,w 8t^3-2t=0

[thorny flame]

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[shell orbit]

Wait sry to interrupt I have a lot of questions

[rigid cloak]

Okok

Ask

[shell orbit]

Was I right about this? That the red line is the vertical tangent to both points?

We need the derivative of something to get the slope of the tangent
And I think we set something equal to 0 cause the slope of the tangent is 0
Im also thinking maybe we ignore y(t) and only do something with x(t)? Because we dont care about the y axis or something? Really not sure
And was I right about these?

Why do we want tan(90)?

[rigid cloak]

The three points we get for

Three values of t

Fr

No

Slope of tangent is undefined

Slope is the tan of angle made with+ve horizontal axis

Y axis makes angle 90 degree

Slope is tan(theta)

Like dy/dx=tant

If line is vertical

t=90

And tan(90)is undefined

That is slope is also undefined

[shell orbit]

Ok I see

1 more thing

[rigid cloak]

That's why I made dy/dx undefined

By finding values for which denominator is 0

Ok

[shell orbit]

The middle point I can clearly see is at (0, 0), can I just say 'I see from the graph that its 0, 0' or do I have to use time proving it? I dont have a lot of time in my exam...

[rigid cloak]

See

U have to prove

Ike see when I set the denominator

8t^3-2t=0

I get t=0

As a solution for which slope is undefined

And so I get x=0 and y=0 by putting

Moreover depends on your exam

If MCQ or choice baswd

Mark directly

Else show

The working

[shell orbit]

Hmm ok

Thats easy to prove at least

I think I know how to solve the rest tho

I just sub in 1/2 for t

And -1/2

And get the coordinates

easy enough

[rigid cloak]

Okok

Welcomd

[shell orbit]

Yeye thank you bro

❤️

.close

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[clear juniper]

Yay finally a trigonometry question

What's the doubt

-_-

[neon iron]

Can u explain me why cot theta has values

When it is 0

Tan theta also becomes 0

[clear juniper]

bruh

[deft holly]

nahh

[neon iron]

Then shouldn't it be pi and 2 pi?

[clear juniper]

Depends on angles

And there are multiple angles

That might be some equation

[deft holly]

you know the tangent graph?

[neon iron]

Tan theta=0 and so theta=0

[clear juniper]

Like as he mentioned in above question? Or typically?

[clear juniper]

Okay and?

[deft holly]

i just wanna talk about asymptotes

[neon iron]

Where does tangent comes from?

[deft holly]

tanx= infinity on those parts

so 1/infinity = 0

[clear juniper]

Asymptotes well like the specific tan graph

[neon iron]

So I will take values of 1st and 3rd quadrants only

But why the heck theta is half pi and 3/2 pi

[clear juniper]

I'm confused

First

Share the question

Like the question of the answer u mentioned

[deft holly]

yea

uh

cottheta=0 does not become tantheta = 0

[clear juniper]

Womp

Tan theta is infinite

[rigid cloak]

Yo

[deft holly]

fr tho

ayo whatsup

[clear juniper]

sup

[neon iron]

Here is the question

[clear juniper]

Which

[clear juniper]

Blud first check ur answer

[neon iron]

Be it's zero...

[clear juniper]

How did u get tan theta zero

Bruh

[deft holly]

bro really said 1/0 is 0

[clear juniper]

Cot inverse is tan

[clear juniper]

Einstein

@neon iron any more doubtss

[neon iron]

Good question 😢 TT

[neon iron]

Then how am I supposed to solve this

[deft holly]

when tanthta is infinity

[clear juniper]

Tan theta is infinity

[neon iron]

Please I am ashamed

[deft holly]

1/infinity is 0

[clear juniper]

Calm down it's fun

No need to be ashamed

[deft holly]

weve all done something like this before lmao

[clear juniper]

@deft holly explain him

No confusion

-_+

[neon iron]

Seems like it's new for me, never done something like this before

[deft holly]

so

you know the tan graph?

you know the asymptotes?

[neon iron]

OK, I get this but later what happen

Like how did the solution comes 1/2 pi?

[neon iron]

Yes

[deft holly]

yea

tantheta = infinity on asymptotes

[neon iron]

Okay then...and asymptotes lines are on x= 1/2 pi, 3/2pi like this?

[deft holly]

yea

[neon iron]

Can u explain a little bit further?

[deft holly]

on what exactly

[neon iron]

If I want my answer while doing on the quadrants?

[deft holly]

i mean

what im saying is just a theory thing

this is also just exact values of cotangent

[vernal ledge]

its not defined

[clear juniper]

Yeah yeah

Same stuff

[deft holly]

it approaches infinity

[clear juniper]

Lol

[clear juniper]

Doubt solved?

[neon iron]

Ig kind of

[clear juniper]

Yeah

Type.close

[neon iron]

Thanks @deft holly @clear juniper

.close

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[clear juniper]

Hmm

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[strong sable]

\left.f(x,y)=\begin{cases}\frac{2x|y|}{\sqrt{x^2+y^2}}&(x,y)\neq(0,0),\0&(x,y)=(0,0).\end{cases}\right.

[thorny flame]

riyobi
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[strong sable]

does f_x(0,0) = 0 = f_y(0,0)?

[ebon thunder]

for all values of 0 it is 0 otherwise it is the top function

[topaz sinew]

@strong sable Has your question been resolved?

[clear juniper]

Hu

Hola

What's the doubt

[strong sable]

This formula seems so complex

why cannot we just use this

[strong sable]

I only want to know this so why do i need the differentiable thing

[golden blade]

Well you wanna know if there is a derivative at (0,0) so the approach I sent above would be to see if it is differentiable at (0,0). If it is differentiable in (0,0) then it has a derivative in (0,0)

[strong sable]

why cannot i just directly check if the drivatives exist? why do i have to check it by checking if it's differentiable?

[golden blade]

Well you can also calculate the partials and then see if f_x and f_y exist at (0,0)

[strong sable]

so is this correct?

[golden blade]

I guess

yea

[strong sable]

okk, thank you, i'll let this channel open to see if others can confirm this maybe

[shadow salmon]

Well in general you could have partials exist but not differentiability

[shadow salmon]

Yeah thats correct, since you seemed to only ask about the partials

[clear juniper]

Hey

Is there still a problem

[golden blade]

#help-18 message

[shadow salmon]

?

theyre asking about the partial derivates, whatever just the word "derivative" means in this case

[golden blade]

how can the partial derivative exist in one point but at the same time you say it's not differentiable there?

[shadow salmon]

By the definitions

theres some classic examples of it

theres even some bizarre ones

https://math.stackexchange.com/questions/3831023/can-a-function-have-partial-derivatives-be-continuous-but-not-be-differentiable

However what you can say is that if your partials are continuous, then yes we certainly have differentiability

[golden blade]

then i dont understand snow's statement

what was wrong with my statement then

[shadow salmon]

Maybe theyre talking about some other notion of the word "derivative"?

and not partials in this case

[golden blade]

idk if i should ping them but anyway

[shadow salmon]

if they dont bother sure

[golden blade]

i mean it's the same situation with |x|

they said that differentiable and derivative exists is synonymous

[shadow salmon]

how is it the same with |x|?

[golden blade]

but that would go against what you sent

[golden blade]

well there exists a derivative by calculation

but it's not differentiable in x = 0

[shadow salmon]

Uh, how exactly?

[golden blade]

x/|x|

[shadow salmon]

What

no

Everywhere aprt from x=0 the derivative exsits

[golden blade]

and what did i say?

[shadow salmon]

yeah but that's different in this case

it's differentiable also aprt from x=0, in the one variable case it doesn't make sense to talk about partials

[golden blade]

i wasn't talking about partials

[shadow salmon]

yeah so how is this the same in that case??

[golden blade]

it's an analogy

[shadow salmon]

Uh how?

[golden blade]

in both cases you are asking if it's differntiable in some point

[shadow salmon]

It would be an anaolgy if you had some kind of notion of "derivative" which existed at x=0

[golden blade]

one case is in one dimensional other in two dimensional

well you have x/|x| but the right and left hand side limit are not equal

[shadow salmon]

for the partials we have a limit at particular point, meanwhile for differentiability this does not work for the same point

so i dont see how this is an analogy

in the one variable case the the notion of derivative and differentiability are the same thing since their def. are the same, but for higher dim. we have a notion of a partial derivatives, and then differentiability again; but this time their def. are different

and maybe snow is using the word "derivative" differently here, but for partials this is most definitely not true unless im missing some context from ur convo with them

[golden blade]

i don't understand why it would be different in higher dimension, the partials derivatives are also just a limit that comes from the definition in one-dimensional case

[shadow salmon]

have u checked the definitions?

you can sorta see how they reduce to the same thing once we're working in one variable

also what's funny is that this can be a wording thing too, in Sweden we dont ever use the word differentiability until we talk about the def. in higher dim. and that was precisely to avoid confusion

[golden blade]

it's not funny cause i am getting crazy lately and feel my whole concepts are based on trash

it's not funny when you are entering your 5th semester knowing shit

[shadow salmon]

I think it's funny in the sense that maybe it's the way the words are badly used is what is causing the confusion

i.e not your fault but the way theyve been taught

like we would use the word "derivable" if i try and translate it for one variable and i thought it was dumb right up until we started using the "right" word in higher dim. and it clicked a bit faster that way imo

[golden blade]

Well we defined in one-dimensional case back then that a function f is differentiable in some x0 in D(f) if 1. the left hand side limit equals the right hand limit of the derivative limit definition with that "h" and 2. f is continuous in x0

[shadow salmon]

notice how that just becomes the same thing if we try and reduce the dim for the partials def. aswell

in reality i think it's sorta weird to talk about partials in one variable, but i would say we are basically treating it as the derivative in that case

[golden blade]

for the 2-dimensional case we didn't even bother to define differentiability other then what partials are and how you calculate them. We only got shown how to investigate continuity

[shadow salmon]

Huh thats weird

Also btw we should prob move this convo away from this help channel

@strong sable i hope u got the help needed btw

[strong sable]

no this is a related topic, i can learn from it also

Thank you btw

.close

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[wise frigate]

I feel like the function is supposed to have a ^2 in it somewhere but I’m not exactly sure where

[outer portal]

the 3 on the left before the ()s isnt correct, youll need to check that

remember that f(0) should be -3

the function doesnt have a ^2 in it because all of the roots cross through the x-axis like lines, so theyre ^1

[wise frigate]

if a graph as x turning points, wouldn’t have a degree that’s is essentially x + 1 ?

that’s how I got three, for context

[outer portal]

if you can see the turning points, then you can see the roots

when you add up the degrees of each root you still get 3

something like this is degree 6

the root on the right is degree 1, it looks like a line
the root near the right is degree 2, it looks like a parabola
the root on the left is degree 3, it looks like a cubic
the 6 is from 1 + 2 + 3

however the graph itself only has up to 4 turning points

so seeing the roots directly will tell you the degree, the turning point idea only sometimes works

[wise frigate]

oohh okay okay

it helps a lot with the example, thank you

[outer portal]

np

[wise frigate]

.close

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[shell orbit]

Part C

c1 is (y + 1)^2 + (x + 0)^2 = 16

c2 is (y - 5)^2 + (x - 8)^2 = 36

So for the question I know we need to use the formula (ax + by + c)/sqrt(a^2 + b^2) = length of radius of circle c2

Im not really sure what to do with it though, or how to get the unknowns

[topaz sinew]

@shell orbit Has your question been resolved?

[shell orbit]

<@&286206848099549185>

[rigid cloak]

Yes

Yo

[shell orbit]

yoyo

[rigid cloak]

Give me a white page

Let's draw the diag

[shell orbit]

Wym sry?

[rigid cloak]

Can u send me a white page

A online photo

Leave it ok

Imma explain

Verbally

@shell orbit

U know end points ?

I mean diametric form oc circle?

See BC is a diameter

So B and C become end points of diameter of

C2

Now the diametric end points form says

A circle with end points of diameter (x1,y1) and (x2,y2) can be written as

(x-x1)(x-x2)+(y-y1)(y-y2)=0

Now solve and tell me c2

@shell orbit

Take your time read understand and tell me the equation of c2

[shell orbit]

ye I need to reread a few times

[shell orbit]

Its not in this form right?

[rigid cloak]

No

Check your equation of c2

[shell orbit]

Its wrong? 👀

[rigid cloak]

Yes

[shell orbit]

Or just not in the right form

Oh

[rigid cloak]

First check what I wrote

Read u will understand

[shell orbit]

(x-2)(x-14)+(y-5)(y-5)=0

[rigid cloak]

Good

Good 👍

Now I hope u can

Do

Next

[shell orbit]

Ummm

Ive never seen a circle in this form before

[rigid cloak]

See u know when do circles touch

[shell orbit]

So not really sure

[rigid cloak]

?

Two circles touch when distance b/w their centres is R1+R2

In this case they touch externally

For the case in which 1st circle is inside 2nd circle

And touches internally

It's different

[shell orbit]

(16/5, 7/5)

[rigid cloak]

What

[shell orbit]

Thats where they touch

[rigid cloak]

You can do b part

Okok

Now for C

Part

[shell orbit]

Yeye im stuck on this

[rigid cloak]

See

[shell orbit]

c1 is (y + 1)^2 + (x + 0)^2 = 16
c2 is (y - 5)^2 + (x - 8)^2 = 36
So for the question I know we need to use the formula (ax + by - c)/sqrt(a^2 + b^2) = length of radius of circle c2
Im not really sure what to do with it though, or how to get the unknowns

[rigid cloak]

Do u know

Ah

I forgot the form

Ula

Maybe I need to derive it

Wait lemme derive

See

Consider a circle

And we have a point outside it

We draw two Tangents from it to the circpe

Circle

[shell orbit]

I drew a little diagram

Very rough

[rigid cloak]

Yes

Good

Now see

I don't remember much

But see

If the lines touch

The circle

We can get the slope of them

By doing derivative as they tangent

@shell orbit

We know one point on both the lines

So we can calculate their slope using that

Then we equate the slopes using both forms

To get x and y

That is where the lines touch circle

[shell orbit]

What about my method? Using (ax + by - c)/sqrt(a^2 + b^2) = length of radius of circle c2

[rigid cloak]

I forgot so I am doing this if your have better approach then good

[shell orbit]

It doesnt use derivatives

[rigid cloak]

Hm i remembered

Correct

Perpendicular dist of a point from line

Hm

Correct

Line is our tangent

And point is centre

Hm correct