help-27

[restive river]

You didn't ping themmm

[zealous ether]

@polar wharf could you help me with this other question

[zealous ether]

lol facts

Adding all gives 4/9 but answer is 5/9 i guess

This is the question for reference

[restive river]

No one came yet?

[zealous ether]

:C no

[restive river]

I would have helped but idk abt this 1/3 method like can you explain what are you trying to do by taking (1/3)^3

[zealous ether]

np i appreciate your will to help

also 1/3^3 is basically

just cuz

if diana, for instnace, has to choose a group of 3 then there's 1/3 chance she'll choose 1 group

nnothing fancy

similarly for the other 2 brothers

so 1/3 * 1/3 * 1/3 = 1/3^3

[restive river]

Oh oh oh okayyy

Got it

Okay so

I got the answer

From this 1/3 method 😭

But my thinking again is ig different may I ?

In the first one

Where all are in same group

Shouldn't it be (1/3)^3 * 3C1

(Selection of one group where all will be together out of three groups)

That would be uhhhhhh 1/9

Then second case,

Where either B or C are with D

That means if D comes, C shoudlnt come in that place

So

[[(1/3)^2 x (3C1) ] *[(1/3)^1 x 2C1] ] into 2 for both cases

Damn

Adding both we get 5/9

@zealous ether

[restive river]

Here the first part is like B or C with D , in one group they get 3C1 chances to select one group of their choice

Then the remaining person who is not with D, gets to choose group in 2C1 ways

Whole into 2 cuz I directly chose C or B and you can do it differently

[restive river]

So here what's wrong is

[zealous ether]

lol bruh i don't get it maybe my brain is not working rn

but thank you!

I'll read it later cuz now i got class

[restive river]

When you multiply by 3! It means you're distributing B C D when you have already taken them in one group

[zealous ether]

okay sure sure

thanks once again

i'll check and then let you know

[restive river]

Aye np

[zealous ether]

cheers

[restive river]

Yeah

[zealous ether]

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[whole iron]

what can i do further to simplify ut

[sonic smelt]

Make common denominators in $1 - \frac{\cos^2\theta}{\sin^2\theta}$

[woven radish]

A Lonely Bean

[whole iron]

thanks

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[gilded wharf]

Hi guys can someone please help me get started with this question?

[supple knot]

draw a picture

[gilded wharf]

A picture?

[supple knot]

of p(x)

[gilded wharf]

I’m not sure what you mean by drawing p(x)

I’ve done something similar to that question but this one is a bit complicated

[supple knot]

the same method works.

drawing a picture isn't absolutely necessary

[supple knot]

just do the integrals like you did in this question

[gilded wharf]

That’s it?!

[supple knot]

the definition of standard deviation involves two integrals

[flat rain]

Here to learn, can I have the definition?

I only know the sum definition for like a list of data points

[gilded wharf]

‘ and “

Excuse me @flat rain

[flat rain]

Erm

Helli

Hello

Sorry I didn't mean to interrupt

[gilded wharf]

@supple knot do I work with 1 and 3

[supple knot]

what do you mean "work with"

both values are important

[gilded wharf]

Like with substituting

[gilded wharf]

Here I used 0 and 1

[supple knot]

yes both endpoints are necessary when evaluating integrals

[gilded wharf]

So far is this correct?

[supple knot]

should be dx

[gilded wharf]

Oh sorry my bad

What’s the next move?

[supple knot]

do you know how to do integrals?

power rule would be useful in this case

actually the left side is wrong

the integral is the expected value

this is p(x)

[gilded wharf]

Oh

The expected value of x

We determine it through integrate

[supple knot]

read this since you keep getting stuck
https://www.probabilitycourse.com/chapter4/4_1_2_expected_val_variance.php

[gilded wharf]

Here is the E(x)

Wait

Correct?

[supple knot]

no?

[gilded wharf]

I think the second last line is incorrect

I fixed it up

<@&286206848099549185>

[devout snow]

@gilded wharf Has your question been resolved?

[gilded wharf]

<@&286206848099549185>

[supple knot]

,w 0.25 * int 1 to 3 of x^2

[woven radish]

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[supple knot]

looks right

you can use wolframalpha to check your work

[supple knot]

read this for the rest of the definition of standard deviation

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[plush knot]

Can you explain better what you mean?

[supple knot]

^

show the entire instruction set

screenshot is better

[devout snow]

@restive river Has your question been resolved?

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@restive river Has your question been resolved?

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@restive river Has your question been resolved?

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[restive river]

How do i solve this?

[restive river]

$$x^2\cos(x) = 1$$

[woven radish]

Brandon H

[pastel pasture]

,w x^2cos(x) = 1

[woven radish]

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[pastel pasture]

Like that

[restive river]

ummm, i can't do that

[pastel pasture]

There's really no other way to solve stuff like this

Unless you're really really good at plotting graphs

[restive river]

Do you know how i'm supposed to solve this then?

[pastel pasture]

think you can use Rolle's theorem

or MVT actually

[restive river]

MVT

make sense

thank you

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[ivory bluff]

in geometry, how do you prove that some AB is longer than say AC

[ivory bluff]

like a reliable tool to do so

[pastel pasture]

triangle inequality mayhaps

Without the exact situation it would be hard to say

[ivory bluff]

anything else

i mean we have ptolmys inequality

or however you spell it

[last hill]

What are you given about A, B and C?

[manic condor]

Ptolemy's

[ivory bluff]

yes

[ivory bluff]

im just talking in general

i guess its just play with simple inequalities we have

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[alpine forge]

How?

[alpine forge]

please reply to my message

to notify me

[manic condor]

Equate their coordinates.

[alpine forge]

their parametric equations?

[versed jetty]

yes

you get a set of three simultaneous equations

[alpine forge]

sorry it's like my first/second time properly studying this

[versed jetty]

it's ok

we all have to start somewhere

if the lines meet then they'll meet at some well-defined point, and consequently the (x,y,z) of line 1 and the (x,y,z) of line 2 will have to coincide

[alpine forge]

thank you!

close

uh

[versed jetty]

thr c9mmsnd is .close

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[zealous ether]

if i have

[woven radish]

hiii

[zealous ether]

can i just add them?

so like

[woven radish]

hiii

[sonic smelt]

Of course you can

[zealous ether]

okay just confirming

one more thing

[sonic smelt]

13^17 is not congruent to 7 though I think

Same thing with the other equation

[woven radish]

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[zealous ether]

oh sure

maybe i just messed something up np

but that aside

if i have:
2^{99} \equiv 6 (mod 11)
and
2^{99} \equiv 8 (mod 9)

i want mod 99 so i'd be sieving right?

,, 2^{99} \equiv \text{something} \pmod{99}

[woven radish]

hiii

[zealous ether]

the intent of sieving is to find that something right?

which is just 17

[sonic smelt]

What's sieving?

[zealous ether]

it's used to find the first solution for n systems of congruences

after which chinese remainder generalizes

this

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[mighty flume]

How do I approach this problem?

[mighty flume]

I first have to find theta but I'm not sure how to unwrap sin θ = 1/4

would 1 be opposite and 4 for the hypotenuse?

[long pasture]

you don't have to find theta, you'll just have to find which quadrant theta is in
that's the first step

[mighty flume]

it would be in quadrant 1

[long pasture]

this would be the second step

[mighty flume]

sine would be in quadrant 1 or 2

since its positive

[long pasture]

but then, since tangent is positive, therefore it's in Q1

[mighty flume]

right right

tan > 0

[long pasture]

yep

next step

cos(theta/2)

what identities can you think of

[mighty flume]

is it a pythagorean identity?

[long pasture]

hmmm

that, we will use it later

something about double angle formula?

[mighty flume]

cos(2θ) = 1-2sin^2(θ) ?

[long pasture]

good, and similarly

cos(theta)=1-2sin²(theta/2)

or

cos(theta)=2cos²(theta/2)-1

[mighty flume]

how would i plug in sin θ as 1/4?

[long pasture]

we would now use this

sin(theta)=1/4 and theta in Q1

what will cos(theta) be

[mighty flume]

it will also be positive if its in Q1

[long pasture]

correct

and the exact number for cos(theta)?

[mighty flume]

let me draw the triangle

[long pasture]

sure

[mighty flume]

my minds blanking, how would i solve for the adjacant angle?

i have opposite = 1, hyp = 4

ohh

nvm

[long pasture]

hehe

you found it

you just solve for cosine theta

not the theta itself

which will be adj/hyp

therefore we just have to solve for the length of adj

[mighty flume]

2/4 = cosθ

[long pasture]

em...

Pythagorean?

[mighty flume]

sqroot3

square root 3

[long pasture]

1²+x²=4²

[mighty flume]

1 * 1 = 1

[long pasture]

good

[mighty flume]

4 * 4 = 16

16 -1 = 15

square root both

x = squareroot 15

[long pasture]

yey

[drifting hearth]

finding the other side won’t help since you need cos (x/2)

you’ll need to solve for theta to get the answer

[long pasture]

now that we have
cos(theta)=(√15)/4

[long pasture]

we plug back into this

[drifting hearth]

oh i see

[long pasture]

good

[mighty flume]

i put that over 2?

[long pasture]

$\cos\theta=\frac{\sqrt{15}}{4}$

[woven radish]

Biscuity

[long pasture]

nope

[long pasture]

.

[mighty flume]

wouldnt it be cos2theta = 2cos^2(theta/2)-1?

since its a double angle identity?

[long pasture]

nah

lemme type again

cos(2x)=2cos²(x)-1

[mighty flume]

oh is it because you changed the inside to /2

[long pasture]

cos(x)=2cos²(x/2)-1

[mighty flume]

i guess thats allowed i never thought about that

then i plug

[long pasture]

hehe

[mighty flume]

square root 15/4 = 2cos^2(x/2)-1

[long pasture]

good

and then simplify

[mighty flume]

I get to here

[long pasture]

opps

[mighty flume]

i feel like i misstepped in the beginning

my first step should be to get rid of the /4 right?

or move over the -1?

[long pasture]

well both works for this q

anyways

sqrt(15)/4+1=2cos²(x/2)

(sqrt(15)/4+1)/2=cos²(x/2)

good?

[mighty flume]

that it?

cant go any further right?

[long pasture]

nah

we still need to find the non-squared one

[mighty flume]

in simplifying? should i sqrt both sides?

[long pasture]

we just take sqrt on both sides of course

[mighty flume]

sqrt(sqrt(15)/4 + 1)/2 = cos(x/2)

[long pasture]

yey

remember to see if it's positive or negative!

sqrt(sqrt(15)/4 + 1)/2) = cos(x/2)

[mighty flume]

sqrt is plus or minus?

it has to be positive right?

since tan is positive

[long pasture]

since theta is in Q1, theta/2 also is in Q1, therefore +ve

[mighty flume]

yes

[long pasture]

got it?

[mighty flume]

how do i deal with the denominators?

[long pasture]

now simplify it to get the final answer

for the numerator of the whole thing

do common denominator thingy first

that is

sqrt(15)/4+4/4

[mighty flume]

huh?

[long pasture]

expand fractions

$\frac{\sqrt{15}}{4}+1=\frac{\sqrt{15}}{4}+\frac44$

[woven radish]

Biscuity

[mighty flume]

oh i see you want to focus on the numerator

okok

sqrt(15)+4 /4

[long pasture]

[mighty flume]

cross multiply instead?

[long pasture]

nah

now that we have

[mighty flume]

not sure

[long pasture]

why is there an equal sign lol

anyways

[mighty flume]

yo uwant me to get rid of the 1 first ?

we do that by having a common denominator

[long pasture]

$\sqrt{\frac{\frac{\sqrt{15}+4}{4}}{2}}$

[woven radish]

Biscuity

[mighty flume]

oh yes yes

then we divide 4 and 2 right?

[long pasture]

nah

we need to be careful with this

[mighty flume]

oh wait no

yes

[long pasture]

it's the numerator part divided by 2

[mighty flume]

the denominator of 4 isnt being divided by 2?

[long pasture]

$\sqrt{\frac{\sqrt{15}+4}{8}}$

[woven radish]

Biscuity

[long pasture]

this

[mighty flume]

oh so we can multiply the denominators

i thought we could but looking at the answer choices i wasnt sure

oh wait

nono i get it

sqrt8

[long pasture]

hehe

now that 8 is NOT a square number, we do something to make it into a square number

i.e. expand this new fraction by 2/2

[mighty flume]

expand 4

right?

no we can

multiply 8

[long pasture]

$\sqrt{\frac{\sqrt{15}+4}{8}\cdot\frac22}$

[mighty flume]

until its squarable

[woven radish]

Biscuity

[mighty flume]

right of course

[long pasture]

so it will become?

your turn~

[mighty flume]

2sqrt(15) + 8/16

sqrt of 16 is 4

you get the answer of

[long pasture]

yesssss

yessssssssss

YESSSSSS

[mighty flume]

sqrt(2sqrt(15) + 8)/4

which is B

[long pasture]

correct!

done!

[mighty flume]

easy!

[long pasture]

nah, it's not easy

but it's good you made it!

[mighty flume]

i get a little confused on the fractions a lot of the time

thanks for the help !

[long pasture]

try writing them in × and ÷ with in case you are stuck with fractions

e.g.
(2/4)/6

we can write it as
(2÷4)÷6

[mighty flume]

i see i see

[long pasture]

and now, given that you are familiar with brackets operations, we can express it into
2÷4÷6
or

$\frac24÷6$

[woven radish]

Biscuity

[long pasture]

or

$\frac{2}{4}×\frac{1}{6}$

[woven radish]

Biscuity

[mighty flume]

oh right right

keep change flip

[long pasture]

yep

that's how i deal with fractions

hope that helps!

[devout snow]

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[rain bluff]

is this enough info or do they need all sides distance

[winter torrent]

seems fine

Word has an equation editor I think if you want to play with that but you don't need to

[rain bluff]

ok ty

but like

its a rhombus

ive done left and right

do i need to do top and bottom too or

[supple knot]

Show the image

We don't know what top/bottom/left/right mean

[winter torrent]

did they tell you it's a rhombus?

[rain bluff]

yeah

sorry hold on

[devout snow]

@rain bluff Has your question been resolved?

[winter torrent]

that does not look like a rhombus to me at all

[rain bluff]

what??

my lines are kind of wonky but

[winter torrent]

,w rhombus

[woven radish]

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[winter torrent]

wow that is no information

a rhombus has all 4 sides the same length

[devout snow]

@rain bluff Has your question been resolved?

[rain bluff]

i am dumb

looking at my assignment

i actually wrote that it was a paralellogram

sigh

[violet wind]

Yeah they would need all sides

An isosceles trapezoid is an example of a shape with only one pair of opposite parallel sides

[devout snow]

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[mild crypt]

So ( ((1)/(x+y)) - ((1)/(x)))/(y)
Can be simplified with some common denominator thing but I don’t rly understand it. Like I get how you can rewrite the top part of the fraction to -y/(x+y) but in the video I’m watching they write (1/y)*((x-(x+y))/((x+y)x)

[mild crypt]

I don’t get what they are doing

I think it has something to do with the fraction they write before it but I don’t get how they just get rid of the top fraction

Oh sorry should I use the help forums ?

[wooden wraith]

$\frac{1}{x+y} - \frac{\frac{1}{x}}{y}$ This is the expression you're starting with?

[woven radish]

tatpoj

[wooden wraith]

@mild crypt

[mild crypt]

Waif I found notes

So ye I guess this is a shortcut then right? I guess it does look easier tho lol

[mild crypt]

I understand the notes a lot better like I understand every step I just don’t get how you can get the value that is infront if the thing in the first step. Is that just gotten by multiplying the denominator of the top parts?

Idk it still feels a bit weird

Like I get the steps but

[wooden wraith]

value in front of the thingi n the first step?

[mild crypt]

Its weird how you write something that’s literally 1 *

[wooden wraith]

oh you mean the (x0)((x0 + deltax) over itself on the third line?

[mild crypt]

This

I do ye

What does that do I don’t get it

[wooden wraith]

Yeah it's so you can cancel the denominator of the numerator

Instead, you could just combine the two fractions in the numerator by making a common denominator

it's the same thing

it's the GCF of the denominators in the numerator

[mild crypt]

Like I could Do the top part seperatly and get something similar but it would be (-dx /(x0+dx) ) / dx

[wooden wraith]

cancel those dx's and you're right at the end

that's equivalent to the last line in your picture

[mild crypt]

How can I cancel them

Its like

A fraction inside a fraction

Idk how to rhen cancel the numerator of the numerator with the denominator of the big fraction

[wooden wraith]

Think back to middle school, how would you simplify any fraction like that?

These are variables but it works the same way

as something like (2/3) / 4

or whatever

[mild crypt]

Im in highschool and I never got that 💀

[mild crypt]

Uhhhh

[wooden wraith]

Haha no worries

[mild crypt]

Like multiply i guess

[wooden wraith]

$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \cdot \frac{d}{c}$

[hot steeple]

look at the left bit

[woven radish]

tatpoj

[mild crypt]

Oh ye I never got that 🤣

[wooden wraith]

Dividing by any number is the same as multiplying by its reciprocal

[mild crypt]

Makes sneee tho I know how dividing is multiplying the opposite

By its what

[wooden wraith]

its inverse, reciprocal, whatever you want to call it

like 2 and 1/2

[mild crypt]

Ah

[wooden wraith]

or 2/5 and 5/2

[mild crypt]

Ye

[wooden wraith]

or in your case, dx and 1/dx

[mild crypt]

Yes wait can you like do that math thing so I can visualize

For like the method I did

Idk how that works

[wooden wraith]

from here, right?

[mild crypt]

Yep

[wooden wraith]

$\frac{\frac{-\Delta x}{x_0 + \Delta x}}{\Delta x}$

eh, it doesn't show up well but the dx at the bottom is the "whole" denominator

also I forgot a negative lol

[woven radish]

tatpoj

[mild crypt]

Ye

[wooden wraith]

You're dividing by dx, which is the same as multiplying by 1/dx

[mild crypt]

Ah so you’ll just do -dx * dx

So -1

Ah and then you get -1/x0+dx?

Wait no I think that’s wrong tho

[wooden wraith]

$\frac{-\Delta x}{x_0 + \Delta x} \cdot \frac{1}{\Delta x}$

[woven radish]

tatpoj

[mild crypt]

Ye what

That’s different

Becuase you can divide the too bu dx to get the -1

But then you have a *dx missing in the bottom

[wooden wraith]

oh you're missing a factor of x0 in the denominator

[mild crypt]

Wait nvm that’s a x0

ye

Where is rhat coming from

[wooden wraith]

can you show your work on how you got here?

[mild crypt]

NVM ye that is what you get if you do the top part properly

I messed that up XD just did it quick on my phone

[wooden wraith]

haha i didnt' notice it was missing either, looked right at first

[mild crypt]

Here it looks like you can also divide the very top with the very bottom and take the middle as the new bottom tho

[wooden wraith]

That is also true, yes

[mild crypt]

Ok well tyvm! I understand now and can probably do it in the future myself now ^^ altho maybe I should practice like 2 more 🤣 imma sleep now tho idk why I’m doing math at 00:30 midnight

[wooden wraith]

$\frac{\frac{a}{b}}{c} = \frac{\frac{a}{c}}{b} = \frac{a}{bc}$

[woven radish]

tatpoj

[wooden wraith]

no problem 👍

[mild crypt]

Uh what

But if I multiply that with opposite

1/c * b/a rhats b/ac not a/bc @wooden wraith

[wooden wraith]

you only flip the denominator

$\frac{\frac{a}{b}}{c} = \frac{a}{b} \cdot \frac{1}{c} = \frac{a}{bc}$

[woven radish]

tatpoj

[mild crypt]

Nothing is flipped there

[wooden wraith]

huh?

[mild crypt]

There is nothing flipped tho

[wooden wraith]

the reciprocal of c is 1/c

[mild crypt]

Yes

OH just

[wooden wraith]

remember that c is equal to c/1

I flipped it to 1/c

[mild crypt]

C on its own is flipped?

Just the bottom?

Ah ok

[wooden wraith]

yes, a more general form would be $\frac{a/b}{c/d} = \frac{a}{b} \cdot \frac{d}{c}$

[woven radish]

tatpoj

[mild crypt]

Oh ye fair that’s what you did here too

Ye in my mind you just took it and put it down a level lower

Its nice to visualize it makes sense like

Damn that’s pretty easy

Ok I’ll practice some more but I think I got it

Thank you again 🙏

[wooden wraith]

np 👍

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[glass plume]

Why is the answer D?

I thought for the formula for inverse trig sin,

its sin^-1 (u/a) + C

but where did the 1/3 come from

[vocal cypress]

because you take out the 9 from the square root, which becomes sqrt(9(4/9-x^2))

and sqrt9 = 3, and its on the bottom so when you take it out you get a 1/3

[glass plume]

hold up

oh rlly?

ok i see

thanks

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[left robin]

hey

[weak cove]

hello

[left robin]

$\text{for small }\frac{s}{r_0}: \ \frac{1}{(r_0+s)^3}\approx\frac{1}{r_0^3}-\frac{3s}{r_0^4}+...$

[woven radish]

Martin

[left robin]

in a previous exam someone had to use this

can someone explain how we can derive this?

my first guess would be taylor

but then i wouldnt know if we would differentiate with respect to r or s

[weak cove]

[r0(1+s/r0)]^3

[left robin]

this shows how we get the first term, but how do we get the second?

nvm got it

thx^^

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[weak cove]

can you show

[left robin]

yes, one sec

using binomial formula

wait

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[left robin]

this is wrong, one sec

@weak cove

[weak cove]

ah nice

ty

[left robin]

.close

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[restive river]

hello

[devout snow]

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[restive river]

Just wanted to make sure if my working is correct

Sorry if handwriting sucky just ask if something is unclear

[zealous ether]

i mean yeah it is correct

laplace transform of cos(at) is a standard result

and you did end up with exactly that

[restive river]

Great nice

[woven radish]

hiii

[restive river]

Another question just for clarification on another question

What do they mean by "use the linearity of the Laplace transform" exactly

[zealous ether]

no clue, idk where you can even use linearity here

[restive river]

Hmmm

I should probably include the full thing here

Do they just mean to like

Split up the integral?

[last hill]

Yes

[zealous ether]

yeah

they want you to use the exponential definition

and then find the laplace transform of each

[restive river]

So the integral for the cos one should match the one I found thru ibp

[zealous ether]

sure

[restive river]

Oh why not multiply g(t) by a constant too?

[zealous ether]

you can lol

idk why the picture didn't include it

but yeah basically they want you to find the laplace transform of e^{ibt}/2 and e^{-ibt}/2

then linearity is trivial

[restive river]

Fair enough let me try it

Hmm so

I get [
\map F s = \f1{2i} \bracks{\int_0^\infty e^{t(ib-s)} \dd t -\int_0^\infty e^{-t(ib+s)}\dd t}
]

[restive river]

I'm guessing that's the point?

[last hill]

Yes, and these are easy integrals to solve

[restive river]

Wait am I tripping but doesn't the first integral just diverge

[last hill]

No, note that the real part is negative

[restive river]

Oh huh, I never really tackled with complex integrals but I assume we ignore the imaginary part when we take the limit?

[last hill]

No, just work with i as if it's a number

But $\lim_{a\to\infty} e^{-a + bi} = \lim_{a\to\infty}e^{-a} \left( \cos(b) + i\sin(b)\right)$

[woven radish]

DAILI

[last hill]

And cos and i*sine are bounded in size by 1 so it goes to zero

since the decaying exponential dominates

[restive river]

Ohh I see this makes a lot of sense ty

Hmmm

So all in all

[
\map F s = \f1{2i} \bracks{\f1{ib-s}+ \f1{ib+s}}
]

[restive river]

Fingers crossed I didn't mess it up lmfao

[last hill]

I think you're off by a minus sign

Since it's - the antideriv at 0

Could be wrong though

[restive river]

Oh right I see why

I think it's like

• for both

And we pull it out

Wait well

,w Laplace transform of sin(at)

[woven radish]

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Requested by lexqa

[restive river]

,w a/(a^2 +s^2) = -1/(2i) (1/(ia-s) + 1/(ia+s))

[woven radish]

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[restive river]

Okay ayee

Ty @last hill @zealous ether

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[untold vector]

Should expected and observed frequencies match in the table ?

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[untold vector]

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[void lance]

how in the world do I find this

[devout snow]

@void lance Has your question been resolved?

[winter torrent]

why not just plug in 100 into the formula?

[void lance]

wdym ?

[winter torrent]

you have a function that gives cost right?

so if i asked you for the cost if t = 50 you could calculate it

[void lance]

Ah I see

Another question rq too

I couldn't understand how to solve this one bc there's no x intercept

[winter torrent]

at the origin?

[devout snow]

@void lance Has your question been resolved?

[void lance]

so 0 ?

I see

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[onyx grove]

I am concerned with part iii)

[pallid rock]

Is there a free tutor here?

[midnight bane]

#❓how-to-get-help

[onyx grove]

this is a video of the creator of the question doing it

He explains how to do the question roughly but doesn't actually do it because he runs out of time

but I don't see how he gets the answer in the show that part of iii)

that makes cos theta = 1/2 / root7/4

which makes 2 arccos 1/2 / root7/4 the angle of the sector, because theta is half of the angle

which is not what he got, he got 1/root7 ,which as he says in the video you should get if you do the question the way I just showed, as the angle.

Also another question I have is I know that 7pi/4 is the area of the entire circle, and that the angle is one I have gotten wrong, but I don't get why the 1/180 degrees, and also that if you calculate it by area of circle * portion of sector proportional to the entire circle you get the area of the sector, not the triangle they have in question

so I was wondering if you did it with 1/2absinc but you get sin as a result so it couldn't be that.
PS. the question is a non-calculator paper

[devout snow]

@onyx grove Has your question been resolved?

[onyx grove]

<@&286206848099549185>

[devout snow]

@onyx grove Has your question been resolved?

[lunar harbor]

$\frac{\frac{1}{2}}{\sqrt{\frac{7}{4}}}=\frac{\frac{1}{2}}{\frac{\sqrt 7}{2}}=\frac{1}{\sqrt 7}$

[woven radish]

Civil Service Pigeon

[lunar harbor]

You could convert between sin and cos by drawing a right triangle and interpreting geometrically tho

[lunar harbor]

Pretty sure the website is just wrong cause here’s the original paper

At least if I’m interpreting you correctly

Run-on sentences are hard to understand

[onyx grove]

ohhh I see

where did the pi go in the answer

How would I use the 1/2 ab sin c rule if I have c in terms of cos c

also when it says it is bounded by the arc, does it mean the straight line from A to B or does it mean the curved circumference of the circle from A to B

[onyx grove]

so I have the cos-1 1/root7 as half of the angle required, so why doesn't it multiply the angle by 2 and where did the pi on 7/4 go.
This is in the case if you think the arc AB is the curved circle line from A to B which I'm not sure it is

[lunar harbor]

Nvm I’m dumb

The idea is to convert 1/180 deg to radians probably

Idk why that’s there but ig that’s a “modification”

[lunar harbor]

Pythagorean

[lunar harbor]

Curved circumference

[onyx grove]

Why do they have 1/180 degrees in the first place

I thought the formula for area of a sector is just =area of circle * proportion of angle compared to full circle

[onyx grove]

Oh yeah, I would just get the angle in terms of sin instead, but the answer is in terms of cos which is my question, if I get some 1/2 ab sin c answer how would I change it to cos without a calculator, it would be something really messy

[onyx grove]

If they were to translate it to radians it would not cancel out the 7pi/4 as radians is simply a measure of degrees and is not actually any value of area, so it would not interact with the 7pi/4 and would only be inside the arccos

Which is why I am confused because why is there a measure of degrees outside of the arccos

[lunar harbor]

https://youtu.be/H7GwDntpGUY

[lunar harbor]

You do you ig

I gave you my interpretation

You’re free to take it or leave it

[onyx grove]

Sorry I wasn't trying to be a dick just confused about the question

thank you for the help though it does clear quite a bit up

i'll try again to figure it out tommorow

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[sharp citrus]

is there some formula im supposed to use for this question

cus i have no idea how to approach it

[potent nebula]

P(A U B) = P(A) + P(B) - P(A and B)

for events that are not mutually exclusive

if they are then P(A U B) = P(A) + P(B)

do you know how to calculate P(A and B)

[strange arch]

@sharp citrus ?

[sharp citrus]

yep

still here

yeah yeah

seems clear to me now

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[stark shuttle]

How do I calculate the horizontal shift for a function like this?

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[cursive totem]

,,13^{2019}\mod 95

[cursive totem]

I've used fermat's to obtain 13^4 = 1 mod 5 and 13^18 = 1 mod 19

but idk where to go from there

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[restive river]

Am i reading this problem correctly?

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@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

I did not read it right

okay

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[wooden axle]

I have to find the equation of a plave which passes through x axis and (3,2,4) point

[wooden axle]

So here i can imagine by+cz=0

2b+4c=0

What should i do next?

@pseudo basin

[pseudo basin]

☠️

[uncut crow]

☠️

[weak cove]

💀

[pseudo basin]

anyway it's 2b + 4c = 0 not 2b + 4z = 0

[wooden axle]

@uncut crow your PFP😂

[wooden axle]

Yes

[uncut crow]

what about it

[wooden axle]

Typo

What next

[pseudo basin]

you can now find b and c

take one of them as whatever you want (except 0) and find the value of the other one

[wooden axle]

U meant 2b=-4c ?

[pseudo basin]

and you are done

sure whatever

[wooden axle]

Can you tell me vector solution?

I want to solve the question by vectors

[pseudo basin]

do the same thing we did yesterday only with (0,0,0) and (3,2,4) as your two points

[wooden axle]

Now it's done🤦

[pseudo basin]

i don't want to repeat myself with different numbers

[wooden axle]

Ohhh i missed origin

.close

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[past laurel]

dont even know how to start this i plug in -1 on top and get 3.45... but idk what to do in the bottom

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[past laurel]

<@&286206848099549185>

[supple knot]

That's very small

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[wooden axle]

How to find the equation of a plane which passes through middle point (3,-2,1)(1,4,-3)

[wooden axle]

@manic condor

Middle point yes

[manic condor]

There are infinite planes passing through a single point.

[wooden axle]

No no

Middle plane

[manic condor]

What do you mean?

[wooden axle]

We need to find the equation of place which is perpendicular to these points and passes through the middle point

Do you understand hindi?

[manic condor]

Plane perpendicular to a point? What does that even mean? 😐

[manic condor]

Sure. Go ahead.

[wooden axle]

I got it

Dot product

[manic condor]

Even now, if they want plane passing through perpendicular bisector of that line segment, there are infinite such planes.
However, if angle between line and plane has to be 90°, then it has a unique plane.

[wooden axle]

Middle point and x,y,z then direction ratio will be 2-x,1-y,-1-z and dot product with normal(1,-3,2) will be zero

[manic condor]

Well, it's even simpler. In the second case that i mentioned, perpendicular vector to plane is in same direction as this line.

Just put value of middle point and solve for constant.

[manic condor]

How is normal (1, -3, 2)?
Check your thinking again.

I'll be back in a few minutes.

[wooden axle]

This is normal direction ratio@manic condor

(3,-2,1) - (2,1,-1)

[manic condor]

Ah. You took with midpoint. That's correct.

[wooden axle]

I am confused about whether I do dot products or cross?

[devout snow]

@wooden axle Has your question been resolved?

[devout snow]

@wooden axle Has your question been resolved?

[vernal monolith]

I think you need 2 vectors for cross product, you only found a single vector that passes through both points

Given vector (a,b,c) that is perpendicular to a plane, can you type the formula for that plane

Hint: take two arbitrary points on the plane, find a vector that passes through both points on the plane, and lastly take dot product of these vectors , since one vector is perpendicular to other, it should be equal to 0

[wooden axle]

See

I guess this is correct

[vernal monolith]

That vector will be our normal vector yes

You can type the formula of plane from that

[manic condor]

This is already done. What are you confused about now?

[wooden axle]

Why i didn't try the cross product

[vernal monolith]

Do you know the purpose of cross product?

Why we use it

Its used to find a normal/perpendicular vector to our current 2 existing vectors

Here we only have 1 vector

Plus our vector is already a normal to the plane we are trying to find

[wooden axle]

So that we can get a perpendicular vector

[vernal monolith]

Exactly

Do we need to do that when we already have one though

[wooden axle]

If it was given two points of plane

[vernal monolith]

You need like at least 3 points to take cross product

[wooden axle]

Yes exactly that is what i wanted to hear

Can we not try dot product?

[vernal monolith]

On 3 points?

[wooden axle]

Yes

[vernal monolith]

It wouldnt be perpendicular to every point then

[wooden axle]

Ohh now i got it the loophole

Here we are taking all the points because it is in x,y,z form so all points included

So dot product is making sense

[vernal monolith]

Lets say you have points P, Q and R
Then to find perpendicular vector you need to take 2 vectors whose union contains all these points, so i will take the vectors PQ and QR,
( but you can also do it like: PR and PQ or PR and QP, however you'd like)

Then i take cross product of PQ and QR => |PQ x QR|
Reminder that cross product is perpendicular to the vectors it consists so
|PQ x QR| . PQ = 0
|PQ x QR| . QR = 0

[vernal monolith]

Actually dot product isnt needed for your question if you know the shortcut

[vernal monolith]

Shortcut i mean is this

But its derived from the dot product way, i think

So far your work is like this @wooden axle

I typed equation of the plane from the shortcut, but longer way to do is by taking points $P=(x_0,y_0,z_0)$ , $Q=(x,y,z)$ and constructing $$\overrightarrow{PQ}$$ which is normal to vector $\vec{n}$

[woven radish]

Cyrenux

[wooden axle]

We can take a point P as middle point

[vernal monolith]

By middle point are you referring the to point between (3,-2,1) and (2,1,-1)

[wooden axle]

Nope

Middle point is (2,1,-1)

[vernal monolith]

Ah i see, my bad

The vector you typed still passes through it, so we are good i think

After typing the equation of plane

We can find term d, by using the fact that middle point should satisfy the plane since plane is passing through it

[wooden axle]

Suppose we have to find a plane equation and we are given a point perpendicular to plane, 2points on plane what will we do cross product?

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[manic condor]

What do you even mean by a point perpendicular to plane ?

[wooden axle]

Like in this

[vernal monolith]

There is no 'point perpendicular to plane'

[wooden axle]

Question

[vernal monolith]

Its a vector that is perpendicular to a plane who consists the point

[wooden axle]

You guys are not understanding my doubt😪🙈

Do you know hindi?

[manic condor]

Write in Hindi.

[wooden axle]

Mera sawal hain
Agar ek plane ka equation nikalna hain and hamko diya hain
Ek point (3,-2,1) jo plane k upr hain aur 2 points plane ka upr hain (2,-1,1) and (4,-5,2)

@manic condor

[manic condor]

plane k upr hain Vs plane ka upr hain ?

Don't they mean the same thing?

[wooden axle]

Nope

Different

Perpendicular point hain

[manic condor]

Perpendicular point makes no sense to me.

Do you mean that (3,-2,1) is lying above the plane ?

Or somewhere out of the plane(not on the plane) ?

[wooden axle]

Yes

[wooden axle]

Out of the plane and above

Give me the idea how will cross/dot product will happen here?

[vernal monolith]

Is this a seperate question?

Cant tell

[wooden axle]

This is my doubt

Different question

[vernal monolith]

That plane you drew seems random, what was your idea with it?

It seems like you want to find a plane which consists these 3 points

[wooden axle]

I guess here we will do cross product

[vernal monolith]

Yeah its forced kinda

[wooden axle]

We will find vector PQ and QR then cross product

And cross product will make all the points of the plane

[vernal monolith]

Cross product, which you will evaluate is the vector that will be perpendicular to plane, if cross product is (a,b,c) then equation of plane is ax + by + cz + d = 0

Since cross product, who is perpendicular to the vectors who consists the points P Q R, the plane who is perpendicular to the cross product, will in fact contain these 3 points

After finding equation of plane you can insert one of the points

To find value; d

[wooden axle]

This is another situation like our previous question

[vernal monolith]

That doesnt seem to contain the point P

[wooden axle]

So here PQ like perpendicular to the plane

[vernal monolith]

Also i can already see that you are trying to say P perpendicular to the plane, but that is like the most illegal thing ever, in a way its like saying two points are perpendicular which makes no sense

[wooden axle]

We will do dot product

[vernal monolith]

You can say that P is a point on a normal vector

[wooden axle]

Yes

[vernal monolith]

There are many planes that include points R and Q

[wooden axle]

In this situation we can do cross dot both right?

That is my doubt

[vernal monolith]

But i believe there is only one plane that contains P Q and R

[vernal monolith]

We need 2 vectors for it, NOT a plane and vector

We can take PQ and QR or QP and QR or ... etc

[vernal monolith]

Oh i see an issue with your take

PQ is not perpendicular to that plane

You already know points P, Q and R, evaluate PQ and QR
Since Q and R are in the plane so is the vector QR, but notice that PQ . QR is not 0

[empty flame]

hello everyone , what are the coordinates of P,Q,R?

[wooden axle]

Both pictures are different

P is perpendicular in last picture

[vernal monolith]

Here are the points P Q and R

[wooden axle]

In the first picture p is not perpendicular so we choose only cross product

[vernal monolith]

Point P cannot be perpendicular

Because points cannot be perpendicular

[wooden axle]

In the second picture it is perpendicular so we can use cross product and dot product both that's my doubt

I meant PQ straight line is perpendicular

[vernal monolith]

To what

[wooden axle]

To the plane

[vernal monolith]

Its not

[wooden axle]

I am stating it is

[vernal monolith]

I just explained why its not

The numbers you picked dont satisfy such conditition

[empty flame]

is this the second picture that you are talking about ?

[wooden axle]

Yes

PQ ×QR only way to derive plane

In the first picture

[empty flame]

then no $\vec{PQ}$ isnt perp to the plane containing Q and R

[woven radish]

calculus is fun

[wooden axle]

But in the second picture PQ.QR and PQ×QR both are possible or not that's my doubt because PQ perpendicular to QR

[empty flame]

try $\vec{PQ}.\vec{QR}$

[woven radish]

calculus is fun

[wooden axle]

You are making things worse to me

@manic condor are you there or not?

[empty flame]

can you try this

[wooden axle]

Seriously i need help in Hindi to be sure when we do cross and dot and in which situation we can use both products

[vernal monolith]

They said its irrevelant from the values they posted earlier

[wooden axle]

Is there no situation where we can do both products?

[empty flame]

when you try this you will see that $\vec{PQ}$ isnt perpendicular to the plane that you showed in the picture

[woven radish]

calculus is fun

[vernal monolith]

The picture he sent is irrevelant to his previous one so not same values

[wooden axle]

I am saying it is perpendicular

[vernal monolith]

He wants to see if there are shortcuts when dot product is 0

An assumption

[wooden axle]

I don't want shortcuts

[vernal monolith]

If you dont want shortcut just do cross product then why bother to see their dot product is 0

[empty flame]

look if the dot product is 0 then you get the equation of the plane directly

if the dot product is not 0 then you use the cross product to get a normal vector of the plane ( a vector perpendicular to the plane) then you get the dot product between a vector in the plane and the result of the cross product to get the equation of the plane

[vernal monolith]

If the dot product is 0, does plane contain all P Q R or only Q and R or only P

[empty flame]

i am saying that if P is a point outside the plane containing Q and R and PQ.QR = 0 then you can directly get the equation of the plane

[vernal monolith]

Yes im aware

This plane consists all the points P,Q and R?

[empty flame]

but no matter what points P Q and R you choose you can always put them in the same plane

so as long as you choose 3 points you can always find a plane that contains all 3 points

[empty flame]

which plane

[vernal monolith]

I see