help-38

[solid kiln]

atif

[shrewd obsidian]

please check that this is equal

to your equation

and you should be able to factor now

but please do learn the underlying lesson of this example

[shrewd pewter]

Of course

Quadratics

[shrewd obsidian]

you had 2x^2 and that's why you needed to multiply 9 with 2

[shrewd pewter]

Ahhh

So whenever in other examples

Like lets say instead of 2x² we have 6x²

[shrewd obsidian]

yes

[shrewd pewter]

Do I multiply the third term by 6?

In other words 9 x 6

?

[shrewd obsidian]

yes, but not in the equation, that's just the trick to help you facto

r

[shrewd pewter]

Ahhh

And then after it should equal the middle term right?

[shrewd obsidian]

ok go easy

[shrewd pewter]

Oh my bad

I meant to say, the sum of the factors should equal to the middle term?

[shrewd obsidian]

not necessarily those factors

so e.g. whats 9 times 6? 54

but 9 and 6 wont be your numbers

but other factors of 54 will be

[shrewd pewter]

Ahhh

[shrewd obsidian]

this was a coincidence

that you got 9 and 2

[shrewd pewter]

Interesting

[shrewd obsidian]

so -9 + 2 = 7

and -9 x 2 = -18

i mean what would you do if you had just x^2

not 2x^2

[shrewd pewter]

this equates to the answer yes? Or is it purely coincidence

[shrewd obsidian]

you know the trick right?

so there's an invisible 1 there

it's actually $1 \cdot x^2$

[solid kiln]

atif

[shrewd pewter]

Oh ..

(X ) (X ) = 0
( X - 3) (X + 3) = 0
?

[shrewd pewter]

Yess x has a coefficient of one

[shrewd obsidian]

yes so whenever the coefficient isn't one, multiply the free term without x with that coefficient

and then you will get a number

that's product of those two, and then you do the OLD trick

you know

just instead with the product, as opposed to the free term as you would do if you had 1 times x^2

were you able to factor the original problem?

[shrewd pewter]

Could you provide an example please? Something different, id want to try and solve the previous question on my own.

[shrewd obsidian]

did you end up factoring your problem?

[shrewd pewter]

Ah.. I haven't, in spite of your great discussion. I'm still really confused lmao, explain it like I'm 5. Sorry if I wasting your time, i find this to be really interesting though

[shrewd obsidian]

you can't factor this?

[shrewd pewter]

I'm not sure how?

[shrewd obsidian]

do the most basic thing

rearrange the terms if that helps

[shrewd pewter]

Ahh

Oh hold on

Ahh

2x² -7x + 9 = 0?

[shrewd pewter]

Okay yeah I understand what you meant noww

[shrewd obsidian]

but can you finish the problem

you need to get ( ) ( )

that form

[shrewd pewter]

Not quite sure, hmm

[shrewd pewter]

Ahh

[shrewd obsidian]

use the pic i sent and just factor whatever you can

there's some common factors there

go step by step

i have another one for you

but do this first

[shrewd pewter]

I think il be wrong but il give it a shot

Alrighty

[shrewd obsidian]

how is it going?

i can give another hint

[shrewd pewter]

Do I need to find the gcf? I think there is none

[shrewd pewter]

I'm not sure what you mean by factor it..

Its been a while since I've done 6th grade math

Actually a long while

[shrewd obsidian]

$2x^2 + 2x - 9x - 9=0$

[solid kiln]

atif

[shrewd obsidian]

if you can't simplify this in ANY way

apply ANY manipulation that you know

[shrewd pewter]

Ohh quadratic formula?

[shrewd obsidian]

except adding 2x and -9x

ofc

no bro

[shrewd pewter]

Ah

[shrewd obsidian]

i think you need to go watch some youtube videos

https://www.youtube.com/watch?v=9MUDAr1PnlE&t

here's one

[shrewd pewter]

Simplified this is

2x² - 7x - 9 = 0?

[shrewd obsidian]

no

[shrewd pewter]

Ahh

[shrewd obsidian]

ok i ll give you the solution

[shrewd pewter]

Thanks

[shrewd obsidian]

$2x(x + 1) -9(x+1) = 0$

[solid kiln]

atif

[shrewd obsidian]

but you need to go watch that video

[shrewd pewter]

OHHHHH

OHH

Okayyy

[shrewd pewter]

Yeah for sure

[shrewd obsidian]

sorry i didnt give the final answer, the above was just another step

let me know if you don;t know how to put into ( ) ( ) form

that last image

[shrewd pewter]

Alrighty

Ty

Okay uh

I learnt a different method

And the answer I got is

Question: 2x² - 9 = 7x

Ans in ()() form: (x + 3) (3x -2)

[shrewd obsidian]

try multiplying that and see if you will get 2x^2 ...

it's not correct

you should get (2x-9)(x+1)=0

this is the final step after the last image i sent

[shrewd pewter]

Oh hold on I made a mistake in the calculation

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[restive prairie]

can anyone help me out?

[restive prairie]

i’m confused

this is the question

[unique minnow]

What confuses you?

[restive prairie]

what i’m supposed to do after this

I don’t think it’s right

[unique minnow]

Well that's half of the integral of 1/u.

Do you know what the integral for 1/u is?

[restive prairie]

No what would it be?

ln(u)

[unique minnow]

Yes

[restive prairie]

but then what do i do with the du/2? isnt it supposed to cancel out

[unique minnow]

Well du stays there, it tells you that you're now integrating with respect to u, not t.

And the 1/2 is just a constant you can bring it to the front of the integral

[restive prairie]

like this?

[unique minnow]

Yep

[restive prairie]

ohhh ok i see thank you!

[unique minnow]

No worries!

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[sage anvil]

guys how is this possible

[sage anvil]

i saw in solutions its 2 ^2/3

but I don't understand

[naive otter]

What is X's value ?

[sage anvil]

it is not given

[rotund grove]

then it's not possible

[wraith hinge]

then u cannot decide

[rotund grove]

you can't just magically remove x (unless there is cancellation

[sage anvil]

hmmm

[naive otter]

We have 2x•x½ = 2x^(3/2)

[sage anvil]

task was to derive square root of x * (x^2-1)

[rotund grove]

x(x^2-1)?

[sage anvil]

so it is like 1/2 x ^-1/2 * (x^2 -1) + square root of x * 2x

[sage anvil]

.close

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[rotund grove]

,selfroles

[solid kiln]

Gave you the she/her selfrole.

[wraith hinge]

,selfroles

[solid kiln]

Gave you the she/her selfrole.

[wraith hinge]

lol

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[fair forge]

is it correct to write it like this?

( ln(x) )' is how i write it

[fair forge]

i do see why it is correct

but is it common in usE?

[spare yarrow]

prime as in derivative?

[grim sparrow]

I've seen it before, yes

[fair forge]

ye

[grim sparrow]

I usually write log'(x) instead of (log(x))', actually

it's just easier haha

[fair forge]

yeah but it looks yk

[spare yarrow]

not very comon, indeed. Just be careful so that you don't get marked for that

[fair forge]

it's my teacher's

anyway thanks

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[grim sparrow]

[kindred kayak]

i opened 2

my misstake

.close

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[quiet remnant]

The graph of f (x) is shown. Make the graph of 3f ((x/2) − 1) indicating each of the transformations carried out.

[quiet remnant]

May someone check if I did it correctly? please

Its in spanish and my letter is pretty bad, so sorry if its hard to understand.

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[quiet remnant]

<@&286206848099549185>

[strange needle]

i SEE

Helo

[quiet remnant]

Hello

[quiet remnant]

<@&286206848099549185>

[tulip hollow]

Please be patient, someone will assist you.

[quiet remnant]

ok

[tulip hollow]

It would also help if you wrote it in English and specify what you need help with.

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[quiet remnant]

Oh

I think I finished the problem

I think the answer is tha graph of bottom

[quiet remnant]

I mean, firstly I expanded the graph, then I moved x to the right and then I expanded on Y

Thats it

But I want to make sure that it is correct

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[quiet remnant]

<@&286206848099549185>

[wise mural]

Seems good to me, you can check at some specific values to validate your answer

For instance, 3f((x/2)-1) at x=4 would be 3*f(1), that is 3, as you wrote in the answer

That is not a proof but it helps being confident in your answer

[quiet remnant]

Thanks

I just wanted that lol

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[sharp gust]

In doing I.V.T., if you have something like cosx=2x^2 --- do you just make it equal to zero?
so 2x^2-cosx = 0.

then plug in values like 0, 1, and -1? what counts as a "solution" exactly, and I don't really get what it means to have a solution in the interval

[serene shale]

solution = value of x that makes the equation, cos(x) = x^2, true

you can indeed subtract one side of the equation over to the other

so for example if f(x) = 2x^2 - cos(x),
NOW, a solution is a value of x that makes f(x)=0

now you try to find two values of x, one that makes f(x) positive, and another that makes f(x) negative

i think x=0 makes f(x) negative and x=pi makes f(x) positive.

proof:
f(0) = 2(0)^2 - cos(0) = 0 - 1 = -1, which is negative
f(pi) = 2(pi)^2 - cos(pi) = 2(pi)^2 + 1, which is positive

now you can use the IVT (because f(x) is continuous) to conclude that there is some value of x on the interval (0, pi) that makes f(x)=0

that value of x, which is between 0 and pi, is a solution of the original equation cos(x) = 2x^2

hope that made sense

[sharp gust]

oooh pi is a good one to remember for trig, I was grimacing at the values 1 and -1 gave me, and 2

lemmie re-read it again

the answer was -1,0 for a solution in the (a,b) interval

but cos(-1) gives me weird numbers that don't match 2(-1)^2

[serene shale]

yeah i mean

there’s always other values of x you can choose

x=-1 is weird here

[sharp gust]

yeah

so in the (a,b) format it's not actually x,y but rather left endpoint and right endpoint

how tf is -1 the left endpoint if algebraically it gives you like 3.4 something.
cos(-1) is .54 and 2x^2 is 4

[serene shale]

yeah correct

( , ) notation is used for 2 different things

one is ordered pairs (x, y) which represents a single point

the other is an open interval (a, b) which is a infinite stretch of real numbers

[serene shale]

and yeah idk. i guess 2(-1)^2 is 2, and cos(x) is at most 1, meaning 2(-1)^2 - cos(-1) is definitely positive

[sharp gust]

ah

[serene shale]

¯_(ツ)_/¯

[sharp gust]

Yeah lol idk I'll email the TA

it doesn't make sense to me yet because it's just intro calc stuff, so it should just be some simple equation with simple answers

[serene shale]

usually you plug in simple values of x like -1, 0, and 1

but with trig functions you can do other stuff like pi and such

[sharp gust]

oooo

[serene shale]

but yeah in general when someone says a “solution on an interval”

they just mean a value of x, that’s in the interval, that makes the equation true

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[wraith hinge]

pls help with the part b

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[wraith hinge]

<@&286206848099549185>

[wraith hinge]

okay sorry

[proven onyx]

Okay it ok

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[viral pumice]

Solve this By factoring x²=16x-63

[burnt mulch]

!occupied

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[burnt mulch]

Look up “projection of point onto line 3D”

https://stackoverflow.com/questions/7023240/projection-of-point-on-line-in-3d

[wraith hinge]

x^2 -16x + 63=0
x^2-7x-9x+63=0
x(x-7)-9(x-7)=0
(x-7)(x-9)=0
x=7 or x=9

[wraith hinge]

Ermm i dont get it

[burnt mulch]

As in how to derive the formula or smt else?

Because the formula is already written there

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[strange hamlet]

Im unsure where to start

[radiant spindle]

Well do you understand all the terms listed

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[faint horizon]

How do I write this in the form “c a^p b^q” if they don’t provide those variables

[radiant spindle]

They're asking you to simplify the expression in the form of ca^p*b^q

So you can figure out was c,p, and q are

[faint horizon]

But how can you have p and q without a and b

Oh wait nvm I figured it out

Thanks

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[radiant spindle]

What part?

[primal spade]

evrything

basically

[radiant spindle]

ok so 11a

[primal spade]

like 12

[radiant spindle]

Oh ok

[primal spade]

no no leave q11

[radiant spindle]

So 12 id look at the graph and write coordinates of everywhere you know

[primal spade]

oh ok

[radiant spindle]

like in 12a it looks like the line goes through (-1,0)

[bright quarry]

that approach is unnecessary for all of these

they all give the y intercept

you don’t need all the points

just look at the slope

[radiant spindle]

Oh sorry I meant to specify, find two

There are multiple ways of doing this

[bright quarry]

first one goes up four over one thus the slope is 4 and the y intercept is 4

you can tell what the slope is just by looking for all of these

[primal spade]

What abt this

<@&286206848099549185>

[radiant spindle]

well what's the question

[primal spade]

Wait hold on

this is all the textbook gave

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[brazen gazelle]

how do i use null factor law on this equation

[marble wharf]

its one thing (x^2-4x+2) multiplied with another e^(-x+4)

so for the product to be zero, one of those things has to be zero

[brazen gazelle]

so what does that mean

@marble wharf

<@&286206848099549185>

[sacred sapphire]

set this equal to 0?

[brazen gazelle]

yeah

[sacred sapphire]

like what he said

[brazen gazelle]

yeah but what next

[sacred sapphire]

for two things multiplied to be 0

either one of them must be 0

[brazen gazelle]

why cant both of them be 0

[sacred sapphire]

they can

[brazen gazelle]

so what should i actually do to the euqation

[sacred sapphire]

well now which one can be equal to 0

can e^(x+4) be 0?

[brazen gazelle]

yeah

but cant the other one also be = to 0?

sorry i had to go to the bathroom

<@&286206848099549185>

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[brazen gazelle]

<@&286206848099549185>

[sacred sapphire]

sadly e^(-x+4) cant be 0

[brazen gazelle]

y not

ohh wait i think i know

is it because e approaches 0 but never reaches it

[sacred sapphire]

yeah

so the only possible 0 that can be is the other one

which is just a quadratic equation

[brazen gazelle]

so what do we do with the e^(-x+4)

[sacred sapphire]

just ignore it

[brazen gazelle]

oh fr

yay

so anything with an e i can usually ignore??

[sacred sapphire]

not really

something like e^x - 5 = 0 is solvable

[brazen gazelle]

how would u solve that

[sacred sapphire]

e^x = 5

x = ln(5)

[brazen gazelle]

ohh okay

do u ln to get rid of e

[sacred sapphire]

yes

[brazen gazelle]

wait im having some trouble doing null factor law on the first part

[sacred sapphire]

what about it

@brazen gazelle

[brazen gazelle]

what adds up to -4 and multiplies to 2

[sacred sapphire]

nothing

quadratic formula time

[brazen gazelle]

so if i cant solve it i just use quadratic formula?

[sacred sapphire]

for quadratics?

[brazen gazelle]

ya

cuz i thought i was doing something wrong

[sacred sapphire]

yeah

[naive otter]

Some ways to solve quadratics include
•Forming square
•Making the x- term vanish
•Quadratic formula
•Manipulation through visualisation using Vieta's formulas

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[brittle root]

hi

[brittle root]

im having some problems with limits

is this correct?

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[pale juniper]

Yes

[brittle root]

ty

.close

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[covert whale]

help with this

[wary basin]

how far have u gotten through so far

[covert whale]

ive filled first question

[wary basin]

yep

so what's confusing

[covert whale]

Last quest

[wary basin]

the part p?

[covert whale]

yes

[wary basin]

well twice as likely would indicate

2:1 ratio

and it looks like it is 28:57 which would be close but slightly higher than twice as likely

[covert whale]

oh

28x2=56

but rocky. road is 57

[covert whale]

can u help me with o

<@&286206848099549185>

[rough goblet]

okay first off

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[covert whale]

plz help

[wary basin]

path of green then blue or blue then green

[wary basin]

in this case would be 4/5x1/5+4/5x1/5

[covert whale]

Oh

[covert whale]

Is the answer next to it right

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@covert whale Has your question been resolved?

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[ornate cape]

Can anyone solve this?

[trim jolt]

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

[ornate cape]

1

[wind cloak]

try to simplify the expression first

[ornate cape]

then?

[wind cloak]

huh?

if you've simplified then just differentiate

is there something else to do?

[ornate cape]

no

[wind cloak]

...

[wind cloak]

then what's the probelm?

[ornate cape]

I don't know if my simplification is correct or not

[wind cloak]

show your work if possible

[ornate cape]

now i have no camera around me?

[wind cloak]

uh, then type what you've got

or latex it

[dull temple]

or mspaint

[ornate cape]

??

[wind cloak]

what did you do in the last step?

[wind cloak]

evrything except that last thing is fine

[ornate cape]

i plugged in the value of theta

[wind cloak]

you could simplify further while keeping the theta

[wind cloak]

what's wrong with my english

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@ornate cape Has your question been resolved?

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@ornate cape Has your question been resolved?

[wind cloak]

@ornate cape are you done with the problem?

[ornate cape]

no

[wind cloak]

what are you even trying?

taylor expansion☠️

[ornate cape]

??

[wind cloak]

uh ignore it

do you need further help?

[ornate cape]

no

[wind cloak]

so if you are done with this channel then remember to close it

[naive otter]

x = cos2ø

[trim jolt]

@ornate cape Has your question been resolved?

[rotund grove]

bro close it

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[brittle gulch]

.create

(x+2) (x+3) +(x-3) (x-2) -2x(x+2) =0

[brittle gulch]

: )

Solve

[tulip sapphire]

Hi please be polite we are volounteers and help on our free time, if noone is helping you, wait 15 minutes and ping the helpers

[tulip sapphire]

what have you tried?

.status

!status

[trim jolt]

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

[brittle gulch]

Yeah, yesterday I solved and I forget it how to solve

Its easy but. I forget

[tulip sapphire]

what step are you on?

[brittle gulch]

How I solved

[tulip sapphire]

have you done any progress so far, could you send what you've done

[brittle gulch]

Wrong ams

My writting is bad

[tulip sapphire]

could you show the steps you've done to solve it so i can find the mistake

no problem

[tulip sapphire]

Expand the paranthesis, multiply everything etc..

[brittle gulch]

The x's are so trifying

[tulip sapphire]

start by expanding the parenthesese, so (x+2)(x+3) becomes x*x + 3x + 2x + 6

[brittle gulch]

Oh

[trim jolt]

@brittle gulch Has your question been resolved?

[rotund grove]

sometimes using brackets like this is good:
༼x+5༼༼x+4༼ = 5

find x

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[tender onyx]

Hi I need help with integartion

[tender onyx]

this is my answer

(i know the sine term is 0 so I canceled it out)

so this is the final answer

(excluding the 1/pi)

why is the t-pi term flipped

like

it screws up my limits

I think

please help

please help

me

I need it

[wind cloak]

wdym?

(t - pi)^2 thing?

[tender onyx]

ye

my integral is 0...

like the ans

idk if thats good or not.

[wind cloak]

a^2 = (-a)^2

[tender onyx]

ye, look at my answer, mine is -(pi - t)^2

[wind cloak]

for a being a real number

[tender onyx]

is (pi - t)^2 even by any chance?

[wind cloak]

that thing is squared

[wind cloak]

so yeah definitely

[tender onyx]

ah

so

hmm

you know how to do fourier transfrom?

sereis*

fourier seres*

series*

[wind cloak]

uhm, no sorry

[tender onyx]

:(

this is so stoopid

my integral answer is 0

but

idk if thats right

💀

[wind cloak]

try to plug it in wolframalpha ig

maybe that'll help

[tender onyx]

👍

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[lusty delta]

depends how quickly and how well you can relearn

[wraith hinge]

not sure

[dusty garnet]

assuming not well at all

[wraith hinge]

very cooked

[dusty garnet]

would it be feasible for me to reteach myself trig while learning calc 1

[wraith hinge]

yes

[dusty garnet]

any good resources? preferablly something that doesnt take eons of time

im aware this question is stupid btw lol so thanks for da help

[prime lynx]

🍳

[grim sparrow]

probably

[grim sparrow]

http://tutorial-math.wip.lamar.edu/Extras/AlgebraTrigReview/TrigIntro.aspx looks like a good one

there's also this cheat sheet that probably contains more trig than you need for calculus lol
https://tutorial.math.lamar.edu/pdf/trig_cheat_sheet.pdf

[dusty garnet]

aight

i am garbage at math so im probably fuckedf but thank yall

[grim sparrow]

you're not bad at math!

[dusty garnet]

professors aint ready for this comeback

[grim sparrow]

it's common to forget stuff

that doesn't make you stupid or bad at math

you can make it

it's certainly feasible

[dusty garnet]

i must presevere...

[grim sparrow]

good luck!

[dusty garnet]

ty <3 ily

[grim sparrow]

❤️

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[dusty garnet]

.close

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[dusty garnet]

o7

[dusty garnet]

time to make it happen

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[grim sparrow]

[grim sparrow]

.close

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[grim sparrow]

sigh

why does it reset right then and there lmao

oh well

[pearl lotus]

Damn that's a hard prob
Try Differentiating wrt to x

[pearl lotus]

That's me in a chemistry class

[grim sparrow]

so true

[pearl lotus]

They be doing bullshit reactions then assuming the temp to be constant to calculate heat released

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[short spear]

How to do question b without using the l'hopital rule?

[ionic pendant]

split tan(3x) into sin & cos, and use the provided limit

[short spear]

so you have (sin3x)/((6x)(cos3x))

but idk how the provided limit would work if there is a 3 in front of the x

[ionic pendant]

we can write the provided limit as [ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 ] so we can do a change of variables and replace some function of $x$ with $u$, as long as $\lim_{x \to 0} u = 0$

[solid kiln]

cloud

[short spear]

so the answer is 1/2?

[ionic pendant]

yes

[short spear]

ohk thanks for the help

.clsoe

.close

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[dim kettle]

We knoe that LHS=(x-a)(x-b)(x-c)(x-d)(x-e)(x-f). If we plug in x=0, then we get acf=abcdef, so acf=0 or bde=1. The latter is obsviously impossible as no 3 distinct integers product equal 1.
For acf=0, if a=0, then x(x^2+bx+c)(x^3+dx^2+ex+f)=x(x-b)(x-c)(x-d)(x-e)(x-f). If we plug in x=0, we get 0=0. Or can we instead divide both sides by x first yielding (x^2+bx+c)(x^3+dx^2+ex+f)=(x-b)(x-c)(x-d)(x-e)(x-f)??

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[frigid lark]

i have no idea how to do this

i wouldve used F = qvB = mv^2/r but we only have v, B, and r

im sure that the photo of the track is supposed to do something to help but i dont really know what

ok i found that its a negatively charged particle because of the right hand rule

but because its a fundamental particle, does that mean that its charge is gonna be the same as an electrons charge? i only have electron charge (and proton bc its just +) on my formula sheet and i dont remember if fundamental particles have different amount of charges

.close

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[hasty yew]

guys how do you type off diagonal ddots $\ddots$ but the other diagonal

[solid kiln]

normalAtmosphericPa=101,325

[hasty yew]

in lartex

with this bot

[vale bobcat]

Oh there's a package that can rotate characters but it may not be included in the bot

I don't think there is a command for that on the vanilla tex.

\rotatebox[origin=c]{90}{$\ddots$}

[nova spire]

\rotatebox[origin=c]{60}{$\ddots$}

damn not far

[solid kiln]

rafilou2003

LucienF

[nova spire]

well approximately 70 degrees maybe

[vale bobcat]

You can still use the rescaling package afterwards ahaha

[marble wharf]

if you google it you find at least one stackexchange post with a few options

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[unkempt wyvern]

can anyone help me understand what angle i should take the make the W force the same "direction" as the Na ?

[unkempt wyvern]

i saw that its 4/5 but im struggling to understand why

is it this?

[wind cloak]

wdym?

[unkempt wyvern]

you see how the Na is tilted compared to the W

if i set my axis in that tilted angle, i would need to also have the W go in that direction, so i would need to multiply it with a factor of 3/5 or 4/5 to take the Y force

i don't think i'm explaining myself well

[wind cloak]

so you want a component of W in the direction opposite to that of N_A right?

[unkempt wyvern]

like i want to find this

actually now that i draw it i think i see it more clearly

[wind cloak]

see, W vector is along the side 3
{if you understand what I mean}

[unkempt wyvern]

like what i want to find is w y

yeah i get you

so then it would be 3 instead of 4

and so Wy = W*3/5 ?

no

its 4/5

wait then what do you mean by its along the side 3

what i did in my head is just take the 345 triangle and rotate it

so i got the 4 at w

5 as wy and 3 as wx

[wind cloak]

see here, you made a little 345 triangle,
so vector W is parellel to the side 3

that's what I meant

[unkempt wyvern]

yeah

but i don't really get how that helps

becasue my axis is tilted as well

this is the full figure with the axis i set

so from what i saw

after redrawing the triangle

W is 5

3 is Wx and 4 is Wy

[wind cloak]

see if you understand in the terms of angles it'll be better

angle psi represents approximately 53 degrees

so 90-psi will represent an angle of approx 37 degrees

[unkempt wyvern]

okay

[wind cloak]

it is easier to think that way

[unkempt wyvern]

wait hold on something else that confuses me with angles

[wind cloak]

what?

[unkempt wyvern]

in this case

where i'm given this 30 degree angle

[wind cloak]

yeah, so what will be the acute angle with W?

[unkempt wyvern]

60?

[wind cloak]

yeah

[unkempt wyvern]

ah

okay

lmao

and then i can apply what you did just before

[wind cloak]

hmph..

[unkempt wyvern]

huh

i think i get it though

thank you

.close

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[unkempt wyvern]

how can i find the movement relation between A and B

[unkempt wyvern]

like A moves 1 m how much does B go up

its not the question exactly but i need to know the relation

[violet gust]

complete the triangle with line of contact, x axis and y axis

so any movement of A is strictly along x axis

and for B it is along y axis

so, you can project those movements along the contact surface

[unkempt wyvern]

so i would have a tan relation?

[violet gust]

yes

[unkempt wyvern]

oo

okay

so when a moves a bit

b moves tan(15) of a

so Sa = Sbtan(15)

or is it the opposite

no yeah its the opposite

tan is opposite/adj

a is the adjacent force

[violet gust]

well, just to visualize, if b moves a small abount, would that require a to move a lot more or less?

[unkempt wyvern]

so its Sa tan(15)

uhhh

a lot?

[violet gust]

exactly

so is tan15 more or less than 1?

[unkempt wyvern]

less

[violet gust]

then you see which side its gonna be on

[unkempt wyvern]

so it should be on b?

[violet gust]

(its just a hack so you can verify stuff)

you are saying sb < sa

[unkempt wyvern]

ahhh

[violet gust]

so is sb * something smallr than 1 gonna be equal to sa?

[unkempt wyvern]

no

sa* something smaller

[violet gust]

mhm

[unkempt wyvern]

okok

thank you

[violet gust]

so its not gonna be on b

[unkempt wyvern]

gonna take some time to really get in there but i get it

but damn thats a nice hack

it will be on A

[violet gust]

[unkempt wyvern]

dope thank you very much

[violet gust]

npnp

[unkempt wyvern]

.close

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[boreal geode]

Hello guys I am lost during solving algebra , should I take option 1 or option 2 (I am not sure which one is correct)

Here is a better quality of the picture

[wind cloak]

none

[boreal geode]

what is the correct option sir

[placid radish]

u made some errors in both 1 and 2

[barren mountain]

?

a huge mistake

[placid radish]

think abt the following: is $$\frac{1}{3}-\frac{1}{6}=\frac{1-1}{3-6}$$?

[solid kiln]

qwertytrewq

[boreal geode]

Oh shoot

I am so sorry guys I should have proof read one more time before I send

I apologise 🙏🙏🙏

[boreal geode]

Oh yes sir ! Huge mistake!

[rotund grove]

💀 🙏

[boreal geode]

Oh wait Nuh uh I accidentally sent the wrong one I forgot to crop away the top

Should be just here

[gilded pulsar]

Yeah but still

[boreal geode]

OPPs

[boreal geode]

wait sir let me send the question too

[barren mountain]

what are you doing?

[gilded pulsar]

What’s funny is, you added correctly at the top

[gilded pulsar]

Both are wrong

[boreal geode]

Idk I was using a method I just learnt but I think I used it absolutely wrongly

[boreal geode]

oh god

[gilded pulsar]

When adding you don’t do anything with the denominator

Just add the numerator

[boreal geode]

But I need to make the demo same

[gilded pulsar]

ofc that only works when the denominators are the same

Yeah

[boreal geode]

Mb I didn’t let u finish

[boreal geode]

Wait sir let me show u

[boreal geode]

like this

[boreal geode]

Applying it to the question

I think I did it correctly

[gilded pulsar]

Look at the final result

[wind cloak]

how would you combine the fractions after this?

[boreal geode]

And add it?

[wind cloak]

can you show it

[boreal geode]

Oh ye man I forgot to finish the demo dumb me

[boreal geode]

Wait sir I will be back in few minutes after I finish my food rq 🙏🙏

[wind cloak]

kk

[boreal geode]

like this?

[wind cloak]

yeah

[wind cloak]

but you didn't do these like that

[boreal geode]

True I am just bit uncautiois

and messed up

Need more practice

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[simple patrol]

This is the question

[simple patrol]

ive reached here but idk how to continue

the soln i saw online used this which i dont want to do (they dont do the last step i did and did this instead of what i did at the end)

[wispy sparrow]

i feel like this is the go-to method for trig integrals like this

[simple patrol]

it is

but its just that i didnt like it here

[wispy sparrow]

other than that, multiply top and bottom by sec^2

[simple patrol]

i want to know if there is another mehtos

[wispy sparrow]

from second last step

[simple patrol]

ill try it

[wispy sparrow]

nvm that doesn't work

[simple patrol]

ya it becomes messy

[wooden remnant]

Other method could be going to complex exponentials

But also not pretty compared to just using the tan(x/2)

[simple patrol]

i dont know what those are

[wooden remnant]

Use eulers formula

$e^{ix} = cos(x) + i \cdot sin(x)$

[solid kiln]

thijs2725

[wooden remnant]

But if youre not familiar with integrating with this youre probably better of not starting with this integral

[simple patrol]

we havent learnt that so we wouldnt be allowed to use it

[wooden remnant]

Then I sadly THINK that you would HAVE to use the tan(x/2)

[simple patrol]

alright

thanks

.close

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[heavy belfry]

What is the probability that the triangle made out of 3 randomly selected points on a circle contains the center of that circle?

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@heavy belfry Has your question been resolved?

[nova spire]

fix the first point on that circle (we can rotate the circle so that this point doesn't move from our referential)

oh and I forgot about something oops

ok wait my approach needs some revision to do

[heavy belfry]

if the are on the same semicircle there is no chance that triangle will contain the center except maybe if the two points make the diameter?

[nova spire]

they would need to be on opposite sides of the circle, so in different semicircles

but again this approach doesn't really work

ok so

[heavy belfry]

why would that mean they are in diferent semicircles basically two points if making a diameter can be in both semicircle right?

[nova spire]

the semicircles we're looking at are delimited by point 1

now place point 2 for example

either it's on the "left" of point 1 or on the "right"

(or at exactly opposite but this has probability of 0)

if you want points 2 and 3 to make a diameter

point 3 is gonna have to be on the other side

so on the other semicircle

anyways so the way to do it is a bit different

[nova spire]

Ok sorry for the delay

so I think the key intuition is noticing that, except for when 2 of the 3 points form a diameter (probability of 0), the center is NOT inside the triangle when there is a semicircle that contains all 3 points

which leads to this well-known problem https://math.stackexchange.com/questions/325141/probability-that-n-points-on-a-circle-are-in-one-semicircle

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[hazy viper]

hi

[hazy viper]

how would i solve it

i tried getting y to one side and x to the other

and then integrate both

but this doesnt work

because i cant integrate e^(-y^2)

[naive otter]

dy/e^y² =dx/(-2)(1+x²)

Were you given any range?

Upper limit , lower limit ?

[hazy viper]

y(0) = 1

thats all

[hazy viper]

yes thats what i got

the right one is

[naive otter]

?

[hazy viper]

what?

[naive otter]

Definite integration or indefinite

[hazy viper]

no this is all i got

[naive otter]

y(0) = 1

[hazy viper]

that doesnt help me integrate e^(-y^2)

[naive otter]

Yeahh true

But the function isn't integrable

[hazy viper]

ye so there has to be another way of solving

[naive otter]

One min I'm sending you https://en.m.wikipedia.org/wiki/Gaussian_integral

Maybe they missed some term ! Or intentionally, according to your grade

Most probably they missed a y on the left side

[hazy viper]

i dont think they missed anything, this question was on my last exam

[naive otter]

Which grade ?

Are you from Russia ?

[hazy viper]

Calculus in one Variable

no sweden

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[proven latch]

does anyone know how to shade in desmos using inequalities?

[proven latch]

I want to shade within these functions

[topaz fog]

This works without the restrictions for x but doesn’t work with those restrictions, I don’t know if it will be useful

bottom line says y> median(f(x),g(x),p(x))

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[sudden seal]

when writing in standard form,

8 x 3 x 10^6 x 10^-2 = 24 x 10^4

but to make a between 1 and 10 it will become 2.4 x 10^n.

I thought we should divide it so it would be 2.4 x 10^4 x 10^-1 = 2.4*10^3. but this is wrong, why?

[dull temple]

you went backwards

think about this -- 24 = 2.4 x 10

so 24 x 10^4
= (2.4 x 10^1) x 10^4

[sudden seal]

hm i see, but in theory, could we also do it in my method, but make it work?

since 24 / 10 = 2.4

[dull temple]

isn't that exactly what i did?

[sudden seal]

yes, but 10^-1 = 1 / 10 correct?

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[barren hearth]

How can I make a picture of what #7 is saying?

[kindred pier]

,rotate 180

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[barren hearth]

How would I do this

The wording is throwing me off

[half pasture]

Did you try breaking it down first

Twice the measure of angle 6 is 12 more than the measure of angle 7

[barren hearth]

I don’t understand what that means

So 2x-12 = x

?

[limber minnow]

I think that's on the right track, though would it be x on both sides?

Also, it should be "12 more than 4 times the measure of angle 7"

[barren hearth]

I don’t understand what it means

It says twice angle 6 is 4 times plus 12 angle 7

That means no sense

So it’s twice and then it’s 4 times???

Like what does that mean

[limber minnow]

How about "twice angle 6 equals four times angle 7 plus 12" ?

[barren hearth]

Nope

So

Angle 6 times 2

So

2x

[limber minnow]

Sure

[barren hearth]

2x+4y+12 = 180

?

Made the x and y different since they’re not congruent so the variables are different values

[limber minnow]

Hmm, you have all the parts but I think you're combining them too much

How about:

2x - 12 = 4y
x + y = 180

[barren hearth]

Lol no

I would’ve never done that

That makes no sense

[limber minnow]

Okay well we know by definition of <6 and <7 being a linear pair that x + y = 180 right?

[barren hearth]

Yeah

[limber minnow]

So "Twice the measure of <6" gives 2x
"Is twelve more" gives - 12
"than four times the measure of <7" gives = 4y

It could also be written as 2x = 4y + 12

Does that make sense at all?

[barren hearth]

Wouldn’t it be 2x = 4y-12

Since it says 6 is twelve more

Meaning 7 would be twelve LESS than 6

[limber minnow]

Hmm, just to be sure you have it that x is <6 and y is <7 right?

[barren hearth]

ya

[limber minnow]

Okay cool cool

So, we can say that "2x is twelve more than 4y"

[barren hearth]

yeah

[limber minnow]

So we would have that 2x > 4y right?

[barren hearth]

Yeah

no

They're equal

They're a linear pair

One isn't greater

[limber minnow]

No I think that in a linear pair the angles can be different? They just need to add up to 180

[barren hearth]

It can be 2

[limber minnow]

Like, 70 and 110

[barren hearth]

We don't know if the linear pairs are 110 and 110 or 110 and 70 for example

So idk how you would solve that in an equation if you dont know if they're equal

[limber minnow]

I think in that example A and B, A and D, C and B, and C and D form all the linear pairs?

Well we don't have that x and y are equal, but we do have that x + y = 180

Do you think you can move terms in that equation to have y alone on one side? y = ?

This is the reference I'm looking at btw:
https://www.cuemath.com/geometry/linear-pair-of-angles/

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[barren hearth]

Yeah

Sorry I was driving

So I can have x and y equal 180?

Or how would I set it up

I get the wording part

[limber minnow]

Yeah, we have that x + y = 180 from them being a linear pair of angles

So you'd want to move the x to the right side in order to have y alone on the left side

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[barren hearth]

So we’d have 2 different equations?

[limber minnow]

Ah, yup!

[limber minnow]

That's what I was referring to here!

Sorry I think I was jumping around a bit with my reasoning and didn't explain why

Let's go back to 2x - 12 = 4y for a moment, are you still unsure about this equation?

[barren hearth]

No

Wait

I didn’t understand the - 12

[limber minnow]

Okay, well what if I rewrote it as 2x = 4y + 12 ?

[barren hearth]

Yeah