help-13

[winter apex]

No but I probably know what it does lol

[split crane]

thoey*

lemme show

[split crane]

what grade are you in?

[winter apex]

I'm a sophomore in high school

[split crane]

ah

Im in eight grade tryna pass to be a freshman

[winter apex]

You're good man

Honestly, I know some Precal and Calculus as well

[split crane]

HTF

[winter apex]

Mainly so I would know how to solve for some problems on my calculator test

I placed 1st today actually

[split crane]

W

[winter apex]

But anyways

[split crane]

so this is triangle sum theory its super duper easy fr

this what Im working on in class

Its like free time to be real with you

[winter apex]

Oh yeah, that's pretty easy

4y-19+55+5y=180

[split crane]

but theres these problems that come up that have like 5 traingles in a big triangle and you have to find the angles

lemme draw a pic

[winter apex]

If you looked at this

Does that make it easier?

The extra angle they gave you honestly just makes it look scary

[split crane]

no really

its so weird hang on lemme draw this out

[winter apex]

You're good

[split crane]

sorry bad at drawing with a mouse

thoes are deff not cerrect angles and its prob impossible to solve

but thoes are what I have a hard time with

[winter apex]

Oh I see, yeah

[split crane]

You dont have to solve that Ill figure out eventually

itsd 1 oclock so ima go to bed

Have a great night bro you helped me alot tonight

❤️

[winter apex]

No worries man, anytime

You get to look forward to this in precal though lol

[split crane]

NO

NO

NO

SLEEPPY TIME

[winter apex]

Anyways, you have a goodnight man

.close

Oh, I can't close it

[split crane]

what about bom

[winter apex]

I think bom's out

[split crane]

yea

.close

lol

[winter apex]

I'll dumb this down lol

[split crane]

You should become a helper

[split crane]

^ 😭

[winter apex]

Do you know the pythagroian theorum?

[split crane]

yes ofc

[winter apex]

It's that but with a kick of drugs

[split crane]

IDK what rad. is

[winter apex]

It's just another unit

[split crane]

oh ok

[winter apex]

Kind of how like we have Feet and Inches

Radians is Degrees

180 degrees = pi radians

$C^2=A^2+B^2-2AB*Cos(c)$

[wraith dagger]

Unparalleled

[split crane]

WTF

[winter apex]

That's the formula

[split crane]

IDK THAT

[winter apex]

Yeah, don't worry about it lol

[split crane]

what is COS

[winter apex]

Cosine

[split crane]

what is (c)

[winter apex]

You have to use a calculator

The angle between A and B

in this case 0.323

[split crane]

oh like that what all thoes buttons at the top do

[winter apex]

Yeah

[split crane]

lol

[winter apex]

It's really simple if you know the formulas and stuff lol

[split crane]

you saw nothing

[winter apex]

But it's learning them, yk?

[split crane]

yea

😭

LMFAo

[winter apex]

Anyways, you have a good night

[split crane]

Aye I mean we all did it once

[winter apex]

@lucid lantern Sorry for all the extra stuff, this was the last problem

[cedar kiln]

@lucid lantern Has your question been resolved?

[cedar kiln]

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[silver idol]

I need help answering this

[lucid lantern]

Oh nah you’re good I was finna text you later Ina bit if that’s Igh

[lucid lantern]

Yes

[silver idol]

ok

[chrome quail]

what have you tried until now?

[silver idol]

me or bom?

[charred scarab]

You, it's your channel

[silver idol]

oh well i just tried to simplify the fraction

[chrome quail]

how many cards below 3 are in a deck of cards?

[silver idol]

52? i think

[chrome quail]

below 3

[silver idol]

oh then 8 i think?

[chrome quail]

8? what are those?

remember aces are considered high

[silver idol]

Then none I think, if im wrong then idk

[chrome quail]

What about 2?

there ar four 2s, one for each suite

so 4 out of the 52 cards are less than 3

[silver idol]

i see

soo

[silver idol]

so now what

[cedar kiln]

@silver idol Has your question been resolved?

[cedar kiln]

@silver idol Has your question been resolved?

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[simple ingot]

okay im definitely missing something very obvious but how do i get the areas of the two slanted triangles ?

[simple ingot]

i can tell you right now that the area is going to be xy

but i dont know how to get it

[mighty shuttle]

You mean the area of the squares?

Which triangles

[simple ingot]

the slanted triangles

[simple ingot]

you see how theres one ttriangle in the centre

then three squares bordering it

then three triangles sandwiched between the squares

[mighty shuttle]

Ah

[mighty shuttle]

0.5 Xy

Not xy

[simple ingot]

*for both

[simple ingot]

how thou

cuz the question gives the area and i know the area for everything except for the two, so if i subtractt i can get the area of the two triangles

[simple ingot]

but i dont know how to do it the other way

[mighty shuttle]

I'm talking about this triangle

[simple ingot]

yeah i have that one

i mean the other two

the one on tthe right and top leftt

[mighty shuttle]

Ah

[simple ingot]

these ttwo

okay i know you can do it with some trigonometry and cosine rule

but theres definitely an easy way

[simple ingot]

cosine rule is not taught in this course so that is definitely not the method

[radiant lance]

This is the route I was going, maybe it can help your brain gears turn until someone knows the proper way

[cedar kiln]

@simple ingot Has your question been resolved?

[simple ingot]

oh tysm I'll have a look at that

okay maybe you can do it with assume x=y

then you'll know the angles

and can cheese with cosine rule

idk bruh

this is making me suffer

[cedar kiln]

@simple ingot Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Suppose U and W are subspaces of V such that V = direct sum of U and W. Suppose also u1, ..., um is a basis of U and w1,..., wn a basis of W. Prove that u1,...,um,...wn is a basis of V.

[crimson sedge]

It's not difficult to do, I already tried but I don't like my solution because it's not really well formalized. I am currenlty in the first year of electrical engineering and I am not that good at writing shit down basically...

Anyway what I tried it's basically to use the lemma that I can extend a linear independet list to a basis. I chose u1,.., um and because w1,...wm is not in the span of the vectors before (the span is U), then i can adjoin the list to w1,...,wm to form a basis of V.

That is possibile of course because cap U W = additive identity

[royal loom]

might I suggest a possibly simpler proof?

[crimson sedge]

And u1,...,um, w1,... wn being basis of U and W are linear independet

[crimson sedge]

Say it

[royal loom]

to show u1, ... um, w1, ... wn is a basis of V

you want to show that they span V, and that they're LI

LI should come from the direct sum easily

[crimson sedge]

yep

[royal loom]

to show they span V

let v in V

since V = direct sum U and W

then

[crimson sedge]

ok

[royal loom]

v=u+w for some u in U some w in W

[crimson sedge]

i get it

[royal loom]

ya

[crimson sedge]

yeah

I did the same

before

lol

So saying that is formally correct

That was my problem, basically my prof want stuff to be abstract or whatever he means

[crimson sedge]

Thank you for your help

[royal loom]

the proof is formally correct if you make it detailed enough, it is also abstract enough to solve your problem in full generality

did he say specifically he had an issue with that method of proof?

[crimson sedge]

We well see

I will write both of them in case

[royal loom]

I believe it should be fine if you flesh out the details properly

[crimson sedge]

Yes I do

[cedar kiln]

@crimson sedge Has your question been resolved?

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[dry fossil]

$y' = \tan(x) * y$

[wraith dagger]

alee

[dry fossil]

[ \frac{dy}{y} = \tan(x) , dx ]

[wraith dagger]

alee

[dry fossil]

do i have to do this ?

[royal loom]

it certainly is the easiest method

well

unless you already know the answer

[dry fossil]

[ \ln|y| = -\ln|\cos(x)| + C ]

[wraith dagger]

alee

[dry fossil]

so i integrate and get this ?

[vagrant elbow]

yeah

now get rid of all the lns

[dry fossil]

how can i do that

$y = e^{-\ln|\cos(x)| + C}$

[wraith dagger]

alee

[royal loom]

yes now simplify

=C/e^(ln|cos(x)|)

=..

[royal loom]

this should be solvable at first glance without any integration or anything if you wanted to be slick with it btw

[dry fossil]

could you show me pls

i would appreciate it

[royal loom]

well

there's really nothing to it, it's just a trig derivative you should know

[dry fossil]

derivative of tan ?

[royal loom]

that is sec^2(x)

so no

[royal loom]

try sec(x), what is its derivative

[marsh lake]

quotient rule?

[dry fossil]

1/cos(x)

[royal loom]

no

[dry fossil]

aaaa

tan(x) sec(x)

[royal loom]

yes

so if y=sec(x) then y'=tan(x)sec(x)=tan(x)y

but

the way you are doing it

is good practice

see if you get the same answer

[dry fossil]

wait

so what i need to solve

[royal loom]

you were doing it like this, right?

just simplify now

you should get the same answer

[dry fossil]

e^C / e^cos(x) ?

im trying simplify

@royal loom

[royal loom]

well, you dropped the ln by accident

[royal loom]

and you can replace e^(C) with a different constant, like k

because after all it is just a constant

[dry fossil]

wait can you write the correct one pls

[royal loom]

No, you got this

you're close

just make those fixes I suggested

[dry fossil]

e^C / e^ln(cos(x))?

$\frac{K}{ e^\ln(\cos(x))}$

[wraith dagger]

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

[royal loom]

sure this fixes one issue

now do the final simplication

what is e^(ln(stuff)=?

[dry fossil]

$\cos(x)$

[wraith dagger]

alee

[royal loom]

yes

so your simplification is

K/cos(x)

or just

K*sec(x)

Right?

[dry fossil]

yes

[royal loom]

same thing as what we got earlier

we just didn't include the constant

[royal loom]

🔠

[dry fossil]

wait so

now it is K * sec(x)

before it was y = sec(x)

right?

[royal loom]

right, like I said, we just left off the constant earlier

[dry fossil]

and nothing changes?

[royal loom]

our case was a special solution where k=1

the most general solution is with the constant included

maybe one last important note

K > 0

[dry fossil]

ok but so before could the constant be included or not?

[royal loom]

it could be

[dry fossil]

if y=sec(x) then y'=tan(x)sec(x)=tan(x)y

so this

[royal loom]

earlier we didn't exactly solve it

we just said

hey this looks like the derivative of sec(x)

[dry fossil]

can be this: if y= K sec(x) then y'=tan(x)sec(x)=tan(x)y

?

[royal loom]

well

now you must include the constant

if y=Ksec(x)

what is y'?

[dry fossil]

0 * tan(x) sec(x) ?

[royal loom]

No, why 0?

y' = K sec(x) tan(x)

[dry fossil]

Is the derivative of a constant 0?

[royal loom]

it is a constant being multiplied, not added

[dry fossil]

aaaa

yes

sorry

right

[royal loom]

so y' = K sec(x) tan(x) = y tan(x)

and we still see that it works

[dry fossil]

yes

[royal loom]

and actually

we just verified that k does not need to be > 0

in this method of solution

we forced K > 0 because K = e^(-C)

but

as we can just see by what we just did above

let K be any number, maybe even negative

if y=Ksec(x)

then

y'=Ksec(x)tan(x)

y'= tan(x) y

is still a solution

so K really is just any constant, not necessarily positive

makes sense?

[dry fossil]

yes

thanks!

[royal loom]

no problem

[dry fossil]

but does the fastest resolution step have a name?

or is it something mechanical?

i mean

does this resolution method have a name?

@royal loom

[royal loom]

Which method?

We did it 2 ways

[dry fossil]

the fastest one

[royal loom]

One way was just getting lucky, this was an easy problem and we already knew the function

In general it won’t be so easy

But the other way will always work for equations of this type

[dry fossil]

ah ok

[royal loom]

It’s called

A separable differential equation

[dry fossil]

ah okok

[royal loom]

And you used the method of separation of variables

[dry fossil]

yes

thanks!

.close

[cedar kiln]

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[cedar kiln]

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[jade ibex]

a question came up in my schoolwork: One of the numbers below was printed incorrectly. Which number is it?
536 563 546 556

[mighty shuttle]

any more information?

[livid hound]

needs more context

the printer supposedly printed the numbers someone told it to print

[jade ibex]

oh okay thanks

[jade ibex]

uh no none

.close

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[left cove]

im confused can anyone help

[cedar kiln]

Please don't occupy multiple help channels.

[left cove]

.close

[cedar kiln]

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[buoyant latch]

is there any reason for the difference in order

i yoinked these from the laplace transform page of wikipedia

[crimson delta]

probably two different people wrote it

[buoyant latch]

ok

.close

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[void dock]

What is hyperbola and parabola

[lyric narwhal]

They are two different conic sections

Parabola is the one with eccentricity = 1 and hyperbola is the one with eccentricity > 1

Do you have a specific question in mind?

[pastel vault]

Google exists for your question

[void dock]

No thank you 😊 my doubt is clear now

[pastel vault]

Npnp

.close

[cedar kiln]

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[mint viper]

I am looking for some intuition as to why the partial derivative of f with resepct to y tells us anything about the uniqueness of a DE?

[cedar kiln]

@mint viper Has your question been resolved?

[pastel vault]

It has to do with Lipschitz continuity: |f(x) - f(y)| <= M |x - y| for some M and for all x, y

Notice that |f(x) - f(y)| / |x - y| looks awfully like the derivative: it becomes the derivative as you let x approach y or vice versa

So if Lipschitz continuity is violated, that basically means the function has an unbounded derivative somewhere

So integrating f(x, y) under Picard iteration may not give you a fixed point

Integrating a function with an unbounded derivative should give you an unbounded function

So you can't contain the function in order to use the Picard-Lindelhof theorem

IDK, tldr the reasons all require results from beyond a first course in DEs

I don't think the proof of Picard-Lindelhof is that bad

You might think otherwise

[cedar kiln]

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[crimson delta]

can you answer it for 8 squares?

[cold wadi]

no...

[stray otter]

alr so what is the next one

[crimson delta]

can you draw the next few squares?

[stray otter]

1 more should be enough

[cold wadi]

It's just the left foot and this will sequence in this image until 48 squares are reached, as if jumping 48 times in this type of sequence

[crimson delta]

well you could also draw all 48 squares but maybe just a few at first and maybe you nottice a pattern

[stray otter]

so something like this right?

[cold wadi]

That's it, just with her left leg, how many times did she jump with just her left leg.

[cold wadi]

ok.

is portuguese...

[cedar kiln]

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[low forge]

How many 9-card hands contain four cards of the same value for a standard deck of cards. (52)?

[cedar kiln]

@low forge Has your question been resolved?

[low forge]

no 😦

[low forge]

i understand the basic premise

but i can't seem to get the solution

so for 5 hands

it would just be 13 x 48 c 1

@toxic thistle

[gray oar]

lol u talking to bot

[low forge]

<@&286206848099549185>

[cedar kiln]

@low forge Has your question been resolved?

[languid canyon]

😂

[worldly chasm]

@low forge #help-36 message

You two asked the same question ^

[cedar kiln]

@low forge Has your question been resolved?

[cedar kiln]

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[vale berry]

Need help writing a generalization for this patern

[weak otter]

I mean just write it as log(10^p * x) and simplify

[vale berry]

thank you

.close

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[crimson sedge]

How do I determine if the function is One to One, Onto or Bijective

[slim gorge]

to determine if it's one to one, you use both the horizontal and vertical line tests

[crimson sedge]

ahh I've done those tests in prev topics but

what are we exactly looking for here by those tests?

[slim gorge]

if there are multiple outputs for an input or if there are multiple inputs for an output

so for every y there's a unique x and for every x there's a unique y

[crimson sedge]

that is true in this grapj

graph*

[slim gorge]

yep

so it's one to one

for it to be onto, for any value belonging to R+/0, let's say z belongs to R+/0, and there must always be a value of x for which f(x) = z

and if it's bijective then it's both one to one and onto

[crimson sedge]

I think it is bijective

but what about the interval notation they have provided

[slim gorge]

basically that's the codomain

so to find if it's onto

there would have to be a value of x for which f(x) is any value in the interval [0, infinity)

so you're right, it is bijective

[crimson sedge]

yeah so every x has to map with atleast one f(x)

ehh

[slim gorge]

it's different

[slim gorge]

wait

oh my bad

been a while since I've done this

misread it

R is all real numbers

so is there a value of x which produces f(x) = -1?

[crimson sedge]

nope

the graph is on the positive quadrant

[slim gorge]

yep

so it'd be only one-one

[crimson sedge]

since there is no solution for f(-1)

right?

[slim gorge]

yes, or any negative values

as in no x exists for f(x) < 0

[crimson sedge]

okay

thanks

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[elder sparrow]

.open

how to ask

[elder sparrow]

ok

now

this question is VERY basic

∫1/2x dx

right

now theres 2 ways u can do this

one is to factor out the 1/2 to simply get 1/2 ln|x|

the other is
let u=2x, so du/2 = dx
hence ∫1/(2x) = ∫1/u du/2 = 1/2 ln(u) = 1/2 ln|2x|

why am i gettin different answers from both methods?

[mighty shuttle]

$u=2x$

[wraith dagger]

ƒ(Why am. I here)=I don't know

[mighty shuttle]

$du=2dx$

[wraith dagger]

ƒ(Why am. I here)=I don't know

[mighty shuttle]

$dx=\frac{du}{2}$

[wraith dagger]

ƒ(Why am. I here)=I don't know

[mighty shuttle]

which is the same thing

[elder sparrow]

isnt it now ∫1/u (du/2)

so 1/2 ∫1/u du

[mighty shuttle]

it is

[elder sparrow]

so 1/2 ln(u)

but u=2x

so 1/2 ln(2x)

[mighty shuttle]

wait

ah

makes sense

it differe by a constant

$ln(2x)=ln(2)+ln(x)$

[wraith dagger]

ƒ(Why am. I here)=I don't know

[elder sparrow]

OHHHH

that makes sense

thankuu

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[slow fiber]

Tom, Todd, Maude, and their mom are all playing poker
and betting for peanuts. They’re allowed to go into peanut debt, and during the
course of the game some of the peanuts get eaten.
If, after the game, there are 10 peanuts left and no one is in debt for more than
5 peanuts, how many possible distributions of peanuts are there?

[slow fiber]

How do I do this with bars and stars if there are negative numbers?

[deft saddle]

whats bar and stars

[slow fiber]

its just a counting method

[deft saddle]

if there's 10 total peanuts left

is it?

[slow fiber]

?

[deft saddle]

is 10 peanuts left totally from 3 people?

4*

my

theory is

if maximum debt is 5

there can be 2 cases

no one has debt

someone has debt

let's finish one case first

if no one has debt and 10 peanuts among 4 people

10 0 0 0
9 1 0 0
9 0 1 0
9 0 0 1
.
.
0 0 0 10

u gon count?

[cedar kiln]

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[crimson sedge]

can someone explain how it got into the last step?

[void glen]

$(\frac{k}{2}+\frac{2}{2})$

[wraith dagger]

y0shi

[void glen]

$\frac{k+2}{2}$

[wraith dagger]

y0shi

[crimson sedge]

thanks

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[meager vapor]

Review the conditional statement: “If you do not sleep, then you will be tired.”

A second statement was derived from the conditional statement above: “If you do sleep, you will not be tired.”

How are these two statements related?

The second statement is the converse of the first statement.
The second statement is the conclusion of the first statement.
The second statement is the contrapositive of the first statement.
The second statement is the inverse of the first statement

IS it c or d?

[runic garnet]

if u do not sleep (p) then you will be tired (q)

now identify what these are in terms of p and q

if you do sleep

you will not be tired

define those clauses in terms of p and q

[meager vapor]

So would the answer be c?

[runic garnet]

do this first, then we will go from there

[meager vapor]

if you do sleep (q) you will not be tired (p)

[runic garnet]

look at this

if "if you do not sleep" is p

then what is "if you do sleep"

[meager vapor]

p?

so its inverse?

[runic garnet]

p: "if you do not sleep"

"if you do sleep"

those are not the same expression

how can they both be p

[meager vapor]

so it has to be q

[runic garnet]

no

if p is "if you do not sleep"

and we want to define "if you do sleep" in terms of p

we notice that "if you do sleep" is the negation of "if you do not sleep"

[meager vapor]

yes

[runic garnet]

so what is this in terms of p

[meager vapor]

not p?

negative

[runic garnet]

correct

~p

now

we have defined q to be "then you will be tired"

[meager vapor]

so the other is not q

[runic garnet]

correct

so what do we end up with

[meager vapor]

inverse

[runic garnet]

~p -> ~q

yes

[meager vapor]

thanks

.close

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[indigo plover]

if n = 8^2022, what is n/4

[glass sky]

4+4 , then binomial expantion?

[indigo plover]

8

[queen estuary]

no just convert everything to a power of 2 and use exponent rules

[indigo plover]

oh okay

ty

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[glass sky]

wait ofc im so tunnel visioned 💀💀

[cedar kiln]

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[green crow]

Need help on this problem

[green crow]

Here’s what I tried

[cedar kiln]

@green crow Has your question been resolved?

[cedar kiln]

@green crow Has your question been resolved?

[dreamy spire]

Call it

[void glen]

the pendulum isn’t really in circular motion so we cannot use that formula

i would suggest breaking the force of gravity up into components (perpendicular and parallel)

and solve for the acceleration on the parallel axis

[green crow]

On part A i got right that the radial acceleration is 4.5 m/s^2 using the v^2/r formula

[void glen]

quick question, is the pendulum going around is circular motion or is it just swinging up and down

basically i’m asking if it’s a conical pendulum or a simple pendulum

[green crow]

The way I see it is the pendulum swinging up and down

[void glen]

yeah then that isn’t circular motion

so we cant relate it to v^2/r

since the force getting it to turn is changing every instant

i believe the question is trying to trick you by giving you all that unnecessary information

[green crow]

Why did the problem say "swings on a 2m long string"

[void glen]

as a trick i assume

[green crow]

Is it common for physics problems to try to trick you

[void glen]

yeah

what unit is this for

cuz i could be interpreting this wrong

[green crow]

Dynamics: Motion in a plane

[void glen]

okay yeah then it’s prob trying to trick you

to get the tangential acceleration

we just need to split up the force of gravity into two components

parallel and perpendicular components specific

try that

[green crow]

Maybe like this

[void glen]

yep

[green crow]

From here is it finding net force

[void glen]

yeah

in the parallel direction

also i’m pretty sure that angle isn’t theta

but 90-theta

[void glen]

yep

we don’t really need y here tho

we just need x

since we can set that equal to ma

and solve for a

[green crow]

Why the x direction

[void glen]

nvm theta is 15 here as well

since it’s with the vertical

[void glen]

well in the y direction

it wouldn’t be accelerating

since we know the rope isn’t going to extend

nor is it going to retract

in the x direction is where the ball is trying to move at that instant

[green crow]

Am I still thinking of the pendulum as swing back and forth or it’s something else

[void glen]

yeah you are

[green crow]

Is solved for a_y but don’t know the tension

Never mind I forgot I’m solving for in the x direction

[void glen]

yeah

since a_y would just be 0

[green crow]

It counted it right but apparently the tangential acceleration is supposed to be positive but the way I wrote it was -gsin(theta)

[void glen]

it just depends on your sign convention

you took down as negative and up as positive

and also the question asked for the magnitude

[green crow]

I also took left as negative and right as positive

[void glen]

so we dont count for direction

[void glen]

yeah

[green crow]

So when a problem asks for magnitude is it like putting absolute value

[void glen]

yeah basically

just ignore direction

in your final answer

[green crow]

I also have to solve for Tension and I do have a tension expression from the net force but the number I got for tension using mgcos(theta) was wrong

[void glen]

so when you submitted mgcos(theta) as tension it said that it was wrong?

[green crow]

Yes I put in (4)(9.8)cos(15) ≈37.864 in degrees

[void glen]

i got this when i subbed it in

did you put 15 in degrees or radians

[void glen]

are you meant to use 9.81 or 9.8

[green crow]

I've always use 9.8 unless I'm supposed to actually use 9.81

[void glen]

hmm

you used 9.8 for the previous answer right

if you did then we prob went wrong somewhere for this question

[green crow]

Yes

Maybe the pendulum is going along a circle because the first question used v^2 /r for radial acceleration

[void glen]

hmm

if it was then it wouldnt make sense why the first question was asked

since if its in uniform circular motion

then the tangential speed remains the same

[green crow]

Could this work

[void glen]

well the force that is centripetal now is the component of tension

not force of gravity

so the first equation is wrong there

it wouldnt make sense if we got the first part right tho

and it was actually going in a circle like that

since then the tangential acceleration would not be gsin(theta)

there would be no tangential acceleration in fact

only centripetal

[green crow]

To find the Tension do you have to find the components of the tension vector

[void glen]

well for a conical pendulum like this

yes

unless the question somehow switches up

i dont see how we got the tangential acceleration to be -gsin(theta)

if it wasn't a normal pendulum

[green crow]

I think it does switch up because it asked to find radial and tangential acceleration

[void glen]

can i see the full question

[void glen]

and all the parts

[void glen]

radial meaning centripetal acceleration right

so then we would just have $T-mg\sin(\theta)=\frac{mv^2}{r}$

[wraith dagger]

y0shi

[void glen]

yeah then what you wrote would be right

oh right because it is turning

so $T\neq mg\sin(\theta)$

[wraith dagger]

y0shi

[void glen]

thats my bad

[green crow]

Wouldn't that mean acceleration in y is v^2/r i just set it equal to that because i found the radial acceleration earlier

[void glen]

yeah

[void glen]

basically just this

[green crow]

Why is it also T-mgsintheta too

[void glen]

since those are the forces present there

[green crow]

I guess that's it for this problem. This was a tricky one I had the right idea at the beginning but keep executing it wrong

[void glen]

yeah it was definitely tricky

i got tricked by it too

[cedar kiln]

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[crimson sedge]

any ideas ?
idk how to start

[coral jewel]

did you try finding the intersections of the two circles?

[crimson sedge]

how to do that ?

[coral jewel]

equate two equations of circle

you should have a line

[crimson sedge]

x^2+y^2+6x-4y-7 = x^2+Y^2-6x-4y+5
?

[coral jewel]

yes

[crimson sedge]

x=1
y= 0,4

so int points are
(1,0)
(1,4) ?

[coral jewel]

yeah

now construct a circle based on those 3 point

[crimson sedge]

how to construct a circle?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

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[wraith dagger]

Ransik

[cedar kiln]

@crimson sedge Has your question been resolved?

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[solemn perch]

hello

[solemn perch]

whos a computer science major here?

i would like to ask for help

its more on a runtime algorithm analysis model

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[strong jacinth]

b please

[cedar kiln]

@strong jacinth Has your question been resolved?

[livid hound]

what have you tried so far?

[strong jacinth]

uhh idk how to do it

[livid hound]

there's probably a typo in the question
a) is most likely supposed to be asking for sin(A)

and then its finding the two solutioins to that for 0 < A < 180°

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[full forum]

Im stuck with Q12..

I dont know where 7 and 25 came from

[cedar kiln]

@full forum Has your question been resolved?

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[dull plover]

how did we solve for x here

[dull plover]

i dont understand ;0

[green hornet]

open the LHS bracket

group all x terms together and simplify

[dull plover]

u mean by distributing the sqrtq2?

..?

[green hornet]

yes

[pastel vault]

The first step is of course to take the square roots of both sides, so you need a plusminus sign

And then you distribute the 2 - x as was mentioned

[dull plover]

why didnt we take the plusminus for (2-x) though

[pastel vault]

There's no plusminus on the left

[dull plover]

so after distributing it we end up with 2sqrtq2 - xsqrtq2 = +-xsqrtq1

[pastel vault]

Yes

[dull plover]

x = (2sqrtq2-xsqerq2)/+-sqrtq1

[pastel vault]

No

You add x sqrt q2 to both sides

And then factor out an x on the right

[dull plover]

2sqrtq2 = +-xsqrtq1 + xsqrtq2?

ahhhh yes it makes sense nwo

[pastel vault]

Yes

[dull plover]

thank you !

[pastel vault]

Then they put the x sqrt q2 first

No worries

[dull plover]

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[gloomy snow]

Wtf is this😭

[gloomy snow]

Is this even math

[coral jewel]

!redir

[cedar kiln]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[coral jewel]

.close

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[slate garden]

can someone explain to me SI units

[slate garden]

What does m^-2

and why do we do 15 divided by 1000/ 1 divided by (100)(100)

like shouldnt it just be over 100

cause cm to metre

is divide by 100

[pastel vault]

m^(-2) = 1/m^2

So it's $15 \frac{g}{cm^2}$ for example

[wraith dagger]

south

[slate garden]

ohhh right yh

thx

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[pastel vault]

So 1 m^2 = 1 m * 1 m = 100 cm * 100 cm

That's why, nw

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[toxic hinge]

Hey guys, can you help wit this task?

[pastel vault]

Write sin^2 x as 1 - cos^2 x

Don't forget the brackets around (1 - cos^2 x)

Then you have a quadratic in cos x

Also your quadratic is concave up so you will have cos x < a and b < cos x

Two distinct regions

[toxic hinge]

Like this?

[pastel vault]

Yes

[toxic hinge]

So is it right?

[pastel vault]

You're missing solutions

Cause cos(-x) = 1/4 and cos(-x) = 1/5 as well

So you need to multiply those two branches by -1

And you'll get 4 sets of solutions

Otherwise that's correct

Oh wait you didn't do the inequality part either

https://www.desmos.com/calculator/auxbz3skah

The inequality is the tiny region between the two x-values

[toxic hinge]

Maybe this?

[pastel vault]

Say arccos(1/5) < x < arccos(1/4) and so on

[toxic hinge]

Ohhh

Okay

Thanks

[pastel vault]

Nearly, so -arccos(1/5) < x < -arccos(1/4)

[toxic hinge]

Thanks for help

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[pastel vault]

No worries

[cedar kiln]

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[crimson sedge]

Why cos (-0) is equal to cos(0) while sin(-0) is equal to -sin(0) and not sin(-0) . "Note that 0 represent THETA"

[pastel vault]

Unit circle

So -theta is the half-triangle below the x-axis

The definition of cos theta is that it is the x-coordinate of the point

Sin theta is the y-coordinate of the point

So -theta has the same x-coordinate as theta

[pastel vault]

-theta has the negative y-coordinate of theta

[warm crescent]

replace x and y with the black equations

[crimson sedge]

OH. So cosin is still positive and now sin is negative. This is what you meant?

[pastel vault]

Yes

The unit circle tells you why

[crimson sedge]

Oh, I got confused for a moment. Let me check If I understood well.

Now if you have a equation cos(-90grade +x) equals to cos(-X) Because we wanna find the small angle and after it equals to cos(X)?

[pastel vault]

Ah that's not correct

cos(x - 90) is the same as shifting the graph of cos(x) 90 degrees to the right

If you do that, you get sin x

There's also a unit circle explanation

cos(x - 90) = cos(-(x - 90)) = cos(90 - x)

[crimson sedge]

Exactly:

[pastel vault]

Ok you were talking about this right

Then yes

Then that shows you cos(-90 + x) = sin(x)

[crimson sedge]

But why?

🤷

[pastel vault]

This is one method

Knowing the graphs of sin and cos

[crimson sedge]

Okay, let me break it down.

[warm crescent]

x-90=-(90-x). So cos(x-90)=cos(-(90-x)). Using cosx=cos(-x) you have cos(90-x)

it absorbes the - sign

cos(a+b)=cos(-(-a-b))=cos(-a-b)

[crimson sedge]

Why in math you say: "using". Do I have to memorize this: Using cosx=cos(-x) you have cos(90-x)

I hate memorizing.

[warm crescent]

ok then dont memorize

[crimson sedge]

I have to understand why.

[pastel vault]

If you have cos(90 - theta), the 'length' of the TOP triangle is the green side, sin theta

And the 'height' of the triangle is cos theta, so sin(90 - theta) = cos theta

[crimson sedge]

.

[pastel vault]

You have to tilt your head 90 degrees to see it

[crimson sedge]

Any video that explains this with examples?

[pastel vault]

https://www.youtube.com/watch?v=Cu0yrPf2660

[crimson sedge]

Very well.

[pastel vault]

The diagram in this one is pretty clear

[crimson sedge]

Now any sheet paper about cos, sin, tg. 12 year of school.

[pastel vault]

But yes pls don't memorise

Anything not involving double or compound angles comes from the unit circle

(There are diagrams for the compound angle theorems also)

[pastel vault]

For example

[crimson sedge]

I will instand memorize it for life if i understand why. Thats why I skiped vectors cuz its hard to image it in my brain.

"Why" for everything.

[crimson sedge]

That's why I am acting dum.

[pastel vault]

Exactly, you have the spirit

[warm crescent]

ok find cos(θ) and sin(90-θ) in terms of ratios (think about 90-θ swapping the adj and opp in general)

[pastel vault]

It's better to spend 2 hours understanding something that you'll then know forever

Than 20 hours trying to re-understand something cause you just memorised it and don't actually know why that's true

[crimson sedge]

I have a dum friend like me here. I am not alone.

😁

I will check the video.

[pastel vault]

Okok nw

[crimson sedge]

Is cos of 90-theta the same as sin of theta cuz we are talking about the right triangle with fliped cos and sin since its position is fliped?

Is this what it means?

[pastel vault]

Yes, it's cause it's flipped

[crimson sedge]

Ok one more moment to do an exercise, so I lock it in.

[crimson sedge]

Good? Before i click save as >> C:Brain/main-cell.

[pastel vault]

Nope, the picture on the left shows you that cos(180 - theta) = -cos(theta)

The x-coordinate is the negative of that of cos(theta)

The other picture is correct yes

[crimson sedge]

Oh thats a little mistake. I got the logic.

Thank you!

[cedar kiln]

@crimson sedge Has your question been resolved?

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[signal ibex]

How to sketch 1/f(x) if this is f(x)

[pastel vault]

1/0 = asymptote
1/(very small positive number) = very large
1/1 = 1
1/(very large positive number) = very small

Also 1/f(x) doesn't change the sign of the function

[signal ibex]

How is it this tho, y is there a closed circle on zero, and the line still gets larger, shouldn't it be smaller?

and y is the curve after 1 look like that

[pastel vault]

Read what I said again

The open circle on zero means that the 1/f(x) is not 0 when x = 0

Cause 1/-1000000 is a tiny number

But 1/anything can never equal 0

[signal ibex]

how do u know it's a closed cirlce and not a root, isn't roots asymptotes and visa versa when u do reciprocals?

[pastel vault]

Yes, that's correct

Also there's an open circle on 0

[signal ibex]

yes but isn't it meant to be closed

cus it's a root

[pastel vault]

No.

[signal ibex]

oh ok

why does it get bigger if the original also got big

[pastel vault]

It doesn't

The rule is that close to 0 becomes far from 0

Far from 0 becomes close to 0

If you check all the parts of the graph, you will see that this is true

So for example, if f(x) gets closer to 0, 1/f(x) gets further from 0

[signal ibex]

but in both cases the y is getting bigger

[pastel vault]

Where?

[signal ibex]

in the question and answer foir the first line y is getting bigger in both

[pastel vault]

No, f(x) is getting closer to y = 0 for 0 < x < 1

And 1/f(x) is getting further from y = 0 for 0 < x < 1

[signal ibex]

no it's getting closer to x=0, the y is going off infinitely into the negatives

[pastel vault]

So it's getting further away from y = 0

[signal ibex]

yea but they both are

if f(x) is going away from y=0 shouldn't 1/f(x) go towards it

[pastel vault]

Yes that's exactly what is happening

[signal ibex]

isn't this going away from 0 (getting bigger)

[pastel vault]

For which x-values?

[signal ibex]

0<x<1

[pastel vault]

Yes

Ah you don't follow the arrows

[signal ibex]

ok, and is this not going away from 0 for the sae values? https://media.discordapp.net/attachments/903430334681608232/1226110379084746842/image.png?ex=66239346&is=66111e46&hm=52addffb3a615c682582a58e7f7b329bd8aaeadc8ae0eec2ac8f51c002c02e3f&=&format=webp&quality=lossless

[hot crag]

...?

dude

[pastel vault]

You always have to read the graph from left to right

[hot crag]

x = 0 and y = 0 are different

;-;

that's what ur confusing

[pastel vault]

Also this

[signal ibex]

yea but in both cases the y is going down

[signal ibex]

wdym

[pastel vault]

Where is the y going down here

[hot crag]

"y=0" not "x=0" here, for example

[pastel vault]

I'm still talking about 0 < x < 1

[pastel vault]

I'm struggling to think of why you think the y is both going down

[pastel vault]

THIS

YOU DONT FOLLOW THE ARROWS

You read the graph from left to right

[hot crag]

^^

[pastel vault]

The arrow is the wrong way as you're going from right to left

[signal ibex]

oh ok so the first is getting smalelr and the 1/f(x) is getting bigger

[pastel vault]

Yes

[signal ibex]

oh ok

do the arrows just mean it's continuous

[hot crag]

yes

[pastel vault]

They mean the graph continues further in that direction

[hot crag]

basically

they mean that it goes on

in the same manner