help-49

[quiet jackal]

oh

the one for max ?

[brazen palm]

yep

and the whole shape is symmetrical along x=y, so we can find some angles are equal

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[olive matrix]

.reopen

[midnight plank]

✅

[brazen palm]

something like that

[quiet jackal]

i found an easier way

[brazen palm]

oh nice

[quiet jackal]

look at triangle 0.0

4,4

4,0

[brazen palm]

i get that

[quiet jackal]

it’s isocoles so the argument of the red line is 45

blue line sorry

[brazen palm]

pi/4-pi/6 right?

[quiet jackal]

that means small angle is pi:12 yes

we can do the same thing but with 0,4

[brazen palm]

yep

[quiet jackal]

but that doesn’t get anywhere

how did u get the max

[brazen palm]

well, you can use addition

[quiet jackal]

how did you work it out to be 30 i mean

[brazen palm]

arcsin(1/2)?

[quiet jackal]

ohh

yes yes

[brazen palm]

i thought you got it lol

[quiet jackal]

idk i confused myself

it’s pi/12 to 5pi/12

[brazen palm]

yep

[quiet jackal]

thank you

[midnight plank]

@quiet jackal Has your question been resolved?

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[midnight plank]

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[flat grove]

Hi again I'm back with the same problem as yesterday

[flat grove]

Here's again the scheme

We know that the radius of the little circle is 1cm

That all four big circles are tangents

And tangent to the two other circles

We're looking for the radius of the biggest circle

And here's what I found

So BO'AO is a square (if we stick other tangents coming from O it makes a square)

So the two triangles are rectangular and isosceles

But I found without any calculs the result

And I can't make it to this end

Even with the Pythagorean theorem

So where am I wrong or what did I miss

[olive matrix]

this is how i would do it

[flat grove]

Mhm?

[olive matrix]

how do you know that three circles make it to the end?

[flat grove]

Just drew it

I assumed

[olive matrix]

yeah that's not really a proof, diagrams can be deceiving

[flat grove]

So my researched answer is different, the drawing is just assuming I wanna find that result I assume but with demonstration

The one thing you drew is the hypotenuse minus the radius of the small circle, and it is also the radius forming the right angle with the tangent that starts from O

That one

[midnight plank]

@flat grove Has your question been resolved?

[hot totem]

Any progress?

[flat grove]

Yes and no—

Trying to think about some other properties

[hot totem]

Are you aware of the equation of the circle

[hot totem]

Actually you should be able to solve the question with just this image

[flat grove]

The equation of the circle

Wait

How can I use it

[hot totem]

Scrap that

That works, but does more algebra than necessary

[hot totem]

Focus on this image

[flat grove]

It doesn't jump in my eyes, I've been on it since yesterday or I'm really stupid—

[hot totem]

Ok

You know that angle OAO' is a right angle

And angle OBO' is a right angle

[flat grove]

Yeah

Yes

[hot totem]

What about the other two angles

[flat grove]

45

Each

[hot totem]

ok

You are claiming that OA = AO'?

[flat grove]

Yes

[hot totem]

ok

What is the quadrilateral OBO'A

[flat grove]

Square

[hot totem]

ok

For any square

What is the ratio of the square's diagonal as compared to its length?

(If you don't know, rely on Pythagoras)

[flat grove]

Um that makes two right triangles ?

[hot totem]

Yes

So

[flat grove]

So

Isn't it right ?

[hot totem]

That's going nowhere

I'm asking you about the ratio of a square's diagonal and its lengths

[flat grove]

Diagonal^2 = 2(one of the length)^2?

[hot totem]

ok

What is OO'?

[flat grove]

Diagonal

[hot totem]

What is OA

[flat grove]

Hypotenuse

Oh

[hot totem]

....

[flat grove]

On lengjt

[hot totem]

You see it?

Do you see it?

[flat grove]

Um—

[hot totem]

Ok

Let's go further

[flat grove]

Please

:')

[hot totem]

We have that OBO'A is a square. So OA = O'A?

[flat grove]

Yes

[hot totem]

We have that OO' has one unknown

[flat grove]

Yea

[hot totem]

And that O'A has one unknown?

But now we have an equation relating OO' and O'A, right?

[flat grove]

Yes

[hot totem]

Write it

[flat grove]

OO'^2= 2O'A^2

[hot totem]

Use R

[flat grove]

OO'^2= 2R^2

Hmmmm

[hot totem]

You know OO'

It's a very simple function of R

It's inside the diagram

[flat grove]

1+R ?

[hot totem]

Yes

Solve the quadratic

[flat grove]

Yeah

[hot totem]

||(There's actually no quadratic if you stated 1+R=sqrt(2)*R) - you can think about what is going on if you'd like||

[flat grove]

1=sqrt(2*R) - R

?

[last slate]

This one fr is useful

2(R+1)² = (2R)²

Find R

@flat grove

[flat grove]

Yeah I'm

Confused now$/$skdkdjs

Wait

[grand pond]

Yuko

[flat grove]

Huh

[last slate]

Yeah

[last slate]

You didnt understand this?

Or that equation i wrote?

[flat grove]

The equation you wrote

[last slate]

Yeah so

You know whats Pythagoras Theorem right?

[flat grove]

Yea

[last slate]

You see the triangle made by the purple and green lines

[flat grove]

Yea-

[last slate]

Apply the theorem THERE

[flat grove]

Oh

[last slate]

What do you get?

[flat grove]

Wait

[last slate]

(R+1)² + (R+1)² = (2R)²

??

[flat grove]

Yes so

[last slate]

Find, R

[last slate]

2R² + 2 + 4R = 4R²

2R² - 4R - 2 = 0

R² - 2R - 1 = 0

(R-1)² -2 = 0

(R-1) = √2

R = 1 + √2

@flat grove

[flat grove]

I'm here

[last slate]

Ye

[flat grove]

You're just really fast

I'm tryna do it on paper

[last slate]

Yeah np

Youll get the same result

Youre welcome

Ping me if needed

[flat grove]

Thanks-

[last slate]

You understood it all?

say the truth

[flat grove]

Not sure I got this

[last slate]

Yeah

[flat grove]

Ok i got iy

It

[last slate]

Yeah

[flat grove]

Damn thank you

[last slate]

I just added and subtracted 2 to make it a perfect square

[flat grove]

Yeah

[last slate]

You could have used the quadratic formula too

[last slate]

Welcome

[midnight plank]

@flat grove Has your question been resolved?

[midnight plank]

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[last slate]

can someone give me an example where acceleration is positive and velocity is negative

[surreal moon]

consider a(t)=1

what is v(t)?

[last slate]

im not sure are u supposed to integrate?

[surreal moon]

yes

[last slate]

t+C?

or smth

[ebon parrot]

@last slate derivative of position is velocity derivative of velcoity is acceleration

[last slate]

yes i know that

[ebon parrot]

yeah so now you do the opposite u start with a(t) = 1

get v(t) = x+c right

[last slate]

yes

[surreal moon]

yes

So ask yourself, what would v(0) be?

[last slate]

it would be C

[surreal moon]

so how can you make v(0)<0?

[last slate]

if C is < 0

[surreal moon]

there you go

[last slate]

can u give me a irl situation?

[surreal moon]

v<0 requires a sense of direction

practically, imagine your car driving in reverse and you apply the brakes

[ebon parrot]

your car is going -30 m/s from your house you apply a force in reverse towards your house even though ur still going -30m/s at +5m/s

[surreal moon]

negative velocity, positive acceleration

[last slate]

oh i see

and the reverse would be like

[ebon parrot]

going towards ur house

[last slate]

driving a car to your house, but slowing down

?

[ebon parrot]

yeah sure if your using the store as a relative since - velocity would require opposite direction

[last slate]

i see

@ebon parrot is accleration just speeding up towards the positive direction?

or am i understanding wrong

[surreal moon]

acceleration is just how velocity is changing with time.

That's the most general, most accurate way to think of it.

likewise, velocity is just how position is changing with time.

[last slate]

ok i see

ty

.close

[midnight plank]

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[midnight plank]

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[keen zephyr]

I'm struggling to find the limit for the following equation

[keen zephyr]

I want to say -1/20

[rose kayak]

can you show your work?

[keen zephyr]

I essentially took the root in the denominator and put it outside the limit but I think that is wrong since the numerator isn't a root either

My algebra isn't the best so that root in the denominator is really holding me up

cause I also want to just square both the denominator and numerator to get rid of the root and simplify the x^2 to 1 which is -1/15 but that was wrong as well

[rose kayak]

so with this one, I don't really see a way to simplify it. If we do an incremental check we'll see that diverges negatively

Let $f(x) = \frac{x}{\sqrt{x^2-16}}.$ Then:
$\\ f(-4.1) = -4.5555555556$
$\\ f(-4.01) = -14.1686383294$
$\\ f(-4.001) = -44.7297442682$
$\\ f(-4.0001) = -141.424007874$
$\\$
And so on and so forth. As we can see it's diverging the closer we get to $-4$

[grand pond]

MellowDramaLlama

[rose kayak]

does that make sense?

[keen zephyr]

Then the limit does not exist right

[rose kayak]

the limit itself does not exist correct

but the left handed limit and the right handed limit exists

but they both diverge

since x = -4 is an asymptote for the function

[keen zephyr]

That was it! I had put - infinity earlier because I saw the asymptote but it should have been DNE

Thank you!

[rose kayak]

wait no it should be -infinity lol

it's asking for hte left hand limit specifically

[keen zephyr]

Thats what I thought

but I put - infinity and it was marked wrong

I just put DNE and it was right

so IDK haha

[rose kayak]

yeah they're trippin'

the LHL absolutely exists and it's -inf

[keen zephyr]

haha at least we on the same page

[rose kayak]

yep lmao

if it was -4^+ then DNE would be valid 🙂

[keen zephyr]

Well, I appreciate the help anyways! At least we got the right answer on the assignment but you and I know the real answer haha.

[rose kayak]

yep for sure

best of luck!

[keen zephyr]

Have a good one!

.close

[midnight plank]

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[midnight plank]

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[harsh egret]

if I have the limit as x approaches 1 from the right side of f(x+1), does that really just mean I'll look at the limit as x approches 2 from the right of f(x), or does it mean I'll take the limit as x approches 1 of f(x) and add one?

[drifting umbra]

This is algebra correct?

[harsh egret]

its AP Calc AB

I can send a pic

one sec

[drifting umbra]

Ok

[drifting umbra]

Nono

Someone else is going to have to help you. I thought it was algebra and I was going to be able to help you with that

I'm close not there. I'm so sorry I shouldn't have come here if I didn't know

Excuse myself. So sorry

But that does seem right

[harsh egret]

yep thats right

.close

[midnight plank]

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[white flicker]

Help

[midnight plank]

Please don't occupy multiple help channels.

[fallow scarab]

!onechannel

[midnight plank]

Please stick to your channel.

[fallow scarab]

.close

[midnight plank]

Channel closed

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[tidal forge]

need help

[fallow scarab]

Factor the numerator

[tidal forge]

isnt just doing the limit so -18?

i need to find f(x) and c

[tidal forge]

would f(x) be -18?

.close

[midnight plank]

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[weary crescent]

mvt not apply bc although continuous on interval -3 to 7, it's not differentiable right?

my brain doesn't rly make sense of the answers tho

like if k the value of tangent line parallel to f'(c)?

[main current]

Why close and open again lol

You're correct, mvt applies if the function is on a closed interval and is differentiable on that interval

.
We know by the question that the function is on a closed interval. We know by your logic mvt doesn't apply. Therefore

[weary crescent]

i half understood it so i closed since no one was there

i need help with this one

[main current]

Did you get the other one lol

[weary crescent]

only time A'(t) = 0 is between 4 and 5

yeah

[main current]

There's also an A'(t) = 0 between t = 0 and t = 2

Any others?

[midnight plank]

@weary crescent Has your question been resolved?

[weary crescent]

and between 7 and 8

[midnight plank]

@weary crescent Has your question been resolved?

[midnight plank]

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[spiral tangle]

period of sin2x and |sin2x|

[midnight plank]

Please don't occupy multiple help channels.

[spiral tangle]

sin2x period is pi

[lyric charm]

pi, not pie.

[spiral tangle]

yes

now

[spiral tangle]

Any idea to get it without desmos?

|2sinx cosx|

2|sinx||cos|

Now i got it why it happens

Thanks ann
Am I right?

[midnight plank]

@spiral tangle Has your question been resolved?

[midnight plank]

@spiral tangle Has your question been resolved?

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[cold jolt]

I know for sure that this is the “right” answer but my intuition tells me it’s not right

[cold jolt]

I think the answer would actually be an association because knowing with response the people gave helps us predict what treatment they received

or are those the same thing

Oh would that be the same thing

I guess whichever treatment they gave you would also help you predict their response… but

Or like help you “simulate a prediction” since more people said yes for smashed into

But idk can someone help or clarify?

<@&286206848099549185> Pleasee don’t let me go to the bottom again

[verbal anvil]

Sorry, I’d help if I knew how

[cold jolt]

👍

[lament linden]

Hmmm,my math skill isn't that good enough to help you out

[cold jolt]

Bro there’s no way you’re an undergraduate and don’t know basic stats 😭

I don’t even think you need to know anything

[lament linden]

😋

Good luck bro,Wish people could help you out cuz im busy right now

[cold jolt]

Yeah no worries

I know exactly what’s going to happen

I’m going to slowly go to the bottom of the list

No one is going to help me

I’m going to sleep

The channel is going to expire

And the next person who types here is immediately going to be helped

[random vortex]

hi im next person

just kidding, lemme look over your thing

@cold jolt definitely 4th choice, trust me

[cold jolt]

explain

Well

Out of all of them that makes the most sense

But i still think it’s a bit off

[random vortex]

its really not

i could explain but i gtg to school in 10 mins sooo

basically "smashed into", "hit" and control treatments reveal how they got hurt by the glass so its easier to get a better understanding of their situation before they got hurt

thats all bye

[cold jolt]

????

Damn

I’m struggling to understand how if you know that you would know to predict yes or no

z cuz wouldn’t you always predict no?

I feel like if you knew if their responde was yes or no, then you can predict the treatment better

.help

[midnight plank]

Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[midnight plank]

@cold jolt Has your question been resolved?

[midnight plank]

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[cedar oxide]

how would I go about solving this?

do I need to use the equation of the tangent line?

[red carbon]

do you know integration or derivatives yet

[cedar oxide]

derivatives

im not allowed to use integration

[red carbon]

a general form you can use for quadratics is $$f(x) = a(x - h)^2 + k$$

[grand pond]

o.O

[cedar oxide]

so id have a(x-5)^2+10

?

[red carbon]

potentially

you have the restriction $f'(5) = 8$

[grand pond]

o.O

[cedar oxide]

what do I need to do with the given restriction? Im trying to think of a way but I honestly have no idea

[verbal anvil]

use the fact that the slope of the tangent line is the derivative

and that the y intercept is the c in ax^2 + bx + c

[cedar oxide]

i got y=2x^2-12x+10

wait can I do the thing $y=2x^2-12x+10$

[grand pond]

Ogre of Smokies

[cedar oxide]

.close

[verbal anvil]

Do what thing

[midnight plank]

Channel closed

Closed by @cedar oxide

Use .reopen if this was a mistake.

[cedar oxide]

.reopen

[midnight plank]

✅

[cedar oxide]

the $$ thing

.close

[midnight plank]

Channel closed

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[verbal anvil]

Yes, you can learn how to do that in #latex-help or #latex-testing

[midnight plank]

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[keen zephyr]

I got -2/7

but not an answer so am I trippin

[nova yoke]

can you show your work?

[keen zephyr]

Top simplifies into -2(x-4) and the absolute value foils into (x-4)(x+3) cancal out the x-4 in numerator and denominator and left with -2/ absolute value of (x+3) fill in 4 for x and get -2/7

hopefully that makes sense haha

[verbal anvil]

there is an absolute value

can you cancel out immediately?

oh, x - 4 is positive

I guess you can

[keen zephyr]

right since it approaches from the right

rti you helped on the last one but the answer wasn't there either so I wonder if we are missing something

Sorta odd there would be 2 without right answers

This has me feeling I'm going crazy haha

[verbal anvil]

maybe they made a typo

[verbal anvil]

That appears to be the case here

[verbal anvil]

you could ask the person who assigned you these problems

I plotted on desmos just to be sure, but it agrees with us

[keen zephyr]

Craziness haha. This was supposed to be a take-home quiz haha so open notes but idk if he would answer specifics off it, but I can try.

Shouldn't have gone to community college haha

Theres no way it can be infinity right

because there is an asymtope and it is not continuous

[verbal anvil]

It’s not infinity

the answer choices are all wrong

[verbal anvil]

answer specifics? The answer choices are wrong

and you can show your calculations as proof

[keen zephyr]

I'll give it a go. Thank you!

.close

[midnight plank]

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[spiral tangle]

Period of |sinx|

[verbal anvil]

try graphing it

[spiral tangle]

@ebon sable

sinx period 2π

[terse spear]

Don't ping specific people

[verbal anvil]

this is a correct statement

|sin x| is different

[spiral tangle]

There is a reason so don't tell me what to do not

Ohh I got it now. Yesterday we talked about it

|sinx| will be π

[terse spear]

Okay and? Still don't ping specific people

It's rude

[spiral tangle]

Okay fine. Not going to ping again

So yesterday they said we should do LCM/2

But here it's only one function so it will be π

[prisma idol]

isn't it pi/2?

because the bottom part is reflected

[spiral tangle]

No

[verbal anvil]

No

[verbal anvil]

Yes, so pi

[spiral tangle]

If it was more than one function

[prisma idol]

oh

yeah sorry

[spiral tangle]

Then it can be π/2

[prisma idol]

pi

[spiral tangle]

Like |sinx|+|cosx|

It will be π/2

[wary thorn]

@spiral tangle

[dreamy thicket]

New @spiral tangle post just dropped

[spiral tangle]

.close

[midnight plank]

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[daring hull]

how can i break sin3x ?

[daring hull]

gotta prove this

[wary thorn]

3x=2x+x

[verbal anvil]

you could have started with sin3x and sin x first instead of sin2x and sinx

[daring hull]

but that would not give fractions

rhs has fractions

[verbal anvil]

oh

[verbal anvil]

then use 3x = 2x + x

[soft token]

Using double angle formula on sin(3x) should work.

[verbal anvil]

or compound angle formula

[daring hull]

cant really take commons

after using the formula

[verbal anvil]

can you show your work

[daring hull]

.rotate

,rotate

[daring hull]

i think 3/2 + 3/2 will work

[verbal anvil]

try that

[daring hull]

🙏

that works

thank tou everyone

you*

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[timid tangle]

Does this mean arcsin or 1/sin?

How do you know?

[potent veldt]

1/sin of 1/2 sounds impossible without a calculator

Also by convention it means arcsin way more often than 1/sin

Like I've just never seen it mean 1/sin

[timid tangle]

Oh okay

Thanks guys

.close

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[olive matrix]

if you want 1/sin just write csc

[midnight plank]

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[bitter tartan]

i need help

[fossil knot]

You can eliminate the first choice because it's an even function

Now think about the differences between the 2nd and 3rd choices

||The 2nd choice is always positive, but what happens when a<x<b?||

[bitter tartan]

b is negative and a is postive?

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[fleet skiff]

<@&286206848099549185>

please help

[midnight plank]

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[open cloud]

Let $$A=\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}}+\sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \cdots + \sqrt{1 + \frac{1}{2019^2} + \frac{1}{2020^2}}$$ Then we can write $A/2019$ as a fraction $pq$ , where $p$ and $q$ are integers, in lowest terms. Find $p+q$.

[grand pond]

donut123

[open cloud]

I know this is a probably some type of telescoping series

but i dont know how

I tried to write it in terms of n and simplify, and got something, but i don't think my factoring was correct

If someone could give me some clues or guidance for this that would be appreciated

[midnight plank]

@open cloud Has your question been resolved?

[midnight plank]

@open cloud Has your question been resolved?

[last slate]

hello

[open cloud]

hi

sry i was away for a couple minutes

<@&286206848099549185>

Let $$A=\sqrt{1 + \frac{1}{1^2} + \frac{1}{2^2}}+\sqrt{1 + \frac{1}{2^2} + \frac{1}{3^2}} + \cdots + \sqrt{1 + \frac{1}{2019^2} + \frac{1}{2020^2}}$$ Then we can write $A/2019$ as a fraction $pq$ , where $p$ and $q$ are integers, in lowest terms. Find $p+q$.

[grand pond]

donut123

[midnight plank]

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[midnight plank]

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[fathom summit]

Hello

[last slate]

Hello

[dull yoke]

@fathom summit

Hello

@last slate

Hello

[midnight plank]

@fathom summit Has your question been resolved?

[fathom summit]

This

<@&286206848099549185>

<@&286206848099549185>

[lyric sundial]

!15m

[midnight plank]

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

[midnight plank]

@fathom summit Has your question been resolved?

[fathom summit]

<@&286206848099549185>

[thin badger]

pls
we can't read

[fathom summit]

Here @thin badger

<@&286206848099549185>

[thin badger]

use the epsilon delta definition of the limit to find delta

For any eps > 0, we can always find a delta > 0 such that
$$ f(B_\delta (x)) \subseteq B_{f(x)} (\epsilon) $$

[grand pond]

Secret

[fathom summit]

The thing is I don't understand what the epsilon is

[thin badger]

they gave $ \epsilon = 0.001 $

Here f(x) = 5x - 9

[fathom summit]

Can you walk me through its step by step the teacher didn't bother to explain it in class he was like you can figure it on your own 😅😂

@thin badger

[thin badger]

aight so
Now u have
$$ \abs { f(x) - 1 } < \epsilon $$

[grand pond]

Secret

[fathom summit]

Ok

[thin badger]

cuz f(x_0) here is 1

anyway

u get
$$ \abs{5x - 10} < \epsilon \implies \abs{x - 2} < \frac{\epsilon}{5} $$

[grand pond]

Secret

[thin badger]

more like

$$ \abs{5x - 10} < \epsilon \iff \abs{x - 2} < \frac{\epsilon}{5} $$****

[grand pond]

Secret

[fathom summit]

Damn I didn't even know you can do that 😭

[thin badger]

So what u can claim is
if $ \abs{x - 2} < \frac{\epsilon}{5} $, then definitely $ \abs{f(x) - 1} < \epsilon $ holds

so our $\delta$ of interest is now
$\frac{\epsilon}{5}$

if $\abs{x - 2} < \frac{\epsilon}{5} , then definitely $\abs{f(x) - 1} < \epsilon$ holds

if $\abs{x - 2} < \frac{\epsilon}{5}$ , then definitely $\abs{f(x) - 1} < \epsilon$ holds

[grand pond]

Secret

[thin badger]

so our $\delta$ of interest is now
$\frac{\epsilon}{5}$

[grand pond]

Secret

[fathom summit]

So he don't need a number

[thin badger]

yeah

[fathom summit]

Oh wait can't we just plug in our 0.001

[thin badger]

now substitute $\epsilon = 0.001$, we get $\delta$

[grand pond]

Secret

[thin badger]

correct

[fathom summit]

Thank you soo much

❤️❤️❤️

[thin badger]

no probs

[midnight plank]

@fathom summit Has your question been resolved?

[midnight plank]

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[light rain]

does someone know how I would use the squeeze method to solve this problem

[midnight plank]

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[jaunty hatch]

how my channel got closed

[jaunty hatch]

burh

@stable halo here u can say please

what were u saying

ah he is gone

how do i repost my question

[stable halo]

send rthe question again

just copy pasta

[stable halo]

ok

so 12 volts across the first line

and 12 volts across the second line

[jaunty hatch]

yes

[stable halo]

now whats the resistance in the first line

[jaunty hatch]

18+6

24 ohms

[stable halo]

Yea

but you need to find voltage across resistor 2

so fuck that line

we go 2nd line

We have 8 and 12 in PARALLEL

so whats rthe total resistance of thart part

[jaunty hatch]

in second line total resistance is 6.8ohms

[stable halo]

Yea

so now

[jaunty hatch]

1/8+1/12= 4.8
4.8+2

=6.8ohms

[stable halo]

2/6.8 * 12

[jaunty hatch]

woah

why that

where th does that 12 come from

🤨

[stable halo]

12 volts al togeather

[jaunty hatch]

bruh

[stable halo]

SO back to the first line

[jaunty hatch]

wth is this explain 😭

ok back to first line

[stable halo]

we have 12 volts going across the line

[jaunty hatch]

yep

[stable halo]

But we have 18 ohm and 6 ohm resistor

[jaunty hatch]

yes

[stable halo]

So the voltage acorss the 18 ohm resistor is 18/(18+6) * 12

and th e6 ohm ressitor

6/(18+6) * 12

[jaunty hatch]

what is this formula

u are dividing voltage current resistance

😭

everything?

[stable halo]

No

Wait

[jaunty hatch]

okok

[stable halo]

So like

If that was One BIG ressitor

it would be 24 ohms

[jaunty hatch]

yessir

[stable halo]

with 12 volts across it

[jaunty hatch]

sum of both resistors yes

[stable halo]

but if it was 2 6 ohms resistors

[jaunty hatch]

its also 12v across

[stable halo]

HM i think i have a better idea

bare with me

[jaunty hatch]

☺️

i hope the bot doesnt close the channel again

im here bot

dont close the channel

[stable halo]

Just type the X when it asked if the questions are u finished or something

[jaunty hatch]

X

it also asks to calculate current through 12ohms resistor

current across series is same, but in parallel its different

X

[stable halo]

Yea

should be fine

do you know how to work out current?

[jaunty hatch]

its 12v across 2ohms?

[jaunty hatch]

yes but its bit complicated, i will try

[stable halo]

YOu see

its just a bit confusing

becuase there are kind of 3 branches coming off the voltage source

not 2

thats where its confusing

[jaunty hatch]

wdym

so its not 12v?

circuits have always been confusing for me 😭😭

bruh u been typing since ages but u dont send anything? 💀

[stable halo]

Bro

idk anymore man

Like

i think its 6.8 ohms togeather for top half so we do 2/6.8 * 12

[jaunty hatch]

😭

[stable halo]

But i than simulated it

and it was wrong

So now im using a different simulator

[jaunty hatch]

😭

wdym

im so confused, im just gonna write 12v

in answer

[stable halo]

https://gyazo.com/6c36d6a906a87543378599eab164feab

You see the top and bottom

theyre not the same

[jaunty hatch]

yes

what is this u are using

anyway

[stable halo]

just whiteboard

Well im trying to find a good one

[jaunty hatch]

i think the answer is 12v across all resistors

so im gonna write it 😭

[stable halo]

no

no

I GOT IT

[jaunty hatch]

is it not 12v?

[stable halo]

https://gyazo.com/c21eea4f04b24366bcdf43eee39990c9

[jaunty hatch]

woah how 3.53v?

[stable halo]

2/6.8 * 12

like i said

Its just confusing

[jaunty hatch]

2/6.8 *12?

but what is this formula

like how did u derive it

i never seen this

[stable halo]

voltage in series

[jaunty hatch]

how did u do it from this website?

[stable halo]

https://phet.colorado.edu/sims/html/circuit-construction-kit-dc/latest/circuit-construction-kit-dc_en.html

its nice and simple easy to use

[jaunty hatch]

wowwww

can i also find out current with this and then check my answer?

[stable halo]

Yeap

[jaunty hatch]

wow lemme try

thanks so much man

this is gem

how did u edit the voltage

because its only showing 100v or above

[stable halo]

CLick on the battery

and at the bottom of your screen should be a scale

[jaunty hatch]

ok

scale minimum is 100v?

[stable halo]

ahh

your using the big battery

on the item thingy

go up one page

and click the battery with the copper tip

https://gyazo.com/54c35bcfcf73ed9036f5739ec2e1633d

[jaunty hatch]

oh ok

thanks

[midnight plank]

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[ornate steeple]

can somepne please walk me through the potential errors i made in my integration, i got an answer but it not on the multiple choice

[velvet oar]

looks right to me

perhaps your answer is equivalent to one of those choices, it might just look different

[ornate steeple]

i was also thinking of that

but i cant see any way to get to any of those answers

[velvet oar]

well the most noticable difference between your answer and any of the choices is the (x+1)^(3/2)

try factoring out (x+1)^(1/2)

[ornate steeple]

okok lemme see

still nothing promising

[velvet oar]

🤔

can you show your work?

one of those is equivalent to what you wrote

[ornate steeple]

yea lemme show you

thats what i got

[velvet oar]

looks good, keep simplifying inside those brackets

(x+1)/3 - 1, I mean

[ornate steeple]

ooooooooo

looks like it might be C

only one with a minus 2

let me finish simplifying

[velvet oar]

👍

[ornate steeple]

🙏

incredibly

it is C

thank you

[velvet oar]

no problem 👍 good work

[ornate steeple]

.close

[midnight plank]

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[midnight plank]

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[lost hamlet]

Hi

[lost hamlet]

Is anyone there?

[chilly sigil]

Wassup

[lost hamlet]

I need help with this constrained optimisation question

Basically I’m trying to optimise the surface area of a frustum connected to a cylinder. Radius of cylinder = larger radius of frustum = 4.14cm.

[midnight plank]

@lost hamlet Has your question been resolved?

[lost hamlet]

Hi, are you still there?

Are you able to answer my question or no

@gray widget

<@&286206848099549185>

.close

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[tawdry fox]

i have already asked about this, but i did some more googling and i dont seem to understand this still

[tawdry fox]

how do i find the asymptote

clearly there are two, i checked in desmos. but i dont understand how do i find those...

like, x^4+4 cant be equal 0. so what do i do

[visual pier]

Where are the asymptote?

[tawdry fox]

oh damn

i used desmos incorrectly lmao. okay my bad thanks

put the wrong function there

[visual pier]

👍

[tawdry fox]

.close

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[waxen island]

How can I do it algebraically?

[ancient warren]

differntiate

i guess

[dense idol]

literally do it

[waxen island]

how does finding the derivative help

[ancient warren]

cos g prime x is the deriviative of gx ???

[waxen island]

hold me shoes

thanks, i thought it meant inverse function since all the others were inverse

.close

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[floral thorn]

Reference ans:

What I got instead:
,$$ W_{2x} : cos(140)=\frac{W_{2x}}{30} $$
$$ W_{2y} : sin(140)=\frac{W_{2y}}{30} $$
$$ W_{2x} = -22.98 $$
$$ W_{2y} = 19.28 $$

[grand pond]

Kai Funaba

[midnight plank]

@floral thorn Has your question been resolved?

[midnight plank]

@floral thorn Has your question been resolved?

[midnight plank]

@floral thorn Has your question been resolved?

[floral thorn]

.close

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[haughty pewter]

Calculus I
I am so lost

[faint cairn]

Does the tangent line you gave intersect f(x) + g(x) at x = 3?

[haughty pewter]

All of the information I am given is in the picture

My understanding is that the derivative of f(3) is y=3x+5?

I think I'm just having trouble transforming the language into something I understand better

[faint cairn]

Do you know how to find the tangent to a curve?

[haughty pewter]

I guess not off the top of my head

[faint cairn]

The tangent to a curve y = f(x) at x = a is the line (y - f(a)) = f'(a)(x - a).

In other words, it is the unique line y = mx + b such that the line intersects the curve at x = a and the line + curve have the same slope at that point.

[haughty pewter]

So the equation given to me is the result of (y - f(a)) = f'(a)(x - a)

and a=3?

[faint cairn]

Yes. The first tangent is for f and the second is for g.

You need to use that information to figure out how to find the tangents for linear combinations of f and g.

[haughty pewter]

so a. y=f(x) +g(x) is just telling me to do this: y= 3x+5 +2x - 2?

[faint cairn]

No. It's telling you to find the slope of f(x) + g(x) and then the point the tangent line intersects at x = 3.

You cannot simply add lines like that.

[haughty pewter]

Okay I am confused because the answer that it gave me to question a. is the result of what I asked above.

[faint cairn]

Ah sorry I made an addition error before

Yeah that's right

You generally can't multiply tangents but I guess adding them is fine

[haughty pewter]

So that tells me that question b. would be
b. f(x)-3(g(x)) = 3x+5 - (3(2x-2))?

[faint cairn]

Yeah

[haughty pewter]

-3x+11

[faint cairn]

Yup

[haughty pewter]

was incorrect

https://tenor.com/view/ashita-no-joe-anger-rage-anime-gif-20587055

[faint cairn]

Pretty sure those are right. If the program says it's wrong, then I might be making a stupid mistake (which hopefully someone else will catch). However, I think those should be right.

[haughty pewter]

the answer for a. was given to me after I got it wrong. So 5x+3 is correct for a.

Okay -3x+11 worked when I typed it again

not sure what I did

maybe I had a space or something

[faint cairn]

Yeah it was probably something like that. When I take calculus in college our grading program also had those types of issues.

[haughty pewter]

Got this the other day

because of a space

[faint cairn]

[haughty pewter]

Okay I redid it with a new question and got em right. Thanks for the help!

.close

[midnight plank]

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[midnight plank]

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[solemn prism]

The 4 x 4 square shown below on the left must be divided into 8 rectangles. The rectangles must have their sides on the lines of the grid and also their vertical sides must have measure 1. The two figures on the right show 2 possibilities. How many possibilities are there in total?

[midnight plank]

@solemn prism Has your question been resolved?

[solemn prism]

<@&286206848099549185>

[wintry bane]

Try assuming the number of rectangles in each of the 4 rows as a,b,c,d

[solemn prism]

ok

[wintry bane]

You have a+b+c+d=8

[solemn prism]

oh

Why 8?

[wintry bane]

Because there are 8 rectangles in total

[solemn prism]

Oh ok

I thought you were saying other thing

ok

[wintry bane]

In first figure, a=2 b=1 c=4 d=1 for example

[wintry bane]

This is the first constraint for a,b,c,d

[solemn prism]

Sorry for asking, but a, b, c and d, what are they?

[wintry bane]

It's oki, don't be sorry :p
a,b,c,d are the number of rectangles in the 4 horizontal stacked rows of the square

For example look at the last figure

[solemn prism]

Ohh

Ok so

[wintry bane]

The topmost row has 1 rectangle

So a=1

Then the one below it has 2 rectangles

b=2

Same for the 3rd row, c=2

And the bottommost has 3 rectangles (6,7 and 8)

So d=3

Do you get what a,b,c and d are now?

[solemn prism]

Yeah

[wintry bane]

Since their vertical dimension is always 1, they are always going to be divided into 4 separate rows

And their addition has to be 8 cos it's given in question

That's the first constraint

Can you think of any other constraint about a,b,c,d?

Or can they take any value?

[solemn prism]

The maximum value they can take is 4?

[wintry bane]

How come

Why not more than that

[solemn prism]

Oh wait

It can technically be more than 4, it will look weird

But yeah

ok damn

[wintry bane]

It can be 1,1,1,5 no?

Oh wait

It's a 4x4 grid

[solemn prism]

yes

[wintry bane]

Ok right

Any other constraint?

[solemn prism]

No that I can think, no

[wintry bane]

Any idea about their minimum value?

[solemn prism]

1

[wintry bane]

Right!

So we have a+b+c+d = 8 where 1 ≤ a,b,c,d ≤ 4

[solemn prism]

So

The maximum is 4?

[wintry bane]

Yes

Cos it's saying the rectangles can only have their sides along the grid lines

[solemn prism]

oh ok

[wintry bane]

So if you want to put 5 squares in one row then you will have to compromise that condition

I suppose this question is related to combinatorics

[solemn prism]

Yes, in fact

[solemn prism]

But ok

[wintry bane]

So do you know anything like Beggar's formula?

[solemn prism]

Not very much

I know it

But I cant apply it

[wintry bane]

Ok do you know the formula?

[solemn prism]

n!/(n-r)!

?

It was something like that

[wintry bane]

Hm no not that one

That's nPr

[solemn prism]

n! / (p! q! r!)

?

If not, I dont know

[wintry bane]

Yeah not that one

Hmm

Does this sound familiar? (n+r-1)! / n! (r-1)!

[solemn prism]

oh

How do I apply that in here?

[wintry bane]

So we had a+b+c+d = 8 where all of them were greater than 1 right?

[solemn prism]

Yep

[wintry bane]

The formula for distributing n things among r variables (where each of them starts from 0) is this

So if we were asked to find the number in ways in which we can distribute a,b,c,d in a+b+c+d = 8 (where all of them were starting from 0)

Then it would be (8+4-1)! / 8! (4-1)!

Since n=8 and r=4 (a,b,c,d)

BUT we cant simply use that here since our a,b,c,d are starting from 1 AND NOT 0

[solemn prism]

Ok

[wintry bane]

So to work around that, we can use new variables

Like p=1+a, q=1+b, r=1+c, s=1+d

So a=p-1, b=q-1, c=r-1, d=s-1

Putting these in a+b+c+d=8

It becomes p+q+r+s = 4

Where p,q,r,s can all vary from 0 to 3

[wintry bane]

Remember a,b,c,d could vary from 1 to 4

So p,q,r,s can vary from 0 to 3 since they're 1 less than the original variables

[wintry bane]

Now we can apply the formula to find all the integer sets here

Here n=4 and r=4

[wintry bane]

Using this, we find it to be (4+4-1)! / 4! (4-1)! = 7! / 4! 3!

So p,q,r,s can take 35 values

[wintry bane]

To meet this condition

But the thing is

It also includes the cases like (4,0,0,0) etc