help-27

[restive river]

If I take the general equation

And make equation with kx. And treat that like phase difference.

Would that work

But that’s not constant is it?

[fallen condor]

yeah, I thought you do gram-schmidt to find the orthonormal basis, but the question said "use QR factorization or gram schmidt_ to find the orthonormal basis" and I did a double take cuz I didn't know you could use QR factorizatoin to find orthonormal basis

ALSOS GUESS WHAT

95 on midterm

🤩

I've been watching the mit lin alg lectures to supplement the book to supplement the community college lin alg class and I think I have a pretty decent understanding of shit

I love Gilbert Strang soooo much 🥹

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[cerulean quarry]

How do I do q6

1/2mu^2 = 1/2mv^2 + mgh
v^2 = u^2 - 2gh
T-0.4g = mv^2/r

Plug v^2 into the T equation but then what do I do

[hushed lance]

I think T-0.4g only works at the instant it receives the impulse

[cerulean quarry]

Oh wait I used the wrong thing

The acc equation I used is

T = mgcos(theta) + (mv^2)/r

[hushed lance]

what does it want you to describe actually

[cerulean quarry]

Motion of the bob

By considering the tension

[hushed lance]

well there's a lot of ways to describe that

I think it just wants you to calculate where the pendulum stops

Because otherwise you get into functions which are a lot more complicated

[cerulean quarry]

What do I do then

[hushed lance]

well you already wrote the energy conservation

just plug in v=0 to see what height it reaches

[cerulean quarry]

h = (u^2)/2g

Wait

I alr have a height tho

Is 0.3(1-cos(theta)) not the height

[hushed lance]

the problem is cos theta varies along the motion

[cerulean quarry]

Yh

[cerulean quarry]

So what do I do with this

[hushed lance]

the question gives you values of u² I think

so just plug them in to solve for the height in each case

[cerulean quarry]

Yh that’s what I thought

But this doesn’t work out

[hushed lance]

oh I see

[cerulean quarry]

Cause in the ms it’s like

When the string makes an angle of 83.6 with the downward vertical, tension will be lost and the bob will fall freely under gravity

[hushed lance]

ok I'm sorry I actually don't know how you would do this without vectors

[cerulean quarry]

Np ty tho

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[severe condor]

Can i get help w this

[severe condor]

The equation

It might seem easy

But it needs quadratics

[frozen aurora]

,rccw

[frozen aurora]

(2x+1)^2 + x^2 = -4 ?

what are you supposed to do, solve for x?

[severe condor]

There are 2 ways

To represent 1 numer as 2 like 5x to represent like 7x - 2x

Or what i did is quadratics

U here?

<@&286206848099549185>

This is what i got

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[twilit comet]

please don't make several help channels for one question.

your occupied channel is #help-10

[timber sorrel]

oh sorry plus it isnt there

[twilit comet]

please close one of your channels.

[twilit comet]

like i said: #help-10

[twilit comet]

click on the right-facing arrow at the top, or you can click on the blue link-thing i sent

[timber sorrel]

ok

should i delte this?

[twilit comet]

no, i'll do it.
you got the original channel, right?

[timber sorrel]

yes plus i need help so see me at my og channel

[twilit comet]

okay

.close

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[mortal ferry]

I corrected the equation, I think. Is it?

[mortal ferry]

The argument is that, according to the original equation, if n=1, then it would get " 1 < 0 "

Therefore, I corrected the equation which is pointed by the arrow. The others are fine, I think.

[prime hollow]

Notice that the sum starts at r=2

[mortal ferry]

Yes

[prime hollow]

So it's an empty sum

[mortal ferry]

Please elaborate

[prime hollow]

If n=1, the you sum from r=2 to r=1

Which is nothing

[mortal ferry]

Shouldn't the "n" above the summation notation be the "number of squares"?

[prime hollow]

The n is the last value of r. Wjhen you write $\sum_{r=a}^{b}$, then it means "compute the summand with r=a, a+1, ... b-1, b and sum the values up."

[woven radish]

Ivar Ängquist

[mortal ferry]

Oh

[prime hollow]

So n=1 doesn't really work. If anything, it would be zero

[mortal ferry]

Thank you

[prime hollow]

No problem. Understandable confusion.

[mortal ferry]

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[restive river]

(Stuff in blue ink) I’m kinda struggling at using trig identities to verify equations 😓 I’m assuming there has to be more simplification since sin^2-cos^2 seems too close to an identity to not be something. Is this an accurate answer and if not how can I work on trig identities?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[bronze badger]

what is exactly your problem

[fervent sundial]

U don't really need an identity for sin^2 (x) - cos^2 (x) in this problem

but cos(2x) = cos^2 (x) - sin^2(x) so -cos(2x) gives sin^2 (x) - cos^2 (x)

(if you do really need it for some reason)

[restive river]

I see

so is what I got fine @vestal pebble?

been hung up on it lol

[stark anvil]

Looks a lot messy

[restive river]

my bad

I'm gonna try and redo it

[stark anvil]

Just try to rewrite the question and send it again if you can

Just the question

Is this like a prove lhs = rhs question?

[obtuse chasm]

probably

[restive river]

The problem says “For the following, verify that each equation is an identity”, so I’m thinking it’s simplifying?

[obtuse chasm]

@restive river Just convert to Sin and Cos and try it again man

[restive river]

I will thank you

[obtuse chasm]

👍 👍 ATB

[restive river]

Hahaha no idea what I was trying right now

[stark anvil]

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[obtuse chasm]

that's scary damn

[obtuse chasm]

Ohh mb, just joined.

[restive river]

It’s not hw thankfully just review problems for our test

Got covid and missed one week of review lectures lmao

[obtuse chasm]

Damn, take care man.

[stark anvil]

You understood the process though right?

[obtuse chasm]

try to write more neat ig so u understand stuff better

and um write a variable or something after the trignometric function- like Sin'x' or Sin 'theta'

[restive river]

Yeah

honestly the way you solved it is awesome, I'm going to try and approach similar problems in a similar way going forward

[obtuse chasm]

hehehe tyty

in 10th trigno was my forte hehe

[restive river]

I used to be good at these problems but never practiced 🥲

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[rose flower]

Hello

[rose flower]

Please explain this part

How did they know p(z greater equal 1) is 0m8413 and same for

The other one

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[high fossil]

Can anyone tell me where's the mistake?

The answer is - infinity and i get 1/4 😭

[stone stump]

what happened to the -4x in the denominator

[high fossil]

oops

Okay, I am blind

sorry 😅

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[valid knot]

Let f(x) = 3x - 4 and g(x) = 5x + c. Find c if (f g)(x) = (g f)(x) for all x.

[valid knot]

I have (fg)(x)=15x^2+3xc-20x-4c right now

[proud bridge]

wait why there is x squared

[valid knot]

in this?

[proud bridge]

yes

you multiply by 3 there

[valid knot]

because when you multiply functions you multiply the terms by eachother right

3x*5x=15x^2

[proud bridge]

oh

okay bro

i misunderstood

like (fog)(x)

yk

[hushed lance]

no I think you're right

[proud bridge]

(f×g)(x) he said

[valid knot]

yea

thats what i meant

[hushed lance]

fg is another notation for fog

[valid knot]

i didn't mean fog

i meant multiplying functions

[proud bridge]

okay maybe

but

[valid knot]

mb

[proud bridge]

f(x) × g(x)

okay

go on

and ther is np actually

[hushed lance]

but the question doesn't make sense

[proud bridge]

yes

actually

oh

[hushed lance]

of course f(x) times g(x) is equal to g(x) times f(x)

[proud bridge]

(fog)(x)

it is

[hushed lance]

so yeah it is fog and gof

[proud bridge]

yeah i've realized now

[valid knot]

not f(x)*g(x) it's (fg)(x)

[proud bridge]

(fog)(x) = f(g(x))

f(5x+c) here

[restive river]

yeah, that symbol means they're wrapped one around the other

[valid knot]

oh

[proud bridge]

if f(x) = 3x-4

then what is f(5x+c)

we have to find this

[valid knot]

ah

i didn't really get what the problem was asking

[proud bridge]

okay let's do it now

[valid knot]

so it's f(g(x))=g(f(x))

?

[proud bridge]

yes

[valid knot]

so its just inverse functions

ok

[proud bridge]

why inverse

[valid knot]

inverse functions are functions that cancel eachother out

[valid knot]

this

[proud bridge]

Are these like that?

[valid knot]

yea

[proud bridge]

because idk about it really

how

[valid knot]

inverse functions cancel eachother out to x

nvm

[proud bridge]

yea i mentioned about that bro

ahaha

but it's fine

let's continue

we will just write the equality

clearly

and solve it

i think so

[valid knot]

i got 15x+3c-4=15x-20+c so then 3c-4=-20+c
2c-4=-20
2c=-16
c=-8

yep its right thanks :)

.close

[devout snow]

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[valid knot]

i just didn't understand the notation it was saying lol

[proud bridge]

lol okay, you are welcome

have a nice day/night whatever it is

[valid knot]

u too

[proud bridge]

thx

[devout snow]

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[vocal trellis]

Yall tryna help me out rq?

[vocal trellis]

I'm a bit confused on what to do

Wait I'm dumb

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[cursive spoke]

We derive several results on European call options written on a dividend paying stock $S$. In what follows, let $0$ denote the time today, and let $C_0$ and $P_0$ denote the premia for the European call and put options, respectively, written on $S$; both options mature at time $T$ and have strike $K$. Let $p$ be the continuously compounded risk-free zero rate, assumed constant and independent of maturity. We denote by $S_t$ the market price of the stock $S$, and we assume that at time $0$, $0 < t < T$, the stock is scheduled to pay a cash dividend $D$. The results will be derived by using simple Absence of Arbitrage Opportunities (AAO) arguments.

\begin{enumerate}
\setcounter{enumi}{2}
\item Using the AAO arguments similar to those used in class, show that the following put-call parity formula needs to hold:
[
C_0 - P_0 = S_0 - \frac{D}{\exp(pt)} - \frac{K}{\exp(pT)}
]
\end{enumerate}

[woven radish]

Hsway

[cursive spoke]

Struggling with this question. I have tried setting it up such that the two strategies are equivalent to the costs in the put-call parity formula. For context, this involves European options.

What I have done so far:

Strat A: Long call and short put at strike K
Strat B: Long underlying S and borrow sum of discounted value of dividend and strike

But when I go to resolve the payoff equality to satisfy AAO, I cant seem to figure it out.

[devout snow]

@cursive spoke Has your question been resolved?

[spare crypt]

not familiar with the relevant equations but if you post them I may be able to help you show the equivalence needed

[cursive spoke]

I think the gist of the question is to look at the two strategy costs, which are the respective sides of the given equation, then to construct their respective payoffs and prove that they are equal. Here is an example from the notes:

Where the payoffs of a long call are max{S_t - K, 0} and a short put are min{S_t - K, 0}

[spare crypt]

mostly spitballing but it looks like your question is the same as the image you posted, just with another cost D/... and evaluated at 0

[devout snow]

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[fresh wasp]

Find the value of the constant "b" for which the equation has two distinct roots that sum to -1
Im absolutely cooked

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[fresh wasp]

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[vivid topaz]

hello

[vivid topaz]

In economics, primitivation makes it possible to express the cost function C(x) of x objects from the marginal cost function C'(x) Thus, for example, the boss of a company manufacturing electronic gadgets expresses, in euros, the marginal cost of making x gadgets by: C'(x) = 30-0.015x. Assuming that the production of a part costs 25 euro. Expresses the cost function C(x).

is it this ?

<@&286206848099549185>

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[hasty flame]

what is a special case?

[quaint citrus]

seems like something u woulda learned in class

like what a "special case" means in this context

but i see a potential answer

so

[hasty flame]

id hope so too but im doing this online so

no teacher on hand for qcc

[quaint citrus]

rip

[hasty flame]

im thinking its like denominators

[quaint citrus]

so how would u go about solving this problem

yea me too

[hasty flame]

since

there are like denominators..

[quaint citrus]

yea lol

thats what id put

[hasty flame]

yeah ill try it

[quaint citrus]

continue at ur own risk ig lol

[hasty flame]

lol, maybe some googling will help

thx

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[tough moss]

I keep getting this wrong, I've solved the problem but it still says wrong

so much work has been done

someone help me please

GCNIG

<@&286206848099549185>

[winter patrol]

show your work

[devout snow]

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[restive river]

<@&286206848099549185>

[restive river]

I need help with math

[devout snow]

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.

[restive river]

how do I line up the graphs to get the right answer

[devout snow]

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.

[devout snow]

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.

[restive river]

oh-

[devout snow]

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.

[restive river]

like-

BRUH

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[cosmic bolt]

hello

<@&286206848099549185>

[devout snow]

@restive river Has your question been resolved?

[restive river]

can you help me

<@&286206848099549185>

it took so long

.close

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[lost junco]

Hi I will help.

[devout snow]

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[austere geyser]

uh hello?

[austere geyser]

i just joined

i heard alot about this server and i have problems with math anybody mind helping me?

alright anybdoy mind helping with

1/5 x+1/3=4/3 whats the value of x

i was absent on my teachers explanation so i dont really understand it

he deosnt explain to me precisely

[quartz sky]

I don't understand.

[austere geyser]

i might need to be afk

for a little

wdy,

wdym

1/3 is 1 over 3 as a fraction

[quartz sky]

you mean (1/5)x+1/3=4/3 right?

[austere geyser]

how

did u get that

i need explanation

im not just gonna put the answer

[quartz sky]

I think x=5

[austere geyser]

how did u get that

im sooooooo confused

[quartz sky]

So am I

why?

[devout snow]

@austere geyser Has your question been resolved?

[austere geyser]

yes

sorry i was afk

[devout snow]

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[languid monolith]

Z

[languid monolith]

How do I solve this?

[weak cove]

Is this a test?

[languid monolith]

One I did previously yes

[deft hemlock]

My friend, what does it mean for two curves to intersect?

@languid monolith

[languid monolith]

The curves will meet at a certain point

[deft hemlock]

How could you write that in equations?

[languid monolith]

Make them equal each other

[deft hemlock]

Indeed.

Now you have reduced the problem from something abstract, confusing, to determining whether a quadratic equation has real roots.

Let me let my sir take it from here @weak cove

[languid monolith]

Ok

[weak cove]

I'll let my sir the quadratic formula take it from here

[languid monolith]

Why are we using quadratic formula

[wheat pawn]

cuz you have a quadratic equation?

[devout snow]

@languid monolith Has your question been resolved?

[languid monolith]

But I’m not finding x?

[deft hemlock]

Try writing it out first.

[devout snow]

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[languid monolith]

.reopen

[devout snow]

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[woeful blade]

i need help with this pls

[safe knoll]

if the lines are perpendicular then $m_{1}m_{2} = -1$

[woven radish]

JustToPro

[safe knoll]

$m_{1}$ being slope of line 1 and $m_{2}$ being slope of line 2

[woven radish]

JustToPro

[woeful blade]

is this the answer?

[safe knoll]

its the hint towards the answer

[woeful blade]

dawg im sleep deprived rn hints won't help me 😭

[safe knoll]

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[safe knoll]

sorry but rules

[woeful blade]

@safe knoll

[weak cove]

!noans

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

@woeful blade Has your question been resolved?

[woeful blade]

yo

[weak cove]

Don’t DM or ping me

Be patient

[woeful blade]

i tried solving it, but i need confirmation if the answer is -4 or not

i believe it is but im not sure

can you lmk yes or no?

for part 1

[devout snow]

@woeful blade Has your question been resolved?

[woeful blade]

its been 40 minutes

shall i wait more?

[weak cove]

Shall you not ping me more?

[weak cove]

As opposed to what?

[woeful blade]

i didnt ping you tho?

wdym

[weak cove]

replying to me, pings me..

[woeful blade]

and does pinging kill you?

[weak cove]

If you don't want to wait for help, feel free to close the channel

No I just specifically asked you not to

[woeful blade]

like wtf is your problem with pinigng

dawg

[weak cove]

I just specifically asked you not to and you keep doing it..

[woeful blade]

its a yes or no question

pinging literally gives you a quiet heads up

its nothing to be annoyed by

[weak cove]

I don't want to help with your question though, so I don't think you should keep pinging me, especially if I asked you not to. Sorry. I'm sure someone will help eventually

[woeful blade]

mf you got involved and pinged me

is it my fault you got involved?

involving yourself obviously means you want to help

smh

[weak cove]

I was telling you the server rules

no it doesn't

[woeful blade]

yeah when someone already fucking did 🤣

[weak cove]

they sent a different message

[woeful blade]

???????

its the same fucking message with different formatting

[weak cove]

Okay?

[woeful blade]

💀

[weak cove]

What do I have to do with your problem? Wait or don't wait, I don't care, you don't need my permission to wait longer for help

[woeful blade]

and turn ping sounds off if you are so distracted by them?

people that get bothered by pings really are annoying

[weak cove]

What does telling me about how you should keep pinging me have to do with your problem?

[woeful blade]

you responded in like 3 seconds, obviously you have nothing better to do

[weak cove]

I have a lot of things better to do than help someone who is being entitled

[woeful blade]

LOL yeah sure you do bud.

you responded in 3 seconds

but sure bro

[weak cove]

What does any of this have to do with your problem?

[weak cove]

because you pinged me so the notification popped up.. that is how that works

[woeful blade]

what does it matter, you're not trying to help anyway

[weak cove]

Exactly my point from the very beginning

yet you think trying to berate me does your problem any good for some reason

[woeful blade]

so you just left whatever "important" task you had at hand, to respond

🤣

[woeful blade]

It has nothing to do with the problem

[dusky void]

lmao

[woeful blade]

you're just annoying lmao, bro doesn't like pings

bro must be fun at parties

[weak cove]

I don't like them from people who I asked multiple times to stop doing them already

anyways, again, has nothing to do with ur problem

[woeful blade]

buddy you're just yappin tryna sound smart

f off or something lmao, if you're busy

why are you talkin to me

[weak cove]

because you keep pinging me and calling me names and trying to make fun of me?

[woeful blade]

my fault bro, you seem sad

i wont ping you anymore

[weak cove]

Okay thanks

[weak cove]

Which part are you working on?

[woeful blade]

first part

im not sure if the answer is -4 or not

i just need clarification

[weak cove]

Show your work of how you got -4

[woeful blade]

https://tenor.com/view/krackpink-gif-22744290

[woeful blade]

😭 dawg issa yes or no question

[weak cove]

If you want help, you have to explain why you're confused and participate in learning

that is why you got the !nosols and !noans commands earlier

[woeful blade]

https://tenor.com/view/kurt-kurt-angle-100-yard-stare-what-are-you-serious-gif-4081464694509837388

[hazy yarrow]

forgot we dont do mutes anymore but timeouts

Please be nicer to helpers in the future and generally follow the rules @woeful blade , this is unacceptable

[devout snow]

@woeful blade Has your question been resolved?

[turbid beacon]

i doubt it

[devout snow]

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[waxen steeple]

how is part b 24

[devout snow]

@waxen steeple Has your question been resolved?

[devout snow]

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[regal condor]

Hello everyone, i just want to share that im having problems understanding certain trigonometric applications on most physics problem like this especially tension, can you recommend a material or book that can help me understand and review this concepts? i dont know where to start, exams are next week

[devout snow]

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[devout snow]

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[eager kite]

I've come across a problem calculating something for a game, need help! I tried my best to make the problem more 'mathematical'

There are 2 values, x and y. Both of them have a maximum, though x can go beyond max(x)
In this example, max(x) = 100 and max(y) = 7

There is also an unknown formula called β which can only be assigned to x
β(10) ≈ 90
β(50) ≈ 75
β(100) ≈ 110
The above happens if max(x) = 100, if max(x) = 75, then:
β(10) ≈ 67,5
β(50) ≈ 56,25
β(100) ≈ 82,5

y is in some way connected to formula β. For example (max(x) = 100):

if y = 1 and β(x) ≈ 90, then y will equal 0,9
if y = 1 and β(x) ≈ 75, then y will equal 0,5
if y = 1 and β(x) ≈ 110, then y will equal 0,25

I need to find formula β and to find a way to make a formula in which the equations above will apply
im fine with anything close enough, i dont have to have all the numbers to match up exactly

[eager kite]

if max(y) = 7...

if y = max(y) and β(x) ≈ 90, then y will equal 6,3
if y = max(y) and β(x) ≈ 75, then y will equal 3,5
if y = max(y) and β(x) ≈ 110, then y will equal 1,75

ok looking at this i think this is just impossible?

ok i think I found something

β(max(x)) = max(x) + max(x)/10
β(max(x)/2) = max(x) - max(x)/4
β(max(x)/10) = max(x) - max(x)/10

formula α = the y formula i put at the bottom of the initial message

if y = 1...
y * α(max(x)) = 0,25
y * α(max(x)/2) = 0,5
y * α(max(x)/10) = 0,9

in this case if max(x) = 75 then

β(75) = 82,5
β(37,5) = 56,25
β(7,5) = 67,5

y * α(75) = 0,25
y * α(37,5) = 0,5
y * α(7,5) = 0,9

ok going back to non-math terms the problem was that I had 2 variables and I wanted to make it so that on signal, 1 of the variables (which i call x here) always goes up and the other variable (y) always goes down

if x is high then itll go up a tiny bit and y will go down a lot
if x is low then itll go up a lot and y will go down a little

if x is 50% of max(x) then x will go up a tiny bit and y will go down a little too

considering that y always goes down idk if it was necessary to mention that y has a maximum but it does ig

in my code max(x) = 4200 and max(y) = 7

[devout snow]

@eager kite Has your question been resolved?

[devout snow]

@eager kite Has your question been resolved?

[devout snow]

@eager kite Has your question been resolved?

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[frank juniper]

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[frank juniper]

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[unique spire]

a function like that is usually in form $y=a*\left| x-h \right|+k$

[woven radish]

nico

[unique spire]

that's the vertex form

there's a another form which is:

$y=a*\left| b*x-c\right|+d$

[woven radish]

nico

[frank juniper]

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[unique spire]

exactly

[frank juniper]

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[unique spire]

wait, i thing maybe they want you to do $f(x)\circ g(x)\text{ so }f(g(x))$

[woven radish]

nico

[frank juniper]

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[unique spire]

so $f(x) = a*\left| g(x)-h_g \right|+k_g$

[frank juniper]

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[woven radish]

nico

[frank juniper]

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[unique spire]

then solve $a_f*\left| x-h_f \right|+k_f=A_1*\left| g(x)-h_g \right|+k_g$ for $A_1$

[woven radish]

nico

[unique spire]

you already have the equation for f(x) so you know $a_f$, $h_f$ and $k_f$,

[woven radish]

nico

[frank juniper]

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[unique spire]

it's called an "indice"

it's used to differenciate variables so when i say $a_f$ i'm talking about the constant "a" in the "f" formula

[woven radish]

nico

[frank juniper]

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[unique spire]

gl

[devout snow]

@frank juniper Has your question been resolved?

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[valid halo]

two points charges, q1 and q2, are placed 15cm apart in vacuum. the magnitude of the electrostatic force between them 6mN. if one of the charges 8mC, determine the magnitude and sign of the other charge

[valid halo]

Cant seem to get the right answer

[devout snow]

@valid halo Has your question been resolved?

[devout snow]

@valid halo Has your question been resolved?

[runic whale]

you have the law of electrostatic force

and the value of the force

[devout snow]

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[covert pier]

how is this the answer

[covert pier]

don't i have to replace every x and y with 0

[hushed lance]

does the question ask for the equation of the tangent?

[covert pier]

yes

[hushed lance]

misread sorry

[covert pier]

np

[hushed lance]

basically it wants you to find the slope of the tangent using implicit differentiation

[covert pier]

yeah but how do u do that

i did implict differiation

how is the answer y=x

[hushed lance]

then solve for dy/dx when x and y are 0

that's only the slope of the tangent

[covert pier]

yeah but u have to replace every y and x with 0?

[wheat gyro]

Could u help me ???

[hushed lance]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hushed lance]

yes

[covert pier]

then how is y and x in the answer

thats what im confused about

[hushed lance]

and then you use that slope and the point 0,0 to deduce the equation of the line

even if you plug in 0,0 dy/dx may not necessarily be 0

what did you get after differentiating both sides

[covert pier]

(cos(y)-ye^sin(x)cos(x))/((e^sin(x)+xsin(y)) = dy/dx

[hushed lance]

yeah so what do you get if you put x=0 and y=0

[covert pier]

1/1

i think

[hushed lance]

yes

So the slope of the required tangent is 1

as the tangent has slope 1 and it passes through 0,0, it would have equation y=x

[covert pier]

ohh i understand it now

i didnt know replacing x and y with 0 is solving for the slope 😭

ty

.close

[devout snow]

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[devout snow]

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[fair thunder]

i want help in how to work out area of compound shapes

[fair thunder]

help

How to work out area of compount shapes

[solid bolt]

do you have an example in mind

[fair thunder]

area of quadrilateral

[solid bolt]

it depends on the quadrilateral

[desert fulcrum]

you have the coordinates?

[fair thunder]

no

[solid bolt]

actually a basic formula for compound shapes would be to split them into triangles and then keep calculating (if they are irregular)

[calm valley]

Equations ?

[fair thunder]

any quadliateral but jst how to do it

yes

[calm valley]

Example any?

[fair thunder]

for my gces

[fair thunder]

no but any will do

just need to see how to do it

[fair thunder]

uh

do you know what gcse are?

[calm valley]

You got equations in ur question?

[solid bolt]

gcse like the board

[fair thunder]

no

[fair thunder]

ye

for yr 11

[calm valley]

No? Full form?

[fair thunder]

no i just wanted the equations

if it is possible

[calm valley]

This will help then

[fair thunder]

alr so what is the equation for rectangal?

cause idk only that one

[solid bolt]

so you can do 0.5 * h1 * d1 + 0.5 *h2 * d1
which is actually 1/2 * d1 (h1 + h2)

[fair thunder]

and a little bit confusing

[fair thunder]

okay

[solid bolt]

where h1 and h2 are heights and d is any diagonal

[calm valley]

It's divided into two triangles , and ig u know area of triangles

[solid bolt]

yea

thats how it works for all irregular shapes

[fair thunder]

yes b*h/2

right?

[fair thunder]

okay

[calm valley]

Yep

[fair thunder]

Thanks guys

very helpful

.close

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[crude tapir]

I have to explain why 2-2²+2³-2⁴+.....-2²⁰²+2²⁰³-2²⁰⁴ is devideable by 3

[crude tapir]

All i tried is that i found the first 2 groups of 3 numbers are devideable by 3

[solid bolt]

what was the first step you did

[crude tapir]

Grouped 2-2²+2³

=6

Ik that in these kind of problems i have to group

But now on idk

[solid bolt]

so what you do is instead

factor out 2

[crude tapir]

3x2

[solid bolt]

from the entire geometric series

[crude tapir]

Oj

Oh

Alr

Than?

[solid bolt]

dyk the formula for sum of finite geometric series

[crude tapir]

No i forgot it

But

Id like to solve it in a way without formulas

[solid bolt]

uh so

[crude tapir]

Also i really appriciate your help

[solid bolt]

so there is another somewhat complicated way

i dont know how to explain it

you start by factoring out 2

then we notice except the first term which is 1 all the others are in terms of 2

then you group them differently

[crude tapir]

Lemme try

[solid bolt]

= $2(1 - (2^2 - 2^3) + (2^4 - 2^5) - \ldots + (2^{202} - 2^{203}))$

[woven radish]

flurry

[solid bolt]

if that makes sense

$= 2(1 - 2(2 - 2^2) + 2^3(1 - 2^2) - \ldots + 2^{201}(1 - 2^2))$

$= 2(1 - 2(2(1 - 2) + 2^3(1 - 2^2) + \ldots + 2^{200}(1 - 2^2)))$

[crude tapir]

If i factorise by 2 its not that

Yrah

Now correct

[woven radish]

flurry

[solid bolt]

then it becomes $= 2(1 - 2(2(1 - 2) + 2^3(1 - 2^2) + \ldots + 2^{200}(1 - 2^2) + 2^{202}))$

[woven radish]

flurry

[crude tapir]

Its 2(1- 2... not 2²

[solid bolt]

bro im writing in a hurry

but are you understanding the concept

[crude tapir]

Not well

[solid bolt]

wait lol

uh i mean thats what i was saying

its easier with formulas

https://k12.libretexts.org/Bookshelves/Mathematics/Analysis/07%3A_Sequences_Series_and_Mathematical_Induction/7.04%3A_Sums_of_Geometric_Series/7.4.01%3A_Sums_of_Finite_Geometric_Series

[crude tapir]

Thanks

.closr

.close

[devout snow]

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[fervent bear]

Help me out

[fervent bear]

Question 7 and 8

[devout snow]

@fervent bear Has your question been resolved?

[fervent bear]

@fervent bear

[devout snow]

@fervent bear Has your question been resolved?

[devout snow]

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[restive river]

yes

that's correct

there are 2 conditions

your function g, has an inverse denoted as (g^-1) if and only if its injective and surjective

otherwise a inverse function does not exist.

yes

your function is not injective, neither surjective

so none

therefore an inverse function does not exist.

[devout snow]

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[zealous bone]

Does anyone know how to solve task 9 here?

[zealous bone]

I finished everything but that one and i really can't seem to see how to solve it

[devout snow]

@zealous bone Has your question been resolved?

[devout snow]

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[prisma frost]

can someone help me out with this question

[unique spire]

what have you tried so far?

[prisma frost]

so i but then tensioon force

at the angle going up

then did moments

about where the dot is

so

(80 x 4) = 100x

however im 99 percent sure thats wrong

and im missing stuff

[unique spire]

yes it is wrong

have you made the diagram yet

[prisma frost]

yea i show it

This is it

Ignore the arrow going down

I’m assuming the tension force is up since it’s holding the rod up or am I wrong?

[unique spire]

the tension force has an "x" and a "y" component

[prisma frost]

?

wdym

[unique spire]

one sec, i'll draw something

[prisma frost]

Kk

[unique spire]

you can decompose the tension in 2 vectors

[prisma frost]

ah ok

would x and y be equal?

[unique spire]

and since the sum of all force in Y is equal to 0, then only $T_y$ is used to balence the weight

[woven radish]

nico

[prisma frost]

wait why is y

0

[unique spire]

$\sum_{}F_y^{}=0$

[woven radish]

nico

[unique spire]

you are working in static right?

[prisma frost]

gonna be honest i havent heard of that term before

but imma assume yes i am

question doesnt say if i am or not

[unique spire]

in the question it does not tell you it's moving, so you are working in static

[prisma frost]

ahh ok

so after know the x and y components how would i use it to help me?

[unique spire]

since you are working in static that means that $\sum_{}F^{}=0$ that means$\\sum_{}F_x^{}=0\\sum_{}F_y^{}=0\$ and$\\sum_{}Moments^{}=0$

[woven radish]

nico

[unique spire]

$T_x=T*\cos(angle)$

[woven radish]

nico

[prisma frost]

oh

ok

i see now how to answer it now

its basically wants me to do

80 x (4 x cos (30)) = x x 100

then i can rearrange it

or no?

[unique spire]

not exactly

[prisma frost]

what part was wrong?

oh

sorry

[unique spire]

$\sum{}Fy^{}=0\
-weigh+T_y=0$

[woven radish]

nico

[prisma frost]

i meant sin

im kinda confused now

what would i sub into that then?

[unique spire]

i was kinda waiting for you to add the last force in the diagram but

[prisma frost]

oh

it think i figured out anyways

[prisma frost]

would it be this

but sin

instead?

with me doing it around the point on the left

[unique spire]

actually yes

the way you wrote it is very confusing but yes

[prisma frost]

sorry about that

anyways thanks for the help bye

[unique spire]

have a nice day

[prisma frost]

thank you. you too

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[strong forge]

Need help on this question, I don't really know how to approach it or what to do.

[uncut crow]

t_n - seems like a typo

should be t_n =

?

[strong forge]

I dont think its a typo

thats how its written in all the other questions he's taken up

[uncut crow]

[heady plinth]

Agree

[uncut crow]

anyway you can plug some numbers in to get relations between b and c and solve for b and c

[strong forge]

nvm ur right it is supposed to be = and not -

and what #s do i plug in for b and c

[uncut crow]

ones from the table pick whatever you want

n = 2 for example

[strong forge]

ok lets say I make n = 2
does that mean t_n = 5, then it would be b (t_2-1) so that would be t_1 = 4 and then c(t_2-2) and that would be t_0 = 1

5 = 4b + 1c

then i'll choose another n value and then just solve for b and c?

@uncut crow

[uncut crow]

[strong forge]

Thanks

@uncut crow so ive got what b and c equal now, it says to put it in terms of n, how would I do that

[devout snow]

@strong forge Has your question been resolved?

[devout snow]

@strong forge Has your question been resolved?

[strong forge]

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[surreal jay]

can someone help me with this?

[near stone]

i'd assume you need to find the derivative of f(x)

then plug f'(16)

[surreal jay]

yeah i got f'(x) = ((6x+7)sqrtx - (3x^2+7x+6)1/2x^-1/2)/x
then substituted x for 16 but it says the answers wrong

[near stone]

did you apply the quotient rule?

i'd probably start with that

[quaint citrus]

ew

[near stone]

why ew

[devout snow]

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[tardy hawk]

where do I start 🙏

[worldly maple]

idk 🙏

yo i might be able to help but i can't read it

[near stone]

press windows + prt sc button to make a screenshot

[worldly maple]

i think its a school chromebook so he cant use discord

[near stone]

is this clear?

[worldly maple]

nvm i am of no use to this operation😭

use pythagorean theorem or smth idk

[tardy hawk]

sorry, It's my friend who needs help, and the picture he sent me suck

[near stone]

I could use pythagorean theorem or sohcahtoa

[tardy hawk]

Its not a right triangle tho

Oh wait

I figured it out

nvm

[near stone]

good job

[devout snow]

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[whole flame]

hi

[whole flame]

for this question should i be doing arctan(1) - arctan(0)?

[lost laurel]

11 times that

[proper mica]

Yes, the answer would be 11 × π/4

[whole flame]

why 11 times?

[mighty galleon]

derivative of arctan is (1)/(1 + p^2)

in order to get 11 on top you would do 11arctan

like how the derivative of 11x = 11

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[devout snow]

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[jaunty abyss]

Is this a graph that opens downward ?

[lunar harbor]

Yes

[jaunty abyss]

Okay

[devout snow]

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[frail sun]

I'm confused on this problem of finding max and mins.

[frail sun]

What I did was take the first derivative which is

[woven radish]

Someone

[frail sun]

Then from there, you set is to 0 and solve for x, which is -5/6, right?

-5/6 is a "critical number", but just want to make sure I'm on the right page

[silent reef]

You're right

[frail sun]

Okay, then what I do is draw a little line and plug in a value that is less than -5/6 (-1) and greater than -5/6 (0) and see if they're positive/negative?

[silent reef]

A hack: when bounded intervals are given such questions, the absolute minima usually lies at x = lowest value (-3 in this case) and the absolute maxima at x = 1 (greatest value)

[silent reef]

Once you find all the critical points, put those values in the function and check for the greatest/smallest value

You need not do the step you just mentioned

[frail sun]

ah

the original function, right?

[silent reef]

That's only for finding which interval the function is increasing/decreasing in

[silent reef]

Yes

[frail sun]

-3 = -20
-5/6 = 0
1 = 22

I guess where I'm confused at is where the hell did they get 27 and -7/6

(that's the answer key)

[silent reef]

Hmm it's probably incorrect in that case

[frail sun]

the answer key?

I swear this happens literally every exam review too 😂

makes me doubt everything i know

[silent reef]

wait these values are incorrect

[frail sun]

oh?

[silent reef]

The maxima is 27

at x = -3

[silent reef]

haha it happens

[frail sun]

12(-3)+10 = -26

dunno how i got -20 the first time

[silent reef]

this works on positive intervals only then i suppose

[silent reef]

No the value must be put in the original function

not its derivative

[frail sun]

ohhhh

that's what i did wrong

[silent reef]

yep

[frail sun]

I even had it in my brain, use original function

ahahah

[turbid tiger]

you have just 5 values of x
if you have less values try to substitute the value of x in function

[silent reef]

😭

[turbid tiger]

if you are stuck anywhere

[frail sun]

Then on the test, do they usually always want the y value or is there a way to determine what they're asking for?

sometimes they can be tricky

[silent reef]

usually, yes

Critical points are rarely asked

if the word "value" is used it's always f(x) = y

[frail sun]

easy, thank you so so much!

[silent reef]

no probs :)

[turbid tiger]

please close this if your doubt is resolved

[frail sun]

.solved

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[reef tundra]

Im trying to derive the moment of inertia of a torus about the axis depicted in the right, but I know I need the volume first. I attempted to set up the volume integral in spherical coordinates, but its incorrect and I don't know why. I do not get the 2pi^2ab^2 volume.

My process was seting up the integral so that the z axis is the axis of rotation depicted to the right. here's what I got so far. Where am I going wrong?
heres the volume integral I got

rho 1 is acos(theta)-sqrt(b^2-a^2sin^2(theta))
rho 2 is acos(theta)+sqrt(b^2-a^2sin^2(theta))

[devout snow]

@reef tundra Has your question been resolved?

[reef tundra]

<@&286206848099549185>

[spare crypt]

uh if you try spherical coordinates you might spontaneously combust so I'd try cylindrical or just the washer method

[reef tundra]

yea i know, but what exactly am i doing wrong in my volume integral?

[spare crypt]

everything looks good so far, haven't looked at the last row yet

hm yea I don't see an issue

[reef tundra]

that so wierd. Whenever I use Desmos to compute this integral its wrong 😭

[spare crypt]

oh the formula for p=.. should be phi

maybe there's like a variable typo

theta is the easy 2pi multiplication that's symmetrical with respect to the function

oops jk you just use different letters from me

er there's something off, like one change is making it p^2sin(theta)

the sine angle in the integral is supposed to be the vertical one

[reef tundra]

the angle is supposed to start from the y axis?

[spare crypt]

er the z axis

like this should work, just wolframalpha is timing out because it's spherical

https://www.wolframalpha.com/input?i=volume+integral+wolfram+alpha&assumption={"F"%2C+"TripleIntegral"%2C+"rangeend2m"}+->"arcsin(b%2Fa)"&assumption={"F"%2C+"TripleIntegral"%2C+"intvariable1i"}+->"x"&assumption={"F"%2C+"TripleIntegral"%2C+"rangestart1i"}+->"a*cos(y)-sqrt(b^2-a^2sin(y)^2)"&assumption={"F"%2C+"TripleIntegral"%2C+"rangestart3"}+->"0"&assumption={"F"%2C+"TripleIntegral"%2C+"rangestart2m"}+->"-arcsin(b%2Fa)"&assumption={"C"%2C+"volume+integral"}+->+{"Calculator"%2C+"dflt"}&assumption={"F"%2C+"TripleIntegral"%2C+"integrand"}+->"x^2*sin(y)"&assumption={"F"%2C+"TripleIntegral"%2C+"rangeend1i"}+->"a*cos(y)%2Bsqrt(b^2-a^2sin(y)^2)"&assumption={"F"%2C+"TripleIntegral"%2C+"intvariable2m"}+->"y"&assumption={"F"%2C+"TripleIntegral"%2C+"intvariable3"}+->"z"&assumption={"F"%2C+"TripleIntegral"%2C+"rangeend3"}+->"2pi"

did you try flipping sin(phi) in the integral with sin(theta)

[reef tundra]

then I get 0

[spare crypt]

uh maybe swap dphi and dtheta

idk if you can link what you have

[reef tundra]

https://www.desmos.com/calculator/u3sqzz6em4

dont worry if u cant figure it out. I dont wanna take up too much of your time

[spare crypt]

I wasn't gonna stop till I tinkered it to work lmao, here it matches https://www.desmos.com/calculator/uqajo0zujr

I can't explain why

[reef tundra]

thankyou!!!

.close

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[twilit granite]

say we have the function f:A->B
how do i prove that the inverse is from B->A if and only if f is bijective?

[devout snow]

@twilit granite Has your question been resolved?

[heady plinth]

Well if it’s not subjective then you can’t use B as the domain, you would have to exclude the elements that don’t have an inverse

If f is not injective then the inverse is just not a function because one element goes to two

[twilit granite]

got it, but how can I show that mathematically, in a general case?

[devout snow]

@twilit granite Has your question been resolved?

[faint zinc]

@twilit granite by assuming it is not surjective and showing it leads to a contradiction, and assuming it's not injective and leads to a contradiction

[devout snow]

@twilit granite Has your question been resolved?

[twilit granite]

working on it

[magic pine]

the question is also a little bit ill posed id say

generally you'd say f has an inverse f^-1 : B -> A iff f is bijective, not that the inverse is from B -> A iff f is bijective