help-39

[mortal sparrow]

Okay got it thanks

I think I've got it then yea thanks very much

.close

[pearl pond]

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[wise rampart]

Could you please help me with the steps of partial fraction decomposing $\frac{−4x−10}{x^2+4x+3}$ using the way they used in the symbolab link below?
​ https://www.symbolab.com/solver/step-by-step/partial fractions \frac{−4x−10​ }{x^{2}%2B4x%2B3}?or=input

[jolly parrot]

kangaroo rat

[plush bramble]

https://www.youtube.com/watch?v=QKkdYW77xNI

[toxic lichen]

your link only takes me to the problem itself but not any steps for it

[midnight haven]

you are better off understanding how to decompose in general than following some "steps" from a website

[toxic lichen]

^

[pearl pond]

@wise rampart Has your question been resolved?

[pearl pond]

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[frozen night]

This is what I have done so far

not sure how I obtain $\frac{\partial f}{\partial u}$

[jolly parrot]

張嘉棋

[frozen night]

and $\frac{\partial f}{\partial v}$

[jolly parrot]

張嘉棋

[lunar trench]

How about just keep it in that form and plug into equation

[frozen night]

erm huh? how?

[lunar trench]

You don’t need to obtain $\frac{\partial f}{\partial v}$ and $\frac{\partial f}{\partial u}$

They just ask you to simplify the equation

Omg how to use latex

[frozen night]

u just forgot the \

before the first frac

[jolly parrot]

asuasu

[frozen night]

hmmm

[frozen night]

then what do I put and where?

because it ain't clear

[lunar trench]

Your method is correct

[frozen night]

I think I need final equation in terms of $u$ and $v$

[jolly parrot]

張嘉棋

[lunar trench]

The final equation will be in the form of $\frac{\partial f}{\partial v}$ and $\frac{\partial f}{\partial u}$

[jolly parrot]

asuasu

[lunar trench]

Instead of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$

[jolly parrot]

asuasu

[frozen night]

oh no then

we can easily get $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ from the given equation?

[jolly parrot]

張嘉棋

[frozen night]

(given is a better word here, rather than initial lmao)

[lunar trench]

Basically, you transform $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$into $\frac{\partial f}{\partial v}$ and $\frac{\partial f}{\partial u}$ from this

[jolly parrot]

asuasu

[frozen night]

hmm

erm

my brain is frying rn, bare with

[leaden cape]

I think I've figured it now.

[frozen night]

okay so, $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ can be obtained where?

[jolly parrot]

張嘉棋

[frozen night]

I see this now, but how do I extract?

just simply by moving things?

[lunar trench]

Just substitute the whole thing into equation

[frozen night]

erm the given equation into the concoction of partials?

[leaden cape]

Assume the regular d is partial d. Using x = x(u, v) = e^u cos(v) and y = y(u, v) = e^u sin(v).
If f(x, y) = f(x(u, v), y(u, v)) then df/du = df/dx (dx/du) + df/dy (dy/du) = df/dx (e^u cos(v)) + df/dy (e^u sin(v))>
In short, df/du = x df/dx + y df/dy.
Perform a similar calculation for df/dv and then notice what df/du - df/dv equals.

[lunar trench]

Now if you simplify this form you will get df/du=df/dv

[lunar trench]

This is also alternative solution

[frozen night]

oh like this 💀 my brain thought the other way to substitute

so is $\frac{\partial f}{\partial u}$ just the derivative of $e^ucos(v)$?

[jolly parrot]

張嘉棋

[frozen night]

oh wait that makes no sense

[leaden cape]

f = f(x(u, v), y(u, v)) so

$\pdv{f}{u} = \pdv{f}{x} \pdv{x}{u} + \pdv{f}{y} \pdv{y}{u}$.

[jolly parrot]

stabulo

[leaden cape]

Since x(u, v) = e^u cos(v) and y(u, v) = e^u sin(v) then

$\pdv{f}{u} = e^u \cos v \pdv{f}{x} + e^u \sin v \pdv{f}{y}$,

[jolly parrot]

stabulo

[leaden cape]

$\pdv{f}{u} = x \pdv{f}{x} + y \pdv{f}{y}$.

[jolly parrot]

stabulo

[frozen night]

why does it seem cyclic? meaning $\pdv{f}{x}$ is used?

[jolly parrot]

張嘉棋

[leaden cape]

I don't think it is. It's just the way you've been around in circles I think.

[frozen night]

well they are quite round 🤣

[leaden cape]

We're assuming, without proof here, that we can solve for x and y in terms of u and v.

Then we simply assume we started off instead with f(x(u, v), y(u, v)) instead of simply f(x, y).

We use the chain rule to calculate df/du and df/dv so really there was no circular logic.

[frozen night]

This is where I’m at

[leaden cape]

Maybe bring the outside brackets inside to form a simplification but I didn't do this process.

[frozen night]

the other guy did

[leaden cape]

Is that the current method you're going to use?

[frozen night]

yessir

oooohhhh okay so i'm taking the $\pdv{f}{u}$ and $\pdv{f}{v}$ out

[jolly parrot]

張嘉棋

[pearl pond]

@frozen night Has your question been resolved?

[pearl pond]

@frozen night Has your question been resolved?

[frozen night]

@lunar trench

(sorry for pinging)

[lunar trench]

Did you find the solution?

[frozen night]

nope

lemme send where I got to tho

[lunar trench]

I think your sign is wrong

[frozen night]

which sign?

[lunar trench]

Plus and minus on last line

[frozen night]

erm

there are bunch of them

[lunar trench]

Yeah i want to mark but i dont know how to

[frozen night]

is it between the (x+y)?

or the bracket from PD?

[lunar trench]

It is the one after $\frac{x}{x^2+y^2}$

[frozen night]

ahhh so they should be flipped?

[lunar trench]

Wrong syntax again?

Yeah

[frozen night]

wait it seems correct tho? the syntax

[lunar trench]

Also the other one is after $\frac{y}{x^2+y^2}$ on second bracket

[frozen night]

$\frac{y}{x^2+y^2}$

hmm

[lunar trench]

In df/dv part

[lunar trench]

Yeah and also the first one

From plus to minus

[frozen night]

ahhh yh got that

that can't be the answer tho

[lunar trench]

Now you simplify the form in each bracket

[frozen night]

$(x+y)(\frac{x}{x^2+y^2})$

bruh

idk how to simply that tho

[lunar trench]

Ok i will show you the first half

[frozen night]

it goes to 1

all this time I thought it didn't

okay I should be good now

all it was, was just the signs

😭

so the answer is just $\pdv{f}{u}+\pdv{f}{v}=0$

[jolly parrot]

張嘉棋

[lunar trench]

Ok it wasn’t supposed to be plus

[frozen night]

so minus?

[lunar trench]

And i just noticed that your calculation in the first partial differentiation was wrong

[lunar trench]

Here

[lunar trench]

Yeah should be minus

[frozen night]

done

phew

thank you very much

.close

[pearl pond]

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[tall flint]

That is indeed what it's asking for

For a transaction that isn't suspicious, what is the sample space of the tagging test?

Every transaction gets tested. What is the sample space of this test?

If something gets tested, it's either tagged or not tagged, right?

So when a transaction gets tested, given that it's not fraudulent, what is the test's sample space?

The test can't make mistakes?

It can't incorrectly tag a legitimate transaction?

Yes the sample space doesn't change

The point I'm trying to make (and perhaps not doing so well on) is that conditioning on events like this doesn't change the sample space, but it does change the assigned probabilities

The sample space has two elements. You know the probability of one of them already

That's not how this works

Given that the transaction isn't fraudulent, it either is tagged or isn't tagged

You know the probability that it isn't tagged

10%?

The question says 0.99

Or 99% I guess

Yes

Np

Where's that coming from?

But they're asking for fraudulent and suspicious. It's no longer conditioning on anything

Nothing is given in b

Yes

That sounds right

Why do you think that?

the conditioning is going the other way

of the legitimate transactions, 99% aren't tagged

but of those not tagged, how many are legitimate?

err, of those tagged

you get the idea

where's that coming from?

and that gives you P(F'|T)?

yes

it does not

you computed P(T) in a previous question

1-P(T'|F')

because when you condition on F (or F') then it's no longer random; it's information, it's given

thus your sample space (the space of possible outcomes) only contains randomness in T

ie whether it's T or T'

Np

[pearl pond]

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[midnight ravine]

Why did they set first and second deeivative equal to each other

<@&286206848099549185>

.close

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[grizzled lodge]

Stuck again on this one

[grizzled lodge]

I got that it’s 2-sec^2(x/2^k) but now what

[bold ledge]

that's cursed

[shrewd basin]

,tex \sec^2{\left(\frac{x}{2^k} \right)} \left(2\cos^2{\left(\frac{x}{2^k}\right)} -1 \right)

[jolly parrot]

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

[shrewd basin]

,tex \sec^2{\left(\frac{x}{2^k} \right)}cos{\left(\frac{x}{2^{k-1}}\right)}

[jolly parrot]

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

[shrewd basin]

$$\int \frac{\cos{x}}{x} \prod_{k=1}^{\infty} \sec{\left(\frac{x}{2^k} \right)} \ dx$$

[glacial sequoia]

\prod and not \Pi I believe

[shrewd basin]

aight ty

you should be able to go from here

@grizzled lodge

[grizzled lodge]

Er

Idk bruh how do you simplify the product?

[jolly parrot]

Pure

[shrewd basin]

typo

i fixed it

[grizzled lodge]

Still don’t know bruh

@shrewd basin

@glacial sequoia

[glacial sequoia]

Don't ask me 🗿 just as clueless as you

But I'd think that you'd wanna evaluate the first few terms and have a look and see how a pattern develops

[grizzled lodge]

Okay I’ll try

[glacial sequoia]

You can write cos x in terms of cos x/2 I mean and stuff

[grizzled lodge]

Bruh I still don’t get it

.close

[pearl pond]

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[grizzled lodge]

.reopen

[pearl pond]

✅

[glacial sequoia]

Usually these problems are magical

With a bit of manipulation you can start a chain reaction the simplifies the whole product

Not sure how you'd do that though

[grizzled lodge]

I’m just gonna wait for pure or someone else

[glacial sequoia]

Yeah sorry I can't help

[grizzled lodge]

Nah dw it’s a tough question

[glacial sequoia]

I'll be looking forward to seeing the solution to this though

Looks fun

[shrewd basin]

im gonna play minecraft. try some trig identities

[grizzled lodge]

[glacial sequoia]

I think

You can split the cos x

And it'll give you a term with cos^2 (x/2)

With one of those cancel with the sec x and the other split it into x/4 to cancel with the next sec x

And so on 🗿

[grizzled lodge]

I’m not following bruh

[glacial sequoia]

Magical chain reaction but I'm unsure of the logistics

[glacial sequoia]

Can I just ask

What level is this problem

Like grade

[grizzled lodge]

Idk I found it online

Instagram

[glacial sequoia]

Damn

Uh

Alright

$\cos 2x = 2\cos^2 x - 1$

[jolly parrot]

NEONPerseus

[glacial sequoia]

Are you aware of this formula

[grizzled lodge]

Yes

[glacial sequoia]

Yeah so

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

[jolly parrot]

NEONPerseus

[glacial sequoia]

Why did that take that long to type out

Anyways

[grizzled lodge]

Okay still following

[glacial sequoia]

The goal is to somehow get the secants in that product to cancel with the cosines in this formula

[grizzled lodge]

Ok

[glacial sequoia]

One of the cos (x/2)s can be turned into 2 cos (x/4)s and one of those can be turned into a cos (x/8) and so on

I'm in Grade 11 though so take whatever I say with a grain of salt 💀

[grizzled lodge]

Grade 11 is like what year in the UK?

[glacial sequoia]

I just have one more year of high school left before I go to college

[grizzled lodge]

I see

[pearl pond]

@grizzled lodge Has your question been resolved?

[pearl pond]

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[modern talon]

its asking for the value associated with the 97.5%

[pearl pond]

@subtle bolt Has your question been resolved?

[modern talon]

np

and yes

its the largest

so you want the values on the right

meaning the 97.5%

yep

[pearl pond]

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[raw zephyr]

Would anyone here be able to help with physics?

[pearl pond]

@raw zephyr Has your question been resolved?

[pearl pond]

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[limber ravine]

Can you exclude any of the answers from the domain of x alone?

[buoyant panther]

you any simply check if answers fit

[pearl pond]

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[midnight haven]

How do i divide 1/5 on paper?

[midnight haven]

Or 1/6, 1/3 etc

[tall flint]

What do you mean "divide 1/5"

[midnight haven]

Like i need to 1:5

[tall flint]

Is there a specific problem you're trying to solve?

[limber ravine]

Do you need to find the decimal representation of 1/5?

[midnight haven]

I just want to know how to do it

[midnight haven]

I guess so

[limber ravine]

Do you need to use a certain method, like long division?

[midnight haven]

I don't know what that is

[fiery vortex]

uh

use calculator

[midnight haven]

I need to do it on paper

[fiery vortex]

but ydk long division

[midnight haven]

Are there ways easier than that?

[limber ravine]

Remember the most common fraction's decimal values

[fiery vortex]

^

[limber ravine]

1/2 = 0.5, 1/3 = 0.333..., 1/4 = 0.25 and 1/5 = 0.2 are good ones to remember

[midnight haven]

But i can't do it if it's 1/43, can i?

I even can't do it after 5 properly

Like 6, 7, 8, 9

[copper kestrel]

yeaa that's why you should learn long division

if you really want to do those without a calculator

[midnight haven]

I once heard of a method that used multiplying

But i can't remember it

[cinder flower]

I can think of a really inefficient way with multiplication lol

[limber ravine]

@midnight haven I tried to write out an explanation of long division, but it's pretty hard to format it properly. Try to watch this video to get a general understanding of the method: https://www.youtube.com/watch?v=eIUoIhfupuA

[midnight haven]

Thanks

[pearl pond]

@midnight haven Has your question been resolved?

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[sturdy bolt]

,rotate

[sturdy bolt]

Someone help 😬

🗿

[pulsar lark]

Find angle C first

[sturdy bolt]

How

Nice prfp😉

[pulsar lark]

Also BC

Pythagoras and trigonometry

[sturdy bolt]

Bc is 13/2

I mean 13

Each segment is 13/2

[pulsar lark]

Yes find angle C

Using trig

[sturdy bolt]

We haven’t done trigonometry the teacher said u wouldn’t have to use it

[pulsar lark]

Ok

[sturdy bolt]

Is there another way

Wouldn’t u have 2 similar triangles if u bisected angle A

I think you’d have to use that? But I’m not sure

@pulsar lark

[pearl pond]

@sturdy bolt Has your question been resolved?

[sturdy bolt]

<@&286206848099549185>

Pls check this out

@buoyant panther I think this is similar to the other one

[buoyant panther]

hyp is 13

you have similar triangles here

[sturdy bolt]

Yes hypotenuse is 13

[buoyant panther]

DC = 6.5

[sturdy bolt]

ABC and edc are similar?

[buoyant panther]

now for example DE/5 = 6.5/12

[buoyant panther]

yes

[sturdy bolt]

I thought 6.5

Since 13/2

[buoyant panther]

a

typo

you're right

[sturdy bolt]

So what else from there?

[buoyant panther]

DE/5 = 6.5/12

[pearl pond]

@sturdy bolt Has your question been resolved?

[sturdy bolt]

That means x is 2.70

2.708

But how is that the answer

And if u make that a mixed number it still doesn’t work

@buoyant panther

[buoyant panther]

how

,calc 6.5/12 * 5

[jolly parrot]

Result:

2.7083333333333

[buoyant panther]

it's not 2.708

,calc 2 + 17/24

[jolly parrot]

Result:

2.7083333333333

[buoyant panther]

advice: work on fractions

[pearl pond]

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[unborn sundial]

.reopen

[pearl pond]

✅

[unborn sundial]

.close

[pearl pond]

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[pearl pond]

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[vernal grove]

Is this correct

"Nico asks us to predict the ocean temperature at the surface using his data. What value would you provide to Nico?" Do I use this equation to predict?

[primal robin]

Hello, the slope is wrong

The dependent variable is not the depth(meters)

It's asking for a linear model of temperature as a function of depth

The slope should be the inverse

I mean you found ∆z/∆T

That would be correct if you consider z as a function of T, but that's not the case here

Yes

The slope is always ∆y/∆x, where y is a function of x

I think you have to find the independent term again

y is a function of x

The problem says that T is a function if z

It should be T2-T1/z2-z1

but you originally found z2-z1/T2-T1

No, a and b are x0 and y0

$y-y_0=m(x-x_0)$

[jolly parrot]

ELeonardo

[primal robin]

where $(x_0,y_0)$ is any known point on the line

[jolly parrot]

ELeonardo

[vernal grove]

oh yes im slow

so basically this

thru the two points

wait is this right

@primal robin

[primal robin]

Yes it looks correct

👍🏽

[vernal grove]

Yes now

Do I use this equation

and predict what the temp will be at the surface level

Surface level meanin depth would b 0

"Nico asks us to predict the ocean temperature at the surface using his data. What value would you provide to Nico?"

says using his data so Im confused what data.

Or do I j straight up predict off this last line

[primal robin]

Yes that's correct

[vernal grove]

Yes but

Do I j plug in 0

in my equation

to find?

because it says predict using Nicos data.

[primal robin]

You used the data to find the linear model

Now use the linear model to predict temperature at depth=0

[vernal grove]

mhm I see

so that would be 9.267

this would be correct right

[primal robin]

Yes 👍🏽

[vernal grove]

for this its p = f(z) n findin the derivative of p with respect to z basically?

m i suppose to find the derivative of both graphs?

or do I find derivative of both graphs and then equal to each other?

[pearl pond]

@vernal grove Has your question been resolved?

[primal robin]

It's asking for a sketch of just dp/dz

[vernal grove]

yes but

findin dp/dz

[primal robin]

It's just a sketch

Look at the slope of p(z) and try to sketch the derivstive

You don't have to find a number, equation or something

[vernal grove]

oh

[vernal grove]

that would be the graph on the right?

[primal robin]

Yes but rotated

The graph given is z(p)

You are asked for the derivstive of p(z)

[vernal grove]

Basically like this

[primal robin]

Almost

[vernal grove]

wait no other way Im slow?

[primal robin]

The y axis should increase when you go up

[vernal grove]

so x is my y and y is my x

[primal robin]

Yes that's great

[primal robin]

Yes 😵‍💫

[vernal grove]

Now I j find the slope on the curve?

[primal robin]

Yes

Draw a sketch, basically when it's increasing and decreasing, and how steep is the curve

[pearl pond]

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[vernal grove]

would this b correct

sketch of

p(z)

I had to sketch because the values were confusing me

[primal robin]

Don't forget that it's flat at the beginning

[vernal grove]

yeh

but other then that

wait does that flat part in the beginning matter?

when Im trying to find the slope

[primal robin]

It does matter because it's part of the given graph

And you have to sketch the derivative using that graph

[vernal grove]

the derivative of this graph right

why dont we care about the graph

in the left

[primal robin]

Because the problem does not ask for that graph

[vernal grove]

also

zeros dont exist for this right

since there cant be no horizontal line drawn for the curve

[primal robin]

Wdym

The graph of p(z) is horizontal st the beginning according to the given graph

[vernal grove]

Oh yeh

So that would be my only 0

for x

and then rest is just increasing

[primal robin]

Yes, but there's a very steep part

so the slope is big

Then it gets small because it's much less steep, almost horizontal

[vernal grove]

I see

sonething like this?

@primal robin

[primal robin]

Mmm I don't think so

Notice that it's very steep next to the beginning

Then it's almost horizontal

but keeps growing

Anyway the slope is small

[vernal grove]

🤔

[primal robin]

The graphing the derivative, think of the y axis as the slope

[vernal grove]

Shit im too slow for this

Imma try again tomorrow

[primal robin]

Don't worry, just take your time later

Don't forget that you are graphing how steep is the curve, not how high it is

[vernal grove]

Bet ty

Its like 1:35 am here so imma get some sleep

and work on it tmrw

[pearl pond]

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[floral gull]

Is it possible to have a continuous function in a closed interval that has a local min but no abs min? I feel like no, because how can you have a local min without an abs min

[plush bramble]

You should try to prove it then

[floral gull]

well

how can you have a local min without a abs min

sorry thats not a proof at all but

I was hoping someone could just confirm my logic

well I suppose thats actually for me to do

nvm then

.close

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[pearl pond]

@elfin falcon Has your question been resolved?

[pearl pond]

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[pearl pond]

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[calm herald]

.close

[pearl pond]

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[calm herald]

<@&286206848099549185>

[abstract locust]

you haven't even asked a quetsion

[toxic lichen]

is there a reason why you think the statement is false?

right...

well

perhaps you could instead try to find some property that is true for Y1+Y2 but false for a Bin(n, p1+p2) distribution

for example, what values can each of these RVs even take?

[calm herald]

.close

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[pearl pond]

Please don't occupy multiple help channels.

[tepid current]

.close

[pearl pond]

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[pallid wasp]

idk where to start

[stark barn]

equate the horizontal forces

But you dont have the horizontal forces from the outset

so to find the Tension force of each string you equate the vertical forces

[pearl pond]

@pallid wasp Has your question been resolved?

[pallid wasp]

why r u here LOL

[stark barn]

Heyyyy

its you!

[pallid wasp]

yeah

[stark barn]

I've met clix here as well lmao

[pallid wasp]

i read ur name and i was like this guy seems familiar

[pallid wasp]

._.

[stark barn]

hehehe

Hows your specialist study going

[pallid wasp]

bad

[stark barn]

same here man

[pallid wasp]

im stuck on that question

[stark barn]

Im considering just skipping all the probability questions

[pallid wasp]

😭

[stark barn]

Look at the vertical components first

[pallid wasp]

yk i skipped the whole last section

in methods e2

bc i didnt see it 💀

oh ok

[stark barn]

oh boy

[pearl pond]

@pallid wasp Has your question been resolved?

[pearl pond]

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[teal flame]

I dont understand 23a it jumped kinda too fast

[manic arrow]

just plug in the value
and see it LHS and RHS is equal.

like if you substitute (1,7)
you get 7=3(1)+4

similarly plug in second point

and see what comes

[pearl pond]

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[teal flame]

Hi I would like to ask 23c I dont understand about it

[pearl pond]

@teal flame Has your question been resolved?

[pearl pond]

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[primal comet]

hello can someone explain what -1/3 means?

[primal comet]

is that 1 / -3?

[river hazel]

Gradient

[kind rampart]

or slope

[primal comet]

no i just mean -1/3

[kind rampart]

wait let me check

[primal comet]

like how do you calculate it

[kind rampart]

1 moment

[kind rampart]

oh it doesnt matter

they both have the same slope

$\frac{-1}{3}=\frac{1}{-3}$

[jolly parrot]

insult

[primal comet]

oh cool

thanks

.close

[pearl pond]

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[fallow yew]

What does it mean when a Probability distribution is normalized that it integrates to 1?
The context is the function of exponential family, a class of probability distribution, which will be used to model the data y

[real jungle]

@fallow yew It means that the sum of all your probabilities is equal to 1

[pearl pond]

@fallow yew Has your question been resolved?

[fallow yew]

Awesome, thanks a lot

[pearl pond]

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[halcyon tulip]

good 1

[abstract locust]

elementary engineering principal

[vivid ginkgo]

tf does that meant, I'm really confused about this. My teacher sent us this, it took me 4h still couldn't get it. My teacher give us 2 weeks to solve it.

Gave

[abstract locust]

wdym you coudln't get it

it's just a fact

[vivid ginkgo]

Whatever

Forget it, I need help on this one

[limber ravine]

You're actually serious that it's supposed to be a question?

[vivid ginkgo]

No I don't think so, is it sort of prank? I message this to my math teacher, and she sent me a laughing emoji.

[pearl pond]

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[pearl pond]

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[cold frigate]

Hi! I’m practicing for my abstract algebra midterm and I’m struggling with these true/false questions that sum up the last chapter:
———

1. If H is a subgroup of G, then <H> = H

2. The group (Q-{0},•) can be generated by a finite number of generators.

3. The subgroup of Sn generated by {(1,2),(2,3),…..,(k-1,k)} with k inferior or equal to n is ISOMORPHIC to the group Sk.

4. If H is a subgroup of G, then H is a subgroup of or equal to Cg(H)

5. The subgroup <1> of the group (Z,+) is Z itself.
   ————-
   I have already been through the course material but since I’ve been sick I can only catch up. Can someone explain to my why these would be either true of false?

[cold frigate]

<@&286206848099549185>

[pearl pond]

@cold frigate Has your question been resolved?

[cold frigate]

.close

[pearl pond]

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[clear ridge]

Hello. How to graph this exponential function? Thank you!

[pearl pond]

@clear ridge Has your question been resolved?

[clear ridge]

<@&286206848099549185>

[feral apex]

Shift the e^x graph above by 3 units and then to the right by 1 unit.

[dense ravine]

are you familiar with how coordinate axis transforms work?

[clear ridge]

Oh. Could you by any chance draw it?

[clear ridge]

Not really. But if it is going to help me to solve it I can do my research

[dense ravine]

so here's the thing

if you have (x-a) in your function

you can setup a new axis, let's call it t which is t = x-a

this means that axis t is shifted to the right by a with respect to x

so in your case t starts at x=1

and you function is now 3+e^t

does that make sense?

[clear ridge]

Well. Kind of. But I guess my teacher would be surprised where I took the t

But I am fascinated

That there is a way like that

[dense ravine]

oh it's just a new axis we defined

we defined a new axis, like x, shifted by a to the right

and next we do the same for y

y=3+e^t

y-3=e^t

and we invent a new axis, s = y-3

which is shifted by 3 on the y axis

and we get a new coordinate system s-t

in which we draw the function s=e^t

which is just the regular exponential function

here, I'll draw it real quick so you understand better

[clear ridge]

Oh. This sounds super interesting. I would like to understand that. Thank you that you will draw it

[dense ravine]

so t lies on top of x but is shifted by 1

and s lies on top of y and is shifted by 3

and now we just move them on top of each other to form a new coordinate system t-x

t-s*

[clear ridge]

Thank you. Teacher will be surprised! I will now study it.

[dense ravine]

and now you just draw the function s=e^t in your new coordinate system

as a shorthand for the future, whenever you have x-a in a function it means that the graph is shifted by a units to the right

and if you have y-b it's shifter b units to the top

[clear ridge]

You are very smart. Thank you.

[pearl pond]

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[nocturne timber]

idk how to solve

[ebon skiff]

Hi lol

Havent seen you in a while

You could complete the square and read off the vertex from there

Or you could find the vertex with some formulas

[nocturne timber]

oh hi lol

ok

[ebon skiff]

Let's try to complete the square

Just a reminder

If the function is

f(x)=a(x-d)^2+e

From here

We can directly read the vertext

It will be

S(d,e)

Ok?

[nocturne timber]

V(x) = 2×325x - 4600

[ebon skiff]

We need to complete the square to bring the function into the form from above and read the vertext from there

[nocturne timber]

for the deriative

[ebon skiff]

?

We dont need the derivative lol

[nocturne timber]

ik but

[ebon skiff]

Complete the square

[nocturne timber]

shouldnt we do all the steps

wtv ok

[ebon skiff]

Do you know what that is?

?

[nocturne timber]

we need to find x

??

[ebon skiff]

No

Complete the square

Means

"Force a square"

[nocturne timber]

where V represents the value of the home and x represents each year after 2020

[ebon skiff]

I suggest watching a video about it

Since it's preety hard from me to explain by typing

[ebon skiff]

That doesnt matter

You you need is the vertex of that function

[nocturne timber]

it does matter

[ebon skiff]

All

[nocturne timber]

if it didnt matter

it wouldnt be in the equation

[ebon skiff]

It's there to throw you off

[ebon skiff]

That describes the function

But all the quesion is asking is the vertex of that function

You dont care abou the rest

[pearl pond]

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[warm horizon]

derivative x exist at 0?

[buoyant panther]

no

[viral kiln]

Absolutely not

[honest bough]

What about conditionally? /s

[warm horizon]

why

it is not 0?

[honest bough]

The derivative is undefined whenever there's a sharp point

[warm horizon]

wait what would be the equation for derivative anyway

of |x|

[honest bough]

There's a few different ways to represent it

[viral kiln]

isnt it abs(x)/x

[honest bough]

|x|/x is my favorite way, yes

[buoyant panther]

the other way around

x/abs(x)

[honest bough]

|x|/x = x/|x|

[buoyant panther]

in fact yep

[warm horizon]

how tf r we supposed to find that

with our rules

[honest bough]

|x| = √(x²)

[buoyant panther]

or sgn(x)

[warm horizon]

Oh shit ur big brain

[honest bough]

You can also find the derivative using the piecewise definition

[viral kiln]

its one of those ones u remember tbh

like d/dx ( lnx ) = 1/x

[warm horizon]

rip

.close

[pearl pond]

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[lime ginkgo]

How do I find the derivative

of this

.close

[pearl pond]

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[strange lantern]

hello

[lyric elbow]

$\int_{3}^{2} = -\int_{2}^{3}$

[jolly parrot]

『Marius』

[lyric elbow]

this is how you change the intervals right?

[strange lantern]

I need help with this

[lyric elbow]

dude

[strange lantern]

yes

[feral sedge]

Yes

[feral sedge]

And this isn't your help channel

[lyric elbow]

we send it in the same time

[feral sedge]

You're the one interrupting

Channel name is not yours

[strange lantern]

I was here before

[feral sedge]

Assuming they're independent

[strange lantern]

yes

[feral sedge]

So you need P(A and B) when A and B are independent

[strange lantern]

correct

[feral sedge]

in that case you can multiply P(A) and P(B)

[strange lantern]

yes

is that it

@feral sedge

[feral sedge]

Yes

[strange lantern]

Find the probability that BOTH alarms will work.
Find the probability that AT LEAST ONE of the alarms work.

[feral sedge]

P(A and B)=P(A)*P(B)

[strange lantern]

there is two parts to the question

[feral sedge]

That's the first part then

[strange lantern]

ok

[feral sedge]

actually

Maybe you should do like a tree diagram thing

[strange lantern]

can you help me solve this question

[feral sedge]

https://youtu.be/mkDzmI7YOx0

Like this kind of thing

There's probably a better video on it somewhere

[strange lantern]

.close

[pearl pond]

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[warm horizon]

i thought the 1st and 4th are true but it said no

[copper kestrel]

this is the graph of f'

so the local mins and maxes are only when the derivative changes signs

[pulsar lark]

Is f(6) absolute max?

[copper kestrel]

uhh no?

I don't think so at least because we don't know the actual graph of f right

[pearl pond]

@warm horizon Has your question been resolved?

[warm horizon]

OHHHHH

Im stupiddd

I didn't see the `

[pearl pond]

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[junior pawn]

$[\frac{u^{-n+1}}{-n+1}]^2_1 = \frac{2^{-n+1}}{-n+1} - \frac{1^{-n+1}}{-n+1}$

[jolly parrot]

Max.cc

[pearl pond]

Please don't occupy multiple help channels.

[junior pawn]

am i right?

[grim patrol]

errr

is this antiderivative

wrt u

#❓how-to-get-help

[junior pawn]

x if you want it doesn't matter here

[grim patrol]

no i mean

idk like

weve 2 vars, u and n

anw

u just replaced all u's with the vals

so it shld b wrt u

looks fine

[junior pawn]

hum

i asked the question because when i calculated this integral it gives me :

the opposite

https://www.integral-calculator.com/

that's the integral

when i'm putting 1 and 0 instead of x i just don't have that so i don't get it

[lunar trench]

Both are equal

[junior pawn]

wait

$\frac{2^{-n+1}}{-n+1} - \frac{1^{-n+1}}{-n+1}$

[jolly parrot]

Max.cc

[junior pawn]

those 2 expression are equal?

[lunar trench]

Yes, and the one you said it was an opposite

[junior pawn]

ok thc

.close

[pearl pond]

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[marsh crater]

How do I prove that Haar's measure on the borel sigma algebra generated by the unit circle with subspace topology is unique?

[pearl pond]

@marsh crater Has your question been resolved?

[pearl pond]

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[pearl pond]

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[mystic pebble]

i need help visualising and doing these questions pls

[pearl pond]

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[mystic pebble]

oop

.close

[pearl pond]

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[vital bluff]

need help with this. complex numbers. I only need the answer because i want to compare it with mine

[glacial sequoia]

What's the question?

[shrewd basin]

,w solve |z+1-i| < sqrt(2)

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Pure#9389

[shrewd basin]

What are you meant to do

Sketch this?

[vital bluff]

yeah

[shrewd basin]

It’s the region inside a circle centred at -1+i with radius sqrt2

Not including the border

[vital bluff]

thanks

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[pearl pond]

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[midnight haven]

i got A'U B' = U

but its wrong

[sharp quest]

there are multiple correct answers

[midnight haven]

i workeed out a and b

but i cant work out c

is c coreect?

[sharp quest]

the last option?

[midnight haven]

yess

.close

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[broken delta]

hey

[broken delta]

is the derivative of f(x) = h(g(x)k(x))

this:

f'(x) = h'(g(x)k(x))(g'(x)k(x)+g(x)k'(x))

[plush bramble]

yes

[broken delta]

okay thanks

.close

[pearl pond]

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[broken delta]

.reopen

[pearl pond]

✅

[broken delta]

then if I was to find f'(1)

i would replace every x with 1

so:

f'(1) = h'(g(1)k(1))(g'(1)k(1)+g(1)k'(1))

from this formula

f'(x) = h'(g(x)k(x))(g'(x)k(x)+g(x)k'(x))

[plush bramble]

yup

[broken delta]

cool thanks again

.close

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[low jungle]

m=p^4 , p^5q=2, express m in terms of q

[low jungle]

I'll try to figure out textit but yeah I don't know how to solve that ^

[sharp quest]

solve your second equality for p
then substitute it into the first equality

[low jungle]

wait hold on i wrote my question wrong

[low jungle]

hi, i don't believe this applies anymore does it?

[sharp quest]

why not? if you know what p equals in terms of q, then sub it into the first equality, you have m in terms of q

[low jungle]

let me try one sec

okay i got it haha

thanks

.close

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[pearl pond]

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[midnight haven]

what i can do after that

to verify if cos2a(1+tana.tan2a)=1

[pearl pond]

@midnight haven Has your question been resolved?

[coarse citrus]

wait, im doing it now

almost there, just need a picture if you still need the answer @midnight haven

[midnight haven]

ok

i need

[coarse citrus]

ok tell me if anything is unclear

i went through a different path because your proof went into a wall

[midnight haven]

its unclear

[coarse citrus]

the picture lol?

[midnight haven]

yes

[coarse citrus]

ok let me retake

my camera is not good

[midnight haven]

no problem

[coarse citrus]

i broke it up into two

upside down though sorry:(

[midnight haven]

;rotate

,rotate

[midnight haven]

,rotate

[midnight haven]

ok thank you

.close

[pearl pond]

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[vapid trench]

In the first picture the first graph is supposed to be the same as the second pne

idk y its not the same since its just a long divided version

i had to long divide since i wasnt getting the correct answer without

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[pearl pond]

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[vivid solar]

i’m stuck proofing this😓

[mental bloom]

What is sin90

Its 1 🙌

[vivid solar]

ah omg i don’t know why i didn’t think to simply them like that

thank you!

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[pseudo gulch]

hi guys, how can i evaluate this limit without using l'hopitals rule?

[pseudo gulch]

i literally dont know where to start. i know lim x-> 0 sinx/x is 1, but this is lim x-> infinity

[midnight haven]

use the substitution u=1/x

[pseudo gulch]

how does that help?

[midnight haven]

you should get something that looks familiar

[pseudo gulch]

yeah but its limit to infinity

not 0

[tacit mulch]

If you make that substitution, then u will approach 0 as x approaches infinity

[pseudo gulch]

oh ok

but our teacher didnt teach us that yet

is there no other way?

[noble sigil]

https://www.youtube.com/watch?v=BKD5Rtylk1A

[tacit mulch]

That's okay, you're allowed to get creative

Substitutions like that are very common, you probably did things like that in earlier classes