help-10

[dry echo]

but I can't seem to figure out how to get -1 from this...?

[wise talon]

chain rule

[unreal musk]

What happens when you differentiate the insides wrt x?

[wise talon]

⛓️

[dry echo]

OH

WAIT

I FORGOT 1

LMAOO

once again

my genius frightens me

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[signal hollow]

Could someone tell me the process of how to solve this? I've watched several videos and still don't completely understand. As you can see in the attachment, I get stuck there.

[signal hollow]

in some examples that I see, they subtract the arccosine from pi then multiply by u and then also multiply arccosine by u without subtracting

[latent walrus]

replace u with what it originally was and rearrange for t
there will be 2 values for t that are less than 365

or you could do it before, say u=arccos(1/2.9), 2pi-arccos(1/2.9), 2pi+arccos(1/2.9),... and then solve for t

[signal hollow]

wait where did you get the 2pi

[latent walrus]

,w graph y=cos(x) and y=0.3 from 0 to 6pi

if theres a solution at say idk x=0.1 then there will also be one at x=2pi-0.1

[warm shale]

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[signal hollow]

ohh okok I kinda get that. But why does it, in some examples, subtract from pi instead of 2pi?

[latent walrus]

that im not sure about - that would generally apply if it was sin rather than cos

,w graph y=sin(x) and y=0.3 from 0 to 6pi

[signal hollow]

ok so for cos it's 2pi cause there's two intercepts in one period but pi for sin cause there's only one?

[latent walrus]

theres still 2 for sin. if theres a solution at x=0.1 then theres another at x=pi-0.1

as you can see the line still intersects at 2 points

[signal hollow]

so I do that for arccos

for arcsin it would be pi

instead of 2pi?

[latent walrus]

basically, it may be clearer if you graph them, i can extend the graphs over a few periods

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[latent walrus]

,w graph y=cos(x) and y=0.3 from 0 to 6pi

[warm shale]

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[latent walrus]

sin and cos are repeating functions, so adding 2pi to any solution is also a solution

but the stuff i mentioned before is what you would look at in a single period

you shouldnt need to go beyond 2pi for this question though (when finding u)

[signal hollow]

okay

i kinda get it now

thank you sm

[latent walrus]

np

it might help to play around with it a bit to get used to it

[signal hollow]

yeah, looking at the graphs helped too

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[frosty spoke]

I'm feeling really lazy right now, but I have a gamer argument with some people, and I need to know:
Suppose we have X_1, X_2, ... iid success/failure random variables with probability p
what is the probability of a run of length M in X_1, ..., X_N?

This almost answers it for me, but it appears to only treat the cases where p = 1/k for k natural
https://www.gregegan.net/QUARANTINE/Runs/Runs.html

[fathom flicker]

Sounds like you could use binom.dist in excel

?

[frosty spoke]

okay yeah turns out the question in the game is simpler than this, so I don't need the answer

[drifting wraith]

the answer is the same, it's only a calculator limitation

[frosty spoke]

but it's not actually a binomial distribution thing, because runs have to be contiguous

[drifting wraith]

T(n) = (1-T(n-m-1)) * (1-p) * p^m + T(n-1)

you just calculate that one by one

sum of (....) no (yes^m) with .... not having a streak or (....) (anything^1) with ... having a streak already

[frosty spoke]

you mean something like this recurrence?

but wiht ds replaced with ps, and no normalization?

[drifting wraith]

i don't know, yes?

[frosty spoke]

ill have to think about this whole thing more lol

ty

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[obtuse pebble]

Please don't occupy multiple help channels.

[timid silo]

I don’t understand what to do from here

[twin sapphire]

6-3=3

then simplify

[timid silo]

How can I simplify it?

Do I divide?

[autumn merlin]

yes divide 12 by 3

[timid silo]

Can I multiply the roots by 3 as well?

DIVIDE*

[twin sapphire]

$12\sqrt{3}/3 = 4\sqrt{3}$

[warm shale]

Benjamin

[timid silo]

My answer is $4sqrt6 + 4sqrt3$

[warm shale]

1_00_52

[timid silo]

@twin sapphire

[twin sapphire]

yeah ok

you can factorize if you want

$4\sqrt{3}(1+\sqrt{2})$

[warm shale]

Benjamin

[timid silo]

So if I have smt like this

Am I supposed to divide the 2 by the 2

Or leave it like that

[twin sapphire]

$\frac{a \times b}{c} = \frac{a}{c} \times b$

[warm shale]

Benjamin

[obtuse pebble]

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[tropic patio]

does someone know how to get the values of a b and c

[trail cloak]

Hmm

[tropic patio]

by making x to something i cant find A

[trail cloak]

The left side is a 3rd degree polynomial

And the right side is a second degree polynomial

[tropic patio]

yes

it comes from this

[trail cloak]

Oh

[tropic patio]

that becomes into this

[trail cloak]

You gotta carry polynomial division first

Both the numerator and denominator have the same degree

[tropic patio]

ok

[trail cloak]

You have to make the numerator at least one degree lower than the denominator's degree

[twin sapphire]

this is equal to $1 - \frac{4}{x^3-2x^2}$

[warm shale]

Benjamin

[trail cloak]

Hmm...

[twin sapphire]

and now do partial fraction on this easier one

[trail cloak]

Yep

[tropic patio]

oh

i was adding them

my bad

thx

[twin sapphire]

wdym?

partial fractions are added to get the original fraction yeah

thats why we use them for integration

[tropic patio]

well i was doing the substraction

[trail cloak]

You didn't carry out the long division first though

[twin sapphire]

yep

[trail cloak]

Which is why you got stuck

[twin sapphire]

essentially i did "short division"

[trail cloak]

Synthetic?

[tropic patio]

you took out the four

[twin sapphire]

because long division is just numerator = 1*denominator - 4

which is short

[tropic patio]

and then cancelled the left side

[trail cloak]

Oh ye

Algebraic manipulation basically

$$\frac{x^3 -2x^2}{x^3-2x^2} -\frac{4}{x^3 -2x^2}$$

[warm shale]

VulcanOne

[tropic patio]

yes

i did that but by mistake instead of separating the integral i did the substraction

thank you so much

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[vapid badger]

i need help with b.iii

This is what I did

[obtuse pebble]

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[steep snow]

I get answer b why answer paper write is d

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[shadow cave]

Have I made any mistakes from a-d?

[obtuse pebble]

@shadow cave Has your question been resolved?

[obtuse pebble]

@shadow cave Has your question been resolved?

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[shadow cave]

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[sage dagger]

when performing completing the square is it recommended to put it into fraction form

[sage dagger]

or does this work fine as welll

[frosty spoke]

what's fraction form

you mean the 3.5 as 7/2?

[sage dagger]

do we do that instead of 3.5

[sage dagger]

yes

[frosty spoke]

3.5 and 7/2 mean the exact same thing

it's like asking whether you can write "automobile" in the place of "car"

they're literally synonyms

[sage dagger]

oh i was just wondering because i was watching a yt video that was using the fraction instead of the 3.5 but makes sense then

[frosty spoke]

and I mean this in the notational sense

not in that it works out mathematically

like 7/2 and 3.5 refer to the same thing

so you will never encounter a situation where writing 3.5 gives you a different answer than writing 7/2

well, okay maybe if some people redefine division or whatever

but not at your level

[sage dagger]

does this look right

[rustic drum]

i think the square must write out of the bracket

[sage dagger]

oh right

yeah

but apart from that

its correct?

[rustic drum]

it this completing the square ?

[sage dagger]

yes

[rustic drum]

correct

[sage dagger]

also when completing the square during the halfing process

do u just half the number only or do you also have to include the negative or positive sign

[timber island]

Well it doesnt matter, cus anyways you are doing both subtraction addition to make sure the value doesnt change

[sage dagger]

oh yeah its squared anyway

it wont change

does this look right?

[timid silo]

welp if you add 3^2 on one side

you'd have to add it to the other

also, would probably recommend you to just divide the whole equation by 3 from the very start

[obtuse pebble]

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[opaque galleon]

can someone explain to me how the bounds was changed?

[unreal musk]

When x=e^4, you get sec(theta)=2, hence cos(theta) = 1/2, corresponding to theta = pi/3

Similarly x = e^2 gives sec(theta)=1, cos(theta)=1 and theta=0

As per your choice of substitution ln(x) = 2 sec(theta)

[obtuse pebble]

@opaque galleon Has your question been resolved?

[opaque galleon]

wait

[opaque galleon]

like this?

[unreal musk]

Yep, basically!

[opaque galleon]

So we are changing the bounds to angles because we are rewriting in terms of theta

I don't understand why we do it this way though

can someone explain to me please

what is this part called so I can search it up

[unreal musk]

Erm, probably something like "u substitution change of bounds" would do? or "u substitution definite integrals"

[opaque galleon]

we are studying trig substitution, I understand that we have to change the bounds to an angle because we rewrite the expression in terms of theta but I don't understand why we are doing it this way

okay so u = lnx

then since we are doing trig sub, let lnx = 2sec(theta)

then get the derivatives of both sides

we have

1/x dx = 2 sec(theta)tan(theta) dtheta

ok now we have to change the bounds

u = lnx = 2sec(theta)

we plug in e^2 to lnx

then solve like I did awhile ago

so when changing the bounds, we plug it in to the u we chose

then solve for theta?

[obtuse pebble]

@opaque galleon Has your question been resolved?

[unreal musk]

In this case "u substitution" is a tiny bit misleading, and in this situation it's rather "theta substitution"

[unreal musk]

Which you're doing as per here

[unreal musk]

You plug the $x$ bounds into the expression you're working with, in this case $\ln(x) = 2\sec(\theta)$, and in this case the $x$ bounds are $x = e^{4}$ and $x = e^{2}$

[warm shale]

@unreal musk

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[worn siren]

can someone explain this step?

i feel like im missing something but i dont get the jump

[high lily]

factored out e^-x

then simplified

[worn siren]

sorry could you explain it again?

you're left with

(Sinx-cosx)-(sinx+cosx)

[high lily]

i said factor out

not erase / wipe from existence

[worn siren]

yeyeye okok

yep got it

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[worn siren]

ty

[obtuse pebble]

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[fast turtle]

If I try to solve this I get 3√3x = -6√2
I'm stuck
How to solve the x?

[naive fossil]

What have you tried so far?

[high lily]

note that 3sqrt(3) is just a number

as is -6qsrt(2)

3√3x = -6√2

that isn't that different from something like

7x = 2

would you be able to solve for x in that simpler equation?

[fast turtle]

yes

[high lily]

how would you get the solution to that?
(I'm not asking for what the solution is)

[fast turtle]

Bring the Coefficient to the other side

[high lily]

bring how?

that's very vague/poor wording that's open to interpretation

[fast turtle]

7.X becomes :7

[high lily]

that's just as bad

wdym by becomes :7

actually, can you tell me what the solution would be?

[fast turtle]

V = 2/7

[high lily]

you mean x?

to concisely unambiguously describe to to get
x = 2/7
the action performed can be described as dividing both sides of the equation by 7

$$\red{7}x = \blue{2} \implies x = \frac{\blue{2}}{\red{7}}$$
$$\red{3\sqrt{3}}x = \blue{-6\sqrt{2}}$$

[warm shale]

ℝamonov

[fast turtle]

Oh so that's how you learned it we don't have to divide both sides

[fast turtle]

so there is no solution for x here?

[high lily]

wdym

Oh so that's how you learned it we don't have to divide both sides
that's what you're supposed to be doing and what you are effectively doing with all that vague
"move", "bring"

i even coloured the coefficients / constants in red and blue for you

in an attempt to indicate that these two equations aren't that different

(red)*x = (blue) → x = (blue)/(red)
in the 7x = 2

is in the exact same form as the equation you have

[obtuse pebble]

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[sacred wolf]

integrating factor of a non-exact de where dM/dy-dN/dx is a function of both x and y

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<@&286206848099549185>

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[timid silo]

hey

[timid silo]

so i'm not getting that part ; C*cos(wt+ p) = Acos(wt+p) + Bsin(wt+p2)

To rewrite a sin(t)+b cos(t) as a single cos function, you can use a trigonometric identity that states:

a sin(t)+b cos(t) = R cos(t-C)

where R = sqrt(a2+b2) and C = arctan(b/a).

i don't know the intuition for this trig thing

https://www.youtube.com/watch?v=a93rk9B0AsU&ab_channel=DanielAn

i'm not getting it from 12:00

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[little shuttle]

hey

I already have the answer but I think I did it a pretty bulky way

if someone could show me the most concise way to complete the problem that would help

[golden oriole]

How did you do it ?

[little shuttle]

chain rule and wayy too much factoring

[golden oriole]

Assume that it is equal to y

Then divide both sides by -2 and then take the secant

Now you can do implicit differentiation

[little shuttle]

so now its secarcsec which cancels out?

I dont understand

[golden oriole]

Yes

It works for regular trig functions im sure
Not 100% sure it will go well with reciprocal functions

[little shuttle]

so now its y=x+1?

if -2 and arcsec are factored out?

[golden oriole]

No no

Its
Sec(-y/2) = x + 1

[little shuttle]

oh ok

where do I go from there?

[golden oriole]

Implicit differentiation

[little shuttle]

ok thanks

-solved

.close

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[obtuse pebble]

Please don't occupy multiple help channels.

[gleaming wagon]

Does anyone know how to answer thiss?

[tranquil sonnet]

Notice that STU is isosceles

[gleaming wagon]

Ohh wait nvm

I see it

since the equalateral triangle has equal sides

[tranquil sonnet]

ST=SU

[gleaming wagon]

so SU is the same as SR

and ST is the same as SU

[tranquil sonnet]

Yup

[gleaming wagon]

thank you!

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[timid silo]

have no idea how to do this

we aren't using trig functions either

[rain forge]

do you know sinus rule or just cos = adjacent / hypothenuse

[timid silo]

she didn't teach it to us like that at all

[rain forge]

what do you try first

[timid silo]

she used pythagoras

i think

im not sure i dont remember too well

[rain forge]

what is your course

[timid silo]

geometry

[sour lance]

think about an equilateral triangle cut in half.

try to see how that relates to this problem and use the pythagorean theorem

[timid silo]

umm

wait nvm

.close

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[frail shell]

how do I start to solve this?

[gleaming wagon]

Are you trying to differentiate it?

[frail shell]

yes

[gleaming wagon]

what do you have so far?

[frail shell]

nothing

I forgot how to differentiate

im starting my calc 2 class

[gleaming wagon]

Do you remember the chain rule?

[frail shell]

I can watch over it again on youtube

[fierce lagoon]

,tex \dxchainrule

[warm shale]

Umbraleviathan

[gleaming wagon]

You would differentiate the whole thing and times it by the derivative of it's parts

Sorry I'm really bad at explaining

[frail shell]

its okay

I can try to figure it out

[obtuse pebble]

@frail shell Has your question been resolved?

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[timid silo]

hello where is the physics channel

[timid silo]

whyd it get pinned 😭

[tardy epoch]

#old-network

[timid silo]

ty

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

ig

[obtuse pebble]

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[next wagon]

Hello, how can i tridiagonalise a 3x3 matrix ?

[next wagon]

.close

[obtuse pebble]

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[next wagon]

Hello in the case of a 2x2 non diagonalisable matrix, lets call it A, my objective is to do the next best thing which is to Tridiagonalise it

[next wagon]

So we have 1 double eigenvalue, lets call it v1, and 1 eigenvector called u1

Lets call the matrix A

So we have A* u1 = v1* u1

We can find a vector u2 such as u1 and u2 are linearly independent, so (u1,u2) is a basis of R^2

[mighty geyser]

what do you mean, tridiagonalise it

[royal shard]

tridiagonalise is basically an upper right triangle

but the diagonal does not have to be 0

[next wagon]

Now we can write A * u2 = a* u1 + b* u2, a and b being reel numbers

Ok finally set up everything, here is my issue now

[royal shard]

you have u2 twice there

[next wagon]

Yep that isn’t a typo

[mighty geyser]

more like schur decomposition but carry on

or jordan normal form

[next wagon]

What i’m saying is we can write u2 as being a linear combination of both vectors of the basis (u1,u2)

[royal shard]

👍

[next wagon]

In my lesson, they state that b must be = v1 otherwise the matrix would be diagonalisable

And i don’t understand why

They say otherwise A would contain b as an other eigenvalue that’s associated to an eigenvector that looks like u2 + c * u1, for a particular c

[mighty geyser]

generalised eigenvalues it seems

[next wagon]

I don’t know what that is ?

[mighty geyser]

hmm, let's suppose it's different and see what happens

$u_2+ku_1=v_1ku_1+au_1+bu_2$\
can you find some k such that $v_1k+a=bk$

[warm shale]

Element118

[mighty geyser]

(basically, what I'm doing is applying the matrix A to get from the left hand side to the right hand side)

[next wagon]

Is that supposed to be u1 or v1 ?

Right side of the =

[royal shard]

honestly, im so confused by this
my intuition tells me that when we look at
u2=a u1 + b u2
then a=0 and b=1 are a must, since they form a basis and are thus linearly independent
b=v1 would either imply that there is a second solution or that v1=1 which is not always the case

[next wagon]

I’m not sure i understand how you got this equation ?

[next wagon]

Ohh sorry i forgot to multiply by A on the left side

Let me correct it

Done

[royal shard]

A * u2 = a* u1 + b* u2

[next wagon]

Did you mean to multiply the left hand side by A too ?

[royal shard]

and then b must be v1 you say

[next wagon]

If the matrix isn’t diagonalisable yes

[next wagon]

Otherwise we apparently get this but i don’t get why

[royal shard]

so

we can form a jordan normal form

you know what that is?

i think that helps here

[next wagon]

No i don’t ?

[royal shard]

hmm

[dark stirrup]

I think I used a Jordan form to solve one of your problems a while ago, @next wagon.

Not 100% sure tho

@royal shard is this an example?

[royal shard]

if 2t is the eigenvalue, then i would say so

[next wagon]

Just looked it up and this shows up

[royal shard]

i have not yet done matrix exponent though

[next wagon]

Welp never mind

Was a very specific question

[dark stirrup]

Okay, what was the original question?

I think, thanks to @royal shard, I may be able to answer now

[next wagon]

This

[dark stirrup]

I do not follow this question

[next wagon]

I’ll try and put it differently

A * u2 = a* u1 + b* u2, a and b being reel numbers

A * u1 = v1 * u1

If b wan’t equal to v1, we could find an other eigenvalue and therefore A would be diagonalisable

So b must be equal to v1

So A * u2 = a* u1 + v1 *u2, a being reel number

Why does b have to be = to v1 ?

[dark stirrup]

How does this work? Isn't lhs vector and rhs scalar?

[next wagon]

Lhs ? Rhs ?

[next wagon]

That’s just the definition of an eigenvalue

U1 being an eigenvector associated to v1

Wait i just figured it out

But only partially

Still don’t get why we have this : They say otherwise A would contain b as an other eigenvalue that’s associated to an eigenvector that looks like u2 + c * u1, for a particular c

I’ll write it down

[dark stirrup]

"lhs" means "left-hand side". In your case: A*u2. This should be a vector, right?
"rhs" means "right-hand-side": v1 * u1. I'm assuming these are both vectors, so their product would be a scalar, right?

[next wagon]

V1 is a reel number

U1 is a vector yep

So the product is a vector

Hope this kind of makes sense

Just for reminders

[dark stirrup]

ah ok

[next wagon]

It’s our eigenvalue

[dark stirrup]

Yeah I'm completely lost, sorry. I'm at work, so I'm useless right now

[next wagon]

Haha no worries just discussing the issue and formulating a question to ask helped me answer part of my question 😇

[dark stirrup]

awesome

That's called "rubber duck" method

[next wagon]

Oh ok ! Didn’t know about it

Good luck at work and thx for the help as always, love you man ❤️ @dark stirrup

@royal shard thx for your help too 🙂

Take care guys !

.close

[obtuse pebble]

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[earnest field]

Helllllo

[earnest field]

Number 5

I’m confused, I tried to do it and it isn’t making since

It’s not supposed to have a decimal and 240/9 is 26.6 or something

So I’m def doing something wrong but idk how to do this haha

[royal shard]

Length/width stays constant
Therefore also width/length stays constant

Lets use width/length

In the first picture we see 9/12

And in the second?

[earnest field]

X/20

[royal shard]

Yes

Since it stays constant, we set them equal

So 9/12=x/20

[earnest field]

And then?

Cross multiply?

[royal shard]

Rearrange go get x

[earnest field]

?

OH WAIT A

[royal shard]

We have
9/12=x/20
We want to find x, that means we want to isolate x
This is also called rearranging for x
In our case the only thing that is left is this "/20" which means divide by 20

[earnest field]

I got it I think

[royal shard]

What is your result?

[earnest field]

15=X

[royal shard]

Great

[earnest field]

12/9 and 20/15 are both 4/3 so that means it’s right, right?

[royal shard]

Yes

[earnest field]

Thank you so so much! I might need some more help. I missed a few days of school due to being sick so I am trying to catch up and I’m missing patches so I’m half getting it haha-

Does this look correct?

[obtuse pebble]

@earnest field Has your question been resolved?

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[marsh sentinel]

So $6^{30}+4=6\cdot (6^{29}) + 4$

[marsh sentinel]

Then for which q and r $6^{30}+13=6\cdot (q) + r$

and r needs to be between 0 and 6

[warm shale]

weegee

weegee

[obtuse pebble]

@marsh sentinel Has your question been resolved?

[obtuse pebble]

@marsh sentinel Has your question been resolved?

[obtuse pebble]

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[fierce lagoon]

Notice how you can factor out x^2 first

You don't even need to do any arduous division after that

[obtuse pebble]

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[sterile bone]

can anyone help me w this please

[nocturne minnow]

Find the roots of that function

You can do factor by grouping

[sterile bone]

.close

[obtuse pebble]

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[nocturne minnow]

Fyi, when the remainder is 0, that means it's a root of the function

[obtuse pebble]

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[brittle dune]

how do I use the vc

[obtuse pebble]

@brittle dune Has your question been resolved?

[obtuse pebble]

@brittle dune Has your question been resolved?

[forest sinew]

you dont, its gone

[brittle dune]

huh

[forest sinew]

there were moderation problems

[brittle dune]

then what do I do now

[forest sinew]

do?

[brittle dune]

yes how do I get help

[forest sinew]

#❓how-to-get-help

just ask

[brittle dune]

but

[forest sinew]

id make a new channel though

[brittle dune]

VC is better

[forest sinew]

well i cant help, sorry

[brittle dune]

sadge

[forest sinew]

isnt up to me

i just post the cats

[brittle dune]

what if I ring up the mods

[forest sinew]

ah you can do this

message modmail

[brittle dune]

you really think so?

how?

[forest sinew]

i dont think theyll change their mind

its top of the list ->

[brittle dune]

I'll form an alliance

[forest sinew]

if you do message modmail, you should make your entire message one line

.close

[obtuse pebble]

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[forest sinew]

gl

[obtuse pebble]

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[near elk]

Ask

[near elk]

A .3, 7,15 ___
B. 9, 7, 5 ___
C. 2, 12, 72 ___
D.3 9 27 ___

hi it says to find the next number as pattern

Hoping for help, I will write my answer down

A. 31

B. 3

C. 432

D. 81

[fierce lagoon]

Yeah they look fine

[obtuse pebble]

@near elk Has your question been resolved?

[near elk]

D. 3 9 27 ___ << 81 was incorrect

33 is correct answer by the teacher but I dont understand

[obtuse pebble]

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[near elk]

.reopen

[obtuse pebble]

✅

[nocturne minnow]

33 isn't even a choice that you gave...

Oh wait, my bad

I thought those were answer choices

[near elk]

🙂

sadly i dont understand why teacher says 33 is the correct answer

D. 3 9 27 ___ << 81 was my answer by teacher marked it as incorrect
33 is correct answer by the teacher but I dont understand

[nocturne minnow]

You could ask your teacher on what crazy logic they had

[near elk]

hehehe

[obtuse pebble]

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[near elk]

thank you\

[obtuse pebble]

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[timid silo]

can someone step me thru this please im having trouble understanding it?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

please

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[mortal sun]

ignore the 2nd one

but why is my 3rd one wrong?

[hallow plaza]

You didn’t use tan

[mortal sun]

isnt anti derivative of cos -sin

[fierce lagoon]

Your answer to the second one is funny

Also the antiderivative of + cos(x) is sin(x)

So it should be + 5sin(t)

[keen gust]

im p sure it's also ct not X lol

[mortal sun]

right right

[fierce lagoon]

You also uh

[mortal sun]

lol let me try that again

[fierce lagoon]

Introduced x

Dt

Instead of Dx

[mortal sun]

now i got it right...lol

thanks guys

i think its enough math for today

.close

[obtuse pebble]

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[hallow plaza]

I need help with questions a and b algebra 2. I don’t know where to start

[verbal pulsar]

it will stop at s = 0

[obtuse pebble]

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[gentle shard]

How do I solve this? I did long division and got x^4 + 4x^4 + 16 + 64/(x^2 - 4) but idk what to do from here

[trail musk]

it's just asking for the form

not for the numerical values of the coefficients

[gentle shard]

ik, how do I find the form?

[fathom flicker]

can the bottom be factored into linear factors?

[trail musk]

even simpler

[gentle shard]

yea but the top isn't linear

[fathom flicker]

you have to continue applying long division until the degree of the top is less than the degree of the bottom

then you can do partial frac setup

[gentle shard]

already did, I got x^4 + 4x^4 + 16

how would I do the partial frac setup though?

[fathom flicker]

x^4 isn't lesser in degree than x^2

[obtuse pebble]

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[wary flame]

integral from -3 to 0 of f(x) = sqrt(9-x^2)

[wary flame]

wait

[devout sable]

[wary flame]

i swear give me 30 seconds

-_-

ok it's just 9pi/4 right

lmfao

[devout sable]

,w integral from -3 to 0 of f(x) = sqrt(9-x^2)

[warm shale]

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[devout sable]

[wary flame]

i forgot about circles :))))

[devout sable]

[wary flame]

thank you anyways

is that ur cat

.close

[obtuse pebble]

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[fiery iron]

Is my proof correct?

[fiery iron]

This is graph theory

<@&286206848099549185>

[obtuse pebble]

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[patent bronze]

I don't understand. My answer is 74cm^2, but it's not one of the options.

[patent bronze]

Can I make a parallelogram without mirroring the shape? If so, how? Maybe that is what the question is asking

If not then maybe I should go with the next largest option that works, which is B I think

What should I do lol 😅

[sonic anchor]

idk how you got an answer that is cm^2

is asking for perimeter

there are only 3 ways to construct it i think you can go over all of them

[obtuse pebble]

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[opaque galleon]

is this allowed?

[timid silo]

no

,w
(5+4x-x^2)^(5/2) = -(x^2-4x-5)^(5/2)

[warm shale]

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[timid silo]

only works for two values

the equality does not hold in general

[obtuse pebble]

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[wicked current]

how to solve these ?

[royal shard]

basically rearranging it

[wicked current]

how ?

[royal shard]

$x^{a\cdot b}=(x^{a})^b$

[warm shale]

~Martin

[royal shard]

this will help us

$x^{-\frac{4}{3}}=(x^{-1})^{\frac{4}{3}}$

[warm shale]

~Martin

[royal shard]

and we know that if we take a fraction to the power of -1, the numerator and denominator change places

so we are left with

$\sqrt{27}^{4/3}$

[warm shale]

~Martin

[wicked current]

let me try

[royal shard]

hint: to solve further, try to write the square root as an exponent instead of a root

[wicked current]

okat

what now ? @royal shard

[royal shard]

1 to the power of anything will still be 1

so we are left with

$\frac{1}{27^{-4/6}}=27^{4/6}=27^{2/3}$

[warm shale]

~Martin

[wicked current]

but theres no option like that

[royal shard]

$a^{b/c}=\sqrt[c]{a}^b$

[warm shale]

~Martin

[royal shard]

applying this, we get

[royal shard]

yes

now, we can also pull out the 2

$27^{2/3}=\sqrt[3]{27^2}=\sqrt[3]{27}^2$

[warm shale]

~Martin

[royal shard]

so we now need the third root of 27

this works, because taking roots is basically the same as multiplying exponents together, and those are commutative

$\sqrt[b]{a}^c=a^{\frac{1}{b}\cdot c}=a^{c\cdot\frac{1}{b}}=\sqrt[b]{a^c}$

[warm shale]

~Martin

[wicked current]

yes

[royal shard]

as we have perfect squares (1, 4, 9, 16 and so on), we also have perfect cubes

27 is one of those

1^3=1
2^3=8
3^3=27

[wicked current]

so,

9 ?

[royal shard]

$\sqrt[3]{27}^2=\sqrt[3]{3^3}^2=(3^3)^{2/3}=3^2=9$

correct

[warm shale]

~Martin

[royal shard]

👍

[royal shard]

this is the math behind it, but normally we skip the third thing there (the (3^3)^2/3)

[wicked current]

let me try some more. Thanks

.close

[obtuse pebble]

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[forest crypt]

Chess is math too right

[forest crypt]

I'm so bad, can you give me tips?

[royal basin]

these stats tell us nothing

[royal shard]

uhm

16 and 37 blunders?

i dont think math is what helps you the most

if you know someone who plays chess, better look at how he plays and let him look how you play

that will be more productive

[royal basin]

^

[fiery birch]

No, it isn't , however the programs used to calculate whether a move is good or not does use maths. To get good at chess, you need to practice and be able to see multiple moves ahead , sometimes in less time if you are playing a shorter version of the game. However, don't think that being good at maths will make you a great chess player. It won't.

@forest crypt

[obtuse pebble]

@forest crypt Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[prisma moat]

i need help finding the expected value of X

[prisma moat]

nvm i figured out my mistake

.close

[obtuse pebble]

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[haughty coyote]

use power series instead if you can

[prisma moat]

oo that seems interesting

how exactly?

havent gone over an example where i would use power series yet

[haughty coyote]

the sum of the (k+1)x^k is the derivative of the sum of the x^k+1

which is known

[prisma moat]

ooh and we have n(2/5)^n

[haughty coyote]

so the sum of the k x^k is that - the the sum of the x^k

or you factor one x out

to say it's x sum of the k x^k-1

which is the same up to an index shift

[prisma moat]

like this?

[haughty coyote]

,rotate

[prisma moat]

i need to review derivatives in infinite series

[haughty coyote]

you mixed the 2 methods together into a non-method I think

[prisma moat]

hmm

[haughty coyote]

the 2nd one is simpler btw

[obtuse pebble]

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[fiery birch]

What is the issue here? @fresh acorn

[fresh acorn]

you have to draw the worksheet

idk how too

[fiery birch]

Do you know how to use Excel?

[fresh acorn]

yes

but they ask us to draw in paper

[fiery birch]

Where? In the above question, it says to enter

[fresh acorn]

isn't the following transaction the pic?

[fiery birch]

No it's telling you to enter these transactions into a worksheet and then calculate profit or loss

[fresh acorn]

the thing is....

i just studied accounting... I don't know shit

💀

which one is supposed to be debit or kredit

[fiery birch]

Neither did I study finance , but let's see this

The first one is telling that capital was injected which means it's a debit as the person started a business using some money

Is the next one which tells purchased office stationery clear? As in whether it's credit or debit?

[fresh acorn]

it just says there's a purchase that cost $1600 on office stationery

[fiery birch]

When you purchase , do you lose money or gain money?

Debit = losing money, credit = gaining money

[fiery birch]

Let me reframe this again. If I go to a shop to purchase something, will the shopkeeper give me money or will it be me paying the shopkeeper some money?

Am I making the slightest sense at all? @fresh acorn

[fresh acorn]

i pay

[fiery birch]

Good , if you pay do you lose money or gain money?

[fresh acorn]

lose

[fiery birch]

So it's a debit or a credit?

[fiery birch]

Remember this

[fresh acorn]

debit

[fiery birch]

Awesome , well done

Can you try the next one now?

Think that the things being mentioned are happening with you

[fresh acorn]

ok

[fiery birch]

Forget about what you might or might not have studied, just think in terms of losing and gaining money

Then convert them to debit or credit at the end as per the definition I gave you earlier

[fresh acorn]

OOOWHH

[fiery birch]

Do share what you are doing with each statement

[patent blaze]

i need help

[fiery birch]

Create a separate help channel #❓how-to-get-help

[obtuse pebble]

@fresh acorn Has your question been resolved?

[fresh acorn]

like this?

[fiery birch]

Yes but why did you put the first one as credit?

Rest of it seems correct

@fresh acorn

[fresh acorn]

250k is injected fund right?

so doesn't the company pocket 250k

[fiery birch]

We are probably not talking about the company here, look at the other statements.

It's probably someone starting a company, don't you think?

Let's say someone like the CEO of a company

[fresh acorn]

yea

[fiery birch]

I know it's confusing but here when they say injecting it means investing

So if you are investing you will actually use your own money but if you get a loan you are not

[fresh acorn]

OOOWHH

I GET IT

omg tyvm

[fiery birch]

No problem

[fresh acorn]

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[north gull]

Not sure how to go about this

It looks simple but

My teacher didn’t provide any information in the notes I think it was a mistake

If somebody could just give me the formula or guide me please that’d be great

[fierce lagoon]

Is this homework

[north gull]

Yes it’s an online class

[fierce lagoon]

Do you know your power theorems

[north gull]

No

[mighty quarry]

how can you do this ?

[fierce lagoon]

https://www.dummies.com/article/academics-the-arts/math/geometry/apply-three-power-theorems-circle-problems-229707/

You wanna use the tangent-secant theorem

[north gull]

Thank you

.close

[obtuse pebble]

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[inland matrix]

can someone explain why total charge is Q1-Q2

[misty gate]

is this capacitators?

@inland matrix ??

q=cv?

[inland matrix]

yes

[inland matrix]

these are

also Q1 > Q2

[misty gate]

i think bcs of sign change the q2 goes negative

and q(total) = q1+(-q2)

[whole dock]

Charges add up

[inland matrix]

wait he draw it wrong

let me redraw it

[inland matrix]

yeah but its supposed to subtract according to the answer

lemme show

[whole dock]

Yeah because Q2 is negative

[misty gate]

hmmm

[inland matrix]

why though? how do you know that

[inland matrix]

also the question has no battery attached

it also gives you this

[whole dock]

This looks like series

[whole dock]

And this one parallel

But Q1 and Q2 should be positive

[inland matrix]

yeah thats what i thought but i only get the answer treating it as parallel

[inland matrix]

basically u have this

[inland matrix]

its disconnected and reattached like this

[whole dock]

So that's a battery right and the positive plate of c1 is connected to positive plate of c2?

[inland matrix]

the answer is 750 and 250 btw

[whole dock]

Negative with negative

For the charges?

[inland matrix]

yep its just a notmal parallel circuit thing

[whole dock]

Q1 = 750?

Q2 = 250

[inland matrix]

yes

[whole dock]

?

For the series one or parallel

[inland matrix]

series

[whole dock]

Can you send a larger picture that has the whole question

[inland matrix]

they are on two different pages

[inland matrix]

does that help?

[whole dock]

Yep

[inland matrix]

the solution is you do Q1-Q2but im not quite sure what the logic for doing that is

[whole dock]

Alr so

I think you drew it incorrect

It is not in series

It remains in parallel but just the connection of the plates have changed

[inland matrix]

ohh

wait but why is it the difference

wait because theyre switches is different

and capacitance is always positive plate

omg youre a genius

tyty

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[obtuse pebble]

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[sinful marsh]

can someone help me in finding the range of this function?

[ruby ravine]

Well, this is a messy function. Finding range of this function is not going to be easy (just saying by looking at it). Usually what you would do is, find the minima/maxima using derivatives, and if that turns out to be the global maxima/minima, then you are a lucky man because then thats your range.

[ancient jacinth]

the sqrt(x^2-1) is a good start

[sinful marsh]

guys i don't know how to use derivatives and minima and maxima

i know limits

[ancient jacinth]

hmm

[ancient jacinth]

this is an even function

[ancient jacinth]

try taking the limit of this function as x approaches infinity

and also try to find when f(x) = 0

[ruby ravine]

well alright, first start off by simplifying this function

[ancient jacinth]

if f(x) can equal 0

[ruby ravine]

the numerator is factoring nicely

[ancient jacinth]

if you do that, you will see that all positive numbers are within the range of f(x)

[ruby ravine]

that would at least get you rid of the denominator, but make sure to add the condition for x != +-1

[ancient jacinth]

yes

[sinful marsh]

ok

[ancient jacinth]

now since there is square root involved, some values of x are undefined

for example

f(0) is undefined

[sinful marsh]

yes

domain is x < -1 U x > 1

[ancient jacinth]

yes

we know that 0 is part of the range of the function

because f(2) = 0

and as x goes to infinity, this function also goes to infinity

so we can say that the range must include all positive numbers and 0

but we need to find out what negative numbers belong to the range of f(x)

[sinful marsh]

how do you know that f(2) = 0?

[ancient jacinth]

when does x^4-5x^2+4 = 0?

[sinful marsh]

ah ok

[sinful marsh]

yes

[ancient jacinth]

can you solve this btw?

[sinful marsh]

yes

[sinful marsh]

how do i find this

[ancient jacinth]

hmm

tell you what

find the minimum of this function

[ruby ravine]

well uh, this function can be reduced to

y = (x + 2)(x - 2) sqrt(x^2 - 1)

x != +- 1

should be easy from here onwards I believe?

[ancient jacinth]

well kaplan knows that all numbers from 0 to positive infinity belong to the range

however they dont know how to find what negative numbers belong to this range

[ancient jacinth]

also this function is a quartic

how could you factor it like this?

[ruby ravine]

what I factored it into is still a qartic

[ancient jacinth]

y = (x + 2)(x - 2) sqrt(x^2 - 1)

this is not the original function

[ruby ravine]

nah wait, its fine

[ancient jacinth]

one sec

this is x^2 -4 / sqrt(x^2-1)

[ruby ravine]

yip

[ancient jacinth]

the original function is x^4-5x^2+4/ sqrt(x^2-1)

[ruby ravine]

(x^2 - 4)*sqrt(x^2-1)

not divide

[ancient jacinth]

https://tenor.com/view/drake-computer-notebook-explaning-gif-21152267