help-13

[sacred grail]

for $\cos(140\degrees)$, you want to use the other equation
[
\cos(180\degrees - \theta) = -\cos(\theta)
]

[orchid oriole]

So cos 180-40 equals -cos40

[sacred grail]

yes

and what about for tan(140)

[orchid oriole]

How about these questions

Question a

[sacred grail]

do you know when sin(θ) = 1/2?

[orchid oriole]

When it’s 180?

30?

[sacred grail]

not 180 but sin(30) = 1/2 yes

can you convert that into an obtuse angle

[orchid oriole]

180-30

[sacred grail]

yes

[orchid oriole]

Do I juts put the inverse in the calculator

For example for tan

[sacred grail]

you shouldnt need to

[orchid oriole]

Will it be -tan (-1)

[sacred grail]

?

[orchid oriole]

Which is -45

[sacred grail]

-tan?

[orchid oriole]

Inverse tan

Tan^-1

[sacred grail]

$\tan^{-1}$

[orchid oriole]

Yes that’s what I meant

[sacred grail]

that is one way of calculating what the angle should be

are these values something you are expected to know from memory though?

[orchid oriole]

Nope

[sacred grail]

hm

[orchid oriole]

Thank you for helping

I understand them now

[sacred grail]

-45 isnt an obtuse angle however

[orchid oriole]

Oh

What if I do 180-45

[sacred grail]

well

yeah that works

[orchid oriole]

Thank you

.close

[cedar kiln]

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[cedar kiln]

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[crimson obsidian]

Is the domain of f(x)=x^x = R-{0}?

[earnest socket]

x = -0.5?

[crimson obsidian]

=(-0.5)^(-0.5)

[mighty drift]

,w domain of x^x

[earnest socket]

is what?

[wraith dagger]

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[crimson obsidian]

Wait

[earnest socket]

are you sure?

[crimson obsidian]

It's complex number

Not defined

[earnest socket]

yes, not allowed

so re-evaluate your answer

[mighty drift]

Arguably you could define over all rationals with odd denominator

[crimson obsidian]

So it's domain is R+?

[earnest socket]

how would you define the denominator of a rational number?

m/n = 2m/2n

[mighty drift]

Obviously in lowest terms

[earnest socket]

any rational with an odd denominator is a rational with an even denominator

but then it would be defined for one, and undefined for the other

even though both quantities are equal

[mighty drift]

{p/q, (p,q) in Z x N*, q odd}

[earnest socket]

so is the function defined there or not?

[crimson obsidian]

You're right

So Tushar is domain x belongs to R+?

[earnest socket]

i would say it's that and all the negative integers

[crimson obsidian]

I see

Is 0 excluded

[mighty drift]

Things get iffy for negative numbers

[iron saffron]

x^x=e^(xlog(x))

[crimson obsidian]

It's defined for negative integers though

[iron saffron]

domain is clear

[mighty drift]

In general

[earnest socket]

but this only holds for x > 0

[iron saffron]

that's why it's clear

R+

[earnest socket]

well

it's defined for the negative integers

the RHS is not

[crimson obsidian]

They are two different functions

I don't think they are identical

Because like Tushar says x belongs to negative integers works in former but in latter x>0

[iron saffron]

right

[earnest socket]

i would say it wouldn't be defined for negative rationals that are not integers

[earnest socket]

because of this

[crimson obsidian]

The iffy stuff about negative numbers is those with odd and even denominatirs

[earnest socket]

it a function is defined somewhere, i think we enforce that it is defined for all "versions" of that number

[crimson obsidian]

Somehow it looks like mathematical loophole to me

At x=-1/3 you said f(x) is defined but at x= -2/6 you say f(x) is undefined

But both x are equal

Wait they are defined for -2/6

.close

[cedar kiln]

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[crimson obsidian]

Is -1/2 and -2/4 same number?

[drifting marlin]

Yep

[crimson obsidian]

Someone told me that they aren't

They are equal but not same

[drifting marlin]

I mean they take on the exact same value, despite not looking the same

[earnest socket]

need context lol

for all intents and purposes, equal means same

[crimson obsidian]

So equal and same are synonyms

Alright

[earnest socket]

need context

[crimson obsidian]

I see

[crimson obsidian]

No this is my confusion created

From previous question I asked

[split harness]

One is reduced to prime factors the other isnt

[fair geyser]

is 2 and "the smallest prime" the same number?

[split harness]

Yes

[crimson obsidian]

So they are equal but not same?

[fair geyser]

or just equal?

[crimson obsidian]

Same and equal both

[split harness]

They are equal

[crimson obsidian]

Oh alright

[split harness]

"the same" is very vague

[tame wraith]

to this day i still think 1 should be a prime number

[crimson obsidian]

Based on this I've a proof that they are not equal. I need to write that proof here

[tropic oxide]

who told you this, where/when did they tell you this, and what are the EXACT WORDS they said?

if you are unable to provide this info, then say "i am unable to provide this info".

[crimson obsidian]

They said me in some other server. They said they are not same number but same fraction

[tropic oxide]

so this was a discord message?

[crimson obsidian]

Yes

[earnest socket]

could be related to this discussion: #help-13 message

[tropic oxide]

take a screenshot of the message and send it here.

[crimson obsidian]

Okay

[tropic oxide]

blur/black out name/pfp if needed

[fair geyser]

it's hard to imagine when it would be wrong tbh
you're not really going to make a mistake because you have a separate notion of sameness for numbers

[tropic oxide]

but i want to see the message and, if relevant, the dialogue leading up to it

[crimson obsidian]

He directly answer nope

[tame wraith]

hmm i guess i could see what hes trying to say

[tropic oxide]

......

did Joncord elaborate on this in any way?

[earnest socket]

i would've answered yes

[fair geyser]

you have to invoke like occam's razor to argue that numbers are never equal but not the same, and that's not like a law, you're allowed to have weird complex notions that people hate

[crimson obsidian]

Not really

[tropic oxide]

okay so like

let's all of us stop here.

[tame wraith]

well, depends on if he means numbers as in, 1 2 , 2 4

or numbers as in 1/2 and 2/4

[crimson obsidian]

I see

[earnest socket]

more context: they asked for the domain of x^x and someone suggested that it could be defined for all negative rationals with odd denominator, and i remarked that every such number is a rational with even denominator so it would be nonsensical to make the function defined for only certain representations of a rational number

[tropic oxide]

oh so there was context?

as in you're on the same server as the two of them?

[earnest socket]

no

that happened here

[crimson obsidian]

Yes they is me

[tropic oxide]

... okay

[earnest socket]

in this channel actually

[tropic oxide]

calamity you should probably go get yourself a pronoun role

anyway like

to say the obvious:

[crimson obsidian]

Alright

[tropic oxide]

-1/2 and -2/4 are the same number

in any reasonable interpretation of the word "same"

[earnest socket]

agreed

[tropic oxide]

if joncord is going to insist that -1/2 and -2/4 are "the same fraction" but NOT "the same number" then she ought to define both concepts.

[crimson obsidian]

That makes sense

[earnest socket]

is 1 + 3 the same as 2 + 2?

[tame wraith]

if anything, id be more likely to argue the opposite, same number but not same fraction (even though it is same fraction)

[earnest socket]

i think you can either view 1/2 and 2/4 as results of a computation of two real numbers, or as rational numbers

[crimson obsidian]

My question arose, because (-1/2)^(-1/2) was undefined and (-2/4)^(-2/4) defined, which wouldn't be possible since both are same numbers

[tropic oxide]

My question arose, because (-1/2)^(-1/2) was undefined and (-2/4)^(-2/4) defined,
WHY IS THIS ONLY COMING UP NOW

THIS IS SUCH A HUGE XY PROBLEM

[crimson obsidian]

I mean I was about to bring it up but I had to reply some messages

[tropic oxide]

you could've said it at the outset no?

[crimson obsidian]

This question is related to context Tushar brought

Yes

[earnest socket]

linked it here. this was the only context as far as i was aware

[tame wraith]

whats that website again @tropic oxide

[tropic oxide]

what website

[earnest socket]

https://xyproblem.info

[tame wraith]

yh

that one

[tropic oxide]

anyway i dont have a good answer for how to tell if a fractional power of a negative number is defined or not

to me the question is why would you want to work with those things itfp

[earnest socket]

.

[crimson obsidian]

(-2/4)^(-2/4)=(-2)^(2/4) which is defined unlike (-1/2)^(-1/2)

[earnest socket]

well

are you sure (-2/4)^(-2/4) is defined

[tame wraith]

,w (-2/4)^(-2/4)

[wraith dagger]

Results provided by WolframAlpha

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Requested by doctor99268#2294

[tame wraith]

,w (-1/2)^(-1/2)

[wraith dagger]

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Requested by doctor99268#2294

[crimson obsidian]

Well it's a chain reaction, all started with me asking domain of x^x and I was curious

[earnest socket]

maybe you want (-1/3)^(-1/3) and (-2/6)^(-2/6)

[crimson obsidian]

But earlier ones also work

[tropic oxide]

as far as i am concerned the domain of x^x is either (0, +∞) or Z ∪ (0, +∞)

depending on whether or not you care about negative integers

[earnest socket]

and i wouldn't include rationals with odd denominators in lowest terms for this reason

[crimson obsidian]

Yes that's what Tushar said as well but then later there was a argument that negative fractions with odd denominator also satisfy the equation

[earnest socket]

it should be representation-independent

[split harness]

Why is that

[crimson obsidian]

Because (-2)^(2/4) would be 4^(1/4)?

[earnest socket]

how

you are squaring first

[crimson obsidian]

Yes

[earnest socket]

,w (-2)^(2/4)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by tushar#0942

[earnest socket]

,calc (-2)^(2/4)

[wraith dagger]

Result:

1.4142135623731i

[crimson obsidian]

Why can't I square first

(a^b)^(1/c)=a^(b/c)

Can't I do this?

[earnest socket]

i don't know if that holds for a < 0

these sort of operations are bound to give you all sorts of problems with negative numbers

[crimson obsidian]

I see

[earnest socket]

(-1)^(2/4)

is that 1 or i?

[crimson obsidian]

So there is a limitation that a must be non negative

[crimson obsidian]

i but if I square first then it's 1

[earnest socket]

exactly

[crimson obsidian]

I think I understand then

[earnest socket]

https://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities

[crimson obsidian]

Thanks for that link

.close

[cedar kiln]

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[zinc vortex]

Where did those partial derivatives come from?

This is Green's Theorem

[upper garnet]

curl

[tame wraith]

@zinc vortex curl

this is 2d version of stokes theorem

if your surface is only on the xy plane

then

the vector dS = (0,0,dA)

so the dot product of the curl and that

will only give you the k component of the curl

[zinc vortex]

I see

I just screenshot this

Thanks

[tame wraith]

if you want some more proof as to why the curl

is even there

well the principle is that

one sec

ill draw it

[zinc vortex]

I'll wait 🙂

[tame wraith]

as you know

curl is a measure of rotation

and basically

all the inner rotations

add up

to be the one big rotation

that your closed line integral is doing

better image

$\nabla \cross F \cdot d\vec{S}$

[wraith dagger]

doctor99268

[tame wraith]

this will give you the rotation

that is on the teeny bit of surface

because you want to know how much of the rotation is actually happening normal to the surface

which is what the dot product to the normal of the surface is for

then you just integrate it

across the entire surface

[zinc vortex]

Thanks for detailed the explanation, doc! I screenshot this one too.

[cedar kiln]

@zinc vortex Has your question been resolved?

[cedar kiln]

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[viral tartan]

Hi can someone take a quick look at these 2 sheets and tell me if the answers look right.

[cedar kiln]

@viral tartan Has your question been resolved?

[cedar kiln]

@viral tartan Has your question been resolved?

[cedar kiln]

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[shrewd sparrow]

Can I get a fact check on this?

[uneven crater]

Yeah, the multiplication is correct

[shrewd sparrow]

Awesomeness

[uneven crater]

Awesomeness indeed 🤘

[shrewd sparrow]

Confused how to do this

[lean thunder]

U can calculate SA for the base n the top using the given radius

As for the cylindrical SA u'll need height and radius, one of which u already know
Use the given volume for the height of cylinder

[shrewd sparrow]

Okay and now

This doesn’t seem right, I didn’t use the length

But I used the formula?

Which didn’t ask for the length

Hm idk

[nova glacier]

the area of the rectangle is (9 * 7)/2

To get the volume you need to multiply by the length

[shrewd sparrow]

So 31.5 x 19?

[nova glacier]

19 * 9 * 7 / 2

[shrewd sparrow]

Ah

[nova glacier]

$598.5 cm^3$

[wraith dagger]

Cain

[nova glacier]

@shrewd sparrow is that clear?

[shrewd sparrow]

Yes it is

Mostly

Well I’m not sure where the

Multiplying the 19 comes in

[nova glacier]

How would you calculate the volume of a box?

[shrewd sparrow]

Multiply the two sides together

[nova glacier]

like this one

[shrewd sparrow]

Multiple the three together

[nova glacier]

Right

Let's call w = width, l=length, h=height

So one way to think about it, is that you calculate the area of the rectangle w*h, and multiply it by the length l times

The same with the prism

You calculate the area of the triangle, and you multiply it by the length

[shrewd sparrow]

Oh that makes much more sense

Thank you!

[nova glacier]

you are welcome

[shrewd sparrow]

Okay

So wtf is the net of a cube. 💀

I know the net of like everything else

We never learned the net of a cube

[nova glacier]

Sorry, can you tell me what the net is? I'm not a native English speaker

[shrewd sparrow]

Oh what are you native?

It’s like, the dimensions unfolded

Like these

[nova glacier]

Like this?

[shrewd sparrow]

Yes that is it I think

Ah good

He never actually told us but like yk okay then😂

[nova glacier]

Just try to unfold it in your imagination 🙂

[shrewd sparrow]

I try but then my brain starts unfolding 😂

[nova glacier]

😂

[shrewd sparrow]

What are you native?

[nova glacier]

Hebrew

[shrewd sparrow]

That’s awesome 😲

[nova glacier]

🫶 Yes

Good luck with your hw

[shrewd sparrow]

Thanks, I def need it. I’m struggling with this packet, and the test is on Tuesday 🥲

[nova glacier]

These are all cubes

[shrewd sparrow]

😃

Let’s take a moment to appreciate my amazing drawing skills

[nova glacier]

Great!

[shrewd sparrow]

No it’s so bad hahahaha

[nova glacier]

If you are good at math you are bad at drawing

That's the way it is

[shrewd sparrow]

Ahem

I’m bad at math

💀

Can I get another fact check here?

It’s cool and all I can try and maybe get it wrong but I can’t get it wrong on the test. If I’m doing this wrong I definitely need to be corrected ahha

And also while I’m here

There is a problem involving the volume is a half cylinder

And idk at all how to do that 🙃

It’s a real struggle out here y’all

[weak heath]

good

[jaunty pumice]

Which question?

[jaunty pumice]

Just use calculator and check yourself

[jaunty pumice]

can you tell me what's the radius of cylinder?

@shrewd sparrow

[shrewd sparrow]

Uhm

10

?

[tawny juniper]

@shrewd sparrow Volume of a cylinder = 𝜋r²h (pi x radius squared x height), therefore volume of half a cylinder = (𝜋r²h)÷2
1 yard = 3 feet, therefore 50 yards = 150 feet
Height = 150 feet, radius = 10 feet, therefore Volume = 7500𝜋 cubic feet ((𝜋 x 10^2 x 150)÷2)

[cedar kiln]

@shrewd sparrow Has your question been resolved?

[shrewd sparrow]

Thank you for the help😄

[tawny juniper]

You're welcome ^^

[cedar kiln]

@shrewd sparrow Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[remote tartan]

not sure how to start this

[royal loom]

maybe polynomial long division?

[remote tartan]

how do i divide that

[lyric narwhal]

just keep r as a variable

and consider it a constant when ur dividing

[remote tartan]

I'm still a lil lost

[earnest socket]

the factor is completely determined from the information you're given

what form must the factor be in?

[cosmic spindle]

U can guess it is x-1

And then do rest

[earnest socket]

yeah, not just guess

but that's the only way the leading term and constant term are ensured

[cosmic spindle]

Divide first 2

And then it’s x-1

[remote tartan]

how is it x-1

[earnest socket]

.

[remote tartan]

dividing by 2 will give me a fraction

[remote tartan]

i dont understand

[earnest socket]

what form must the other factor be in?

[remote tartan]

i don't know

[earnest socket]

what degree must it have?

[remote tartan]

to the first degree?

[earnest socket]

yes

all such polynomials are of the form ax+b

[remote tartan]

i was thinking of factorising the quadratic but the r is got me stuck

[earnest socket]

can you determine a and b from this equation?

[wraith dagger]

tushar

[remote tartan]

no?

[earnest socket]

look at the constant term

of both sides

[remote tartan]

-r and +r?

[earnest socket]

the constant term on the right is not r

it's a product

you can expand if you want, but the constant term will be r*b

[remote tartan]

why is it r*b

[cursive narwhal]

You can fix it by identification

[earnest socket]

every other term has x as a factor

expand it and see if you want

[remote tartan]

whats identification

[remote tartan]

so to multiply the right with the (ax+b)?

[earnest socket]

sure

the only important information you need is that the x^3 term will be 3ax^3 and the constant term will be rb

which you can see directly from the form of the product

but if you need to expand to confirm this, then that's fine

[cursive narwhal]

Expand it and you resolve a system of coefficients

[earnest socket]

yes, even though two equations are in one variable alone and give you the solution immediately

[cursive narwhal]

Yes

[remote tartan]

so r=-b?

[earnest socket]

no

how are you getting that?

[remote tartan]

nvm

[earnest socket]

-r = rb

[remote tartan]

uh

still lost

[earnest socket]

your goal is to find a and b

equate the coefficients of x^3 on both sides

and equate the constant terms on both sides

[remote tartan]

do i equate 3bx^2 to - 14x^2 or 3bx^2 - 11ax^2 to - 14x^2 ?

having trouble finding b

a=3

b= -17/11?

or b = 19/3

[earnest socket]

just look at the first and last terms

forget the middle terms

what constant term did you get?

[remote tartan]

so is it - 11ax^2+3bx^2?

[earnest socket]

how is that a constant term

do you know what a constant term is

the constant term is everything without an x

[remote tartan]

br then?

[earnest socket]

yes

and the constant term on the left?

[remote tartan]

-r

[earnest socket]

yes

those should be equal

solve that equation

[remote tartan]

b=-1?

[earnest socket]

yes

do the same for the term with x^3

[remote tartan]

a=3?

[earnest socket]

no

how did you get that

what is the x^3 term on the right

[remote tartan]

i forgot abt the 3

a=1?

[earnest socket]

yes

3 = 3a

so you have

$3x^2 - 14x^2 + 17x - r = (3x^2 - 11x + r)(x - 1)$

[wraith dagger]

tushar

[earnest socket]

(x - 1) is a factor of your polynomial

can you use this information to determine r?

[remote tartan]

aaaahhh

[earnest socket]

if you are stuck: ||think about roots||

[remote tartan]

x=1 and sub in?

[earnest socket]

exactly

[remote tartan]

both sides?

[earnest socket]

x = 1 makes the right-hand side 0, so it must also make the left-hand side 0

i trust you can proceed from here

[remote tartan]

i hope i can

[earnest socket]

having (x-1) as a factor implies that x=1 is a root of your polynomial

[remote tartan]

I'm somehow off by 1

i got r=7 and it should be r=6

[earnest socket]

how did you get r=7

[remote tartan]

uh

[earnest socket]

check your work

,calc 3 - 14 + 17

[wraith dagger]

Result:

6

[remote tartan]

i have - 8 on the left side

[earnest socket]

this is just basic arithmetic

[remote tartan]

i moved things around

[earnest socket]

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[earnest socket]

why is your first line true

[earnest socket]

that is not what we have here

[remote tartan]

do i include the (x-1)?

[earnest socket]

of course lol

can't just remove it

otherwise that says a degree 2 polynomial is equal to a degree 3 polynomial

false

[remote tartan]

i thought it was the two equations from the start to be equaled

[earnest socket]

A is a factor of B does not mean A = B

it means A x (something) = B

[remote tartan]

so i need to multiply the line by x-1?

[earnest socket]

just copy that equation

and substitute x=1

[remote tartan]

but if i sub in 1 for x doesn't that bracket become 0?

[earnest socket]

the right-hand side should vanish

yes

that's the point

having (x-1) as a factor means the polynomial has x=1 as a root

[remote tartan]

so if its sypposed to all vanish what's the point?

[earnest socket]

the point is to solve for r

[remote tartan]

and i do multiply 3x^2-11x+r by x-1?

[earnest socket]

$$3x^2 - 14x^2 + 17x - r = (3x^2 - 11x + r)(x - 1)$$
substitute $x = 1$:
\begin{align*}
3(1)^2 - 14(1)^2 + 17(1) - r &= (3(1)^2 - 11(1) + r)(1 - 1)\
3(1)^2 - 14(1)^2 + 17(1) - r &= (3(1)^2 - 11(1) + r)(0)\
3(1)^2 - 14(1)^2 + 17(1) - r &= 0
\end{align*}

[wraith dagger]

tushar

[remote tartan]

ah

i see now thank you

one more question

[earnest socket]

what do those being factors of that polynomial tell you about its roots?

[remote tartan]

x=3 x=-4?

[earnest socket]

yes x=3 is a root and x=-4 is a root

so formulate those statements into equations

[remote tartan]

then simultaneous equations?

[earnest socket]

yes

[remote tartan]

okay

i got it thank you

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[fossil thorn]

can someone help me with this, this is what I have tried:

[vague rapids]

picture on way?

[fossil thorn]

,rotate

answer doesn't make any sense tho

[fossil thorn]

,rotate

[uneven iron]

I think it can be solved using pyhsics

[fossil thorn]

im studying length of arc atm

radians etc.

[uneven iron]

Then forget what I said

[tribal kite]

I think ik how.

Find the circumference

Of the larger circle

Pi is the ratio of the circumference to diameter

So just find the diameter

Then u know that's in 2h

[vague rapids]

why 2pi/180
isnt perimeter of semicircle equal to pi*r

[tribal kite]

So divide by 2

[tribal kite]

It's pi*diameter I believe

[vague rapids]

that's for circle not semicircle

[fossil thorn]

nvm, i was trying to convert it into radians, but realized that doesn't make any sense

[fossil thorn]

what does that mean?

[vague rapids]

you don't need that stuff rn

[fossil thorn]

this is what the answer key says but i don't get how they got 2pi

[vague rapids]

they further divided by 2 also

2pir/2=pir

[tribal kite]

Wait brb I gotta try my method first

Yeah it works

It's a different method but same ans

[fossil thorn]

but how do they know that θ is going to be 2pi in the first place

[vague rapids]

Because complete circle is 2pi

[tribal kite]

Cuz d=2r

[fossil thorn]

ahhh

[tribal kite]

Diameter is 2x of radius

So 9000 is radius

18000 is diameter

Then x pi to get distance in 2h

Then get 1h

Same answer

Try it first

[vague rapids]

you should better mention what the variables mean

[tribal kite]

It's not a variablr

It's multiply

[vague rapids]

Be clear man

[fossil thorn]

why are we multiplying it with pi?

[vague rapids]

360 degrees is 2pi

180 degrees is pi

[tribal kite]

It can be solved without radians

[tribal kite]

Since circumference is pi times diameter

[vague rapids]

Yes but that's how the solution is done

[tribal kite]

Indeed but there are multiple methods to a question

[fossil thorn]

both these methods look good?

[vague rapids]

,rccw

[vague rapids]

yeah fine

[fossil thorn]

Okay thank u both very much!!

[tribal kite]

Yeah there's multiple methods to a question

[fossil thorn]

Yep both useful

Ty!

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[distant cosmos]

https://youtu.be/mBCiKUzwdMs

[distant cosmos]

I don't get what they are graphing tbh

Liek at 2:44 what's sample value and what's "probabilty of density "

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[spice light]

solve for x?

[flint plinth]

maybe start by taking reciprocals, and you can write (x+R)/R = x/R + 1

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[simple heron]

how do I solve this? the value of A is -1/2 and the value of B is 3/2

[radiant topaz]

you know the series expansion of 1/(z+1)?

[simple heron]

yeah

[simple heron]

It depends on the region

[radiant topaz]

you should first write the function in terms of z-1 first

$f(z) = \frac{1}{(z+1)(z+3)}$ can be written in terms of $z-1$ like $f(z) = \frac{1}{((z-1)+2)((z-1)+4)}= \frac{-1}{2((z-1)+2)} + \frac3{2((z-1)+4)}$

[simple heron]

There's no use of partial fraction ?

[radiant topaz]

there is

[simple heron]

alright I got it

Thank you

[wraith dagger]

fäf

[muted tendon]

Check this again

[radiant topaz]

what's the right partial fraction?

[muted tendon]

A= 1/2 B = -1/2

[simple heron]

uh

[dire geode]

,w partial fractions 1/(z+1)/(z+3)

[wraith dagger]

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[simple heron]

thank you so much

[radiant topaz]

So the thing I wrote is wrong

[simple heron]

It's okay. I got the idea. you guys helped me a lot

[radiant topaz]

you're welcome dear

[cedar kiln]

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[crimson sedge]

I know what the question is asking, but am I doing the work correctly?

[radiant topaz]

yes

[crimson sedge]

Alright, but from here I don’t really know how to find the expression for the inequality of n

Especially with factorials

[radiant topaz]

when does some natural number's factorial crosses 100?

[crimson sedge]

Oh, I think I get it now

So I have to basically find the smallest value of N that is greater than 100 from the inequality?

[radiant topaz]

yes

[crimson sedge]

Got it, thank you

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[young totem]

I was able to 'fix' the line by changing the length of it to get all the stars

by changing the [x<_] section to [x<7]

but i dont understand how it actually works

[long swan]

The stars don't even look like they're on a linear path

[young totem]

and why the length of the line changed

yes

[radiant topaz]

how cute

[young totem]

its suppose to extend to a part and then drop down to get the last star

[long swan]

So you need another piecewise conditional

[radiant topaz]

whats the actual question?

[young totem]

to change one number in the row to fix the line

[radiant topaz]

what?

what row?

[young totem]

the equation under the text on the left

[radiant topaz]

x<11 prolly

what does fixing marble slide mean?

[livid hound]

the braced part is the domain restriction

[radiant topaz]

yes

[livid hound]

by having {x<7}
it only graphs for values of x below 7
and nothing after that

[young totem]

oh i see

but why does that change the length of the line though?

[radiant topaz]

you remove the restriction

[young totem]

the restriction of what?

[radiant topaz]

domain

the x values

[young totem]

but isnt it the other way round?

didnt i add a restriction that makes it graph x values less than 7?

im really confused rn

[radiant topaz]

you added the restriction?

[idle folio]

Lol

[young totem]

like i added {x<7}

so isnt that a restriction?

[manic scaffold]

yes it is

what's the problem?

[young totem]

mhm

so why does my line change in length when i add this restriction?

i dont rlly understand how that works

[livid hound]

by having {x<7}
it only graphs for values of x below 7
and nothing after that

[manic scaffold]

ok, when you say (equation) {x<7}, that means that equation is valid for all values of x less than 7

but any other value of x, the equation is not valid, so it doesnt get drawn

[livid hound]

if you have some other value
{x<0}
the line would stop at x=0

[manic scaffold]

so if you change it from 7 to 11, the equation can now be drawn for more values of x up until 11

[young totem]

ok that makes more sense

lemme change it to 11 to see what u guys mean

it gets longer??

[young totem]

but when its {x<7} , it didnt stop at 7?

[livid hound]

it did?

[radiant topaz]

it did

at x=7

[young totem]

OH WAIT

yes

[radiant topaz]

not y=7

[young totem]

sry

[livid hound]

to make things clearer

[young totem]

kinda braindead

rn

[livid hound]

make two new lines / equations
x = k
k = 7
change the stuff in the brace to {x<k}

and use the slider to change the value of k

[young totem]

Ok

Oh i see the logic now

tysm for all ur help 🙂

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[frigid vector]

I am trying to understand chain differentiation

[frigid vector]

using same method I tried to compute gradients for my backprop implementation, and it just don't work.

I thought maybe I misunderstood something, and come up with this sample...

maybe there is a problem with recursive definition of x

[cedar kiln]

@frigid vector Has your question been resolved?

[mighty drift]

When you write dy/db = 1, x isn't a function of b
If you let x = y-b / a, then that means y is constant because x will always have the value that makes y(a, b, x(a, b)) the same value y0. Or you define x as a function of y to find what y is, and then you've got a definition problem

The chain rule is fine. The way you create dependencies is not

[cedar kiln]

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[rough canopy]

Does this approximation for a square root of 3 work using binomial expansion?
(1+x)^(1/2) =
1 + (1/2)x + [(1/2)((1/2) - 1) x^2)]/2! + [(1/2)((1/2) - 1)*((1/2-2)) x^3)]/4! ........ I'm not sure if this is mathematically incorrect or it is just a very inefficient series that converges slowly.

[rough canopy]

using x = 2, this should give square root of 3 right ?

[muted bear]

if i recall correctly, it should work

it converges relatively quick because of the factorial too i think

[rough canopy]

I have tried it untill the 9th term and it did not approach anywhere near the square root of 3.

[muted bear]

hmm

i dont quite know how to make this a series

did you forget the binomial coefficient or am i going insane

[rough canopy]

the binomal coefficent is the 1/2 * ((1/2)-1)/2! * ((1/2)-2)/3! * ... ((1/2) * ...((1/2)-n)/(n-1!) parts in each of the terms

I think the regular coefficients of a binomial expansion dont apply here as our power is a non integer, these ones are derived from the combinatorics formula.

[cedar kiln]

@rough canopy Has your question been resolved?

[cedar kiln]

@rough canopy Has your question been resolved?

[mighty drift]

,w taylor series (1+x)^(1/2)

[wraith dagger]

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[mighty drift]

(1/2)Cn x^n
By the ratio test,
(1/2)C(n+1) / (1/2)Cn = (1/2 - n-1) / (n+1) ~ 1
So the radius of convergence is 1

[mighty drift]

2 > 1 so the series should diverge as the term get exponentially bigger

[rough canopy]

so because it is a divergent series, it doesn't work to estimate the squareroot of 3?

[mighty drift]

Can't estimate when you don't converge

[rough canopy]

but surely (1+2)^1/2 is the square root of 3, so does this mean that applying the binomial expansion in this case is not appropriate?

[mighty drift]

Power series have a radius of convergence

Outside of it they diverge

Inside of it they might converge to the function you generated it from

[chilly shuttle]

diving straight down at a steep angle can provide the fastest possible descent and acceleration for an aircraft, which could lead to the highest possible speed

[rough canopy]

I'm not familiar with the concept of a radius of convergence, I shall look into it, thanks for the explanation. 👍

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[covert aspen]

Help

[covert aspen]

2^2145 mod 2143

I can't do this

I want an actual response, not someone using a bot

[tropic oxide]

@covert aspen last time i asked you whether you're able to state fermat's little theorem

since you mentioned it

but then you didn't respond

[covert aspen]

2^2143-1 = 1 (mod 2143)
2^2142 = 1 (mod 2143)

I don't know if this is correct

[tropic oxide]

it would've been better if you could state the theorem in general before applying it to your problem specifically

also just to verify

,w is 2143 prime?

[wraith dagger]

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[tropic oxide]

ok cool

yes, so you have 2^2142 = 1 (mod 2143), that's correct

are you able to continue from here?

[covert aspen]

Nope

[tropic oxide]

well

you can reduce 2^2142 modulo 2143

but you're asked for 2^2145

how do you think you could get from one to the other?

[covert aspen]

By adding 3?

I'm not sure

[tropic oxide]

... gonna need you to elaborate on this.

$2^{2145} = 2^{2142} + 3$ is not true even if you consider this modulo 2143.

[wraith dagger]

Ann

[tropic oxide]

so maybe write out what you want to do in a little more detail?

[covert aspen]

I don't know how else to get from 2142 to 2145

[tropic oxide]

how do you get from **2^**2142 to **2^**2145?

there's only so much i can say here without, like, explicitly saying 2^2145 = 2^2142 * 2^3. you overthought it all.

[covert aspen]

Oh, so it was 2^2142 * 2^3?

If so, I guess I did

[tropic oxide]

that's what i was hoping you'd come to, yes.

the same relationship holds modulo n, of course. whatever it is that n may be.

[covert aspen]

What would the next step be?

[tropic oxide]

well you know what 2^2142 is mod 2143

and you also ought to know what 2^3 is

[covert aspen]

Yes

Do I break down the 2142?

[tropic oxide]

you are overthinking it again

[covert aspen]

Hmm

I'm confused

[cedar kiln]

@covert aspen Has your question been resolved?

[covert aspen]

I don't know the next steps

[covert aspen]

Can anyone help me?

Please, thanks

Hello?

Someone 😢

[deep plover]

gm

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[tepid elbow]

Anyone can help me on quadrileteral

[long swan]

i can help you spell it

it's 'quadrilateral'

[tepid elbow]

Damn alr

How

Ok thank yoy

Can anyone help me on quadrilateral

[long swan]

no

[tepid elbow]

Bruh

[muted tendon]

Post your question

[tepid elbow]

Why

Ping me when ur back

[muted tendon]

6x+8x = 56

Solve for x

Then i will tell further steps

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[deft sigil]

Question 4

[deft sigil]

Need help bruh

[hushed spoke]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[deft sigil]

1

[bright bridge]

everything on the perimeter of that triangle is when added up is46

[deft sigil]

Ye

[bright bridge]

so $2x + x + (x-2) = 46$

[deft sigil]

Ye

[wraith dagger]

bacchess

[deft sigil]

But what is x

Gotta find x

[deft sigil]

Ye I know that bit. I need to find X. It’s actually making mad 😭

[bright bridge]

combine like terms?

[deft sigil]

Omg

[bright bridge]

lol

[deft sigil]

I’m actually so dumb

Thank u bro

[bright bridge]

np

[deft sigil]

😘

[bright bridge]

thats all?

[tepid elbow]

How do you get 56 i dont understand

[muted tendon]

It is given

[tepid elbow]

Oh alr

Whatd the rera

Rest

[cedar kiln]

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[cedar kiln]

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[icy pilot]

Triangle ABC has 4 lines parallel with BC, 5 lines parallel with AC, and 6 lines parallel with AB. How many trapezoids are formed from those 15 lines (exclude parallelogram), knowing that no lines in those 15 lines are concurrent.

[icy pilot]

How do I solve this with partial permutation? The answer for this quesrion is 720. However, my answer was 900, which is wrong

I looked up for the solution, but it was hardly understandable, as no explaination was given.

Nvm, i figured it out

.close

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[icy pilot]

.close

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[strange shoal]

guys can u help me w this

[strange shoal]

i js wanna know the missing value

wait nvm i got it

[vague rapids]

You may close the channel then

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[meager trellis]

I’ve got stuck resolving a problem of a finite difference method. The problem is defined in the following image. I’ve expanded the first term in Taylor series and calculated the truncation error subtracting the r(x) from the numerical method. I get stuck because the q(x) functions is evaluated in xj+1 and xj-1. Should I expand the q(x) terms in Taylor series too?

[cedar kiln]

@meager trellis Has your question been resolved?

[cedar kiln]

@meager trellis Has your question been resolved?

[crimson delta]

yes

[cedar kiln]

@meager trellis Has your question been resolved?

[meager trellis]

If I do that I’ll come to that

For the method to be 4th order I have to eliminate the term $\frac{h^2}{2} u^{(4)} (x_j)$

[wraith dagger]

danimasa

[meager trellis]

$\tau_j = \frac{h^2}{12} u^{(4)}(x_j)+\frac{h^4}{360} u^{(6)}(x_j)+O(h^8) - (\alpha_2 [q(x_j) + hq’(x_j)+\frac{h^2}{2}q^{(2)}(x_j)+O(h^3)][u(x_j)+hu’(x_j)+\frac{h^2}{2}u^{(2)}(x_j)+O(h^3)]+ \alpha_1 q(x_j) u(x_j) + \alpha_0… )$

[wraith dagger]

danimasa

[meager trellis]

But all the terms with the alpha is multiplied by q(x), that’s is my doubt how to eliminate that term

[meager trellis]

<@&286206848099549185>

[cedar kiln]

@meager trellis Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[viral tartan]

Hi guys quick question. Can someone help me find practice sheets for similar question to this for my non calculator paper that’s coming up

[viral tartan]

Not sure how to identify the exact area of algebra/what to write on Google

[stone marsh]

Itll probably be hard to find questions exactly like that. A good place to start would be quadratic factorisations problems.

[carmine bronze]

Try googling --> "college algebra" filetype:pdf. You should find one or two textbooks that you can download that will have math exercises in it.

[viral tartan]

I will look in to both of those things. Thanks guys!

.close

[cedar kiln]

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[gusty rain]

How can I simplify this?

[crimson sedge]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[grand forge]

x^2 -4 is the difference of 2 squares remember

[gusty rain]

I have no idea what to do tobegin

[crimson sedge]

factor the denominator using what Mortta said

[gritty viper]

also expand the numerator

[gusty rain]

How can I do that?

[crimson sedge]

read

use

[grand forge]

AW HELL NAw

he got light mode

[crimson sedge]

light mode > dark mode

[grand forge]

THAT IS CAP

Eyes that dont hurt > eyes that hurt

[crimson sedge]

lol

[gusty rain]

Honestly I don't get it

I have to factor x^2 - 4?

[crimson sedge]

yes, thats the first step

[gusty rain]

(x+2)(x-2)

[crimson sedge]

yes

now expand then factor the numerator as eggnote said

[gusty rain]

x^2 -5x -14 = (x+7) (x-2)

Wait

2 and -7*

[crimson sedge]

yes

now cancel

[gusty rain]

Cancel what?

[crimson sedge]

cancel the common factor from the numerator and denominator

[gusty rain]

So x+2

I'm left with x-7 and x-2 now is that the answer?

[crimson sedge]

thats the answer

[gusty rain]

Awesome, what is this type of problem called? I wanna practice it more

[crimson sedge]

factoring

[gusty rain]

Alright thank you

.close

[cedar kiln]

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[teal magnet]

If https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:standard-form-quadratic/v/ex3-completing-the-square?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJBCxIIVXNlckRhdGEiHmthaWRfMTA0MTAxOTkyMDMyMzYzMjE5MzE0MDE1OQwLEghGZWVkYmFjaxiAgICApNeHCgw&qa_expand_type=question is correct, I don't understand the section starting with "By factoring out 𝑎"; I though to balance taking out 𝑏 ∕ (2𝑎) that you'd have to substract b²/4a² yet its "− 𝑏² ∕ (4𝑎)". Is the denominator right, should it be as I suspected, or should it be something else?

[drifting marlin]

you multiply by a when you remove that term from the bracket

[teal magnet]

Is "− 𝑏² ∕ (4𝑎)" correct for that section?

[drifting marlin]

yep

[teal magnet]

Well... I have no idea how to get there. In the replies of that comments T.Angelo have the same issue as I have.

[drifting marlin]

so start by looking at just what's within the parentheses

what do you have to do to complete the square on x^2 + (b/a)x?

[teal magnet]

add & remove b²/a².

ooops

[drifting marlin]

not quite

[teal magnet]

add & remove b²/4a².

[drifting marlin]

better

so now you have a(x^2 + (b/a)x + b^2/4a^2 - b^2/4a^2) + c

if you want to remove that final term from the parentheses, you have to multiply it by a as you do

[teal magnet]

oooh! thanks 😄

[drifting marlin]

leaving you with a(x^2 + (b/a)x + b^2/4a^2) - b^2/4a + c

[surreal cave]

which of course can be factored as a(x+b/(2a))^2

[drifting marlin]

• but yes

[surreal cave]

oh dang

good catch my b

[teal magnet]

I see as I'm having the same problem in maths as in programming, some mistakes I can't see no matter how many times I look at it.

Is there a way to make a post as the answer to my question and leave the channel for someone else?

!answer so now you have a(x^2 + (b/a)x + b^2/4a^2 - b^2/4a^2) + c if you want to remove that final term from the parentheses, you have to multiply it by a as you do

@answer so now you have a(x^2 + (b/a)x + b^2/4a^2 - b^2/4a^2) + c if you want to remove that final term from the parentheses, you have to multiply it by a as you do

!close

!abandon

[drifting marlin]

.close

[cedar kiln]

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[sleek condor]

isnit already given?

[quasi jungle]

This is the problem: Hi I am having trouble setting the domaine in y for this problem. I found that the domaine in x is -2>x>7

[sleek condor]

lol

[cedar kiln]

Channel closed

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[sleek condor]

oops

[quasi jungle]

.reopen

lol

nooo

[sleek condor]

it wont work

open a new one

[quasi jungle]

yess

[cedar kiln]

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[frail dagger]

I think it might be called regression analysis

<@&286206848099549185>

[cedar kiln]

@frail dagger Has your question been resolved?

[frail dagger]

anyone?

[crimson sedge]

You can ping the helpers if more than 15 minutes pass (:

[frail dagger]

I did didn't I?

[crimson sedge]

You don’t ping them immediately

[frail dagger]

what?

[crimson sedge]

Try pinging again.

Is what I meant

Or you can close this and ask the question in a newer channel.

[frail dagger]

I'll try pinging again first. Thanks

[crimson sedge]

Welcome

[frail dagger]

<@&286206848099549185>

welp

.close

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[limpid jasper]

Hello! I’m having a hard time in this question.

[limpid jasper]

I already answered it to some extent but I’m having a hard time making the curve

<@&286206848099549185>

Not sure if this is right

This is what I did

I’m not sure if I should use the z-score formula and z table

[astral bay]

,rccw

[cedar kiln]

@limpid jasper Has your question been resolved?