help-26

[clever lichen]

Hi. I'm struggling with reverse FOIL factoring of a polynomial 5w squared - 29w - 42. I have done a factor tree for 42 but am unable to come up with two factors that will equal -29 when doing the FOIL check, particularly when accounting for the positive 5w from the initial FOIL

[knotty ledge]

Factoring quadratics where the leading term is not 1 is more complicated

[true minnow]

not really, OP should find two numbers m, n such that m+n = -29 and m*n = (-42)*5

[clever lichen]

I can find -21 and -8, but they don't multiply to -42. That's where my check is failing.

[true minnow]

they need not multiply to -42

they have to multiply to -210

[clever lichen]

So I get 15 and -14 for that, but how do I account for the final -42 in the equation? The two factors that equal the -210 still have to multiply to equal -42 at the end.

[true minnow]

wdym?

generally, if you have $ax^2 + bx + c$, you can factor if there are $m$ and $n$ such that $m+n=b$ and $mn = ac$

[clever lichen]

So in the original equation of 5w squared - 29w - 42, I can come up with -15 and -14 for two factors that equal -210, but they don't multiply to equal -42

[thorny flame]

musava_ribica

[true minnow]

first, -15 and -14 when multiplied do not give -210, but 210 (positive), second why do you want them to multiply to 42?

[clever lichen]

Sorry, added an extra negative there. I'm doing a FOIL check on my work to see if it equals the original equation and it doesn't. When you do FOIL, you multiply the last two factors, yes/

[true minnow]

$5w^2 - 29w - 42 = 5w^2 - 35w + 6w - 42 = 5w(w-7) + 6(w-7) = (w-7)(5w+6)$

[thorny flame]

musava_ribica

[true minnow]

where exactly do you "fail" in FOIL check?

[clever lichen]

Oh! I see. I missed a step in the middle. Ok. I understand now. I missed the factoring to remove the (5w-6).

Thank you.

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[trail heath]

.reopen

[sweet shard]

!msgdel

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[trail heath]

im unsure how to do this

i understand polar just not this process

[uncut crow]

The process of finding a 3d volume?

[trail heath]

somewhat

i havent delt with a constraint in xy but the problem in z,x

[uncut crow]

It becomes pretty nice in polar coords

Try writing D and z under that transformation

[knotty ledge]

read this

[trail heath]

do we swap channels

[knotty ledge]

please open a new channel

[uncut crow]

sorry

[trail heath]

channel 15

[uncut crow]

close this channel

[trail heath]

.close

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[little venture]

What is inverse trig functions? Is it just cscx = 1/sinx

[little venture]

Like how the inverse of 5 is 1/5

My friends tells me it’s wrong but they won’t tell me the correct answer

So I come here to seek the answer

[pearl peak]

thats the reciprocal

the inverse is sin^-1(x) or arcsin(x)

[little venture]

Multiplicative inverse basically

[pearl peak]

yes

[little venture]

Yeah sin to the negative one power

Is 1/sin

[pearl peak]

no

[little venture]

What

[pearl peak]

no sin^-1(x) is not 1/sin(x)

[little venture]

How?

[little venture]

How tho

[wooden osprey]

yeah the notation is a bit confusing

sin^-1(x) is arcsin(x), not the reciprocal of sin

[little venture]

It’s like 5^-1 being 1/5

[wooden osprey]

yes, but not for trig func

[little venture]

Sin^-1 has to be 1/sin

[little venture]

So trig functions get special tra Yemeni?

Treatment

[wooden osprey]

pretty much

[little venture]

So what is sin^-1 then

[wooden osprey]

in short, inverse trig func is the opposite of trig func

[little venture]

?

Opposite?

[wooden osprey]

cos0 = 1 => arccos1 = 0

[little venture]

Of a trig function?

[wooden osprey]

arccos is the inverse trig func of cos

[little venture]

Oh you could have just said f^-1(x)

Like That Inverse

Not that inverse

But like THAT inverse

Ahh

[plucky token]

An inverse trig function swaps the x values and y values around

[little venture]

Yessir

Wait

[wooden osprey]

no, inverse trig func is not reciprocal

[little venture]

so ln and exp are opposites

and they cancel out each other

so does sin^-1 and sin cancel out each other

[plucky token]

Yeah

[wooden osprey]

athough sin^-1() and f^-1() are of the same notation, they sre very much different

[little venture]

And this goes for any trig function?

[little venture]

Ye I know

[plucky token]

Yep

[little venture]

So I have can 1/csc^-1

Which is sin^-1

Wait

[plucky token]

Well, be careful

[little venture]

Wtf am I talking about

[plucky token]

I think u assumed that the inverse trig function was the reciprocal

[little venture]

Used to

[plucky token]

Nice job catching yourself tho

[little venture]

Wait can you do trig identities like

Tan^-1 = sin^-1/cos^-1

And also would it be sin^-1y or sin^-1x

[plucky token]

It would be sin^-1(x)

[little venture]

O ok

[plucky token]

And I think this is true

[little venture]

Yay

When do people learn this stuff

[plucky token]

Yea, I don't see why not

[little venture]

It seems so confusing

[plucky token]

You see, I actually don't know

[little venture]

O

[plucky token]

I've been randomly self-studying

[little venture]

O lol

I’m in 8th is this math too early

[plucky token]

Apparently, you learn some of it in algebra 2, than you do more stuff in pre-calc. Idk where to get good pre-calc stuff tho

[little venture]

O

Algebra 2 hmm

Pre calc is like before calc so

[plucky token]

Yeah, not having a good background in trig can make this feel real confusing

[little venture]

wait then calc has this stuff

[little venture]

I think I need to strengthen my trig base

What is the best way to introduce trig

[plucky token]

Good idea

[little venture]

Cuz I got introduced with the unit circle

And I recently learn about SOHCAHTOA

And my friend wants to learn a bit of trig too

*at 8th grade how tf

[plucky token]

What course are you in now?

[little venture]

Int 1 (Accelerated maybe)

[plucky token]

https://youtu.be/I3jyBUyjg48?si=py6XmPldY08hKHrO

This should give a good intro to SOHCAHTOA

[little venture]

Oh no I already learned SOHCAHTOA a bit

It is in relationship with the right triangle

[plucky token]

Unit circle right after is actually pretty easy, knowing that the hypotenuse is 1

[little venture]

And sohcahtoa also led me to a new way to prove sin^2x * cos^2x = 1

[plucky token]

Yea, do u know the other Pythagorean identities?

[little venture]

Uhh the tan squared plus one is secant

Secant squared

And the cosec with cot

[plucky token]

Yea

[little venture]

I don’t really study the double angle or triple angle ones

Or the half angle ones

I kind of remember the sum ones tho

[plucky token]

Yeah, you don't rly need those when you have a calculator

[little venture]

Sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

I think

[plucky token]

Yep

[little venture]

And then cosine one is

Cos(a+b) = sin(a)cos(a) + cos(b)sin(b)

I think I’m wrong

But I’m not too sure

[plucky token]

I think it's cos(a+b)=cos(a)cos(b)+sin(a)sin(b)

[little venture]

Oh yeah that makes more sense

And then the other sim identities are kind of chill there

[plucky token]

Yea, that's it

[little venture]

Cuz of sine cosine

And the other identities

[plucky token]

Yeah, you can derive most of the other stuff just from these

[little venture]

Btw In the complex world I can’t wrap my head around why sine definition is being divided by I while cosine isn’t

[plucky token]

I haven't got to parametrics yet

[little venture]

Complex world isn’t parametrics I don’t think so

[plucky token]

Oh

[little venture]

Idk I call it the complex world

I don’t know if it has a formal name

[plucky token]

Yeah, it involves i

[little venture]

Imaginary numbers

Ye

Or “Lateral” numbers

[plucky token]

Eh, idk. I'll get to it soon, thats all I know

[little venture]

Wait parametrics is imaginary stuff?

Or unless I got mixed

[plucky token]

I think polar coordinates involves transferring equations to cos(x)+isin(x) sometimes

[little venture]

Ye

Cis(x)

[plucky token]

Ye

[little venture]

Or cis z

[plucky token]

Oof, I gotta go now

[little venture]

Oop

Aight I’ll see you then

Wanna firend rq if you want

[plucky token]

Sure

[little venture]

Aight I sent one

[plucky token]

I accepted it

[little venture]

Aight see ya then

[plucky token]

Make sure to close the channel btw

[little venture]

Maybe we can math study lol

[plucky token]

Yea

[little venture]

Aight thanks for that reminder

Alright see ya

[plucky token]

Cya

[little venture]

.close

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[red thunder]

please help

[red thunder]

Can someone please help

[topaz sinew]

@red thunder Has your question been resolved?

[vivid spire]

Have you tried anything?

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[shut violet]

How do I find the equation of the locus?

[frosty belfry]

let z = x + yi, then remove the modulus

[shut violet]

$$x^2+y^2=4\left(x^2-4x+4+y^2\right)$$

[thorny flame]

Lex1729

[frosty belfry]

yeah

[shut violet]

$$4x^2-16x+16+4y^2=x^2+y^2$$

[thorny flame]

Lex1729

[shut violet]

What do I do now?

Do I divide both sides by 4?

[frosty belfry]

no you just simplify from there to make it in the form of a circle

[shut violet]

But I need to complete the square right?

[frosty belfry]

yeah

[shut violet]

But doesn't the x^2 term and y^2 term need to be 1 in order to do that?

We would need to factor out a 3 right?

@frosty belfry

[frosty belfry]

lemme show u

[shut violet]

Sure

<@&286206848099549185>

[frosty belfry]

@shut violet do you understand my working?

[shut violet]

Yeah, that makes sense.

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@shut violet Has your question been resolved?

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[spice flicker]

Can someone help me understand this

[spice flicker]

Trig identities make no sense to me

[half edge]

Do u know what csc(x) can be written as in terms of sin and cos

[spice flicker]

Like 1/cos

[half edge]

No

csc x = 1/sinx

[spice flicker]

Oops

[half edge]

Do u know how to write tan in terms of sin and cos

[spice flicker]

And tank is Sinx/cosx

tan*

[half edge]

Ye

[spice flicker]

Omfg

Is it just that

[half edge]

So do multiplication

[spice flicker]

Multiplying those togethe

r

[half edge]

Yea

[spice flicker]

Okay

I understand this

Thank you

[half edge]

Np

[spice flicker]

This makes way more sense

.close

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[mild reef]

How do you solve this?

[mild reef]

I don't get how to integrate this

[neon iron]

use u-substitution and take x² + 3 as derivate of x² + 3 is 2x, so after the u-substitution the integral would look like du/√u which can be solved using power rule

you're facing issue while integrating 1/√u?

[mild reef]

Yeah it's the last solving using power rule part I'm confused about. There wasn't a power rule in my textbook

I canceled out 2x and got the 1/√u, not sure where to go from there

[distant nymph]

Are you sure that you haven't been taught power rule? Seems odd.

$$\int x^n \dd{x} = \frac{x^{n+1}}{n+1} + C$$

[thorny flame]

Enemagneto

[mild reef]

I missed a ton of class so actually we might have gone through it at some point

[distant nymph]

This?

[mild reef]

OH

[distant nymph]

It's arguably one of the most fundamental and useful rule you learn.

In integration, i mean.

[mild reef]

I GOT IT - THANK YOUU!!!!

.close

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[mossy nest]

On a set of n vertices, we construct a random undirected graph without loops by connecting each pair of vertices with probability p. Let the random variable X represent the number of isolated vertices of the resulting graph.

[mossy nest]

can someone help me with calculating E(X)?

[topaz sinew]

@mossy nest Has your question been resolved?

[distant nymph]

Do you know how many undirected graphs without rings can be formed with n vertices?

[topaz sinew]

@mossy nest Has your question been resolved?

[mossy nest]

what do you mean by rings?

[distant nymph]

Loops

[mossy nest]

i think they do not mean loops but edges from vertex v to itself

[distant nymph]

Oh.

That makes the question way easier.

Lol. I ended up finding Catalan numbers and what not.

[mossy nest]

then how to caluclate it?

i found that if we have n vertices, there is n(n-1)/2 possible edges

[distant nymph]

Anyway, so how many undirected graphs are possible with n vertices such that no vertices are isolated?

[mossy nest]

uff

dont know how to calculate that

im not sure whether we need that

[distant nymph]

Yeah. I think that there might be another way to go about it.

[mossy nest]

let X is discrete random variable that denotes number of isolated vertices:

if X=k, then

$P(X=k)=\left( \frac{n(n-1)}{2} - k \left( n-1 \right) \right) p \left( n-1 \right) \left( 1-p \right)$

[thorny flame]

Michal

[mossy nest]

im not sure whether its correct

first half describes that all vertices except k has edged and second part describes that k vertices are isolated

[distant nymph]

If k vertices are isolated, then the remaining (n-k) vertices can have (n-k)c2 edges. How do we consider possibility of there being an edge between any two vertices among those (n-k) vertices?

[mossy nest]

hmm

its wrong

[distant nymph]

Because all (n-k)c2 edges don't need to be there for those vertices to not be isolated.

[mossy nest]

the solution from my lesson is way simpler but i dont get it

[distant nymph]

Let me think for a bit more time. I wanna try it properly.

[mossy nest]

So don't look at the photo

It's not in English...

[distant nymph]

Okay. I don't think my way works because apparently there is no clean formula for "number of unlabelled undirected graphs with n vertices such that no vertices are isolated".

Try asking in #probability-statistics once.

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[edgy quail]

!status

[topaz sinew]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[edgy quail]

4

<@&286206848099549185>

[sonic escarp]

to be surjective, there must exist an x for which f(x)=0. which x is this?

[edgy quail]

shit

sorry mate

wait can i say this tho

like there is a number d such that

there is no a s.t. f(a)<d, and

there is a for all f(a)>=d

then can i proceed with the sol

[sonic escarp]

how do you argue this?

i get f(f(f(f(n))))=f(f(n)+1)+1=f(f(n+1)+1) which is different to your statement.

[topaz sinew]

@edgy quail Has your question been resolved?

[edgy quail]

ah shit

sprry mate

thanks anyway

.close

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[neon iron]

Yo

[foggy bronze]

yo

nice qs i see

[neon iron]

Yo

Can anyone help me

[pearl peak]

send the question

[lusty yacht]

Resent the question

[neon iron]

Number 15

[foggy bronze]

,rotate

[foggy bronze]

Ans given in terms of log?

[neon iron]

Yes

Expand

The log

[pearl peak]

use log rules: log(ab) = log(a)+log(b)

[neon iron]

Can you show me

[foggy bronze]

First the exponent one?

[neon iron]

Ye

[pearl peak]

theyve already used the exponent one thats why i didnt mention but yes that one also

[neon iron]

Can you show me what the answer would b

[foggy bronze]

oo didnt see that

[neon iron]

Can you show me plz

[neon iron]

Where would the 1/3 go

From the rad

[pearl peak]

okay sure

are you sure you have attempted the question properly though?

[neon iron]

No

[pearl peak]

the 1/3 will be distributed itll be 1/3 (log base 3 of (2x^5))

[pearl peak]

try the question using this rule

try to split up the 2 and x^5

[neon iron]

log3(2x^5)=log3(2)+log3(x^5)

[pearl peak]

can bring the down as well

and dont forget the 1/3

[neon iron]

Where would the 1/3 go

[pearl peak]

final answer: ||1/3log3(2)+5/3log3(x)||

[pearl peak]

itll be 1/3 [log3(2x^5)] right, so the 1/3 goes in front of each term of the LHS

[neon iron]

Oh thanks

Can you help me on one more

[pearl peak]

ye sure

[neon iron]

#13

,rotate

[neon iron]

#13

[pearl peak]

bring the 3 down

note that log(a/b) = log(a) - log (b)

[neon iron]

Answer?

@pearl peak

<@&286206848099549185>

@foggy bronze

[topaz sinew]

@neon iron Has your question been resolved?

[pearl peak]

Don’t forget to distribute the 3

There should be a 3 in front of the log2(y)

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@neon iron Has your question been resolved?

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[fair pulsar]

how would you get the values of x from this as my answers were wrong

[restive inlet]

what's part a),
and show your work

[fair pulsar]

part a is irrelevant as its just proving that 11cos^2-10cos=-3sintan is equivalent to 8cos^2-10cos+3=0

[fair pulsar]

.

<@&286206848099549185>

[pearl cove]

t = cos x

8t^2 - 10t + 3 = 0

[vale furnace]

What is x?

[fair pulsar]

in the range of 0<=x<90

[vale furnace]

And x is?

[pearl cove]

(4t-3)(2t-1) = 0

t = 1/2

t = 3/4

[fair pulsar]

cool!

thats where i was getting stuck 🙂

i couldnt find out where they got 1/2 and 3/4 as i hadnt subbed in t

[pearl cove]

=> x = pi/3 + k2pi

[pearl cove]

=> x = arccos(3/4) + k2pi

[fair pulsar]

thank you 🙂

imma close

.close

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[pearl cove]

what grade are you in ?

[fair pulsar]

A level but smoothbrained this

[topaz sinew]

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[plush glacier]

Does anyone know how to solve #1?

[dense ridge]

Formula for the difference of square

[plush glacier]

I tried it but i have no idea how it can be all canceled out

[past knoll]

first question right?

[plush glacier]

Yess

[past knoll]

1-(1/(n+1)^2) formula then

[dense ridge]

For example
In first term
1-(1/2²)=(1+(1/2))(1-(1/2))=(3/2)(1/2)

Then doing the same thing with the other terms

[past knoll]

after doing the same thing

it will show that u got more than half every time u times it

so it answer 1

u will got like (3/4)*(8/9)..... in the pattern of ((n+1)^2 -1 )/(n+1)^2

[plush glacier]

Do you have to cancel out?

This is what I did so far but I am stuck. Do i do more terms than the given?

[dense ridge]

If you do more terms you will find
For(1-1/2²)(1-1/3²)...(1-1/n²)
[k/(k+1)] and [(k+1)/k] cancel together and k is from 2 to(n-1)
the remaining part is in the first and the last paratheness
You can list the case when n is small to check this result

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[neon iron]

Why doesn't this work?

[acoustic pecan]

if x was -40 then the left would be -40*(-52)

which isnt -40

you can do stuff like this if x(x-12)=0

but with a different constant it doesnt make much sense

[neon iron]

Hmm okay it's just a property of zero to do that, got it

Thanks

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[fleet imp]

https://math.stackexchange.com/questions/2989242/verifying-an-increasing-function-on-a-closed-interval-is-riemann-integrable

[fleet imp]

In this proof he says $x_j-x_{j-1}<\varepsilon/(f(b)-f(a))$. why?

[thorny flame]

jsidind810

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[fleet imp]

<@&286206848099549185>

[fleet imp]

<@&286206848099549185>

[zealous wigeon]

think f(b) - f(a) = m is constant

He chose e/m because in further down the proof it get cancels out by m*e/m = e which essentially completes the proof

Since epsilon is arbitrary it doesn't make a different to choose epsilon or epsilon*constant

So it was to make proof much more straightforward

[topaz sinew]

@fleet imp Has your question been resolved?

[fleet imp]

but how can he just say $x_j - x_{j-1} < \epsilon$

[thorny flame]

jsidind810

[fleet imp]

for any epsilon > 0?

oh

because you can pick $n(\varepsilon)$?

[thorny flame]

jsidind810

[zealous wigeon]

You make partitions like really small, so supf and inff are close, for reimann integrability

[fleet imp]

i.e. you make n really large

[zealous wigeon]

We chose n based on e so yup

e=epsilon

[fleet imp]

ye

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[calm gyro]

Can someone help with this <@&286206848099549185>

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[neon iron]

are my steps incorrect?

[topaz sinew]

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 was told that it's wrong.
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7. None of the above

[neon iron]

I'm getting the wrong answer

[loud ingot]

What is the question

[neon iron]

this right here

it's asking me to evaluate the trig ratio

[loud ingot]

sin - 210?

[neon iron]

yes

[loud ingot]

Sin of what?

[neon iron]

-210

[loud ingot]

Should be sin (-210)

[neon iron]

I guess so

it has -210 ^ degree symbol

but that's besides the point

[loud ingot]

Explain your diagram

[neon iron]

I drew the degree, found that -180-(-210) was 30, (my reference angle), from there determined the rest of the info

[ruby tree]

Why is the hypotenuse 2?

[neon iron]

is it not supposed to be?

corresponding to the 90

[loud ingot]

How do you come up with hyp

[neon iron]

just out of memorization of the special triangle

[loud ingot]

Cool

[ruby tree]

I see

Well, your answer is 1/2, right?

[neon iron]

I'm not concerned about my diagram, nor sin being opposite over hypotenuse

yes

[ruby tree]

It's correct

[loud ingot]

Look up the trig cycle

Very useful

[neon iron]

Oh I see what I did wrong

I mistyped the question into a calculator

instead of putting 210 i put 120

[ruby tree]

Haha nice

[neon iron]

thanks for verifying

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[fierce sand]

okay no one using this?

[fierce sand]

question #49

that's my work above

this is the answer

pretty close

my question is, why does the left side of the graph to the left of the -2 VA cross above the HA?

when can i tell when the graph will pass the horizontal asymptote?

<@&286206848099549185>

[keen venture]

Don't expand everything just to find leading coefficients. You can tell the leading coefficient is 1, just by multiplying the leading coefficient of every factor

I don't think you can tell without calculus. I'll think on it, though.

@fierce sand

[fierce sand]

okay ty

[fierce sand]

oh gotcha okay thanks

@keen venture

I figured it out

If I plug in y = 1

Which is the horizontal asymptote

And it return an x value

Then it crosses

Otherwise it's undefined and never crosses the HA

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[late flame]

I thought the correct answer would be just plugging in the bounds into FTC like f(x^6)-f(x^2), but why are they multiplied by the derivatives of the bounds?

[elfin halo]

First integrate that shit

And you’ll get a function. Of X

[late flame]

but isn't the expression in the integral already f(x), or F'(x)? why would we need to integrate it?

[hollow swallow]

The first thing that came to my mind was the chain rule.

[late flame]

yeah, but i'm not sure how they reached a step where they could apply it

[strange whale]

derivative of F(x^6) - F(x^2)

F(x^6) is a composite function

the composition of F and x^6

[late flame]

ohhhh

got it!

thanks guys

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[wet pecan]

1d

[wet pecan]

This is what ive tried

[hollow drum]

,w simplify (2ysqrt(1+x^2y^2)-xy^2)/(x^2y-2xsqrt(1+x^2y^2))

[thorny flame]

Results provided by WolframAlpha

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

Requested by captainnova22

[hollow drum]

Looks like your answer is the same as the answer key of y' = -y/x

You just need to simplify your answer

[wet pecan]

My teacher said that once weve solved for y’ we dont need to simplify algebraically

So i don’t know if he would count that as a correct answer

[hollow drum]

Well your answer is the same as the key, if you simplified it. So if you don't need to simplify it, then what you have is correct

[wet pecan]

I see

Ok thank you

I appreciate it

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Please don't occupy multiple help channels.

[frosty belfry]

you need to close one of your channels

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[ivory wave]

they left the server

<@&268886789983436800>

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[shrewd shore]

Did I at least do this right?

[vale furnace]

Yeah

[shrewd shore]

Okay, so what would be my next step then since the website isn't accepting my answer

[vale furnace]

Ah

[gleaming reef]

xybe

[vale furnace]

should be + 63/64

[gleaming reef]

ur back?

^^

[shrewd shore]

Wait what

How +

the - has to be distributed

[gleaming reef]

what is 1 - 1/64

[vale furnace]

1 - 1/64 = 64/64 - 1/64

[gleaming reef]

^

esta persona es correto

[shrewd shore]

oh

it still didn't take it, unfortunately

[vale furnace]

Idk I don’t think I’m missing anything

Maybe times everything by 64

[shrewd shore]

I normally don't have to rationalize the denominator... maybe I have to here?

[vale furnace]

Oh wait

Why has it become x^2

Should be -x^2

[gleaming reef]

mhm

[shrewd shore]

It worked

I need coffee

I am making dumb mistakes

Thank you

[vale furnace]

Np

[shrewd shore]

.close

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[plush otter]

is this set countable? I think yes but I dont know how to explain it

[opal vault]

First of, are the Z^i countable?

[plush otter]

Isn't it?

At least for i=1 and i=2 it is

[opal vault]

Well I'm asking if you can explain why

Maybe a little induction wouldn't hurt

[plush otter]

Hmm

I'm not sure how haha, for i=1 and i=2 i was thinking in terms of Hilbert's hotel, I'm not sure how to generalise

[odd pagoda]

have you seen the proof that Q is countable?

[plush otter]

Yes

In the context of Hilbert's hotel infinite buses

[opal vault]

Here's a hint on how you might continue for higher powers of i

Z² ≈ Z (as it is countable)

So let's write Z³ = Z² × Z

[plush otter]

Ohhhh

I get it now thanks

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[leaden rock]

What is i^5

[ivory wave]

in what context

[leaden rock]

i is sqrt-1

Idk how to do the visual

[willow bone]

i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
you just need to follow the pattern

[leaden rock]

Pattern?

[strange topaz]

i'd start from ^0

[willow bone]

yeah

since sqrt(-1) = i, that meants i^2 = -1

[strange topaz]

1,i,-1,-i

0,1,2,3

[leaden rock]

So it repeats every 4?

[strange topaz]

yes it does

[willow bone]

yea

[leaden rock]

Why is that

Guys can u teach me how to do the visuals

{i}

Idk

[meager dawn]

yes

[strange topaz]

https://www.youtube.com/watch?v=CgztSanrYf0

[leaden rock]

Thanks

[meager dawn]

actually yeah that video is probably better than a discord text message lol

[hollow drum]

Wdym visuals? Latex? Because #latex-help

[leaden rock]

Oh ok

$ I know how to type now $

Or not

$hello$

$hello$

[thorny flame]

Edge

[leaden rock]

Why does $e^(i*pi) = -1$

I can’t type

U know what I mean

[bitter hemlock]

$e^{i\pi}=-1$

[thorny flame]

Soosh

[leaden rock]

Oh

Yeah

[bitter hemlock]

it's a complex topic

i.e. it has to do with complex numbers : )

[leaden rock]

I know that $e^{i\theta} = \cos(theta) + i\sin (theta)$

Oops

Almost worked

[bitter hemlock]

use \cos \theta \ sin

[leaden rock]

Ok

[bitter hemlock]

you can just edit past msgs btw and the bot will refresh the tex rendering

[thorny flame]

Edge

[leaden rock]

Good enough

And there is a Maclauren expansion series for them

[bitter hemlock]

watch this
https://www.youtube.com/watch?v=v0YEaeIClKY&ab_channel=3Blue1Brown

[grave widget]

3 blue one brown is goated

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[earnest wagon]

Why is this neither continuous nor discrete?

[earnest wagon]

X is the beta distribution here

Y=g(X)

[potent silo]

We say a random variable is continuous if its cdf is continuous on R

As you can see, F is not continuous in 0

So Y is not continuous

[earnest wagon]

But why isn’t this just considered a discrete rv?

[potent silo]

Finally it is not discrete anymore because the image of CDF of a discrete random variable is finite, i.e., |F(R)| < infinity

or if you prefer, card(F(R)) < infinity

And here, this is not the case

[earnest wagon]

I’m a bit confused on that, could you point out why Y doesn’t satisfy that?

[potent silo]

Ok, because actually $F(x)$ can take any values in \
${0} \cup [1 - (1/2)^\theta, 1]$

[thorny flame]

wlmmm

[earnest wagon]

I tried working out the pdf for the case when Y=x but I get a constant value rather than the interval you have?

P(Y<x)=P(Y=0)+P(X>1/2)=1

[potent silo]

What is x?

[earnest wagon]

My bad I see where I went wrong

[potent silo]

You see, it could be simpler to look at the possible values for Y instead of working directly with the CDF

X takes value in [0,1]

The possible values for Y are 0 (if X <= 1/2), and the probability P(Y=0) is non zero
and if X > 1/2, then Y = X, so Y can take any values in the interval [1/2, 1]

Hence, it cannot be discrete

Otherwise, the possible values for Y would be a finite list of numbers (and all of them would have non zero probability)

[earnest wagon]

That makes sense, but if I was to go with deriving it through the cdf how would it be

For y>0.5, P(Y<=y)=P(Y=0)+P(0.5<X<y)=P(X<y)=1-(1-x)^theta?

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[junior nimbus]

Hello

im using the work energy theorem

and I know that Wnet = 0

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[tardy cove]

anyone knows how to do this question？

[tardy cove]

You have 200 meters of fencing to enclosed two rectangular areas (as in the figure), both areas
are the same size. The total area of the enclosed region is 1400 square meters. What is the dimension
of each area?

[wispy tiger]

Put

Variables

For sides

@tardy cove

U there

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[candid knot]

Can someone check my answers to these problems

[hollow drum]

You can graph it to check your work instead of getting people to check it for you

Even if it says no calculator, you're just graphing it to confirm your answers

[candid knot]

Ok

Will do

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[jagged meadow]

I'm not in school but if this is considered cheating please let me know, and I will remove my question. I bought a TI Nspire CX II calculator (thinking it had CAS) from Office Depot to help me with my independent calculus studies. I didn't get a chance to use it for a couple weeks, and now that I've started using it and realized it cant do simplification like 4x/2 -> 2x, I was wondering if there's a way to put CAS software onto my non-cas calculator

[gleaming reef]

I wouldn't do that

adding CAS software to an non-cas calculator could allow you to gain unfair advantages on assessments

that require non-cas calculators

[jagged meadow]

I'm not in school, does that still apply generally?

am I just out $165? They wouldn't let me return it

or do you have any possible suggestions on where I might be able to trade it in?

[gleaming reef]

https://education.ti.com/en/software/details/en/3dc447eb50e0466697761768fe3b6391/ student-nspirecxcas

@jagged meadow try this

[jagged meadow]

I'll check it out, but if that doesnt end up working, do you have any suggestions on places I might be able to to trade it in, if such a thing exists?

[long dawn]

It would likely apply if you do plan on taking any assessments I guess

[jagged meadow]

I dont 😅

[long dawn]

But if you're not in school or using that calculator on assessments then it's your calculator, you do you

[jagged meadow]

not currently, but if I do, I plan on only using school provided calculators on tests. Not planning on carrying around a $165 calculator just to do the basic stuff allowed for tests

[jagged meadow]

ideally I could just get the cas version, but given office depot wont take it back, I'm thinking it's just not going to happen

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@jagged meadow Has your question been resolved?

[jagged meadow]

how does one manually flash it? I'm only finding things about nspire connect

it shows screenshots of stuff but not what the software is called or where it can be downloaded

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@jagged meadow Has your question been resolved?

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@jagged meadow Has your question been resolved?

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@jagged meadow Has your question been resolved?

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[cinder oxide]

why is the orange line equal to (red line - green line)?

if i were to eyeball it, orange looks much larger than that

[long beacon]

looks fine to mee , also its vector subtraction

[cinder oxide]

meaning, not magnitude?

we are not getting the length of the orange line?

what are we getting instead?

[knotty ledge]

Another vector

[cinder oxide]

with the origin at proj tip? and the tip meeting with u tip?

[long beacon]

sounds good (idk about saying it origin)

[cinder oxide]

origin is where it starts

[long beacon]

fair

[cinder oxide]

for orange vector it starts here

[long beacon]

yeah

[knotty ledge]

Really all vectors start at the origin

[long beacon]

i guesed he was reffering origin as like the tail of the vector not the (0,0,0)

[knotty ledge]

Yeah I guess i should clarify that you can translate a vector whereever you want and it is still the same vector, we just care about magnitude and direction

[cinder oxide]

meaning all vectors have origin (0,0,0)?

or in the case of orange vector, it's origin is not (0,0,0)?

it's here instead

in other words...
the origin for a 2d graph refers only to the coordinate (0,0)?
but the origin for a 2d vector can refer to a different coordinate?

[long beacon]

fair

[long beacon]

vectors have a freedom of being picked up and placed and still having the same significance , if it carried the same magnitude and direction

[long beacon]

you can translate it to origin and it will still say the same

[long beacon]

no?

[cinder oxide]

translate to graph origin you mean

[long beacon]

i mean 2d only has 2 dimentions

[cinder oxide]

i am talking about vector origin

[long beacon]

if you say (0,0,0 ) <-- reffering to 3d

[long beacon]

tail?

[cinder oxide]

is it called tail and point?

start and end of vector

[long beacon]

i mean vectors initial point is tail

(some reffer to it as tail i should say)

expresses what you are telling more clearly

[long beacon]

and the arrow part as head

[cinder oxide]

ahhh, OK

it's tail and tip

[long beacon]

fair from where i am its reffered as head(tip)

[cinder oxide]

tail is starting
tip is arrow (end)

OK

tyvm

.close

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[long beacon]

np

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[neon iron]

I can find only one solution for each question

In answer, for 0 to π/2 there is different answer and for π/2 to π there's different answer

And I have no idea first why there's different answers and I also can't find

The different answer which is given

[keen raptor]

?

That shouldn't be true

[neon iron]

Yeah

I can send ss

[keen raptor]

Sure

[neon iron]

Answer for i)

I founded the 0 to π/2 one but for π/2 to π I have no idea

For every one there's answer like that except fourth one and third one have three

@keen raptor what do you think 🤔

[keen raptor]

Oh I see, weird

Wtf kind of class is this

[neon iron]

Oh it's 11th

[keen raptor]

Indian?

[neon iron]

Yes

[keen raptor]

They need to do quadrant by quadrant because of things like sec(a) being positive or negative

the r should be positive

but sec(a) changes signs at pi/2

Things like that

be back in a bit

[neon iron]

Ok

I thought the same for a moment but then there should be answer for every quadrant right?

[keen raptor]

Sometimes they happen to be the same

[neon iron]

I dont how should I proceed for that

And how will I know for which quadrants I had to calculate

See the answer for third one it's same for 0 to 90 and then same for 90 to 270 and then for 270 to 360

And for second one there's answer just like first one

[keen raptor]

Well, when they're the same you combine them

If it's the same answer for two quadrants

[neon iron]

Ok so I should replace cos 90+ theta and sin 90+theta and then 180 and then 270

Like that

?

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@neon iron Has your question been resolved?

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[copper valve]

Hello

[copper valve]

Ive calculated that b= 38

[cobalt dawn]

thats correct

[copper valve]

What do I do now?

Is C = D?

And B = E?

[cobalt dawn]

angle <AED is the same as <ABC

[copper valve]

Mhm

[cobalt dawn]

then <AEB = 180 because its a straight line

then you can find <BED

[copper valve]

What do you mean straight line?

Every triangle has one

Oh I’m stupid

I get what your trying to do

Then 180 - 38

[cobalt dawn]

yep

[copper valve]

Ima take a random guess

8/12 * 20

= DE

<@&286206848099549185>

[sweet shard]

!occupied

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

[copper valve]

I haven’t figured it out yet

[sweet shard]

@terse acorn read this

[jagged valley]

question?

[copper valve]

.

[sweet shard]

<@&268886789983436800>

tried warning

[copper valve]

There’s math help available

Help 10

[knotty ledge]

go to any channel in "Math help available"

[copper valve]

19

24

Ok back to my question

[copper valve]

.

[cobalt dawn]

AE = 12 - 8 = 4

then you get 20/x = 12/4

[copper valve]

Why do I do AB / AE?

[topaz sinew]

@copper valve Has your question been resolved?

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@copper valve Has your question been resolved?

[cobalt dawn]

well when you have two similar triangles with sides a, b, c and a',b',c' you have the following
$a/{a'} = b/{b'} = c/{c'} = k$

[thorny flame]

Scythed

[cobalt dawn]

in this case AB is a and AE is a'

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[quasi sedge]

what is speed of sound in air at -20 degree celsius

[ruby tree]

,w what is speed of sound in air at -20 degree celsius

[thorny flame]

Results provided by WolframAlpha

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Requested by nelfeal

[ruby tree]

Did you want to add any context to that?

[quasi sedge]

what formula did we utilise here

[ruby tree]

Probably something like this:

,w (331.3+(-20)0.606) from m/s to km/h

[thorny flame]

Results provided by WolframAlpha

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Requested by nelfeal

[ruby tree]

There you go

(source: https://en.wikipedia.org/wiki/Speed_of_sound)

[quasi sedge]

okay thanks

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[tender girder]

tan doesn't make sense

tan a is 4/3 and tan b is -15/8???

[knotty tusk]

did u do this tan?

[tender girder]

I tried the tan identity and got 181/24

I got 77/24 in the numerator

[knotty tusk]

1 + at the bottom

[tender girder]

yes

[knotty tusk]

its sus that it so small

but i believe u

[tender girder]

so ummm I have 4/3+ 15/8

which is 32/24+45/24 right?

[knotty tusk]

maybe error in simplify, because denom is fraction

why not leave fraction in denom? Maybe website isn't picky

[tender girder]

I have 1 + (4/3*15/8) = 1 + 60/24 = 1 + 10/3

in the denominator

[tender girder]

I think it is

I tried turning the +1 into a part of the fraction and that didn't help much, this is so hard

I hate this so much

how can this possibly be negative

I'm so upset

I've spent so many fucking hours on this problem and nothing fucking works

math is a fucking sham

[knotty tusk]

tan addition identity is pain

[tender girder]

it is

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[earnest wagon]

Can I just use rank nullity, we know eigenvectors are distinct so create a basis with eigenvectors and extend it to V. The ker of T is greater than 0 by condition i thus dim(v)>=2+1

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[scenic brook]

What do I do for this?

I get to this

[sweet shard]

missing square on the constants

[scenic brook]

wdym

[sweet shard]

Look at the formula for arc length again

Oh you're missing squares on everything

[scenic brook]

oh

lmao

I see

.close

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[pearl pulsar]

For this problem, my work is this but I'm having trouble finding a bound for the "less than" part of the inequality

$B^ \frac{1}{n}$ I can bound it easily because $B > 0$

[thorny flame]

Jesses