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[naive sorrel]

How would you go about taking the reciprocal of

[naive sorrel]

$\color[rgb]{0.35,0.35,0.35}x+(y/a)$

[ocean seal]

WOWWA

[naive sorrel]

can you say

$1/(x+y/a)$

[ocean seal]

WOWWA

[naive sorrel]

then multiply by $a/a$ on top and bottom

[ocean seal]

WOWWA

[naive sorrel]

is that valid

[minor needle]

multiplying by this rather wouldn't get anything

try to add x + y/a making common denominator

then simplify

[naive sorrel]

oh i see

but that yields the same solution as my method i believe

so my method would be correct then?

[minor needle]

but your method gives

ah in fact it would be the same

ye

[naive sorrel]

alright

thanks

.close

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[jaunty socket]

how do i answer this?

[minor needle]

use fundamental theorem of algebra

I mean factored form of a polynomial

Assume:
$$f(x)=a(x-x_{1})(x-x_{2})(x-x_{3})$$

[ocean seal]

Modus

[minor needle]

where x1, x2, x3 - roots

to find leading coefficient "a" substitute point

[placid zinc]

Note the strategy here is to construct the cubic. Once we do, we can just plug in to find the answer.

[jaunty socket]

so would i just plug the point in to the equation you gave me?

[jaunty socket]

plus the point in here?

plug*

[minor needle]

Have you plugged roots into?

[jaunty socket]

no

[minor needle]

so let's do it

[jaunty socket]

okay wait so what should i write first?

[jaunty socket]

should i write this first?

[minor needle]

assume x1 = 2, x2 = 3, x3 = -5

then write this

[median oar]

When x = 2 it equals 0

You know at least this cubic can be factored into some form (x-2)(quadratic)

Same for x=3

[dry ridge]

Sorry what do they mean by “has a value of 120”?

[minor needle]

f(x) = 120

[median oar]

(x-2)(x-3)(linear)

[dry ridge]

At which point?

[minor needle]

this is the question

[median oar]

That’s the question

[dry ridge]

Ok.

[jaunty socket]

ok i have written this down andthen wrote down the equation you gave me

[jaunty socket]

✅

[median oar]

What did you write down

[jaunty socket]

here lemme send what i have so far

,rotate

[minor needle]

now

[median oar]

Never mind I’m blind

[jaunty socket]

gotcha imma do that now

ok just subbed them in

[median oar]

Now can you solve for a

[jaunty socket]

how do i solve for a?

[median oar]

What information from the question haven’t you used yet?

[jaunty socket]

the points it passes thru

[median oar]

Right

A straight line is defined by 2 (unique) points

A parabola 3

A cubic 4

You have 3 roots that’s 3 points

[jaunty socket]

gotchu

[median oar]

Your 4th point will define a unique cubic

[jaunty socket]

ok so what do i put for a?

[median oar]

Well it goes through what point

[jaunty socket]

4,36

so would 4 be subbed in for x and for f(x)

and a be subbed in for 120?

oh wait

where do i put the 36?

[median oar]

Why are you subbing in 120

We don’t know what a is right now

We are trying to find it

[jaunty socket]

nvm i just sub 4 in cus its the x value no?

[median oar]

Yes and what for y

You also know the y value because the curve passes through this point

[jaunty socket]

i know the y value is 36 but idk where itshould go

[median oar]

What is x

[jaunty socket]

4

[median oar]

Ok what is y

[jaunty socket]

36

[median oar]

Ok how do you find y for some value of x

[jaunty socket]

you plug x into the equation and solve

[median oar]

Right

So your equation is something something = f(x)

And you know f(4) = 36

[jaunty socket]

yea

[median oar]

So sub in x = 4 for you equation and make it equal to 36

[jaunty socket]

gotchu

[median oar]

Say I have some function f(x) = x+ 1

When x = 2, y = 3

f(2) = 2 + 1 = 3

Now if I said f(x) = x + a

And when x = 2, y = 3

[jaunty socket]

a is 1?

[median oar]

Then you write f(2) = 3 = 2 + a, a must be 1

So use this method to find a for your cubic

[jaunty socket]

ok just a sec

ok lemme send what i got

,rotate

[median oar]

Ok so you found a

[jaunty socket]

perfect

[median oar]

So now you know exactly what f(x) is

Now find f(x) = 120

Find x

[jaunty socket]

okay imma try that rn

so rn i have 120= (x-2)(x-3)[x-(-5)]

what should i do from here? expand?

[median oar]

Where did the 2 go

Kinda sus

[jaunty socket]

ohhhh omg

ye ok so including that should i expand?

[median oar]

I mean

That wouldn’t be my first thought

I’d try like x=7

[jaunty socket]

oh just guess and check type thing?

[median oar]

5*4*2 = 40 ok not quite

Maybe 8?

6*5*3 = 90 ok too much

[jaunty socket]

i guess ill do 6

[median oar]

Wait

That’s wrong lmao

[jaunty socket]

owh

[median oar]

It says +5 at the end

[jaunty socket]

wdym

ye it is adding 5 not subtract

[median oar]

7 would be 5 * 4 * 12

Way too much maybe 6?

[jaunty socket]

thats just 88

so weird

so its like 6 point something?

nono

oops

1 sec

6 is 264

ill do 5

got it

its 5

[median oar]

See that’s a bit easier than expanding

In this factored form it’s easy to guess and check

[jaunty socket]

so now what is the next step?

[median oar]

Nothing

You found it

[minor needle]

Fun thing is there are also two more, you can see that 60 = (4)(3)(5) (also possibilities with negative signs)
so you need (x-2)(x-3)(x+5) = 60

[median oar]

x=5

Well

Yeah that’s possible too

[jaunty socket]

what would my therefore statement be? therefore the cubic function with zeros at 2, 3, and -5 that pass through the point 4,36 and have a value of 120 is 5

but like shouldnt i write a full cubic function?

[median oar]

Also since you know the graph is - + - +

You know that if there is another sol it has to be between -5 and 2

Since the one you had was above 3

Could be 2 more sols even

[jaunty socket]

uhh what is a sol

[median oar]

Solution

[jaunty socket]

oh lmao

but my friend did it and got one solution, with a cubic function

[median oar]

You know there’s at least 1 sol

You know know there’s only 1 yet

[jaunty socket]

what should i write for my therefore statement?

<@&286206848099549185>

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[limpid moat]

can someone help me solve b)?

[raven dagger]

1 ?

[fallen verge]

I may be wring but i think its ||22C18||

[raven dagger]

Oh wait no

Didn't read full combination

[limpid moat]

is it 22C18?

[raven dagger]

Yes

[limpid moat]

okay thank u

.close

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[edgy flare]

Complex Variables: Laurent Series and Residues

I’m looking at other examples it looks like you are supposed to find a way to manipulate the equations by factoring out and keep doing this until you have a couple terms then a term that turns into a power series.

[fallen verge]

Ah more cv

[edgy flare]

I’m confused on how to determine what to factor out and how I know I’ve reached that term to turn it into a power series.

I tried this but didn’t really know where I was going with it.

[fallen verge]

Im gonna try to learn laurent series for you

[edgy flare]

Good luck.

I tried using this (formula for a geometric series) but I’m not sure if this is correct or valid to do.

[merry depot]

depends on z

that works only when |1/z^2| < 1

[fallen verge]

Yeah i have no clue

I do know however that z^2/(z^2-1)=1/(z^2-1)

[edgy flare]

Is there a way to tell if it is less than 1?

Or do I just write that and say valid for |1/z^2|<1

?

[fallen verge]

|z^2| > 1 i believe

Which in turn means |z| > 1

[fallen verge]

I also dont think this is true bc the function is defined and we're using what is essentially analytic continuation

Assume i replied to the next message

[edgy flare]

<@&286206848099549185>

I also tried this. I don’t know what is right and what is wrong though.

[lone heart]

@edgy flare Has your question been resolved?

[lone heart]

@edgy flare Has your question been resolved?

[naive valley]

the problem asks for the series expanded around the point z=1

your series is around z=0

you want powers of (z-1), not powers of z

[edgy flare]

So just input z-1 where I have z?

[naive valley]

well you can't just replace z with z-1, that won't be the same function

but maybe you could do a similar manipulation to get something of the form 1/(1 - (z-1)^2) (or some other power of (z-1)) ?

[placid zinc]

Something I like to do is substitute:
u+1 = x

Then write a series in terms of u, and then substituting back gets it in terms of (x-1)

[edgy flare]

You mean u+1 = z?

[placid zinc]

Yeah that

[edgy flare]

I’ll try that in the morning. Probably post it again tomorrow to see if I did it right.

Thanks

.close

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[alpine sable]

Hi can someone help me with my set theory work?

[alpine sable]

Im having trouble decoding these 4 questions

[placid zinc]

I think "PrOvE iN sEt ThEoRy" just means "prove" here

Just, be sure to use your known definitions, as you would with any other proof.

[alpine sable]

I have recently started taking this course, and am having trouble catching on with the proofs :(

[placid zinc]

That's fair, it can be hard to start

Trust me when I say that you're not alone

Unfortunately, since I'm not certain what definitions you're expected to use, I can't help with all of them

Uh, 2 seems possible though

What is a total order?

[alpine sable]

If it helps, we’re using the ZFC set theory

total order requires us to be able to order all elements in a set

[vale wigeon]

youll probably want to have a formal definition of Z here that youre expected to work with

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@alpine sable Has your question been resolved?

[lone heart]

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[gray ingot]

You should ask this in the physics server #old-network

[vale wigeon]

or perhaps tell us what the relationship is between surface temp and wavelength (or freq) of light emitted

@rich basin

[lone heart]

@rich basin Has your question been resolved?

[willow canopy]

Wien's law is

$\lambda_{max} = \frac{b}{T}$ where $b$ is the wien's constant which i forgot what is it.... nice

[ocean seal]

Azzurala

[rich basin]

yeah

so I used lambda = c /f

in which f = 750 * 1 * 10^12

and thn plugged into wein's law

[vale wigeon]

,w wien's law

[ocean seal]

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Requested by Ann#0413

[rich basin]

so how would we find max wavelenftyh

[willow canopy]

the star is red

[rich basin]

so how would a star be red explain the maximum wavelength?

[vale wigeon]

the peak wavelength determines which color you see the star as, no?

[rich basin]

oh yeah

thank you so much

.close

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[obtuse hatch]

hmm

[harsh girder]

https://en.wikipedia.org/wiki/Rolle's_theorem

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@obtuse hatch Has your question been resolved?

[obtuse hatch]

.close

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[quartz cave]

Hello

[quartz cave]

Understanding the Problem: This is a "prove" type problem.
The hypothesis is that a, b, and c are odd integers, and the conclusion is that equation ax2 + bx + c = 0 can not have a rational root.
The hypothesis is straightforward. In the conclusion, "rational root" means a root, that is, the value of x that satisfies the equation, and that can be expressed as m/n, where m and n are integers. So the conclusion means that there is no number of the form m/n that satisfies the equation under the hypothesis.```

``the value of x that satisfies the equation, and that can be expressed as m/n`

How can I infer that x can be expressed as m/n

?

Are there any rules for this, is this just something I should know?

[vale wigeon]

"can be expressed as m/n, where m and n are integers" is the definition of rational. does this answer your question?

also you're missing a ^ in ax**^**2 + bx + c = 0

[vale wigeon]

yes

[quartz cave]

Its how they wrote it

[vale wigeon]

they wrote it as an exponent

in plaintext to write exponents you use this symbol: ^

ax^2

[quartz cave]

I think so yeah, a rational number can be expressed as m/n where m and n are integers. Where as irrational numbers cannot be expressed in the same way?

[vale wigeon]

that is what it means for a number to be rational.

[quartz cave]

Ah I see what you mean

Ok got it

So next time I read rational number in a problem, I should immediately think m/n

[vale wigeon]

i don't think so

you should not attach yourself to the letters m and n

but "rational number" means "ratio of two integers" and it always pays to be aware of that, yes.

[quartz cave]

Ok thank you, that answers my question 😁

.close

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[long kayak]

$\left( 1-\prod_{k=0}^\infty(1-(1-\rho)^{(1+\delta)}\rho^{x+k})^{\Delta_k} \right) \leq \left( \sum_{k=0}^\infty \Delta_k(1-\rho)^{(1+\delta)}\rho^{x+k} \right)$, $\rho \in (0,1), \Delta_k \geq 0, \delta >0$ why does this inequality hold, on the text is says something about expanding the infinite product up to the first-order terms, but it doesn't help me

[ocean seal]

Annie.

[carmine bronze]

fuck me is this calc 2?

[long kayak]

Its a line from a proof in some paper

[carmine bronze]

what math level is the paper

[long kayak]

I would say post graduate

[carmine bronze]

ok good

[long kayak]

But I feel like this is not that difficult I just can't figure it out

[carmine bronze]

I hope to never encounter this again

[long kayak]

Hahaha

[carmine bronze]

It's not difficult but you can't figure it out

That's me most days

[long kayak]

Me too

[carmine bronze]

yeah but you're doing post graduate work

I'm trying to find simple stuff like f''

[long kayak]

That would be nice to do again 😂

[carmine bronze]

well I wish you luck on your math journey

[long kayak]

Thx, you too

[carmine bronze]

would it though?

Well yeah it must seem easy once you finish all of math

[long kayak]

Yeah, that's the point

Once you see much worse stuff, you are going to miss this

[carmine bronze]

😐

[long kayak]

.close

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[keen python]

how can this be surjektive ?

[vale wigeon]

the codomain is the set of all odd numbers

would you say you believe that this function isn't surjective or you have no idea what's happening?

[keen python]

let me send you waht I got

according to my solution its not surjetive

but the prof said its surjective

[vale wigeon]

ok then show your solution and i will point out where your mistake is

[vale wigeon]

e.g. y = 4

4 ∉ 2Z+1, you know?

[keen python]

ahh

I see

[vale wigeon]

in fact if y is an odd number then y+3 will be even

so there's no contradiction at all

[keen python]

I see ye

if the co domain was for example all Z then it would not be surjective

thats why he made the co domain all the odd numbers

[vale wigeon]

yes

[keen python]

oh so what I did on the right is actually the right ?

bc the definition says

if I find an x

and put x into f

i should get y

[lone heart]

@keen python Has your question been resolved?

[keen python]

how can I prove that this is not injective ?

ive examples such as

f(0) = f(-2) = 4 => x1 != x2

but i want to prove it formally

[weary wyvern]

Wdym formally

[keen python]

like you would normally do it by

x1^2 +2x1 + 4 = x2^2+2x2 + 4
x1^2 + 2x1 = x2^2+2x2

now what do I do ?

[weary wyvern]

No you just need a counterexample to the statement f(x1)=f(x2) => x1=x2, which you provided

[keen python]

but what if I dont find them on the fly ?

[weary wyvern]

Graph the function, then it should be easy to find examples

[lone heart]

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[heady void]

how do I do questions like this

[heady void]

Hey this channel is occupied you have to go to another one for help

[spice mesa]

yea my bad

i did alr i thought i delelted this msg

[heady void]

oh ok

[heady void]

So I got the workings

explain why

is it 28

[prime badge]

10^26 has 27 digits, times 16 it gains 1 digit

[heady void]

oh

that makes sense

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[errant cedar]

i used the result given and cos 2theta = 2cos^2theta-1 to get

$4cos^3\theta+2cos^2\theta-2cos\theta=1$

[ocean seal]

reiikoo

[errant cedar]

i tried solving the cubic by letting cos = x but there are no nice solutions

my other option was factoring out the cos theta

but then it could be +- 1

and same with the terms in the brackets soo a lot of solutions to consider?

[harsh girder]

,w factor 4 x^3 + 2 x^2 - 2 x - 1

[ocean seal]

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[errant cedar]

ah

seems hard to recognise

alright thanks

.close

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[potent garnet]

My work so far: I split this congruence into two congruences (Chinese Remainder Theorem): x^2 + 6x-31 == 0 (mod 8) and x^2 + 6x-31 == 0 (mod 9). After some manipulation I factored these two congruences to (x-1)(x+7) == 0 (mod 8) and (x-4)(x+10) == 0 (mod 9). Since 8 and 9 aren't prime ( I cant use the fact that if ab == 0 (mod n) then either a or b must be == to 0 (mod n)), how do I continue?

[vale wigeon]

think you may be better off completing the square first

[alpine sable]

there is none

[potent garnet]

So the first one would be x^2 + 6x == 4 (mod 9) and the second would be x^2 + 6x == 7 (mod 8). Completing the square gives us (x+3)^2 == 4 (mod 9) and (x+3)^2 == 0 (mod 8)

[little drum]

As Ann suggested and FireBlazer did, you could make up a mod chart (mod 9) after concluding that (x + 3) is of the form 4k for some 'k' integer and that (x + 3)^2 == 4 (mod 9) and matching to check another form for (x + 3) and merge the two

[potent garnet]

what do you mean by mod chart?

[vale wigeon]

so we have (x+3)^2 - 40 == 0 (mod 72) right

i.e. (x+3)^2 == 40 (mod 72) which reduces to (x+3)^2 == 0 mod 8 and (x+3)^2 == 4 (mod 9) by the looks of it

then write out all the squares mod 9 to see which ones fit the bill

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[lament basalt]

This is about logarithms

Why is 7^log of 38 with base of 7 = 38

[lament basalt]

I thought of 7^7^1 = 7^38^x where the x is exponent of an exponent

[wary stream]

You mean $7^{\log_7(38)}$?

[alpine sable]

log_7(38)=log_7(38)

[ocean seal]

dldh06

[thick star]

?!

[alpine sable]

thats the easiest way to understand imo

[thick star]

that's the equivalent of saying a = a

[lament basalt]

I thought this

[alpine sable]

ye but its 7^log_7 38=38

[wary stream]

What is it, mine or yours?

[thick star]

the general rule is that $x^{\log_c y} = y^{\log_c x}$

[ocean seal]

Yash

[alpine sable]

cuz log_a(b)=c is just saying a^c=b

[thick star]

for c > 0

[lament basalt]

Instead of this sry

[lament basalt]

Wdym?

[wary stream]

What is the actual question?

[thick star]

you can very conveniently derive this from calculus. either way, if you use this you find that what you said is indeed true

[wary stream]

The sentences you are saying makes no sense

[lament basalt]

This @wary stream

[wary stream]

That is just log properties

[thick star]

you can't do that

[wary stream]

This thing

[lament basalt]

Why is 7 log 7 (38) = 38

[gray isle]

Why is log 7 (38) = 38
its not

[lament basalt]

This

[gray isle]

that's very different from what you said

[lament basalt]

Yes sry I forgot

[gray isle]

log_7(38)
(by definition of log) is the value that when 7 is raised to that power gives 38

$\log_7(38) = x \iff 7^x = 38$

[ocean seal]

ℝamonov

[gray isle]

#❓how-to-get-help wdym and generate your own channel if you want help

[lone heart]

@lament basalt Has your question been resolved?

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[silent sage]

What formula / methodology has been applied here?

[thick star]

x - 2 = (2x-3) - (x-1)

[silent sage]

I was referring to the step beneath the obvious. (My apologies for not specifying)

[thick star]

oh

[thick star]

that's just squaring both sides

[silent sage]

Could you elucidate?

[thick star]

i would but im too tired to latex this

it'll become apparent

square both sides. you'll find that a lot of the terms cancel out. you'll get something in the end which looks like

|(2x-3)(x-1)| = -(2x-3)(x-1)

since LHS must be positive, RHS must be positive too.

you have a negative sign to deal with so (2x-3)(x-1) must be less than 0

[slender gull]

$|a| + |b| \geq |a - b|$\
The equality holds true when a and b are of opposite signs. That means ab $ \leq 0$\
Now use 2x-3 and x-1 in place of a and b that's all.

[silent sage]

Does this happen to be a general property? A lot of questions in the same row directly use what seems to be a propertly along these lines: If |x|+|y|=|x+y|; x.y>=0 , If |x|+|y|=|x-y|; x.y<=0.

[slender gull]

That is right.

It happens to be a general property.

[ocean seal]

What the hell am I doing here?

[slender gull]

|x| + |y| is the sum of two non negative real numbers.
While |x +- y| could be the sum or the difference.

Even if you have |x+y|

[silent sage]

Either x or y could be negative.. correct?

[slender gull]

Yes and that would make

|x| + |y| greater than |x+y|

So only when both of them are of same signs will the equality hold true here.

Or one of them is zero.

[silent sage]

The book did mention these sets of properties.. I suppose the conclusion used in the questions wasn't mentioned because it is directly derivable from here.

(As you had mentioned)

[slender gull]

That's right.

[silent sage]

That's right.

Thank you for your assistance and time.

[slender gull]

You're welcome!

[silent sage]

.close

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[open grail]

After calculating this integral

I've managed to get to this

And so does this calculator

But after i substitute x for 4,-4 i get 0

But why does the calculator add these stuff

Where this - 9 came from??

[frigid crest]

can anyone help me with this queation

A riding lawn mower has wheels that are 15 inches in diameter, which are turning at 2.5 revolutions per minute. How fast is the lawn mower traveling in miles per hour?

[last ether]

Looks like it tried to do a trig sub

[last ether]

#❓how-to-get-help dude read rules

I've had to deal with like 3 of you guys in the last hour

[last ether]

@open grail

That's my guess

[open grail]

Um

Even so shouldn't the answer be the same

[last ether]

It should but idk

[open grail]

If trig sub or not

[last ether]

The |x| is suggesting they did some trig sub for whatever reason

[open grail]

Hmm lemme try

[last ether]

I kinda see the issue though

f(x) [the antiderivative] only works for $x\geq 0$

[ocean seal]

Umbraleviathan

[last ether]

Desmos seem to agree with this

But if I'm gonna be honest, the -9 is useless

@open grail I see what's it's doing now

The -9 is useless

[open grail]

This is after I've done trig

[last ether]

It's multiplying the antiderivative by $\frac{x}{|x|}$ because it's trying to mimic the behavior of $\sqrt{9x^2+x^4}$, where it acts like an absolute value.

[ocean seal]

Umbraleviathan

[open grail]

U mean when we

Sqrt(x^2)

The made it |x|

[last ether]

Yes

Because of the function behavior

Where it got -9, that's useless

[open grail]

Ok even if i solve it in that state

-9

Well

See

No.9

[last ether]

Yeah I see

[open grail]

The - 9 is important here

[last ether]

Really

[open grail]

If i neglect it, it will be c
If not it will be b

[last ether]

Oh

Hm

Hold on

In your calculator, try changing the 9 to a 4 @open grail

[last ether]

Why is it 8/3 now

Wait that implies that the -9 was -27/3

Where did the -27 come from then

Hm

Well lemme try this

$$\sqrt{x^4+9x^2} \geq 0, x\in \mathbb{R}$$

So if we factor this out, where $x$ is an unknown variable:

$$\sqrt{x^4+9x^2} = |x|\sqrt{x^2+9}$$

[ocean seal]

Umbraleviathan

[last ether]

Try IBP with that

Just note that the derivative of $|x|$ is $\frac{x}{|x|}$

[ocean seal]

Umbraleviathan

[open grail]

Ye

[last ether]

I think that's possibly where the -27/3 comes from

Maybe ping a helper because from here, I'm stuck

[last ether]

I just know that this is true, and it gives an iota of where they get the -9

OH WAIT

@open grail one last thing, try 5

Instead of 9

[open grail]

Ok

[last ether]

Something just clicked

[last ether]

I knew it

It was taking 9 and raising it to 3/2

For whatever reason, idk

$$\sqrt{x^4+ax^2} \rightarrow \frac{x\left((x^2+a)^\frac{3}{2}-a^{\frac{3}{2}}\right)}{3|x|}$$

[ocean seal]

Umbraleviathan

[last ether]

How to prove why it's subtracting by that value, idk

[open grail]

<@&286206848099549185>

[last ether]

Lol yeah

Riemann my G explain

Fuck hes not online

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@open grail Has your question been resolved?

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[lone heart]

@slow fiber Has your question been resolved?

[lone heart]

@slow fiber Has your question been resolved?

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[pearl wraith]

how tf do you even start this question

[pearl wraith]

this is due tomorrow

[vale wigeon]

try making such colorings for small sets

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@pearl wraith Has your question been resolved?

[lone heart]

@pearl wraith Has your question been resolved?

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[stark island]

On one of my homework’s the problem asks, if (alpha) is in quadrant 2 and tan (alpha) = -4/3 and (beta) is in quadrant 3 and cosine (beta) = -15/17 then sin(alpha-beta) =

[stark island]

I tried graphing it with the unit circle but I’m just totally confused

On what the proper steps to a problem like this are

[real gazelle]

do you know the formula for sin(a-b)

[stark island]

Is it also referred to as an identity?

Like the sum and difference identity

[fluid basin]

yes, these formulas are often referred to as identities

there should be one you can use to rewrite sin(alpha-beta) as some combination of things on alpha and on beta

[stark island]

Oh I think, is it the one where you take the sin and cos and subtract them

Of alpha and beta

wait let me try it

[fluid basin]

you should do that, yes

[stark island]

I’ll tell you if I get stuck on something

Okay so I encountered a problem

It’s sin(a-b) = sin(a)cos(b) - cos(a)sin(b)

Since tan(a)= -4/3 does that mean sin(a)= 4?

Since tan = sin/cos

[fluid basin]

can sin ever be 4?

[stark island]

ohhh

no

Bc unit circle

[fluid basin]

not for real values at least

[stark island]

How do I find sin of alpha then

[fluid basin]

you're getting closer

so you have sin(a-b) = sin(a)cos(b)-cos(a)sin(b)

[stark island]

Yes

[fluid basin]

you know cos(b)

[stark island]

Yes

How do I find the rest of them

[fluid basin]

and you know tan(a), right?

[stark island]

yes

[fluid basin]

so ther emust be something you can do to what you have to get rid of the unknowns

[stark island]

yea

Is it another identity?

[fluid basin]

so look at yhou ridentities and see if there are any that have tangent in them that might be useful

[stark island]

Hm okay

[fluid basin]

remember that sin a / cos a = tan a

[stark island]

Quotient identities

[fluid basin]

honestly i don't see an obvious course on th is one, but i haven't done one of these in decades

[stark island]

yea like I’m really lost

What do I do then

@fluid basin do I ping a helper?

[fluid basin]

might as well, i'm stuck here too

[stark island]

Alright

[fluid basin]

at least not without pulling out the big guns, which won't help you at all

[stark island]

Calculators?

[fluid basin]

software

[stark island]

oh

Yeah I’m not allowed calculators

[fluid basin]

you need to learn how to do this, not have it done for you

[stark island]

<@&286206848099549185>

yea I do

Need to learn it

[fluid basin]

and i'm afraid i don't know how to teach it to you

[stark island]

alright that’s fine thank you for your help

[fluid basin]

sorry i wans't more help

[stark island]

No problem

[fluid basin]

i kinda know how to do this in another way but it involves math you're not ready for

[stark island]

Yeah I’m just taking pre calc right now

[open grail]

.reopen

[stark island]

Do I type that?

.close

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[open grail]

May someone reopen that

[open grail]

<@&286206848099549185>

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[modest trout]

I think type .reopen @open grail

[modest trout]

what...

.close

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[open grail]

.

Hh .reopen

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[strong wyvern]

I'm a little stuck on this problem just nkt sure where to start

[strong wyvern]

find (a + b + c) given that 2a - b + 5c = 13, and 2a +3b + c = 75

[tacit arch]

isn't it infinitely many possibilities

2 equations 3 unknowns

[minor needle]

a + b + c is constant in this case, for any solutions

[minor needle]

solve for (b,c) in terms of a, then u'll see something

[tacit arch]

OOOh nice

[strong wyvern]

alright ill be back in a little

yeah im stuck

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@strong wyvern Has your question been resolved?

[real gazelle]

try subtracting the first equation minus the second equation

it gets rid of the a

[strong wyvern]

yeah I got that part, which gives me 4b - 4c = 62

or 2b -2c = 31

[real gazelle]

hm wait lemme work this out real fast

[strong wyvern]

I have b - c = 31/2

[real gazelle]

okay so

yep that's right

so you can solve b = 31/2 + c

and then substitute this back into one of your original equations, let's say 2a+3b+c=75

then 2a+3(31/2 + c)+c=75

so now can you solve for a

[strong wyvern]

a = (57 - 8c)/4

[tacit arch]

then you can plug in your equations for a, b into a + b + c

[real gazelle]

^^^

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[fluid raven]

Find the area enclosed by the graph r = 3 sin theta + 3
I'm not sure how to do this question and im really lost

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[mystic yoke]

How would I complete parts a b and c?

[mystic yoke]

.close

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[swift shore]

@neat sierra what is proposition 6.3(5)

[neat sierra]

It just says either two cosets of a subgroup equal or are disjoint

[swift shore]

sorry

[neat sierra]

we havent gotten to normal subgroup yet

ok, but I was thinking for this question, the example I have is G = S3 and H = {identity, (12)}. So then take a = (13), then aH = {(13), (123)} and Ha = {(13)(132)}. So (13) is in both aH and Ha.

So i was thinking, this only happens when H contains the identity?

wait, nvm do a and b have to be distinct?

So like if a = (13), b cant be (13)

[lone heart]

@neat sierra Has your question been resolved?

[placid zinc]

Proposition 6.3(5) basically says that if aH ≠ bH, then aH ∩ bH = ∅

What we are showing does not contradict 6.3(5), because bH need not be Hb

You are interested in a right coset that isn't equal to the corresponding left coset

@neat sierra

Oh wait, you literally found it already

[placid zinc]

What only happens?

[neat sierra]

THe situation described in the qeustion

[placid zinc]

Oh I see. aH and Hb may overlap. They definitely won't if Hb = bH, though.

[neat sierra]

oh i see

so a = b

and H contains the identity?

[placid zinc]

No conditions on a,b. H is a subgroup and always contains the identity.

[neat sierra]

Ohh. i see what you mean

I got it. Having a dumb moment...

thanks

.close

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[bitter violet]

proof x^n +y^n=z^n n cannot be larger or equals to 3

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[weary belfry]

question

[weary belfry]

apparently lim (a * b) = lim a * lim b

both to positive infinity

and a, b is function (x)

lets say a =x, b = 1/x

lim (a*b) = 1,

lim a = positive inf, lim b = 0, hence, lim a * lim b = 0;

meaning that the statement is false?

[fallow vale]

Usually arithmetic like infinity times 0 does not work well with limits

That first result holds always if the limits exist and are real-valued

[weary belfry]

ah thanks

quite reassuring to hear that

oh, real valued

could u elaborate?

like is 0* inf not real

[fallow vale]

Like the limit is a real number

Well it is not defined in general

It will depend on the context

[weary belfry]

0 * anythin = 0?

[fallow vale]

Yes but these are for real numbers

Infinity is not a real number

But sometimes

[weary belfry]

good to know

[fallow vale]

It is useful to add infinity in to account for some other things

But you don't have all the properties that the other numbers have

[weary belfry]

also i like ur pfp alot

[fallow vale]

with infinity

Thanks!

[weary belfry]

yes for sure

thanks man ima do sum other problems

[fallow vale]

Ok good luck!

[weary belfry]

how do i terminate this channel?

like

end this conbo

combo

comvo

[fallow vale]

I think you are supposed to type .close but I'm not sure

[weary belfry]

convo*

.close

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[cyan mountain]

They would be equal right ? Cus complementary angles?

[timber sparrow]

yes.

[last ether]

Yes

[cyan mountain]

Okok

Thank you ^_^

.close

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[burnt galleon]

Are these the right transformations

😢

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@burnt galleon Has your question been resolved?

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[livid sail]

To measure the height of the cloud cover at an airport, a worker shines a spotlight upward at an angle 75° from the horizontal. An observer D = 575 m away measures the angle of elevation to the spot of light to be 45°. Find the height h of the cloud cover. (Round your answer to the nearest meter.)

[last ether]

Split D up into h + (575-h)

Then use trig ratios

[livid sail]

hmmm

h +?

[last ether]

D = h+ (575-h)

left segment + right segment

[livid sail]

then what do i do

[last ether]

Find h with the 75° angle

[livid sail]

i understand that bc with 45 degree angle the sides have to be the same

[last ether]

$$\tan{(75)} = \frac{h}{575-h}$$

[ocean seal]

Umbraleviathan

[last ether]

You know how I got this?

[livid sail]

yes

then do you factor out an h

?

[last ether]

Well no, it just becomes a linear problem

[lone heart]

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[umbral herald]

hi, I'm kind of an idiot sometimes but is this correct? speed / distance = time I'm just looking for a way to calculate time to target for a system in a game I am building.

[normal ingot]

speed = distance / time

[umbral herald]

thanks, as i said i can be an idiot at times.

.close

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[carmine bronze]

so any x we see is 0?

[remote heron]

what?

what do you mean?

[carmine bronze]

I have no idea

[remote heron]

what do you mean any x you see?

[fallow vale]

That expression describes the limit as x tends to infinity

[carmine bronze]

well all the fractions

[carmine bronze]

yes

[remote heron]

they are cancelling out the fractions which tend towards 0 along the limit

its just a trick

[carmine bronze]

so cancel out all the fractions

[fallow vale]

Because $\lim_{x\to\infty}x^{r}=\infty$ for any $r\in\mathbb{Q}$ with $r>0$

[ocean seal]

ninjahuman

[remote heron]

not all the fractions

$\lim _{x \to \infty} \frac{1}{x^p} = 0$ provided that $p > 0$

[ocean seal]

jan Niku

[remote heron]

this isnt really a rule i guess

[carmine bronze]

1/x has a 0 exponent though

[remote heron]

it has exponent 1

[fallow vale]

What?

[carmine bronze]

oh nvm

yeah 1

[remote heron]

if you dont write the exponent, you assume its 1

[carmine bronze]

yeah ok

[carmine bronze]

alright that makes it easier

I didn't see that written down anywhere on the explanation so I don't get when to cancel out

like this then

[remote heron]

yea, youre just missing the sign and the simplification

[carmine bronze]

missing what? they're right there though

[remote heron]

oh, i mean in the screenshot

[carmine bronze]

oh yeah I got that one wrong I need to redo

but the work is right?

[gray isle]

notation was abused

but the end result of -3/4 is correct

[carmine bronze]

you're accusing me of abuse?

where

[tacit arch]

The "going to 0"

[gray isle]

not writing the limits in your work

[tacit arch]

You're missing a limit statement

[carmine bronze]

should be good?

[tacit arch]

Yup

[carmine bronze]

nice

thanks

hey ramonov has honorable

I forgot

Only honorable that has actually helped me

[carmine bronze]

am I wrong if I put lims here?

[remote heron]

you should remove the limit once you apply it

[silent sage]

There is no term of "x" anymore.. its quite meaningless.

[remote heron]

youre applying the limit to get those rational terms to 0

[carmine bronze]

oh yeah

there is no x

mmm alright

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[lone heart]

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[lost coral]

idk how to strt

[lost coral]

whats the 1st step?

[naive valley]

if v points southeast (more or less), which way does -5v point?

[lost coral]

less?

-5 means it decreases?

[naive valley]

it means it points in the opposite direction

[lost coral]

yep

[naive valley]

so which points does that rule out

[lost coral]

well i dont have the coordinate of E

soo idk what to do

[naive valley]

well you know it can't be D E or F, correct?

[lost coral]

B?

[naive valley]

because they are all in the same direction as E

[lost coral]

yes

so its B and C

[naive valley]

also, how does the length of -5v compare with the length of v

[lost coral]

u just take the coordinates of v and multiply the x and y by -5

so its in the negative side

[naive valley]

and how do the lengths compare?

[lost coral]

what do u mean?

by compare

[naive valley]

-5v is _____ times as long as v

[lost coral]

1/5 v?

[naive valley]

no...

[lost coral]

-5 times as long as v

[naive valley]

well lengths are always positive, so 5 times

and in the opposite direction

that should be enough to deduce which point it must be

even allowing for imperfect drawing

[lost coral]

i dont understand what ur saying

is there a way u can rephrase it?

[lost coral]

also how is it 5 times longer?

it aint longer

its shorter

does origin mean starting point at 0,0 or the point A?

.close

[lone heart]

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[flint raven]

Whats the difference between AAS, ASA
SAS, SSA

in terms of congruency of triangles

[mortal trellis]

what do you mean with "difference"

they have different conditions (which sides/angles have to be the same) but they all show the same thing (that two triangles are congruent)

[flint raven]

The conditions i guess

[high rapids]

Hopefully you see the pattern

[flint raven]

Thanks

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[uncut depot]

https://youtu.be/7JuPEnK4vKQ?t=142 they say that the slope is 2 over 1 here

[uncut depot]

what do they mean by 2/1

what's that 1

[keen pasture]

2/1 = 2

[uncut depot]

i know that but, does that 1 change

like, for this equation its 1

for some other equation its not 1 for some reason

like, a slope formula or something

[keen pasture]

They probably say 2 / 1 as slope= Delta y/ Delta x

And to point out where this slope comes from they probably write 2/1 instead of 2. But both options are fine. However, you would usually just write the 2

[uncut depot]

oh

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[thick lynx]

$$\sqrt{a^2} = \frac{b}{c}$$
$$\text{First way: } a = \frac{b}{c}$$
$$\text{Second way: }a^2 = \frac{b^2}{c^2}$$
$$a^2 = \frac{b^2}{c^2}$$
$$a = \pm \frac{b}{c}$$

[ocean seal]

Kepe

[alpine sable]

$|a|=\sqrt{a^2}$

[ocean seal]

Toby

[thick lynx]

$$\sqrt{a^2} = \frac{b}{c}$$
$$\text{Canceling the square root with the exponent should work too, no?}$$
$$a = \frac{b}{c}$$

[ocean seal]

Kepe

[pliant dune]

typically (like in the first way) $sqrt(a)$ denotes the positive square root

[ocean seal]

Tom

[alpine sable]

the squareroot is the inverse of the square only on $[0,\infty)$ as $x^2$ is not injective

[thick lynx]

But if you square the whole equation, you get quadratics on both sides, then take the square root, you will have plus minus

[ocean seal]

Toby

[thick lynx]

For negatives, you can't cancel out the root with the exponent?

[alpine sable]

yeah, you can't: eg $(-1)^2=1$ but $\sqrt1:=1$

[ocean seal]

Toby

[pliant dune]

by squaring you lose the sign

so $x^2=4$ isn't equivalent to $x=2$

[ocean seal]

Tom

[pliant dune]

for example

[thick lynx]

yeah, $x^2 = 4 \implies x = \pm 2$ so I just shouldn't square it, right?

[ocean seal]

Kepe

[thick lynx]

Cancel out the exponent with the root right away if I know that a is nonnegative

right?

in $$\sqrt{a^2} = \frac{b}{c}$$

[ocean seal]

Kepe

[pliant dune]

that would be one way to avoid the problem

[thick lynx]

Alr, thx

.close

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[thick lynx]

Determine if $\sqrt[3]{0.8}$ is rational or irrational.

[ocean seal]

Kepe

[thick lynx]

Let's try by contradiction

[ocean seal]

Kepe

Kepe

[thick lynx]

How to continue?

[pliant dune]

I'd suggest something like replacing 0.8 with 4/5 and multiplying out the fraction