help-33

[lethal bridge]

what the blud @summer trench

[jagged relic]

The question says "show [this] is not integrable over D", so yes, it diverges

[lethal bridge]

i see

.solved

[marsh citrus]

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[marsh citrus]

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[alpine hull]

in this question, the model answer plotted the circle with equation x^2 + y^2 = 1, and then set a value for 2x + 3y (= k) and made this line tangent to the circle, then found k through geometry. why does this method work for maximizing 2x + 3y?

[obtuse umbra]

aight

[cunning fiber]

Cauchy

[obtuse umbra]

ill use pure algebra and inequalities for this

[spark otter]

Lines 2x+3y = k are all the lines that are parallel to the one you drew

For there to even exist (x,y) on the circle such that 2x+3y = k, the line and the circle must intersect

(because intersect means there's a point that's on both the circle and the line)

If you want k to be maximized

you need the line to be as high as possible

The extremum just so happens geometrically when the circle and the line are tangent

• The tangent does "intersect" the circle at one point

• If you're a parallel line above the tangent, you don't intersect the circle and so no solution to "2x+3y = k" on the circle exists

• if you're a parallel line below the tangent, then you're not a maximum

[alpine hull]

ohhhh

that makes a lot of sense

is there an alternative non geometric method for this?

[spark otter]

Well, personally I thought of (2x+3y)^2 = ...

You get a function in either x or y, whichever you decide to not remove using "x^2+y^2 = 1"

[alpine hull]

hmm okay

thanks for your help anyways

.close

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[scarlet quartz]

How many digits does the natural number 3²⁰²³ have

[scarlet quartz]

i'm so stuck

i'm learning logarithm

[jagged relic]

How would you use logarithms to find the number of digits of, say, 16?

[harsh scroll]

Ello my Vietnamese fella

[scarlet quartz]

yê

[scarlet quartz]

erm i dont think that i know the answer

[jagged relic]

Ok let me ask something else then

What would be the value of b in b^x = y such that 1 <= x < 2 and 10 <= y < 100 ?

[scarlet quartz]

b would be 10

[jagged relic]

Right, and do you see how y is a 2-digit number and if you round x down and add 1, that gives you 2?

To elaborate, same question but with b^x = y such that 2 <= x < 3 and 100 <= y < 1000

[scarlet quartz]

hmmmmm

b would be 10 again

[jagged relic]

Right, and here y has 3 digits and floor(x+1) = 3

(floor is "round down")

So now instead of 10^x = y, you want x = something

[scarlet quartz]

complicated, i still feel kinda vague

oh im onto something

[wide sequoia]

you take the logarithm of base 10, and then you round your number up, to the nearest integer

and the great part is that $\log(a^{b}) = b \log(a)$

[elfin berry]

DM ModMail for new nickname

[scarlet quartz]

like i wanna understand deeply

[wide sequoia]

so it means that what power of 10 gives that number

for example $\log(100) = 2$

[elfin berry]

DM ModMail for new nickname

[wide sequoia]

because 10^2 gives 100

and mb, I meant the number of digits $n$ of any number x is given by $n = \floor{\log(x)} +1$

[elfin berry]

DM ModMail for new nickname

[wide sequoia]

btw the $\floor x$ symbol is the greatest integer less than or equall to x

[elfin berry]

DM ModMail for new nickname

[wide sequoia]

so if we need to find the number of digits of 100

the floor of the log of the number is 2, since the log of 100 is 2

then when we add 1, it becomes 3

so it means that 100 is a 3 digit number

likewise the number of digits of 3^2023

[scarlet quartz]

hmmmm let me see

[wide sequoia]

you can use this law $\log(a^{b}) = b \log(a)$

[elfin berry]

DM ModMail for new nickname

[scarlet quartz]

yea yea i know this one

[wide sequoia]

so the number of digits of digits of $3^{2023}$ is $\floor{\log(3^{2023})}+1$ which is $\floor{2023 \log(3)}+1$

[elfin berry]

DM ModMail for new nickname

[wide sequoia]

,w calculate 2023*log(3)

[elfin berry]

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[scarlet quartz]

wow

[wide sequoia]

so it is 2222,49, whose floor is just 2222, and then when we add 1 it becomes 2223

so 3^2023 has 2223 digits

,w how many digits does 3^2023 have

[elfin berry]

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[delicate moat]

Wolfram's log is log base e

[wide sequoia]

wait

[delicate moat]

you need to specify log_10

[wide sequoia]

oh ok

,w log_10(3)*2023

[elfin berry]

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[wide sequoia]

yess, it is 965,21 whose floor is 965

and when we add 1 it is 966, and that is the actual answer

@scarlet quartz question answered?

[scarlet quartz]

yea yea

it might a little time to fully understand

but yours explanation will definitely help me

[wide sequoia]

yea it doesnt sink in the first time, thats natural

[wide sequoia]

glad i helped

more questions?

[scarlet quartz]

let me see in my homework

[wide sequoia]

alright, ill be waiting

[scarlet quartz]

question 18

do you need me to translate it

a,b,c ∈ R+

to satisfy that equation

and just find T

[wide sequoia]

i am not really following

is that spanish?

[scarlet quartz]

vietnamese

Let a, b, c be positive real numbers satisfying a^...= 9, b^...=7, c^... = 11. Calculate the value of the expression T=

[wide sequoia]

alright

the internet just came back

so if you have noticed, the expression of T is basically a^2+b^2+c^2

[scarlet quartz]

oh=)

nah ưay

really

[granite valve]

Uh no it is a^log^2 type

[wide sequoia]

let us give the terms names. let $x = a^{log_{3}7} = 9$, then Y the next term and Z the other one

[elfin berry]

DM ModMail for new nickname

[wide sequoia]

yea I meant that, I just gave the varaibles names in my head

[granite valve]

Alr

[wide sequoia]

so T is defined like $T = x^2 + y^2 + z^2$ where $x = a^{log_{3}7} $ and same goes for y and z

[elfin berry]

DM ModMail for new nickname

[scarlet quartz]

yess

[granite valve]

Uh still no $x^2 = a^{2log_{3}7}$

[elfin berry]

scoob

[granite valve]

and not $a^{(log_{3}7)^2}$

[elfin berry]

scoob

[wide sequoia]

doesnt $a^{\log^2}$ mean $(a^{\log})^2$

[elfin berry]

DM ModMail for new nickname

[granite valve]

No

,tex .exp rules

[elfin berry]

scoob

[wide sequoia]

oh okk

[hardy sonnet]

No, you wanna exponentiate by log to get log²

$a^{log^2_{3}7} = (a^{log_{3}7})^{log_{3}7}$

[elfin berry]

a handsome russian dude

[hardy sonnet]

So you just exponentiate 9, 7, 11 by their respective logarithms and sum that

[scarlet quartz]

ohhhhhhh i know how to do it now

thanks you guys so so much

appreciate all of yall

🫶🏻🫶🏻

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[marsh citrus]

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[marsh citrus]

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[still temple]

Anyone there

[tight furnace]

No, sorry

[still temple]

Huh

[obtuse umbra]

oh what the hell

[still temple]

Ikr this is concerning

I just want to know where to start

I don’t think s

[obtuse umbra]

SR is the tangent of (P) and (Q)

try to think that

[still temple]

Umm so forms t a right angle/

Right? So pqr is right angled /?

[obtuse umbra]

SP and RQ both perpendicular with SR

[still temple]

So just use simple cosin for first part

Right

Let’s move on

How to find r/s

Wait

I’ll just use Pythagorean

Ok ur

Wit guys there a problem

Is anyone even here

lol I’ll just close it

.close

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[simple reef]

I have a differential equation problem i was hoping someone could help clarify some things

[simple reef]

i solved through it and found a implicit solution

but im stuck at the part where i verify that it is in fact a solution

Any help is appricated

[main idol]

!show

[paper raptor]

what does your working through look like?

[marsh citrus]

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

[simple reef]

Sure let me attach everything here

one sec

[gloomy merlin]

you gotta check the total differential and see if it matches your original DE

[simple reef]

@paper raptor

.close

[marsh citrus]

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[main idol]

sign error here

[marsh citrus]

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[sturdy python]

if log(a) 3 = 0,47, log(a) 5 = 0,69 and log(a)20 = 1,28 determine the value of each expression

[sturdy python]

log(a) 75

[marsh citrus]

@sturdy python Has your question been resolved?

[cyan quail]

remember that addition in log makes the number multiplied

[sturdy python]

yeah but what do i multiply because none of those multiplied make 75

[cyan quail]

3×5×5

[elfin berry]

Xavier 🌺

[sturdy python]

log(a) 75 = log(a)3 + log(a)5 + log(a)5

?

alr

wait thats it? @cyan quail

[cyan quail]

doesnt it?

[sturdy python]

oh alr ty

but what if its log(a)4

how am i supposed to do it cause its smaller

wait i divide 20 by 5?

[cyan quail]

yeah

youre good

[sturdy python]

ty

[cyan quail]

that would be subtraction

[sturdy python]

yeah my teacher never taught us this thats why im having trouble a bit

[sturdy python]

log(a)4 = log(a)20 - log(a) 5

?

[cyan quail]

yes

[sturdy python]

first i have to do this tho right

then i have to do the subtraction?

or only the subtraction

[marsh citrus]

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[cyan quail]

just subtraction not dividing

[marsh citrus]

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[sturdy python]

what if its log(a) 12

[sturdy python]

because here i cant use any

[cyan quail]

you got log(a) 4
and log(a) 3

[cyan quail]

here youve got log(a) 4

and log(a) 3 is given

[sturdy python]

oh log(a)12 = log(a) 4 + log(a) 3

[cyan quail]

yeah

[sturdy python]

gotcha

ok ok for the last one i think i know how to do it?

its log(a) 36/a^4

wait nvm

[cyan quail]

log(a) 36-log(a) a⁴

[sturdy python]

its log(a) 36 - log(a)a^4

[cyan quail]

36=3×3×4
log(a) a⁴=?

[sturdy python]

axaxaxa

[cyan quail]

so what is log(a) a

[sturdy python]

1

[cyan quail]

so then

[sturdy python]

log(a)36 + 1

[cyan quail]

uhh think again

log(a) a⁴={log(a) a}×4=1×4

[sturdy python]

im not sure

so log(a)36 = log(a) 3 + log(a) 4 + log(a) 3 + 1

?

[cyan quail]

so log(a)36 = log(a) 3 + log(a) 4 + log(a) 3 + 4

[sturdy python]

oh 4

because of the exponent

[cyan quail]

yeah

[cyan quail]

can you do with this now

[sturdy python]

wait

its not finished?

[cyan quail]

so just calculate it now

[sturdy python]

log(a) 14

[cyan quail]

?

how to get u saying?

[sturdy python]

wait what do we calculate

[cyan quail]

so log(a)36 = log(a) 3 + log(a) 4 + log(a) 3 + 4

did it?

[sturdy python]

yes i wrote it

[cyan quail]

the value?

[sturdy python]

wait how do i calculate the value because theres no base

its only argument

[cyan quail]

log(a) 3= 0.47
log(a) 4=1.28-0.69=0.59

some logs were given

[sturdy python]

oh

[sturdy python]

log(a) 20 is 1.28

log(a)4 isnt there

[cyan quail]

look at the second one

[cyan quail]

you got it already

[sturdy python]

1.28 - 0,69

[cyan quail]

here

[sturdy python]

oh

gotcha

[cyan quail]

now good?

[sturdy python]

yeah ty

log(a) 36 = 0,47 + 0,47 + 0,59 + 4

log(a) 36 = 5.53

[cyan quail]

log(a) 36= 0.47+0.47+0.59

[sturdy python]

what happend to the 4

[cyan quail]

log(a) 36/a⁴= log(a) 36-4

[cyan quail]

you were asking this

[sturdy python]

wait

i dont get how the +4 got removed

[cyan quail]

it was -4

i misread sorry for that

[sturdy python]

oh -4

because -a^4

?

[cyan quail]

it was division

so you have to -4

[sturdy python]

ohh

okok

so log(a) 36/a^4 = 0,47 + 0,47 + 0,59 -4

[cyan quail]

log(a) 36/a⁴

[sturdy python]

like that?

[cyan quail]

yeah

sorry i went somewhere

[sturdy python]

alr ty

[sturdy python]

np

[marsh citrus]

@sturdy python Has your question been resolved?

[sturdy python]

.close

[marsh citrus]

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[marsh citrus]

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[kindred hinge]

I need help figuring out where I went wrong

[late geode]

can you show your work?
getting a negative value is immediately a red flag

[kindred hinge]

sure here give me one second

I used u sub

[queen osprey]

when you substituted

ln 3

and 0

why did you not include 0

you have to add pi/2 there also

[kindred hinge]

😭

[queen osprey]

then you get the right answer

it won't become 0 here

it'lll become pi/2

unlike polynomials

[kindred hinge]

why didn't I catch that

Thank you!

.close

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[tawdry crescent]

An ellipse is drawn with the major and minor axes of
lengths 10 and 8, respectively. Using one focus as center a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse. Then find the radius of the circle.

[tawdry crescent]

I tried this but am getting the following result pls help-
what does this yield the wrong result -)
let, r²=y²+(x-3)² be the required circle
y²=r²-(x-3)²
putting this into our ellipse
x²/25 + [r²-(x-3)²]/16 = 1
9x² -150x +625 -25r² =0
now putting D=0 as if it is tangent to ellipse then x coordinates will be equal
150² -4(9)(625-25r²)=0
150² =4(9)(625-25r²)
150²=(36)(25)(625-25r²)
r=0 ??

[cunning fiber]

Consider what happens when your value of $r$ makes $y^2<0$. \ \

I'd instead parameterise the ellipse by
$$(x,y)=(5 \cos \theta, 4 \sin \theta).$$
and minimise the square of the distance from this to $(3,0)$.

[elfin berry]

Civil Service Pigeon

[tawdry crescent]

So we cant solve quadratic equations in x,y by just substitution ?

If we want x,y belong to R

[marsh citrus]

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[marsh citrus]

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[tawny pewter]

how do i do this idk what to do for 4c

[tawny pewter]

accidentally put the same image twice smh

[marsh citrus]

@tawny pewter Has your question been resolved?

[cunning fiber]

discord.gg/physics

wrong server

<@&268886789983436800>

[marsh citrus]

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[dense surge]

Confused here because I got KL and JK but when i multiply them together and divide by 2 I don’t get an integer??

[waxen dust]

Can you show your work

[dense surge]

yes here

[waxen dust]

,rccw

[marsh citrus]

@dense surge Has your question been resolved?

[dense surge]

<@&286206848099549185>

[wooden shard]

try using the geometric mean theorem

[dense surge]

that’s how i figured out KL and JK

[wooden shard]

mb

[wooden shard]

oh okay try using JL and KM as base and height

[wooden shard]

okay you really didn’t need this

cuz area is 1/2(base)(height)

[dense surge]

I figured I needed them because I thought JK was the base and KL was the height of JKL??

but i see how it’s 125, just don’t get why the other lengths don’t work the same way

[wooden shard]

KL is not the height of JKL

it’s a base also

height is a perpendicular distance from any point to the opposite side

[dense surge]

yeah I thought KL was also a height cause it was perpendicular

[wooden shard]

well what the opposite side of L?

[dense surge]

JK?

[wooden shard]

well yes but that’s a base

wait i might be saying this wrong

for right triangles you can only have a height coming from the 90° to the hypotenuse

@dense surge forget what i just said

[dense surge]

okay

[wooden shard]

you evaluated JK and KL wrong

[wooden shard]

your proportions are wrong

actually i don’t think you need proportions for this

[dense surge]

i used geometric mean leg theorem

[wooden shard]

i used this

[dense surge]

Yes that’s the quicker way of what i did

[wooden shard]

here’s what i did

x is JK and y is KL

[dense surge]

Ohh I see where I messed up now. I got the same numbers as you in my head, but I took root 500 and thought of 5x10

Thank you

it should be 10 root5

[wooden shard]

👍

[dense surge]

.close

[marsh citrus]

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[gentle siren]

i tried working this through and i got stuck simplifying it, how would i work around the (n!)^3?

[high dock]

We're using d'alembert's ratio test right

[gentle siren]

yes that's right

[high dock]

so we take the limit of ratios of successive terms

a_2/a_1, a_3/a_2, etc.

[gentle siren]

a_n+1 / a_n

[high dock]

1/6, 8/6!, 216/9!, etc.

[gentle siren]

oh so rather than trying to do it with n itself just plug in numbers

[high dock]

wait scratch that I did something wrong i think

technically you could do that i think but here the terms blow up so its not practical

we have $a_n+1 / a_n = (n+1)!^3 / (3n+3)! * (3n)!/(n!)^3$

we can cancel some stuff out to get

$\frac{a_{n+1}}{a_n} = \frac{(n+1)^3}{(3n+3)(3n+2)(3n+1)}$

[elfin berry]

nc129

[high dock]

to get the denominator i used the fact that

$(3n+3)! = (3n+3)(3n+2)(3n+1)*(3n)!$

[elfin berry]

nc129

[high dock]

taking the limit as x approaches infinity, we apply l'hopital's rule 4 times and i think r = 1/27 unless i did my derivatives wrong somewhere

[gentle siren]

how did you get to this

[high dock]

this is our ratio right

$a_n+1 / a_n = (n+1)!^3 / (3n+3)! * (3n)!/(n!)^3$

[elfin berry]

nc129

[high dock]

we can factor (n+1)!^3 to be (n+1)^3 times n! ^3

[high dock]

and then the (3n+3)! factors to this

so we have

$n!^3 * (n+1)^3 * (3n)!$

[elfin berry]

nc129

[high dock]

as our numerator

and

$(3n+3)(3n+2)(3n+1)(3n)!(n!)^3$ as our denominator

[elfin berry]

nc129

[high dock]

and I cancel to get this

[gentle siren]

oh ok i was doing the denominator for a_n+1 wrong

[high dock]

oh rip

yeah the factorials are rly annoying

[gentle siren]

.close

[marsh citrus]

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[marsh citrus]

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[left quest]

yolo

[wintry hare]

How does this not become 1^8 since the 1 has an exponent of 1

[jagged relic]

It does

1^8 = 1

[left quest]

it does

1^8 = 1

because

1^8 can be thought of as 1 11111111

sry

1 x 1 x 1 x 1 x 1 x 1 x 1 x 1]

and it will result 1

[brave bluff]

1^8 = 1

[left quest]

therefore

(1/y)^8 = 1/y^8

[wintry hare]

It's Just clicked I'm sorry that was so dumb

Thank you

[left quest]

np

dont forget to close

[wintry hare]

How

[brave bluff]

Use .close

[wintry hare]

thankss

.close

[marsh citrus]

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[faint breach]

Q intersection (0, infinity)

[faint breach]

Does it make R+

How will it include irrational numbers?

[crystal lintel]

the question is cut off?

if the question is which one is equal to R^+, then the answer is none of those

so it's probably not that

[marsh citrus]

@faint breach Has your question been resolved?

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[novel juniper]

Kind of confused here

[novel juniper]

I want to show if (x) in R[x] is a prime ideal, R is an integral domain

[static quarry]

ok, any thoughts?

[novel juniper]

Contreadiction is probably the way to go

[static quarry]

yea i think so

(or contrapositive)

[novel juniper]

p(x)q(x) \in R[X] would mean p(x) or q(x). \in (x) , and both have to have their constant term as 0

as no non-zero element is in (x)

oh

it's so simple

Let $R$ not be an integral domain, then exists elements p,q in R such that pq=0. (x) being a prime ideal must have p or q,but this isn't possible as (x) has no non-zero element of R

[elfin berry]

wai

[static quarry]

yea that's valid

(x) is an ideal and therefore an additive group so it must contain 0 = pq
and it's a prime ideal so as you said...

[novel juniper]

mhm

[static quarry]

looks good to me

[novel juniper]

I can't believe it took that long for this to strike

TYSM

[static quarry]

yw!

[novel juniper]

.close

[marsh citrus]

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[kindred locust]

Let $f:\mathbb{R} \to \mathbb{R}$ such that $f(x^{2}-y^{2})=(x-y)(f(x)+f(y))$. Find all such functions $f$. $\newline$

$f(a^{2}-b^{2})=f((a-b)(a+b)) \newline$
Let $\lambda=a-b$. Then $f(a^{2}-b^{2})=f(\lambda a + \lambda b)=\lambda[f(a)+f(b)]=\lambda f(a)+ \lambda f(b) \newline$
This implies $f$ is a linear transformation. The only linear transformations from $\mathbb R$ to $\mathbb R$ are of the form $\phi (x) = kx$ for some $k \in \mathbb R$, so $f$ must also be of this form.

[kindred locust]

is my solution correct?

(i also got f(x)=kx in another way, but i wanted to see if this works too)

[elfin berry]

Green

[indigo nest]

Can't you just set y to be 0

f(x²) = xf(x) + xf(0)

[kindred locust]

yeah i did that for the first solution i had
i got f(x)=xf(1) and let f(1)=k

[marsh citrus]

@kindred locust Has your question been resolved?

[knotty trellis]

It feels wrong to have lambda depedent on a and b

though the conclusion seems correct

[indigo nest]

Yeah I've been trying to convince myself for a while about that and I'm not convinced yet

[knotty trellis]

Linear transformations should satisfy
f(c1a + c2b) = c1f(a) + c2f(b) for all c1, c2, a, b

You only proved it satisfies it for all a,b and c1 = c2 = a - b

[kindred locust]

okay I'll try to see if i can manage to show that

ty

.close

[marsh citrus]

Channel closed

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[indigo nest]

I think you can actually derive f(0) by plugging in x = 1 and y = 0

Wrong tag

.reopen

[marsh citrus]

✅ Original question: #help-33 message

[kindred locust]

i did f(0-0)=(0-0)(f(0)+f(0))=0

[indigo nest]

Right that works too lol

[knotty trellis]

so then we'd have just f(x^2) = xf(x)

It's also obviously odd

[kindred locust]

i had gotten f(n)= n^(1/2)f(n^(1/2))
you can simplify f(n^(1/2)) to n^(1/4)f(n^(1/4))
this way you get a limit as the exponents of the n outside are the sum 1/2^k, and the n in the argument of f goes to n^0=1

[indigo nest]

Interesting

[knotty trellis]

It assumes continuity at 1

[kindred locust]

shit 😔

[knotty trellis]

it's still interesting though

[kindred locust]

seems like that wasn't completely right either then

this was a thumbnail of a yt vid i saw a couple of days back which i decided to work on now, unfortunately i dont remember the title

[knotty trellis]

I wonder if we could get some discontinuous sol by defining f on each branch x^(2^z) with z in Z

[knotty trellis]

f(x^2) / x^2 = f(x) / x
Define g(x) = f(x) / x
So g(x) = g(x^2)

this might also get us sth

if g is continuous, it should be constant

if there exists discontinuous non-constant g, then it might transfer into a discontinuous f

Define x ~ y iff there is z in Z s.t. y = x^(2^z)

now this is an equivalence relation

take the partition on R

So that should partition R into sets of form {x^(2^z) | z in Z}

Now I summon the axiom of choice to pick a representative from each such set

And define g on those representatives arbitrarily

Use g(x) = g(x^2) to extend g on each set from the partition, using the values on representatives

oh, so actually each parittion will have g constant on them

Now define f(x) = g(x) x

To verify f(x^2 - y^2) = (x-y)(f(x) + f(y)
we need to verify
(x^2 - y^2)g(x^2 - y^2) = (x-y)(xg(x) + yg(y))

or g(x^2 - y^2) = (xg(x) + yg(y)) / (x+y)

Oh, this probably wont be necessary true

[kindred locust]

ouch

[knotty trellis]

that's not an ouch

That means that there will probably be no discontinuous sol

[kindred locust]

oh wow

[knotty trellis]

We just need to use something else than f(x^2) = xf(x)

[knotty trellis]

My construction only broke at this step, and it broke because x^2 - y^2 is not nice, exponents dont interact with + or - nicely

[kindred locust]

also, i found another solution given x,y =/= 0

it's simple enough i think

[knotty trellis]

What is it?

[kindred locust]

right

[kindred locust]

give me a min to type

you have y²=(-y)²
then you can derive (x+y)(f(x)-f(y))=(x-y)(f(x)+f(y)), you have to use the fact the function is odd
you can simplify it to 2x f(y)= 2y f(x) => f(x)/x=f(y)/y
x, y are arbitrary so it must be of the form mx for some m in R

[knotty trellis]

,w expand (x+y)(f(x)-f(y))=(x-y)(f(x)+f(y))

[elfin berry]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by methisalwaysright

[knotty trellis]

oh yeah, xf(y) = yf(x)

f(y) / y = f(x) / x

[kindred locust]

finally we got it

[knotty trellis]

Wait how did you get the first line?

[kindred locust]

consider f(x²-y²) and f(x²-(-y²))

[knotty trellis]

ohh

they must be equal

[kindred locust]

I'm on my tab so I'm too lazy to type it out entirely 😔

[knotty trellis]

That's a nice sol

[kindred locust]

and also much simpler than the others 😭

[knotty trellis]

yeah

[kindred locust]

anyway thanks for your time

.close

[marsh citrus]

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[marsh citrus]

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[finite forum]

how do i do b?

[wind whale]

can you calculate the perimeter for an arc for which I give you the angle it subtends?

[finite forum]

r(theta)?

[wind whale]

precisely

[dim forge]

l=rthetha

[wind whale]

you can calculate the perimeter by adding the length of two arcs, one in the left and the one on the right

separately

and add them

however you have to be careful while considering the angle

[finite forum]

which two arcs?

i thought r(t) only works for sectors

[wind whale]

it works for arcs, sector is the area which is subtended by that arc

perimeter is a length not area

[finite forum]

well yeah

but you need a theta

idk what to use for that

that's what i meant

[wind whale]

but you calcualted something in the previous part

that is an angle which will help you arrive at theta

try joining O with A and B then see which angle you are required

and then use basic circle theorems to actually find the one you need

[marsh citrus]

@finite forum Has your question been resolved?

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[south cradle]

Angel C is 140 degree and B 40 degree a is 6cm I have to find out c anyone can help me I trying all stratigies

[still temple]

angle C cannot be 140 degrees

[south cradle]

180-40? No?

[still temple]

well

Thats for the sum of interior angles of the triangle

Whats the interior angles here?

[south cradle]

Like B and C?

[still temple]

so whats A?

[south cradle]

Not listed

So 0?

[still temple]

what

A+B+C=180 degrees

Since the sum of interior angles of the triangle = 180 degrees

you cant just exclude it

[south cradle]

So I have to do 140 :2?

To find out A

[still temple]

no

[stoic saddle]

Angel C is 140 degree
are you sure about that

[south cradle]

In school we did tha Stratigies

[still temple]

Do you know what the dot in angle C means?

[south cradle]

It’s an angle?

[still temple]

it looks like you gotta use trig here

havent you guys learned that that dots means the angle is 90 degree?

[stoic saddle]

that doesn't answer my question either way

[still temple]

Or am I wrong

nvm I am right

[south cradle]

This is how I did in school

[lime rampart]

,rccw

[south cradle]

Sorry my bad writing

[still temple]

Not being rude

But I cant understand

lol

you guys have learned trig identities right?

[south cradle]

C is 125 c 5cm

Sin (125) * 5cm

Is a in length

[south cradle]

Kinda

[still temple]

So here, the angle C is 90 degrees

so what is AB and BC?

[south cradle]

It’s not 90 degrees

[still temple]

It is infact 90 degrees

Otherwise you can't solve this question

[south cradle]

Wait it really is

[still temple]

uh huh

the dot means its a right angle

afaik

[south cradle]

so it's a right triangle

[still temple]

yup

then finally you can use trig identities

whats AB and BC then?

[south cradle]

You mean the small b?

The capitalised is Angle

[still temple]

the side AB

and the side BC

[south cradle]

AB is 6cm

[still temple]

not quite

were solving for AB

that is c

[south cradle]

I don’t get what you mean

[still temple]

when I say side AB I mean the line that has the point A and point B

[south cradle]

That’s 6cm

That’s c AB?

[still temple]

yes

We are solving for it

Thats not 6cm

[south cradle]

Oh wait a is 6cm

[still temple]

yes

and now

in terms of perpendicular, base and height, what is AB and BC?

the trigonometric thingy

p, b and h

[south cradle]

To solve AB we need angle sin right

[still temple]

not quite

[south cradle]

And the angle A?

[still temple]

first answer my question

in terms of p, b and h, what is AB and BC?

<@&286206848099549185> I need a bit of helping 😅

[south cradle]

AB is c and BC a

[still temple]

https://tenor.com/view/how-can-i-help-dr-max-goodwin-ryan-eggold-new-amsterdam-can-i-do-anything-gif-16988221

[still temple]

can you explain this guy

how to use trig identity to solve this problem

I think I am a bit too tired to explain properly

so where are we

I just explained to him how angle C is 90 degrees

and now were moving on to what AB and BC is in terms of p, b or h

[south cradle]

I have to find out length of c

[still temple]

ok so angle C is 90 degree what makes 90 degree special about this triangle

we can apply trigonometric ratio deriectly here

[still temple]

do you know what is cos

[south cradle]

yes

thats our topic sin cos tan

[still temple]

yeah so cos = adjacent side/hypotenuse

hypotenuse is side opposite to right angle which is C here which side is opposite to C

[south cradle]

yupp

[south cradle]

AB

Which is small c

[still temple]

yeah

if you look at angle B what is side CB

[south cradle]

a

[still temple]

yeah but it is adjacent to angle B right

[south cradle]

Umm I guess

It’s adjacent because we know B is 40 degree right?

And we wanna find out AB

And it’s opposite of C

[still temple]

adjacent means the side beside angle B

CA is the side opposite to angle B

[south cradle]

Well if we have the adjacent I can do the equation right? Cause I have CB and I have B

[still temple]

yeah

[south cradle]

6/ cos(40)?

[still temple]

yeah

[south cradle]

7.83

[still temple]

,w 6/ cos(40degrees)

[elfin berry]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by hi016662

[still temple]

right

[south cradle]

Nice

So can you help me with one more?

[still temple]

sure i'll try

[south cradle]

Soo

By my experience this is also a right triangle

I have to know AB

[still temple]

i dont know does the dot mean 90?

[south cradle]

Yes

Always

[still temple]

yes

[still temple]

ok then do same

[south cradle]

AC is the adjustant?

[still temple]

yh but here you don't need to find AC, you need AB

AB is opposite to $\gamma$

[elfin berry]

firestepper

[south cradle]

So it’s cos again?

Nö wait

It’s sinus

Right?

[still temple]

yes sin

[south cradle]

Cause we have hypotenuse

[still temple]

what language is this

[still temple]

yh

[south cradle]

German

[still temple]

ok

[south cradle]

So it’s 4,5: sin (45)?

Or we don’t use the same ?

[still temple]

do you mean 4,5 x sin 45

[south cradle]

Unfortunately

[still temple]

or 4,5/sin45

[south cradle]

┬─┬ノ( º _ ºノ)

Divided

[still temple]

,, \sin45 = \frac ca

[elfin berry]

firestepper

[still temple]

now see what c =

multiply both side by a

[south cradle]

Umm

Kinda confused

[still temple]

if $x = \frac {y}{z} \implies z \cdot x =y$

[south cradle]

We don’t have c so how we divide it by a?

[elfin berry]

firestepper

[south cradle]

or do you want me to set it

Oh I get it

[still temple]

you have a

[south cradle]

Sin(45) x 4,5 =3,829

[still temple]

right

,calc 4.5/sqrt(2)

[elfin berry]

Result:

3.1819805153395

[south cradle]

What’s this?

[still temple]

that is what you should get

[south cradle]

What did you do on 3.829

Cause I thought 3.82 is the final solution

[still temple]

,,\sin45 = \frac{1}{\sqrt2}

[elfin berry]

firestepper

[still temple]

so $\sin45\cdot 4.5 = \frac{1}{\sqrt2}\cdot 4.5$

[south cradle]

Did my calculated do it automatic ?

[elfin berry]

firestepper

[still temple]

yh this is it

[south cradle]

I’m just confused with the 1/V2

[still temple]

how did you get this

[south cradle]

My keyboard calculator

But it’s bad

[still temple]

it is a common angle dw you can do calculator

[still temple]

ic

so yeah 3.182

[south cradle]

So one question

When do I multiplie when dividing the angle with the length

Cause with cos we divided length with angle

[still temple]

its about what fraction you get and what information you have
eg
you have adjacent value and want to find adjacent
then cos = adjacent /hypotenuse
so hypotenuse = adjacent /cos
similarly if you have hypotenuse but you want adjacent
cos = adjacent /hypotenuse
adjacent = hypotenuse x cos

if you have $ x= \frac yz$ by multiply both sides by z \
$\implies z \cdot x = y$ then divide both side by x \
$\implies z = \frac yx$

[elfin berry]

firestepper

[south cradle]

Can you give me a triangle equation similar to what we have done ? And I will try to solve it on paper and send it

[still temple]

hm.. idk

you can search

[south cradle]

Alright

Thank you

Teaches me a lot

.close

[marsh citrus]

Channel closed

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[still temple]

https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/trig-solve-for-a-side/e/trigonometry_2

np

[marsh citrus]

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[turbid willow]

im not sure where i am going wrong, but i cant seem to get the right answer

[peak cobalt]

its right, i used calc and got v=11.699

[turbid willow]

huh really?

i keep getting negative, and u cant square root a negative

[peak cobalt]

nahh

check again bro

[turbid willow]

but how tho i did it like 10 times

[peak cobalt]

do it 11 times

[turbid willow]

11th time no help

[peak cobalt]

show me the equation that u put

[turbid willow]

i did (10tan(30)-1960)/3 and then sqrt that

[peak cobalt]

ma guy whats that

Wheres the 1?

[turbid willow]

i times up by v^2

[peak cobalt]

no no no

[turbid willow]

v^2 = 10tan30 -1960/3

[peak cobalt]

10tan30 -1 = 1960/3v² right?

Where the 1 go buddy

[turbid willow]

ohhh

yeah u right

[peak cobalt]

lol u did everything correct and u got stuck here

it happens :(

[turbid willow]

wait but im still not getting the right answer

[peak cobalt]

What r u gettin

[turbid willow]

when u sqrt that

u get 20

not 12

yoo is my brain not functioning today

[peak cobalt]

ma guy

where the 3 go

[turbid willow]

ohhh

[peak cobalt]

divide by 3 right?

[turbid willow]

YEAH

THANKS

I GOT IT

broo my brain is not working today idk why

[peak cobalt]

bro u need to sleep lmao

put that head to rest

have a good day/night/evening/whatever

[turbid willow]

lmaoo thanks alot

.close

[marsh citrus]

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[marsh citrus]

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[elfin badger]

can someone help with what step to do after this

[lime rampart]

partial fractions, probably.

[static quarry]

well the blue is wrong already

[lime rampart]

oh right I didn't even see that, sorry.

[static quarry]

if you factor out sqrt(3) you'd have sqrt(x^2 - 1), not x^2 - 1

[lime rampart]

okay now this looks like a trig integral of some sort.

[elfin badger]

not sure why they use x=coshu

[static quarry]

that's the usual trick with sqrt(x^2 - a^2) forms
you use it because it works

there's also a clever substitution for this integral that lets you get an answer in terms of log

instead of hyperbolic trig functions

(or equivalently, you can use this funky identity):

[elfin badger]

so would you just remember it?

[static quarry]

yea for the various trig substitutions it's probably best to memorize them unless you are able to remember how to derive them under pressure (like on an exam)

[gloomy merlin]

What i do is try to make sense of what substitution works based on the unit circle (hyperbola) equation

[elfin badger]

okay ill try do the full working for it

[elfin badger]

i havnt learnt the uni circle

[gloomy merlin]

You are doing trigonometric substitutions and you don't know the unit circle?

[elfin badger]

im doing integration by substitution and by parts I know some tri identities

[static quarry]

how were the trig substitutions taught to you? didn't they draw a picture involving the unit circle to explain how / why they work?

[elfin badger]

no I was just given the identities

[lime rampart]

oh goodness...

[gloomy merlin]

If you dont know/understand the unit circle you don't understand trigonometry in general

[elfin badger]

Im thinking I should learn the unit circle now...

[static quarry]

maybe you could watch the relevant khan academy videos to get a better explanation

(or check your book if you have one)

[elfin badger]

Ill watch the video thanks

[gloomy merlin]

I mean nonetheless you can still make it work with only

cos^2(x) + sin^2(x) = 1
cosh^2(x) - sinh^2(x) = 1

For the sake of your problem (and most trig sub problems)

[elfin badger]

these were the trig identitie i knew

.close

[marsh citrus]

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[dim forge]

@elfin badger you can remember this too

[marsh citrus]

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[noble geyser]

I dont think you are supposed to solve it like this

I think you should first simplify the term that you get at dy/dx

You can simplify by 2xy-1

So that xy=1/2 shouldnt be a solution

5th line

Ok, here is another way to explain this

You got xy=1/2 as one of the answer, right?

Bruh

Nevermind, i have another way to explain this

At the end you got xy=1/2

Right sorry

So 2xy+1=0, right?

Now look at the 9th and 10th line

In the denominator, there is 2(x^2)y+x

Which is x(2xy+1)

[noble geyser]

Which is 0x

So you essentially divided by 0

Yep

Because x^2=1

So x = 1 OR -1

[stray orchid]

im sorry

[noble geyser]

!occupied

[marsh citrus]

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).

[noble geyser]

I guess youre right

I dont really understand what you are saying

Oh wait yea

Now i understand

Yea youre totally right

Sorry im chinese

It takes a bit of time to realise

Sometimes i forget too

[marsh citrus]

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[noble geyser]

Np

[marsh citrus]

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[hot rivet]

<@&268886789983436800>

[marsh citrus]

Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[marsh citrus]

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[zealous halo]

How to proof ,

If a sequence is bounded , then each of its subsequences is bounded.

[vernal forge]

you can go by contraposition

if any subsequence is unbounded, then the whole sequence is unbounded

[zealous halo]

Can you guide , how to start the proof

Exact words .

[zealous halo]

.

[lime rampart]

that sounds like a stronger form of Bolzano-Weierstrass?

[zealous halo]

I would say , a part of its proof

Which they have used as a result in theorem

[vernal forge]

no, weaker

it's one of those things that are so obvious it's hard to prove lol

[jagged relic]

Just write down the definition for an unbounded subsequence, no?

[stoic saddle]

isn't it quite clear that any bound which works for the original sequence also works for any subsequence of it

[zealous halo]

Now write a formal proof for it

[stoic saddle]

let M be a real number s.t. for all n ∈ N we have |a_n| ≤ M

[zealous halo]

Yes

[stoic saddle]

let (a_n_k) be a subsequence of (a_n)

since n_k is always a natural (by definition) we have |a_n_k| ≤ M for all k ∈ N

therefore the subsequence (a_n_k) is bounded by M

this formal enough for you?

[lime rampart]

ah my bad, sorry.

[zealous halo]

Welllll

[stoic saddle]

clear answer not given

explain yourself

[zealous halo]

No , all good

.close

[marsh citrus]

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[marsh citrus]

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[zealous halo]

Suppose we have an interval I , and we bisect I into two parts I1, I2 .

Does it mean , we divided the interval I into two equal parts , "left part" and "right part ".

Or does it mean we divided it into two parts ,which are equal but those two parts contains random numbers from the interval I .

[stoic saddle]

i would understand "bisect" to mean splitting into left and right parts.

[zealous halo]

Equal parts ..?

[stoic saddle]

yes, unless stated otherwise.

[zealous halo]

Let me send u the image

Proof of Balzano Weierstrass Thm

[stoic saddle]

bolzano

also it says subintervals btw

[zealous halo]

Bolzano

[stoic saddle]

and it does say equal subintervals, here meaning that they are of equal length

[zealous halo]

Understood

We are driving the interval into two equal parts

But subparts can have different no. Of term of my sequence

Maybe the density of terms is more in one part

[stoic saddle]

"density" is overkill

it's more like, there's infinitely many terms, so at least one of A1 and B1 has to be infinite.

[zealous halo]

The way I m thinking it is like this

I have natural numbers marked on x axis and the value at each index is the the height (corresponding to y axis )

In that sense , if I divide the interval into two parts ...then the idea of being dense in one sub part comes

[stoic saddle]

sure

[zealous halo]

Thank u once again

.close

[marsh citrus]

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[marsh citrus]

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[normal fog]

TRUE OR FALSE, these were questions from my quiz i took yesterday

[static quarry]

what is a in the first one?

[normal fog]

That is all the information given

[main idol]

then you were taught what a represented elsewhere and aren't telling us

[static quarry]

i find that hard to believe

[normal fog]

@halcyon elk

[main idol]

uh

!noping

[marsh citrus]

Please do not ping individual helpers unprompted.

[normal fog]

Thats all the quiz says

[lime rampart]

if you are looking for a particular helper why not just DM them?

[static quarry]

show the rest of the quiz

[main idol]

elsewhere

[lime rampart]

but I suppose you are working with the assumption that a is the exponent?

[normal fog]

A is the base

[lime rampart]

in short, are your exponentials x^a or a^x?

[normal fog]

I think

2nd one

Exponential functions

[main idol]

that would be the answer to this question

[lime rampart]

guess my role here is done then.

[tight furnace]

It's probably like a*b^x

negative base does not work well with non-integer exponents

[normal fog]

So what do you guys think the answers are

[main idol]

yea or this one