help-28

[fast peak]

we have enough channels

no need to micro mod

so?

[obsidian torrent]

[hot herald]

nothing disruptive enough going on for urgent action

it'll close naturally if they've abandoned. channel supply isn't an issue

deleting mod pings is worse

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@rancid gate Has your question been resolved?

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[viral pelican]

Hey, I'm trying to create a RSA encryption using Mathematica, I've currently come this far, however I'm having trouble with the encryption part of it.

But how do make sure that all the numbers in the list gets encrypted with the formula I've got within the Solve?

I've written this code
https://gyazo.com/c0507621fae4ad0fec4dc1165f6f23d4
https://gyazo.com/cbe1370674c8b61b838d5e39af6627e0

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[paper kelp]

help please

[paper kelp]

are the two triangles similar?

is there any trick or formula for this question?

[full forum]

@paper kelp Has your question been resolved?

[paper kelp]

<@&286206848099549185>

[rough plaza]

hello there

i think they would like you to first express
OT in terms of t
and
OL in terms of t and p

[paper kelp]

ot is t but for ol how?

[rough plaza]

using the 2:1

oh, i might have mistyped

well, we can start from here

PT

PT, we can find PT

[paper kelp]

is PT = OT

[rough plaza]

in terms of t and p

nah

remember tip to tail?

[paper kelp]

vector stuff?

[rough plaza]

yep

[paper kelp]

yea i know that

so i draw a line that connects the two vectors?

[rough plaza]

PT will be from Point P to O and then Point O to T

[rough plaza]

so, that we can make a vector from P to T

so, what is PO?

[paper kelp]

Po is -p?

[rough plaza]

correct!

now OT , as you've said it's just t

so by tip to tail, PT will be?

[polar valve]

does the notation $\overrightarrow{OP}$ means t is vector?

[glossy valve]

ThM

[paper kelp]

t-p?

[rough plaza]

i guess so

at least in my convention

[polar valve]

isnt it then 1/3 ( -p+t)?

[rough plaza]

there we go, PT is t-p

[paper kelp]

so PL would be 1/3 of that right?

[rough plaza]

there we go

[paper kelp]

so thus it'd be (t-p)/3

[rough plaza]

correct, that's the answer

[paper kelp]

Oh wow thanks

I thought it was something to do with similarity

didn't expect vectors

anyways thanks for the help 🙂

[rough plaza]

nah, just boring tip to tail

no problem👍

[paper kelp]

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[fading cargo]

The GCD of the two polynomials, X^3 - 2 and X^2 - 1 is 3, why are they co primes?

[fading cargo]

mustn't the gcd be 1 ?

[fast peak]

in the polynomial ring 1 and 3 are units

[onyx glen]

in what ring? R[x]?

[fading cargo]

yes in R[x]

[onyx glen]

3 is invertible

gcd is only defined up to multiplication by an invertible

[fading cargo]

then all the constants are invertibles ? which means whenever i find the gcd of two polynomials in R a constant then they are co primes

?

@onyx glen

[fast peak]

all nonzero constants

[fading cargo]

thank you so much

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[stable plover]

,w periodic function

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by tushar#0942

[indigo ore]

@torn jolt What is your first guess?

Ok, what would be the period?

Do you know what is the period of a periodic function?

A periodic function is basically a function that has a repetitive cycle

Or a pattern if you prefer

If you see the function f(x) = sin(x) graph, you'll notice a pattern being used over and over

Another example could be the function
f(1) = 1
f(2) = 4
f(3) = 1
f(4) = 4
f(5) = 1
f(6) = 4
..................... do you see the pattern?

Ok so this is what we call a periodic function

The period is the ''distance'' for which it takes to complete one full pattern/cycle

Here you see that after 2 units, the pattern is over and repeats itself

The period is also the smallest number T>0 for which f(x+T) = f(x)

Here, for example, f(x+2) = f(x)

It is true (as you said) that f(x+6) = f(x) too, but it is not the smallest number

So back to your example, I would suggest you to write explicitly what is f for the first values

and see if you recognize a pattern

It will be

Write me a list:
f(0) =
f(1) =
f(2) =
f(3) =
....

until f(9) say

Yes, what is the period?

Look back at our discussion above

Of what would be the period

[indigo ore]

We deduced that the period for that example would be 2

Yes

And also because f(1) = f(4)

f(2) = f(5)

and so on

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[regal aurora]

hi so just a quick question

[regal aurora]

how do you simplify a^-1 with only positive exponents

sorry for the dumb question just cant really find an answer

[clear lily]

$a^{-1} = \frac{1}{a}$

[glossy valve]

Gijs

[clear lily]

and more generally

$a^{-n} = \frac{1}{a^n}$

[glossy valve]

Gijs

[regal aurora]

ok and another one

(a^-2)^3

!close

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[restive hawk]

could anyone help me through the start of this integral

[restive hawk]

im not sure whether i would want trig substitution or if something like integration by parts is better

[velvet spoke]

trig sub sounds good

[restive hawk]

so im going to have to deal with the square root first?

i was deterred from trying trig sub because of the x^2

[velvet spoke]

try it

trig sub, simplify and then maybe some trig identities

[restive hawk]

so just deal with the square root as usual?

alright will be back in a few lol

@velvet spoke kind of ended up stuck

noticed i miswrote the sin under the sqrt

should be 361 but wouldnt change anything

alternatively should i have replaced that x^2 earlier on in my substitution>?

[velvet spoke]

note that the 19 is also squared

[restive hawk]

i ended up rewriting as this

[velvet spoke]

one sec

well you can simplify the bottommost expression more

1/tan = cos/sin right

[restive hawk]

right

[velvet spoke]

so the cos terms cancel

[restive hawk]

so now im back to 1/sin@

[velvet spoke]

and you're left with 1/sin^2

[restive hawk]

sin^2

[velvet spoke]

can you integrate that?

[restive hawk]

and i assume

i can u sub

??

actually nop

[velvet spoke]

hmm usually those are integrals with a well known solution

[restive hawk]

i know theres identities for them

[velvet spoke]

integral of csc^2(u) is just -cot(u)

[restive hawk]

but since it is on the denominator is it different?

oh

interesting

did not know that

[velvet spoke]

but you can probably derive it using identities as well

[restive hawk]

so if i were to not know that identity

i wouldnt be able to u sub anyways

right?

[velvet spoke]

yes, probably not using u-sub

[restive hawk]

so knowin the identity would be the important part here

so ive gotten to =-cotu

and u would be inverse sin of x

?

/19

[velvet spoke]

yes

[restive hawk]

so my final answer would be

-cot(arcsin(x)/19)

+c

no that is wrong

but not sure how

ohhh

is it arcsin(x/19)

by order of operations

unfortunately also wrong

nvm messed up a fraction

thanks

[velvet spoke]

what did you get?

[restive hawk]

i just forgot to put the 1/361 outside

[velvet spoke]

good

[velvet spoke]

great :3

[restive hawk]

cool

thank you

[velvet spoke]

but you can simplify it even more

[restive hawk]

oh

[velvet spoke]

but dunno if thats needed

[restive hawk]

true

invere sin

uhh

oh

it ends up wit he same

sqrt(361-x^2)

or something like that

not needed for this problem set though

[velvet spoke]

alright then you're good

[restive hawk]

thank yu

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[gray nacelle]

hi can someone help me with this?

for a) idk if i should do a proof by contradiction or something

@topaz valley hello

[normal tree]

write out the definitions of f(n) = O(g(n)) and g(n) = O(h(n))

[torn jolt]

whats your definition?

[gray nacelle]

not sure how to continue from here

[gray nacelle]

i have to do something to c_0 and c_1 right

[torn jolt]

well, your goal is now to find a c_2 and n_2 such that
f(n) ≤ c_2 h(n) for all n ≥ n_2
right?

[gray nacelle]

yes

wait

f(n) <= c_2(n)?

oh

yes

[torn jolt]

ok one hint, what if you multiply the second inequality by c_0?

[gray nacelle]

$c_0g(n) <= c_0c_1h(n)$?

[glossy valve]

WolframAlpha

[torn jolt]

yeah, now you have

[gray nacelle]

$f(n) <= c_0g(n) <= c_0c_1h(n)$

[glossy valve]

WolframAlpha

[gray nacelle]

wait where did urs go

[torn jolt]

na its not needed anymore

[gray nacelle]

o

so my thingy is correcto

[torn jolt]

almost

[gray nacelle]

o

[torn jolt]

so what do you choose c_2 as?

[gray nacelle]

c_0 * c_1

i think

[torn jolt]

yes, but you still gotta be careful with the n

what would you choose n_2 as?

[gray nacelle]

hmm

n_0?

[torn jolt]

but then the second inequality would not always work

[gray nacelle]

oh right

hmm

n_0 + n_1?

but thats quite loose i think

[torn jolt]

that would work yes

[gray nacelle]

ohhhh

[torn jolt]

alternatively

max(n0, n1)

[gray nacelle]

max hmm

OH

okay

i get

lemme write that down

[torn jolt]

you just want to make sure that both
n ≥ n_0
n ≥ n_1

[gray nacelle]

yes

i understand

[gray nacelle]

could u help me with another one

[torn jolt]

which

[gray nacelle]

part b

[torn jolt]

whats the definition

[torn jolt]

for all n?

[gray nacelle]

oh right

yea

wait no

n_0

[torn jolt]

so there exist c1, c2, n0 such that for all n ≥ n0?

[gray nacelle]

yea

[torn jolt]

well id start with one direction

=>

[gray nacelle]

hmm actually lemme write it out differently

more pleasing 🙂

okay one direction hmm

[torn jolt]

think of how to get rid of the c in front of the g in the top one

[gray nacelle]

multiply top by c_2?

1/c_2 i mean

hmm but theres c_1

[torn jolt]

might want to split the top into two seperate inequalities

[gray nacelle]

can i ignore the 0

[torn jolt]

at first yeah

its a little detail you could fill out at the end, not that important

[gray nacelle]

okay

ill multiply both equations with 1/c_1 and 1/c_2 respectively?

[torn jolt]

yeah

[gray nacelle]

since i wanna rid of the c

hmm

so c_3 = 1/c_1?

and c_4 = 1/c_2

[torn jolt]

careful, can you recombine them first?

[gray nacelle]

hmm

not sure where to put 1/c_2

[torn jolt]

i meant like putting the g in the middle

[gray nacelle]

oh so recombine into the second equation

is this what u meant

[torn jolt]

yeah but thats not quite correct

[gray nacelle]

hm

wait this si getting confusing lol

one sec

hmm why isnt it correct

[torn jolt]

i think you can ignore the second inequality and any substitution for now, just look at
g(n) ≤ 1/c1 f(n) and 1/c2 f(n) ≤ g(n)

if we combine them by putting the common g(n) in the middle we get
1/c2 f(n) ≤ g(n) ≤ 1/c1 f(n)

[gray nacelle]

yes

i gets

[torn jolt]

and now comparing that to the second inequality it fits

you just had the c3 and c4 swapped

[gray nacelle]

OH LMA

[torn jolt]

now what should n1 be?

[gray nacelle]

n_1 hmm

it would make sense to be n_0 i think

[torn jolt]

yeah

and any justification for why 0 ≤ 1/c2 f(n) holds?

[gray nacelle]

i would think its by definition

[torn jolt]

it follows from the first inequality
0 ≤ ... ≤ f(n)

and because c2 is positive

[gray nacelle]

oh i see

but cant f(n) return a negative

[torn jolt]

the first inequality does not allow that

n0 is chosen to be large enough so that doesnt happen

[gray nacelle]

oh right

i see

okay

makes sense
lemme write that down

lemme do the other direction

should be similiar i think

[torn jolt]

btw the other direction now is trivial and doesnt require any work

https://media.discordapp.net/attachments/903486480985489478/1063998878074019840/SmartSelect_20230115_095134_Samsung_Notes.jpg

[gray nacelle]

oh

i should have uh seen ur msg

i just did it

lolol

i need to use the toilet when i return could we do part c XD

[torn jolt]

i gtg, you can close and make new channel or try alone

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@gray nacelle Has your question been resolved?

[gray nacelle]

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[paper kelp]

is a correct?

[brave basin]

@paper kelp Where's the line for the upper quartile?

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@paper kelp Has your question been resolved?

[paper kelp]

it's there no? at 149?

148 is the median and 149 is the upper quartile

[main elm]

what is the question

send ur question

[paper kelp]

my question is that did i draw my box and whisker plot correctly according to the given information?

<@&286206848099549185> please tell me if i'm correct

[brave basin]

@paper kelp 159 is upper quartile

[paper kelp]

would it be like this?

[brave basin]

Yes

[paper kelp]

Thanks for the help!

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[novel sparrow]

Can somebody tell me why simplifying the term inside the log like that is allowed?

[vivid aurora]

dude

if youhave taken LCM

[vast fossil]

$-\ln{\abs{x - \frac{7}{2}}} + C_1 = -\ln{(\frac{1}{2}\cdot{\abs{2x - 7}})} + C_1 = -\ln{\abs{2x - 7}} - \ln{\frac{1}{2}} + C_1 = -\ln{\abs{2x - 7}} + C_2$

[glossy valve]

A Lonely Bean

[vivid aurora]

then in denominator 2

should be there

[vast fossil]

Basically ln(1/2) gets devoured by + C

[vivid aurora]

ok

now i see

[novel sparrow]

Thanks

[novel sparrow]

what is lcm?

[vast fossil]

Lowest common multiple

[vivid aurora]

lowest common multiple

[novel sparrow]

ahh

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[nocturne lance]

how did it end up with sin²(theta) + cos²(theta) / sin²(theta)? Right before 1/sin²(theta)

[hollow grove]

sin²(theta) + cos²(theta) is 1

[hot herald]

common denominator? combine fractions?

[torn jolt]

$a/b + c/b = (a+c)/b$

[glossy valve]

starlight

[nocturne lance]

Ah so it just combined fractions?

I understand now thank you

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[finite bluff]

anybody know matlab and willing to help . paid

[torn jolt]

[spice orchid]

Help here is free, just post your question

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[torn jolt]

is the vector B negative even tho its pointing in a positive direction (for x)

i.e. Bx = -3 and By = -3

[torn jolt]

channel is taken

[brave basin]

@torn jolt Doesn't matter if the vector is pointing in a positive x/y direction or negative x/y direction

[torn jolt]

kk

just matters what quadrant its in?

[brave basin]

Nope

B is just B, you don't call it negative or positive

[torn jolt]

but whats the magnitude of Bx

[brave basin]

You mean how much it travels in the x-direction?

[torn jolt]

bruh

yes, the 'distance'

[brave basin]

You talk about the magnitude of a vector, idk what you mean by magnitude of Bx

[torn jolt]

its physics 1

Bx is the x-component of the vector B

[brave basin]

Oh you mean B_x

Okay okay

yes its 3

[torn jolt]

ok coolio gangstas paradise thanks

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[faint dock]

why c is wrong?

my book says that the answer is e

i am stuck between c and e

but i think it's c

since dom of f i [1:+inf)

[brave basin]

both functions are the same

[faint dock]

and g is -inf;-3)U[1;+inf

but why is c wrong

[brave basin]

Because c only says that dom g is a subset of dom f, which is true but it is not the whole thing

you also have dom f subset dom g

[faint dock]

oh i see

so it;s the vice versa

g is subset of f

i mean dom

or is it equaly correct to say that domg is subset of domf

[brave basin]

both are subsets of each other

in other words, the domains are the same

[faint dock]

no i don't think so

the dom of first is smaller than the second

dom f(x) = [1;+inf)

[brave basin]

both functions are the same

how can their domains differ?

[faint dock]

in think the second one is not defined for x in (-3;1)

[brave basin]

also for first

basically sqrt[f(x) / g(x)] = sqrt[f(x)] / sqrt[g(x)]

[faint dock]

but if you put 10 for example

you don't get the same numbers

you get 2.5 for f and 0.77 for g

so f(x) != g(x)

no wait

yeah it's true

i think

e

but still the domain thing i think i'm correct

[brave basin]

You are half-correct

Since both functions are the same, you have that domg = domf which implies domg is a subset of domf and domf is a subset of domg

Only domg is a subset of domf is not correct

[faint dock]

no dom is not the same

function look the same

but domain is different

[brave basin]

How?

[faint dock]

i told you

[woeful dawn]

Domf=[-3;inf) ?

[faint dock]

.

[faint dock]

1*

[brave basin]

My my, you are correct

[faint dock]

don't worry

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[brave basin]

@faint dock Yes mb for confusing you. But actually domg = (-inf,-3) U [1,+inf) and domf = [1,+inf)

[brave basin]

So its domf subset domg

[faint dock]

yes

that's what i'm saying

[brave basin]

c is the other way around

it says domg subset domf

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[vestal perch]

Okay this is more of a general understandig question. i am currentely learning chain rule ( i think thats the english name ) and what i dont fully understnad is what to use as the " core " and what to be the outer part of the derivation. like is there a general rule to what the core is supposed to be.

[vestal perch]

when i have problems that includes this like logartimes square roots and other harder things to derive i kinda get stuck and dont know what to use as the core in the chain rule

[clear lily]

so for example g(f(x))

by core you mean f(x) right?

[vestal perch]

yes

[clear lily]

ideally you choose the core as something you know the derivative of

for example

$f(x) = 2x + 3 \newline g(x) = x^5 \newline g(f(x)) = (2x+3)^5$

[glossy valve]

Gijs

[clear lily]

then $\frac{\dd}{\dd x} g(f(x)) = g'(f(x))f'(x) = 5(2x + 3)^4 \cdot(2)$

[glossy valve]

Gijs

[vestal perch]

i understand that

if i have sqrt of something i know that the inside of the sqrt should be the core

but if i have somthing like e^sqrt(2x+1)

then i get kinda confused becuase what is supposed to be the core then

the whole exponent or the problem inside of the sqrt

[clear lily]

sometimes you need to choose the core twice

for $e^{\sqrt{2x+1}$ first choose your core as $\sqrt{2x+1}$ then you need to find the core of that so you choose the core $2x+1$

[glossy valve]

Gijs
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vestal perch]

oh so like doing the chain rule thing twice

[clear lily]

exactly

[vestal perch]

i see this makes more sense than trying to brute force trough questions hehe, thank you!

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@exotic epoch Has your question been resolved?

[exotic epoch]

No

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[worn breach]

If it says b^2-4(a)(c) > 0 are real roots, why would you then use k = -6, if its not valid?

[worn breach]

(please correct me, im unsure)

after getting k= 2, k= -6, i thought only k=2 fits the criteria? as its greater than 0

[devout mason]

Are you looking for values of k such that the quadratic has real roots

[worn breach]

yes

[devout mason]

It should be greater than or equal to

[vast fossil]

k < -6 is valid

Because we have (k - 2)(k + 6) > 0

And when k is less than -6, then k - 2 is negative and so is k + 6

Negative times negative gives positive

[worn breach]

Oh i see

Thanks !

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[idle shoal]

Could someone please explain why the z bounds of the bottom integral are 0 and 2

[idle shoal]

Here is a clearer picture

is anyone there?

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[torn jolt]

Hi can someone please help me with this: Prove that if a < 0 then there exists n ∈ N such that a + n > 0

[fast peak]

I mean this is either trivial or not, depending on what you are allowed to use

[torn jolt]

Okay, I made this proof

Take n = [1-a]. Note that -a>0. So 1-a > 1. Since n = 1 -a , then n + 1 > 1 -a . So n+1 > 1 -a, then n > -a, then n+a > 0

How do you see it?

[fast peak]

whats the notation [1-a] ?

floor? ceil?

[torn jolt]

No, sorry, just n = 1 -a

[fast peak]

well thats not necessarily a natural number

[torn jolt]

I see

Can you please give me a hint?

[fast peak]

I implicitly did already

[torn jolt]

This?

[fast peak]

.

[torn jolt]

Absolute value should be? I'm not sure about that, sorry

[fast peak]

have you not seen floor or ceil before?

[torn jolt]

No, we are not learning that topic for now

[fast peak]

ok what is a?

a real number or an integer

[torn jolt]

Real number

[fast peak]

hmm

have you proved something like "for every real number there is a natural number that is bigger" ?

[torn jolt]

Yes, Archimedeanity, right?

[fast peak]

ok

try to apply that

[torn jolt]

Alright, I'll see, thank you

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[devout valley]

Goated pen, but that's like illegible

I can't read the text in your picture

Try and send a better one if you can please?

[dim compass]

Use CamScanner from playstore to get clear images

[tough sandal]

im glad i came to watch this

[devout valley]

Still unclear, can't make it out

[sonic sparrow]

wtf

[devout valley]

Can you take the picture again maybe?

...one time is bad luck they say

[tough sandal]

clearly a troll

[limber pier]

alr so basically

what part do you need help on

a or b

a is pretty easy if you know law of cosines

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[visual salmon]

How do I do this problem?

[tidal citrus]

you need to start with known sum formulas

show us the formula for the first n consecutive integers.

[visual salmon]

(n1 + n2)/2 ?

[tidal citrus]

kinda, notation is iffy

(1 + n)/2

[visual salmon]

ok

[tidal citrus]

because we are starting from 1 and ending at n

ok, how about this now:

$\sum_{i=1}^n c \cdot i = ?$

hang on

[glossy valve]

Disorganized

[tidal citrus]

where c is a constant

also,

$\sum_{i=1}^n c = ?$

[glossy valve]

Disorganized

[visual salmon]

Ok

[tidal citrus]

ok what

@visual salmon

[visual salmon]

I am unsure what to do

[tidal citrus]

I'm asking you to fill in the ?

do you know how to rewrite LHS?

[visual salmon]

nope

[tidal citrus]

how about you look it up for us real quick

[visual salmon]

Alright

[tidal citrus]

@visual salmon https://ms.uky.edu/~123/lecturenotes/Chapter9_answers.pdf

see "summation rules" (first 3 examples)

[visual salmon]

Ok

[visual salmon]

nc

[visual salmon]

$\nsum_{i=1}^n i \

you would use the second example

not sure how to latex it

[tidal citrus]

You dont need it here, just tell me what the sec9nd one comes out to

We have more to do so I'll just say it

It's just n times c

[visual salmon]

Ok

[tidal citrus]

Or rather you typed the answer to the second one

The first one is c times the sum-of-consecutive-integers

c*n(1+n)/2

[visual salmon]

I see

[tidal citrus]

Ok so let's apply these rules

$\sum_{i=1}^n 5i - 1 = ?$

[glossy valve]

Disorganized

[visual salmon]

5* sum-of-consecutive-integers - n

[tidal citrus]

$\frac{5n(1+n)}{2} - n$

[glossy valve]

Disorganized

[tidal citrus]

Rewrite it

[visual salmon]

(5n^2 + 5n) / 2 - n

[tidal citrus]

...so that it is in the form of the RHS

[visual salmon]

Ok

So is this the proof to show it equals?

since its (5n + n^2)/2 - n = (5n^2 + 3n) /2

[tidal citrus]

Ok, almost there

Still one more thing to factor out

[tidal citrus]

Yes, provided you have already proved sum of consecutive ints (an easy proof)

However

This is not proof by induction

😬

[visual salmon]

ah

[tidal citrus]

Gimme a second

[visual salmon]

ok

[tidal citrus]

I'll remind you of the steps

Step 0: show this works for i=n=1

Step 1: show this holds for an arbitrary value k > 1

Step 2: show this holds for the next arbitrary value k+1

Step 2 here is called the inductive step

Ok, so Step 0

If n = 1, then there's just one term

So i=n

Just plug 1 into LHS and RHS, show they match

[visual salmon]

Ok I will

[tidal citrus]

To do step 1, you do what we just did

We let i be an arbitrary number

Represented by k (or just i)

[visual salmon]

Yes so I just showed that if you plug in 1 for n then its going to be 4=4

[tidal citrus]

And use rules we already proved

[visual salmon]

Ok

[tidal citrus]

The last step is to plug in k+1 for i (or i+1 for i) and show that you can basically get the same formula

[visual salmon]

Ok for k >=1 do I just plug in 1

[tidal citrus]

No, that was step 0

Also, let me be more clear

i is just an increment. We are counting up to n

n is what is changing

[visual salmon]

I see

[tidal citrus]

Not really i, since we need to count up all the terms up to n anyway

So it's more like "let n=1, then let n=k, then let n=k+1"

Because we are trying to show it doesn't matter what size n is, it will always work

[visual salmon]

Oh ok

For k > 0 would I just plug in any number greater than 0

for example 2, 5(2)(1 + 2)/2 - 2 = 2(5(2) + 3)/2

which is 13=13

[tidal citrus]

Step 2 is ALSO the same work we just did, but now, instead of letting n=k, we let n=k+1 and do exactly the same thing we just did.
(also there is a typo in the last formula, my bad, watch out for that. 5 should be inside square brackets on the k+1 term)

[visual salmon]

Got it just looked over it all

[tidal citrus]

QED is Latin for "and now it's proven"...maybe

[visual salmon]

Ok

[velvet marlin]

Quad Erat Demonstrandum iirc

That which is to be proven

[tidal citrus]

Ooh, fancy!😗

[velvet marlin]

that's why I learnt it 😄

[tidal citrus]

@visual salmon some formulas, like recursive formulas, may require you to use other numbers than 1 as your base cases (like 1 AND 2, i.e. the first and second term)

That should just make sense on a case-by-case basis

[visual salmon]

Got it. Thank you for all the help

[tidal citrus]

No problem

[visual salmon]

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[dark escarp]

1 was moved to RHS in the equation marked in green, then why later (marked red) they mentioned in LHS while substituting k to k+1 ?

[rose rain]

notice for k+1 you get

2^(3k+3) = 8*2^3k

[dark escarp]

yeah, it is what we substituted there, k+1, -1 was already taken to RHS(green) right?

[rose rain]

sure

[dark escarp]

there was no point to show it again in LHS

[rose rain]

er

no its for rigour

[dark escarp]

Okay thanks @rose rain 👍

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[torn jolt]

I already have the half of it but I need to know if the proof is good enough
Here the question :In the figure, the point S lies inside the line FE.
Starting from S, as indicated in the figure, six circular arcs are drawn step by step around
arcs around A, C, B, A, C, B are drawn.
Show that the sixth arc leads to S again and that the six arcs together are then exactly as long as the two dotted arcs together.
I need help by showing that the six arcs together are then exactly as long as the two dotted at s together.
I’ve already got an answer to showing that the sixth arc leads to S again here a photo
Immagine
my answer:
Immagine
I need to know if the proof is good enough and I need help with the other proof(Show:that the six arcs together are then exactly as long as the two dotted arcs together)

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@torn jolt Has your question been resolved?

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

[regal ibex]

Hi

[torn jolt]

Hi can you help me

[regal ibex]

What seems to be the problem here

[torn jolt]

the problem stands abovwe

above*

could you help me with that

[regal ibex]

I..

Maybe?

[torn jolt]

do you think it's proof enough

?

[regal ibex]

Ma shite, lad, I have no idea

If it's all in indisputable problems, with no other valid possibility, then probably?

[torn jolt]

<@&286206848099549185>

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[muted vigil]

any theorems that can point me in the right direction?

[quiet stirrup]

Not sure what kind of theorem you could possibly be looking for. Being told to use the fundamental theorem of calculus or the residue theorem is not useful.

[gritty rose]

What's the question?

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[muted vigil]

Determine whether it converges/diverges

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[wild tundra]

I have "X=arccos(-0.2) x∈[0, 2𝜋]"

[golden furnace]

yeah?

[wild tundra]

I have to find all solutions to X/theta

This is X

[golden furnace]

well since -0.2 falls under the interval [-1, 1], we know that it is reversible so that's out of the way

so just cosine both sides

[wild tundra]

cosx=-0.2...

[golden furnace]

yup

[wild tundra]

I have to find x not prove it

[golden furnace]

wait what does it mean by finding all solutions?

by exact value or something else

[wild tundra]

x∈[0, 2𝜋] in this range

,w X=arccos(-0.2) x∈[0, 2𝜋]

[golden furnace]

hm forgot how to graph it but I think cosx hits -0.2 multiple times in that interval

[glossy valve]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Yoshi#4432

[golden furnace]

and at irrational points

[wild tundra]

,w Solve for x, x=arccos(-0.2) x∈[0, 2𝜋]

[glossy valve]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Yoshi#4432

[golden furnace]

,w x = arccos(-0.2), x ∈[0, 2π]

[glossy valve]

Results provided by WolframAlpha

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Requested by biggboy#8536

[golden furnace]

no?

oops

[wild tundra]

,w cos(x) = y, y = -0.2

[glossy valve]

Results provided by WolframAlpha

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Requested by Yoshi#4432

[wild tundra]

2pin+arccos(-0.2)?

[golden furnace]

hm it does intersect at 2 points

idk what the question really wants

approximations?

[wild tundra]

Find the solution for theta or x in the given domain

[golden furnace]

well you got all the information you can

better off working out those values I guess

[torn jolt]

dont ask me questions

i feel dumb today

[golden furnace]

!help

[full forum]

Please read #❓how-to-get-help

[torn jolt]

!help

[full forum]

Please read #❓how-to-get-help

[wild tundra]

x = 1, y = 2

[torn jolt]

thx u

[torn jolt]

is worng

wrong*

question refreshed

[wild tundra]

10x+6y=34, 5x+3y=17

[torn jolt]

what x and y

[snow obsidian]

you could solve by substitution method or eliminating method

[wild tundra]

7x+4y=24, -6x-4y=-20,

x=4

y= -1

You're welcome

[torn jolt]

thx

this my last question if you mind helping

[wild tundra]

-6x-9y=-66

-2y=-12

y=6

2x+18=22

2x=4

x=2

[torn jolt]

thx u

[wild tundra]

Back to business

Find the general solutions for theta or x

theta = arctan(-4)

[torn jolt]

ok

alr im go practice

cya

[wild tundra]

ok

[wild tundra]

Can you solve this

theta = arctan(-4).....

I'll do some later today

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[tired bison]

I dont understand how it went from the 2nd to the 3rd step (the light blue part)

[tired bison]

I know this was used but I fail to understand how

would be thankful if someone could explain it to me

[viral jasper]

Do you know how chain rule works?

[tired bison]

if its this thing then yes, if not then no

[viral jasper]

Yeah it's that thing

[shrewd hamlet]

Also, $e^{x\ln{x}} = e^{\ln{x^x}} = x^x$

[glossy valve]

Stephen

[golden furnace]

I think he understands that but yeah

[viral jasper]

In your problem, do you know what f(x) and g(x) would be?

[tired bison]

g(x) = xInx
f(x)= e^x?

[viral jasper]

Other way around

[tired bison]

no its the other way around

yeah

[viral jasper]

Okay so f'(g(x)) is taking the derivative of e^x, then substituting x->g(x)

I assume you know derivative of e^x

[tired bison]

I don't 😦

[viral jasper]

ah

That may be the issue then

e^x has probably the easiest derivative for you to learn

It's itself

[golden furnace]

the most famous

[viral jasper]

Derivative of e^x is e^x

[tired bison]

okkkkkkkkkkk

that what I didnt understand

thanks a lot @viral jasper 🙂

[viral jasper]

np

[tired bison]

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[robust depot]

for this cylinder i want the mass moment of inertia

along the z axis

is it correct that i can break the Integral into this

The density i assume is constant so this holds true

now im wondering how exactly i would solve the Integral

just assuming it has standard radius r

im guessing i would have to transform it into cylinder coordinates for ease of use?

considering if i were to define the bounds right now of the integrals for dx and dy it would be a very annoying sqrt term for the circular boundaries

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[turbid schooner]

I’m on 14

See how I changed the bases for the rest? How do I do that for 14?

[compact comet]

1/2^2

[rustic hamlet]

Well, 32 is 2 ^ 5

[turbid schooner]

Yeah but what about 1/4?

[glossy valve]

Mehdi_Moulati

[turbid schooner]

Oh! Ok I see thanks

[worthy tree]

np

[atomic blade]

Why lime

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[cursive badger]

how to do i find this

?

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[cursive badger]

how do i find this

[cursive badger]

🙂

.close

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[hot wadi]

I was going to answer but ok

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[leaden forge]

where can i find some good resources on calculating probability, specifically whatever this is?

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[leaden forge]

<@&286206848099549185>

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[icy horizon]

im not sure how to solve this without the sine and cosine rules

[rough plaza]

On first thought:
Let CE = x feet and CA = y feet.

[snow obsidian]

you are given the measures of angles so where are you stuck exactly?

[icy horizon]

how can i use soh cah toa when the only side measurement i got is in a non-rightangle triangle?

[snow obsidian]

you can take BC as x so AC will be x - 100

[icy horizon]

im not sure how to proceed if i have an unknown?

i think im way off

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[icy horizon]

😔

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[icy horizon]

any help guys? <@&286206848099549185>

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[icy horizon]

<@&286206848099549185>

still stuck 😦

[barren heron]

hi!

alright, im looking at the picture, and you want to know how to solve based on the line?

the picture is talking about two triangles, with the smaller one being the dotted line and the bigger one, the entire solution

you can find the top inside angle, for because with the angle being 25 and it being a 180 degree line, you can subtract 180-25 to get your remaining angle

from there you can use your soh cah toa to figure out the remaining angles and lines

[barren heron]

does this make sense?

if not i can re- explain

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@leaden phoenix Has your question been resolved?

[brittle steeple]

show that if a vector is in the left set it has to be in the right set

and then show if its in the right set it has to be in the left set

it is true, yes

idk

what does it mean if a vector is in the left set

(or at least, what does it mean if a vector is in A+B)

yeah that makes sense

thats good

(technically you wouldn't use the same n for both since they're different numbers but regardless)

yeah probably

yeah exactly

you got it

np!

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[ashen breach]

I’m not sure why is x =4 when undefined?

[atomic blade]

Wdym undefined

Every triangle has a centroid

[ashen breach]

i mean

why is the x =4

where did they get that from

in segment WX work

[atomic blade]

Oh because the slope of one of the medians is undefined. It's a vertical line. That's useful since the equation for that median is x = 4, and it means the centroid's abscissa is also 4 (since it has to lie on the line x = 4)

[ashen breach]

i’m not sure i get it

a vertical line means 4? sorry i lm not following

[atomic blade]

So the purple median, it's a vertical line

It's slope is undefined

That means the equation for that line is x = 4

[ashen breach]

are you getting the 4 from the vertex?

[atomic blade]

I mean I'm just getting two points: the vertex, Y, and the midpoint between WX

But like just visually, it's gonna be x = 4

[ashen breach]

so basically

because it is an undefined slope

x=4 but why did you use 4 is where i’m confused

[atomic blade]

The centroid has to lie on the median. If the median's equation is x=4, the x-value/abscissa of the centroid must also be 4

It would make no sense if the centroid did not lie on a median

[ashen breach]

oh wait

this is the homework key. in the actual question the graph is empty, how would i figure out x=4 then?

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[vernal dock]

Csn someone help me with the question number 2?

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[modern hazel]

No problem!

What's up?

[vernal dock]

I cant do the question 2

And i need to do this paper u til tomorrow morning, i am new in calculus thats why

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[torn jolt]

(a) (f(3)-f(3.5))/0.5 @vernal dock

[vernal dock]

Why /0.5?

Is there any formula?

[torn jolt]

|3-3.5|=0.5

[vernal dock]

So there is a formula then, right? Like f(x)-f(y)/|x-y|

[torn jolt]

Yes
Technically

|f(x)-f(y)|/|x-y|

Except for cases when the interval include peak moment @vernal dock

[vernal dock]

Alright, thanks you man appreciated it

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