help-19

[wintry temple]

key

[undone mauve]

what's the issue

[wintry temple]

i don't have answer key

so i don't know

if i got it right

[undone mauve]

show the steps

[zenith tartan]

also we dont tell the answer here, if you've solved it maybe then you can show your work here

[undone mauve]

that you followed

[wintry temple]

fine

0=n/2(2(161)(n-1)-7)

then

[undone mauve]

wrong

[wintry temple]

how?

[undone mauve]

check it again

[wintry temple]

sum is 0

and

d= -7

[undone mauve]

n/2*(2a+(n-1)d)

[wintry temple]

yea

[undone mauve]

you multiplied 2a and n-1

[wintry temple]

yea

2*161

[undone mauve]

substitute the values, you'll get a linear in terms of n

[wintry temple]

...

huh?

[undone mauve]

?

[wintry temple]

i don't get what u mean

[undone mauve]

substitute the values in 0 = n/2(2a+(n-1)d) and you'll get a linear equation

since its 0 you can just do 0 = 2a+(n-1)d

[wintry temple]

u get 0= (315n+7)/2

...

[undone mauve]

no?

you don't know the value of n, how did you substitute it?

[wintry temple]

mb

its 315 n

[undone mauve]

no

[wintry temple]

why?

[undone mauve]

0 = 2a + (n-1)d

=> 0 = 161*2 + (n-1)*(-7)

[wintry temple]

what happened to n/2?

[undone mauve]

you can cancel it since its a common term

[wintry temple]

waaaa

fine

ok

.close

[odd edge]

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[vagrant haven]

Helooooo

[vagrant haven]

Need help with num 10 only

Im confused

<@&286206848099549185>

Anyone free?😭

Hellooo

[vagrant haven]

Num 10 on this one

So like 26²=(4x+4)²+(2x)²?

And i solve it?

Okay uhm lemme see if i can😭

Yes

Semicircle theorem i think

I believe i am now stuck again

Where do i go from here? Or is this entirely wrong?😭

Wait wha?

Where did 32x come from TT

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[vagrant haven]

.close

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[halcyon quail]

Let f(x) = x/(x+a), where x > 0 and a > 0. If f(f(x)) = x /(3x+4) , then find the
value of a. Pls solve

[odd edge]

Please don't occupy multiple help channels.

[halcyon quail]

Ho w is it done

[odd edge]

@halcyon quail Has your question been resolved?

[odd edge]

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

@halcyon quail Has your question been resolved?

[quasi sparrow]

Compute f(f(x))

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[hybrid spade]

how can i know X and Y? i thought the triangles were similar but they arent, the ABC perimeter is 36

[whole vine]

no other info?

[hybrid spade]

AD is dividing A angle in two equal halfs

AD is bisector of A

[whole vine]

can you like send the pic of the entire question?

!original

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[nimble blaze]

apply angle bisector theorem

[hybrid spade]

its in another language brb

In the triangle
ABC in the figure below,
AD is the bisector of the angle
A. If its perimeter is 36 cm, then the values, in centimeters, of X and Y
It is
they are:

In the triangle
ABC in the figure below, AD is the bisector of the angle A. If its perimeter is 36 cm, then the values, in centimeters, of X and Y is:

[whole vine]

yeah. you can use the angle bisector theorem for this

so you get

$\frac{2x}{y} = \frac{x}{2x}$

[clever fjord]

Galactic Fur

[hybrid spade]

oh

i didnt know that

[whole vine]

additionally, you also know that 2x + x + 2x + y = 36

you can use these 2 eq to solve for x and y

[hybrid spade]

ok let me try that

uhh yx = 4x²

and 5x + y = 36

wait

[whole vine]

you can cancel out the x here

[hybrid spade]

oh

y = 4x

[whole vine]

yep

[hybrid spade]

5x + 4x = 36

oh

[whole vine]

now it should be easy to solve

[hybrid spade]

yep

x = 4

i had thought of something wrong but it was correct gms

[whole vine]

ggs

[hybrid spade]

i figured DC was double of BD

so AC would be double of AB

so y would be 4x

because i thought they were similar

but they arent

.end

.cancel

.close

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[odd edge]

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[mystic saffron]

How should i go about starting with this?

Since the time for one revolution is constant, this is uniform circular motion i suppose

so [
a_{\t{rad.}} = \f{4\pi^2R}{T^2}
]

[mystic saffron]

smaller values of a correspond to bigger sizes of T

so T_min corresponds to when the acceleration is maximum as well i suppose

T_max corresponds to when a is minimum as well

[zenith tartan]

look i think i can help, i am not sure that's gonna work but maybe that would provide some hints ig?

you know it take T time to complete one revolution, that means a angular displacement of 2pi

and $\omega=\frac{\theta}{t}$
$\omega=\frac{2\pi}{T}$

[clever fjord]

ഴajat

[zenith tartan]

you got the value of omega from here right

now we have to make thier FBD's, from which we have to use the condition F_static leq F_max

but the only problem is that we would also have the friction coefficient mu in our answer

and this omega that we calculated is gonna be used in the centripetal acc= R(omega)^2, so we can make an equation with T in it

should i elaborate myself?

"from which we have to use the condition F_static leq F_max" : and this because we have to maintain the height

@mystic saffron hello?

[mystic saffron]

yes i was working on it mysaelf

[zenith tartan]

and what did you tried?

[mystic saffron]

sadly we havent covered angular motion yet

[zenith tartan]

😭

[mystic saffron]

im still thinking

[zenith tartan]

if you havent covered circular motion yet, then from which topic does this question belong to?

[mystic saffron]

i have

i just havent had it in terms of its angular terms

its okay i think tho i got an idea ill work it out

.close

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[spiral laurel]

How do you go about approximating this (conditionally) convergent series? Hint: This should have to do with finding the remainder to a series.

[odd edge]

Please don't occupy multiple help channels.

@spiral laurel Has your question been resolved?

[spiral laurel]

<@&286206848099549185>

[odd edge]

@spiral laurel Has your question been resolved?

[quasi sparrow]

Did you use the hint

[spiral laurel]

I would if I knew how to, but my class skipped remainders for series’ for some reason.

[quasi sparrow]

https://youtu.be/FkUrAgBzAZo?si=5jZBzIXfI2Jgy3J1

Or read your book

[spiral laurel]

Our class puts empty variables in our books to teach us how to do things intuitively, but I would prefer not to multiply this problem into a series of problems that I also need to solve just to comprehend what’s going on.

Thanks for the ‘help’ though.

.close

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[forest marlin]

sorry i know this isnt hard but my head isnt working right now

[spiral laurel]

@forest marlin Subtract the value opposite of the variable's branch from the 'root' of the branch to get the variable.

For instance, A is calculated by subtracting 14 from 23, and so forth.

[forest marlin]

so b or c is 1?

@spiral laurel

[spiral laurel]

A should be 23 - 14.
B should be (23 - 14) - 2.
C should be 14 - 3.

[forest marlin]

@spiral laurel thanks bro

.close

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[tight notch]

i need to find x but i have no clue how

i got 48 but it was wrong

i have no clue how to do this

ping me if anyone is here to help me

[rich night]

im putting money on 74

personally

[tight notch]

hello

you got 74?

howd u get it

[rich night]

MOJ = 160 cuz angles half at circumference

[tight notch]

yh

[rich night]

JOL because = 126 because 🌟 vibes🌟

[tight notch]

...

so you jus guessin that JOL is126

i get one attempt at this question i need a definite explanation

lmaooo this is some weird question

no one in my calss has been able to do this

[rich night]

aight fine hmmm, what circle theorems do you need to know

[tight notch]

i have no clue

[rich night]

(there are a bunch out there)

[tight notch]

the quadilateral isnt a cyclicl

so idk how to do this

[tight notch]

only thing i know is that MOJ is 160

@crisp pasture

[rich night]

listen, i probably cant help with your problem, but can you help with my problem?

[tight notch]

im 15 i dont think im a good source of help

but ig i can try ahaha

[rich night]

bro same whats your point lol

[tight notch]

ohh lmao

wats the question

[rich night]

its help-27

[tight notch]

i dont think any helpers are coming

sad

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[oak arrow]

I am supposed to check these things on reflexivity, symetry, asymetry, antisymetry and transivity.
I haven't been to the lecture to be honest, because I literally don't understand a single thing.
How difficult is this exercise ? What am I even looking at, cause I don't understand nothing ? What are the steps to solve this riddle ?

[odd edge]

@oak arrow Has your question been resolved?

[oak arrow]

Help me onegaishimasu

[waxen jackal]

@oak arrow pls if you have source to study this lmk , i have same your problems 🥲

<@&286206848099549185>

[oak arrow]

buddy, if I find good sources, I will dm you

[waxen jackal]

I will send msg to your dm 👀

[native ruin]

do you know what a relation is?

[native ruin]

?

@oak arrow

but for the difficulty these are rather easy problems which are very straightforward once you see what you have to do

ping me when you are on again i might be able to help

[oak arrow]

Yes sir, a relation is a subset of the cartesian product XxX

[native ruin]

ok

so lets start with the first example

$mR_1 n \iff 2n \mid m$

[clever fjord]

Homelama

[native ruin]

can you come up with a pair $(m,n)$ such that $mR_1 n$

[clever fjord]

Homelama

[oak arrow]

(2,4)

[native ruin]

yes exactly

do you know how reflexivity is defined?

[oak arrow]

I think if the relation is reflexive, then x = y means y = x

[native ruin]

no, a relation is reflexive iff $nRn$

[clever fjord]

Homelama

[native ruin]

so is this relation reflexive?

[oak arrow]

No

[native ruin]

yeah correct why?

[oak arrow]

Because, I think, there must be a divider m for 2m for it to be reflexive which is not possible for natural numbers

[native ruin]

exactly because 2m does not divide m

[oak arrow]

I am a genious

[native ruin]

ok what about symmetric?

[oak arrow]

Symmetric was mR1n => nR1m, right ?

Okay. Let me see

If 2m divides by n, than n cannot divide by 2m for the natural numbers. So not symmetric

[native ruin]

yes

[oak arrow]

Ok, nice

[native ruin]

so now to asymmetric

[oak arrow]

What is asymmetric and anti symmetric ?

[native ruin]

it means that
$$mR_1 n \implies \neg nR_1 m$$

[clever fjord]

Homelama

[native ruin]

like mRn and nRm cannot hold at the same time

[oak arrow]

Oooooooooh, then it is asymmetric

Answer locked

[native ruin]

think again

[oak arrow]

Okay, wait...

[native ruin]

you just have to find one counterexample

[oak arrow]

Is 0/0 = 0 ?

[native ruin]

yeah 0|0, but 0/0 is not defined

this is like an edge case

but there are also other pairs that work

but m=n=0 would be a counterexample

[oak arrow]

Oh, okay. Thank you, Homelander. You helped a lot.

[native ruin]

sorry yeah 0 0 is the only counterexample

[oak arrow]

Oh, nice

Np

[native ruin]

so yeah, what about antisymmetric?

[oak arrow]

Imma close now and watch an episode of Narcos, because I really need to

I think I got the hang of it

You my hero

[native ruin]

where are you in narcos?

[oak arrow]

I watched the whole show a year ago. I want to start over again so s01 e01

[native ruin]

oh very nice, i think i watched it like in a week all seasons

[oak arrow]

Wooow, impressive

[native ruin]

so have fun bye!

[oak arrow]

Thanks, bud. Bb

.close

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[waxen jackal]

I found your

[waxen jackal]

Ops

.close

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[sudden ledge]

hey guys
for this question since one of the constraints is not convex
KKT does not gaurantee that the solution is globally optimal correct?
so how would I prove that the solution here is optimal

[odd edge]

@sudden ledge Has your question been resolved?

[odd edge]

@sudden ledge Has your question been resolved?

[sudden ledge]

<@&286206848099549185> any help?

[glossy summit]

Well if we know all of the x's

no matter 1, 2 or 3

are 1

we know 1 to the power of 4

is 1

idk im waffling. Im new

I tried.

[sudden ledge]

np lol

[sudden ledge]

.closee

.close

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[hearty delta]

i dont know what to do after I find the 23

what is the area?

[vocal granite]

depends on your bounds of integration

the double integral over the area form dA should just give you the signed area of whatever region you are integrating over

[hearty delta]

I figured it out i think thanks!

.close

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[lime prawn]

so by letting $f(x)=y$ we can then say that $g(y)=x$

[clever fjord]

mygenderisanonnewtonianfluid

[lime prawn]

and we also know that $f'(4)=-\frac{3}{4}$

[clever fjord]

mygenderisanonnewtonianfluid

[lime prawn]

one of my favorite properties of derivatives is that $(f^{-1})'(x)=\frac{1}{f'(y)}$

[clever fjord]

mygenderisanonnewtonianfluid

[lime prawn]

and using this property, we see that $g'(y)=\frac{1}{f'(x)}$

[clever fjord]

mygenderisanonnewtonianfluid

[lime prawn]

plugging in $y$ at the point of $f$ for which the derivative is shown ($4$) gives us $g'(6)=-\frac{4}{3}$

[clever fjord]

mygenderisanonnewtonianfluid

[lime prawn]

hope this helped

if it did, please close the question

[static rapids]

ty

and one more

i say it's B because the tangent slope in B should be the greatest?

not sure

[grizzled tide]

yea

[static rapids]

ah

[lime prawn]

yes

[static rapids]

i almost choose D because i saw that the slope is more closer to a vertical line, but i realized it's negative

alr ty

[lime prawn]

np

[static rapids]

ima keep this open for now

if i dont have any questions ill close it

[grizzled tide]

just close it and make a new channel if you have questions

!1q

[odd edge]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[static rapids]

alr

.close

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[ember badge]

how to evaluate sum

[south plume]

do you see 'i' anywhere in the sum

[ember badge]

wdym

[south plume]

your sum is in terms if a variable i (because you have i=5 written below)

but so you, literally, see 'i' anywhere in the body of the sum?

[ember badge]

ohhhhhhhhhhhhh

wait i get it

so u just add it 5 times

4

[south plume]

yep

[ember badge]

+4

[south plume]

4*5

[ember badge]

ok

yeah yeah

ok

[odd edge]

@ember badge Has your question been resolved?

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[mystic saffron]

@odd edge

[odd edge]

Please don't occupy multiple help channels.

[mystic saffron]

help

[mystic saffron]

ok

help

@odd edge help

@odd edge

HELP

[brittle beacon]

#help-35 stay in that one and ask a question there. The bot doesn’t answer questions

.close

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[kindred tartan]

hi! im trying to figure out how to calculate effect size (cohen's d and r squared) while only knowing one standard deviation

[kindred tartan]

this is basic college stats btw

<@&286206848099549185>

pls help test tomorrow 😦

[sudden ledge]

did you try something?

[kindred tartan]

no i dont know how to start because our formulas need both standard deviations

and also every online calculator needs both standard deviations

lol

[sudden ledge]

not sure what formulas you guys using but for the 1st question you can easily find the t-statistic with the given info

i assume thats what theyre askin

[kindred tartan]

im looking for cohen's d and r squared rn

i just sent our formulas

[sudden ledge]

i think here since you have just 1 group u dont need the pooled

u just use SD of the sample which is gven

[kindred tartan]

i tried that with d = (Msample - µpopulation) ⁄ σ and i got 0.3 and it said it was wrong

so im really confused

[odd edge]

@kindred tartan Has your question been resolved?

[odd edge]

@kindred tartan Has your question been resolved?

[tame comet]

i can help you cheat

i’ve helped some people cheat on diff serves

servers

what’s the test on,?

[tame comet]

ah this

np, i can help.

[kindred tartan]

i just wanna know what i’m supposed to do

real talk i do not know anything about statistics

[odd edge]

@kindred tartan Has your question been resolved?

[odd edge]

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[modest plinth]

Can someone proofread?

The text in the middle says "so we can remove the absolute value"

for lack of better terminology lol

I put it in wolfram and got a different answer so

<@&286206848099549185>

[odd edge]

@modest plinth Has your question been resolved?

[odd edge]

@modest plinth Has your question been resolved?

[odd edge]

@modest plinth Has your question been resolved?

[modest plinth]

Can someone help me understand what I am doing wrong? <@&286206848099549185>

[mystic saffron]

hi guys

can someone help me solve fratctions put it has pi in it

[mystic saffron]

What answer did you get from Wolfram?

@modest plinth

[modest plinth]

(-∞, -3)

not sure what that means

[mystic saffron]

hm

[modest plinth]

https://www.wolframalpha.com/input?i=|6x%2B5|>%7C2x-7%7C+

[mystic saffron]

did you try to find for other several cases?

[modest plinth]

the number line is 1/4 and up and also -3 and down

[mystic saffron]

show me the number line u drew

[modest plinth]

check the link

the wolfram line is there

my line is on the page up there

[mystic saffron]

oh so it means

x < -3

[modest plinth]

yea

[mystic saffron]

hold on

[modest plinth]

x>-3

more than -3

lol

[bold summit]

what is your problem bro

[modest plinth]

this

I don't know where I went wrong

[mystic saffron]

*the union

not the inequality

[modest plinth]

how did you come to the conclusion?

[mystic saffron]

you did not make possible cases

[modest plinth]

wdym

[mystic saffron]

i think u should watch the stuff online

organic chemistry guy i think is the name

[modest plinth]

This just doesn't make any sense to me

I've not learned this much

[mystic saffron]

i only know half of this stuff because im still in 8th grade, the additional lessons i get outside school is the gist of it

[modest plinth]

that or I'm not understanding what you are doing there

[mystic saffron]

well idek, i got the idea when I tried to perform the operations on our given homeworks

[modest plinth]

if abs(6x+5) > abs(2x-7)

then both abs(6x+5) and abs(2x-7) is equal or more than 0

which should mean I can just remove the abs modifier?

Then there is just one case?

that's what I gather

but obviously I'm wrong

and I don't see how you can get the other result

[mystic saffron]

i don't know dawg, I haven't took the abs lesson yet

[modest plinth]

rip

[mystic saffron]

but i think I tried my best to give a reason why Wolfram gave that answer

[modest plinth]

Thank you for trying

[mystic saffron]

you're welcome homie

[odd edge]

@modest plinth Has your question been resolved?

[cosmic sandal]

You can only remove the absolute value when both sides are positive.

If only one side is positive and the other is negative, then you need to multiply one of the sides by -1. (Potentially reversing the inequality if you pick the positive side.)

If both sides are negative, then you need to multiply both sides by negative 1 which reverses the inequality.

For checking your answer, the easiest method is usually to plug in a few values.

For example, try checking it at -10, 0 and 10.

[cosmic sandal]

For example, -3 < -2 but |-3| = 3 > 2 = |-2|

[mystic saffron]

oh that's what abs meant bruv

[odd edge]

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[mellow axle]

”Solve for z, one of the roots is a pure imaginary number”

[mellow axle]

So i got two of the roots right, but one is incorrect and i am missing two additional roots

What am i doing wrong?

<@&286206848099549185>

Here is the corrects answer:

[modest plinth]

Yeah but since there's an abs value on both sides then neither can be negative?

Can you pm me ?

[odd edge]

@mellow axle Has your question been resolved?

[mellow axle]

https://tenor.com/view/twitching-eye-gif-26162222

[odd edge]

@mellow axle Has your question been resolved?

[heady badger]

wait is
20 cm^2 = 20 * 10^-4 m^2

$20 cm^2 = 20 * 10^-^4 m^2$

[mellow axle]

https://youtu.be/QMQ1qGNSVIA?si=S042WzJKffES_abP

[clever fjord]

bob420
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[mellow axle]

Watch this^^

[heady badger]

tnx I will

[mellow axle]

.close

[odd edge]

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[rocky edge]

S is an equivalence relation on ${a, b} \text{ and } |S| = 3$

[clever fjord]

luke1

[rocky edge]

!status

[odd edge]

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

[rocky edge]

1

my knowledge on this topic isnt the biggest

[viscid flint]

is there a question here?

[rocky edge]

oh

find set S, or argue why set S doesn't exist

[viscid flint]

ok well a good place to start would be to look up the definition of an equivalence relation

[rocky edge]

transitivity, reflexive and symmetric

but what does it mean for it to be "on" {a,b}

[viscid flint]

that's a list of properties

can you please go find your book's definition of an equivalence relation and post it here?

[static totem]

there are 4 thngs that can be in S, and you need 3, you can examine all 4 subsets

a relation on a,b is a susbset of {a,b} × {a,b}

[rocky edge]

Reflexive $\forall \in S,$ $xRx$

[clever fjord]

luke1

[rocky edge]

Symmetric $\forall a,b \in S$ . $aRb \iff bRa$

[clever fjord]

luke1

[rocky edge]

Transitive $\forall a,b,c \in S$ . $aRb \land bRc \rightarrow aRc$

[clever fjord]

luke1

[rocky edge]

$aRb = (a,b) \in R$

[clever fjord]

luke1

[viscid flint]

what's S here?

i suppose R is the relation

[rocky edge]

$R \subseteq S \times S$

[clever fjord]

luke1

[rocky edge]

sry thought it was implied

[viscid flint]

well you seemed to be struggling with it

R is a relation on S if it's a subset of S x S

[rocky edge]

oh i see

we get the set {(a,a), (a,b), (b,a), (b,b)} and we have to find the pairs that have an equivalence relation

[rocky edge]

|S| = 4 right?

[vocal granite]

The space of all possible pairs has size 4, S should be some subset of this space which satisfies the definition of an equivalence relation

[rocky edge]

{(a,a), (a,b), (b,a), (b,b)} these are the possible pairs

and it is reflexive and symmetric no?

[vocal granite]

Yeah, the space of all pairs is an equivalence relation of size 4

But your question is to determine whether or not you can make a smaller equivalence relation with only 3 elements

[rocky edge]

cant see anyway to make this smaller

[vocal granite]

One way to make it smaller would be to just take the set {(a,a)}

Which is still an equivalence relation but now has size 1

But, are there any other possible equivalence relations with size between 1 and 4? And what about specifically size 3?

[rocky edge]

{(a,a), (b,b), (a.b)} ?

@vocal granite Thoughts?

[vocal granite]

Doing this kind of construction by adding and removing elements is a good way to approach this

But your new set is no longer an equivalence relation!

(What does symmetry say)

[rocky edge]

oh (b,a) not there

i dont think this is possible

@vocal granite ?

sry for the ping

s

[vocal granite]

Nah, the pings are helpful easy to lose tabs in discord

But yeah

If (a,b) is in there you also need (b,a)

[vocal granite]

Yep! It’s not possible, can you give an argument for why?

[rocky edge]

wait can u explain why (a,b), (b,a), (a,a) is not an equivalence relation

@vocal granite

[vocal granite]

What does reflexivity require?

[rocky edge]

$(x,x) \in R$ doesn't that satisfy it?

[clever fjord]

luke1

[vocal granite]

In particular $(x,x)in R$ for all $x\in S$

[clever fjord]

Willow

[vocal granite]

Which is to say that every element must relate to itself

[vocal granite]

And in this example b doesn’t relate to itself

[rocky edge]

ohh

ive been thinking in a different way this whole time

[rocky edge]

to show this can I just say in order for the set to be an equivalence relation a pair relies on another pair (except a pair on its own) so we cant make 3 pairs in the set?

[vocal granite]

Yep, that’s should work

[rocky edge]

for this your allowed to do this as you never introduced b ?

[vocal granite]

Yeah

If b was included

Then the smallest possible equivalence relation would be {(a,a),(b,b)}

[rocky edge]

.close

[odd edge]

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[ember oak]

,rotate

[odd edge]

@vernal island Has your question been resolved?

[odd edge]

@vernal island Has your question been resolved?

[amber niche]

,rotate

[amber niche]

,rotate

[amber niche]

,rotate

[odd edge]

@vernal island Has your question been resolved?

[deep arrow]

,rotate

[amber niche]

what even is that

[odd edge]

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[mystic saffron]

How do i simplify [
x'y'z' + yz' + xy
]
without using a K-map into its minterms and maxterms

[mystic saffron]

.close

[odd edge]

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[cold urchin]

boolean

[odd edge]

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[pliant crest]

Hi can someone help identify the problem? The answer is 41/6. But I get -3/2

[latent scaffold]

You have to flip the sign of the first integral.

[pliant crest]

Why?

[latent scaffold]

You split it, but if you do nothing, then that is just the integral of 3t-5 from 0 to 3.

The reason why you split it is that the graph is negative from 0 to 5/3.

But that will yield negative area if you just compute the integral. You want the sum of unsigned area to get the total distance travelled.

[pliant crest]

how do i know the graph is negative in that region?

[latent scaffold]

If you want (3t-5) >= 0, you can follow the steps you did at the very beginning and you wil get t >= 5/3.

So that means between 0 and 5/3 it must be negative.

It helps to draw it out in situations like this,

[odd edge]

@pliant crest Has your question been resolved?

[odd edge]

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[neat pagoda]

In sequences/series , for the Arithmetic formula, can An=a1+d(n-1) be used instead of An=a1+(n-1)d ?

[mystic saffron]

multiplication is commutative, so yes

[neat pagoda]

Thank you

[odd edge]

@neat pagoda Has your question been resolved?

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[toxic sigil]

Could someone tell me if my logic is on the right track for this problem?

Here, I'd need to determine the values of two unknowns, a and b. Each plaintext-ciphertext pairing provides a linear equation in terms of a and b. Therefore, two distinct pairings are required to set up a solvable system of two linear equations for these two unknowns, right?

[odd edge]

@toxic sigil Has your question been resolved?

[odd edge]

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[mystic saffron]

I'm trying to represent this function as a power series, center being 0, I'm not exactly sure what I'm doing wrong though, the resulting series I'm getting isn't correct

[toxic monolith]

$\sum_{n=0}^{\infty }\left( -1 \right)^{n}\frac{x^{n+1}}{n+1}=\sum_{n=1}^{\infty }\left( -1 \right)^{n+1}\frac{x^{n}}{n}\text{ for }-1<x\le 1$

[clever fjord]

Joanna Angel

[toxic monolith]

I hope you undersatnd it, you did well, just you wrote in diff way

[mystic saffron]

whats wrong with my answer though

its not correct?

[toxic monolith]

it is correct

bu tother books show it in my way

[mystic saffron]

why doesnt it map on on desmos?

[toxic monolith]

on interval i wrote, desmos shwos ok

but ouside interval, you cant compare them

Maclaurin expansion of ln(1+x) works only in interval (-1,1]

= convergence interval

[mystic saffron]

ok, gotcha, I appreciate your help

thankyou

.close

[odd edge]

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[kindred cosmos]

Is this the right place to ask for help?

[mint mirage]

Depends, what kind of help are you looking for?

[ember mulch]

!da2a

[odd edge]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[cold sage]

[kindred cosmos]

Great.
I need help with this I just don't know how to do the problem.

Is using a formula.

And 2439 is confusing me.
Because is either 9 to the power of 4 since the compliment is that it has no zeroes and there is one containing one zero so it would be 10 and the rest 9, but I can't seem to get an answer close to it.

[odd edge]

@kindred cosmos Has your question been resolved?

[kindred cosmos]

Also having trouble here as well.

<@&286206848099549185>

[ember oak]

How many 4 digit numbers have no zeros?

[kindred cosmos]

1

The correct answer is 2439 and for b 1512.

Problem is I have no idea how to get to those answers.

[ember oak]

1? How're you getting 1?

[kindred cosmos]

I am responding how many 4 digit numbers have no zeroes and is one of the digits.

Containing at least 1 zero so it would be the first digit.

[ember oak]

one of what digits?

[kindred cosmos]

0-9 1-9 1-9 1-9
10 9 9 9

Or 10 times 9 to the power of 3.

[ember oak]

why is highest digit 0-9?

If first digit is 0, how is that a 4 digit number?

[kindred cosmos]

Count 0 to 9 would make it 10.

Each digit contains

4 digits with one of them containing a zero

First digit Second Digit
is 10 Is 9 and so on and so on.

The only time it is not repeatable is in question B.

Which would go from 10 9 8 7.

[kindred cosmos]

Hello?
Or are you still trying to figure out the problem?

[odd edge]

@kindred cosmos Has your question been resolved?

[kindred cosmos]

<@&286206848099549185>

[tight warren]

In "at least ~" problems it's easier if we look at the complementary cases.

(4 digit numbers containing at least 1 zero) = (All 4 digit numbers) - (4 digits numbers that** DO NOT** contain ANY zeroes)

[kindred cosmos]

Would it be 9 times 10 power to the 3- 9 to the power of 4?

Because it's the only way to get to 2439.

[tight warren]

Yep

thats it

[kindred cosmos]

OHHHHH

[kindred cosmos]

B is a bit confusing.

[tight warren]

It's structually the same

[kindred cosmos]

I got 1521 I guess this a typo in my proffesor's part? Because it's 1512 instead.

[tight warren]

(All 4 digit numbers with no repeating digits) - (4 digits numbers that DO NOT contain ANY zeroes AND with no repeating digits)

[kindred cosmos]

Answer is 1512, but the calculation is getting me 1521.

I don't know this is either a typo on his part or I am doing something wrong.

[kindred cosmos]

Would be 10 * 9 87 -9 to the power of 4=1521.

[tight warren]

9 * 9 * 8 * 7 = 4536.

9 * 8 * 7 * 6 = 3024.

4536 - 3024 = 1512 is our final answer.

[kindred cosmos]

Why is it 9 times 9 times 8 times 7 instead of 10 times 9 times 8 time 7?

[tight warren]

because the 1000s place cannot be 0.

so there are 9 choices from 1 ~ 9

[kindred cosmos]

Oh. That makes sense.

[kindred cosmos]

Ok one more thing can you help me with the monogram?

[tight warren]

what monogram?

[kindred cosmos]

This

[kindred cosmos]

^

Is a combination like (insert) NCR insert.

And that should be all everything else I can understand.

[kindred cosmos]

<@&286206848099549185> A little help with the monogram and that's it.
Is combination or ncr.

[odd edge]

@kindred cosmos Has your question been resolved?

[odd edge]

@kindred cosmos Has your question been resolved?

[kindred cosmos]

<@&286206848099549185>

[odd edge]

@kindred cosmos Has your question been resolved?

[odd edge]

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[sour geyser]

is that simply just the linear function y=2+3x???

[open onyx]

ye no need to overthink

[sour geyser]

ok but my teacher put this in the answer

which is not just simply the function y=2+3x which is what i initially thought to do

could someone explain???

[open onyx]

it is simply just the function y=2+3x acting on a random variable X with uniform distribution in [0,1] it looks like

im not sure how you interpreted it when it asked for a pdf and presumably the question defined X for you in parts (a) or (b)

[sour geyser]

oh right so if i subbed x=0 then i get 2 and if i subbed x=1 i get 5 which is see gets me the length of the uniform distribution but what about how they get hieght 1/3???

or the E(y)

[open onyx]

by definition of probability a pdf has to have area 1

[sour geyser]

or nvm thats cos pdf has area 1

but the E(y) tho?

or is that always the middle?

[open onyx]

linearity of expectations

E[aX+b]=aE[X]+b

(or in this case eyeballing also works)

[sour geyser]

what is that and what is a and b and x in this?

[open onyx]

X is a random variable and a and b arbitrary constants

in this case E[3X+2] = 3E[X]+2

[sour geyser]

right so how does E[3X+2] give 3.5 tho?

[open onyx]

find E[X]

[sour geyser]

thats the integral of x(2+3x) right?

[open onyx]

that’s E[Y]

i just need E[X]

[sour geyser]

is that the integral of just x?

[open onyx]

i mean you can also just integrate Y but sure whichever way you want

[sour geyser]

but what are the bounds like 2 and 5???

or 0 and 1?

[open onyx]

depends if you’re integrating f(X) or f(Y)

[sour geyser]

well f(Y) i think

which was x(2+3x) right

[open onyx]

no

[sour geyser]

...

[open onyx]

it’s just x/3

[sour geyser]

wait how did u get that?

[open onyx]

state f(X)

[sour geyser]

2+3x?

[open onyx]

no, the density for X

[sour geyser]

sry dunno what that is

[open onyx]

function for the pdf of X

(also i say density because pdf= probability density function)

[sour geyser]

idk what other functions there are tho

[open onyx]

i just want you to state the pdf of X

[sour geyser]

you've stumped me

[open onyx]

eh, i mean that in someway explains why you struggled with this

the pdf of X is just 1

[sour geyser]

wait as in the area is 1?

[open onyx]

no, the probability density function of X is given by 1 when 0 ≤ x ≤ 1

( and 0 elsewhere)

[sour geyser]

ok

[open onyx]

now find E[X]

[sour geyser]

integral of x(1) with bounds 0 and 1 right?

which is 1/2

or no?

[open onyx]

ye that’s it

[open onyx]

then we come back to this

[sour geyser]

oh right so 3*0.5 + 2 = 3.5 right

also with this question is that P(-1 < X < 0.5) asking what is the probability that x will be between thoose 2 numbers given it is uniformly distributed between 0 and 1?

[open onyx]

ye

[sour geyser]

and if so i just find the cardinality of the X being in that bound -1 and 0.5 and divide it by the cardinality of the bounds 0 and 1?

which is like 0.5/1 right?

[open onyx]

ye or just shoving it into an integral works once you know the pdf

[sour geyser]

ok thanks

.close

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[gleaming hedge]

could someone help me understand this?

[gleaming hedge]

I don't know how it got to cubic root of a b and c

[faint oak]

Have you tried to visualise it? It might become apparent.

Or looking at simpler forms to see how it works?

[gleaming hedge]

no man that will take time

just tell me how it happened

like what formula

or something

[faint oak]

Sure

[gleaming hedge]

<@&286206848099549185>

[odd edge]

@gleaming hedge Has your question been resolved?

[faint oak]

$\lim_{{n \to \infty}}(\frac{1}{3}(a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}))^n=a^\frac{1}{3}⋅b^\frac{1}{3}⋅c^\frac{1}{3}$

This is a slightly different way to write it

[gleaming hedge]

hm alright

[clever fjord]

kytsu1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[faint oak]

I don't know how to solve $(a^{1/n}+b^{1/n})^n$ Do you?

[clever fjord]

kytsu1

[gleaming hedge]

man

[gleaming hedge]

my lecturer showed a way using natural logarithm

[faint oak]

As n approaches ∞ 1/n gets nearer zero, so we will have a^0 + b^0 + c^0.

Numbers raised to the power of 0 result in 1, so we have ((1/3)(1+1+1))^n which is 1^n = 1

[gleaming hedge]

right

[toxic monolith]

$\lim_{x \to 0^{+}} \left( \frac{a^{x}+b^{x}+c^{x}}{3} \right)^{\frac{1}{x}}=\\exp\left[ \lim_{x \to 0^{+}}\frac{\ln\left( \frac{a^{z}+b^{x}+c^{x}}{3} \right)}{x} \right]\overset{H}{=}\\exp\left[ \lim_{x \to 0^{+}}\frac{3}{a^{x}+b^{x}+c^{x}}\cdot \frac{1}{3}\cdot \left( a^{x}lna+b^{x}lnb+c^{x}lncx \right) \right]=\\=exp\left[ \frac{1}{3}\cdot \left( lna+lnb+lnc \right) \right]=\\=\left( abc \right)^{\frac{1}{3}}=\sqrt[a]{a}\sqrt[3]{b}\sqrt[3]{c}$

[clever fjord]

Joanna Angel

[gleaming hedge]

ohhh

wait

I'll try doing this myself

can I ping you, if I have questions, Joanna?

[toxic monolith]

ok)) sure )

[gleaming hedge]

can you solve it without applying n->inf to x->0+

[toxic monolith]

I'm back, ok, give me 5 - 10 minutes for editing LATex

[gleaming hedge]

alright

[toxic monolith]

We will use here:

• inequality between the arithmetic and geometric mean
• monotonicity of the logarithmic function
• inequality of the form:

$\forall _{x>0}\left( ln\left( x+1 \right)<
x \right)$

[clever fjord]

Joanna Angel

[toxic monolith]

then we get:

$\frac{1}{n}\ln\sqrt[3]{abc}\le \ln\left[ \frac{1}{3}\left( \sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c} \right) \right]=\\\ln\left[ \frac{1}{3}\left( \sqrt[n]{a}-1 \right)+\frac{1}{3}\left( \sqrt[n]{b}-1 \right)+\frac{1}{3}\left( \sqrt[n]{c}-1 \right)+1 \right]<\frac{1}{3}\left[ \left( \sqrt[n]{a}-1 \right)+\left( \sqrt[n]{b}-1 \right)+\left( \sqrt[n]{c}-1 \right) \right]$

[clever fjord]

Joanna Angel

[toxic monolith]

Now just multiply the above inequality by n and use:

$\lim_{n \to \infty } n\left( \sqrt[n]{a}-1 \right)=\ln
a\text{ for }a>0$

[clever fjord]

Joanna Angel

[toxic monolith]

please write it down and then it's good 🙂

Naturally, the second method is not obvious, either you know this trick before or you will struggle all day, unless you use the trick in the first method that I showed earlier

[odd edge]

@gleaming hedge Has your question been resolved?

[gleaming hedge]

ohh right

tysm

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[mystic saffron]

just a general question, can you do difference of cubes

[mystic saffron]

with ^6's?

[sullen ferry]

yea

$(x^2)^{3}-(y^2)^{3}$

[clever fjord]

Moosey

[mystic saffron]

so would it be two times the power of everything?

oh

so they just go on the outside?

[sullen ferry]

just basically replace everywhere you see an 'x' or a 'y' with x^2 and y^2

because that is what you are cubing in this case

[mystic saffron]

what if there's a ^2

[sullen ferry]

wydm

[mystic saffron]

replace it with ^5?

because I have x^6-y^6, would I do the difference of square equation

[sullen ferry]

you could do difference of squares or difference of cubes

[mystic saffron]

with 2 times the power?

oh sorry I meant

different of cubes

lol

[sullen ferry]

$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$

[clever fjord]

Moosey

[mystic saffron]

yeah that's it

would it be $x^2 y^2 * x^4 +xy^2 + y^4$

[clever fjord]

mj

[sullen ferry]

$x^{6}-y^{6}=(x^2)^{3}-(y^2)^{3}=(x^{2}-y^{2})(x^{4}+x^{2}y^{2}+y^{4})$

[clever fjord]

Moosey

[mystic saffron]

hm close enough

so it's just difference of squares

*cubes

sorry

with powers

multiplied by 2

[sullen ferry]

yes, applying difference of cubes to x^6 and y^6

[mystic saffron]

fair enough

thanks for your help

[sullen ferry]

yes, np :)

[mystic saffron]

.close

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[jagged cloud]

can i get a wee reminder how to solve

for pa,pb,pc and lamba values

[sullen ferry]

multiply ways, you could get the augmented matrix into RREF, find its inverse and multiply over (this way is bad), etc.

the first way is most typical

augmented matrix being your coefficient matrix with your solution as an extra column attached on

then the procedure for getting to RREF mainly just involves getting your main diagonal (from left to right down) being all 1s, and entries below the diagonal all zeros. This may not be possible. You do this using row operations.

the process to getting to RREF/REF form is gaussian elimination

[jagged cloud]

can you help me build inital step. im sure ive done this in past. maths bit rusty

[sullen ferry]

$\left[\begin{array}{cccc|c}
0.08 & 0 & 0 & -1 & -1.4\
3 & 7 & 3 & -1 & -1.6\
-2 & -3 & -4 & -1 & -1.8\
1 & 1 & 1 & 0 & 350
\end{array}\right]$

[clever fjord]

Moosey

[sullen ferry]

this is the matrix you start with. We want to make the first entry (the 0.08) into a 1.

[jagged cloud]

ok

[sullen ferry]

How can we do this? well, we have row operations we can work with. We can scale an entire row by any amount (like multiply), or scale and add it to another row.

[jagged cloud]

times by 1/0.08?

[sullen ferry]

OH right, and we can also switch the order of the rows

https://www.khanacademy.org/math/algebra-home/alg-matrices

[sullen ferry]

yes, that will do

[jagged cloud]

of course bottom goes top

im tryna find my notes on this

[sullen ferry]

$\left[\begin{array}{cccc|c}
1 & 0 & 0 & \frac{-1}{0.08} & \frac{-1.4}{0.08}\
3 & 7 & 3 & -1 & -1.6\
-2 & -3 & -4 & -1 & -1.8\
1 & 1 & 1 & 0 & 350
\end{array}\right]$

[clever fjord]

Moosey

[sullen ferry]

Row-echelon form & Gaussian elimination section is probably what you want to look at

[jagged cloud]

cant find notes sadly.

im in fifth. last had maths class in 3rd

[sullen ferry]

https://www.youtube.com/watch?v=eDb6iugi6Uk

[jagged cloud]

cheers. that should be enough

[sullen ferry]

https://www.youtube.com/watch?v=eYSASx8_nyg&t

yea, these should help and show you :)

[jagged cloud]

is that 1st vid u sent

essentially playing abou to get triangle of zeros bottom left

thanks @sullen ferry

[sullen ferry]

yeee

:)

[jagged cloud]

nice nice

[jagged cloud]

im a final year masters elec n mech eng student. sometimes takes a bit to remember all stuff ive learnt

[sullen ferry]

understandable :)

[jagged cloud]

and u want a 1 top left to make row operations easy as fuck right

[sullen ferry]

yea yea

[jagged cloud]

then use any one in middle to remove the other 2 zeros

or is it row 2 column 2 i use to get the 2 below zero

those ones to be zeros

@sullen ferry thanks

[sullen ferry]

np :)

[jagged cloud]

sorry one more thing.. if i was to move row 4 to top for ease, does the 350 also move.. top right

@sullen ferry

[sullen ferry]

yes

[jagged cloud]

ta

where u from lad

[sullen ferry]

US

[jagged cloud]

im glasgow

nice nice

love usa

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@jagged cloud Has your question been resolved?

[odd edge]

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[mystic nebula]

how do I know where the boundary points are?

[mystic nebula]

I know that I'm supposed to examine the critical points & and the boundary points then compare them, but idk how to find the boundary points ngl

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@mystic nebula Has your question been resolved?

[light violet]

Pre sure you find the extreme along the boundary lines of the square so Lagrange multipliers I think

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[hoary jungle]

what does ln^2 mean?

[bright cobalt]

ln*ln ig

Idk I'm dumb compared to most of this server

[hoary jungle]

me too

[weary pelican]

Yep (ln(x))²

[hoary jungle]

ah

what do i even subsititute here

[weary pelican]

Not even a slightest intuition?

[hoary jungle]

1+ln^2x?

oh wait

[weary pelican]

Works

[hoary jungle]

derivative of that is 2 • ln(x) • 1/x right?

[weary pelican]

Yes

[hoary jungle]

yeah i think i solved it

also another question

what happens if i write d(1/x) instead of dx

does my e become x?

and do i get left with x^-2 * x

or is it d(lnx) that way

[weary pelican]

If there was written d(1/x) instead then your silent variable is just t = 1/x
So 1/x³ e^(1/x) = t³e^t

Are you asking about a variable change?

[hoary jungle]

thats my problem

i dont really know what im asking

my lecturer did something like that

and he talked about noticing the derivative or something while doing it

[weary pelican]

Uh maybe in some cases it can happen

Like dx/x

You know that this is d(ln(x))

[hoary jungle]

I just took down the notes

But cant really comprehend the change

[weary pelican]

Sure but it's like delaying the inevitable variable change

d(x²) = 2xdx

[hoary jungle]

but doesnt it make it easier?

[weary pelican]

I'm not arguing for it tho

like just let u = x²

And write du = ...dx

(Here du = 2xdx)

[hoary jungle]

i see wht you mean

[hoary jungle]

had a feeling it would make it easier doing it that way here

[toxic monolith]

i do not want to interrupt, but the most important is understanding the formula: