help-49

[modern bolt]

?

[spiral osprey]

there's only part a and b that i see

[modern bolt]

Eh?

[spiral osprey]

you gave an answer to 4 parts when there are only 2...

[west iron]

he analyzed 4 of the 5 propositions given but answered neither of the questions

which is fine, since

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[modern bolt]

Oh yeah I left out the 4th one mb

It fails whenever x or y is 3

[spiral osprey]

you're looking at the wrong thing lmfao

[modern bolt]

Oh

[spiral osprey]

look at parts a and b

not the facts that were given

[modern bolt]

Ohhh

[midnight plank]

@pale barn Has your question been resolved?

[pale barn]

ok i have made tables for all values

now what

and this

[pale barn]

?

[midnight plank]

@pale barn Has your question been resolved?

[midnight plank]

@pale barn Has your question been resolved?

[pale barn]

.close

[midnight plank]

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[last slate]

SOLVE the equation 5X-6 = 4 (X+2) in the set of rational numbers

[grim vector]

What have you tried ?

[last slate]

nothing..

what am i suppsoed to do

i know nothing

teach me

[grim vector]

Try to develop the right side

[last slate]

huh

5X−6=4X+8 ?

[grave sonnet]

@last slate it smells like poo

[grim vector]

Sure

[last slate]

mb

[grim vector]

And now send the 4x to the right side

[last slate]

5X - 4x - 6x = 8

?

is it correct

is the anwer 14

[modern bolt]

How you got 14?

[modern bolt]

-5x = 8

[last slate]

5x−6=4x+8
5x - 4x - 6x = 8
x−6=8
x=8+6
x = 14

o

[modern bolt]

X = -5/8

[grim vector]

Its not -6x its -6

[modern bolt]

X-6x

[grim vector]

In the original problems

[grim vector]

Stop

[last slate]

omg

[grim vector]

Its not

-6x

[modern bolt]

How old are you?

[modern bolt]

Oh

[last slate]

STOP WHY AREY OU ASKONG

IS 14 CORRECT

[last slate]

no, not really

[modern bolt]

I gave a solution for his simplified form

[last slate]

yes, but can you show your working out?

[modern bolt]

He sinplied it wrong 😔

[modern bolt]

Any 13 year old should be able to do this

[grim vector]

He did

[modern bolt]

And its against discord tos for a people aged less than 13 to use discord

[last slate]

im not less than 13

wtf

[modern bolt]

Okay

[quiet hinge]

Please refrain from making such comparisons when helping people. Everyone has their own path in life

[last slate]

5x−6=4x+8
5x - 4x - 6 = 8
x−6=8
x=8+6
x = 14

[modern bolt]

Was just checking if he was underaged for discord

[quiet hinge]

If you cannot resist making such comments please dont help

[last slate]

i mean, you got -5/8

guys i jut need help 😭

[grim vector]

Just becarful in second line its not -6x its jsut -6

[quiet hinge]

Your method of "checking" is flawed, not scientific in any way and just rude

[last slate]

Ok sorry

[grim vector]

Otherwise its perfect

[last slate]

you have fixed all the errors that i can see

yes

[modern bolt]

Ok chill

[last slate]

can u guys help me with another problem again

sure

send it

[grim vector]

Ok

So

[modern bolt]

Simplify the ratio?

[grim vector]

Try to simplify the top of A

[last slate]

what yakubros said, try to collect like terms on the left hand side of the ratio first

[modern bolt]

Take root 3 out and then do algebra on the top term

[last slate]

(8rt(3) + 14rt(3) - 12rt(3))

[grim vector]

You can also write 75 as 3*25

[modern bolt]

After taking about root3 outside you will get (8+14-12)root3 at top

After solving the top you can check for same factors on top and bottom

[last slate]

i would suggest not giving out the whole answer

[modern bolt]

And eliminate them

[midnight plank]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[last slate]

thank you; i did not know the command for it

[last slate]

8, 14, and -12?

[grim vector]

What happen if you sum them up ?

[modern bolt]

I didn't tho? I just showed how to take root3 outside

[modern bolt]

Try my method , take the root3 out and then perform the algebra

[last slate]

yes, and i was worried that you would continue

thank you for not doing so

[modern bolt]

After you get better you can do it directly

[last slate]

10?

[grim vector]

Yes

[modern bolt]

Yes

[modern bolt]

You got pen and paper ?

[grim vector]

Now how can you rewrite 75 as 3*smth ?

[modern bolt]

🥺 how you got 3*something

[last slate]

YE

[modern bolt]

Is it always 3*something?

[modern bolt]

Write the ratio in division first

The left side at top

The right side at bottom

[grim vector]

Well since you want sqrt(3)

Bot

[modern bolt]

Why you want it ?

Explain properly

[grim vector]

To simplify and show its natural number ?

[modern bolt]

Yes

You didn't explain this at first

[modern bolt]

• there is no need to show its natural number ?

Its called simplified form

[grim vector]

Read the question ig

[modern bolt]

It can be natural , or rational doesn't matter

[last slate]

gyts

guys

[modern bolt]

Eh its not in English 😭

[grim vector]

Este natural

[modern bolt]

Yes?

[modern bolt]

What if natural in other language means something else?

[grim vector]

Not when you talk about math

[last slate]

i got 10 what do i do with the ten

[modern bolt]

Did you write the fraction?

[grim vector]

So 10*sqrt(3) / sqrt(75) right ?

[modern bolt]

Send me the pic

[last slate]

remind me what this server does with unhelpful helpers?

[last slate]

10√3 : √75

what do i do next?

[modern bolt]

Now we need to simplify it

[grim vector]

How can you can get rid of the sqrt(3) ?

[modern bolt]

Btw ratio is just / so you can just divide and then rewrite as : afterwards

[last slate]

ohhhh

[last slate]

uhhh

30?

[modern bolt]

Yah that's why I said to rewrite the term in fraction

[last slate]

10/3?

[modern bolt]

Ok so we got two terms right ?

We need to see if anything cancels out

[grim vector]

No i mean there is a sqrt(3) on top and you want it to disapear so you need it on bot also right ?

[modern bolt]

How to see if it cancels out? There are some ways

First to get factors of the number

Ours already looks not that bad so we can go to the step 2

Step 2 to Checkif we can eliminate the root

The term which got root is @last slate ?

[modern bolt]

Idk not the place to ask

#discussion

[last slate]

Ohhhhhh

[modern bolt]

What's the root we got?

What term has the root ?

Is it 10root3 or root75?

[last slate]

THE ANSWER

2

??

am i correct?

[grim vector]

Yes

[modern bolt]

Yeah

[modern bolt]

Good

[last slate]

YES

[modern bolt]

Explain

How you got 2

[prime hornet]

hard question. there’s no policy for if it’s a one time thing

[last slate]

prime factors

[prime hornet]

but if it’s repeated then sure

[last slate]

75

[modern bolt]

#discussion

[modern bolt]

What

[modern bolt]

Oh okay

True true you can go like that too

Countinue

[prime hornet]

also technically not the place to go chat about it, but you are right, and I’ll stop interrupting

[lapis dew]

unuseful helpers 👀 I wonder who was been offline for 6 months by now

aight, time to disappear from internet for one year once again

[prime hornet]

LMAO

[midnight plank]

@last slate Has your question been resolved?

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[ruby gate]

On number 2, why is the bounds from 0 to 1. Like how would I determine that? someone said so but I dont understand where the 1 bound comes from

[midnight plank]

@ruby gate Has your question been resolved?

[cedar coral]

Compute the intersection of the function with y=0.
You will obtain a closed region, from x=0 to x=1

[midnight plank]

@ruby gate Has your question been resolved?

[ruby gate]

What’s wrong with my shell method here? The hand pic is the answer key

[cedar coral]

you asked why the integration bounds were 0 and 1

[ruby gate]

Ik it’s a different question can u help real quick

[cedar coral]

no, sorry

[ruby gate]

.cloze

.close

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[pallid kestrel]

Problem: In the square ABCD, the side AB has a common equation of 4x-3y-12=0. Find the equations of AD and BC if the side of the square is 5 cm and the middle of CD is M with coordinates (3,5 ; 9). This is what I have tried but at the moment I am stuck. Any help would be appreciated!

[lethal path]

seems like you are getting stuck after finding 4x - 3y + 13 = 0

[pallid kestrel]

Exactly

I don't know how to continue

[lethal path]

one way is to solve 4x - 3y + 13 = 0 and (x - 3.5)^2 + (y - 9)^2 = 2.5^2

[pallid kestrel]

Should I try it?

[lethal path]

there is a simpler way though

rearrange this as y = (4/3)x + 13/3

so for every 1 unit change in x, y changes by 4/3

and the length of the hypotenuse is sqrt(1^2 + (4/3)^2) = 5/3

so by proportionality
1 unit of x : 5/3 units of hypotenuse
3/2 units of x : 5/3 * 3/2 units of hypotenuse

hence the x coordinates of C and D must be 3.5 + 1.5 and 3.5 - 1.5

[pallid kestrel]

Isn't 5/3*3/2=2.5 though

[lethal path]

exactly

we want the hypotenuse to be 2.5 or 5/2

[pallid kestrel]

What do you mean by hypotenuse?

[lethal path]

cause MC = MD = 5/2 by definition of midpoint

[pallid kestrel]

Why would we take it like that though

[lethal path]

what do you mean?

[pallid kestrel]

The side of the square itself

I have never seen anything like that yet

[lethal path]

oh, well I find this the neatest way

you do have to use your brain a bit instead of using algebra autopilot

like if you can find the coordinates of C and D

given that M = (3.5, 9), MC = MD = 5/2, and that M, C, D lie on the line 4x - 3y + 13 = 0

you're pretty much there

[lethal path]

the key is that the line is the hypotenuse

[pallid kestrel]

By the line you mean MD and MC?

[lethal path]

whereas the length of the legs of the right triangle tell you the change in x and change in y from point M

[lethal path]

yes, 4x - 3y + 13 = 0 passes through D, M, then C

[pallid kestrel]

Okay I kinda get it and then we make triangles with a 90* angle

Where MC and MD are hypotenuses

But then what

We have that MD and MC are 2,5 cm

[lethal path]

try using vector rotation

this will be another new technique

[pallid kestrel]

I don't think I have studied that

[lethal path]

so we have that the height of the right triangles is 2 by Pythagoras

so DC = (1.5, 2) + (1.5, 2) = (3, 4)

perpendicular vectors with the same length will be (4, -3) and (-4, 3)

we choose (4, -3) since we want point A to be to the right and below point D

if you draw it out

so A = (2, 7) + (4, -3) and D = (5, 11) + (4, -3) if you work it out

[pallid kestrel]

Where did you get that from

[lethal path]

from this diagram

DC = DM + MC but DM = MC by definition of the midpoint

[pallid kestrel]

Then A would be (6,4) and D would be (9,8)

[lethal path]

correct!

but you should really understand why

[lethal path]

check that (4, -3) and (-4, 3) have the same magnitude as (3, 4) by Pythagoras

then check that the slopes of (4, -3) and (3, 4) multiply to -1

same for (-4, 3) and (3, 4)

[pallid kestrel]

I'm trying to figure it out since English isn't my first language

That's why I answer so late

[lethal path]

magnitude = length

[pallid kestrel]

Isn't that k?

[lethal path]

what is k?

[pallid kestrel]

That's how we name the slope

With the letter K

Let's start back from where I found the common equation of CD

And also I find that AD: y=-3/4 + b from AB being perpendicular to AD

Which means their slopes multiplied equal -1

What do i do from here?

[lethal path]

this step

[pallid kestrel]

From there?

What do I do from there

[lethal path]

just read what I wrote from that message

instead of having to ask me over again

[pallid kestrel]

This is what I don't get

[lethal path]

Pythagoras

[pallid kestrel]

a²+b²=c²?

[lethal path]

so for every 1 unit change in x, y changes by 4/3

this is what slope = 4/3 means

[lethal path]

yes

a = 1 and b = 4/3

[pallid kestrel]

And why is it ±1.5 and not 2.5 as 5/3*3/2 is 2.5

[lethal path]

well, a length can never be negative

[lethal path]

cause you want the hypotenuse to be 5/2

and you want to get the change in x from hypotenuse = 5/2

[pallid kestrel]

I meant why aren't the x units of C 3.5+ 2.5 and D 3.5-2.5

[lethal path]

if the hypotenuse is 2.5, the base length = change in x can absolutely not be 2.5

the hypotenuse is the longest side of the triangle after all

[pallid kestrel]

Ohhhhhh

I got it

But where does 1.5 even come feom

R

[lethal path]

it comes from the ratio being the same

[pallid kestrel]

Okay, I think I understand it

And we have the X of C and D

Now what

[lethal path]

then the vector rotation thingy

[lethal path]

start reading from here

[pallid kestrel]

.close

[midnight plank]

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[ionic magnet]

how to solve

[ionic magnet]

in (….)

[brazen shuttle]

Add them

Find a common denominator

Make 3 pi have a denominator of 2 and add

[ionic magnet]

wait

[ionic magnet]

why 3 pi

[midnight plank]

@ionic magnet Has your question been resolved?

[ionic magnet]

.close

[midnight plank]

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[fleet flume]

Hi, can someone explain me this question

[fleet flume]

I thought that there were "n" trials were the probability of both Prue and Frida finding a typo being p1 * p2 (p1*p2 meaning the probability of both of them occurring)

which was Bin(n, p1 * p2)

but it isn't

[forest jackal]

its not that both of them are occuring simultaniously

it means that one of them got a few and the other got the remaining

it can also mean that a few typos were caught by both

it contains all 3 cases

[fleet flume]

oh, so I should look the prob of 1 of them occurring, not both

at least 1

[forest jackal]

yeah

[fleet flume]

thx

got it, thank you 😄

.close

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[noble lichen]

I get 0,589 cm^2 21.
A steel plate is 20 mm long and 18 mm wide. It is drilled with three holes, 5 mm in diameter. How big is the surface left? What is its size in square centimeters?

[fallow scarab]

,rotate

[midnight plank]

@noble lichen Has your question been resolved?

[noble lichen]

Nvm

I found it out

.close

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[sick lodge]

im trying to find the limit where (x,y) tend to (0,0) but i get nowhere i tried the "majoration" its a french work idk how they say it in english

[sharp coral]

the first thing i would try is different paths to see if you can get them to disagree (then the limit would not exist). only if they all seem to agree then you might start trying to prove it

[sick lodge]

so first thing i need to do is to know if it actually exists or not?

that makes sence

is there a way to know if it exists or not?

[fallow scarab]

Where's the original question

[sick lodge]

calculate the limite in (0,0) of the g function : R^2 -> R

the function in question is the one i wrote above

g(x,y)

[sharp coral]

if you can find two different paths which result in different limits then it definitely doesn't exist. if you can find it using some strategy like you're doing then it does. proving it doesn't exist is easier so that's usually what i would try first

[sick lodge]

do i have to randomly choose 2 paths? as long as they converge to 0 and that they are different from each other?

but cloud based on the question that im sorry i gave late it exists

[sharp coral]

no, if they both converged to 0 that would tell you nothing. but if one of them converged to 0 and the other one converged to 1 (for example), then the limit would not exist

[midnight plank]

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[frank osprey]

Can i have a little help proving that d^2 = R(R-2r)
Where:
d is the distance between circumscribed and inscribed (to a triangle) circle centers
R is the radius of the circumscribed circle
r is the radius of the inscribed circle

[frank osprey]

Also i just started geometry on my book so i think i need to use not too hard tools

If its too hard with elementary tools i would accept a help proving that R >= 2r, but of course without using d^2 = R(R-2r)

[steel kayak]

d is BC?

[frank osprey]

In this case d is DF, the distance between the centers of the circles

[steel kayak]

oh yeah sorry

[frank osprey]

Just ignore these letters lol

Sorry idk how to remove them

[frank osprey]

It was my fault

[steel kayak]

no I should have read the English properly...

[frank osprey]

Its fine

[steel kayak]

R is for the smaller circle?

or r

[frank osprey]

R is for the big circle and r for the small, but all information is here if needed

[frank osprey]

<@&286206848099549185>

[midnight plank]

@frank osprey Has your question been resolved?

[frank osprey]

No sir

<@&286206848099549185>

[midnight plank]

@frank osprey Has your question been resolved?

[midnight plank]

@frank osprey Has your question been resolved?

[midnight plank]

@frank osprey Has your question been resolved?

[midnight plank]

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[obtuse totem]

i don’t know what’s wrong

[fallow scarab]

That's quite hard to follow

[obtuse totem]

yea

ill rewrite?

yo i mightve found the probem

my handwriting is too messy

or like its too disorganized

[fallow scarab]

Handwriting is fine

Logical ordering is not

[obtuse totem]

got it now

when I reindexed the sum I made a small error

i did another question, this one is muchhh more neat

what should you do if you have a form like this

you can’t set bn to 0 cause it’s not solely a taylor series

@fallow scarab

oh I think what you do is you set it to be taylor series

so you set this part to be 0

then eval

[midnight plank]

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Please don't occupy multiple help channels.

[rain wasp]

go back to your #help-28

.close

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[rain wasp]

brother its been 3min

[brazen shuttle]

a number mutlipled by 0 is 0 since you have zero groups of that number

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[safe onyx]

I don't understand this

[safe onyx]

I at first thought 1

because 3 / infinity --> 0
n ^ 0 = 1

but that wasn't an option so I was confused

[sharp coral]

infinity^0 is an indeterminate form

[safe onyx]

ah ok
3 + 3 e^ infinity is infinity

[violet storm]

that only works for constant n

[safe onyx]

so how do I get from
infinity ^ 0 to e^3

Ig it's like in the
lim x x-3 / x ^2 - 9 --> 0 = 0 / 0 so you have to factor kinda thing

[violet storm]

a lot of times when you have x in both the base and exponent of a limit you're taking you can try rewriting it using e^ln(...)

$\lim_{x\to_\infty}(3+3e^x)^{\f3x}=\lim_{x\to\infty}e^{\ln(3+3e^x)^{\f3x}}$

[grand pond]

from math import sqrt

[violet storm]

the idea is to take the 3/x exponent in front of the ln then see if it simplifies from there

so for example that e^{3\x} part will simplify to 1 if you separate it from the rest

[safe onyx]

,, \lim_{x\to\infty}e^{\ln(3+3e^x)^{\frac3x}} = \lim_{x\to\infty}e^{\frac3x\cdot\ln(3+3e^x)} ?

[violet storm]

\f wont work for you i think i defined that custom command

[safe onyx]

oooh

[violet storm]

replace it with \frac

[safe onyx]

didn't know that fraction works nicely with two at once thank you :)

[violet storm]

yes

[grand pond]

smeagol

[violet storm]

if they're just one character each i think, otherwise need { }

or maybe one command each too, not sure

yeah just 1 char i guess? dunno :p

[safe onyx]

¯_(ツ)_/¯

[violet storm]

so now you have e^(3/x) * e^ln(....) you can separate into product of two limits and the first kind of goes away since its 1

i think?

[safe onyx]

i don't think we can

:/

[violet storm]

=\ my brain is shortcircuiting yeah

[safe onyx]

i looked up exponent rules because I wanted to be sure for moving the ^3/x to the front

[violet storm]

well try doing lim ln(3+3e^x)/x now

you can move the limit into the exponent

[cedar pawn]

I was really hoping Smeagol's question was going to be about rings >.<

[elder quiver]

what you can do is

$\ln{(3+3e^x)} = \ln{(3e^x)} + \ln{(1+\frac{1}{e^x})}$

[safe onyx]

sorry i do try to stay on topic in the help channel XD

[cedar pawn]

Rings are a mathematical structure, so it could have been

[grand pond]

Smiley ッ

[safe onyx]

beyond my level

[safe onyx]

I don't understand sorry

[lament basin]

$e^3$

[grand pond]

purplehex

[safe onyx]

yes, i know the right answer, but i have no idea why

[elder quiver]

yeah so let's just look at the exponent part for this here

so if we apply what I just did

we have

$\frac{3}{x}\ln{(3e^x)} +\frac{3}{x}\ln{(1+\frac{1}{e^x})}$

[safe onyx]

right now if we plug in infinity it would be
e ^ 0 * infinity
which is still messy

[grand pond]

Smiley ッ

[elder quiver]

from here we can use log rules again

[violet storm]

3+3^e^x in the first log

[elder quiver]

yep

we split ln(3e^x) into ln3 + x

which when multiplied we end up with 3+ 3ln3/x

[violet storm]

i think you should them them the properly of logs you're using since its not an often used one

for that rewrite of the log

[safe onyx]

i need to practice log rules 🏃

[elder quiver]

yeah its just the addition one

log(A) + log(B) = log(AB)

[violet storm]

for this one i mean

[safe onyx]

what about log(A + B)

[safe onyx]

I don't see how it applies since it is two things added, A + B

[elder quiver]

ln(A) + ln(B) = ln(AB)

[safe onyx]

yes I get log and ln have the same rules, but in this case I don't is it
A = 3e^x
B = 1 + 1/e^x

is this it?

[elder quiver]

yep

[violet storm]

if you multiply them together you get 3e^x + 3 back

[safe onyx]

this makes sense

[elder quiver]

so we move the limit to the exponent

and just solve the limit from there

[safe onyx]

but now we have
e ^ 0 * infinity + 0 * 0

[elder quiver]

apply log rule once more to first term

$\frac{3}{x}*(\ln(3)+\ln(e^x))$

[safe onyx]

,,lim e^{\frac3x\cdot\ln(3e^x)+\frac3x\cdot\ln(1+\frac{1}{e^x}}

[grand pond]

Smiley ッ

[safe onyx]

this is 0 * 0 so we ignore it now?

[elder quiver]

yep

[safe onyx]

e^ 3ln(3)/x + 3

which when limits
is just
e^3

since n / x when x to infinity is 0

e^ 0 + 3 = e^3

yay

[cedar pawn]

🎉

[safe onyx]

thank you all so much!

this was a doozy

.close

[midnight plank]

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[midnight plank]

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[opaque igloo]

I know that one point is 1-2i

and another is 3-2i

and i think it is a major arc

but idk how big the radius is

[weak nebula]

Mind explaining your approach a little perhaps

[opaque igloo]

tbh i dont even know how to do this

[weak nebula]

Try writing them in parametric form using eulers thereom?

[opaque igloo]

we havent done eulers theroem tho

[weak nebula]

Do u know parametric form?

[opaque igloo]

no

[weak nebula]

What formulas do u know bro for complex numbers

[opaque igloo]

like de moviers and shit

[weak nebula]

You don't know eulers formula but u know de moivres?

[opaque igloo]

yh

teacher said we will learn it after my exam

[weak nebula]

Icl I'm just as lost as you rn and i gotta run bro someone else oughta help tho good luck 😭

[opaque igloo]

yeah allg

[midnight plank]

@opaque igloo Has your question been resolved?

[opaque igloo]

no

[midnight plank]

@opaque igloo Has your question been resolved?

[opaque igloo]

no

[radiant roost]

i have kind of a janky argument

if angle AOB is to be pi/4 then angle ACB must be pi/2

and you can find coordinates of C in terms of z

ah, better yet - the angle between two vectors is arccos((a dot b)/(|a|*|b|))

set that to pi/4

although the algebra is hard that way

[midnight plank]

@opaque igloo Has your question been resolved?

[midnight plank]

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[twilit field]

Prove $\lim_{x \to a} f(x) = \lim_{h \to 0} f(a+h)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$\abs{x-a} < \delta_1 \implies |f(x)-L_1| < \varepsilon 1
\
\abs{h-0} < \delta_2 \implies |f(a+h) - L_2| < \varepsilon_2$

[runic hamlet]

no

check again

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

Unwinding the definition even more might be helpful if youre uncomfortable using dummy variables which you’ll sort have to

E.g. notice if |h| = |(h + a) - a| < delta_1 then |f(h+a) - L| < epsilon_1

Oh and u should probably use different symbols for L

[twilit field]

huh?

[humble torrent]

U want to show they have the same limit

You’re assuming they already have

[twilit field]

oh right

$\abs{x-a} < \delta_1 \implies |f(x)-L_1| < \varepsilon 1
\
\abs{h-0} < \delta_2 \implies |f(a+h) - L_2| < \varepsilon_2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

We have $\abs{f(x)-L_1} + \abs{f(a+h)-L_2} < \varepsilon_1 + \varepsilon _2$

[humble torrent]

Idk how ur original question was phrased but u can actually show something stronger that if one of these has a limit iff the other has one

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$\abs{f(x)+ f(a+h) -(L_1+L_2)}<\varepsilon_1+ \varepsilon_2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

Uh

?

[twilit field]

Is that wrong?

[humble torrent]

I mean what is ur goal with this? x and h here are approaching different things, so ull have to be precise what you’re looking at

[twilit field]

hmm

[humble torrent]

Here’s otherwise a hint

For an approach

[twilit field]

Hmm, I 'm trying to combine the implications somehow

that won't work very likely though

[humble torrent]

This is why unwinding the definitions a bit more might be helpful, the key here if you’re “rigorous” is how h and x are quantified

[humble torrent]

So you’re allowed to use dummy variables like I did above in the hint

[twilit field]

it's for all h and x

[humble torrent]

I suggest taking a look at my hint if it feels unclear

[twilit field]

I hae

[humble torrent]

Is it still a bit unclear what I meant?

[twilit field]

no, like I'm just trying to figure out how to apply it

[humble torrent]

Well do you see what I arrived at in my hint?

There L is L1

[twilit field]

yeah, added and subracted a

[humble torrent]

So for clarity as that was before u changed the L’s

Let me rewrite it with L_1

If |h| = |(h + a) - a| < delta_1 then |f(h+a) - L_1| < epsilon_1

[twilit field]

mhm

Similarly if $|x-a| < \delta_2$, then $\abs{f(x)-L_2} < \varepsilon_2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

Yes!

So what does this tell u?

This is before “formalizing” it really

But you’ve basically got the meat of the idea down

[humble torrent]

Also it could of been fine if you stopped here if I’m not missing anything

It highly depends on how the original question was asked

[twilit field]

We let $x = h+a$?

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

Before u do any of that, what does this suggest for example?

It’s good if you’re aware of what this accomplishes

[twilit field]

The limit of f(h+a) at h=0 is L_2

[humble torrent]

You already have that by assumption for L_2

What about L_1?

[twilit field]

same thin

[humble torrent]

Yes!

So what does this accomplish?

[twilit field]

It proves the defns are equivalent

*limits

[humble torrent]

Yes! Atleast the way u phrased the question this is indeed what we get

But this is almost a complete proof

You’ll probably want to formalize this a bit more

E.g. let epsilon > 0 etc

But the main idea that we discussed is the meat of the proof

[twilit field]

So in short is this proof fine:
Formally $\lim_{x \to a} f(x) =L_1$, can be stated as follows\
If $|x-a| < \delta_1 \implies |f(a) -L_1| < \varepsilon_1$
\
\
Formally, $\lim_{h\to 0} f(a+h) = L_2$ can be stated as follows
\
If $ |a+h -a| <.\delta _2 \implies |f(a) - L_2| < \varepsilon_2$

[humble torrent]

A bit unclear what is going at the second paragraph

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

Second formulation of the limit is still not right

[twilit field]

hmm

[humble torrent]

When you say that the limit formally is that, are you unwinding the definition? Cuz it doesn’t like it

Oh and I missed that u did something similar in the first part aswell

What’s with f(a) being there?

[twilit field]

yes

[humble torrent]

Ok so what happened? U wrote the definitions of the limit from before correct, but here it’s different

[twilit field]

Got confused

So in short is this proof fine:
Formally $\lim_{x \to a} f(x) =L_1$, can be stated as follows\
If $|x-a| < \delta_1 \implies |f(x) -L_1| < \varepsilon_1$
\
\
Formally, $\lim_{h\to 0} f(a+h) = L_2$ can be stated as follows
\
If $ |a+h -a| <.\delta _2 \implies |f(a+h) - L_2| < \varepsilon_2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

You’re only stating the definitions (albeit still not correctly) here, this is not a complete proof

[twilit field]

I know

[humble torrent]

So why did you ask if that was a fine proof?

[twilit field]

my bad

This is so much harder than LA

[humble torrent]

No worries, take ur time and slowly go over what we went through

[humble torrent]

I can relate, i found both hard and still do at times

But this struggle you’re having might be a good thing in some sense, it means you’re learning

[twilit field]

Can I now let x=a+h?

[humble torrent]

Well I was a bit fast judging your formulation of the definition, but you probably meant |h| < delta_2 in the second formulation right?

[twilit field]

I mean $a+h-a =h$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

But yeah

that's what I meant

[humble torrent]

I mean okay sure, I usually present the proofs in a way that sort of starts from scratch, so this might be confusing is all

[pastel sequoia]

use first principle

[humble torrent]

But yeah it’s fine

Okay so yes, now let in particular x = a + h

[twilit field]

I'm using the epsilon-delta defn

[humble torrent]

And introduce an epsilon > 0

[humble torrent]

Actually don’t do this yet

We want to do this in the right order

[twilit field]

So what do I do instead

[humble torrent]

Well start in the right order, so first let epsilon > 0

In the back of our minds we already know what we’re trying to do when writing this

I.e the idea we went through from before

Now what you can say is that we pick delta = delta_1 and let a+h = x

In this case delta_1 corresponds to the existence of a delta for the first limit

[twilit field]

So in short is this proof fine:
Formally $\lim_{x \to a} f(x) =L_1$, can be stated as follows\
If $|x-a| < \delta_1 \implies |f(x) -L_1| < \varepsilon_1$
\
\
Formally, $\lim_{h\to 0} f(a+h) = L_2$ can be stated as follows
\
If $ |a+h -a| <.\delta _2 \implies |f(a+h) - L_2| < \varepsilon_2$
\
\
Let $\varepsilon>0$
\
Let $\delta =. \min{\delta_1,\delta_2}$
\
Let $x=a+h$
\
We then have
\
$|x-a| < \delta \implies |f(x) - L_2| < \varepsilon$
\
We then have $\abs{f(x)-L_2}+ \abs{L_1 - f(x)} < \varepsilon_1+ \varepsilon_2$
\
\
$\abs{L_1-L_2}< \varepsilon_2+ \varepsilon_2}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

Okay so when I was talking about this I thought we were both talking about showing that f(a+h) goes to L1

But that’s fine too of course, it just wasn’t clear

[humble torrent]

1. Make ur epsilon > 0
2. Why are you choosing delta that way?

[twilit field]

1), will mention that
\
2) Just thought it's better that way

[humble torrent]

Why do you think it’s better that way? You already know that it has a limit of L1

It doesn’t change the validity of the proof, but it’s not needed is all I’m trying to say

[twilit field]

So in short is this proof fine:
Formally $\lim_{x \to a} f(x) =L_1$, can be stated as follows\
If $|x-a| < \delta_1 \implies |f(x) -L_1| < \varepsilon_1$
\
\
Formally, $\lim_{h\to 0} f(a+h) = L_2$ can be stated as follows
\
If $ |a+h -a| <.\delta _2 \implies |f(a+h) - L_2| < \varepsilon_2$
\
\
Let $\varepsilon>0$
\
Let $\delta =. \min{\delta_1,\delta_2}$
\
Let $x=a+h$
\
We then have
\
$|x-a| < \delta \implies |f(x) - L_2| < \varepsilon$
\
We then have $\abs{f(x)-L_2}+ \abs{L_1 - f(x)} < \varepsilon_1+ \varepsilon_2}$
\
\
$\abs{L_1-L_2}< \varepsilon_2+ \varepsilon_2}$

[grand pond]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[humble torrent]

By the uniqueness of the limit you know L_1 = L_2

[twilit field]

Formally, $\lim_{x \to a} f(x) = L_1$ can be stated as follows:
If $|x - a| < \delta_1 \implies |f(x) - L_1| < \varepsilon_1$.

Formally, $\lim_{h \to 0} f(a + h) = L_2$ can be stated as follows:
If $|a + h - a| < \delta_2 \implies |f(a + h) - L_2| < \varepsilon_2$.

Let $\varepsilon > 0$.
Let $\delta = \min{\delta_1, \delta_2}$.
Let $x = a + h$.

We then have:
[
|x - a| < \delta \implies |f(x) - L_2| < \varepsilon.
]

We then have:
[
|f(x) - L_2| + |L_1 - f(x)| < \varepsilon_1 + \varepsilon_2.
]

[
|L_1 - L_2| < \varepsilon_1 + \varepsilon_2.
]
\
choosing $\varepsilon_1 =\varepsilon_2 =0$, we see that $\abs{L_1-L_2}=0$
\
This proves the desired result

[humble torrent]

Unless that is something you haven’t proved

[twilit field]

Yeah, I've proven that

oh, so all this is unnecessary

the last bit

[humble torrent]

So use it, the point of proofs is to make it concise and as readable as possible, so if you already have a result at ur disposal just reference it instead of cluttering the main idea of the proof

[humble torrent]

Yes

Also I would not choose delta like that, it’s cluttering

[twilit field]

I mean I'm very new to analysis , no harm in doing the same stuff multiple times

[humble torrent]

That’s perfectly fine in that case, no harm indeed

But then I would also expect you to be more precise in your wording, if you’re already being overly detailed about reproving previous results

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

Fair, I'll work on ot

[humble torrent]

What happened at the end there, choosing 0?

They’re strictly bigger than zero

[twilit field]

oh right

hmm

[humble torrent]

I still don’t see why it’s better to choose delta the way you did it

[twilit field]

I'll edit it out once I figure out this last bit

[humble torrent]

Didn’t you previously prove this?

[twilit field]

I did, but I don't remember what I did

[humble torrent]

Oh

Well are you familiar with the idea of showing that if |C| < epsilon for all epsilon, then C=0

Where C is some constant

[twilit field]

No

[humble torrent]

Well that might be a good exercise then, it’s precisely what you’ll need to prove

[twilit field]

Well, I do have one idea

This makes C the greatest lower bound

and the least upper bound

[humble torrent]

With respect to what?

[twilit field]

The set of positive reals

[humble torrent]

Yes sure

[twilit field]

The set of positive reals is (0 ,\infty)

so this makes C=0

[humble torrent]

But uh not the least upper bound

[twilit field]

oops

[humble torrent]

Yeah sure, so why is that true?

[twilit field]

yeah, my bad

[twilit field]

Can I take that for granted for now, I'll prove it in a bit

[humble torrent]

Notice how you’re using a result

You can do this for your original proof as well

[twilit field]

yeah, the greatest lower bound of (a,b) is a

Formally, $\lim_{x \to a} f(x) = L_1$ can be stated as follows:
If $|x - a| < \delta_1 \implies |f(x) - L_1| < \varepsilon_1$.

Formally, $\lim_{h \to 0} f(a + h) = L_2$ can be stated as follows:
If $|a + h - a| < \delta_2 \implies |f(a + h) - L_2| < \varepsilon_2$.

Let $\varepsilon > 0$.
Let $\delta =\delta_1$
Let $x = a + h$.

We then have:
[
|x - a| < \delta \implies |f(x) - L_2| < \varepsilon.
]

We then have:
[
|f(x) - L_2| + |L_1 - f(x)| < \varepsilon_1 + \varepsilon_2.
]

[
|L_1 - L_2| < \varepsilon_1 + \varepsilon_2.
]
\
choosing $\varepsilon_1 =\varepsilon_2 =0$, we see that $\abs{L_1-L_2}=0$ , this follows from the fact that $\abs{L_1-L_2}<\varepsilon_1+ \varepsilon_2$ for all $\varepsilon_1, \varepsilon_2>0$, making it a greatest lower bound of the positive reals, which is $0$
\
This proves the desired result

[grand pond]

ƒ(Why am. I here)=I don't Know

[humble torrent]

Alternatively for intuition, |C| is a non-negative number that is simultaneously less than all positive numbers

So C must be 0

[twilit field]

Got it

That's enough analysis for now 🙏

Thanks a lot of the help!

.close

[midnight plank]

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[last slate]

I require assistance with the first part of this question

[last slate]

I understand that

For binomial distributions:

$$E(x) = np$$

[grand pond]

Edmund Cloudsley

[last slate]

rearranging this equation gives

$$n = \frac{E(x)}{p}$$

[grand pond]

Edmund Cloudsley

[last slate]

n is minized when p is maximised and the maximum possible value that p can have is 1 since probability can never be greater than 1

this gives the minum value of n as 4.5/1 = 4.5 which is also the answer in my book

However, since n is technically the number of trials, shouldn't it be an integer value?

therefore we must find such a p where

$$E(x) \text{ mod } p = 0$$

[grand pond]

Edmund Cloudsley

[last slate]

Feel free to ping me, I would be on another tab. Thanks in advance!

<@&286206848099549185>

[midnight plank]

@last slate Has your question been resolved?

[last slate]

<@&286206848099549185>

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[rigid fern]

idk

[lethal path]

you need to revise your circle theorems before attempting these problems

AC is a diameter of the circle

so hence what must angle ABC be automatically?

[rigid fern]

90 degrees

[lethal path]

great

and then if angle ADB = x, what must angle ACB be?

[rigid fern]

x

[lethal path]

nice

alright so arc AB = arc BC means that angle AOB = angle BOC

[rigid fern]

ok

[lethal path]

great so now look at triangle BOC, and also where OB = OC

that's enough info to find x

and yes we didn't actually need the fact that angle ABC = 90 degrees but like it helps just to write it in case

[rigid fern]

why ob = oc

[lethal path]

OB and OC are both radii

[rigid fern]

oh ok

.close

[midnight plank]

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[grand pond]

xbz

[tidal turret]

my idea was to show that the function is monotonically decreasing/increasing to argue that is injective

,, f(x) = x^3 + 5x + 1 \ f'(x) = 3x^ 2 + 5

[grand pond]

xbz

[midnight plank]

@tidal turret Has your question been resolved?

[limber brook]

that's correct

we know that the function is strictly increasing cuz derivative is always >0

[tidal turret]

yeah

[limber brook]

so there is one solution for sure

for positive or negative, use the fundamental theorem, f(0) = 1 and f(-1)=-5 and f(1) is positive too

from that, we can say that the root lies in (-1,0)

so its one negative root

that is a real number

rest would be complex conjugates

so i think one negative solution would be the right answer

[tidal turret]

yeah

I understood, yeah I see

thank u

.solved

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[midnight plank]

@nocturne silo Has your question been resolved?

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@nocturne silo Has your question been resolved?

[nocturne silo]

<@&286206848099549185>

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@nocturne silo Has your question been resolved?

[brittle steeple]

You got an approach so far?

[nocturne silo]

not really im still a little confused

i know its a 3d shape though

[brittle steeple]

Do you have the diagram?

Or a sketch

Or do you want me to help you with the sketch

[nocturne silo]

can you help me with the sketch please thanks

[nocturne silo]

oo thank you

[brazen shuttle]

if your question has been answered you can close the chat with .close have a good day ❤️

[nocturne silo]

thanks

.close

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[midnight plank]

@gaunt gust Has your question been resolved?

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[umbral sparrow]

am i right with D here?

[umbral sparrow]

am i right with D here?

<@&286206848099549185>

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[barren stag]

you are given 3 sides and need to find an angle

which trigonometric law is able to solve this given 3 sides

[nocturne silo]

cosine

idk why but if i use the cosine law i keep getting 0

[midnight plank]

@nocturne silo Has your question been resolved?

[mental tiger]

Yes. this formula is the one you need to use ~

[mental tiger]

I tried solving for the angle and I got 3 degrees

Just need to make sure that the values are correctly inputted in and that the arccosine is found 👍🏼

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[midnight plank]

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[upbeat plinth]

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[upbeat plinth]

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[sharp coral]

area

[upbeat plinth]

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[sharp coral]

you should consider the side lengths to be fixed

[upbeat plinth]

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[sharp coral]

yes

[grand pond]

Kenzo

[upbeat plinth]

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[grand pond]

Kenzo

[upbeat plinth]

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[sharp coral]

i'm not entirely sure why you solved for b in the first place

[upbeat plinth]

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[sharp coral]

i think you are overcomplicating it

[upbeat plinth]

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[sharp coral]

you are given a formula for A in terms of one variable, theta, and two constants, a and b, and you have to minimize A

[upbeat plinth]

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[sharp coral]

if you were given a formula y = f(x), which is a formula which gives y in terms of a variable x, how would you minimize y?

[upbeat plinth]

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[sharp coral]

but you already have it in the form y = f(x), so all of the steps up to here are already done for you

[upbeat plinth]

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[sharp coral]

yes

[upbeat plinth]

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[sharp coral]

a and b are constants

[upbeat plinth]

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[sharp coral]

so you apply the rule for constant multiples

[upbeat plinth]

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[grand pond]

Kenzo

[upbeat plinth]

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[sharp coral]

where did a and b go?

[upbeat plinth]

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[sharp coral]

yes

[upbeat plinth]

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[sharp coral]

the derivative of y = 5x^2 is 5(2x)

there is a difference between added constants and multiplied constants

[upbeat plinth]

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[grand pond]

Kenzo

[sharp coral]

also no

[upbeat plinth]

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[sharp coral]

you know that the 1/2 stays in front of your derivative unchanged... a and b are constants just the same as 1/2

[upbeat plinth]

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[sharp coral]

you are taking the hint in the completely wrong direction

what you did with 1/2 is correct, what you did with a and b is not

[upbeat plinth]

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[grand pond]

Kenzo

[sharp coral]

yes

[upbeat plinth]

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[sharp coral]

it's a constant which is not 0

[upbeat plinth]

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[sharp coral]

the value of theta such that this expression is 0

[upbeat plinth]

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[sharp coral]

we have a rule which says that if you have two things that multiply to 0, then one of them must be 0

[upbeat plinth]

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[sharp coral]

either 1/2 ab = 0 or cos theta = 0

but is it possible for 1/2 ab = 0?

[upbeat plinth]

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[sharp coral]

correct

[upbeat plinth]

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[sharp coral]

no

we have [ \frac 12 ab \cos \theta = 0]

[grand pond]

cloud

[upbeat plinth]

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[sharp coral]

you can divide both sides by any constants if you want

[upbeat plinth]

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[sharp coral]

yes

[upbeat plinth]

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[sharp coral]

yes, now our equation is simpler

[upbeat plinth]

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[sharp coral]

yes

[upbeat plinth]

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[midnight plank]

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[slim eagle]

a question about triangles

[slim eagle]

so centroid divides orthocenter and circumcenter in 2:1 in euler's line right

and it also divides each median in ratio 2:1 but it is generally that Orthocenter and circumcenter are not on the same line

in equilateral triangle they are same point and in isoceles triangle they lie on the same line for the median through vertex opposite to the unequal side

am i wrong?

[midnight plank]

@slim eagle Has your question been resolved?

[midnight plank]

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[twilit field]

Find an efficient proof for all cases at once, by first demonstrating $(a+b)^2 \leq (|a|+|b|)^2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$a \leq |a|; b \leq |b| \implies a+b \leq |a|+|b| \implies (a+b)^2 \leq (\abs{a}+ \abs{b})^2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

Does this work?

I think I also have to show $a \leq |a|$ and $b \leq |b|$?

[grand pond]

ƒ(Why am. I here)=I don't Know

[nova yoke]

the squaring function is only monotonically increasing on [0,infty), so you have to justify that last implication

[twilit field]

Until there it's all good?

[nova yoke]

yea

[twilit field]

Hmm, then this proof isn't ideal, the book asks for a proof for all cases at once

[nova yoke]

that's definitely possible

but you need a slightly different approach

[twilit field]

Let me think a bit

[slim eagle]

i think u can do this using definite integration too

[twilit field]

I'm studying analysis , integration before I've even reached the completeness axiom is

[slim eagle]

oh shit my bad gang

[nova yoke]

yea that sounds highly sus haha

[twilit field]

some inequality may help

hmm

[nova yoke]

try expanding (a+b)^2 and (|a| + |b|)^2

see how they differ

that might give you an idea what you need to do

[twilit field]

alternatively, I know $a^2=|a|^2$(This can be proven in one line if required, then I have to show $2ab<2|a||b|$

[grand pond]

ƒ(Why am. I here)=I don't Know

[nova yoke]

yea

that's exactly what you need to show

[twilit field]

well, $2ab\leq2 |a||b|$

[nova yoke]

which is basically the same as showing that x <= |x| in general

where did the 2 go on the RHS

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

This will require cases, pretty sure

[nova yoke]

nah

do you have to use cases to show that x <= |x|?

|x| is either x or -x, whichever is larger

[twilit field]

yeah

[nova yoke]

so both x and -x are <= |x|

but if you want to be more formal you can certainly do it with cases

[twilit field]

Got it