help-10

[candid yarrow]

let’s answer this question first

[native grail]

ou

okayy

the answer is +/-(4i+7j-6k)/sqrroot(4^2+7^2+(-6)^2)

[candid yarrow]

that should be right

[native grail]

ye

so about the origin question

[candid yarrow]

equation of what?

[native grail]

this is the equation

for example

[native grail]

of PQR

then i get this 4i+7j-6k

[candid yarrow]

the equation of PQR, or the equation of a line perpendicular to PQR?

[native grail]

i dont have the exact question so i want to use this question to subsitude the values

line perpendicular

the first part was about finding the line perpendicular

then they asked about what is the shortest distance from the origin to the plane

[candid yarrow]

Ok

[candid yarrow]

I’ll assume that means you want to find the line perpendicular to PQR through the origin

[native grail]

ye

[candid yarrow]

you can use the normal vector to do this

[native grail]

thats the shortest distance

wat

[candid yarrow]

or you can use the cross product

[candid yarrow]

no, finding the line and finding the shortest distance are slightly different but related

the shortest distance is a number

[native grail]

ou

[candid yarrow]

do you know how to find the equation of a line in 3D?

[native grail]

i think so

[candid yarrow]

how?

[native grail]

uh

the x axis is j the y axis is k

the i is the

idk about i

i is the one that can move the plane

[candid yarrow]

Isn’t the x axis i and the y axis j

[native grail]

no

ou

yea

it is

so the k is the unit vector in positive Z-axis direction

idk what that means but

i kinda get it

i need to form plane equations to find shortest distance

i kinda get it

ye

thank

.close

[obtuse pebble]

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[torn vine]

how do i find the width and length of a number if only the perimeter is given?

[trim portal]

of a number?

[torn vine]

i mean rectangle my bad.

[trim portal]

you can't generally do that, there are infinitely many solutions

[obtuse pebble]

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.

[torn vine]

so would this not be possible?

[gilded needle]

rectangular triangle?

[trim portal]

it's right triangle i think

[torn vine]

yep right triangle pretty sure

i dont get how to solve it at all 😭

[trim portal]

is the length integer?

or does it not have to be

[torn vine]

im pretty sure not

theres multiple choices do you want me to show that?

[trim portal]

yeah...

[trim portal]

okay

the cruicial part is that the length is larger than the width

[gilded needle]

there must be a piece of information missing

[trim portal]

oh wait

its still unsolvable

yeah

[gilded needle]

e.g.:
35, 84, 91
is a valid solution but doesn't match any of your options

[trim portal]

why 3 numbers?

[gilded needle]

those are the three sides

[trim portal]

oh, okay

[gilded needle]

they're a pythagorean triple and they sum to 210

[trim portal]

it doesnt even have to be integers though

so there are infinitely many solutions

and all of the options given have a solution associated with them

[gilded needle]

yea my example shows that even if you require an integer solution, you can find one that doesn't match any of those options

if you don't require integers then all bets are off

[trim portal]

none of the options match an integer solutions anyway

[gilded needle]

wait why not?

why can't i take a 3,4,5 triangle and scale it by 10?

oh wait

that won't sum to 210

nevmind

[trim portal]

$f\left(x\right)=\frac{210\left(x-105\right)}{x-210}$

[warm shale]

MæthIsAlwaysRight

[trim portal]

this function gives length given width

or vice versa

these are the solution that wolfram alpha gives

over positive integers

[gilded needle]

ha i had to do an annoying amount of algebra but i agree with the formula

[obtuse pebble]

@torn vine Has your question been resolved?

[obtuse pebble]

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[fathom flicker]

I don't understand the definition of Lambda here

shouldn't it be Lambda(v) = something in V''

phi(v) isn't even in V'' its in F

[gilded needle]

yes

[fathom flicker]

Bungo you're a quick one

[gilded needle]

Lambda(v) is an element of V''

[wild swallow]

[fathom flicker]

barely even could get my question off

[gilded needle]

i.e. a linear functional on V'

and phi is an element of V'

so the left hand side makes sense?

(and is a scalar)

[fathom flicker]

Yeah both sides are in F

I'm just confused why it looks like they're evaluating (Lambda v) instead of Lambda

[gilded needle]

well you can't apply Lambda to phi

[fathom flicker]

Like, what if I want to know what is Lambda(v) for arbitrary v in V

[wild swallow]

(Λ(v))(φ)

[gilded needle]

Lambda's domain is V

phi is not in V

[fathom flicker]

yes

yes

This is why I am confused

[wild swallow]

Λ(v) is an element of V''

[fathom flicker]

V''

[gilded needle]

Lambda(v) is a linear functional on V' that sends the element phi of V' to phi(v)

[wild swallow]

Λ(v) is then a function V' -> F

[fathom flicker]

Lambda(v) should be a linear functional psi:V' to F

how is psi defined in terms of v our input

[gilded needle]

v is not the input to what you're calling psi

[fathom flicker]

Yes I get that

v is the input to Lambda

but psi our output should depend on v in some way

and I'm not seeing how

[gilded needle]

each v begets a different psi

[fathom flicker]

yes

in what way?

[gilded needle]

the psi associated with v accepts an arbitrary phi in V' and maps it to the scalar phi(v)

you might call it the "evaluate at v" map

[fathom flicker]

why arbitrary phi in V'

how is this well-defined then

[gilded needle]

the input to phi is any element of V'

phi is some arbitrary element of V'

i.e. take any phi in V' and evaluate it at v

that's what the psi associated with v does

[fathom flicker]

oh

psi takes element in V'

and maps it to F

[gilded needle]

yep

[fathom flicker]

and it does it like you described

I see

🔠

So can we say

Lambda(v) = psi
where psi(phi)=phi(v)
for all phi in V'

[gilded needle]

yep

[fathom flicker]

😓

[gilded needle]

implicitly psi depends on v

you could even give it a subscript v if you want to make it clearer

[fathom flicker]

🫡

this is good

my artificial slave help me solve it

Cause I’m too lazy to have my palm on the calculator

[gilded needle]

[fathom flicker]

😂

thanks Sir

.close

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[dim jasper]

help me i dont really understand this or i just suck but this here say that the function only continuous at x=0 but what if x is an irrational number?

[zenith raft]

continuous at 0 means continuous at the input value 0

not every point whose image is 0

[upper shuttle]

if x is irrational f(x) is x which approaches 0 as x approaches 0

[dim jasper]

if x is irrational f(x) is 0 no?

[zenith raft]

read what i said

[upper shuttle]

sorry. i meant rational

[zenith raft]

is that your confusion?

[dim jasper]

im confused in the last line where it say f is continuous at precisely one point that is 0 is 0 the input or output?

[upper shuttle]

input

[river valley]

doğuş hocam burdaysanız selam verirmşsniasz

[dim jasper]

what language is that

[timid silo]

turkish i think

[dim jasper]

im confused lets say a is irrational number so does limit x approach a is defined or no?

[zenith raft]

it does not exist

[gilded needle]

if you approach a via rational values of x then the limit is a
if you approach a via irrational values of x then the limit is 0

so the overall limit cannot exist unless those two limits are equal, i.e. a=0

[zenith raft]

thank you sir

[dim jasper]

ohh

[zenith raft]

snoseph

[gilded needle]

thank you ma'am

[wild swallow]

slayyloseph

[zenith raft]

change name pls

WTF

change back

thank you

i was thinking about changing my name to that yesterday when i said i'm back and you made a joke

but i didn't want to sacrifice any devotion to simping for sharp

simping for you was fun times

[wild swallow]

sotru

[upper shuttle]

wtf

nice name

[dim jasper]

but here on the next example it says that if a is irrational the function is continuous

[zenith raft]

well it is a different function so that sounds plausible

[gilded needle]

the celebrated popcorn function

[zenith raft]

celebrated by who?

[wild swallow]

thomae is a popcorn

[gilded needle]

by goodmorning, among others

[wild swallow]

among us

[dim jasper]

what

[wild swallow]

https://en.wikipedia.org/wiki/Thomae's_function

this is the function

[dim jasper]

looks skibidi af

[upper shuttle]

skibidi skibidi skibidi

[zenith raft]

the only time i've ever heard of something being 'celebrated' in math was for the prime number theorem. i think in burton's elementary number theory book

the celebrated prime number theorem

[wild swallow]

not celebrating enough clearly

[upper shuttle]

I've never heard of it

[zenith raft]

yep i was right

but also :kekehands:

not the only celebrated result

ok i'm done spamming goodmorning's channel sorry

[gilded needle]

very first book i tried:

[zenith raft]

doesn't count

[dim jasper]

gyat

oh i fkn understand now

so i have to know that between 2 rational number there is a irrational number

[obtuse pebble]

@dim jasper Has your question been resolved?

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[brisk arrow]

I need someone to check on some equations

[opaque dome]

Yes Show them

[brisk arrow]

I got -1/2 as the coefficient of y^2 term in z

But wolfram shows a different number

[opaque dome]

6*-3/4 = -18/4 = -9/2

Where that -1/2 comes from ?

[brisk arrow]

That doesn’t not equal 1

[opaque dome]

No it doesnt

[brisk arrow]

but the result from wolfram is 1

it differs from the result from wolfram

why

[opaque dome]

No he say -9/2 too

That y^2 isint impacted by the multiplication

[brisk arrow]

check it

it indicate that the y^2 term has a coeffiecient of 1

[opaque dome]

Yeah that y^2 is chilling since he is not multiplied by anything so he dont change his coefficient

How do you get that -1/2 in front of y^2

He comes from nowhere

[opaque dome]

It is

[opaque dome]

And it doenst have to

[brisk arrow]

*he

why you assume y^2 is a male

im curious

or you do not and simply said that because of convenience it brings.

whatever.

[brisk arrow]

it is confusing.

I cannot understand.

[opaque dome]

Where does the -1/2 comes from in your work ?

[brisk arrow]

it is a calculation mistake which I have just noticed

[opaque dome]

So you see that the coefficient is 1 now ?

[brisk arrow]

no

[brisk arrow]

1 =/= -9/2

[opaque dome]

Y^2 dont change between the first and second line

Wait why do you have a time y on your work and not in wolfram input?

[brisk arrow]

I think the solution of my question is

f

I love wolfram

.close

[obtuse pebble]

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[brisk arrow]

why you assume y^2 as a male

does it relates to chromosome

[obtuse pebble]

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[brisk arrow]

ahhhhhhhhhhhhhhhh

[brisk arrow]

Is she the prettiest creature in the universe

[brisk arrow]

Can someone answer it

I need to boost my mood

I wonder if I should send and explain the whole process

maybe i need a break

.close

[obtuse pebble]

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[glacial lynx]

this is the only given, and the piece wise equations doesnt add up

[azure wraith]

u want to make this continuous

?

orrr

?

[glacial lynx]

i want to solve so i can graph

but theres no given

idk if the next step is hidden in the conditionals but

-2 less than or equal to x but x must be less than -1

that doesnt make sense?

[obtuse pebble]

@glacial lynx Has your question been resolved?

[glacial lynx]

hello???

.close

[obtuse pebble]

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[brave burrow]

$$Dom(f) = (0, 1) \cup (1, +\infty)$$

[warm shale]

938c2cc0dcc05f2b68c4287040cfcf71

[obtuse pebble]

@brave burrow Has your question been resolved?

[fossil crag]

well now you have to find the image

so study the function on its domain

[brave burrow]

How

[fossil crag]

its derivative can give you monotonicity, limits will do the rest

[brave burrow]

Which theorem is that

[fossil crag]

?

if f'(x) > 0 then f is increasing around x

etc

[brave burrow]

Limits for finding the asymptotes

We take the limit as x approaches the endpoints of the domain

???

[restive gorge]

And what happens at x = 1

[brave burrow]

Thats one endpoint of the domain

[opaque dome]

He means in term of asymptote and stuff

[brave burrow]

Well we need the limit as x approaches 0, 1, and +infty so we find the vertical asymptotes

Iirc

Other than dat we can take the limit as x approaches -infty and. +infty for the horizontal asymptotes

IDK how to find oblique asymptotes doe

[opaque dome]

-infinity of lnx ?

[brave burrow]

I forgot domain of lnx is > 0

My bad it was just a general comment about asymptotes

[opaque dome]

Nah you just scared me a little bit

Do you make sign tables of derivative and then the arrows under ?

[brave burrow]

Sure, bolzano theorem u mean

A corollary of ivt

[fossil crag]

the reason I'm making you do stuff like that to find the image

is to make a variation table

[opaque dome]

Yes! This

[fossil crag]

variation table = image given

so if you find the sign of f'

and the limits at the bounds of your domain

(left and right limits)

[fossil crag]

you can make the last row of this table

[opaque dome]

When does it is greater or equal to zero

[brave burrow]

Two cases
2lnx-2 >=0 and <0

No?

[opaque dome]

They are complementary

[brave burrow]

Wdym

[opaque dome]

Solve youll see

[brave burrow]

Ok a moment

[opaque dome]

Like if something is positive on [1,2] only its negative all the others times so on (-inf, 1)U(2, +inf)

[brave burrow]

How do i do the table

[opaque dome]

Like this but on top top u only write the e

[brave burrow]

When is it zero

[opaque dome]

you found it

Its e

[brave burrow]

S0 confusing

[opaque dome]

What is confusing ? All ?

[brave burrow]

I am trying but i don't get it

[opaque dome]

No only one e

And put two lines under the 0 all the way long to the bottom

And add 1 and put it two lines till the bottom too

Those are forbidden values

So we represent them with two bars

[brave burrow]

Broo u draw it

[opaque dome]

Ok ok

[brave burrow]

I don't understand the instructions

[fossil crag]

this double bar

for showing the function doesn't exist at that point

[brave burrow]

I didn't knew that was a thing

Ok

[opaque dome]

,rotate

[opaque dome]

So

Some explanations

I put the interesting values of x in the order on the x-line

Such like when the derivative will be zero, the forbidden values.

In f'x line i put the sign of it

[opaque dome]

So you follow what u solve here

And after i put the variations of the function f with the limits and images at point where
f'(x)= 0

Then i look at the least value and the greatest value which are the answers to the question

With care of the jumping interval, i put an union

[brave burrow]

I understand the table thingy

Has a lot of condensed info but i got it

How is this table thing related to finding the image of the function

[opaque dome]

Be careful e is not a forbidden value, so its just a image calcul and there must not be two bars

And there are two bars at 0 and 1

[brave burrow]

My bad

[opaque dome]

You wrote the interval of the image for values of x

[brave burrow]

Ahh bolzano

[opaque dome]

Look for x in (0,1) the image interval is (-inf, 0)

[opaque dome]

Hum, not quite

But the idea is the same

Or maybe its bolzano, but i would just say what we are doing

[brave burrow]

Well we used bolzano to find e, if it changes sign it must be at a root i think

Let me fix the table

[opaque dome]

No the derivative change sign

Not the function f

And moreover its not a root since f(e) = 2e

[opaque dome]

Actually that goofy function has no root by bolzano

[brave burrow]

Whats the image then

(-infty, 0) u [2e, +infty)

[opaque dome]

Indeed

[brave burrow]

How can

It be -tive

[opaque dome]

Ln(0.5) ?

[brave burrow]

,calc ln(0.5)

[warm shale]

The following error occured while calculating:
Error: Undefined function ln

[opaque dome]

,calc log_e(1/2)

[warm shale]

The following error occured while calculating:
Error: Undefined function log_e

[brave burrow]

,w ln(1/2)

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[opaque dome]

See

[brave burrow]

Fair exponent is -1

[opaque dome]

Just a fast remainder: ln(x) is negative on (0,1]

[brave burrow]

1 included?

[opaque dome]

0 is negative and positive

The only one to be like this

Its neutral number

(Ok i admit its more complicated than that but for us it doesnt change anything since 1 isint in the interval of the function)

So all good ?

[brave burrow]

I was reading this https://math.stackexchange.com/a/26708

[opaque dome]

Yeah i was on wikipedia and they say it depends from where you look 💀

[brave burrow]

I still need to justify which theorems i used, but maybe for this one I'll skip those points tbh this exercise was a hefty one

[opaque dome]

Ok

Hum

[brave burrow]

Anyways, tysm for the help

😂

[opaque dome]

Honestly

[brave burrow]

I just hope i don't get this one on the exam

[opaque dome]

I clearly dont understand why they want you to cite theorems name, many just have no name

Like this is derivative purpose to find this

So

[brave burrow]

Imma be honest this is not calculus class is like intro to real analysis

But I'm not even in university yet so idk why they make us namedrop theorems

[opaque dome]

Great musical taste btw

[opaque dome]

Even when they dont have one kekw

[brave burrow]

Yeah, how was the name of the table thing btw

[opaque dome]

Ive just ended hs and the only i know are pythagore, bolzano, thales, bachet- bezout and gauss

[opaque dome]

Table sign

Xd

[brave burrow]

Well, i know very little euclidian geo, so idk about that

[opaque dome]

Oh and btw, naming a theorem is not the justifying part, the real part is the given statement and assumptions required

[brave burrow]

U know a lot to be in HS

[opaque dome]

Depends hs

Ive learnt topo basis and triple integral there

[brave burrow]

You are like strongest hs student and I'm weakest uni warrior

Lmaoo

[opaque dome]

Nah bro dont say this please

Youre not weak, clearly not

You have good thinking, and most importantly you dont seem to give up a problem

[opaque dome]

Is it late hs ?

[opaque dome]

And i can assure since i have talked with exams inspector and done some oral exam that the most important part is to not give up

[brave burrow]

Everyone can enter university here, its open to everyone and completely free, but it lasts 6 years in total and first year is a year in which you have to pass this kinda filters to see if u make it to uni(this is in university but not really) (argentina)

I am doing first year, not hs but also not uni per se

[opaque dome]

Ok so you have to pass it, im sure youll get it

[brave burrow]

Thank u for the kind words, i will try to live up to the expectations

Have a good one sir

[opaque dome]

You too hf with maths

[brave burrow]

.solved

[obtuse pebble]

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[obtuse pebble]

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[safe ravine]

Chain rule guy

Z

[opaque dome]

True

[safe ravine]

f(ax+b)=f'(ax+b).(ax+b)'

[opaque dome]

(U^2)' = u'u

[arctic swan]

can't he just develop and divide by -2 ?

[opaque dome]

Put it in factor in front

[safe ravine]

Fr

[opaque dome]

Dont touch him in derivation process

[safe ravine]

The 🐐 s are coming

[opaque dome]

-(U^2)'/2 = -(u'u)/2

[safe ravine]

Yo

[opaque dome]

Oh wait im numb

Its (u^2)' = 2u'u

Mb

Then the 2 cancel each others

Ggwp

[obtuse pebble]

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[timid silo]

on this video he says i should know 2^10 = 1024 and 2^5 = 32 without a calculator

how do I do this mental math

[kind hawk]

well its a good idea to know the first few powers of 2

and like 3 and 5 or something

[wooden cipher]

have you played the game 2048? its a fun-ish way to remember powers of 2

[timid silo]

really i need to download it rn

[kind hawk]

oh god it came out 2014

[timid silo]

lol

[kind hawk]

I feel old

[timid silo]

ok should i just play that game for a month then ill be good

?

[kind hawk]

you wont need that long

[high lily]

just calculate the powers of 2 every few days

[kind hawk]

couple rounds should be more than enough

[high lily]

and pay attention to tech specs, usb stick/sd card capacity

[kind hawk]

I mean its only like 10 numbers

[timid silo]

ok wow

i will do these tricks thanks guys

.close

[obtuse pebble]

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[obtuse pebble]

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[brisk arrow]

I do not know how to do chain rule in a reverse order

Fk

Hello

[viral blade]

Then do it in the normal order by deriving the antiderivative

[brisk arrow]

Ohhh

Big brain

[viral blade]

Chain rule in reverse is called u-substitution and it's treated as a separate technique that you'll learn about at some point

[brisk arrow]

I did learned that tho

[viral blade]

Oh well then you can do that too

Also b/c of your name: did learn instead of did learned

[brisk arrow]

.close

[obtuse pebble]

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[obtuse pebble]

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[honest rune]

Help with this question

[honest rune]

Here's the work

And this is the question

[restive jungle]

Be careful, it’s <CAB = 37º, not <CBA

[honest rune]

OH ur right

That's my mistake right der

So it should looks like this right?

[restive jungle]

Yeah

[honest rune]

So give me a minute let me re do it and show it to u

[restive jungle]

Sure

[honest rune]

Wait just to make sure

I have to get the values of angle first right?

[restive jungle]

You don’t need value of other angles

Just use law of cosines

[restive jungle]

Aka this formula

[honest rune]

After calculation, I got c^2 = 1 cos C

[restive jungle]

No

Law of cosines suggests that

Let me just screenshot from wikipedia gimme a sec

[honest rune]

ok

[restive jungle]

Remember the cos(angle) has to be the angle being sandwiched between the two sides

[honest rune]

8^2 = 7^2 - c^2 -2(7*c) Cos 37?

[restive jungle]

Ahhh check the signs

Very very close

[honest rune]

Re draw the triangle

[restive jungle]

Yes yes yes

[honest rune]

Then I get 64 = b^2 +49 -2b -14 Cos 37

[restive jungle]

Did you expand $2(b*7)cos(37^{\circ})$ right

[warm shale]

candies

[honest rune]

Wait so does it should be 2b +14 cos 37

idk if i gotta multiply cos too

[restive jungle]

Is use of calculator allowed?

[honest rune]

yeah

I do not know

[restive jungle]

Do you know the quadratic formula?

[honest rune]

y= a^2x + bx + c?

I mean y = ax^2 +bx +c

[restive jungle]

Yeah do you know how to find the roots of this equation by the qudrtic formula

[honest rune]

roots are x intercepts right?

So isn't it by facotring the equation?

[restive jungle]

Uhhh

Dont think so..

Do you know the quadratic formula though

[honest rune]

I learned it during this year summer school

[honest rune]

I wouldn't say 100% but like 80 or 70%

[restive jungle]

This guy

[honest rune]

Oh yes

Yeah ik that

[restive jungle]

So when ax^2+bx+c=0, you can find x by this equation right?

[honest rune]

Yeah

[restive jungle]

Try to convert this equation into the form of ax^2+bx+c=0

[honest rune]

How do I deal with the cos 37

That seems only problem for me to convert it

Is it b^2 - 14b - 12 = 0?

Not confident at all

[restive jungle]

Treat cos37 like a normal number

like just like 7 and 2.5 or pi

it's just a constant number

[honest rune]

Ok give me a sec

I left with b^2 -14b -15 cos 37 = 0

[restive jungle]

check if you expanded this right btw

[honest rune]

Idk what it mean by expanded

[restive jungle]

like

[restive jungle]

this equation is a bit wrong

[honest rune]

is the answer 12.3

I approached it with law of sine as well and got this

[restive jungle]

closer to 12.4

[honest rune]

after rounding up right

it was 12.36...

[restive jungle]

36...?

i think its 12.39 sth

[honest rune]

Hold on

12.36931687...

[restive jungle]

Weird

[honest rune]

Could you show me ur step by step equation

to see how did u solve

[restive jungle]

okay gimme a sec

[molten stirrup]

ok here -)

u know the law of cosine right?

[honest rune]

Yeah?

c^2 = b^2+a^2 - 2(ba) cos C

[molten stirrup]

the cos(theta) in the formula, is the angle between the 2 sides known

here it gives u angle BAC

u cant do this bcuz u cant find cos C

[molten stirrup]

.

[restive jungle]

from this, we get:
b^2-(2*7*cos37)b-15=0

[honest rune]

Oh ok I didn't use the roots equation thats' why

[restive jungle]

i see

[honest rune]

So give me like 5 min then I'll re do it and show it to u guys

.close

[obtuse pebble]

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[shell gale]

I am struggling solving this question

[junior granite]

Have you ever heard of Heron's Formula?

[shell gale]

i haven't

[junior granite]

So does the question setter wants you to calculate area by basic geometry?

[shell gale]

yes

[junior granite]

okay
what have you tried so far?

[shell gale]

so i have drawn a triangle bisect into 2 right trangles, showing the hypothenuse as 30 and the adjacent as 12

[junior granite]

perfect

[shell gale]

wait is it just pythagorean theorem

[junior granite]

yes

[shell gale]

oh

😅 oops

thx for the help 🙂

[flint dome]

if i’m trying to find the iroc for a composite function like f(g(x)) and im using the formula f(x+0.0001) -f(x)/ 0.0001, would i add 0.0001 into only g(x + 0.0001) or would it be f(g(x+0.0001)+0.0001)?

[obtuse pebble]

@shell gale Has your question been resolved?

[junior granite]

It will be $\frac{f(g(x+0.0001))-f(g(x))}{0.0001}$

[warm shale]

77²

[junior granite]

!done

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[flint dome]

thank you!

[obtuse pebble]

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[lament ledge]

can someone help me solve this

[lament ledge]

i got 1.380 as my original answer but its wrong

also having a issue with this problem

[obtuse pebble]

@lament ledge Has your question been resolved?

[swift hull]

youll use the standard deviation formula which is
\sigma={\sqrt {\frac {\sum(x_{i}-{\mu})^{2}}{N}}}

[lament ledge]

i dont think your command worked

[swift hull]

weird

youll use this formula

[lament ledge]

how do i translate that into text form

[swift hull]

why do you want to translate it into text form?

[swift hull]

in your case the population mean is 154.2, the size of the population is 10

[lament ledge]

because this assignment is on a website

[swift hull]

the formula isnt the answer

its just the formula you use

you need to plug in the values

[lament ledge]

i got 1.812

is that correct

[swift hull]

sadly no that is incorrect

can you show me what steps/calculations you did?

[lament ledge]

First, I found the midpoints for each time interval: 0.5, 1.5, 2.5, 3.5, and 4.5.

Then, I calculated the mean time spent in line using these midpoints and their frequencies.

Then, I determined the squared differences from the mean for each midpoint, weighted by their frequencies, summed them up, and divided by the total number of customers to find the variance.

Then, I took the square root of the variance to get the standard deviation, which according to my calculator is approximately 1.812.

its possible that im just not using the right calculator

[swift hull]

wait i thought you were talking about the acrobat problem

let me check for the time problem

[lament ledge]

talking about this one

[swift hull]

yeah i know now, i just thought you were talking about the acrobat one before you clarified

[swift hull]

well all of your steps seem to be correct, but the answer is supposed to be 1.380 i think

lets go through this step-by-step, what mean did you get for the time spent in line

[lament ledge]

i get a x for 1.380

[swift hull]

thats weird

[swift hull]

.

[lament ledge]

mean = total/number of customers = 605.5/279 = 2.170

the / means divided

[swift hull]

ok thats correct

[lament ledge]

whats next step

[swift hull]

what did you get for the sum of the weighted squared differences?

[lament ledge]

Total = 214.783 + 25.5663 + 7.0835 + 72.9279 + 211.7471 = 531.1078
So, the sum of the weighted squared differences is approximately 531.108 i believe

may need to recheck

[swift hull]

ok thats also correct

its 531.534 to be exact

but the calculations are correct regardless

so the next step is you divide by the total number of observations

[lament ledge]

this is what i have

variance = sum of weighted squared differences/total number of observations = 531.534/279 = 1.906

[swift hull]

ok then you take the square root

and get approximately 1.380

i wonder why the test is marking it wrong, everything seems to line up

[lament ledge]

standard deviation = checkmark 1.906 = 1.380

[lament ledge]

i am unsure too

double check?

[swift hull]

try 1.381

[lament ledge]

unfortunately not 1.381

[swift hull]

somethings wrong

ask a teacher perhaps? maybe the question was talking about something else

[lament ledge]

i have tutoring tomorrow so i can check there

but im just so confused with this math stuff nowadays

[lament ledge]

ok, you wanna atleast try this one

not try but help me step by step

[swift hull]

for this one the standard deviation is approximately 30.5lbs

[swift hull]

sorry i didnt read that message before sending

i just gave you the answer since clearly you already knew how to calculate standard deviation/had no problems with it

[lament ledge]

this website my teacher gave us to do homework is a bit iffy sometimes

i can confirm

im suprised 30.5 was correct was expecting it to be wrong not saying your math is wrong

this is the big bad of my assignment, ive tried this 2 times with a tutor and i still got it wrong

[swift hull]

wait afk

[lament ledge]

no biggy

[swift hull]

back sorry for the hold up

[swift hull]

what do you find confusing/not understand in the assigment so i can help you with it?

[lament ledge]

everything really,

i did all the correct calculations and still got it wrong

[swift hull]

ok well lets go through it question by question

first the mean

you add up all of the values and divide by 10, which gets you 67.5

so i think you just made a small error in calculation on the first one

[lament ledge]

i think so too

[swift hull]

the standard deviation rounded to one decimal place is 6.9 not 7.0

i think you did the calculations right but the incorrect mean made the whole answer incorrect

since you calculated using the wrong mean

for the next question, the intreval of the mumbers one standard deviation from the mean is 60.1 to 74.4

since you do mean - s.d for the first one and mean + s.d for the second one

and then within two standard deviations you get

53.2 to 81.3

the mistake with the upper bound is just caused by the standard deviation calculation being incorrect, and the mistake with the lower bound seems to be a calculation error

anyways now the percentage of data within one standard deviation of the mean is 8/10 so 80%

and the percentage of data within two standard deviations of the mean is 9/10 so 90%

does that all make sense?

[lament ledge]

yes

[swift hull]

okay good

[lament ledge]

wait i got jumbled a bit, is this for C

[swift hull]

oh i deeply apologize, i made a mistake for the lower bound

its 53.7

not 53.2

the upper bound 81.3 is supposed to be correct though

[lament ledge]

im gonna double check 81.3 rq

[swift hull]

ok

[lament ledge]

still says both are wrong

[swift hull]

weird

i think somethings wrong with the grading, i mean i literally just substracted a standard deviation from the lower bound for b

and added a standard deviation to the upper bound for b

theres no room for errors in calculation since its ltierally just + and -

and theyve already said that the standard deviation calculation is correct

[lament ledge]

maybe lets double check again

try something other than subtracting maybe

[swift hull]

thats the only way to calculate it i think

[lament ledge]

i think we need a third eye on this

[swift hull]

maybe yeah

but i really feel like theres something wrong with the grading scheme itsself

because i cant find any room for a mistake in the first place

[lament ledge]

ya

im rechecking everything

i cant find the mistake either

[obtuse pebble]

@lament ledge Has your question been resolved?

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[reef pike]

hello

[reef pike]

can someone help me spot my mistakes for questions 347 and 351?

i got both the answers wrong

[full maple]

you should take 1/2 out of the integral for 347)

[reef pike]

ohh okay i’ll try that

and i’ll rewrite my answers so they’re legible

this is what i got

on my second attempt i still didn’t arrive at the correct answer

[full maple]

you have to change the bounds of the integral

[reef pike]

sorry what does that mean

[full maple]

your current bounds are from 2 to 4

[reef pike]

right

[full maple]

but since you changed the function to terms of u

you need to change the bounds as well

you know that u = x - 3

so you can substitute x = 2, 4, and find your new bounds for u

[reef pike]

when i find new bounds they replace the subtraction part?

[full maple]

yes, you would substitute your new bounds into the definite integral

so the subtraction part would change

[reef pike]

oh right i think i learned about that

ok i’ll fix thank you

elp

the answer should be divergent. i have no clue how to get there

[full maple]

oh okay, so we should first consider where the function does not exist

[reef pike]

so set i set the denominator to zero?

[full maple]

yes

[reef pike]

am i supposed to foil it out

[full maple]

well, you don't need to

2(x-3)^2 = 0

it's already factorised for you

[full maple]

this is just (x-3)(x-3) = 0

[reef pike]

oh got it got it

i can do (x-3)^2 = 2 right?

[full maple]

no

[reef pike]

uh sorry what do i do

[full maple]

(x-3)(x-3) = 0

so x must be 3

[reef pike]

oh

uh that was easier than i thought

[full maple]

yeah, that's all good

so the function does not exist at x = 3

[reef pike]

makes sense

[full maple]

so we need to evaluate the integral from x = 2 to x = t from the left

as t approaches 3

[reef pike]

ohh because the original bounds include 3

wait so if that happens then would the answer always be divergent

[full maple]

yeah

[full maple]

did you find the integral from 2 to t?

essentially, we’re going to do 2 definite integrals

one from 2 to t from the left as t approaches 3

and one from t to 4 from the right as t approaches 3

if both of the definite integrals give a number, its convergent

[reef pike]

is this how i would write it?

[full maple]

yes, but not infinity

its 3

[reef pike]

oh right

wait was i supposed to stop at some point?

oh nvm

where am i making a mistake?

is it the algebra?

wait how am i going to get infinity like this

[full maple]

also you can’t use u in the integral

you have to change u back to x - 3

and when you try to sub in 3

it’ll be -1/0 which doesn’t exist

meaning the integral is divergent

[reef pike]

ohhhh

thank you so much

i’ll fix that

[obtuse pebble]

@reef pike Has your question been resolved?

Channel closed

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[reef pike]

thanks for your help

[obtuse pebble]

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[karmic hedge]

I know we ignore x(t) cause that doesnt affect the height

I think that we find the extreme values?

Also I think I remember something about we can ignore the square root inside the cos() no? But I dont remember the reason why, it has been a long time since I did these

[obtuse pebble]

@karmic hedge Has your question been resolved?

[weary reef]

To maximise y(t), cos(...) has to be big as possible (you can see that by inspection)

Upper bound of cos(x) is 1

that's when x = 0, 2pi, 4pi,...

So if what's inside the expression inside the cos(...) is equal to 0 anywhere in the range of t, then cos(...) can equal 1

If not, you could just differentiate

[karmic hedge]

Wym by upper bound sry

[cinder kestrel]

maximum value that cos(x) can be

[karmic hedge]

Why is the maximum value 1?

I mean like why is the maximum value of cos(sqrt(t^2 - 9t + 20)) = 1

[cinder kestrel]

well not maximum

but it definetely can't be bigger than that

(but yes also maximum in this case)

and the answer is because sqrt(t²-9t+20) is a number

real number*

actually

is it?

yeah there is a bit where it becomes complex actually

idk then

[karmic hedge]

<@&286206848099549185>

[paper cape]

yes?

[paper cape]

u need help on this?

[karmic hedge]

Yes please 😇

[paper cape]

what u need help in?

like what u need to do?

[karmic hedge]

I know we ignore x(t) cause that doesnt affect the height
I think that we find the extreme values?
Also I think I remember something about we can ignore the square root inside the cos() no? But I dont remember the reason why, it has been a long time since I did these

[paper cape]

yes

we do find extreme values

do u want the answer?

[karmic hedge]

First can you explain about this?

[paper cape]

sorry tho i cant help in explaning, mainly because im not good in geometry

i do pre-calc, calc, trig,

sorry

[karmic hedge]

I mean the bit about why we ignore the square root inside the cos

Oh

[paper cape]

ok wait

lemme check

i can help

oh nvm

<@&286206848099549185>

please report

[junior granite]

?

[karmic hedge]

I dont know how to solve the question

[junior granite]

so you want to maximize y(t) right?

do you know about the range of cosine function?

[karmic hedge]

Not really

[junior granite]

see $cos(x)\in{[-1,1]},\forall{x\in{\mathbb{R}}}$

[warm shale]

77²

[junior granite]

SO, the maximum value the cosine function have is 1

[karmic hedge]

I dont really know what much of this notation means

Whats the upsidown Ax and E

[junior granite]

it mean that for all angles cosine funtcion returns values between -1 and 1

including both -1 and 1 too

[karmic hedge]

Just when its cos(x)? Or no matter what is inside the cos()?

[junior granite]

yeah for any value

[karmic hedge]

Ah cool

[junior granite]

okay so now since in the question argument of cosine is dependent on t parameter

we have to check if cosine attains value of 1 or not

since cos(0) = 1

[karmic hedge]

You mean we need to check if cos ever =1?

[junior granite]

yes

[karmic hedge]

Because that would be the local max, right?

[junior granite]

yup

[karmic hedge]

OK I'm with u so far

[junior granite]

not even local but universal

[karmic hedge]

Ah ye

[junior granite]

so now we will put argument of cosine equal to 0 and find value of t

if the any value of t we find falls between the given range of t

we can say that cosine will attain its maximum value at that particular value of t

and so we will put 1 at the place of cosine(blahblahblah)

so did you got the plan?

[karmic hedge]

By the argument of cosine do you mean 6cos(sqrt(t^2 -9t+20))-2? Or just the bit inside the cosine?

[junior granite]

yeah

just inside cosine

[karmic hedge]

Ah ok

But I can ignore the square root right?

How come?

[junior granite]

bruh
you'll set the whole thing in cosine equal to 0

[karmic hedge]

I'm still a bit confused sry, I thought we were checking if cosine ever =1?

[junior granite]

yes
that's what we are checking

so for cos(x) = 1, x has to be equal to 0

that's why we are doing all that stuff

[karmic hedge]

Ohhhhh

Okokok

I get you now

But wait

What about the 6 and -2?

6cos(sqrt(t^2 -9t+20))-2

[junior granite]

we can do nothing about it,

they will remain unchanged

when we verify that cos(something) = 1
we will substitute 1 at the place of cos(something)

[karmic hedge]

Hmmm I still don't understand 100%, the 6 and -2 will both affect the height of the graph, why are we ignoring them

[junior granite]

we are not ignoring them

[junior granite]

read this thing

[karmic hedge]

Wait, I think I get it

OK yeah I think so

But were doing all of this assuming we get a value of 0 right? (So that cos will =1) what if we don't get a value of 0?

[junior granite]

That depends on the question

If you try you'll still get something important

[karmic hedge]

Surely if there's no value where t^2 -9t +20 = 0 we won't get anything then right?

[junior granite]

yeah

[karmic hedge]

Either there's a value of 0 and we get it or there isn't and we dont

[karmic hedge]

What other important thing could we get

[junior granite]

I mean that depends on the question

not every question is same

[junior granite]

I think you should try to solve and understand about this one first

[karmic hedge]

OK I'll solve it

t^2 -9t+20 = 0

(t - 4)(t - 5) = 0

t = 4 and 5

[karmic hedge]

So we have 2 global maxes for t

[junior granite]

no, it is better to say that we have 1 global maxima for 2 values of t

[karmic hedge]

Ah ok

[junior granite]

did you verified that these values lie in the given interval for t?

[karmic hedge]

Yep they both do

Oh wait, do I plug them in to the equation

Or just check if 4 and 5 are within 8 and -9

[karmic hedge]

They both give 4

[junior granite]

yeah

[karmic hedge]

So what do I do now? Which is the answer?

[junior granite]

obviously that will be 4

[karmic hedge]

Why?

They both have the same height no?

[junior granite]

you just calculated it

[junior granite]

both?

[karmic hedge]

I thought thats what we just worked out?

cos() = 1 at both t=4 and t=5?

[junior granite]

uh I mean yeah
so the maxima is 4

[karmic hedge]

I dont understand

[junior granite]

yeah

The question only wants the maximum height

not the values of t

[karmic hedge]

I thought it meant the graph was like this

Where the first max is t=4 and the second is t=5

[junior granite]

so what?

[junior granite]

we don't need to care about it
it is same at both

the question wants the maximum of y(t)

not t

[karmic hedge]

Im sooooooooooo lost

I thought there was 2 maximums of t