help-38

[elder field]

Does the book you're taking it from expect you to do it without?

[wraith hinge]

it just tells me good luck

not literally

but kind of

no

what if i use

sqrt(x) + sqrt(y) + sqrt(z) <= sqrt(3(x+y+z))

[worldly wing]

...

alright then..?

[wraith hinge]

tried squaring?

no

might try

try it but ig it will goo too far but there's nothing bad to it

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[heavy belfry]

They guy on the image did just sum up the areas and did not take into account that some areas are below the X-axis. Is this correct??

[steel bobcat]

it is used to find the point of intersections

now you have to calculate the area using integration

[heavy belfry]

yes i know sorry i didnt sent entire image but he integrated it as everything would be above the x axis

[worldly wing]

When you subtract the top graph from the bottom graph it all comes above the axis

[heavy belfry]

how?

[worldly wing]

well

(something larger) - (something smaller) > 0

[steel bobcat]

look what he did is just what you are saying

[heavy belfry]

yea thats corect but that doesnt mean you get the right area

[worldly wing]

We do

[steel bobcat]

he just did integrated (f(x)-g(x))

[worldly wing]

You don't care about if the area is under or over the x axis

You can shift both graphs equally as much as you want up or down and the area bounded by them remains the same

[heavy belfry]

but in integration the area that is under the x axis will subtract from the the area above?????

[steel bobcat]

yup

[green needle]

Help me

[worldly wing]

That's only when we're talking about area under a single curve

[worldly wing]

#❓how-to-get-help

[steel bobcat]

thats exactly what we are saying

[steel bobcat]

not here buddy

[worldly wing]

^

We're talking about the area bounded by both the curves here, not the area under the curve

So we only care about the difference

[heavy belfry]

but why would that be diferent if you have more curves it is the same as you would first calculate the first area under the curve and then second but in this case the second area wont be corect because part of the graph is under the axis

[green needle]

@steel bobcat Where?

[worldly wing]

Which graph is under the axis?

[steel bobcat]

ask in some other help channel

[steel bobcat]

#❓how-to-get-help use this

[green needle]

Ok

[heavy belfry]

the x^2+2x has some area under the x axis

[worldly wing]

You only care about the positive difference between the two curves, not whether the actual curves go above or under the x axis

You can shift the graphs above the x axis appropriately so that all the area is above the x axis

It doesn't make a difference

Because you can shift both graphs by the same amount either up or down without changing the area bounded by them

[heavy belfry]

okay but then why you can say so for the area under 1 curve

[worldly wing]

You can't shift a single curve without changing the area bounded by it

[heavy belfry]

lol sorry that was really stupid i get it now

Thanks for the help

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[worldly wing]

In general you only care about the positive difference between the two curves

You can think of it as making tiny rectangles inside the area in question, where the height is the difference between the values of the curves

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[heavy belfry]

but dont the curves determine where the area will be and by integrating it will automatically subtract the areas below the curve from above so even if you can shift it without chaning area this will still be a problem?

[heavy belfry]

.reopen

[worldly wing]

What?

I didn't understand your question

You do need to check which curve is greater than the other one in the region(s) in question, if that's what you're asking

[heavy belfry]

when you will integrate it will automaticly subtrack the area under the axis from above so it doesnt matter if you can shift the graph to be positive if you dont take that anywhere into account

@worldly wing

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[jolly thorn]

could I simplify the "cube root 81x^4" using prime factorization

[marble wharf]

yes you can

[jolly thorn]

Ig I kinda messed up, didn't I

I mean, what now

[marble wharf]

well now 81=3^4

it might help to write 3^4x^4 as (3^3x^3)*(3x)

[jolly thorn]

oh, k

thx

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[verbal spindle]

The answer is 19^~1959*

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[winter egret]

Domain of 1/√(x^2-36)

[winter egret]

How do we show it?

I know it is not exist at -6,+6

But technically how we show

[velvet coral]

its a function right?

if yes, @winter egret

then

[winter egret]

Yeah

[velvet coral]

$\sqrt{ (x^2-36)}$ should be a positive integer

you agree?

[solid kiln]

Greydawn Dewer

[velvet coral]

then x^2 - 36 is positive

thus you can write it as

$$x^2 - 36 > 0$$

[solid kiln]

Greydawn Dewer

[velvet coral]

take out the domain from this

and write it in interval notaion if you want like

this is better

[solid kiln]

Greydawn Dewer

[velvet coral]

where x is the limit value

[winter egret]

Yes

[winter egret]

equal 0 no?

[velvet coral]

um, you would need to be knowing how to solve that inequality

[velvet coral]

we need this inequality to be true for god's sake, and the values which dont allow that come under the domain

the values from 0 to 6 dont allow that

and 0 to -6 for the other side what so ever

and every value to till infinity does allow us to prove that inequality

[velvet coral]

this is the interval notation so yeah

you mainly asked for representation, I gave you the whole context

[winter egret]

Tq very much

I understand now

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[undone dagger]

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[undone dagger]

A group of kids want to buy a gift for their mothers. If every kid gives 10 dollars from his savings, they will need 12 more dollars to buy the gift. If every kid gives 15 dollars, they will go over the price of the gift by 18 dollars. Whats the price of the gift and whats the total number of the kids?

[main sigil]

.reopen

use .reopen

[undone dagger]

.reopen

[main sigil]

Huh

bot?

[undone dagger]

.reopen

bot broke

[main sigil]

Interesting

Claim a new channel for now

I'll inform mods about this

[undone dagger]

it says im in

math help occupied

im good tho

[main sigil]

this channel will close soon

[undone dagger]

o

[main sigil]

It's better to claim a new one

.reopen

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[finite tusk]

in this question (b), can i do 1- the integral from 1 to 5 for 2x^-3?

[magic bloom]

yeah

or from 5 to inf

[finite tusk]

how can i solve the integral from 5 to inf

[magic bloom]

using limits

let inf be a variable t and take the limit of t to infinity after computing the definite integral

[finite tusk]

ohhh yes i remember it now

alright i appreciate it

[magic bloom]

npp

[finite tusk]

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[finite tusk]

.reopen

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✅

[finite tusk]

how can i solve question (e) in this pic?

[magic bloom]

∫f(x)dx from 1 to x = 0.95

[finite tusk]

the answer is gon be in negative should i take the absolute value?

[magic bloom]

why is it negative?

[finite tusk]

cuz the intergral of f(x) is -1/x^2

[magic bloom]

u should get two solutions

reject the negative one

this is the equation u r solving

[finite tusk]

(-1/x^2)+1=0.95
-1/x^2=0.05
-1=0.05x^2

OHHH

OKAY

SORRY

[magic bloom]

-1+0.95 = -0.05

[finite tusk]

lol i made a mistake

yes i realized

im sorry

[magic bloom]

issoke

[finite tusk]

ok last question as you can say in part (d) its X<4 or X>8

should the intergral be from 1 to 4 + 4 to 8?

or should it be 1 to 4 + 1 to 8?

[magic bloom]

its not saying this

its saying x is between 4 and 8

so u integrate from 4 to 8 simply

x<8 means x is anything below 8
4<x means x is anything above 4

[finite tusk]

its C that says X is between 4 and 8

[magic bloom]

oh wait

yeah

i misread

for D, u integrate from 1 to 4
then u 1-(integral from 1 to 8)
then u add

[finite tusk]

okay yeah i understand now its the same as say from 8 to inf right?

[magic bloom]

yeah

[finite tusk]

alright thanks alot!

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[shrewd obsidian]

i need to calculate the following series $$\sum_{k=0}^\infty (k+5)0.6^k 0.4 =0.4 (\sum_{k=0}^\infty k 0.6^k + 5\sum_{k=0}^\infty 0.6^k) $$

[shrewd obsidian]

so the far right sum is geometric, that's easy

but what about $\sum_{k=0}^\infty k 0.6^k$

[sleek canopy]

mhhh seems like a known one

[shrewd obsidian]

it reminds me of power series

or generating functions

where $a_n = 1, 2, 3, ...$

[solid kiln]

atif

[shrewd obsidian]

oops i messed up the indices

[solid kiln]

atif

atif

[sleek canopy]

https://en.wikipedia.org/wiki/List_of_mathematical_series

[low gazelle]

k0.6^k looks like what you would get if u did the power rule

Try differentiating $\frac{1}{1-x}=\sum_{k=0}^\infty x^k$

[solid kiln]

Nekoyashiki Yuki

[shrewd obsidian]

ok thanksi think i used to see this trick when doing generating function problems

[shrewd obsidian]

nice ty

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[keen void]

$2019^{2019^3} \mod 11$

[solid kiln]

Merineth

[keen void]

I got a quick question about congruency

Fermats little theorem say that

$2019^{10} = 1 \mod 11$

[solid kiln]

Merineth

[keen void]

$2019^{10} = 1 \mod 11 \$
$[2019^{10}]{11} = [1]{11}\$
$2019^{10} \equiv 1 \mod 11$

[solid kiln]

Merineth

[keen void]

There we go..

Are all of these saying the same thing? And are they accruate?

@whole coral my love

[nova spire]

yeah they're the same thing

1st line is least preferred compared to the other two

[keen void]

And they are all just saying:

When 2019^10 is divided by 11 we get a remainder of 1?

[nova spire]

yep

[keen void]

Okii so since fermats says that

$2019^{10} = 1 \mod 11$

[solid kiln]

Merineth

[keen void]

I should be able to simplify, right?

$2019^{201920192019}$

[solid kiln]

Merineth

[keen void]

$2019^{999}$

[solid kiln]

Merineth

[keen void]

Like so maybe?

9*9 = 81

so we get 2019^9

But then what do i do D:

I can't simplify further with fermats

because the exponent is smaller than the p-1 in fermats

[keen void]

[nova spire]

you said ...^10 = 1

[keen void]

Right, yea

[nova spire]

so maybe it's good to look at 2019^3 mod 10

and find the remainder

[keen void]

hmm

I'm not sure i follow

Why are we going back to 2019^3?

When i had 2019^9 simplified down

[nova spire]

2019^3 is the power

and it wasn't simplified well

[keen void]

Okay let me retry

$2019^{2019^3} = 2019^{201920192019}$

[solid kiln]

Merineth

[keen void]

This is what i'm starting with, right?

[nova spire]

yep

[keen void]

With fermats little theorem

$2019^{10} \equiv 1 \mod 11$

[solid kiln]

Merineth

[keen void]

This is what fermats says

Meaning i can simplify as many mutliples of 10 in the exponent as i can, right?

and that would be 201 of them

$2019^{999}$

[solid kiln]

Merineth

[keen void]

This should be correct

[nova spire]

no you could have simplified if it were 2019+2019+2019

[keen void]

ooh okay

since we have * that isn't allowed?

[nova spire]

hold on I'm thinking

oh wait maybe it is allowed

yeah ok

also

note that you're allowed to simplify more than the power

[keen void]

2019?

oh GOOOOOOOOD NOOOOOOOOOOOOOOOOO

i'm supposed to do 2019/11

to find the remainder..

aren't i?

[nova spire]

yeah

and then simplify the power

[keen void]

Okay

I have relearned long division

I got a remainder of 6

Sooo

$6^{666}$

[solid kiln]

Merineth

[keen void]

$6^{36*6}$

[solid kiln]

Merineth

[keen void]

$6^6$

[solid kiln]

Merineth

[keen void]

Hm

doesn't seem right

$6^{201920192019}$

[solid kiln]

Merineth

[keen void]

$6^{999$

[solid kiln]

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

[nova spire]

yes this is better

[keen void]

oki good

If i continue from there

i go

$6^{81*9}$

[solid kiln]

Merineth

[keen void]

Which becomes

$6^9$

[solid kiln]

Merineth

[nova spire]

nice

[keen void]

nice

https://tenor.com/view/kevin-the-office-smirk-gif-5248324

OH i know

what to do

i can expand the base

with 6*6 and reduce the exponent with 1

$36^8$

[solid kiln]

Merineth

[nova spire]

uh no

[keen void]

no?

[nova spire]

2^4 = 4^2

not 4^3

if you wanted 6^2 to show up

you'd need to divide the exponent by 2

but from here you can just go 1 by 1

start with 6

then 6^2

then 6^3

do at least those first three and you'll see

we can use 6^9 = (6^3)^3

[keen void]

$6^9 = 666666666$

[solid kiln]

Merineth

[keen void]

Right?

[nova spire]

yep

[keen void]

$6^9 = 666666666 = 36^7$

[solid kiln]

Merineth

[nova spire]

wut?

[nova spire]

.

[keen void]

never mind

[nova spire]

maybe you can compute 6^3 mod 11

and then ^3 again

that'll give you 6^9

[keen void]

$6^9$

[solid kiln]

Merineth

[keen void]

I fail to see how i can use this

[nova spire]

$6^9 = (6^3)^3$

[solid kiln]

rafilou2003

[nova spire]

just like last time

you were allowed to simplify the base

2019 became 6

[keen void]

oh

$108^3$

[solid kiln]

Merineth

[nova spire]

not quite

some bad math happened here

[keen void]

6*6 = 36?

Oh

LOL

$216^3$

[solid kiln]

Merineth

[nova spire]

yep great

[keen void]

Yeah idk what’s wrong with me lol

[nova spire]

now we can maybe reduce 216 mod 11

[keen void]

,w 216 mod 11

[solid kiln]

Results provided by WolframAlpha

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

Requested by merineth

[nova spire]

we all have brain farts, not too long ago I computed 3*3 = 6

[keen void]

$7^3$

[solid kiln]

Merineth

[nova spire]

yep

should be easy from now on right?

[keen void]

$7^3=777=49*7=343$

[solid kiln]

Merineth

[keen void]

,w 343 mod 11

[solid kiln]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by merineth

[keen void]

Final answer

$2019^{2019^3} \equiv 2 mod 11$

[solid kiln]

Merineth

[nova spire]

yippee !

[keen void]

https://tenor.com/view/cat-cat-jumping-cat-excited-excited-dance-gif-19354605

Can i have some prayers that the exam tomorrow will have some easy questions for once?

[nova spire]

You're gonna crush every single question, down to the last bonus point

Stay strong and hopes high

you're gonna crush that exam :)

[keen void]

HAHAHAH

okay i needed a prayer

not a cope

[nova spire]

I gave you a vision xd

[keen void]

I truly pray it goes well

I'll return tomorrow crying or happy

we'll see D:

thanks

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[red sentinel]

Hiya, how would you go about solving this question? I'm a bit confused

[ionic pendant]

we want to solve the equation
[\mathrm{pH} = 0 = \log_{10}\ab(\frac 1{H^+})] for $H^+$

[solid kiln]

cloud

[red sentinel]

yes

[ionic pendant]

how can we eliminate the log?

[red sentinel]

im not really sure

we only just learned logs

[ionic pendant]

what is the opposite of taking a logarithm?

[red sentinel]

an exponent?

[ionic pendant]

yes

[red sentinel]

oh!

okay

is it something to the power of 10?

would it be (1/h^+)^10?

[ionic pendant]

no, the opposite of x^10 is 10√x. the log_10(x) is the opposite of 10^x

[red sentinel]

oh

so 10^(1/h^+)?

[ionic pendant]

we want to apply the function 10^(x) on both sides of the equation

[red sentinel]

how do you do that?

[ionic pendant]

same as applying anything else on both sides of an equation. you have 10^(left side) = 10^(right side)

[red sentinel]

oh okay

so it's just 10^x on one side because it's 0 right? so it would be 10^x=1/h^+?

[ionic pendant]

no... the left side is 0, so it is 10^0

[red sentinel]

so x is 0?

[ionic pendant]

that is the input of the function in this case

[red sentinel]

how does that work though

[ionic pendant]

if i said apply the function log(x) to both sides of the equation z = 10, x is just standing in for the input of log. you would get log(z) = log(10)

[red sentinel]

im sorry

i dont really know what a function is

i google it

okay so

i understand you need to add 10 to both sides

but not really from there?

[trim jolt]

@red sentinel Has your question been resolved?

[ionic pendant]

well taking 10^(something) is the opposite action of taking log_10(something)

[trim jolt]

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[naive solstice]

How would I go about finding the zeros of the polynomial

[earnest nymph]

i mean, they kinda already factorised it for you

in P(x)=(x-a)(x-b), a and b are the roots, since subbing it in will make P(x)=0

[naive solstice]

So do I do P(0)=(0-2)(0-2)

[earnest nymph]

do

each bracket =0

eg. (x+2)=0

x=-2

[naive solstice]

Ahhh THANKS

So when it comes to the outside numbers such as -2 what do I do with it

[earnest nymph]

they dont matter

[naive solstice]

Just the parentheses got it

[earnest nymph]

if you put it im the beacket or anything, its the same as -2x-4=0

its still equal to -2

[naive solstice]

So when it asks the smallest medium and largest zeros is it based on how many of the same number you get or the smallest to largest based off of its negative or positive sign

[earnest nymph]

if you have an x on the outside tho, that means x=0 tho

i mean

id assume its the latter

multiplicity is something else

[naive solstice]

Ok ^^

How do I find that?

[earnest nymph]

multiplicity is

what power the parenthesis is to

all of these dont have a power

so its single roots

if i had P(x)=(x-a)^2 tho, thats a double root

[naive solstice]

So for example if it's F(x)=-4(x + 3) (x + 3) x + 3)(x-2)

The zeros are 2 -3

And the smaller zero has a multiplicity of 3

So is it because the three is in the parentheses?

And then the largest hero has a multiplicity of one, I just don't understand how to get three and one

Sorry that's the last question I really have

[earnest nymph]

?

[earnest nymph]

(x+3)(x+3)=(x+3)^2

[naive solstice]

This for an example, how do I find the smallest zero

I got the first and last one right but the middle one wrong

[earnest nymph]

so multplicity is 2

the zeros are -2 and 4

there are 3 brackets of (x+2)

[naive solstice]

OOOH

[earnest nymph]

so you can group it into (x+2)^3

[naive solstice]

Woww

XD

Thank you! Haha

[earnest nymph]

np

[naive solstice]

.close

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[naive solstice]

Hey I need help for finding the equation for b and c

[naive solstice]

Is b) 6.73e^0.0376(6)?

@ me when u see this ty!

[mild oxide]

@naive solstice

[dry copper]

lollllll

[naive solstice]

Hola

[dry copper]

yes it is

[mild oxide]

He said to ping when any helper sees it

What you said is correct my guy

[mild oxide]

Hi sis

[viscid flower]

but lol

[naive solstice]

Lol ty! So for the last one, section c how would I do it and what's the equation:D pls

[mild oxide]

Okok

[naive solstice]

Triple the help XD

[mild oxide]

It says

Read the first part

It is written number of years after 2012

I hope you will get it now

[naive solstice]

Hmmm

Nope

[mild oxide]

And your are asked to find population 13.1 years after

2012

My msg did not get sent

[naive solstice]

13.1 is answer

It tells u if u get it wrong

[mild oxide]

Oh ok i see

See they are asking

[naive solstice]

I just need to figure out how to get 13.1

[mild oxide]

How many years after 2012

[naive solstice]

Or any other value if it asks question like this

[mild oxide]

Will the population become 11 million

See above P(t)

[naive solstice]

11-6.67?

[mild oxide]

p(t) is population in millions after 2012

And t is years after 2012

I hope u will be able to do it now

[naive solstice]

Uhh... o.o

Is there a formula or how would I set it up I'm confused

@mild oxide

Halp

Pweaseee

@dry copper could u help with section c I am confused

How to write it in a equation or what steps to take

[limpid dawn]

[mild oxide]

Ok

I came back

[naive solstice]

I understand A. And B. I don't understand C my mind goes blank

[mild oxide]

P(t)=0.73e^(0.0376t)

[limpid dawn]

I have a deja vu

[mild oxide]

What do u feel

[naive solstice]

LIsten

[mild oxide]

Bout this equation

[naive solstice]

I have short term memory

That was like 8 days ago DX

[limpid dawn]

buy some ram

[mild oxide]

Lol

[naive solstice]

My mind absorbs and spits out

[limpid dawn]

oh i know that feeling

[mild oxide]

You can buy from adonis

[naive solstice]

It's fried have to get a whole new system now

[mild oxide]

Bro has loads and loads of ram in his mind

[limpid dawn]

it seems like you dont like math at all

[mild oxide]

Like if bro remembers he remembers

[naive solstice]

I do but I zone out easily lol...

[limpid dawn]

c) is your question?

[mild oxide]

Yes

[naive solstice]

I always ended up doing really good in the very beginning of math and then towards the middle and the end of the course and I start drawing

[limpid dawn]

Can you explain what it is you don't understand?

[mild oxide]

Imma explain very simply now

[limpid dawn]

vibe

[naive solstice]

And I look up and I just become lost ones and I can never get back on the train

[mild oxide]

Listen carefully

[naive solstice]

[mild oxide]

P(t) spits out the population in millions t years after 2012

Now when u put t

You get population t years after 2012 in millions

Now they say when the population be 11 million

So they are simply asking after how many years after 2012

Did u understand now?

[limpid dawn]

I think it's better to imagine that t are the years passed since 2012

So t = 0 implies the population in 2012

t = 1 in 2013 (one year later) and so on

and in total we are witnessing an exponential growth which isn't surprising

that's like the idea

So oppa is saying so right

P(t) = 11 mio

[naive solstice]

P(t)=0.73e^(0.0376(11)?

[limpid dawn]

they are asking you to find how many years have passed before the population reached 11 million

[naive solstice]

Oh

[limpid dawn]

nope

[naive solstice]

You explained it so perfectly to me about 8 days ago

I think now I'm just blanking

[limpid dawn]

P is your population

that depends on time since it increases

with years

passing

so we also write P(t) P of t

population with respect to time

now we wanna know how many years have passed since 2012

where the population P = 11 mio

So we do P = 11

P(t) = 11

and solve for t

which gives us the time in years passed since 2012

[naive solstice]

P(11)=0.73e^(0.0376t)?

[limpid dawn]

Why P(11)?

[naive solstice]

11 million?

[limpid dawn]

that means Population after 11 years since 2012

but you say P(t)

P(11) means P at t = 11

but it's the other way

P = 11 at what t = ?

P(t) = 11 is the notation

[naive solstice]

What is the arrow thingy

[limpid dawn]

ok let me delete

[naive solstice]

Sorry ty for helping me

[limpid dawn]

oh no

[naive solstice]

Hm?

[limpid dawn]

the arrow means equivalence

but i deleted it because i thought maybe it's too complicated

[naive solstice]

For my brain yesh

[limpid dawn]

I have a question

are these correct?

[naive solstice]

Yes

[limpid dawn]

Ok weird

[naive solstice]

It's marked red because I left it empty submitted it and it gives you the correct answer

Like basically letting you know what it should have been

[limpid dawn]

I checked the function P(t) and it doesnt make sense

[naive solstice]

So all of those are the correct answer I know how to get A n B of course

[limpid dawn]

,w plot 0.73e^(0.0376t) between 0 and 14

[solid kiln]

Results provided by WolframAlpha

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

Requested by adoniszgks

[limpid dawn]

Like for b)

2018

implies it's been 6 years since 2012

so we would calculate P(6), no?

[naive solstice]

Mhm! That's super easy

[limpid dawn]

,w 0.73e^(0.0376*6)

[solid kiln]

Results provided by WolframAlpha

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

Requested by adoniszgks

[limpid dawn]

so like

that's not even close to 1 mio

but it says 8.4 mio

so i suspect somethings off

[naive solstice]

b

[limpid dawn]

and there is some other function

[naive solstice]

U miss 6

[limpid dawn]

Oh

[naive solstice]

And put zero on accident

[limpid dawn]

well that explains it

ok thx!

[naive solstice]

Np haha Cx I'm a math genius now

[limpid dawn]

Ok let me explain c) from scratch

You dont understand what to do?

What it is about?

delete it please haha

[naive solstice]

Wouldn't this be correct tho

[limpid dawn]

I will retype it

[naive solstice]

OvO

[limpid dawn]

$P = P(t)$ denotes the population
$\ t$ denotes the time in years (passed since 2012)$\$
Now we wanna figure out how many years passed since 2012, where the population $P(t)$ is at 11 million.$\$
Mathematically it means $$P(t) = 11$$ that's what is to figure out.$\$
We know $$P(t) = 6.73e^{0.0376t}$$ so we do $$6.73e^{0.0376t} = 11$$ and the rest is algebra.

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

read it at your pace

brb

[naive solstice]

Ok!

Tyvm!

[limpid dawn]

back

[naive solstice]

That was fast

[limpid dawn]

got some water

do you also need the algebra part?

[naive solstice]

Stay hydrated

[limpid dawn]

yes

[naive solstice]

Nope

I just didn't know what formula to do is all :3

[limpid dawn]

hehe

it's not a formula really

[naive solstice]

Operation?

[limpid dawn]

it's as stupid as it sounds reasoning

[naive solstice]

What would be the proper name

[limpid dawn]

P(t) gives back population (in million) and you want to know when the function reaches 11

[naive solstice]

Yeah I guess a formula is set in stone

Like y=mx+b never changes so is it called an operation? Or just an equation

[limpid dawn]

It's an equation

[naive solstice]

thank you!

[limpid dawn]

anything with ... = ... is equation

operations are like

addition, subtraction, dividing, multiplication

basically what you apply on both sides of an equation

[naive solstice]

I see, Duly noted ty!

[limpid dawn]

any more doubts?

[naive solstice]

Before I close what language do you speak

[limpid dawn]

English?

[naive solstice]

No all cleared thank you thank you thank you

[limpid dawn]

haha you are welcome

[naive solstice]

Music?

[limpid dawn]

Greek

[naive solstice]

Ooo

[limpid dawn]

Ah I see

I speak German, Greek and English

[naive solstice]

Trilingual:D

[limpid dawn]

haha

[naive solstice]

I speak Russian n English

[limpid dawn]

oh

russian

nice

[naive solstice]

Yeppers xD

Anyways imma close for next person

THANKS a bunch

[limpid dawn]

haha

np

[naive solstice]

.close

[limpid dawn]

🔠

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[jovial wing]

Heres my working so far

Im having a bit of trouble changing dx to dy.

From the original equation, I am missing a 2 on the denominator.

Could I simply add this in and add a 2 on the left side of the integrand?

[trim jolt]

@jovial wing Has your question been resolved?

[jovial wing]

<@&286206848099549185>

[wind cloak]

substitution?

just change the bounds and plug in the values of different things involving x in terms of u

[jovial wing]

what im used to is matching the integral to du

to get the dx to the du

so in other words, get the function to look like this or have featues of 1/2sqrtx so I can sub du in and take away the values related to dx

[wind cloak]

yeah but what's the problem?

you're on the right track

[jovial wing]

Thats the part I am a bit rusty on with this topic

Do I have to change up the original function to try to replicate a 1/2sqrt x

and by doing this do I also have to balance it out on the outside of the integral

[whole coral]

(you could also write that, if it makes it easier, to write what you have as $2\sqrt{x} \dd u = \dd x$ and manipulate that way)

[solid kiln]

@whole coral

[wind cloak]

your original function already have a sqrtx in denominator
you can use that too

[whole coral]

(so you can also multiply by 2/2, so that you get the required 2sqrt{x} in the denominator)

[jovial wing]

Yeah I noticed that part, now to add a 2 in. Can I simply add 2 to the denominator and then have the outside multiplied by 2?

would that work out?

[jovial wing]

great, thank you

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[earnest nymph]

help with part a pls

[earnest nymph]

i have no idea what it wants

[whole coral]

Have you tried setting $x = k\cos(\theta)$ and putting that into $27x^3 - 9x$? The idea is that, from that point, there's a choice of $k$ that makes it easier to solve for $\theta$, given the identity $\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$...

[solid kiln]

@whole coral

[earnest nymph]

i have no clue how to execute that idea tho

[whole coral]

That's fair, did you try the substitution or not though?

[earnest nymph]

yea

[whole coral]

And what did you get?

[earnest nymph]

idk what from here

if i try say that
27k^3=4, i get the wrong answer

idk what to do

[whole coral]

Well, we could also have that 27k^3 be some multiple of 4, but then we want 9k to end up being a "similar" multiple of 3

The choice isn't immediately obvious, I will say (that said, to clarify the above point, by "some multiple of 4" I'm meaning you find something like 27k^3 = 4m for some integer m, but by the "similar multiple" part, I'm hinting you find that 9k = 3m as well)

[earnest nymph]

mmmm

should i try find for that integer m and sub it into the other part or wrong way

aaa

i got it

like this?

[whole coral]

That works, yep

[earnest nymph]

and then cause k>0 its 2/3

icic

tysm

[whole coral]

Yep, k is 2/3 pretty hard to hint at without giving it all away, I shall say

[earnest nymph]

got some good cats there

very cute ^^

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[earnest nymph]

.reopen

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✅

[earnest nymph]

i cant think of a legal way to do b lmao

gives right ans if i say theta =… and then say x=kcostheta

but idk why it works

nvmmm

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[unkempt wyvern]

what can i do from here

[unkempt wyvern]

is it just 1?

in the answer key is says limit x -> 0- is -1, and limit x -> 0+ is 1

[nova spire]

simplifying sqrt((dx)^2) into dx is wrong

the answer depends on the sign of dx

just like sqrt((-3)^2) = sqrt(9) = 3, and not -3

[unkempt wyvern]

oh yeah

but how do i find the limit then

if i do have sqrt((dx)^2) / dx

[nova spire]

well it depends on the sign of dx

so when dx-> 0+ for example

how does sqrt((dx)^2) get simplified?

(dx > 0)

[unkempt wyvern]

ohhhh bruh

it would just be the value

we're taking the absolute value of dx

lmao aight

i see

[nova spire]

always good to remember |x| = sqrt(x^2)

[unkempt wyvern]

thank you

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[unreal lark]

what does the underlied sebntence mean

[vagrant marsh]

you have a landmark to make drawing the curve easier is my best guess

[unreal lark]

ohh

ic

ty

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[vagrant marsh]

youre welcome

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[unreal lark]

what does this mean

[placid radish]

i think it means that, for what k do we have that when sin(x) is reflected through the vertical line at k, it turns into cos x

[unreal lark]

oh how do i figure that out

[placid radish]

well, what is sin(x) after it gets reflected by k?

[violet gust]

Suppose you have a point x0,y0. How would that change when you reflect it in a line say x = x1

Use that to evaluate the function at that new point

and that is supposed to be the cos for new point

[unreal lark]

how do i figure out what k is

[placid radish]

first figure out what sin(x) is after it gets reflected by k.

then use this to figure out what k is

you will get sin(something in terms of x and k)=cos(x) [since reflection by k turns sin(x) into cos(x)]

and you can solve for k that way

[unreal lark]

isnt that dependent on what k is

[placid radish]

yeah, exactly, and we will get sin(somthing in terms of x and k)

[unreal lark]

sin(-x-k) ?

[steel glade]

maybe it'd more helpful if you visualize it on a graph

[placid radish]

well almost! what is x after getting reflected by k?

say if k=1, then 2 gets reflected to 0

but -2-1 is -3, certainly different from 0

[unreal lark]

idk im trying very hard to procews what ur saying

sin(-x+k)

[placid radish]

thats close, but not quite. if x gets reflected to another point l, then their average should be k

that is, (x+l)/2=k

[unreal lark]

why is there another point l

wait

let me think

[placid radish]

oh i just created a name for the reflection

[unreal lark]

i kind of get it

[placid radish]

its sort of like this

hopefully this gives some intuition

[unreal lark]

ye ty i get it

[placid radish]

so what is sin(x) after reflection by k?

[placid radish]

what this graph sort of says is that it is important to know what x after getting reflected by k is

it means that our new reflected function evaluated at x = sin evaluated at x after reflection

[unreal lark]

what does evaluated mean

[unreal lark]

idk ;-;

[placid radish]

oh sin(x) is the equivalent of saying sin evaluated at x

”evaluated at“ is "taking the input“

[unreal lark]

sin(x+k/2)?

[placid radish]

still off

let's do a simpler case first: what is x after it is reflected by k?

[unreal lark]

so r u trying to say our new reflected function at x is sin(x) after reflection

[placid radish]

yeah!

it is important fact: if the reflection of a function $f(x)$ about $k$ is $g(x)$

[solid kiln]

qwertytrewq

[placid radish]

then $f(x \text{ reflected by }k)=g(x)$

[solid kiln]

qwertytrewq

[unreal lark]

bro why am i so stupid

[unreal lark]

wdym reflected by k

is k like an imaginary straight line

[placid radish]

yeah

[placid radish]

reflected by the green line

[unreal lark]

its 2 times distance betgween x and k

wait that didnt make sense

[placid radish]

i am asking what is this point in terms of x and k

[unreal lark]

x+2k

[placid radish]

-x+2k?

[unreal lark]

yes

[placid radish]

notice that this fact basically tells us that our reflected function is sin(x reflected by k) which is sin(-x+2k)

[unreal lark]

uh huh

[placid radish]

if you want me to explain why that fact is true i could try

[unreal lark]

yea pls

[placid radish]

ok, but first i need to ask: do you know what i mean by f(x) or g(x)?

some school don't teach this symbol

[unreal lark]

doesnt that just mean different functions

[placid radish]

yes

[unreal lark]

f(x)=y

[placid radish]

its just that some people only write y=stuff

and not f(x)=stuff

[unreal lark]

ah yeah

[placid radish]

OK, lets say we have some function f (in red) and we reflect by the green line (x=k),

let us call the new reflected function g

[unreal lark]

yes

[placid radish]

🤔 then we ask the question: what is g(x)?

[unreal lark]

y=(x-11)^2

[placid radish]

oh f and g are arbitrary functions (i just picked a nice one to explain the intuition

[unreal lark]

oh ok

[placid radish]

well, we know that x after it gets reflected by the green line (x=k), has the formula -x+2k, that you calculated before

[unreal lark]

yes

[placid radish]

now what is f evaluated at -x+2k?

[unreal lark]

g(x)

[placid radish]

notice that the four points i drew forms a rectangle

this means that (x,g(x)) reflected by the green line is exactly (-x+2k,f(-x+2k))

(x,g(x)) is the point on our graph

and (-x+2k, f(-x+2k)) is also a point on the graph

[unreal lark]

yes ic

[placid radish]

but, of course, reflection by vertical line dont change the "vertical" value

[unreal lark]

f(-x+2k)=g(x)

[placid radish]

it means that, the second coordinate of (x,g(x)) is equal to the second coordinate of (-x+2k,f(-x+2k))

[placid radish]

alright, now you get it

[unreal lark]

yes your graphs r helpful

[placid radish]

so, in other words, f(x reflected by k)= g(x)

pretty neat fact

[unreal lark]

yea

[placid radish]

ok back to our question

let us define g(x) to be sin(x) reflected by k

so, we have the formula g(x)=sin(-x+2k)

now the question is: when is our g(x)=cos(x)?

the key to solving question like this, is to plug in special values, like x=0 (this is special because cos(0)=1)

so we have sin(2k)=g(0)=cos(0)=1, what must k be?

[unreal lark]

pi/4

[placid radish]

well notice that sin is periodic (it reaches 1 at multiple values)

so pi/4 is not the only solution

[unreal lark]

is it npi/4

wait

[placid radish]

um not quite

[unreal lark]

3pi/4

pi/4

[placid radish]

pi/4 and 3pi/4 both works yes!

but there are even more

[unreal lark]

9pi/4

[placid radish]

wait my back 3pi/4 doesn't work

5pi/4 works

[unreal lark]

oh rlly

isnt it negative there

[placid radish]

well sin(x) reaches 1 every 2pi cycles

so sin(2x) reaches 1 every pi cycle

we know that sin( 2* pi/4)=0

so k must be pi/4 +n pi (for integer n)

[unreal lark]

sorry does the * mean times or to the power

[placid radish]

times

^ is to the power usually

[unreal lark]

yes ic

[placid radish]

now here is the tricky part: we derived that k must be these values, but does all of these values work?

[unreal lark]

yes right

[placid radish]

lets check: if k=pi/4 + n pi, then g(x)=sin(-x+2k)= sin(-x+pi/2+2n pi)

now is the right most term equal to cos(x)?

can you try and figure it out?

[unreal lark]

ok

[placid radish]

(use some trig identities)

like sin(x)=cos(x-pi/2)

sin(x)=sin(x+2 n pi) for all integer n

sin(x)=-sin(-x)

etc

[unreal lark]

when n=0 it is

[placid radish]

that is correct, how about when n is not zero but some random integer?

[placid radish]

this identity may help

[unreal lark]

am i meant to figure out what x is? when i did sin(-x+pi/2+2npi)=sin(x+2npi) i onyl could get the value of x

[placid radish]

no, because it is said in the question that "the reflected function is cos(x)"
so this is equivalent to the claim "g(x)=cos(x) for all x"
basically we want to see if indeed sin(-x+pi/2+2npi)=cos(x) for all x

[placid radish]

did you mean sin(-x+pi/2+2npi)=cos(x-2n pi)?

[unreal lark]

yea im doing that rn

[unreal lark]

do i just prove it

i got it that way

[placid radish]

sure

if you think all of the k=pi/2 + 2n pi are valid then prove it is all valid

[unreal lark]

it is

[placid radish]

👍 can you show the steps?

[unreal lark]

olk sorry i did do it

but then when i tried to rewrite it to make it look neat i got confused again and now im redoing the question

ok i finished ill upload a pic

[placid radish]

correct!

so we deduced that if k works, then k must be pi/4 + n pi

and all of pi/4+n pi works

and we are done, these are precisely the solutions

🔥

so the key steps are:
figuring out what the reflected funciton is
figuring out what the possible k's are
verfying our values

[unreal lark]

I made it slightly neater

[placid radish]

great work

[unreal lark]

yes thank you i love you qwerty

TY

i got the qeustion now

i think i can close it now ur patience is godly

.close

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Use .reopen if this was a mistake.

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[sick pier]

question about quadratic functions (?): Can't find the point F.

Problem is, I tried using the quadratic function. first, I did using 3 points I have. then I figured its just flipped, and used the point A as point (h;k). doing so, I came to an answer of point F, f(x) being higher or lower than what I expected, doesn't match visually.

Thanks in advance.

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@sick pier Has your question been resolved?

[sick pier]

<@&286206848099549185>

[sick pier]

.close

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[feral dragon]

i do not get what I'm doing wrong?

[nova spire]

!show

[trim jolt]

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

[feral dragon]

give me one sec

it says to round to 4 decimal places?

[honest scroll]

u should probably round after adding

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@feral dragon Has your question been resolved?

[feral dragon]

@honest scroll so how many decimal places do I go then for the square roots?

[honest scroll]

i think add them first in ur calculator

like 5sqrt1 + 5sqrt2 + 5sqrt3 +5sqrt4

then round that answer to 4 decimal places

if u need to show that in between step

ig 5 or 6 decimal places would prevent an error up to 4 decimal places

[feral dragon]

your right