help-13

[fallen moat]

pretty

now the final step, we write them back in terms of x

(this gonna be long)

$$\frac{\sqrt[3]{x+1}+\sqrt[3]{x}+\sqrt[3]{x+1}}{\left((\sqrt[3]{x+1})^2+\sqrt[3]{x+1}\sqrt[3]{x}+(\sqrt[3]{x})^2\right)\left((\sqrt[3]{x})^2+\sqrt[3]{x}\sqrt[3]{x-1}+(\sqrt[3]{x-1})^2\right)\left((\sqrt[3]{x-1})^2+\sqrt[3]{x-1}\sqrt[3]{x+1}+(\sqrt[3]{x+1})^2\right)}$$

[wraith dagger]

Biscuity

[fallen moat]

lol

not enough space

I'll try to write

or do you understand, so i can skip writing? XD

,rccw

[fallen moat]

not enough space....

anyways

the main point is not this

Now, let $$a'=\frac{a}{\sqrt[3]x}=\sqrt[3]{1+\frac1x}$$, $$b'=\frac{b}{\sqrt[3]x}=1$$,
$$c'=\frac{c}{\sqrt[3]x}=\sqrt[3]{1-\frac1x}$$

[wraith dagger]

Biscuity

[fallen moat]

which is similar to the form of your last line

[halcyon granite]

oh my, so much text 🙂
i'm so sorry I left for a moment

lemme skim through it

[fallen moat]

we have $a-2b+c$ equals to
$$\frac{x^{\frac13}(a'+b'+c')}{(x^{\frac23})^3(a'^2+a'b'+b'^2)(b'^2+b'c'+c'^2)(c'^2+c'a'+a'^2)}$$

[wraith dagger]

Biscuity

[fallen moat]

which technically i just did what you did in line 3

but i pulled three (x^⅔) out, thats why it's (x^⅔)^3

finally

\begin{align}
&\lim_{x\to\infty}x^{\frac54}(\sqrt[3]{x+1}+\sqrt[3]{x-1}-2\sqrt[3]{x})\
=&\lim_{x\to\infty}x^{\frac54}(a-2b+c)\
=&\lim_{x\to\infty}x^{\frac54}\left(\frac{(c-a)(a+b+c)}{(a^2+ab+b^2)(b^2+bc+c^2)}\right)\
=&\lim_{x\to\infty}x^{\frac54}
\left(\frac{(-2)(a+b+c)}{(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)}\right)\
=&\lim_{x\to\infty}x^{\frac54}\left(\frac{x^{\frac13}(-2)(a'+b'+c')}{(x^{\frac23})^3(a'^2+a'b'+b'^2)(b'^2+b'c'+c'^2)(c'^2+c'a'+a'^2)}\right)\
=&\lim_{x\to\infty}x^{\frac54}\frac{x^{\frac13}}{x^2}\left(\frac{(-2)(a'+b'+c')}{(a'^2+a'b'+b'^2)(b'^2+b'c'+c'^2)(c'^2+c'a'+a'^2)}\right)\
=&\lim_{x\to\infty}\frac1{x^{\frac5{12}}}\cancelto{-\frac6{27}}{\left(\frac{(-2)(a'+b'+c')}{(a'^2+a'b'+b'^2)(b'^2+b'c'+c'^2)(c'^2+c'a'+a'^2)}\right)}\
=&\lim_{x\to\infty}-\frac2{9x^{\frac5{12}}}\
=0
\end{align}

[wraith dagger]

Biscuity

[fallen moat]

which tends to 0 as x tends to infinity

oooo, typo

\begin{align*}
&\lim_{x\to\infty}x^{\frac54}(\sqrt[3]{x+1}+\sqrt[3]{x-1}-2\sqrt[3]{x})\
=&\lim_{x\to\infty}x^{\frac54}(a-2b+c)\
=&\lim_{x\to\infty}x^{\frac54}\left(\frac{(c-a)(a+b+c)}{(a^2+ab+b^2)(b^2+bc+c^2)}\right)\
=&\lim_{x\to\infty}x^{\frac54}
\left(\frac{(-2)(a+b+c)}{(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)}\right)\
=&\lim_{x\to\infty}x^{\frac54}\left(\frac{x^{\frac13}(-2)(a'+b'+c')}{(x^{\frac23})^3(a'^2+a'b'+b'^2)(b'^2+b'c'+c'^2)(c'^2+c'a'+a'^2)}\right)\
=&\lim_{x\to\infty}x^{\frac54}\frac{x^{\frac13}}{x^2}\left(\frac{(-2)(a'+b'+c')}{(a'^2+a'b'+b'^2)(b'^2+b'c'+c'^2)(c'^2+c'a'+a'^2)}\right)\
=&\lim_{x\to\infty}\frac1{x^{\frac5{12}}}\cancelto{-\frac6{27}}{\left(\frac{(-2)(a'+b'+c')}{(a'^2+a'b'+b'^2)(b'^2+b'c'+c'^2)(c'^2+c'a'+a'^2)}\right)}\
=&\lim_{x\to\infty}-\frac2{9x^{\frac5{12}}}\
=&0
\end{align*}

[wraith dagger]

Biscuity

[halcyon granite]

whooooaa

one hour of work 🙂

[fallen moat]

lol

[halcyon granite]

I'm sooooo thankful ☺️☺️☺️

[marble lantern]

lol

[fallen moat]

but do you understand? it's kinda messy if it's this big

[halcyon granite]

ye, I hope so 😇

[fallen moat]

that's good!

[halcyon granite]

I think I should close this channel so others can get help as well.

Again, thank you so much! ☺️ Have a nice day 🙂

.close

[cedar kiln]

Channel closed

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[fallen moat]

You too! Have a nice day

[cedar kiln]

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[ancient ether]

Can someone please help me with the 2nd one

I don't know where to start

[kindred storm]

The easiest way is to notice that the potential answers are all quite similar, so differentiate answer a and see if you get what's in the integral. Then, note that answer c will be half of answer a, so if answer a was twice what it should be, then the answer is c. Then, try to reuse some of your work for answer a to see if answer b works.

[ancient ether]

Ah

I'll do that

Okay I'm sorry I don't get it

You want me to differentiate option (a) ?

[kindred storm]

Well, an indefinite integral is an antiderivative.

So, if you take the antiderivative of what you start with to get one of the answers, you can take the derivative of one of the answers to see if you get what you start with.

[ancient ether]

Oh

Okay

[raw gulch]

wait, you can check the derivative, but the purpose of this exercise is that you have to evaluate the integral

[ancient ether]

Uhh

Is there another approach to solving this?

[kindred storm]

Yes, you can do the integral.

[ancient ether]

Okay

I'll try again

. close

.close

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[arctic condor]

I need help figuring out how to calculate Sine without actually using the Sine function

[arctic condor]

https://www.desmos.com/calculator/ury0h31rqk

currently trying to compute Sine() In minecraft, but I want to figure out how to calculate it in desmos first.

[harsh ginkgo]

There are different ways to do so, but the simplest way would be to use the power series of sin(x). Do you know what that is @arctic condor ?

[arctic condor]

No

[gritty galleon]

why do u even need this...

[arctic condor]

to understand trig more

[gritty galleon]

https://en.wikipedia.org/wiki/Sine_and_cosine#Definitions

[harsh ginkgo]

I can garantuee you that using the power series will help with understanding the trigonometry.

[arctic condor]

mk

[gritty galleon]

ur joking right?

[arctic condor]

https://www.youtube.com/watch?v=bre7MVlxq7o was using this, very confusing

[gritty galleon]

this has many definitions of sine which dont use sine (obviously)

[harsh ginkgo]

This feel unnecessarily harsh.

[tropic oxide]

in minecraft? with what tools and for what purpose?

[harsh ginkgo]

but Ill let you do your thing

[gritty galleon]

WAIT UR-

[arctic condor]

ignore the minecraft part, i can do that myself

that was just context

[gritty galleon]

bro then CORDIC is the best way

[gritty galleon]

redstone ig

[arctic condor]

also i feel like ive seen you before somewhere but 🤷

[gritty galleon]

lol https://discord.com/channels/1066065273158910012/1154077580673175602

[arctic condor]

oh LOL

[gritty galleon]

why do u even need sine Yayzay

[arctic condor]

wanna make

spinning dot

[opaque lotus]

hello

i need help

in my math hw

[scenic brook]

!help

[cedar kiln]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[scenic brook]

! occupied

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[scenic brook]

!chill

@chill

[gritty galleon]

huh?

mind splaining me

[scenic brook]

#chill

[arctic condor]

a dot on

[gritty galleon]

shut up

[arctic condor]

a

circle in a graph

[gritty galleon]

u have an animation?

[gritty galleon]

oh do u mean like the radius spinning around the circle...?

like a clock?

[arctic condor]

here

[gritty galleon]

some video coming?

[arctic condor]

https://www.desmos.com/calculator/2tekzbdruc

[gritty galleon]

you know whats a simple and non-elegant way to do this

[arctic condor]

?

[gritty galleon]

use mc-schematic and python to write the required positions in memory (barrel/disk shulker ROM) and then read the ROM through an assembler and then just render the dot at that place

[arctic condor]

ouch

that's

what's the word for it

ambitious

[gritty galleon]

trust me, its 2000x easier than what ur trying

[arctic condor]

i want to be able to change radius

and o

[gritty galleon]

u can do that with 1 click

python is ez

[arctic condor]

i don't know python

[gritty galleon]

i could've written it for u but me is busy with school

[gritty galleon]

bruh

u can learn WHILE doing this

https://github.com/Sloimayyy/mcschematic

https://www.python.org/

[arctic condor]

im also

doing it for sth else

instead of just animations

[gritty galleon]

hmm

but trust me sin is crazy impossibaly tough and stupidly slow to implement

[arctic condor]

would rather have is stupidly slow

https://www.desmos.com/calculator/ury0h31rqk

im already there half of thee way

this doesnt have anything to do with minecraft bro

im gonna close this

/close

.close

[cedar kiln]

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[sharp sequoia]

what does the rxn?

stand for

thats beneath G, H & S

[void sand]

"reaction"

[sharp sequoia]

thanks

.close

[cedar kiln]

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[lone tapir]

Is this: GJ(35sin)=55/9 on the right track or is this off

[cedar kiln]

@lone tapir Has your question been resolved?

[lone tapir]

<@&286206848099549185>

[ocean flame]

okay

hi

[lone tapir]

Hii

[ocean flame]

we need to find?

[lone tapir]

The side lengths

[ocean flame]

ic

one moment

im helping someone else

[lone tapir]

Okay 👍🏾

Thank you for your support

[ocean flame]

do u know trig

[ivory cobalt]

for JH -> sin35/9

[ocean flame]

trigonometry?

[ivory cobalt]

for GJ -> cos35/9

[lone tapir]

A little

[ocean flame]

okay

[lone tapir]

Probably not enough for finals tomorrow

[ocean flame]

what the person said above it correct

[ivory cobalt]

For JH -> sin35/9
For GJ -> cos35/9

[lone tapir]

Wait okay

[lone tapir]

I need you to explain why 35 is opposite though

[ocean flame]

alr

soo

in the given triangle

[potent mesa]

I badly wanna study rn goshh

[ivory cobalt]

Sin theta = Side oppposite to the angle / Hypotenuse of the triangle

[potent mesa]

Give me study tips plss

[lone tapir]

Theta is for the sides right?

[ivory cobalt]

Theta is how we represent the angle to be

The side opposite to the given angle is what goes in the numerator

The hypotenuse of the triangle goes in the denominator

[lone tapir]

Ohh okay

[ivory cobalt]

So it becomes sin35 = JH / 9

On cross multiplying

9(sin(35)) = JH

[lone tapir]

5.16

[ivory cobalt]

And then the same procedure for the 2nd part of the question but taking the trigonometric function as cos

[lone tapir]

Ah okay

[ivory cobalt]

Yes

Btw which country are you guys from?

[lone tapir]

Me?

[ivory cobalt]

Yeah

[lone tapir]

estados unidos

U.S.

What about this one

[ivory cobalt]

Okk

Andre is correct

As 9, 12 and 15 are pythagorean triplets

[lone tapir]

Ya lo sé

I know

But I can’t explain Clare

[ocean flame]

?

[lone tapir]

This was my visual but that’s as far as I got

[ivory cobalt]

same here

What I think is that Clare is wrong

[ocean flame]

need help or you got it?

[ivory cobalt]

As theta is 20 degrees

[lone tapir]

Isn’t it cosine

cos20 14.1/15

[ivory cobalt]

Cos(20degree) = 0.4

so

cos(20 degree) = AB/BC

BC = 15/0.4

[lone tapir]

Isn’t it Adjacent over Hypotenuse?

AH

[ivory cobalt]

Yes

[lone tapir]

BC/AB?

[ivory cobalt]

ahh

correct

wait

then

cos(20 degree) = BC/AB

AB x cos(20 degree) = BC

[lone tapir]

I got 19.9

20

[ivory cobalt]

15 x 0.9 = 13.5 =BC

Since given value of BC is not equal to the obtained value, Clare was wrong

[lone tapir]

Ohhh okay

[ivory cobalt]

Yessirr

[ivory cobalt]

Btw I am from India

[lone tapir]

Oh wow

I had a teacher who lived in India, she said overall that it was a pleasant experience

[ivory cobalt]

Ohh

It is a pleasant experience living in India but the winters are a bit out of hand

cya guys

good night from my side

its 12:30 in India

[lone tapir]

Cya ! Thanks for your help

Oh wow

[ivory cobalt]

Np

[lone tapir]

Sleep well

[cedar kiln]

@lone tapir Has your question been resolved?

[cedar kiln]

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[compact hinge]

There are two boxes with oranges and mandarins. The first box contains 4 oranges  and  5 mandarins. The second box has 7 oranges  and 1 mandarin. Three random fruits are moved from the first box to the  second. Then, two random fruits are taken from the second box. Find the probability that they are mandarins.

[cedar kiln]

@compact hinge Has your question been resolved?

[compact hinge]

No

[cedar kiln]

@compact hinge Has your question been resolved?

[cedar kiln]

Channel closed

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[marble ermine]

Given $A \subseteq \mathbb{R}^m$, $x \in \mathbb{R}^m$; how can I show that $d(x, A) = 0$ does not imply $x \in A$?

[wraith dagger]

lilisworld.

[marble ermine]

i need to find a counterex

[kindred storm]

What is d?

[marble ermine]

i think

distance

[crystal raptor]

Just go simple with m=1

And then if you need to show it for all m then it shouldn't be too bad to generalise

[muted bear]

isnt d=0 only when xis in A part of the rules of a metric space?

Maybe this is too out of my league

[crystal raptor]

d(x,A) involves an infimum

In its definition

Is why this is possible

[muted bear]

So arbitrarily close, and in the limit, 0

Got it

[marble ermine]

so you mean d(x, m) or what?

[dusky peak]

you could choose A to be an open set, where x is on the edge of that set

[crystal raptor]

No, R^1

[marble ermine]

ah ok

[crystal raptor]

Ah come on you could let them try

[dusky peak]

ahh sry

[marble ermine]

so d(x, A) = inf(|a|, a in ]0, 1]) = 0? why?

if x=0 and A = ]0, 1]

[crystal raptor]

You would have the unpack the definition of infimum

[marble ermine]

still don't get it, 0 < a <= 1, so why 0 <= |a| ?

i mean why is 0 the infinumum

[crystal raptor]

What is the definition of infimum?

[marble ermine]

uhm

it is the biggest lower bound of a set

@crystal raptor

uhm ok i just got it thanks

.close

[cedar kiln]

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[crystal raptor]

Nice

[cedar kiln]

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[warped coyote]

Uhhh you sure about that

[modest lily]

I think you can make an iscoceles triangle if you where to draw a radius from O to 70

and then you can find y

[polar holly]

It is actually half twice the measure

[modest lily]

then since its a quadralateral you can find O

[warped coyote]

Th this but from the 21 to y with o

[modest lily]

yeah that makes sense because I'm assuming it cuts perfectly mb

[polar holly]

Yeah, angle x is twice the measure of the inscribed 70 degree angle

[warped coyote]

How is this the case?

[cedar kiln]

@trail sonnet Has your question been resolved?

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[trail sonnet]

.reopen

[cedar kiln]

✅

[trail sonnet]

What is the solution for y tho

[visual mulch]

49

[cedar kiln]

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[cloud creek]

A hen knows how to count. We put it in front of a packet of 2024 eggs. She counts the eggs from the first packet and places them in a second packet as she goes. Each time she has counted 4 eggs, she lays an egg which she places in the packet of those she still has to count.

How many eggs will she have to count in total?

[cloud creek]

my answer is 2698

but im getting a possibility where it is 2695

<@&286206848099549185>

someone

plz

<@&286206848099549185>

[cedar kiln]

@cloud creek Has your question been resolved?

[cloud creek]

no

A hen knows how to count. We put it in front of a packet of 2024 eggs. She counts the eggs from the first packet and places them in a second packet as she goes. Each time she has counted 4 eggs, she lays an egg which she places in the packet of those she still has to count.

How many eggs will she have to count in total?

[cloud creek]

<@&286206848099549185>

someone srlsy

.close

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[acoustic quartz]

Would i apply log properities here and then int by part?

[jaunty plume]

integrate by parts

directly

use liate

[acoustic quartz]

Would i be able to apply log rules tho?

[humble karma]

It doesn't apply here.

[acoustic quartz]

when does the log rule apply?

[humble karma]

There's no log rules for $ln(x)^n$

[wraith dagger]

Azyrashacorki

[acoustic quartz]

i thought you would bring the n in front if you hadthis

[humble karma]

There is for $ln(x^n)$

[acoustic quartz]

so it would be like n *ln(x)

[wraith dagger]

Azyrashacorki

[jaunty plume]

no its integrals

i dont think u bring out log rules

do you

?

[acoustic quartz]

Oh i understand this now

[acoustic quartz]

No not in the problem i just sent

But if u had like ∫logx^7

Then i assume you would

[jaunty plume]

nah

i think it would always be integration by parts

[humble karma]

If it was say the integral of log(x^7), you could bring it outside and then you'd just have to integrate log(x) which will require IBP.

[acoustic quartz]

whats ibp?

[humble karma]

If the exponent is outside, however, then I guess it would just be IBP (integration by parts) all the way.

[acoustic quartz]

makes sense

Thanks guys

Are these the same?

[wanton sail]

It's a little ambiguous, but yeah ln(x)^4 and ln^4(x) are probably the same

(ln(x))^4 is most clear

ln^4(x) could mean ln applied four times, like ln(ln(ln(ln(x)))), but it probably doesn't here

[cedar kiln]

@acoustic quartz Has your question been resolved?

[cedar kiln]

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[quasi lantern]

If a, b and c are positive real numbers then a/b + c/b + c/a is greater than or equal to.

[quasi lantern]

I am unable to understand how to use A.M. >= G.M, how is it used?

can someone give an example?

[tepid sail]

What’s the problem statement

[quasi lantern]

oops I forgot to give the MCQ options

a) 3
b) 6
c) 27
d) None of the above

[faint dirge]

what method are you trying to use to solve this?

only one I can think of is optimising it as a multivar function, ie finding critical points in quandrant 1, but that's because I've just done too much calc recently

[quasi lantern]

Arithmetic mean >= Geometric mean?

Idk if its >= or >

but that should apply

[tepid sail]

What do you think

[faint dirge]

3 and 3, A.M. is 3, G.M. is 3, so >=, as an example

[tepid sail]

Yes ofc

[quasi lantern]

wait how?

[faint dirge]

ie if your set of numbers are all the same, their arithmetic mean and geometric mean are the same (it's whatever the number is), ergo they are >= (= in this case)

I was using {3, 3} as an example

[quasi lantern]

oh ic

but how are they same in this case?
a/b b/c?

[faint dirge]

hmm ok just tried taking partial derivatives they have no critical points for a,b,c > 0

since the partial by c requires 1/a + 1/b to = 0, which is impossible if neither is negative

yeah idk how to solve this, it's late and it is clear I should not have approached this problem without an idea in mind. Very sorry!

[quasi lantern]

np

<@&286206848099549185>

[sturdy rose]

well

the very first thing you should try doing is making the denominator equal or same

there needs to be some form of symmtery

oh or else

just use AM > GM

[quasi lantern]

ya I thought about that

but confused if >= or >

[sturdy rose]

a/b + c/b + c/a / 3 >= 1

[sturdy rose]

=

[quasi lantern]

why tho?

[sturdy rose]

AM can be equal to GM

[quasi lantern]

cuz I didn't get why they are equal in this case

[sturdy rose]

AM = GM = HM = RMS

[sturdy rose]

there is always an equality wdym

[sturdy rose]

do you know the circle proof?

[quasi lantern]

no? am > gm when both are unequal?

[quasi lantern]

no idk

[sturdy rose]

you cant know that unless your given the numbers

[sturdy rose]

are u preparing for JEE by any chance?

[quasi lantern]

yes. so

[sturdy rose]

Can u hop on VC?

it would be better to explain it there

[quasi lantern]

but isn't it better to assume they are unequal when they are +Ve real numbers?

sure

[sturdy rose]

wait i will explain

give me 10 min and i will dm

[quasi lantern]

k np

[minor crystal]

i mean you have equality when a = b =c

[quasi lantern]

ya and in this case it isn't given to be equal.

so isn't it better to take > instead of >=

[minor crystal]

u can do > yeah but

where does it say they're not equal?

it doesn't say that in your actual question

[quasi lantern]

oh yes

but it doesn't state it's equality either?

[minor crystal]

i'm saying you can have any instances where a = b = c

where a = b = c and are all >= 0

this would work according to your question

then your AM >= GM

[sturdy rose]

i think his question is why they are equal at the first place

[quasi lantern]

but when a b c are real numbers(+ve) then they could be unequal too?

[sturdy rose]

which is why >=

[minor crystal]

yeah and the >

accounts for that

= accounts for when they're all equal

[quasi lantern]

ohh ic

[minor crystal]

btw is your question right?

[quasi lantern]

yes it is.

[minor crystal]

a/b + c/b + c/a ?

[quasi lantern]

yup.

[minor crystal]

the gm isn't so nice?

[quasi lantern]

b/c

wait no it's b/c instead of c/b

[minor crystal]

okay there u go

[quasi lantern]

a/b + b/c + c/a

[sturdy rose]

it is

it becomes one

[minor crystal]

no it doesn't

it does right now

not in his original question

[sturdy rose]

typo

[quasi lantern]

ya mb

[minor crystal]

we can never know if he made a typo but sure lol

[sturdy rose]

he did

[minor crystal]

yeah now that he said it

anyway

[sturdy rose]

its like the very first question everyone solves when ur preparing for a shit exam

[minor crystal]

idk what that shit exam or whatever the context is lol so i just asked

[sturdy rose]

even ask you to memorise it

😭

[minor crystal]

anyway you're done with the problem right?

also why do you have to memorize am-gm?

there's am-gm-hm-qm

[sturdy rose]

i dont

[quasi lantern]

qm?

[sturdy rose]

but its a thing

[minor crystal]

if you're memorizing am-gm then why not memorizing hm-qm and the others too lol

[sturdy rose]

rms

[sturdy rose]

they do all

in a line too

just show them the circle proof man

literally no one does that here and its so frustrating

[minor crystal]

[sturdy rose]

Also, close the channel if ur doubt has been cleared

[quasi lantern]

ya, doing it now

Thanks for the help, both of you!

.close

[cedar kiln]

Channel closed

Closed by @quasi lantern

Use .reopen if this was a mistake.

[cedar kiln]

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[zealous torrent]

What is x in the sample standard deviation formula

[upper garnet]

theyre just your sample values

[zealous torrent]

I know what xbar is

[upper garnet]

53.2, 57.8 etc

[zealous torrent]

but how do I write those down in the formula

Do I add them?

[upper garnet]

do u know what a Sum is

that big sigma

[zealous torrent]

Yeah

yea ik

I can't do sum (x-xbar)² as sum(x)²-sum(xbar)² tho

[upper garnet]

(x_1 - mean) ² + (x_2 - mean)² + etc

[zealous torrent]

yhyh

Idk I haven't done stats in like a year

[upper garnet]

no

[zealous torrent]

Its easy and im bad at it

[upper garnet]

(x+a)² isnt x² + a² to begin with lol

[zealous torrent]

Yeah ik ik I know I can't do that

I know

I said that I can't do that

[upper garnet]

yeah

[upper garnet]

its just this

[zealous torrent]

I swear there's a quicker way to do it

[upper garnet]

no

[zealous torrent]

💀

[neat dune]

I think $\sum x^2 - n\overline{x}^2$ for the numerator is less operations

[tropic oxide]

\overline or \bar

[dark birch]

How do you factor a cubic polynomial who's constant term is very large say 120 without using a calculator

[wraith dagger]

quantifier #2 lover

[zealous torrent]

Fuck

I have an answer

The question is one mark though

I did too much working out

There has to be a faster method

[neat dune]

Ok yeah cus for the first version, assuming sample mean has been calculated, you do n substractions, n squares, and then (n - 1) additions. Second version you do n squares, (n - 1)additions, and then 1 substraction plus 1 multiplication

[zealous torrent]

This seems wrong

Wrong

I'm just horrific at stats honestly I have no clue whats going on most of the time

[neat dune]

So can do about 33% less work i guess

[zealous torrent]

What's the fast method

The question doesn't even say find b

It says "state"

[neat dune]

.

[zealous torrent]

oh

Can u check my answer tho pls

[neat dune]

hmmmm

Try excel

[zealous torrent]

?

[neat dune]

I don't feel like calculating all of that

[zealous torrent]

How do I calculate it with excel tho lol

[neat dune]

https://www.ablebits.com/office-addins-blog/calculate-standard-deviation-excel/#formula-examples-standard-deviation

It has a function for sample standard deviation

[cedar kiln]

@zealous torrent Has your question been resolved?

[cedar kiln]

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[livid swan]

answers are dodgy i think

[livid swan]

apparently its 269.4cm^2

I got 237.3cm^2

[tropic oxide]

!show

[cedar kiln]

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

[livid swan]

ive basically split it up like this

and calculated the triangle, both sectors and the semi circle

and added them

idk what to do, its 'wrong'

the green dot resembles the angle: arccos(1.6/2) which is around 36.87

then I found the area of those sectors by getting π(r)^2 and multiplying it by 36.87/360

(multiplied by two since i'm trying to find two sectors

then i found the big triangle by going 16x6/2 (the 6 is from 10sin(36.87) which is 6)

so I added that to the two sectors

[harsh zephyr]

is 6 the height?

of the triangle

[livid swan]

ye

[harsh zephyr]

ight

[livid swan]

then i added the semi circle which was obviously (π(r)^2)/2

and its still fucking wrong

fuck

oh shit

I re did it and i'm right

idk what changed

.close

[cedar kiln]

Channel closed

Closed by @livid swan

Use .reopen if this was a mistake.

[livid swan]

thanks ig

[cedar kiln]

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[quasi plover]

so the period is basically the 3tan ( deta/2)?

i understand yellow, but for blue and green. all i understand is like. rules of some sort

[cedar kiln]

@quasi plover Has your question been resolved?

[quasi plover]

<@&286206848099549185>

pls teach me how to do this question please

we can start from the question

[tawdry otter]

The period can be acquired by using the general formula in green on the left

T = pi/b

[quasi plover]

so basically what we do is pi/(3tan1/2deta)?

[tawdry otter]

Not like that no, let me explain

[quasi plover]

ok

[tawdry otter]

$f(x)=atan(\theta\times b - h)+k$

[wraith dagger]

Cewkins

[tawdry otter]

@quasi plover This is the formula, i like to write it this way but you will find it written differently sometimes depending on the source/text book

[quasi plover]

;-;

b-h?

[tawdry otter]

Point is, the coefficient that is multiplied by theta 'b' determines the frequency and the period of the function

[tawdry otter]

Over here you have theta/2 or theta * 1/2, which means the value of b is equal to 1/2

Using the formula T = pi/b you get pi/1/2 which equals 2pi

So the period of the function is 2pi

[quasi plover]

does this one that you show, you mean theta divided by b? not theda times b? or it doesnt matter that ever sign it is. aslong as it somehow related to theta

[tawdry otter]

No theta is just multiplied by b whatever the value of b is

theta/2 is the same as theta * 1/2, b = 1/2

b is just whatever is multiplied by theta

[quasi plover]

oh

this helps

what does atan mean?

and r there any more formula that i should know? cuz i see theres h and k variable

[tawdry otter]

a is for "amplitude", though its worth mentioning that tan functions dont have amplitude so it only expresses how stretched the function is

[tawdry otter]

Are you familiar with transformations?

[quasi plover]

i heard the term before form the math class

but it like before christmas, after christmas break nearly everything i learn are like washed away

[tawdry otter]

Okay i will give you a quick recap

A: a is used to express the amplitude of the wave, in sin and cos functions it changes the height and the depth of crests and troughs, in a tan function it either stretches the line to be more straight or bends it to curve like a cubic

B: the b that is multiplied by theta changes the frequency and period(wavelength in physics term)

h is for shifts on the x axis, a positive h will move the graph to the left and a negative h will move it to the right, the reason why they sound inverted is because there is a negative sign right before the h in the formula so whenever h is positive it turns into negative when you plug it in and thus changes the shift

k is for the shift on the y axis, positive k will push the graph above and a negative k will push it down, it can also represent the middle line for trig functions

[quasi plover]

crest and troughs>

i didnt learn this from physic ;-;

i learn this from igcse additional math

[tawdry otter]

Crest and troughs is a valid term for math as well iirc

[quasi plover]

iirc ;-;

what is valid term

why this is like learning not revising

[tawdry otter]

What i meant is that crest and troughs are correct words to be used for math

[quasi plover]

oh sorry

[quasi plover]

wait im feeling familiar with this

[tawdry otter]

I suggest you watch some youtube videos on the topic, messing around in desmos with trig functions will also help you picture these transformations

[quasi plover]

ah ok thank you alot

.close

[cedar kiln]

Channel closed

Closed by @quasi plover

Use .reopen if this was a mistake.

[tawdry otter]

Glad i could help

[cedar kiln]

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[crimson sedge]

im not sure how to complete these and am asking for any help please

[crimson sedge]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[hazy cargo]

number 4

is uh

so u know exterior angles of a triangle right

[crimson sedge]

the goal is to find x

if that helps

[crimson sedge]

like what all the angles add up to

[hazy cargo]

mk so

[hazy cargo]

this should help for question 4

[crimson sedge]

so with that how do i solve (8x+40)

[hazy cargo]

so u know that 8x+40 is the sum of the two opposite interior angles

so each angle must be that divided by two

[crimson sedge]

yea as it shows in the picture

[hazy cargo]

then just take the divided by two angle and multiply by three

[crimson sedge]

so 8x+40/2

[hazy cargo]

and that must equal 180

u try

[crimson sedge]

ok

[hazy cargo]

hows it going

[crimson sedge]

i subbed x for 2 and 3

then btoh times divied by 2 and muliped by 3

but still cant get 1-

180

[hazy cargo]

what

why did u sub x

lmao

[plucky owl]

You don't need to do all that, what kind of triangle is given?

[crimson sedge]

i though that was suppose to do

[crimson sedge]

equalateral

i spelt tat wrong

[plucky owl]

So then each interior angle is what measure?

[hazy cargo]

its just a solve for x

[crimson sedge]

60?

[plucky owl]

That's what I'm doing

[plucky owl]

Yes

[hazy cargo]

im saying this method words if u want to do the point of the exercise

u can skip the question but like

whats the point lol

[crimson sedge]

Which one where you having difficulties with?

4

[plucky owl]

Now you know that angle is 60

That's on a straight line

[crimson sedge]

8x+40=60?

[plucky owl]

What's the measure of a line?

[hazy cargo]

close

[plucky owl]

Not quite

[hazy cargo]

remember the exterior angles diagram

[plucky owl]

That does not help

[hazy cargo]

what

[plucky owl]

.

[crimson sedge]

8x+40 no?

[plucky owl]

No

[crimson sedge]

60?

[plucky owl]

The measure of a straight line

[hazy cargo]

cant u just equate it to 120

doesnt that also work

[crimson sedge]

180?

[plucky owl]

Yes 180 is the measure of a line

[plucky owl]

You get the same thing at the end but your method is basically memorizing the formula

Mine actually goes through how it was came upon

[earnest radish]

maaat

math is bad

[hazy cargo]

???

wrong place to be buddy

[crimson sedge]

whos manz

[plucky owl]

So because the purple and (8 + 40) is on a line that means it sums to 180

[earnest radish]

joke bro

[plucky owl]

Can you write the equation? @crimson sedge

[crimson sedge]

60+(8x+40)=180

[earnest radish]

What do you think about mathematics?

[plucky owl]

Almost

It sums to 180

What does sum mean?

[crimson sedge]

addition

like the 2 added up

that right?

[plucky owl]

Yes that's the proper equation now

[crimson sedge]

okok

[plucky owl]

Can you solve for x now?

[crimson sedge]

10?

[plucky owl]

Yes

[hazy cargo]

mk now the other ones

[hazy cargo]

question 1 everything just adds to 180

[crimson sedge]

is 3 correct

[earnest radish]

I don't understand any of the problems, I'm leaving bye

[hazy cargo]

same for everything else actually

[crimson sedge]

i tried doing that one

[hazy cargo]

bye

[hazy cargo]

no unfortunately

[crimson sedge]

damn

[hazy cargo]

just think abt it

interior angle sum of triangle =?

[crimson sedge]

180

?

[hazy cargo]

and conveniently you have all three angles

so what must they all add up to

[crimson sedge]

so each is 60

[hazy cargo]

huh no

[crimson sedge]

oh no

[hazy cargo]

theyre only 60 for an equilateral

[crimson sedge]

they js all add up to 180

[hazy cargo]

exactly

[crimson sedge]

ohhh

[hazy cargo]

very nice

[crimson sedge]

i see

so one angle is 40

we already know

so the other 2 add up to 140

do i have to sub a number in for x so they both add up to 140?

[hazy cargo]

just solve the equation

do u know how to solve equations

[crimson sedge]

ik how to solve it where there a number in front

and use distributive property

when it js like that nha

[hazy cargo]

so write out for me what you have right now

[crimson sedge]

(4x-5)

(3x+5)

[hazy cargo]

so whats the equation

[crimson sedge]

(4x-5)(3x+5)=140

Yea?

[hazy cargo]

mhm

but dont forget

u need to add everything

if theres no sign that means multiply

[crimson sedge]

ohh ok

[hazy cargo]

so now write the equation again

[crimson sedge]

(4x-5)+(3x+5)=140

i got 20

[hazy cargo]

very good

now the rest of the questions are basically the same

[crimson sedge]

ight il js do em and can you lmk if i got it right or nah?

[hazy cargo]

oh alr

just dm me

[crimson sedge]

also the triangal for number 3 is named isocoleis yea?

[hazy cargo]

i might be asleep

[crimson sedge]

Okay

[hazy cargo]

no-

[crimson sedge]

damn

[hazy cargo]

only isosceles if two sides are the same length

[crimson sedge]

ohh they looked the same to me

😭

[hazy cargo]

it has to specify

[crimson sedge]

number 2 adds up to 180 also yea

[hazy cargo]

every triangle angle does

[crimson sedge]

ohhh

for 2 i got 16

and 1 i got 65

im not sure how to do 5 6 and 7 could u help me w those

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

.close

[cedar kiln]

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[quasi plover]

i already got (x+1)(2x-1)(x-2)

in the markscheme, its that with 5

how do i get 5

[vague rapids]

y intercept is equal to 10

so f(0)=10

[quasi plover]

?

[quasi plover]

what do i do with that information?

oh

i got it

thanks

.close

[cedar kiln]

Channel closed

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[tacit elk]

In the playoffs, the Leafs played 13 games before being eliminated. If they won 3 more games than
they lost, how many games were won and lost? How would I put these into X and Y equations. (System of equations application)

[crimson sedge]

take W to be the number of games won and L to be the number of games lost

[tacit elk]

x + y = 13

so is it 3x + y or 3x - y?

for the second

[crimson sedge]

3x+y ?

what is that

[tacit elk]

if they won 3 more games

so 3x

[crimson sedge]

3 more not 3 times

[tacit elk]

so 3 + x

[crimson sedge]

3 more than what ?

[tacit elk]

lost

[crimson sedge]

so why 3 + x

[tacit elk]

oh

so what would be the question cause im confused

[crimson sedge]

they have won 3 more than lost

so won is 3 more than lost

[tacit elk]

y = 3x?

[crimson sedge]

3 more not 3 times

[tacit elk]

y = 3 + x

[crimson sedge]

and 3 more than lost

not 3 more than win

[tacit elk]

x = 3 - y?

[crimson sedge]

no

[crimson sedge]

that would be x+y = 3

you already have x+y = 13

[tacit elk]

ohh

[crimson sedge]

so thats wrong

[tacit elk]

Thank you!

wait how does that work tho?

[crimson sedge]

you don't have the 2nd equation yet

[tacit elk]

whats the second?

I thought the first was x + y = 13

[crimson sedge]

yeah

and ?

[tacit elk]

then the second one

[crimson sedge]

which ?

[tacit elk]

If they won 3 more games than
they lost

how would that go into a equation?>

[crimson sedge]

won 3 more than lost

x = y+3

[tacit elk]

ohh

ty

.close

[cedar kiln]

Channel closed

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Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[river sphinx]

.close

[cedar kiln]

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[lyric solar]

I need a bit of help proving this

[lyric solar]

I tired doing

R3 <- R3 - R2
R2 <- R2 - R1

or R3 <- R3 - R1
R2 <- R2 - R1

and I got to the same exact point

[cedar kiln]

@lyric solar Has your question been resolved?

[lyric solar]

<@&286206848099549185>

[cedar kiln]

@lyric solar Has your question been resolved?

[royal finch]

I haven't tried this problem but are you sure that's true? It seems like you need an additional condition to get anywhere

[lyric solar]

well that's what our teacher gave us, I got no idea xD

all the exercises I did until now were correctly given so I assume this one is too

there might be a longer chains of linear combinations

or some logic that I can't figure

[royal finch]

If you multiply row 1 by a, row 2 by b and row 3 by c, and factor out abc, you get the right handed side multiplied by abc which suggests this is not true at a glance

I would at least do a dirty sanity check and make up 3 numbers for a b c such that abc does does not equal 1 and throw it into any calculator to make sure

[lyric solar]

let me see

[royal finch]

oh wait, no that works. It's too early in the morning

Those multiplication steps will cancel when you factor

[lyric solar]

2 = 2

apprently it is true

and I have 5 more worse than this as homework

[cedar kiln]

@lyric solar Has your question been resolved?

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[crimson sedge]

When you define a function, how do you know which properties you need to prove that satisfy well defineness

[crystal raptor]

Really only two things:
Does the function map things to where you say it does?
If a = b, does f(a)=f(b)?

[crimson sedge]

what if it is really obvious, then do you need to prove it?

[eternal ingot]

how obvious are we talkin here

[crimson sedge]

Sometimes in answers, only some parts of well defineness are proved and not all since some are obvious so you can omit the proof. But when is obvious?

[crystal raptor]

depends what the mapping is and who you ask

[crimson sedge]

Also, is proving uniqueness the same as proving injectivity?

[cedar kiln]

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[small gyro]

It is given that f,g are continuous at [0,1] and differentiable at (0,1) with f(0)=f(1)=0 and f(x)≠0 when x∈(0,1)

I showed that the function h(x)=f^2(x)•e^g(x) satisfies the conditions of Rolles theorem and then have to show that a c∈(0,1) exists so that f’(c)/f(c)=-g(c)/2

[small gyro]

I assume that I somehow have to use the rolles theorem again

But I don’t understand how

What i only assume is that f is either + or - because it’s continuous and fx≠0 at the same domain

I don’t know the English term for this but I hope you get it

I cant see how is that relevant tho

[cedar kiln]

@small gyro Has your question been resolved?

[small gyro]

nobody can help me with that😭

[cedar kiln]

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[small gyro]

.close

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[tidal leaf]

so I got to 3 - 2x= x. But i dont know how to solve it from there

[hazy terrace]

Perhaps add 2x to both sides?

[tidal leaf]

to x and 2x?

[hazy terrace]

Yes

[opaque root]

Are you familiar with equations?

[hazy terrace]

I think so, yes

Oh, sorry, you meant mari

@tidal leaf

[tidal leaf]

a little bit

[crimson sedge]

[
3 - 2x = x \
3 = {???}
]

[chrome flax]

7x

[opaque root]

That’s skipping one step

[crimson sedge]

what do you think it should be if you made 3 on the left side?

[opaque root]

That maybe he is not able to see

I would put everything so he can visualize better

[crystal raptor]

so i got to 3 - 2x = x

[crimson sedge]

hint: consider what makes -2x disappear from the left, what do you add with to remove -2x?

[tidal leaf]

2?

[crimson sedge]

-2x + 2 = 0?

is that true do you think

[tidal leaf]

yes

[crimson sedge]

well ill tell you that you are wrong

[tidal leaf]

oh

[crimson sedge]

[
ax + bx = (a+b)x
]
this is called adding "like terms"

[crimson sedge]

meaning like if you have two terms with like both sharing the same variable

then you can just add the coefficients

does thst make sense?

[tidal leaf]

a little bit

[crimson sedge]

like for example

2x + 2x = (2+2)x = 4x

4y + 6y = (4+6)y = 10y

this is how its done

as an example

what would
3z + 4z
be?

[tidal leaf]

7z?

[crimson sedge]

yeah

amazing

ok lets step up the game and involve minus

so

3x - x = (-1 + 3)x = 2x

wherever you just see a - with NO coefficient you can think of it as -1

another example,

2y - 5y = (2-5)y = -3y

does this make sense?

[tidal leaf]

a little bit

[crimson sedge]

ok here is an exercise

how do you do
x - 2x

remember the process you can just totally ignore the x

focus on the coefficients

[tidal leaf]

would it be -x?

[crimson sedge]

yepp

great

okay two more just to solidify your understanding

13x + 6x - 8x

this is a combination

can you do it

remember the order doesnt matter

do it however as long as u get it right

[tidal leaf]

11x?

[crimson sedge]

yes! good job

okay and the last one

3x - 3x

what is this

[tidal leaf]

0?

[crimson sedge]

yeah

good job you have improved

[crimson sedge]

do you see why you were wrong here now

take it as a rule. If they dont share the same variable and degree then you cant add them

if it is x^2 + 3x you cant add them

if it is x + y you cant add them

id it is 3 - x you cant subtract/add them

if it is 3x + 4x or such then You CAN add them

does all of this make sense

[tidal leaf]

yes

[crimson sedge]

okay great

so back to what we had

[
3 - 2x = x \
3 = {???}
]

[crimson sedge]

we want to add something to both sides

but we want to get rid of -2x

what do you add to -2x to make it 0?

[tidal leaf]

x?

[crimson sedge]

remember what we have been saying

[tidal leaf]

can we not do it?

[crimson sedge]

-2x + ??? = 0

like

if you had x there

it'd be -2x + x = (-2+1)x = -x

so, not zero.

so again, what must it be

[tidal leaf]

3x?

[crimson sedge]

read what we said here

again

slowly

[tidal leaf]

2?

[crimson sedge]

no

imma give you time to think about it because i dont think anything else will help you otherwise

[tidal leaf]

i don't know, im sorry

[crimson sedge]

you said it yourself here