help-23

[desert pasture]

yes?

[brazen parrot]

yes

[desert pasture]

now let 3x=u

so you have f(u)

now for 3x=-6

what should x be

[brazen parrot]

-2

[desert pasture]

cool

does it make more sense now?

[brazen parrot]

no what do i put on ma graph

oh

-2

-2 is x

[desert pasture]

yes

[brazen parrot]

how about the y intercept

the same because trheres no +?

[desert pasture]

you know f(3)

yes

now use the same logic

[brazen parrot]

when x=-2 y=-3?

[desert pasture]

y=3

not -3

[brazen parrot]

oh

y=3

[desert pasture]

yeah

[brazen parrot]

how come in the graph its -3

-2,-3

[desert pasture]

where?

Mind taking a screenshot of that portion?

[desert pasture]

oh, it starts at -2,-3

as f(-6)=-3

makes sense ?

[split fulcrum]

Electric fields and circuits when

[brazen parrot]

uh

if i get that in my exam im skippign it

its 1 mark

[desert pasture]

you both are writing the same exam?

good luck !

[brazen parrot]

wdym

noo hes 1 year ahead of me

[desert pasture]

anyway, do you understand this?

[brazen parrot]

guys how come i got this wrong

thats the answer

[desert pasture]

which part?

[brazen parrot]

part C

[desert pasture]

,w slope of x/5+y/3=1

[flat frigate]

Results provided by WolframAlpha

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

Requested by physicsrocks

[desert pasture]

so the other line should have a slope 5/3

if two lines are perpendicualr $m_1 m_2=-1$

[flat frigate]

ƒ(Why am. I here)=I don't Know

[brazen parrot]

oh u are rite

wait now i got dis

[desert pasture]

instead of this

use the slope-point form of a line

[brazen parrot]

slope form?

[desert pasture]

$\frac{\left(y-y_1\right)}{x-x_1}=m$

[flat frigate]

ƒ(Why am. I here)=I don't Know

[brazen parrot]

oh but dis is easier

[desert pasture]

maybe

now you need help with part (d)?

[brazen parrot]

nope i got it

I AM A MATHEMATICAL GENIUS

[desert pasture]

👍

[brazen parrot]

wats da difference

between inverse proprtional and directly proportional

[desert pasture]

if a is directly proprtional to b a=kb

if a is inversely proportional to b ab=k

[brazen parrot]

;-;

[desert pasture]

if I remember right

[brazen parrot]

wut

[desert pasture]

https://www.ncert.nic.in/pdf/publication/exemplarproblem/classVIII/mathematics/heep210.pdf

this may help

[brazen parrot]

?

[desert pasture]

looks right

[brazen parrot]

ohh and then i calculate the constant

[desert pasture]

yes

[brazen parrot]

and the. I can use it for the next one

OK MAKES SENSE

and the other one is k/x

[desert pasture]

yes, I think so

[split fulcrum]

Aren’t u like 10th

[brazen parrot]

ye

but im doing 11 methods

[split fulcrum]

still 2 years above then

[brazen parrot]

oh

what are u

12?

[split fulcrum]

yes

[brazen parrot]

sry cant math

IM DOING 11 SPECS TOO

probs gonna drop

and do it next year

[split fulcrum]

Physicsssssss

[brazen parrot]

yo...

im doing 11 physics too

im doing 11 english atar and 10 literature

[desert pasture]

physics is awesome

but I digress

[brazen parrot]

same i love physics

it is very easy just plug into multiple equations

dont need to study it a lit

lot

[split fulcrum]

what even are those 😭

[brazen parrot]

like 1 day before the test 💀💀

[desert pasture]

that's now how physics works as it gets more advanced though

[brazen parrot]

ummm so one is analysing and one is legit.... comprehension composing and stuff 😔😔

[desert pasture]

you'll need to derive stuff on the spot

or atleast you do here

sorry

[brazen parrot]

oh right now its free

im like 2nd in my class and i legit skipped year 10 science

[split fulcrum]

bro just call it english or be cool and do extension English

[brazen parrot]

NO CUZ ITS LEGIT.

literature

poems and stuff idk

[brazen parrot]

it makes more sense to call it literature cuz it is literature 😭😭

[split fulcrum]

bro literature IS English

💀

[brazen parrot]

NO BUT U DONT LEARN LITERATURE IN ENG

ITS DIFF IM TELLING U

one is english language

one is english literature

[split fulcrum]

go back and do ur nuclear physics

muahaha

[brazen parrot]

no.

[hazy elbow]

bum chicken

[brazen parrot]

Hi beard

um whyd i get a domain error

[split fulcrum]

Cheater

imagine having a graphing calc

[brazen parrot]

wtf

do u not have one in hsc

ive had one since year 9

[split fulcrum]

no…?

[brazen parrot]

they have in VCE and WACE and QCE and SACR

[split fulcrum]

That’s cheating if you use a graphing calculator here

Makes everything too easy

[brazen parrot]

how would u do this

[desert pasture]

do you know calc?

[split fulcrum]

With calculus

[brazen parrot]

no.

[desert pasture]

then it's going to be hard

[brazen parrot]

no i just use my graphing calculator

its to easy

[desert pasture]

you'll need to find the turning points

[split fulcrum]

noob stop using that thing

[brazen parrot]

UR SUPPOSED TO USE IT

[split fulcrum]

Booooo

[brazen parrot]

nsw sucks ive been doing these qs since yr 9 😡😡

WITHOUT CALCULUS

[split fulcrum]

This is why nsw education is superior

Do things the harder way and gain more knowledge smh might as well have desmos in the exam

[brazen parrot]

true my predicted is 99.5 and i legit study 1 day before a test

[split fulcrum]

bro when do u start calc

Hurry up

[brazen parrot]

term 3

next term

but he said only the basics

[split fulcrum]

Wow

Too long

[brazen parrot]

WHEN DOCU START IT IN NSW

[split fulcrum]

It starts in eleventh grade term 2 where I am at

[brazen parrot]

breh 1 term

ok and u guys do matrices in uni

[split fulcrum]

shut up

[brazen parrot]

YK OUR GENERAL KIDS DO MATRICES

k im not gonna say anything else cuz ive seen 4u math

[split fulcrum]

yeah don’t make me pull up the ext 2 syllabus content

[brazen parrot]

😭😭

rmb when they thought it was uni work 😭😭

[split fulcrum]

Actually we do a bit for 3d vectors

[hazy elbow]

🔠

⚔️

[safe radish]

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[knotty perch]

wait i need to confirm if my teacher made a mistake or not

so I think it should be 2x^2 + 4x-5x^2 +5 instead?

am i trippin?

please somebody

[neat fable]

yes it should be +5 at the end

[knotty perch]

yup

symbolab agreed with me too

thx!

.close

[safe radish]

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[safe radish]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[austere brook]

what test for iii?

[austere brook]

ratio?

[hushed tendon]

u can just view it asymptotically

as n approaches inf, the +2 becomes negligible

so its essentially √n - √n

[austere brook]

so 0 so it converges

[hushed tendon]

yes

[austere brook]

how do i do this

i simplified 3^n+1/6^n to 3/2^n

nvm ic an do this

just realised its geometric

[safe radish]

@austere brook Has your question been resolved?

[austere brook]

.close

[safe radish]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[grand scaffold]

What is the solution for this?

[grand scaffold]

(x
3x
0)

?

[safe radish]

@grand scaffold Has your question been resolved?

[tame oak]

it's just the zero vector

Whoops my bad

Yep that's right

Not quite, you should write it as $$ B(Ker(A)) = [\colvec{1}{0}{0}, \colvec{0}{3}{0}] $$

Apologies, B(Ker(A)) = {[1,0,0], [0,3,0]}

[grand scaffold]

Basis of N(A) is
(1
3
0)

Thats right?

[tame oak]

It's actually got 2 basis vectors

[grand scaffold]

Its 3x, not 3y to have 2

[tame oak]

That's a linear combination of the basis vectors

The basis is the set of linearly independent vectors that span that entire vector space

[tame oak]

What you wrote here is just a vector that's in the null space, not the basis

The basis is a set of vectors

A good example would be the bases for R^n

For R^2 a basis would be something like {[1,0],[0,1]}

In R^3, {[1,0,0],[0,1,0],[0,0,1]}

And so on

@grand scaffold ping me if you have more questions

[grand scaffold]

@tame oak from here we get
From Ax=O we get
x -3y = 0
So the solution are infinite in form of (x,3x,0)
So a basis of N(A) is <(1,3,0)>
Is that right?

[tame oak]

no! the basis is a set

[tame oak]

This is your basis

Read through the examples I gave you for R^2 and R^3

You solved the equations correctly but you forgot the fact that the basis is a set of vectors

[safe radish]

@grand scaffold Has your question been resolved?

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scarlet flower]

can anyone help

[obtuse plover]

what have your tried

[scarlet flower]

nothing idk how to start

[sterile cloak]

Do you know greatest common denominators?

[scarlet flower]

no

[spice wing]

@obtuse plover Sir Stephen , such a complex problem is unsolvable for commoners

[sterile cloak]

Ok so you need to multiply each fraction until the denominators are the same.

Then you can move one to the other side of the equation so you can subtract them

[obtuse plover]

Sir what

[spice wing]

Stephen ?

[obtuse plover]

who is commoner

[spice wing]

I am , Ángel is

[scarlet flower]

yes

so wait

i got the solution from a another task

[spice wing]

Very Good

[scarlet flower]

but

wait

idk what i should do

how to start

[fast gazelle]

well it's still the same basic idea

you want to get rid of the fractions by multiplying by things

so for instance if you multiply both sides by 4

$4 \cdot \frac{8+17a}{10} = 4 \cdot \frac{6a-3}4$

[flat frigate]

bee [it/its]

[fast gazelle]

now the two 4s on the right-hand side cancel

$4 \cdot \frac{8+17a}{10} = 6a-3$

[flat frigate]

bee [it/its]

[fast gazelle]

then if we multiply both sides by 10

$4 \cdot 10 \cdot \frac{8+17a}{10} = 10(6a-3)$

[flat frigate]

bee [it/its]

[fast gazelle]

and the 10s cancel

and then it's just $4(8+17a) = 10(6a-3)$

[flat frigate]

bee [it/its]

[scarlet flower]

and then?

i just dont understand it its so hard

i did the beggin in school i wrote

8+17a
_____ x 10 x 4 = 6a - 3 x 10 x 4
10 -------
4

wait i will just send a pic

[sterile cloak]

It’s easier if you cross multiply

[scarlet flower]

help me with it

pls

[sterile cloak]

You take the numerator of the left fraction, multiply it by the denominator of the right. And you take the numerator of the right, multiply, by the left

[scarlet flower]

can you write it pls like this

[sterile cloak]

And you bring it all down

Ok

One second I write on paper

[scarlet flower]

its easier to understand my englishj isnt that good

[scarlet flower]

thanks thats way better

[sterile cloak]

Then you distribute

[scarlet flower]

wait so

check if i have it right i will write it in here

[sterile cloak]

Ok

[scarlet flower]

i did it wrong omd

i think its wrong

its not the solution

i think it needs to be a = ...

[sterile cloak]

When you distribute you need to multiply EVERYTHING in the bracket

The arrows mean to multiply

[scarlet flower]

ahhh

okay wait let me finish it

[sterile cloak]

And then you can just isolate a

Ok

[scarlet flower]

i need help

so i want to do

-32 to remove it from the left side

and i did on my calculator 32- (-30) = 62

is it right?

[sterile cloak]

Yes

That’s right

[sterile cloak]

Yes!

That’s the answer

[scarlet flower]

finally

ty

[sterile cloak]

Yea

[scarlet flower]

but wait i got moire

more

[sterile cloak]

Some people complicate it too much

[scarlet flower]

thats so easy

but i need help here now

[sterile cloak]

Ok for this one you need to find the least common multiple for all of them

Then you can bring them all down and do what we did before

[scarlet flower]

can you write it on a paper?

[sterile cloak]

Ok

[scarlet flower]

i wrote in school this:

[sterile cloak]

What was your answer?

[scarlet flower]

nothing

i didnt finish it

[sterile cloak]

Oh

[scarlet flower]

but thats how my teacher wrote it

[sterile cloak]

I don’t know what method that is, I’ve never used it. But as long as they both get the same answer, use what you know is easier.

But the way I was taught, you need to make the bottom of all the fractions the same, so that you can bring it all down and just isolate

Just use whatever you’re more comfortable with

If you want to know more about this method you can look up “least common multiples for fractions”

[scarlet flower]

ty so much

i need to learn more tommorow is my test but my teachter doesnt answer me

i will be back in like 20 min

[sterile cloak]

Ok

[safe radish]

@scarlet flower Has your question been resolved?

[scarlet flower]

okay i am back

yk how to use log?

@sterile cloak

[sterile cloak]

i did them a while ago, i dont remember much from logarithms. I can try to help though.

if not jsut ask someone else

[sterile cloak]

is this logarithms or finding x?

[scarlet flower]

yes

[sterile cloak]

logarithms?

[scarlet flower]

i think so

should be

its like Kn Ko P n

idk

ah yes it is

i know this one d

but not c

[sterile cloak]

ohh

you need to find exponential form of the logarithm?

[scarlet flower]

2. is we need to solve die follow equation after x

[scarlet flower]

3 Calculate the value for x in the following equations

[sterile cloak]

like this?

[scarlet flower]

wait i will check

[sterile cloak]

i dont remember logs very well

[scarlet flower]

to be honest

idk what i need to do

ah wait

expo9nential form

[sterile cloak]

which question do you need?

[sterile cloak]

this one right>

[scarlet flower]

both of them

[sterile cloak]

ok

to get the log form you just use the formula

[scarlet flower]

idk what the formula is

ah

[sterile cloak]

this is it

[scarlet flower]

what the hell

[sterile cloak]

i write it out

I don’t know if that’s what you’re looking for or what

[scarlet flower]

my teacher didnt send it

wait i got some more stuff we can skip log for now

[sterile cloak]

ok

[scarlet flower]

yk y=3x - 15x -42

[sterile cloak]

y=3x - 15x -42

[scarlet flower]

yes

[sterile cloak]

is it quadratics?

[scarlet flower]

yes

[sterile cloak]

3x^2-15x-42 ?

[scarlet flower]

yes

[sterile cloak]

what do you need to do?

simplify or convert to another form?

[scarlet flower]

idk there just stands

where does it cut

wait look

or wait yk what

yk how to find x1 and x2?

13,5x2 + 10,8 - 153,9=0

abc fomula or something like that

thats how its called in germany

[sterile cloak]

oh yea

thats just to find the x interceots

you need to know the quadratic formula too

[scarlet flower]

idk how

forgot it

i think i got a A on this but i forgot everybtikg

[sterile cloak]

you just take the coefficient of each term and you plug it into the bottom formula

[scarlet flower]

i rememver a little bit

just the beginning

wait i will try

ís on the bottom one a and c x

oder normal numbers without x

[sterile cloak]

yes

just coefficients

no x

[scarlet flower]

so 13 x 153,9

wait i will show you

is i right?

[sterile cloak]

yes

then you just simplify and add eveyrthing

then after that you break the equation into 2

one with + and one with -

the answers to those are your x1 and x2

this is an example i found

[scarlet flower]

how did he get the 1:2

i got -0,5

[sterile cloak]

how he did this step?

[scarlet flower]

yes

[sterile cloak]

because both are the same

he simplified

he divided the top and bottom by 2 to get a more simple fraction

[scarlet flower]

is the answer -0,5

?

[sterile cloak]

for his, yes

[scarlet flower]

okay i understand this now

[sterile cloak]

do you understand yours?

[scarlet flower]

its so hard to remember all of it

[sterile cloak]

youll get it

its just about practice

it looks hard but its just 1 formula

[scarlet flower]

we used to have this little book where we had every formula

[sterile cloak]

we have the same thing in canada

we get a paper with every formula we need

[scarlet flower]

i write a test tommorow with a lot of stuff like this +- and the other things we did

[sterile cloak]

goodluck with your test

[scarlet flower]

bro now way

look

what i need to know

[sterile cloak]

the hardest thing on it is probably the logarithsm

[scarlet flower]

we never did this before

in school

wait i can show you everything i need to do

thats the first paper i need to know

wait now the other one

this too

[sterile cloak]

what grade is this lol

[scarlet flower]

11

[sterile cloak]

its all over the place

really??

[scarlet flower]

ah no its 10

but i go to 11 but its 10 grade stuff

or 11

its mixed

this too

but this is so easy

[sterile cloak]

oh i see

when i took gradde 11 math it was much more straight and even

we learnt mostly quadratics

and trigonometry

[scarlet flower]

its 10 and 11 mixed i will have my final exam in 2 weeks and he wants to prepare us with this shit test

[sterile cloak]

is he grading the test?

[scarlet flower]

i started lerning today its so dumb

[sterile cloak]

if its just for practice

[scarlet flower]

he is grading it

[sterile cloak]

oh

[scarlet flower]

but i think i will make it

i think i will get a B

[sterile cloak]

yea

exams are usually easier than tests, they jsut have more questions

[scarlet flower]

i saw a exam and there is nothing hard its so easy

[sterile cloak]

you just need to know the basics really well. almost everything uses them

[scarlet flower]

the exam from the last class that graduated

we will get the formula so its gonna be easy

in the final exam

[sterile cloak]

yea

simplifying is the hardest part

in my opinion

i didnt do very good in simplifying XD

[scarlet flower]

wait i will go get ready for bed i will be back in like 20-30 min and then i need to show you more stuff that i dont understand if you can

[scarlet flower]

XD

[sterile cloak]

i have plans to go out in a little bit. its only 3:30pm where i live

[scarlet flower]

ahh np than i will just look at the solution

[sterile cloak]

so its better if you jsut make a new help channel

[scarlet flower]

yeah

u helped me so much

[sterile cloak]

ok have a goodnight, goodluck with yoru test and exam

[scarlet flower]

ty for helping me and ty for wishing me luck

[sterile cloak]

you helped me too lol, i needed to review this stuff for summer

[scarlet flower]

xd

[sterile cloak]

no worries, anytime

.close

[scarlet flower]

.close

[safe radish]

Channel closed

Closed by @scarlet flower

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[safe radish]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[terse ether]

how does this work

[river field]

This, fellow math nerds, is notation hell

[dark cargo]

307 $\equiv 1$ (mod 102) since 307 = $102\cdot3 + 1$

[flat frigate]

SirGareth

[terse ether]

what the fuck

[lofty jetty]

yeah its binomial expansion (306 + 1)^1001

expand that and only the last summand remains mod 102

why the reactions bru

[junior smelt]

Why would you binomially expand it when you can just-

[dark cargo]

just use the fact that $a \equiv b$ mod m $\implies $a^n \equiv b^n$ (mod m)

[flat frigate]

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

[dark cargo]

fr

[lofty jetty]

its literally the proof of that fact

[terse ether]

how do i get b though

the question is

[terse ether]

am i supposed to find a random num that satisfies this?

[lofty jetty]

bro dont listen to them just look at this

[dark cargo]

$307 $ \equiv 1$ (mod 102) $\implies 307^{1001} \equiv 1^{1001} \equiv 1$ (mod 102)

[terse ether]

ig i meant a*

[flat frigate]

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

[terse ether]

oooh ok i think i got it

damn you broke it

[lofty jetty]

nice proof by repeating the thesis absolute sigma

[dark cargo]

do you proof a theorem everytime you use it

[junior smelt]

Well, not a good one at that - see the previous comment, which literally follows from the fact that "multiplication works" modulo a number

[lofty jetty]

this guy clearly asked for proof because he's clueless

instead you just keep reciting the same thing

help channel guru

[terse ether]

i am not sure if i did that, i was just confused about how the formula works, not to put you down but no one cares about proof

[lofty jetty]

the proof literally explains how formula works sure thing bro dont ever read proofs they're useless

[dark cargo]

an easier proof would be proving if a = b mod m and c = d mod m then ac = bd mod m

[junior smelt]

If you're gonna give proofs, at least give good ones

[dark cargo]

eitherway, unless there are more qs, consider closing the channel

[lofty jetty]

bro is judging proofs by how good they are is this really biggest math discord server

[terse ether]

yeah gimme a minute i am still processing in my head

[dark cargo]

aight sure take your time

[lofty jetty]

i have a question

related to modular equations

can i ask while bro thinks or do I wait for when he closes channel

[dark cargo]

think you should take one of the available help channels.

[lofty jetty]

but this is related

[terse ether]

does this mean anything

[terse ether]

i mean go ahead ig

[dark cargo]

im not sure why they wrote the same statement twice pretty sure you'd be fine having written it just once

[terse ether]

damn, starting to think the proff is making it harder than it should be

[lofty jetty]

i have following equation $10^{23} - 10^j - 12 \equiv 0 \mod 17$ and next line is $1 \equiv 10^j + 12 \mod 17$. I don't get what happens in between

[flat frigate]

ta

[terse ether]

sorry i really cant figure out how the identities for mods work
how did we get 69(nice)

[lofty jetty]

i think your professor is using bold mod as an operator and regular mod as to indicate that the product is in mod 17 group

[river field]

notation. hell. af.

[terse ether]

my head

[terse ether]

what does the latter mean

[lofty jetty]

mod 17 group means that result of any operation on its elements is taken mod 17. so mod 17 group only has 17 elements: (0,1, ..., 16)

its better to open your professor's notes and look for definitions i'd expect them to define the difference between bold mod and regular mod

[terse ether]

i think i will just skip the mod stuff, i have to worry about rsa, eigenvectors and need to figure out the laplace transform in 2 hours for my exam, i am very sorry and thanks for your help so far

.close

[safe radish]

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[lofty jetty]

np

[safe radish]

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[sudden hill]

can someone help me understand the steps when integrating sec(x)tan(x)???

[devout shale]

What's confusing you?

Do you know the derivative of sec(x) with respect to x?

[sudden hill]

no

[devout shale]

Okay, sec(x)=1/cos(x)

[sudden hill]

is it just a thing i need to know

[devout shale]

You can just know it, or find it using the quotient rule

since sec(x)=1/cos(x)

[sudden hill]

yeH i know that

[devout shale]

Mhm so using the quotient rule you should be able to find the derivative

[sudden hill]

what’s quotient rule again

[devout shale]

the derivative of u(x)/v(x) is ....

quotient rule tells you the formula for that

[sudden hill]

probably why i’m struggling bc i don’t know that rule well

[sudden hill]

photo is black for meeee

dunno if bc i’m on my phone

[devout shale]

is this better

[sudden hill]

yeah thank you

i think it’s just the sec x bit

to get v’

that’s mainly confusing me

idk why

[devout shale]

you don't have to solve it like that

find the derivative of sec(x) like I suggested

(using the quotient rule, since sec(x)=1/cos(x))

[sudden hill]

i think i’m just getting confused bc im differentiating but integrating

i just got sec x = secxtanx

but like no

[summer crow]

derivative of sec x is in fact secx tanx

thus what is the integral of secx tanx?

[sudden hill]

but what do i do with it

i’m blank

oh

wait

so how do i show working

idkkkkk

[summer crow]

before you continue your integration by parts, just read the question again, what is it that you're integrating

[sudden hill]

huh

i swear i used to know this know suddenly im just confuzzled

the meanings of stuff aren’t going in

[summer crow]

it's ok, I'll break it down for you: we are integrating secx tanx

that means, find the anti-derivative of secx tanx, i.e. what differentiates to secx tanx?

and what did we find secx differentiates to?

[sudden hill]

sec x tan x

so is it just have to try ?

[summer crow]

yh it's just inspection

like you just kinda have to know secx's derivative and notice it here

[sudden hill]

so quotient rule is just part of differentiating

not integrating

[summer crow]

I'm not sure why quotient rule is relevant here. But yes it is a method for differentiation

[sudden hill]

ok

i used quotient to show derivative that’s why

[summer crow]

oh

you could've just directly write down

$\int \sec(x) \tan(x) \ dx = \sec(x) + c$

[flat frigate]

lgkoo

[summer crow]

by showing that $\frac{d}{dx} \sec(x) = \sec(x) \tan(x)$

[flat frigate]

lgkoo

[sudden hill]

so i just need to remember sec x differentiates to sec x tan x basically

[summer crow]

yup

[sudden hill]

i wrote this down in a lesson

is that a one way thing then

or unrelated completely

[summer crow]

not really related to the thing we have here yh

[summer crow]

not sure what you mean by one way thing

[sudden hill]

idk 😭

so do i use the previous photo only when it’s a question where its integrate tan x + a and it will be sec ^2x +ax +c

and say if i had to integrate sec^2x +a would that get me tan x +c

or not

[summer crow]

you mean *differentiate * (not integrate) tan x + a gives you sec^2x + ax +c?

[sudden hill]

so only differentiating tan x gets me sec^2x

[summer crow]

yes

[sudden hill]

and only integrating sex^2x gets me tan x

[summer crow]

yes (with constants)

[sudden hill]

ok

so sec x tan x is not related because i can’t use that as i can only integrate sec^2x

[summer crow]

yes

[sudden hill]

i think i just keep thinking about them multiplied together

so i am trying to make sec x =u

and have u tan x

but that gets me nowhere

that’s why i’m just being stupid for some reason

not going in my brain 🤪

[summer crow]

it's not being stupid, it's called process of learning

[safe radish]

@sudden hill Has your question been resolved?

[safe radish]

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[light root]

how do horizontal shifts affect a cotangent function and does it also affect the asymptotes

[light root]

if it does how will the points be also affected

?

[safe radish]

@light root Has your question been resolved?

[light root]

<@&286206848099549185>

[safe radish]

@light root Has your question been resolved?

[remote verge]

Shifting the function also shifts the asymptote

For example, if you have $\cot(x-\frac{\pi}{8})$, your asymptotes will also be shifted $\frac{\pi}{8}$ units to the right

[flat frigate]

otheol

[safe radish]

@light root Has your question been resolved?

[light root]

what are some tricks to graphing tan functions

i struggle with where to put the asymptotes

[light root]

<@&286206848099549185>

[devout shale]

that's exactly why quotient rule was relevant, since they did not know d/dx secx = secxtanx I told them to derive it using quotient rule

[safe radish]

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[tropic tundra]

<@&286206848099549185> find the absolute minima and maxima of f(x,y)= 2x^2 - 6y^2 + 6y Domain: (x,y) all reall numbers^2 : x^2 +y^2 Less or equal 16 . Hint find critical points, derivatives, cosine properties.

[safe radish]

Please don't occupy multiple help channels.

[tropic tundra]

find the absolute minima and maxima of f(x,y)= 2x^2 - 6y^2 + 6y Domain: (x,y) all reall numbers^2 : x^2 +y^2 Less or equal 16 . Hint find critical points, derivatives, cosine properties.

[hard crest]

you have multiple help channels open with the same question. i'm going to close the older one.
please don't ping helpers right away.
there is no need to repeat your question multiple times.
show your work, and if possible, explain where you are stuck.

[tropic tundra]

.CLOSE

[delicate patio]

@tropic tundra your question will be answered here

[hard crest]

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

you need to put in some effort here.

[tropic tundra]

WHEN SOLVING USING THE COSINE PROPERTIES

-64 SIN θ^2 + 24 SINθ + 32 = 0

FROM THERE

CAN YOU HELP ME?

??

[safe radish]

@tropic tundra Has your question been resolved?

[tropic tundra]

NO

find the absolute minima and maxima of f(x,y)= 2x^2 - 6y^2 + 6y Domain: (x,y) all reall numbers^2 : x^2 +y^2 Less or equal 16 . Hint find critical points, derivatives, cosine properties.

[safe radish]

@tropic tundra Has your question been resolved?

[tropic tundra]

X

NO

❌

[safe radish]

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[junior pebble]

I don't know where to even begin to solve 😟

[frigid locust]

tan(A-B) formula

and then use a triangle to solve further

[junior pebble]

I'm not sure how to even set that up

Would it be Tan(sin(1/2))-cos(5/13)) over 1+tan(sin(1/2)cos(5/13)?

[frigid locust]

sin^-1 and cos^-1

$$ \frac{\tan(\sin^{-1}(1/2))-\tan(\cos^{-1}(5/13)))}{1+\tan(\sin^{-1}(1/2)\tan(\cos^{-1}(5/13))}$$

it would be something like this

[flat frigate]

JustToPro

$$ \frac{\tan(\sin^{-1}(1/2))-\tan(\cos^{-1}(5/13)))}{1+\tan(\sin^{-1}(1/2)\tan(\cos^{-1}(5/13))}$$
```Compilation error:```! Argument of \trigbraces  has an extra }.
<inserted text> 
                \par 
l.49 ...+\tan(\sin^{-1}(1/2)\tan(\cos^{-1}(5/13))}
                                                  $$
I've run across a `}' that doesn't seem to match anything.
For example, `\def\a#1{...}' and `\a}' would produce
this error. If you simply proceed now, the `\par' that
I've just inserted will cause me to report a runaway
argument that might be the root of the problem. But if
your `}' was spurious, just type `2' and it will go away.```

[junior pebble]

Like this?

[frigid locust]

yes like that

[junior pebble]

Okay and then can I cross cancel?

[frigid locust]

u cant

[junior pebble]

Crap of course not

[frigid locust]

uh

if u are given that question

do u know how to solve
$$\tan(\sin^{-1}(\frac{-1}{2}))$$

[flat frigate]

JustToPro

[junior pebble]

Not really I'm not doing well in this class at all

[frigid locust]

k

the space provided is extremely small for this question

or maybe they want us to do another way

[junior pebble]

I could use a piece of scratch paper I believe

[frigid locust]

ok ig then we can do it this way

so first take all thats inside as y (this will be helpful)

$$y = \sin^{-1}(\frac{-1}{2})$$

[flat frigate]

JustToPro

[frigid locust]

applying sin on both sides we get

$$ \sin y = \frac{-1}{2}$$

[flat frigate]

JustToPro

[junior pebble]

okay following

[frigid locust]

$$\sin = \frac{Perpendicular}{Hypothenuse}$$

[flat frigate]

JustToPro

[frigid locust]

Use Pythagoras to find the Base of the triangle

[junior pebble]

-sqrt3?

[frigid locust]

$$(perpendicular)^2 + (Base)^2 = (Hypothenuse)^2$$
$$\implies (-1)^2 + (B)^2 = 2^2$$
$$ 1 + b^2 = 4$$
$$ b = \sqrt{3}$$

[flat frigate]

JustToPro

[frigid locust]

how u getting -sqrt3?

[junior pebble]

oh okay not negative

[frigid locust]

its basic algebra :/

do u know what tan is equal to?

[junior pebble]

opposite over adjacent

or sin over cos

[frigid locust]

oh u learnt those terms

[junior pebble]

yeah I know sohcahtoa

[frigid locust]

ah sorry im using wrong , mightve confused u

opposite = perpendicular
adjacent = base
hypothenuse = hypothenuse

my bad

[junior pebble]

It's okay, I'm already confused haha

[frigid locust]

let me know which step confuses u

[junior pebble]

Okay following still

[frigid locust]

yeah

we have opposite and we found adjacent

so for this one , tan become
$$\tan y = \frac{-1}{\sqrt{3}}$$

[flat frigate]

JustToPro

[junior pebble]

okay makes sense

[frigid locust]

As
$$y = \sin^{-1}(\frac{-1}{2})$$
and our question was
$$\tan(\sin^{-1}(\frac{-1}{2}))$$
Using the substitution
$$\tan(y)$$

[flat frigate]

JustToPro

[frigid locust]

and we just found that $$\tan y = \frac{-1}{\sqrt{3}}$$

[flat frigate]

JustToPro

[frigid locust]

so that means

$$\tan(\sin^{-1}(\frac{-1}{2})) = \tan y = \frac{-1}{\sqrt{3}}$$
$$\implies \tan(\sin^{-1}(\frac{-1}{2})) = \frac{-1}{\sqrt{3}}$$

[flat frigate]

JustToPro

[junior pebble]

okay

[frigid locust]

u can use a calculator to confirm the answer , it will be the same (if u want to verify)

[junior pebble]

It makes sense, I am still following

[frigid locust]

do the same with cos^-1(5/13)

use both values , simplify and u get ur answer

[junior pebble]

okay I will give it a shot one moment

[frigid locust]

maybe use another variable like x for cos^-1(5/13)

so u dont get confused

[junior pebble]

would it be -cos or no because that's just part of the equation

[frigid locust]

just find for cos

cuz there is a + cos in the bottom

[junior pebble]

is cos just 5/13?

[frigid locust]

?

yeah cos (x) is just 5/13

do u even know what to find?

[junior pebble]

Doesn't that contradict our original triangle? Or would we use 2 seperate triangles here

[frigid locust]

we using 2 seperate triangles

[junior pebble]

ok

[frigid locust]

thats why i said use 2 viarable

[junior pebble]

gotcha

[frigid locust]

.

cuz it will contradict and u will be confused and stuff

[junior pebble]

ok i think I know how to get to the next step one sec

a^2+5^2=13^2?

[frigid locust]

yep

[junior pebble]

I'm getting 13.93

for a

[frigid locust]

hm?

try again

a is not 13.98

[junior pebble]

12?

[frigid locust]

ye

[junior pebble]

so tan (x)=12/13

[frigid locust]

yeah

[junior pebble]

awesome so I then do $$\tan(-1/sqrt3)-tan(12/13)/1+tan(-1sqrt3)*tan(12/13) ?

$$\tan(-1/sqrt3)-tan(12/13)/1+tan(-1sqrt3)*tan(12/13) ?

[frigid locust]

$$ at the end

also no

[junior pebble]

$$\tan(-1/sqrt3)-tan(12/13)/1+tan(-1sqrt3)*tan(12/13) ?$$

[flat frigate]

𝕿𝖜𝖎𝖘𝖙𝖊𝖉

[junior pebble]

hmm

[frigid locust]

there is no tan

[junior pebble]

really? that's confusing

Ohhh I get it

Because tan(x)= that

[frigid locust]

yes

$$\frac{(\frac{-1}{\sqrt3})-(\frac{12}{13)}}{1+(\frac{-1}{\sqrt3})*(\frac{12}{13})} ?$$

[flat frigate]

JustToPro

[frigid locust]

this is what u get , yeah?

[junior pebble]

yes exactly

[frigid locust]

solve it a bit and thats ur answer

do not use a calculator to bring the answer into decimals

keep in fractional form

[junior pebble]

This is where I struggle the most

[frigid locust]

why?

$${(\frac{-1}{\sqrt{3})-(\frac{12}{13}) \div (1+(\frac{-1}{\sqrt3})*(\frac{12}{13})) ?$$

that is basically this

[junior pebble]

yes it's the square roots that throw me off

[frigid locust]

oh , u will get use to it

ping me if need more help

other wise ".close" if ur question is resolved :)

[junior pebble]

yeah I kinda need help already XD I'm not sure how to solve that equation

I'm really bad at math lol

[frigid locust]

k

[junior pebble]

sorry, I appreciate your help a lot

[frigid locust]

lets first seperate those fractions for easier understanding

[junior pebble]

okay

[frigid locust]

$$\big{(}(\frac{-1}{\sqrt3})-(\frac{12}{13})\big{)} \div \big{(}1+(\frac{-1}{\sqrt3})*(\frac{12}{13})\big{)} $$

[flat frigate]

JustToPro

[frigid locust]

yeah , so we have this

lets start with the top (first 2 fractions)

[junior pebble]

ok

[frigid locust]

do u know how to take sum/subtraction of fractions?

[junior pebble]

I do but it confuses me when there is a sqrt in there

[frigid locust]

keep it the saem

$$\big{(}(\frac{-1}{\sqrt3})-(\frac{12}{13})\big{)}$$

is equal to

[flat frigate]

JustToPro

[frigid locust]

is equal to

$$\big{(}\frac{-13-12\sqrt{3}}{13\sqrt{3}}\big{)}$$

[flat frigate]

JustToPro

[frigid locust]

yeah that

now the bottom side

$$\big{(}1+(\frac{-1}{\sqrt3})*(\frac{12}{13})\big{)} $$

[junior pebble]

I see because we need the same denominator so we have to multiply -1 by 13

[frigid locust]

yes

[flat frigate]

JustToPro

[frigid locust]

first we multiply

$$\big{(}1+(\frac{-12}{13\sqrt{3}}\big{)} $$

[junior pebble]

so -12/13sqrt3 right

[frigid locust]

yes

[flat frigate]

JustToPro

[frigid locust]

now we do the samething as we did before

[junior pebble]

1/13sqrt3

so it becomes -11/13sqrt3?

[frigid locust]

$$\big{(}\frac{13\sqrt{3}-12}{13\sqrt{3}}\big{)} $$

[flat frigate]

JustToPro

[frigid locust]

no?

[junior pebble]

ohhhh i see

[frigid locust]

multiply and divide the 1 with 13sqrt3

[junior pebble]

oops

[frigid locust]

to make the denominator same

u cant just make the denominator 13sqrt3 magically

combining both we get

$$\big{(}\frac{-13-12\sqrt{3}}{13\sqrt{3}}\big{)} \div \big{(}\frac{13\sqrt{3}-12}{13\sqrt{3}}\big{)} $$

[flat frigate]

JustToPro

[frigid locust]

change the div into multiply

[junior pebble]

so we flip the right one and cross cancel

[frigid locust]

$$\big{(}\frac{-13-12\sqrt{3}}{13\sqrt{3}}\big{)} \big{(}\frac{13\sqrt{3}}{13\sqrt{3}-12}\big{)} $$

yes exactly

[junior pebble]

and then final answer is -13-12sqrt3 over 13sqrt3 -12?

[frigid locust]

yeah

[flat frigate]

JustToPro

[frigid locust]

i think thats the final answer

[junior pebble]

You're amazing

[frigid locust]

do u have any key or final answer written?

[frigid locust]

if any more doubts , u can ask

[junior pebble]

You're a life saver my friend

[frigid locust]

other wise ".close" to close the channel :)

[junior pebble]

Thanks a lot!!!

[frigid locust]

np

[junior pebble]

.close

[safe radish]

Channel closed

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[safe radish]

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[winter quiver]

Can somebody help me with simulations, I just have a few questions about titles for the trial sheets that you make

[winter quiver]

ill send a poicture in a second i just have to download discord on my phone

i can show an example

for example the trial sheet is the second picture but i dint get what i title the boxes

like the blue boxes are the titles

i dont really get where it ssays 90% of cruises bc i dont get how to apply that

like would i calculate the totals of how many passengers got the gift they wanted and then if 90% of my 30 trials got what they wanted then that is correct?

and if they didnt then an assumption/recommendation for the company is that they carry more gifts?

<@&286206848099549185>

also i dont know if this has a diffetrent name in other countries but if it looks similar to anything anyones ever done pls help

rlly need an excellence

[safe radish]

@winter quiver Has your question been resolved?

[winter quiver]

no

hep me

.close

[safe radish]

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[sweet mica]

What is the question?

What have you tried?

What is the question?

That is simply a graph

There's no question there

[iron flare]

This is the equivalent of showing us the number 4 and going 'where do I start'

Do you mean like, the equation of each straight line

[sweet mica]

You have to calculate the equations of the lines to get the slopes

Or simply calculate the slope

[iron flare]

When you say it's a new field do you mean the idea of functions being graphical equations

[sweet mica]

m = (y2-y1) / (x2-x1) that is the slope of a line

[iron flare]

Or what h'(x) means etc

[sweet mica]

For each line segment, take two points (x1, y1), (x2, y2)

Of course the slope is constant, it is the tangent of the angle with OY. Have you studied anything of derivatives or slope before trying to solve exercises?

[iron flare]

h'(3) means the derivative of function h at x = 3, because you don't have a fully defined function that's why we're asking you to start finding gradients

nothing of slope?

im a little confused

do you mean you haven't seen the term slope/gradient at all

in class i mean

fwiw i don't think this is 9th grade material

i was actually thinking 9th grade isn't too far off when people learn graphs but learning basic calc in 9th grade seems a little weird but if this is classwork I'll see what I can explain

have you seen graphs in maths much at all

im not trying to; but i do need to understand roughly where you're at

i was going to explain any fundamentals you need for this that you don't have

f(x) and g(x) are functions meaning that when we put an input X in each of them we will get an output, the output is shown on the graph in the question

you can see f(x) crosses the point (-4,2)

as coordinates are (x,y) this means if you input -4 into f you get 2

f(-4)=2

you can follow the line to at least make a very good estimate as to what each output is

h(x) is defined as those two functions added together

so for example if we went to -4 again we see f(-4) is 2 and g(-4) is 4

so h(-4) = f(-4) + g(-4) = 2+4=6

does this all make some kinda sense so far

Something else to note is how steep the functions are at different points

These are all straight lines which makes it easier to see

Yep, we can describe it with the slope

A slope of 0 means it's flat, greater than 0 means it goes up, less than 0 means it goes down

when there is an apostrophe in a function like there is in h'(x) you can while beginning this think of it as "the slope of h(x)"

or rather because it is h'(3), the slope of h(x) at X is 3

h(x) = f(x) + g(x)
and i can tell you that it is also fine to say
h'(x) = f'(x) + g'(x)

so let's find f'(3) and g'(3) so we can find h'(3)

f(3) is the easier of the two since we see it is on the flat line at the top

[iron flare]

so using what i said here what is the slope

yes, so f'(3) is 0

g'(3) is not flat so we are required to use the equation for slope

slope = change in y / change in x
or in coordinates, given two points (X1, y1) (X2, y2) on the line, y1-y2/x1-x2

looks a little weird but it's not so bad

first try and find two exact coordinates on that part of the line

No

They are "coordinate 1's X and Y" and "coordinate 2's X and Y"

For example

(3,5) and (6,7)

X2=6

Y1=5

etc

Take a look at the part of the line g(3) is on and see if you can notice two exact coordinates it passes through