help-39

[gleaming musk]

[-7,9]

if you are not very convinced, draw our the vector and shift it to fit to see if it joints the tip of v and w

i dont use ipad but im assuming thats possible to do without much trouble

[ruby wadi]

But computing if I just graphed v-w, from connecting v to w, as you said. Id get a totally different vector point to a different column vector no?

[gleaming musk]

wdym

u would get the same vector

shift the vector u graphed from joining w to v to the origin

and u woudl realise its [-7,9]

well ur w vector is wrong

on the diagram

[ruby wadi]

Oh, alright. Yeah I think I see it now. Thank you.

[pearl pond]

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[gaunt kite]

This is a pretty straightforward question: Do you calculate magnitude with your calculator in radians or degrees

[heady finch]

depends on your calculator settings

I believe for normal calculators (non scientific/graphic) they default to degrees

but otherwise you should be able to change to either

[gaunt kite]

no ik but like which are you supposed to use because they're gonna change the answers you receive

like which am i supposed to use

for magnitude

[heady finch]

it won't change the answer you recieve

[gaunt kite]

yes

[heady finch]

radians and degrees are just different forms of the same system

[gaunt kite]

it did

[heady finch]

it depends what the question specifies

[heady finch]

then you made a mistake

or

[gaunt kite]

how tho

[heady finch]

your calculator is in degrees/radians

[sharp vigil]

you really should post the specific question you are doing and the context

[heady finch]

you should determine which to use based on your calculator

[gleaming musk]

it hsouldn't result in an incorrect answer

[heady finch]

^

[gleaming musk]

u probably unknowingly switched inbetween the 2 at some point

[gaunt kite]

this is the Q

[gleaming musk]

without realising

[gaunt kite]

lemme post my work

[heady finch]

the question says degrees

so if your calculator is in radians

you use degrees

while if your calculator is in radians

you should convert degrees to radians

and then perform the same calculations

[gaunt kite]

Ok but I did the exact same thing in the other mode and I still didn’t get an of the choices there

The written work is when my calc was in radians

I redid it with my calc in degrees and I still didn’t get one of those options

[sharp vigil]

rewrite your work using degrees mode and then maybe we can see the problem

[gleaming musk]

wait

[gleaming musk]

how do u have 2 different

Ax and Ay value

[gaunt kite]

That’s how I was taught to do it…?

Wait what

[gleaming musk]

u should break it down yes

[gaunt kite]

Wdym

I only have one

[gleaming musk]

but how are the values different

they should be the same value

[gaunt kite]

They are what

[gleaming musk]

27.68 and 44.84?

[gaunt kite]

Am I dumb

[gaunt kite]

Oh

Idk

[gleaming musk]

Ax should equal to Ay

[gaunt kite]

I’ve never seen that in examples in my book

[gleaming musk]

because its at 45 degrees

[gaunt kite]

Oh

Uh

Lemme rework it hold on

Gimme a sec

Does this look right

Yall

[timber stirrup]

itsright

[gaunt kite]

Bet

Alr thanks

Dats it

How do I close this

[timber stirrup]

.close

[gaunt kite]

.close

[pearl pond]

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[pearl pond]

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[wintry oracle]

how do i rotate?

,rotate

[wintry oracle]

so im stuck on 28b, i attempted like 7 times but everytime im concluded with 0=0, i dont know where to start

[pearl pond]

@wintry oracle Has your question been resolved?

[wintry oracle]

<@&286206848099549185>

[timber stirrup]

did u use congruency between ABF and ECF....?

[wintry oracle]

do u mean similar triangles?

[timber stirrup]

yeah

[wintry oracle]

oh my god

i did it

tysm

how did u think of this lmao i tried like 2 hours

i havent used similar triangles in ages

[pearl pond]

@wintry oracle Has your question been resolved?

[pearl pond]

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[sharp smelt]

Prove that if $ac \mid bc \land c \neq 0 \implies a \mid b$

[jolly parrot]

Veni, vidi, perii is not f(wai)

[sharp smelt]

$ac \mid bc \implies \exists k \in \Z ; bc=ack$.; We now multiply both sides by the multiplicative inverse of $c$, which exists as $c \neq 0$. so $b=ak$. Which proves $a \mid b$

QED

[tropic saddle]

what is k

you mean multiplicative inverse

[timber stirrup]

no

[tropic saddle]

what quantifier belongs to k

[sharp smelt]

what do you mean?

[tropic saddle]

bc=ack for all k?

write the exists k in Z part before

[jolly parrot]

Veni, vidi, perii is not f(wai)

[tropic saddle]

ok good enough

[sharp smelt]

Another question

Prove for all real numbers $x,y$ there exists a real number $z$ such that $x+z=y-z$

[jolly parrot]

Veni, vidi, perii is not f(wai)

[sharp smelt]

Solving for $z$ we obtain $z =\frac{y-x}{2}$. As that's a real numbere, we have proven that there does exist a real number $z$ for all $x,y$ , such that $x+z=y-z$

[jolly parrot]

Veni, vidi, perii is not f(wai)

[timber stirrup]

you are ansering your own questions....lmao

[sharp smelt]

Yes? I need my proofs verified

[tropic saddle]

that works

[sharp smelt]

Thanks!

if $\forall x \in A( x \neq 0) \land A \subseteq B \implies x \in B( x \neq 0)$

[jolly parrot]

Veni, vidi, perii is not f(wai)

[tropic saddle]

thats not readable

can you say in words what you mean

[sharp smelt]

Incorrect Theorem. If ∀x ∈ A(x ≠ 0) and A ⊆ B then ∀x ∈ B(x ≠ 0). ( copied from my book)

[tropic saddle]

that is how they wrote it in the book?

[sharp smelt]

Yes

And then They attached a "proof"

Proof. Suppose that ∀x ∈ A(x ≠ 0) and A ⊆ B. Let x be an arbitrary element of A. Since ∀x ∈ A(x ≠ 0), we can conclude that x ≠ 0. Also, since A ⊆ B, x ∈ B. Since x ∈ B, x ≠ 0, and x was arbitrary, we can conclude that ∀x ∈ B(x ≠ 0).

I've been asked to identify what's wrong with this

[tropic saddle]

and what do you think is wrong with it

[sharp smelt]

${1} \subseteq {1,0}$. is a counter example . if $A \subseteq B $, then $x \in A \implies x \in B$ and not $ x \notin A \implies x \notin B$

[tropic saddle]

ok those are some random symbols

[jolly parrot]

Veni, vidi, perii is not f(wai)

[tropic saddle]

ok

but what was wrong with the proof

[sharp smelt]

Since ∀x ∈ A(x ≠ 0), we can conclude that x ≠ 0. Also, since A ⊆ B, x ∈ B. Since x ∈ B, x ≠ 0, a

this part

[pearl pond]

@sharp smelt Has your question been resolved?

[tropic saddle]

that part is correct

[pearl pond]

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[echo basin]

Can I get help with derivatives?

[calm wing]

ask your question

[echo basin]

I have this homework that I'm stuck.

Details: A curve is graphed on the x y coordinate plane. The x axis is displayed from negative 1 to 8 in increments of 1 unit. The y axis is displayed from negative 2 to 4 in increments of 1 unit. The curve y = f(x) enters at approximately (negative 1, 1.6), goes up and to the right, concave down, reaches its high point at approximately (1, 2.5), goes down and to the right, crosses the x axis at (3.5, 0), now concave up, reaches its low point at approximately (5, negative 1.2), again goes up and to the right through (6.5, 0), again concave down, and exits the viewing window at approximately (8, 1.2).

Do I have to find the equation of f(x)?

and then find the derivative

[calm wing]

no

just approximate

[echo basin]

or do I use intervals and calculate the slope?

[calm wing]

yes, do this

[echo basin]

I did for the first interval and got 0.45, but f'(0) is 0.5

from (-1,1.6) -> (1, 2.5)

hello are you there?

I am going to move into another channel

.close

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[short arch]

Hi there, i got this challenging question from my professor, can anyone help me to prove this? I stuck at some steps

[cosmic charm]

progress?

[pearl pond]

@short arch Has your question been resolved?

[pearl pond]

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[opaque raven]

How do I do this question?

[gleaming musk]

hsc?

[opaque raven]

yes

[gleaming musk]

if so, theres online solutions for them on NESA's website but regardless

[opaque raven]

I know but I don't get how they did it

[gleaming musk]

so the goal of the question really is just evaluating which would depreciate more by the end of 3 years

[opaque raven]

what does depreciate mean?

[gleaming musk]

lose in value

[opaque raven]

ohh

[gleaming musk]

so if it says it is depreciating by 3500 per year ==> the salvage value is losing 3500 in value per year

[opaque raven]

and then its the same thing for the 12% one

[gleaming musk]

yes

it would be losing 12% value per year

or equivalent to multiply by (1-0.12)

[opaque raven]

so for the 12% one its (100-12-12-12)% x 60,000

[gleaming musk]

no not quite

[opaque raven]

and for the 3500 one its 60000 - 3500 - 3500 -3500?

[opaque raven]

oh

[gleaming musk]

u are correct for 3500 depreciation per year

[gleaming musk]

for 12% it is different

it is depreciating by 12% per year of its current value

it should actually be $60000*(1-0.12)^{3}$

[jolly parrot]

Dootud

[opaque raven]

ohh

and then I punch both into my calculator and just state which one is lower?

[opaque raven]

also, if I make this mistake on the exam, will I still get 2 marks for it?

[gleaming musk]

im not sure

is this math standard?

[opaque raven]

yes

[gleaming musk]

im not familiar with math standard marking guideline unfortunately

[opaque raven]

that's alr, I'll look into it

[gleaming musk]

my current highschool teacher only marked math advanced and extension 1 so im not familiar with mat hstandard marking guideline

for advanced definitel no

maybe for standard they might be slightly more lenient

best of luck for ur hsc starting next week

[opaque raven]

alr

thank u very much, I'm so cooked bro

[gleaming musk]

its ok me too

[opaque raven]

Have a great rest of your night!

[gleaming musk]

ive done like 0 english practise

[opaque raven]

oh shiii

.close

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[kind kindle]

hi

[kind kindle]

how do ik which way to draw the graph

for the absolute value

ik this case it's probably 2

but how do I normally tell ig

[twilit lion]

You imagine what the limitations of it are.

|x+1| means the absolute minimum value it can be is 0
so if y = 0 -1 then the smallest y you can have is -1

You usually just reflect it through your minimum point

You basically draw the line as if it didn't have absolute, then reflect any parts that would be below your minimum with absolutes in it

[kind kindle]

Is the minimum the minimum uh

Coordinate based on just the absolute value

Like you change it to y = | x+1 |

[twilit lion]

By minimum I mean the smallest possible value your function can take given the absolute

And that's -1 as I mentioned above

[kind kindle]

Oh, so the coordinate (0, -1)

By turning x into 0?

[twilit lion]

No, it would be (-1,-1)

Remember, the smallest value |x-1| can have is 0

so if x = -1 then |x-1| = 0

[kind kindle]

Ohhh

Okay I think I better understand, thanks

[twilit lion]

np

[kind kindle]

So if it was like y = 3 | x -2 | + 4 or something

The value would be (2, -2)?

[twilit lion]

No

if x = 2 then what is |x-2|

@kind kindle

You are right that at x = 2, |x-2| will be the smallest

But are you sure y would be -2?

[kind kindle]

Back sorry

Ohh, so we turn y into 0 for that?

[twilit lion]

No?

Okay, what is |x-2|

if x = 2

Ping me when you're back

[pearl pond]

@kind kindle Has your question been resolved?

[pearl pond]

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[pearl pond]

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[compact ridge]

yes that's the right approach, https://math.stackexchange.com/questions/320348/inductive-proof-that-2n-choose-n-sumn-choose-i2

so then E[X^2] = (2n choose n) / (n + 1)

which is the same as the nth Catalan number

depends on what your teacher or book wants

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[rigid escarp]

How to evaluate this function?

[midnight haven]

the cubic root is a continuous function

so you can use the property that [
\lim_{x\to a} \6f{\6gx} = \6f{\lim_{x\to a}\6gx}
]

[rigid escarp]

I see, thanks!!

[compact ridge]

nice strategy btw

[pearl pond]

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[tender gate]

can anyone help me for h?

[tender gate]

I found that the function is not injective by a counter example

but i'm not sure my way of showing its surjective is correct

[compact ridge]

oh cool, so what did you do for surjectivity?

[tender gate]

thats what i did

[compact ridge]

uhhh that doesn't show surjectivity

you can actually just let say y = 0, so then x - 1 is a surjective function

hence f(x, y) must be surjective

[tender gate]

i don't get it

it would then be a function in function of x only?

[compact ridge]

oh actually your way works, my bad

[compact ridge]

yeah

[tender gate]

and the function f(x)=x-1

is surjective

becasue it has at least 1 f(x) for every x

and i can do the same with x=0

[compact ridge]

yeah so this is fine, let there be $z \in \mathbb R$ such that $z = f(x, y)$ from the definition

then from what you did, there always exist at least one $(x, y)$

[jolly parrot]

south's secret twin brother

[tender gate]

R or R^2?

[compact ridge]

what you're doing is using the original function $\mathbb R^2 \to R$

[jolly parrot]

south's secret twin brother

[tender gate]

becasue proving a surjective function in R^2 is hard/not possible?

[compact ridge]

oh wait I mean it's still $\mathbb R^2 \to \mathbb R$ but $y = 0$

[jolly parrot]

south's secret twin brother

[compact ridge]

corrected

[tender gate]

wait i'm lost so my way works and your way works

and my way works becasue i end up with z=z which is a way of proving surjectivity

and your way is beacsue x-1 is a surjective function

so for the next time i need to prove if a function is injective/surjective that is defined in R^2 i need to try and "delete" one of the variables

to make it into a function in R

[pearl pond]

@tender gate Has your question been resolved?

[pearl pond]

@tender gate Has your question been resolved?

[tender gate]

.close

[pearl pond]

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[twilit pecan]

this is pretty much what i did, my answer isnt wrong but its not finished because there are two answers which are n=1/2 and n=1. i dont really know how to do it

well technically my answer isnt wrong but idk if the work i did is correct lmao

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[compact ridge]

next time, Desmos it btw

[twilit pecan]

i dont really know how to use desmos with those variables n

[compact ridge]

I mean, if you think the answer is n = 1/2, plot those points in Desmos

you can type in (n, 2)

it will ask you to set a value of n, which you can change at any time

[twilit pecan]

oh alr thanks

[pearl pond]

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[pliant shuttle]

1*15

[pliant shuttle]

1*15

[icy narwhal]

?

[pliant shuttle]

3*5

[icy narwhal]

use delta

what is your question

delta = 23 squared -4x4x15

[pliant shuttle]

from those combinations I dont get how I can get +15 and +23?

[icy narwhal]

use delta

[pliant shuttle]

what is delta my tutor never told me about this

[icy narwhal]

yes a sec

it is a formula you use to facorise polynomial formulas

place this

and then from delta you place x with the first formula of the pic

[verbal zodiac]

use taylor series for this question

[midnight haven]

delta?

[verbal zodiac]

fr

[icy narwhal]

in french we call delta the little triangle 😅

idk how to say it in english

[wet swallow]

Delta ?

[midnight haven]

oh

[jolly parrot]

David

[main oxide]

it's a greek letter

[midnight haven]

whatever

it's just a variable

[icy narwhal]

um

why not

it is also used to define a difference

[midnight haven]

yes

[wet swallow]

Can we get back to the question

[midnight haven]

E = delta m * c^2

[wet swallow]

🙃

[icy narwhal]

blud is calculating delta

let him do his math

[wet swallow]

Okay

[random ermine]

what

[midnight haven]

nothing

[latent flare]

hey this is my first time using the server so just tell me I'm supposed to ask this in another channel.

We've started learning about differential equations I was wondering if anyone had a good website or links to math videos for introduction to differential equations.

[random ermine]

this channel is taken

but maybe paul's notes

[latent flare]

oh okay, thank you

[pearl pond]

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[vague stone]

i’ve just felt all my previous knowledge leave me 😭 after spending so much time on other stuff i’ve completely forgotten how this works

[midnight haven]

what is n?

oh

ok

[vague stone]

hahaha dw lmao

[midnight haven]

wait it's easy

Calculate reflex BCD

then you should be able to get it

[vague stone]

OHHH WAIT IM STUPID LMAOO

thanks my brain was on trigonometry mode lmao

[verbal zodiac]

use the formula 180(n-1)

[midnight haven]

180(n-2) *

[vague stone]

thank u ^^

[midnight haven]

welcome

[vague stone]

yeah thought so haha

[verbal zodiac]

mb sleep deprivation

[vague stone]

ur all good man

.close

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[west verge]

Is there a way to prove that the above limit is ln(a) without using L'Hospitals Rule, or substitution of u = a^h - 1. I know how to do both of those methods, but I wonder if there's a third method

[cursive wraith]

I mean $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$ is kinda the definition of something...

[jolly parrot]

rafilou is not not born in 2003

[west verge]

Right, for context, I'm trying to find another way to solve d/dx a^x by definition

[cursive wraith]

ok

[feral sedge]

How do you define e?

and could you do this with a=e?

[west verge]

I define e = lim x approaches infinity (1 + 1/x)^x

I know how to do this method with substitution of u = a^h - 1. You eventually get e in the denominator

[cursive wraith]

how do you define (1 + 1/x)^x

especially ...^x

[feral sedge]

I was gonna suggest a = e^ln(a)

[cursive wraith]

did you mean lim(n-> infinity) instead?

[feral sedge]

ye x varies over integers probably

[west verge]

Sure, we can use n

[west verge]

Not sure what you mean here

[cursive wraith]

i mean it's not "sure we can do that"

[cursive wraith]

have you defined what is b^x with x any real number?

[charred shoal]

can yall help with mine please?

[west verge]

Ok I think I see what you mean. I know how to define it with natural numbers, yes, it's repeated multiplication. Not sure how to with reals though

[feral sedge]

no

[charred shoal]

okay ty lyt

[cursive wraith]

ok so we need to approach infinity from the integers only

welp, only thing I can tell you to do

is to show exp(x) = 1 + x + x^2/2 + ... + x^n/n! + ...

satisfies exp' = exp

[west verge]

Oh right! Taylor Series, ok

[cursive wraith]

exp(1) = e

and then

[west verge]

I think I know how to do this

[cursive wraith]

waitwaitwait I just realized

how can you define this

mostly a^h

[cursive wraith]

if you haven't gone through this...

[feral sedge]

$\lim_{n\to\infty} n(\sqrt[n]a-1)$

[jolly parrot]

mommymorphism aficionado

[feral sedge]

[west verge]

Oh ok, we switch h to n by letting n = 1/h

[cursive wraith]

I think that was meant as a joke

but

how can you try to find this limit

if you don't have the definition of a^h

[west verge]

Besides L'Hospitals? Not sure

[cursive wraith]

no really I just mean the definition of a^h

[feral sedge]

Binomial theorem maybe

[west verge]

Ooh

[cursive wraith]

yeah so it's impossible to find the limit of something

if you don't even know what that thing is to begin with

[west verge]

Ok. To go back to one of your earlier questions, How do we define b^x with x any real number?

[cursive wraith]

yep

(and b > 0)

[west verge]

I can intuitively understand how what b^x is, you just connect the dots (the integers (x, b^(x)) in an exponential way

Not sure how to define it though

[feral sedge]

well, we have more than just integers

what's n^(1/2)

[cursive wraith]

what textbook did you go through/lesson?

[west verge]

Good point

If n^1/2 = u, then u^2 = n

[west verge]

Not using a book. Just trying to remember what I learned in calculus

[cursive wraith]

ok, you might need some book/lesson support then

there are two ways I know of defining b^x

either

$b^x = \exp(x\ln(b))$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

or

$b^x = y \Longleftrightarrow x = \log_b(y)$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

with $\log_b(y) = \frac{\ln(y)}{\ln(b)}$

[west verge]

Then we have to define log first right?

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

yes what I was getting at

and $\ln(t) = \int_1^t \frac 1u du$

[jolly parrot]

rafilou is not not born in 2003

[feral sedge]

$b^x = \lim_{n\to\infty}b^{x_n}$

[west verge]

Oh cool! Yes, I remember this definition of ln(t) from my calc prof

[jolly parrot]

mommymorphism aficionado

[cursive wraith]

this is more of a property rather than a definition

[feral sedge]

where x_n is a sequence of rationals approaching x

[west verge]

Interesting!

[cursive wraith]

(the property being: x -> b^x is continuous)

[west verge]

Ok

[cursive wraith]

(and b^(p/q) coincides with the rational definition)

[west verge]

Alright

[cursive wraith]

so two choices

[cursive wraith]

mostly equivalent

[west verge]

I see. Well, I thank you greatly. You've helped me a lot, and I think I can self-study from here

.close

[pearl pond]

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[pearl pond]

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[twilit pecan]

how do i even do question 7a... ? I cant get the correct answer

[twilit pecan]

this is what i did, and the answer is not even close. final answer is (4, -3)

[formal canopy]

have you tried using something other than coordinate geometry...?

[twilit pecan]

no

[formal canopy]

try

[twilit pecan]

like wat?

[formal canopy]

you're given enough information to isolate the rest of the diagram from the part you need

I hope you remember how to calculate the area of a triangle given base and height

[twilit pecan]

bh/2

[twilit pecan]

how do i exactly get the height tho

[formal canopy]

what type of triangle is ABC

[twilit pecan]

isosceles

[twilit pecan]

i dont really understand. point c is not laid on x-axes so how do u get the height?

[formal canopy]

what's this

[twilit pecan]

triangle area formula

[formal canopy]

do you have the area?

[twilit pecan]

oh yea

oh

so

i am finding

[formal canopy]

yes

[twilit pecan]

the height

[formal canopy]

good day

[twilit pecan]

and thus

i get the

coord

hold awn

WAIT

I STILL DONT UNDERSTAND HOW TO GET

THE COORDINATE

HELP ME 😭

I GOT h=5

OH

5-2

2-5

oKAY

I GOT IT

THANKS

.close

[pearl pond]

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[pearl pond]

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[lament parrot]

Pleaseee 😭

[lament parrot]

I tried using Cauchy Schwarz and Chebyshev many times but i did not get to anything

[pearl pond]

@lament parrot Has your question been resolved?

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[acoustic valve]

does anyone know the process by which i can solve this?

this is the process at which i was referring to for a smiliar problem

[heady finch]

I would probably find the intercept and then consider each as a separate case without the absolute value

[acoustic valve]

that’s what i did

not sure where i went wrong tho

[heady finch]

Why did you integrate it and then leave in integral form

In the working

[acoustic valve]

i wanted to set apart the different intervals at which i have to evaluate

[heady finch]

Yeah but your formatting is weird

Why would you integrate and leave the integrand on the function

You should use square brackets

[acoustic valve]

sorry im new to the notation part but ill keep that in mind

but when solving, what part did i do wrong? or up to what point?

i was suggested to do the one in purple

to break it up into integration by parts

[heady finch]

Integration by parts?

Are you sure

[heady finch]

I think you might not be considering that when the function is negative because of the absolute value function you want to multiply by -1

The whole point is eliminating the absolute value

[acoustic valve]

i have the negative here tho

[heady finch]

And idt you would use integration by parts

[acoustic valve]

bc since the absolute value inflects the values that would’ve considered points below x it’s inflected

therefore you put the negative into consideration

[heady finch]

Well you shouldn't get a negative area for either result

[acoustic valve]

right

[heady finch]

Considering it's an absolute value function

[acoustic valve]

yeah but the negative behind it makes it positive anyways

[heady finch]

So I'm guessing where you got a negative value there might be a mistake

[acoustic valve]

anyways *

that’s what i considered

and if by doing that i got 119/12 but i still got it wrong

[heady finch]

I'm sorry I'm just struggling to interpret your working

Are you subtracting the area from 1/6 to 2?

Because you should add that area

Since the absolute value function won't change the function as f(x) >0

[acoustic valve]

don’t you subtract the area from 2 to. 1/6?

[heady finch]

Why?

Can you explain your thought process

[acoustic valve]

because two is greater than 1/6

[heady finch]

you subtract the area of 1/6 from the area of 2 no?

fundamental theorem of calculus

f(b)-f(a) for an integral from a to b

[acoustic valve]

yes so wouldn’t 2 be “b” and 1/6 be. “a”?

[heady finch]

yes

[heady finch]

but in the bottom step here you're subtacting 2 from 1/6

instead of 1/6 from 2?

[acoustic valve]

oh sorry that’s my older work

i recognize that but still, i’m getting 119/12

i got ant wrong as well

[heady finch]

what is the answer then

if you have the solution

[acoustic valve]

i don’t know yet

i searched it up not sure if it’s right but online gave me 10

[heady finch]

I reckon you aren't too far off

it's just I think you got your integrals the wrong way around

subtracting the upper bound from the lower bound

then you want to do two separate cases where you subtract the integral from 0 to 1/6 and add the integral from 1/6 to 2

based on the function 6x-1

to account for the absolute value

I would also recommend looking at some problems for notation as well 👍

[acoustic valve]

so instead of taking the antiderivative of 6x-1 i keep it as is with the absoulte vallue

?*

[heady finch]

no

basically

the issue of this problem is the absolute value

so we are taking the two cases for this integral

where the absolute value has an effect

to give us two integrals without the absolute value sign

allowing us to find its elementary antiderivative

[acoustic valve]

i dont know what elementary antiderivative means

[heady finch]

basically

there's no straight antiderivative of |6x-1|

at least not what you are usually taught

but if we separate it into the cases

1. function is negative so absolute value makes function positive (0 to 1/6)
   --> if we use regular function must subtract this to find the positive area given by absolute value
2. Function is positive so absolute value has no effect (1/6 to 2)
   --> use regular function and find its antiderivative
   This allows us to remove the absolute value function

and find the area

[acoustic valve]

is it helpful to refer to this to solve my problem?

[pearl pond]

@acoustic valve Has your question been resolved?

[midnight haven]

Yes

[pearl pond]

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[pearl pond]

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[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

from rank nullity

dim(R4) = dim(ker(f)) + dim(Im(f))

4 = 1 + dim(Im(f))

dim(Im(f)) = 3

like if f(S1) is containted in T1 and f(S2) needs to be contained in T2 aswell

we need to find the intersection between S1 and S2

S1 = {x in R4 : 2x1 - x2 + x3 - x4 = 0 ; x1 - 3x3 + x4 = 0}
S2 = {x in R4 : 2x1 -x2 -x3 + x4 = 0}

the intersection of two planes is just solving the system between the equations of the planes

i) 2x1-x2+x3-x4=0
ii) x1 - 3x3 + x4 = 0
iii) 2x1 -x2 -x3 + x4 = 0

,w rref {{2,-1,1,-1,0},{1,0,-3,1,0},{2,-1,-1,1,0}}

[jolly parrot]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

i) x1-2x4=0 ==> x1 = 2x4
ii) x2 -4x4 = 0 ==> x2 = 4x4
iii) x3 -x4 = 0 ==> x3 = x4

(x1,x2,x3,x4)=(2x4,4x4,x4,x4)=x4(2,4,1,1)

S1 n S2 = <(2,4,1,1)>

https://youtu.be/NG-EJ-mHtXs

no idea why he is finding the intersection of S1 and S2 and T1 and T2

I think he is trying to extend a basis of some sort

ping me please

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

https://media.discordapp.net/attachments/268882317391429632/1223598233763319819/1192815796490080286.gif

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

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[sharp smelt]

so as $V_1,,V_2 \subseteq V \implies V_1 \cap V_2 \subseteq V$
\
$0_v \in V_1, V_2 \implies 0_v \in V_1 \cap V_2$
\
let $v \in V_1 \land V_2 \implies \alpha v \in V_1, V_2 \implies \alpha v \in V_1 \cap V_2$\
\
let $v,u \in V_1 , V_2 \implies v+u \in V_1,V_2 \implies v+u \in V_1 \cap V_2$
\
Therefore $V_1 \cap V_2$ is a subspace of $V$

[jolly parrot]

Veni, vidi, perii is $\R - \Q$

[sharp smelt]

Is this fine?

<@&286206848099549185>

[pearl pond]

@sharp smelt Has your question been resolved?

[sharp smelt]

<@&286206848099549185>

.close

[pearl pond]

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[signal scaffold]

i need help to find x and y

[pearl pond]

Please don't occupy multiple help channels.

[midnight haven]

!1c

[pearl pond]

Please stick to your channel.

[signal scaffold]

I Don't see my name occupied so idk

[midnight haven]

#help-26

[signal scaffold]

i see my name was in an occupied channel and suddenly it's gone, so I repeat the question 💀

[midnight haven]

im closing this

.close

[pearl pond]

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[midnight haven]

:O

[pearl pond]

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[muted bloom]

I need help solving this integral using by using parts by integration

[midnight haven]

have you tried anything?

[muted bloom]

yeah should i show what i did

[midnight haven]

yeah

[muted bloom]

i'm pretty sure it's just wrong

[pseudo oxide]

also integration by parts not parts by integration

[muted bloom]

yeah sorry that's what i mean

[glacial sequoia]

Hint: ||You can make a substitution before integrating by parts to make it simpler||

[muted bloom]

that's actually kind of mind boggling

if you see on the paper on the left, i rewrite the problem midway through the page, that's because I end up switching the variables

i got further along with the problem by doing that

and in the very bottom of the page on the right side, that's what I got for my answer

also I know the variables are very different from normal (not u and v) but this is what the teacher used for variables and I've stuck with it

[muted bloom]

so would you set for example u = 2x-1 , and then would you try manipulate it so that I can get rid of x^3 and replace with some u stuff or

[glacial sequoia]

Wait that's not the integral you gave us initially

[ \int \f{x^3}{\sqrt{x^2 - 1}}\dd{x} ]

[jolly parrot]

krypton

[muted bloom]

oh sorry that's what i mea

oh....

[glacial sequoia]

Lmao

[muted bloom]

yeah, may have miswrote the problem

[glacial sequoia]

So what is it

is it x^2 or 2x

[muted bloom]

it's this, cause i legit sent ss of the homework

[glacial sequoia]

Alright

[muted bloom]

clearly i miswrote on my paper and then proceeded to bang my head against a terrible problem

i'll just try to resolve on my own quickly

by quickly it takes me at least 5-8mins

[glacial sequoia]

the one you wrote it also pretty simple btw you might as well try it out for practice

the lesser the power the lesser the burden

[muted bloom]

true

[muted bloom]

ok yeah this is problem is much harder

i've been stuck trying to anti derive 1/root(x^2-1)

i close this for now. just gonna try and watch some more videos

.close

[pearl pond]

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[minor sable]

Yo

[minor sable]

Prove that 1234567890=0987654321

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[midnight haven]

!lol.

[pearl pond]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[oak quiver]

is that an allias for da2a?

nice

[midnight haven]

allias?

!lol.

[oak quiver]

oh

[pearl pond]

@minor sable Has your question been resolved?

[pearl pond]

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[wispy depot]

WHYYY IS ACCELERATION SAME PLEASE EXPLAIN!!

part (a)

[timber stirrup]

u mean on both?

[wispy depot]

yes

[pearl pond]

@wispy depot Has your question been resolved?

[wispy depot]

<@&286206848099549185>

[oak quiver]

@wispy depot if your doubt is that

why the 4kg block and 6kg block have same mass

then that is because the string that ties them

it is inextensible (like the string cant stretch)

so they both will have same velocity and same acceleration
till the string is taut

[ancient lotus]

I think the person is moving in the frictionless surface, so acceleration may bs same

[wispy depot]

How do you deduce the string is inextensible and taut

Just bcz it's light

Or do u make the assumption bcz it says T over there so must be taught

TautZ

[oak quiver]

its understood
its what makes a string a string

[wispy depot]

No but a string can go slack

[oak quiver]

the taut part can be figured from the part that one block is trying to stretch it

[wispy depot]

Well if both have different forces no they won't have the same acceleration or even if they do have same forces the mass of 4 kg lesser hence

[ancient lotus]

It can't slack I think

[wispy depot]

It has more acceleration

[oak quiver]

if the blocks had been coming towards eachother then the string will no longer be taut

[wispy depot]

Why not and what in the question made you come to that conclusion

[oak quiver]

tie a small object with a string and pull it

the string remains taut

[wispy depot]

No like broski imagine this right

Two blocks connected by a rope

[ancient lotus]

Hmm .. that also right

[wispy depot]

Somehow the block in the back moves faster then the front block

Even though I "pulled" the front block

[oak quiver]

that isnt possible in this scenario

[wispy depot]

Then the rope will go slack

[wispy depot]

Why not

[oak quiver]

because the string will pull enough to only give that block to match their velocity

[wispy depot]

Why

[oak quiver]

wdym why

[wispy depot]

Ohh nvm I'm starting to see your point

Oh yeah yeah yeah

[oak quiver]

lol

[wispy depot]

Omg that makes sense

[oak quiver]

gl bro

[wispy depot]

I kinda am getting it

[oak quiver]

this is only the start

[wispy depot]

But wait just one last thing

[oak quiver]

ask

[wispy depot]

Ik this kinda similar to the question above

But ugh

Would it be possible like if I give like some jerk force

That object A goes and hits B

Again friction less

[oak quiver]

lmao

[wispy depot]

But B is really heavy

A is very light let's say

So it wouldn't move as far

And B gon catch up

[oak quiver]

bro you are doing Newton's Laws of Motions right?

[wispy depot]

😭😭😭

Yea

[oak quiver]

stick to it

[wispy depot]

😭😭😭😭😭😭😭

[oak quiver]

collisions is a chapter that comes after a few chaps

you need some more knowledge like Work Energy Theorem

Conservation of Linear Momentum

[wispy depot]

Well I mean I am in uni and in college I did like study conservation for linear momentum and conservation of energy

Like Ik that

[oak quiver]

damn im in 11th

[wispy depot]

💀

[oak quiver]

ok ill try to give an idea

lets assume the block B will attain certain velocity such that it will collide with A

now we have to deicde whteher this collision is elastic, partially elastic or completely inelastic

this is usually given in the question

[wispy depot]

I was kinda talking about A hitting B

[oak quiver]

yeah collision

[wispy depot]

No but I am just asking is that possible

[oak quiver]

yeah if you give a jerk to B

only B

not A

[wispy depot]

hmmmmmmm

[oak quiver]

or give A a jerk in the opposite direction

[wispy depot]

Well if u give a jerk to A it's defo gon hit B wdym it's a string not a steel rod

[oak quiver]

that might work too

[wispy depot]

Anyways leave it

Hank you very much though

I kinda got your point

🫡

[oak quiver]

we cant say definitely depends on stuff like jerk, velocity and length of string

but yeah

nicee

[wispy depot]

.close

[pearl pond]

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[rare fjord]

I am doing radical work for math and am very confused on the topic of subtracting and adding in brackets then multiplying radicals like this: (9√ 11-6√ 5)*(3√ 22+7√ 15)

[oak quiver]

!noclopen

[pearl pond]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[oak quiver]

replace the numbers by a, b, c and d

then exapnd it

and replace the values of abcd

do you know about FOIL?

[rare fjord]

I dont know about foil

[oak quiver]

np i dont either its some american thing lol

[rare fjord]

what do you mean by that?

[oak quiver]

(a+b)(c+d) = ?

[rare fjord]

where a,b,c,d is all equal to the integer?

[oak quiver]

doesnt matter

a-d can be any real numbers

[rare fjord]

ok but if I do that I think I would get a decimal

[oak quiver]

tell me the expansion for

(9a - 6b)(3c + 7d)

in terms of a, b, c, d

[rare fjord]

Im sorry if I seem really stupid, but I don't know what expansion is. I just signed up for AP gr9 and im really confused im sorry

[oak quiver]

ohh np

do you know what (a+b)(a+b) is?

by expanding i mean opening the brackets lol

[rare fjord]

What is a/b in this situation?

[oak quiver]

leave it

tell me what (1+2)(3+4) is

dont add multiply first

[rare fjord]

21?

[oak quiver]

yeah

we do this by 1*3 + 1*4 + 2*3 + 2*4

k?

[rare fjord]

i understand

ohhhhh

i get it

you multiply the first number of the first bracket by the first number of the second bracket and the second number of the second bracket and same with the second number of the first bracket

[oak quiver]

yesss

nice bro

[rare fjord]

wait so do you do that with 9√11 and 6√5?

[oak quiver]

yeah

[rare fjord]

So it's 9√11 * 3√22 + 9√11 * 7√15 + -6√5 * 3√22 + -6√5 *7√15

?

[oak quiver]

Yes

[rare fjord]

How would that go about?

[oak quiver]

√11 * √22 = √(11*22)

(only works for when both numbers are not negative)

[rare fjord]

Ok, so there is now no more √11 and √22 and its now √242

but wouldnt you want to simplify √242?

[oak quiver]

Yeah sure

[rare fjord]

so 11√2

[naive edge]

.

[oak quiver]

Lol

Yeah

[rare fjord]

then after that we would focus on √15 and √5?

[oak quiver]

Yeah

[rare fjord]

so now its √15 * √5 = √(15*5)?

[oak quiver]

Yeah

[rare fjord]

So now we end up with 11√2 and 5√3

[oak quiver]

That is incomplete

Do it on paper and send me

[oak quiver]

Start from here

[rare fjord]

alright

Thank you, here is the equation I got. Let me know if I am correct. If I am then I can continue to do this equation myself. Thank you

[oak quiver]

Your handwriting is very neat

[oak quiver]

I don't think this is correct

[rare fjord]

how can I correct it?

[oak quiver]

Let's look at the first term

What is it?

[rare fjord]

9√11 * 3√22

[oak quiver]

Yes

Then?

[rare fjord]

• 9√11 * 7√15

[oak quiver]

I mean whats this in terms of a single root

[rare fjord]

√11 *√22

[oak quiver]

You forgot the 3 and 9

They get multiplied too

[rare fjord]

9*3 + √11 * √22

[oak quiver]

No

That is multi too

[rare fjord]

9 x 3 + √11 x √22

[oak quiver]

No add

Btw good you know about \*

[rare fjord]

so 9x3 √11x√22

[oak quiver]

Yes

Simplify it

[rare fjord]

297√2

[oak quiver]

Yes correct

Now the next term

[oak quiver]

This

[rare fjord]

63√165

[oak quiver]

Yes

Next term?

[rare fjord]

• -6√5 * 3√22

-18√110

[oak quiver]

Yes

Then?

[rare fjord]

lastly -6√5 * 7√15

which is 210√3

i think

[oak quiver]

Yup

Now add all these

And you have your answer

[rare fjord]

but how do you add if they dont have the same radical?

297√2 + 63√165 - 18√110 -210√3

?

[oak quiver]

This is the simplest form

[rare fjord]

So thats the answer

?

[oak quiver]

If they had given us better values, then we may have simplified it

[oak quiver]

Depends does your teacher expects us to substitute the values of root2 and root165 and all?

[rare fjord]

nah, I think the answer you gave me was good

Thank you so much

oml i wouldve been cooked if it werent for you

it now makes sense

[oak quiver]

Thnxx bro

Gl

[rare fjord]

thanks cya bro

[pearl pond]

@rare fjord Has your question been resolved?

[pearl pond]

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[lucid magnet]

hi, i need help with separating this equation

[lucid magnet]

I got this far

[timber stirrup]

check if the equation is exact

[lucid magnet]

this paper comes from before learning about exact/non-exact solutions so it shouldn't be need to know if exact or not

afaik

.close

[pearl pond]

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[pearl pond]

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[mighty basalt]

$\Vec{a}$ Implies magnitude right?

[jolly parrot]

Ousel

[sharp vigil]

what do you mean by "implies magnitude"?

[mighty basalt]

Im just trying to learn the symbols better. Ive seen magnitude as |a|, but when working with vectors the vector symbol means it as well right?

"World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 m/s2."
Like I could just write that as :

$\Vec{a}_x$

[jolly parrot]

Ousel

[sharp vigil]

a vector like $\vec a$ by itself is just a vector (magnitude and direction). the magnitude of that vector is $|\vec a|$ or sometimes $a$

[jolly parrot]

cloud

[mighty basalt]

so this is just $|\Vec{a}| \text{ not } a_x$

[jolly parrot]

Ousel

[sharp vigil]

well $a_x$ would be the component of $\vec a$ in the $x$-direction

[jolly parrot]

cloud

[mighty basalt]

I think this is where im getting confused: the magnitude of

$\Vec{a} = \sqrt{a_{x}^2 + a_{y}^2$

[jolly parrot]

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

[sharp vigil]

we have [ |\vec a| = \sqrt{a_x^2 + a_y^2} ]

[jolly parrot]

cloud

[mighty basalt]

Ok I think im good now. I just get lost with all the symbols. xP

[pearl pond]

@mighty basalt Has your question been resolved?

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[solemn meadow]

why can't x=0?

[solemn meadow]

is it bc at the original function f can't be differentiated at x=0 since it is a vertical tangent?

[restive kelp]

I don’t know that level of math

[solemn meadow]

wow thx king!

[dense jasper]

correct - f is not continuous at x = 0

[solemn meadow]

gracias

.close

[pearl pond]

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[pearl pond]

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[atomic pendant]

can someone give the concrete example? idk how to formulate it

[atomic pendant]

specifically a transpose x = b

i know that i need to set d = 2

[fluid axle]

an hyperplane in R^2 is a line (for example set a = [2 ; 1] b = 3, you get the line 2x+y = 3)

[atomic pendant]

how do we mutliply it out

a^t x = b

how did you turn it into 2x+y=3

[fluid axle]

right I'll write it in tex

[atomic pendant]

[2 1] x = 3

[2x 1x] = 3 ???

[fluid axle]

when I say [2 ; 1] I mean a column vector

[atomic pendant]

2x + 1x = 3???

[fluid axle]

it's matlab notation

[fluid axle]

x is a vector already mate