help-13

[fringe slate]

i was on a zoom call

[sour mesa]

You're underestimating yourself

Don't hold yourself in the bubble of your negativity

Anyways, for your case this should help you

[fringe slate]

i can tell you im not lol ive looked through it for 30 minutes and only managed to understand 1/3 of it.

anyways i gtg

.close

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[rapid nova]

Good evening, how are you?

I need help classifying this series. I was able to write it this way, but I don't know how to continue. What criteria or what strategy could you use?

[rapid nova]

Given the function $f(x)=e^{x}-\displaystyle\frac{x+1}{x-1}$\
Classify $\sum(g\left(\frac{1}{n}\right)-2)$
[\sum(g\left(\frac{1}{n}\right)-2)=\sum \left( e^{\frac{1}{n}}-\displaystyle\frac{\frac{1}{n}+1}{\frac{1}{n}-1}-2 \right)=\sum \left( e^{\frac{1}{n}}-\displaystyle\frac{1+n}{1-n}-2 \right)]

[wraith dagger]

Isa

[dire geode]

what does f(x) have anything to do with your sum

and what's the index on the sum

[rapid nova]

Sorry, f is g really haha

And the index is n=2, it doesn't say the index, but I imagine it could be 2

[dire geode]

what is "it"

can you just take a picture

[rapid nova]

I meant that the index of the sum is 2.

[=\sum_{n=2} \left( e^{\frac{1}{n}}-\displaystyle\frac{1+n}{1-n}-2 \right)]

[wraith dagger]

Isa

[rapid nova]

That's the series I have to classify.

If I try to use the quotient criterion, it doesn't work.

[dire geode]

did you try integral test

[rapid nova]

No, I've never used it, haha. I'm going to try. Thanks you

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[dire geode]

.reopen

[cedar kiln]

✅

[dire geode]

a simpler suggestion is the limit test. if the terms don't go to zero as n goes to inf, then the series diverges

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[ancient cove]

.reopen

What I am saying is just speculation

[ancient cove]

it approaches 0

The individual series themselves are divergent, e^(1/n) is essentially just 1+1+1 for large values of n,

(1+n)/(1-n) is similarly just -1 -1 -1 for large values of n, which implies divergence

But since its e^1/n - {(1+n)/(1-n) } -2 = 1 - (-1) -2 = 0. Perhaps the series cancel themselves out ? A little mathematical manipulatation will show that a_n+1 < a_n, but again this does not prove that the series is convergent. A counter example is 1/n (which approaches 0 for large values of n, and a_n+1 < a_n)

Ratio test seems to be inconclusive on this though

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[quick latch]

Any hint on how to get the bijective function?

[muted bear]

floor

[quick latch]

$f: \mathbb{R} \to [0, 1) \times \mathbb{Z}$ defined as $f(x) = \left(x - \lfloor x \rfloor, \lfloor x \rfloor\right)$ for all $x \in \mathbb{R}$

[wraith dagger]

Halex

[quick latch]

is this right?

[cedar kiln]

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[rose cliff]

yo im having trouble with a, im getting 4n^2 + 14n + 9 which isnt nicely factorable

[south tundra]

,w polynomial division 4x^3 + 18x^2 + 23x + 9 by x+1

[wraith dagger]

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[south tundra]

The quotient is correct

[rose cliff]

yeah both roots are not nice

[south tundra]

Well it technically doesn't ask you to factor into linear factors

[rose cliff]

but im guessing that you need nice linear factors for the induction in b to work out

[cedar kiln]

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[rose cliff]

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[gilded remnant]

solving X1 > 0.5 how do you get -1.25 < Z ≤ -1

[gilded remnant]

<@&286206848099549185>

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[shut otter]

Hello

[shut otter]

Pls explain me the result because i dont understand all step to obtain that

<@&286206848099549185>

[queen marlin]

I=sqrt(x^2+y^2), also known as (x^2+y^2)^0.5

so first, the partial derivative in respect to x is the derivative of the outer function times the derivative of the inner function as per the chain rule

so power rule on (x^2+y^2)^0.5 gives us 0.5*(x^2+y^2)^-0.5, and then the product rule multiplies that result by the derivative of (x^2+y^2) in respect to x

but notice that since we're differentiating in respect to x, the y^2 term is just a constant, so the derivative just becomes 2x

so di/dx = 0.5(x^2+y^2)^-0.5 * 2x

since the equation is symmetrical, you can tell that for a similair reason you get that di/dy = 0.5(x^2+y^2)^-0.5 * 2y

only thing left now is to substitute x=4 and y=3 to get ur numerical values

[shut otter]

Ok ty

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[crimson sedge]

\textbf{Question:} Consider a transformation which is known to be linear, but not necessarily time-invariant. Its response to three impulse inputs is as shown below:
\e{alignat*}{{2}
\7xn &= \7\delta n &\q \7yn &= \7\delta n + 2 \7\delta{n-1} + 3\7\delta{n-2} \\ 
\7xn &= \7\delta{n-1}  &\q \7yn &= 3\7\delta{n-1} + 3 \7\delta{n-2} + 2\7\delta{n-3} \\ 
\7xn &= \7\delta{n-2}  &\q \7yn &= 3\7\delta{n-2} + 3 \7\delta{n-3} + \7\delta{n-4} 
}
\emph{Note:} Linearity is defined such that for an input $\alpha \7{x_1}n + \beta \7{x_2}n$ the output under the transformation is $\alpha\7{y_1}n + \beta\7{y_2}n$. 

\vs{3 mm}
Time-invariance is defined such that for an input $\7x{n-n_0}$ the output is also $\7y{n-n_0}$.

\vs{3 mm}
Is it possible to know the response of the transformation for the input \[
\7xn = \7un-\7u{n-4}
\]
where $u$ is the heaviside function

[crimson sedge]

(accidentally pressed enter; will finish typing soon)

[crimson sedge]

I think not. As we do not know the response of the transformation for delta[n-3]

Or is it possible to extrapolate the response from the 3 responses we have been given?

[sacred grail]

probably not

[crimson sedge]

ok ty g

u a real one

not when u dont play chess with me

[sacred grail]

okay so $\conj {x[n]}$ gives $\conj {y[n]}$?

[crimson sedge]

uh

what is overline x[n]

[sacred grail]

conjugate

[crimson sedge]

denoting

of what there is no complex numbers

[sacred grail]

thats when you have a real transformation

[crimson sedge]

oh

BRO

[sacred grail]

[crimson sedge]

why did that joke fly over my head

[sacred grail]

[crimson sedge]

time to hide my crimes

.close

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[cedar kiln]

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[cosmic cargo]

how to solve questions like these? after rearranging and putting into calculator it still isnt right according to the mark scheme

[cosmic cargo]

the values in the mark scheme are

bruh

[crimson sedge]

Rearrange the equation to make x the subject of the equation

[cosmic cargo]

yes that was done

what next

[crimson sedge]

$$x = \arcsin (-0.5)$$

[wraith dagger]

Edmund Cloudsley

[cosmic cargo]

plugging sin^-1(0.5) does not give the right answers

what is arcsin?

[crimson sedge]

Sin^-1

Same thing

Anyways

[cosmic cargo]

oh

[crimson sedge]

Have a look at the unit circle

[cosmic cargo]

?

[crimson sedge]

Where does the value of sin x = -1/2 in the unit circle?

[cosmic cargo]

what is the unit circle/

[crimson sedge]

Nvm

Do you know about the periodicity of sin?

[cosmic cargo]

no

[crimson sedge]

Okay so basically the sine function repeats itself

[cosmic cargo]

oh the graph thing?

[crimson sedge]

Therefore it has infinitely many x values for a single y value

[cosmic cargo]

oh yes ok

[crimson sedge]

Unless a domain is specified

[cosmic cargo]

i see

so does it repeat itself every??

[crimson sedge]

So the generalised form is (answer) + 2(pi)(n )where n belongs to Z

[cosmic cargo]

?

could u explain the parts please

[crimson sedge]

https://youtu.be/LBhRiTw_uXw?si=MfD-pDn1hjXICbHc

You might wanna watch this video

[cosmic cargo]

okay

[crimson sedge]

Periodicity is a bit too long to explain

[cosmic cargo]

i see thanks

[crimson sedge]

And i probably won’t do an extremely good job

Good luck

Do come back if you have any doubts you can ping me anytime

[cosmic cargo]

ok

@crimson sedge is this the only way?

do we have to memorise the circle

[crimson sedge]

Nope nope not at all

Unless ofc you are doing a no-calc course

Are you doing a no-calc course?

@cosmic cargo

[cosmic cargo]

i can use a caculator

[crimson sedge]

Oh great then no need to remember

[cosmic cargo]

$$x = \arcsin (0.5)$$

[wraith dagger]

ajaman

[cosmic cargo]

so after this

my calculator gives me 30 as a answe

r

and my current self deduction is that it will be 360-30

so 330 and 30 will be the snwers

but i dont think that works

the markscheme says it is 210 and 330

[crimson sedge]

Remember that if $\sin \theta = \sin x$

Then $x = \theta \pm 2\pi k$

[cosmic cargo]

seen here

what is k

[crimson sedge]

K can have any integer value

[wraith dagger]

Edmund Cloudsley

[cosmic cargo]

what

so what does it do

[crimson sedge]

if $\sin(x) = k$ then this would be true

[wraith dagger]

Edmund Cloudsley

[cosmic cargo]

im not following rn

[crimson sedge]

ok

sin

is a periodic function

it repeats itself

that is why x can have multiple values

how do we find those multiple values?

by adding or subtracting 2pi

[cosmic cargo]

so 30+2pi would be my answer?

[crimson sedge]

,w solve arcsin(-0.5) in degrees

[wraith dagger]

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[crimson sedge]

-30 + 360 = 330

180 - -30 = 210

[crimson sedge]

this is the first rule

[crimson sedge]

and this is from the second

[crimson sedge]

you can't add degrees and radians together mate

[cosmic cargo]

is pi not 3.14....

??

or am i crazy

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

How do I convolve [
\7\delta n + \7\delta{n+1}
]
with itself

[crimson sedge]

So my progress is currently something like:
\begin{align*}
\8{\7\delta n + \7\delta{n+1}}\ast\8{\7\delta n + \7\delta{n+1}} &= \7\delta n \ast \7\delta n + \7\delta n \ast \7\delta{n+1}\ &\q+ \7\delta{n+1}\ast\7\delta n + \7\delta{n-1}\ast\7\delta{n-1}
\end{align*}

[cedar kiln]

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[vernal kite]

where did i go wrong for q1 ex 2

[green plaza]

You messed up the -ve sign in 3rd step

[vernal kite]

wdym -ve sign

[green plaza]

There is a -ve outside (1-tanalpha)²

[vernal kite]

could you show me? im still a bit lost

[green plaza]

Ah the denominator will be (1+tanalpha)(1-tanalpha)

Nvm

[vernal kite]

yea i thought so

but i get the numerator part now

[green plaza]

[vernal kite]

so it's actually -1 + 2tanalpha - tan^2 alpha?

wait no

[vernal kite]

yep gotcha

[green plaza]

Ignore the Halloween doodle 💀

[vernal kite]

lol it's nice tho

[green plaza]

Tyty

Stopped midway though

Alright now that your doubt is solved

You can close this channel

[vernal kite]

should've continued it and hung it up on ur wall or smth

ight thanks

.close

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[high pumice]

hello

[high pumice]

so

can anyone please help me out

im so cooked for my assignment

this is math induction

🙏 😭 much appreciated

[twilit escarp]

You need to prove this ?

[high pumice]

yes

[twilit escarp]

Induction ?

[high pumice]

yessir

[twilit escarp]

Ok so

[high pumice]

its our first topic for our semester

[twilit escarp]

Calculate a_2

[high pumice]

can u pls show it

😭

[twilit escarp]

Let me latex it

[high pumice]

oki

take ur time

[twilit escarp]

$a_2 = \sqrt{2 \cdot a_1} = \sqrt{2 \cdot \sqrt{2}}$

[wraith dagger]

YakuBros

[twilit escarp]

So do you have that sqrt(2) < sqrt(2*sqrt(2)) ?

[high pumice]

im not sure

[twilit escarp]

Protips : sqrt(2*sqrt(2)) = sqrt(2) * sqrt(sqrt(2))

[high pumice]

i dont get what u mean

either im sorry

[twilit escarp]

I check if this is true for the first value of n, that is the first step of induction

[high pumice]

ohhh okay okay

[twilit escarp]

Imagine dominos, you need to push one so that all fall

Thats what we're doing

We pushing the first domino

[high pumice]

alright i get the point to prove that the other domies fall too

right?

[twilit escarp]

But you need a starting point and thats what we are doing

[high pumice]

okay okay

[twilit escarp]

We are checking if its right for n = 2

[high pumice]

i see

i mean i get the basics but i dont know how to solve it

because the term confuses me

[twilit escarp]

This ?

[high pumice]

the question in general

the pic i sent

[twilit escarp]

Ok but do you know how induction work so ?

Like its the step after

Where we are going to prove it

[high pumice]

alright please

[twilit escarp]

For now we are initializing it

[twilit escarp]

But for this now, we need to know if its right or not ?

[high pumice]

okay i get now

[twilit escarp]

So is it ?

[cedar kiln]

@high pumice Has your question been resolved?

[cedar kiln]

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[silent cove]

Need some help

[silent cove]

I think for some its quicl

quick*

[primal vigil]

say

[silent cove]

if

[versed jasper]

!dontasktoask

what is it

[silent cove]

if every 160 seconds I win 20 bucks

In 5 days Can I get 3000

[versed jasper]

speed = 20/160 = 1 buck per 8 seconds
5 days = 24hrs = 24 * 3600 = 86400
money = 1/8 * 86400 = 10800 bucks

[silent cove]

Thanks 😄

@cedar kiln

.close

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[arctic gull]

what is the answer🥲

[arctic gull]

how to do it

<@&286206848099549185>

[primal vigil]

what's an endpoint of latera recta?

[arctic gull]

passing through focus right

[primal vigil]

sorry my english is not too good
let me search and I'll come back

[arctic gull]

ok no worries!

[primal vigil]

hmm I have no idea about that

sorry

[cedar kiln]

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[arctic gull]

.close

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[jolly crow]

This is implicit differentiation, screenshotted from a 3Blue1Brown video. I don't understand why dy/dx = -x/y, in particular why is x negative?

[pastel vault]

tangent and radius are peprpendicular

what is the slope of the radius then?

[bleak viper]

2xdx + 2ydy = 0 -> 2ydy = -2xdx -> dy/dx = -x/y

[pastel vault]

it also follows algebraically from differentiation yes

[jolly crow]

Shit I'm stupid, thanks.

.closed

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[tiny carbon]

Hello, I am a high school student. I need resources and a map that will enable me to specialize in mathematics.

[crimson sedge]

khan academy

[pastel vault]

.close

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[tiny carbon]

Is this a YouTube channel?

[pastel vault]

it's a website but they also have a YouTube channel

https://www.khanacademy.org/math/precalculus

[tiny carbon]

Thanks bro

[cedar kiln]

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[crisp halo]

Hello!

[crisp halo]

Am I able to simplify this further?

x = sqrt2/2

Also I get the feeling I've done something wrong

f = sin(x)

so I assume f' = cos(x)

<@&286206848099549185>

[quick topaz]

hi

which is the exercise?

[crisp halo]

hi

This is the c) part of a question

a) was to describe why a function is invertible

[quick topaz]

can u send the all text?

[dire geode]

!15m

[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.

[crisp halo]

I can try

[crisp halo]

15 mins have passed no?

[dire geode]

:X:

[crisp halo]

Huh

Oh wait

Oh I thought it was 15 minutes between Helper ping

My bad

I just misunderstood what 15 min meant

15 min after question

Sorry.

[crisp halo]

The question is not in English

I was given this function and interval

then I was asked to solve

[crisp halo]

Now I am asked to use a formula and use the result from this

to calculate

[crisp halo]

The solution to this was 9pi/4

[crisp halo]

I don't know how to continue/if I can continue

[crisp halo]

Sorry, and the formula was the first part of this

With a calculator this is probably a useful answer

but could I simplify that more?

oh yeah

lol

thank you

.close

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[dawn mesa]

Find an example of a function that is differentiable but not Lipschitz ($|f(x_1)-f(x_2)| \leq C|x_1-x_2|$ for some constant $C > 0$). I have found $f(x) = \sqrt{x}$, $\forall x \geq 0$ is differentiable on $(0, \infty)$. I have shown for some $x_1 = 0$ and $x_2 = \epsilon > 0$, we have the constant $C \geq \frac{\sqrt{\epsilon}}{\epsilon}$. Thus as $\epsilon \rightarrow 0$ there is no finitie constant $C$ that satisfies the Lipschitz inequality. Is this sufficient evidence?

[cedar kiln]

Please don't occupy multiple help channels.

[wraith dagger]

Stephanial

[cedar kiln]

@dawn mesa Has your question been resolved?

[heady berry]

do you guys help with science work?

<@&286206848099549185>

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[rancid flax]

I need help at number 2: in english: Let alpha and beta be mappings from Z×Z to Z and let them be given as follows: For all let x,y to the power of alpha be elements of z cross z defined as x+y and (x,y) to the power of beta defined as y to the power of 3.
A) alpha is surjective
B) beta is surjective
C) beta is not injective
D) for all z element Z, the primal set alpha across z is infinite.

[rancid flax]

I need to Show A) B) C) D) and i dont know how I should do that

[cedar kiln]

@rancid flax Has your question been resolved?

[rancid flax]

<@&286206848099549185>

[dire geode]

,rotate

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[pliant aspen]

Can you show the solution to the problem? The parentheses denotes GCD.

[quiet delta]

Just consider possible maximum powers of p in factorization of a and b. There are 2 cases: p in a, p^k in b, with k>=1 and the 2nd case to swap them.

So, for (a^2,b) it is p and p^2. For (a^2,b^2) the same thing. The rest is similar.

[cedar kiln]

@pliant aspen Has your question been resolved?

[pliant aspen]

no. I am completely lost?

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[tribal onyx]

Hi i have some questions about triangles and hw i need help with. Gr 9 geometry honors math

[cedar kiln]

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[silver monolith]

ngl

i cant see shit

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[slow hatch]

I'm way off from the answer key when I do this, I'm quite stuck, I tried to just input one of the x values but that didn't work either, I'm not quite sure what its asking

[runic garnet]

@slow hatch still need help?

[slow hatch]

yea

[runic garnet]

ok so can u tell me the properties of a pmf

[slow hatch]

I actually have no clue

imma be honest I was absent this whole unit

tryna find notes and shit but I can't find shit

[runic garnet]

ok so heres the thing

pmf probabilities must sum to 1

here we have a discrete pmf

so find each probability in terms of a, and sum them and set equal to 1

essentially: P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1

got it @slow hatch ?

[slow hatch]

oh

so x= 0, 1,2,3

and plug each one in for the function

wait but how does that give me a tho

wouldn't A be constantly changing

[runic garnet]

what is P(X=0)

[slow hatch]

this is what I was tryna do by inputting x and substituting

[slow hatch]

P(X=3)

[slow hatch]

umm

[runic garnet]

no this is wrong

doesnt make sense that P(X=3) = 3

thats what u did on the left side

[slow hatch]

im confused again

[runic garnet]

ok

[runic garnet]

answer this

[slow hatch]

idk what it is

I don't even know what P(X=x) means

[runic garnet]

ah ok

so then i think u should go to khanacademy and watch some videos before u attempt this

[slow hatch]

wait is P(X=0) = 0

[runic garnet]

no

[slow hatch]

https://tenor.com/view/controlmypc-cat-screaming-discord-gif-20582295

[runic garnet]

🤷‍♂️ what do u want me to tell u

u need to know the basics before u can move on

[slow hatch]

yeah ik

its just stressful

[runic garnet]

before u go and watch, think about this

if f(x) = 4x^2

whats f(3)

[slow hatch]

144

[runic garnet]

ok and howd u get that

[slow hatch]

I plugged in 3 for x

[runic garnet]

4 * (3^2) ?

[slow hatch]

yea

[runic garnet]

and what is 3^2

[slow hatch]

wait

I did it wrong

36

[runic garnet]

yes

similarly lets say P(X=x) = c/(4* x^2) (or smth like this)

P(X=10) = c/(4 * ( 10^2)) = c/(4 * 100) = c/400

[slow hatch]

right

[runic garnet]

ok now consider ur example

what will u do

[slow hatch]

$P(X=0) = \frac{a}{0^2+1}$

[wraith dagger]

Raeed

[runic garnet]

ok why stop there

just keep going

[slow hatch]

$P(X=0) = \frac{a}{1}$

[wraith dagger]

Raeed

[slow hatch]

$P(X=0) *1= a$

[wraith dagger]

Raeed

[runic garnet]

bruh wat

if P(X=0) = a/1 just simplify that

what does it become

[slow hatch]

oh my god

🤦‍♂️

i had no sleep last night

[runic garnet]

rip

[slow hatch]

$P(X=0) = a$

[wraith dagger]

Raeed

[runic garnet]

go through all of them

[slow hatch]

im actually losing braincells

I know this is not right

[runic garnet]

try P(X=1) again

[runic garnet]

maybe itd be best to do this after u get some sleep

[slow hatch]

can't lmao

tested on this tomorrow

[runic garnet]

i c

[runic garnet]

ok well do this

[slow hatch]

answer key says 5/9 tf am I doing wrong

it IS supposed to add up to 1

[runic garnet]

a + 0.5a + 0.2a + 0.1a = 1

that part is correct

u just did the addition wrong

[slow hatch]

wait stupid question

how did you know to convert them to those decimals

[runic garnet]

i couldve done fractions if i wanted , its just i didnt feel like using latex

easier to read decimals in plaintext

[slow hatch]

because I actually get stuck doing it using fractions

[runic garnet]

ok so

$a + \frac a2 + \frac a4 + \frac a{10} = 1$

[runic garnet]

lets make common denominator 20

$\frac {20a}{20} + \frac {10a}{20} + \frac {5a}{20} + \frac {2a}{20} = 1$

[runic garnet]

now adding we get

[slow hatch]

$\frac{a}{1} + \frac a2 + \frac a4 + \frac a{10} = 1$

[wraith dagger]

Raeed

[slow hatch]

sorry

ocd

[slow hatch]

yeah

so now we don't have to worry about the denominator

and we can just add them freely

[runic garnet]

hmm wait

we messed up somewhere i think

[slow hatch]

,calc 20+10+5+2

[wraith dagger]

Result:

37

[runic garnet]

ah yes i see the error

find P(X=2)

[slow hatch]

oh?

actually lemme just do this since calculators are allowed

[runic garnet]

first answer my questio

[slow hatch]

a/4 ❌
a/5 ✅

[runic garnet]

yes

[slow hatch]

yeah

[runic garnet]

$a + \frac a2 + \frac a5 + \frac a{10} = 1$

[runic garnet]

now common denominators

$\frac {20a}{20} + \frac {10a}{20} + \frac {4a}{20} + \frac {2a}{20} = 1$

[runic garnet]

$\frac {36a}{20} = 1$

[runic garnet]

$a = \frac 59$

[runic garnet]

got it?

[slow hatch]

yess this makes so much more sense now

tysm

,close

.close

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[grand arch]

I would just like someone to explain the results of RREF form, in linear algebra, to me. For example, an augmented matrix reduced to have a row of 0's and what it means

[slow hatch]

Hi there, this is specifically for specific problems and stuff you have I think your question is more of a discussion best suited for #linear-algebra

[grand arch]

alr

.close

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[trail dune]

What m I doing?😭

[trail dune]

I wanted to factor then find x

[digital cliff]

uh
so you had (x-9)/[x(x-3)]=-12/(x^2-9)
fine, i would factorise more here before doing anything else, because it just got messy

[trail dune]

Ohhh I see

Okay one sec

Got it ty

.solved

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[oblique prawn]

what does f_xx mean

[digital cliff]

second partial x derivative

[oblique prawn]

why doesnt second partial y derivative work too

[digital cliff]

wdym by work

[oblique prawn]

why is it f_xx and not f_yy

[digital cliff]

at (1,1) x=y
so 6y=6x>0

fxx=fyy at (1,1)

so theres no need to do both explicitly

[oblique prawn]

so in general when ur doing second derivative test both of them would give the same value?

[digital cliff]

not necessarily no

you just happen to have fxx and fyy being of the same form

6x and 6y

theyre the same only if x=y

[oblique prawn]

so in general which one would you use, f_xx or f_yy

[cedar kiln]

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[severe bison]

Can anyone help proving this?

[cedar kiln]

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[pliant talon]

"Prove that if the columns of B are linearly dependent, then so are the columns of AB." - linear algebra

[pliant talon]

my thought process is that the columns of AB are the columns of B multiplied by matrix A, and somehow the linear transformation preserves the linear dependence of the columns of B

but i am not sure how to prove that a linear transformation preserves linear dependence and i don't think i can just write it like that straight up

.close

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[severe bison]

.reopen

.close

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[crimson sedge]

Lemme get the photos here

[vivid wadi]

let $y = (3)^{x + 1}$, you need to solve for x (input) in terms of y (output)

[wraith dagger]

kaue

[vivid wadi]

hint: log base 3

[crimson sedge]

Oh no... idk Logarithm

[vivid wadi]

for this case all you need to know is that log base 3 cancels with $3^n$, so $\log_3(3^n) = n$.

From $y = 3^{x + 1}$ you can take log base 3 both sides and get $\log_3(y) = \log_3(3^{x + 1})$

[wraith dagger]

kaue

[crimson sedge]

I don't get it

[lean lotus]

U need to know logs before u can do these questions

[crimson sedge]

Can.. you guys teach me?

[glacial field]

Kaue just did

$\log_3(3^n) = n$

[wraith dagger]

mari

[crimson sedge]

Well I still didn't understand...

[vivid wadi]

$\log_3(3^{x + 1})$ simplifies to $x + 1$

[wraith dagger]

kaue

[crimson sedge]

uh... how about let's take it to the top...

so we have 3^x+1 as the question

Is it something like... putting log and...
logn(3^x+1)

$\3^x+1
\log_n(3^{x + 1})$

[wraith dagger]

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

[crimson sedge]

WHAT NO.

just..

3^x+1 <- question
logn 3^x + 1

[glacial field]

$3^{x+1}//
\log_n(3^{x + 1})$

[crimson sedge]

$3^{x+1} to
\log_n(3^{x + 1})$

[wraith dagger]

キオ

mari

[crimson sedge]

YEAHH

[crimson sedge]

Just like that?

[glacial field]

What

[vivid wadi]

n is 3

[crimson sedge]

why?

[vivid wadi]

because you want to cancel the power of 3

[crimson sedge]

You mean the base 3?

[vivid wadi]

$\log_n(n^\text{something})$ = something

[wraith dagger]

kaue

[crimson sedge]

I don't get it still...

[glacial field]

Do you know exponentiation and if so how much about it

[vivid wadi]

if the power is 3, you want the base of the log to be 3 so it cancels

[crimson sedge]

Just the laws of exponents..

[glacial field]

What is e^(lnx)?

[crimson sedge]

Where... from the question is the power of 3?

[crimson sedge]

Okay- not like that-

[vivid wadi]

$3^{x + 1}$

[wraith dagger]

kaue

[glacial field]

You should take some time to go learn how logarithms work before doing these questions

[crimson sedge]

You mean the base 3?

[glacial field]

What I asked is entry level exponentiation understanding

[vivid wadi]

yes

[crimson sedge]

OHHHH

3^x+1
logn 3^x + 1
log3 3^x + 1

[vivid wadi]

you can think of it as how subtraction cancels addition

log base n is just the operation that cancels exponentiation with base n

[crimson sedge]

okay??...

What do I do now?

[vivid wadi]

.

take log both sides and simplify

[crimson sedge]

something like...
3^x+1
logn 3^x + 1
(3) log3 3^x + 1
log(y) = 3^x+1

[vivid wadi]

the starting point is $y = 3^{x + 1}$

[wraith dagger]

kaue

[crimson sedge]

okay...?

[vivid wadi]

you're not using both sides on your first steps here

[crimson sedge]

Okay... so.. I take log both sides and simplify but I'm not going to use both sides on my first steps

got it

[vivid wadi]

$y = 3^{x + 1}$ then $\log_3(y) = \log_3(3^{x + 1})$ then $\log_3(y) = x + 1$

[wraith dagger]

kaue

[crimson sedge]

So.. that's the answer?

[vivid wadi]

no

you need to solve for x

now the last step is basically just subtracting the 1

[crimson sedge]

so.. log3 (y) = x?

[vivid wadi]

you need to subtract from both sides, not just one

[vivid wadi]

you need to use both sides every step btw

[crimson sedge]

so.. log-3 (-y) = -x?

[vivid wadi]

$log_3(y) = x + 1$ then $log_3(y) - 1 = x$ then $x = log_3(y) - 1$

[wraith dagger]

kaue

[crimson sedge]

OHHH

3^x+1
logn 3^x + 1
(3) log3 3^x + 1
log(y) = 3^x+1
log3 (y) = x+1
log3 (y) - 1 = x

[vivid wadi]

theres no equation on your first three steps

[crimson sedge]

3^x+1
logn 3^x + 1
log(y) = 3^x+1
log3 (y) = x+1
log3 (y) - 1 = x

[vivid wadi]

theres still no equation on your first two steps

the starting point should be $y = 3^{x + 1}$

[wraith dagger]

kaue

[crimson sedge]

3^x+1
y = 3^x+1
log(y) = 3^x+1
log3 (y) = x+1
log3 (y) - 1 = x

[vivid wadi]

$y = 3^{x + 1} \
\log_3(y) = \log_3(3^{x + 1})$ \
\log_3(y) = x + 1 \
\log_3(y) - 1 = x \
x = \log_3(y) - 1 \
i^{-1}(n) = \log_3(n) - 1$

[wraith dagger]

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

[crimson sedge]

That's.. the answer..?

[vivid wadi]

yeah

[crimson sedge]

Oh okay

[vivid wadi]

$i^{-1}(n)$ is just a notation for the inverse function of $i(n)$

[wraith dagger]

kaue

[crimson sedge]

Oh okay

Nexttt

[void sand]

the exponential function 5^x is the inverse function of log_5(x)

[crimson sedge]

I don't get it

[void sand]

you're essentially just doing what you did in the previous problems, but instead of cancelling out an exponential with a logarithm, you'll be cancelling out a logarithm with an exponential

[crimson sedge]

I see...

So how do you.. do them..?

[void sand]

well, you start by writing $$y = \log_5(x + 5)$ and interchanging $x$ with $y$ to get $$x = \log_5(y + 5)$$

[wraith dagger]

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

[void sand]

this is the same process you were doing earlier

[void sand]

now you just gotta solve for y

[crimson sedge]

How do you.. do it with the log?

[void sand]

by remembering that exponents and logs with the same base cancel each other out!

[crimson sedge]

So.. how does that work?

[void sand]

namely, $5^{\log_5(x)} = x$ and $\log_5(5^x) = x$

[wraith dagger]

higher!

[void sand]

in our case, you want to use the former to cancel out the log_5 on the right hand side

[crimson sedge]

uhh

So..

j (x) = log5 (x + 5)
x = log5 (x + 5)
x = 5^log5 (x)

[void sand]

x = log_5(x + 5)?

where did y go?

[crimson sedge]

j (x) = log5 (x + 5)
x = log5 (y + 5)
x = 5^log5 (y)

[void sand]

ah, not quite

for one, when you apply 5^ to the right hand side, you gotta also apply it to the left hand side to keep the equality

also, the argument of log5 is still y + 5; the 5 doesn't suddenly disappear

[crimson sedge]

j (x) = log5 (x + 5)
x = log5 (x + 5)
5x = 5^log5 (x)

[void sand]

where's y?

[crimson sedge]

Idk where to put it

[void sand]

this was good, minus my two comments above

[crimson sedge]

I don't get it

[void sand]

j (x) = log5 (x + 5)
x = log5 (y + 5)
x = 5^log5 (y)

is correct, besides the fact that you should've applied 5^ to both sides of the equation, not just the right hand side, and that the argument of log5 is still y + 5, not y

[crimson sedge]

j (x) = log5 (x + 5)
x = log5 (y + 5)
5x = log5 (y)

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[opal sparrow]

on 16 and 8 wouldnt the domain be -inf to inf? i font get why my teacher put only positives and on 16 when one the answers has a negative

[solid juniper]

that's not relevant

the solution would make a little more sense if D: (0, \infty) wasn't written

[opal sparrow]

also on the number line test idk why -1/sqrt2 wasnt tested too?

[solid juniper]

well you don't really need to

the function is symmetric along the y axis

so like

[opal sparrow]

oh

[solid juniper]

you can just look at the points with x > 0

[opal sparrow]

oh is that why the domain is only pos

[solid juniper]

that might have been what she was trying to do with this

but it doesn't really make sense to write that

[opal sparrow]

then you just find both bc its the same but negative?

[solid juniper]

yea. if you find the closest point (x,y) to (0,-1) with x > 0, the closest one on the other side of the graph will be (-x,y)

and the distances will be the same

[opal sparrow]

is that the same case for number 8? bc she put the same only postive domain

[solid juniper]

can i see the problem like it was written in the book?

[opal sparrow]

yeah

[solid juniper]

yea this is similar. if (x,y) is an optimal solution, so is (-x,y). since x^2 = (-x)^2 for any x

[opal sparrow]

ohh okay, i was so confused when i intially saw it lol

thank you sm!

[solid juniper]

oh also

[opal sparrow]

hm

[solid juniper]

here it says specifically positive numbers

[opal sparrow]

oops

did not read that 😭

[solid juniper]

😭

[solid juniper]

but even if it didn't you could just look for solutions with x > 0 if that was more convenient for whatever reason because of this

and then you know the negative ones from that

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[mellow meteor]

I have a question

[mellow meteor]

If you prove discriminant of discriminant less than or equal to zero, then how does this support the original quadratic will have real roots?

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.reopen

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✅

[mellow meteor]

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[marble oasis]

how to do this?

[marble oasis]

i dont have any clue to do this i wasnt at school

For 10 I can probably say 170

inches

for 9 is x is 2

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[tawny drum]

Context:

so i have equations of s+f+l=p (p=extra total amount)

15s+5d+20L=500

[tawny drum]

s-F+2L=0 (from this)

so i solved that p=50

[tawny drum]

then to solve this i used 2F-L=p

which is now 2F-L=50

and since L must be greater or equal to 0

i know that F must be greater or equal to 25

i cant think of what else to use to form more inequatities

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[balmy flint]

Had completely no idea why answer is e 🥲

[dull oxide]

The domain of the inverse function is simply the range of the original function

[balmy flint]

So like this? But then what do I do🥲

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[latent smelt]

which of the products is equal to 8^x

A 16 16
B 16 32
C 32 32
D 32 64

[latent smelt]

16 = 2^4
32 = 2^5
64 = 2^6

what can I do from here?

[clear umbra]

is x an integer?

[latent smelt]

or yes I think it mean integer

[clear umbra]

then just convert 8^x to a power of 2

[latent smelt]

(2^3)^x?

[clear umbra]

yes

[latent smelt]

ok so I need to find which integer is equal to (2^3)^x

[clear umbra]

you can simplify (2^3)^x

[latent smelt]

2^3x

[clear umbra]

and so since x is an integer the power must be a multiple of 3

[latent smelt]

A 2^8
B 2^9
C 2^10
D 2^11

So it must be B

or do you not need to calculate them?

[clear umbra]

those powers?

[latent smelt]

yea

[clear umbra]

nah

[latent smelt]

so you just find their answers in power and compare to 3x

9 = 3x, x is an integer

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[wary coral]

the question is if f′(3)<f′(4), like true or false, and idk what decides it, im assuming it has something with like which of them has the highest rate of increase. So im guessing f'(3) idk (btw i do not have the function itself only the picture of the graph)

[upper abyss]

It's pretty unclear. f'(3) looks slightly larger, maybe?

[wary coral]

maybe xd

but if like

if it was f′(3)<f′(5)

then f'(5) would be going downwards

makin it false correct or?

[upper abyss]

f'(3) is positive, f'(5) is negative.

[wary coral]

yeah so then f′(3)<f′(5) would be incorrect?

[upper abyss]

ye

[wary coral]

ok so if i think it seems like f'(3) is sligtly more positive than f'(4) i shud just answer that right?

(there are no conclusions on this task btw, it just says "discuss with ur classmates")

cooked

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[hybrid gyro]

Can someone help me with this, Im completely blank:
John cut 2/9ths of a roll of tape, then 55cm more. When he finished that, he was left with exactly 7cm less than 4/9ths if the roll of tape pre cutting. How long is the tape pre and post cut?

[hybrid gyro]

I got to here:

X-tape length pre cut

X-(2/9x+55)=4/9x-7=

And now im stuck

[wary coral]

shudnt it be 4/9x+55 in the ( ) (i aint no helper jus askin)

[cedar kiln]

@hybrid gyro Has your question been resolved?

[hybrid gyro]

Yeah

<@&286206848099549185>

[hybrid gyro]

Wait no

[lofty kraken]

drawing it out we get something like this

and we know between the 2/9 and 4/9 are 3/9 of the roll

[lofty kraken]

do you think you can work out from the drawing what those 3/9 of the roll would be

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[mint moth]

im not sure how to solve this? i know that square root is the same as ^1/2 but im not sure what to do next

[mint moth]

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[fringe mauve]

,rotate

[mortal grove]

convert 10 +/- sqrt(5) into decimal form, and use the fact that X is discrete

[fringe mauve]

do i round it

[mortal grove]

no, keep that in decimal form

if i tell you i flipped heads more than 1.842 times, what does that mean to you?

[fringe mauve]

2

or more

oh oik

that makes sens

[mortal grove]

👍

[fringe mauve]

thx

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[sour surge]

.

[sour surge]

Is this the correct answer to the problem

[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).

[queen ice]

Notice how it says pingu in the name

You can’t open a channel with .

[sour surge]

ok i will try again

[fervent depot]

someone gel

someone help

[azure swift]

Have you gone over the mean value theorem?

[fervent depot]

yes

it’s been a while so i’m a little rusty

[azure swift]

See if you can recall the statement of the theorem by checking your notes or book

@fervent depot

[cedar kiln]

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[fringe mauve]

,rotate

[wraith dagger]

Couldn't find an attached image in the last 10 messages.

Oct 24, 2024, 2:52 PM

[fringe mauve]

Is this correct I t his case

,rotate

[dusk coral]

@fringe mauve 12 people possible times 10 possible times 8 possible = 8(10)(12)

[fringe mauve]

Also what about this one

[fringe mauve]

Huh yeah I guess that works

But what’s wrong with mine

I choose a pair

Take 1

Choose another pair

Take 1

Choose the last pair take 1

[dusk coral]

Wouldn’t this be 2 choose 1

Inside the pair

[fringe mauve]

Oh yeah

U get same answe

Mb

[dusk coral]

Over three iteration 2^3=8

Is what you’re off by

[fringe mauve]

Wdym

[dusk coral]

Hmm?

[fringe mauve]

This

[dusk coral]

Your answer was off by a factor of 8 since you didn’t have the 2’s

[fringe mauve]

Oh ok

Thanks

.clore

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[crimson sedge]

Guys

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

No

It never will

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[lone lagoon]

how would you determine the term of the secong grade of this?

[grand shale]

segundo grado de

[lone lagoon]

as far as I know, we wanna have (27 k)

yeah

[static willow]

what is secong grade?

[lone lagoon]

so pretty much

like

when the X^2?

[static willow]

coefficient of x^2?

[lone lagoon]

Like

to find a term where

the term is x^2

[static willow]

so when the thing is 5x^2

u want the 5

[lone lagoon]

yeah, pretty much

[lone lagoon]

why

[static willow]

no thats an example

[lone lagoon]

I want to find that, yeah

[static willow]

so the whole thing is full of terms, and one of the term is kx^2, and you want to find “k”

[lone lagoon]

so we want to find which term is ^2

yeah

[static willow]

is my statement right

alright

lets call k “coefficient of x^2”