help-17

[lusty briar]

Okay wait let me chase back to the step and fix it

[paper depot]

the upshot is that the 119 should actually be 115.

i suppose a bit more white-out can't hurt.

[lusty briar]

So it shld be 119

Shldnt

Oops

Im making more thana rthimetic mistakes

Spelling mistakes too

-7.2

Instead of 3

[paper depot]

yeah you're swallowing vowels like you're eating them for breakfast

"shldnt"

[lusty briar]

Simplified words

[paper depot]

d y wnn by mr vwls r smthng lk wh tps lk ths

[lusty briar]

ts pmo type beat

3

3=a

So answers 3

Wow

[paper depot]

ok FINALLY yes

[lusty briar]

HELP

AND THATS JUST PART A

😭😭😭😭

AND MY EXAMS TMR😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭

[paper depot]

the conclusion is that you need to brush up HARD on your equation solving skills..

ah fuck.

[lusty briar]

Im naturally bad at math

[paper depot]

i think you're kind of cooked for the exam

though i would hesitate to write that off as "naturally bad"

[lusty briar]

Im sctually pretty good at factorizing

As in

Its one of the only kind of questions i cna solve without hiccups

[paper depot]

well ok you are good at factorizing but you fumble the simple shit hard

incl. arithmetic

so kinda uhhhh

good luck

[lusty briar]

Ok thats cause im like cluksy n studd

Mode is easy

Mode is 2

Aka 16

Median is

Oh fuck

Wait

[paper depot]

Mode is 2
Aka 16

the mode is 2, the mode is not 16

don't confuse mode with modal frequency

[lusty briar]

im trying media mn😭😭😭😭😭😭

[paper depot]

??

[lusty briar]

Median

Rn

https://tenor.com/view/orange-cat-gets-flung-and-explodes-orange-cat-funny-cat-meme-explodes-gif-10706110874965244466

Can i bet u that every other persons question here is tenntimes harder than mine but im struggling the most

Okay ykw im mcoing on to another question

Thanks for the help😭😭😭and being pateint twhen im being an idiot

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[cold locust]

dumb question, but i genuinely forgot. unlike when you eliminate the denominator in a equation by adding or substracting, when you are doing that by multiplication or division, you have to multiply / divide all terms of the side of the equation, right ?

[civic otter]

Always do things on both sides of the equation or inequality

Both if you wanna multiply/divide by something and if you wanna add/subtract something

[cold locust]

im asking if i multiply all the terms of the side or just one element ?

[civic otter]

Everything, the whole equation, i.e. all the elements

You have to imagine it as keeping the LHS and RHS on a scale

[cold locust]

thanks

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[vale tulip]

guys how do i do this using delta epsilon definition of limit

[pure marten]

Have you gotten started?

[vale tulip]

yeah, ive written 0<x-a<delta => I f(x)-L I<epsilon and 0<x-a<delta => Ig(x)-M I<epsilon

how do i proceed?

[pure marten]

you need to show that |fg - LM| < ε

[vale tulip]

yes

[pure marten]

I suggest you work in reverse form this inequality and then just reverse the steps

Let A be a bound for f, i.e. |f| < A (choose an actual number for A if you want here)

You can add and subtract inside the absolute value

|fg - Lg + Lg - LM|
|g(f - L) + L(g - M)|

Triangle inequality helps from here

[pure marten]

you can retrieve the inequalities from here and use them

[vale tulip]

oh okay

wait i'll try further calculations

[vale tulip]

oh

i understand

wait i dont understand, i took my two inequalities

what do i do with them now

[meager shoal]

ok we use archimedean principle to do the trick

we will show that that ${g(x)(f(x) - L) < \epsilon/2}$ and ${L(g(x)- M) < \epsilon/2}$

[twin meteor]

k

[pure marten]

What do you have so far?

(I assume you're going to flesh out the details of the proof yourself)

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@vale tulip Has your question been resolved?

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[frail violet]

for the first half of the question can someone please explain whats wrong with my logic

[frail violet]

I take a random vector in U call it u
u = a(x1) + b(x2)
now if this vector is closet to x
(x-u).u = 0 (orthogonal)

I dont get how this is wrong

[vocal sleet]

@frail violet Has your question been resolved?

[vocal sleet]

@frail violet Has your question been resolved?

[frail violet]

@reef grove

[vocal sleet]

@frail violet Has your question been resolved?

[flat whale]

why do you think it's wrong

[frail violet]

Bcz the answer i get from this is not the same as the answer given in the book

[vocal sleet]

@frail violet Has your question been resolved?

[languid hare]

remember that solving this gives you only u

they want you to find u, and they want you to express x = u + x-u (u is in U and x-u is orthogonal to U, as asked)

your method is correct, just make sure at the end you have answered what they have asked for

😄

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[tight delta]

im given a sine function 3sin(pi/5x + 2pi/5)-3 how would i be able to find the min/max points?
i know how to find the amp, mid, and period but when plotting the min/max points. i cant figure how to find the points?
please advise
thanks

[flat python]

do you know what derivatives are?

[tight delta]

no

this is a trig question

[flat python]

well you know how to find the amplitude right

[wary mantle]

Do you want the values of the min/max or the x-coordinates to it?

[tight delta]

yEA its 3

mid is -3

period is 2pi divide by pi/5

[flat python]

so you need to find the x values such that the function equals the amplitude

as the amplitude is the highest value the functions can obtain

(or the lowest)

[tight delta]

yea so sine(0) = 0 right?

so that would be the original midpoint minus -3 would be shifted down by 3

amp would be -7 and 1

is that right?

[tight delta]

hey sorry, i need to figure out the min/max points coordinates

in order to plot

is it possible to figure out if i can find out the perioud?

cause i can do that

period should be 10 right?

so basically

amp = 3
mid = -3
period = 10

phase shift = 2pi/5

is that right?

but i dont think the phase shift alone is enough info to figure min/max points

or is it? im not sure

actually it should be enough

cause it's basically telling you how far to move left/right

[tight delta]

and that alone should already give you the min max points right?

yea i understand that

[flat python]

So what part do you not understand about solving for the maximum and minimum points?

Where are you stuck

[tight delta]

phase shift is 2pi/5? is that correct?

[wary mantle]

The values of the min and max are clear, sin will be maximal with $1$, so you get $3(1) - 3 = 0$ as maximal value, minimal with $-1$, so you get $3(-1) - 3 = -6$ as the minimal value. You just want the $x$-coordinates now. You know that $sin(x)$ is maximal at $x = \frac{\pi}{2} + 2k \pi$ with $k \in \mathbb Z$. So: \begin{align*} \frac{\pi}{5}x + \frac{2\pi}{5} &\overset != \frac{\pi}{2} + 2k \pi \ \pi x + 2 \pi &= 5 \frac{\pi}{2} + 10k \pi \ \pi x &= 5 \frac{\pi}{2} + 10k \pi - 2 \pi \ \pi x &= \frac{\pi}{2} + 10k \pi \ x &= \frac 12 + 10k \end{align*}

[tight delta]

i need to graph that equation and im unsure how to graph the min/max points

want to seee my graph? and you can tell me if its correct?

[flat python]

👍

[wary mantle]

This means the first maximum is at x = 1/2

The second at x = 10.5

And so on

[tight delta]

cause all the yt videos on graphing, only provide a general graph of how the funciton should look like....it doesnt provide exact points

anyways...give me a sec let me take a screenshot

[wary mantle]

Do the same and set the LHS equal to 3pi/2 + 2k*pi and you will get the coordinates of the minimum points

You can also leave off the 2k pi because you know the period is 10

[tight delta]

https://i.imgur.com/BQHcpkU.png

[wary mantle]

Looks right

[flat python]

looks great

[tight delta]

yea but thats only because i used demos to graph the function

[flat python]

ah

[tight delta]

i dont really understand why its supposed to look like that

[wary mantle]

[You can also skip this step, you know the min points will be in the middle of the max]

[flat python]

calm down and listen to this yea

[tight delta]

@wary mantle im going to look at your explanation...thanks

so to find max, general formula is basically amp(1)-mid?

or amp-mid?

[wary mantle]

Yes, almost

[wary mantle]

amp(1) + mid it should be though

[tight delta]

and min is amp(-1)+mid

[wary mantle]

Yes

[wary mantle]

You don't really need to remember this, just know that sin(x) has maximal value 1 and minimal value -1

So something like 500*sin(x + 5 + x/1000) + 6 - 1 will be maximal at 500 * 1 + 6 - 1

[tight delta]

yea so its
max = amp+mid
min = -amp + mid

[wary mantle]

Yes

[flat python]

I really think you're going about understanding this the wrong way, but maybe this works better for you

[tight delta]

but this gives the y value only right?

[wary mantle]

Yes

For the x-value see the second step of my explanation

[tight delta]

how should i be understanding it?

i want to follow best practices

[wary mantle]

He must be referring to what I wrote here

[flat python]

Yes

sin(whatever) is maximized if sin(whatever) = 1

and when is sin(whatever) = 1?

[tight delta]

i dont get how you're solving for x

[wary mantle]

$sin(\text{whatever}) = 1$ when $\text{whatever} = \frac{\pi}{2}$ or $\text{whatever} = \frac{\pi}{2} + 2 \pi$ or $\text{whatever} = \frac{\pi}{2} + 4 \pi$ and so on

[twin meteor]

Kepe

[tight delta]

what??

[wary mantle]

This is just directly from the unit circle, because sin takes on 1 at x = pi/2

[flat python]

Yes. Do you remember the unit circle?

[tight delta]

in what sense...yea i know

[wary mantle]

This is why I set whatever to pi/2

[tight delta]

yea 90

why you chose 90

[wary mantle]

Because there the value of sin is maximal

[flat python]

because that's sin(x) highest value

every other value of sin(x) is lower than that

so the function you have is maximized if sin(x) is 1,

[tight delta]

ok makes sense

[flat python]

so we need to find an x such that sin(x) = 1

which is pi/2 + k2pi where k is an integer

[wary mantle]

2pi is going a full circle around, you will land at the same spot, that's why you want the 2k*pi to get all solutions

[tight delta]

hmm

[flat python]

now look at this again

[tight delta]

we're trying to find the x value right? want to make sure we're on the same page

[flat python]

yes

[tight delta]

shouldnt the phase shift tell us?

which is 2pi/5?

cause i thought the phase shift will tell us how fat to move either to right or left

so not sure why we need to reference unit circle to solve x

[wary mantle]

Well that's basically what we are doing

We want whatever is inside of sin, together with the phase shift, to be pi/2

[flat python]

the sin function doesn't change, if it has sin(2392pi/23290 + 23290x), it will still be at it's highest point if the thing inside the sin = pi/2 + 2kpi

[tight delta]

im even more confused now... where did you get those values from?

[flat python]

it doesn't matter what those values are, this works for any function of sin

alright lets restart

what do you think of when you hear sin(x)

[tight delta]

sin(x)= 0

i think

[flat python]

not the unit circle or trig or the graph of sin(x)?

[tight delta]

sin(x) will tell you the amount of spread of the wave

amount of spread along the x-axis

[flat python]

alright imma re-introduce the sin function to u

imagine a circle

[tight delta]

wait

sine function will pass through 0,0

i get that

cos amp will be (0,1)

cprrect?

[flat python]

sure

[tight delta]

but you knno what im trying to say?

[wary mantle]

Did your class cover sin as the graph or did you also do the unit circle?

If you didn't do the latter it might not make sense to talk about that

[flat python]

did you only learn about the graph y = sin(x) and the phase shift, amplitude and period?

[wary mantle]

Have you ever seen this thing?

[tight delta]

yo man, im self learning al lthis

[flat python]

and all that stuff

[flat python]

oh fr? respect

[tight delta]

yes mostly

do you know KA? i'm basically using that

[wary mantle]

Khan Academy, yes

[tight delta]

to be honest, he glosses over a lot of the material which is probably why there is gaps in my knowledge

[flat python]

damn

[tight delta]

but im pretty sure i understand the fundamentals of sin/cosine

and unit circle stuff

i just cant figure out how to find x,y max min points

well i think i know how to find max/min y points now

but finding max/min x is still tripping me up

can you explain one more time why we're using the unit circle to find the x value?

if i still dont understand, then i will just bypass this topic and move on

trig is one of those subjects where i cant just learn everything in one sitting i've learned

i really have to understand wtf is going on

unlike other math topics i was learning previpously

[wary mantle]

It's funny, once you advance to more advanced things, everything before that will feel trivial because you have used it a lot

[tight delta]

ya

hopefully i can reach that advanced level

but im starting to lose hope

[wary mantle]

Math is all about being stuck

[tight delta]

if i cant figure out basic trig, not sure how i can learn the stuff that comes afterwards

[flat python]

You'll figure it out, there's no need to worry

No one gets it on their first try

[wary mantle]

I will TeX a kind of summary up, then I'll go to sleep. Read it slowly and carefully and surely you'll get it

[tight delta]

whats tex

[flat python]

thanks Kepe, I don't think my bad discord chat explanation is working XD

[tight delta]

ok thanks anyways guys

[flat python]

a program where you can type math

[tight delta]

ill try to figure it on my own

ive been stuck on this for a few days now

[flat python]

Why? We're here to help

[tight delta]

oh i thought you were going to sleep

[flat python]

he's typing an explanation for u to make it more clear

in the math typing program

[tight delta]

ok

[wary mantle]

You can imagine $\sin(x)$ either as the \underline{vertical} length on the unit circle or as the graph, both are equivalent. The latter might sometimes be more convenient, but I will use the former now. \ \hr \ Given $f(x) = \sin(2x + 1)$, let's find the \underline{$x$-coordinates of the maxima}. Imagine the graph of $\sin(x)$ instead, you know that the first maximum is at $x = \frac{\pi}{2}$. \[10pt] Now instead of $sin(x)$ you got something else instead of $x$, but that something else still needs to be $\frac{\pi}{2}$ as before for $f$ to reach the first maximum! \[10pt] So we set $2x + 1 = \frac{\pi}{2}$ and solve for $x$. \ \hr \ This gives you the $x$-coordinate for the \underline{first} maximum. Now since you know the period, you can juts add that to what you found, and you will get the \underline{second} maximum. And so on.

[twin meteor]

Kepe

[wary mantle]

Graph of sin(x):

[tight delta]

ok let me break it down

[wary mantle]

You want whatever is inside of sin to be pi/2

[tight delta]

you're using pi/2 because that would be 90deg, which is 0,1 which is the max point for sine?

[wary mantle]

cross the 0, 1 out

[wary mantle]

It's because you know this graph by heart and you see that at pi/2, it has the biggest value

Now instead of as in this graph, sin(x), you got something else instead of x. So you want that something else to be pi/2 because then, sin(something) = sin(pi/2) = max

[tight delta]

ok

can you use my function as an example

the sine function i provided above

3sin(pi/5 + 2pi/5) -3

[wary mantle]

Given sin(500000x + 1 - 6000 + 2), if you want to find the x-coordinate where it's maximal, you just need to set 500000x + 1 - 6000 + 2 = pi/2. Because if you find that x, then you know that sin(500000x + 1 - 6000 + 2) = sin(pi/2) = maximal

[wary mantle]

If you want to find where that's maximal, you want to find the x for which pi/5 x + 2pi/5 = pi/2

[tight delta]

ok so doesnt matter whatever value it is....it sohuld always equal to pi/2?

[wary mantle]

Yes

[tight delta]

and then you just solve for x?

[wary mantle]

Because there, the sin(x) function is maximal

Yes

[tight delta]

so in my example

[wary mantle]

pi/5 x + 2 pi/5 = pi/2

Gives you ...

[tight delta]

wait...let me do it

1/2

[wary mantle]

Yes

[tight delta]

x=1/2

[wary mantle]

Ok and now every sin function is periodic, right

So there is not only one maximum but infinitely many

But you know the period

[tight delta]

so this is that max x point so max would be 1/2, 0?

[wary mantle]

What is it?

[tight delta]

correct?

[wary mantle]

Yes

[tight delta]

period = 2pi/pi/5

[wary mantle]

Yes

Which is ...

[tight delta]

which is 10

[wary mantle]

Yes

Ok so where is the second maximum?

If the first is at x = 0.5

If you know everything repeats after 10 units

[tight delta]

so multiple 1/2 by 10

5

[wary mantle]

Not quite

Everything repeats after 10 units

If you go 10 to the right, everything will repeat

[tight delta]

10.5

[wary mantle]

Yes

The third is 20.5 and so on

Same happens to the left

[flat python]

30.5

40.5 ...

[wary mantle]

If you go 10 to the left from 0.5 you get another one

-9.5

[tight delta]

ok and then to solve for min y= its basicaly kx=3pi/2?

[wary mantle]

Yes but now that you know where the maxima are you don't need that anymore

[tight delta]

i mean min x

oh min x should already be given

if you know the mix and perioud

[tight delta]

is that correct?

[wary mantle]

Let's look at this

Here, the period is 2pi

Say we are at the first maximum, pi/2

How much to the right do we need to go to reach the first mimum?

[tight delta]

pi/2

[wary mantle]

That's where the graph is 0

Not the minimum

[tight delta]

3pi/2

[wary mantle]

Yes, that's where the mimimum is

[tight delta]

ya

[wary mantle]

So from pi/2, the maximum, we go half a period to the right to reach 3pi/2

Same thing here:

In our case, 10 is the period

First maximum is 0.5

So we go half the period to the right

0.5 + 5

5.5

That's the first min!

[tight delta]

yea cant we also multiply pi/2*4

to get perioud?

[wary mantle]

Sure..

[tight delta]

because from 0 to pi/2 is quarter of a period?

or is that wrong

[wary mantle]

Yeah

No you can do that

If you know the point where the graph is at the mid

And you know where it's at the max

You know the distance between them is a quarter of a period

So you can multiply that distance by 4 to get the period

[tight delta]

ok i think i got it

i will continue to pracitice and review this stuff

[wary mantle]

You can also set whatever is inside of sin to 3pi/2 to get the first mimimum, but that's longer if you already know the max

[tight delta]

ok.....its slowly starting to make sense

my brain needs time to process all this new information

but i tihnk i understand it

[wary mantle]

Make sure to do exercises

[tight delta]

yea i've been doing those KA practice questions....

its super annoying cause he doesnt explain any of the stuff that you just explained to me

like i said, he glosses over a lot of info

[wary mantle]

Tbf once you get to uni level or calc, I've heard and seen that KA's problems are very elementary

You won't get really tricky ones from those automatically generated ones

[tight delta]

well yea...i think KA stuff is mainly elementrary, high school stuff

[wary mantle]

And it can be beneficial to think about difficult stuff

[wary mantle]

Yeah but you can still find tricky problems to everything

It's just that KA doesn't have them

[tight delta]

the question i was doing is tricky to me

but yea it's probably nothing for you

[wary mantle]

Yeah for the beginning the questions they give are good

[tight delta]

what are you? highschool student?

[wary mantle]

But once you practice enough of them, you will be able to solve these in under a minute

[tight delta]

well i thats the plan

[wary mantle]

Just finished high school but I was doing uni in parallel to it

[tight delta]

holy f.....yea some people just get math.

ok.....i gotta go

thanks for your help....

[wary mantle]

np

[tight delta]

i'll review your notes again...dont delete them

or ill just save them now

ok thanks

@nimble heath thanks

@flat python thanks

[vocal sleet]

@tight delta Has your question been resolved?

[vocal sleet]

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[marble tinsel]

this is adding the two vectors right

[flat whale]

Find the vector first and show it here

[marble tinsel]

nvm i figured it out

.close

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[tame frigate]

hello!!

[tame frigate]

asking for opinion here

Study #1. Automated Discovery of Algorithm Optimality Proofs Using Program Synthesis and Formal Methods

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Study #2. Pharmacokinetics and Pharmacodynamics Modeling, Density and Wave Functional Theories, and Molecular Dynamics Simulations for Dual-Stage Inhibition of HIV-1 Infection: Novel Bio-Computational Analyses of Stephania tetrandra on CD4-gp120 Interaction (Entry) and Mutant PRF53L (Replication), with F24-DR11-RQ13 Peptide Complex Binding**

description: This study used advanced computer-aided drug design (CADD) to evaluate Stephania tetrandra alkaloids as potential inhibitors targeting both HIV-1 viral entry (CD4-gp120 interaction) and replication (Mutant PRF53L), addressing drug resistance. The selected compounds demonstrated strong binding affinities, favorable pharmacokinetic and toxicity profiles, and stable protein-ligand interactions, indicating their potential to enhance CD4+ T cell responses and improve immune defense against HIV.

Study #3. Deep Learning Discovery of Mathematical Patterns in Viral Evolution for Real-Time Pandemic Prediction Using Novel Topological Data Analysis and Stochastic Differential Equations

description: "AI-Driven Mathematical Discovery for Pandemic Prediction and Prevention" uses advanced topological data analysis, stochastic differential equations, and deep learning to discover novel mathematical patterns in viral evolution, creating the first real-time global pandemic prediction system. This breakthrough combines cutting-edge mathematics with artificial intelligence to prevent future pandemics by predicting viral mutations and transmission dynamics before they occur, potentially saving millions of lives and trillions in economic damage.

What if this three compared, which do you think will win in a competition in a science fair? under the category of mathematics and computational science, just asking for opinion, tysm! :>

[vocal sleet]

@tame frigate Has your question been resolved?

[summer orbit]

Study 1 is something that is impossible to verify.
Study 2 addresses a problem that actually is there.
Study 3 is just plague inc.

you should dig into the criteria that the project will be evaluated

if it's just sounding flashy, you can picky anything

again, everything is dependent on the criteria

[tame frigate]

here's the criterai if you want to know

[summer orbit]

since you're aiming for mathematics and computational science

2, even if it's a good idea, doesn't relate to the topic

1 is something that's very hard to verify.

3 really seems like the best option, even if it's a plague inc ahh

[tame frigate]

alright, thank you.

[vocal sleet]

@tame frigate Has your question been resolved?

[tame frigate]

uhh I wanted more opinions to it sooo :<

[vocal sleet]

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[terse fiber]

I just had my math exam, and saw this question. Solve the equation.
ln(x)-ln(x+1)=1. I came down to the conclusion that there is no valid solution, I was wondering if anyone else also comes down to the same conclusion?

[terse fiber]

I first of used the log ruled to make the left part into ln(x/(x+1)), where I after used e^ on both sides

[woeful igloo]

You are right. You have an expression with logarithm, so that obv means that x>0

And that makes x/(x+1) < 1, which doesnt follow the given expression

[signal pendant]

well, you can see this simply by noting that log x is strictly increasing

so there can never be a solution

infact there can never be a solution for ln x - ln (x+1) = c for any c \geq 0 by this same logic

[terse fiber]

Ah right, thanks for confirming 😄

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[tropic minnow]

x³ −x−5 = 0 ? I-...

[paper depot]

any context for this question?

and also progress

[tropic minnow]

I need to do a complete study of the function to sketch the graph and I forgot how to take the roots of the function

What is the value of x when the function is equal to 0? I don't know if there is an easy way to solve it.

[chilly widget]

U need exact value?

[tropic minnow]

Can be approximate

[chilly widget]

U can find the points where the slope is 0

And see the number of roots first(1 or 3)

[tropic minnow]

oh i dont known how to do this

[chilly widget]

Find the first derivative

And equate it to 0

[tropic minnow]

Oh but

This wont give me the number where the function cut the abscissa axis

will give me the maximum and minimum values

relative

[chilly widget]

Yeah but to draw the graph u need the turning points too.

I think it's more important than the roots

[tropic minnow]

i see

well im gonna do this then

How can u tell it?

[chilly widget]

Differentiate

Yea sorry made a mistake

[tropic minnow]

ah yes i got u

[chilly widget]

±1/√3

Now I think the y value at both these value of x is -ve

[tropic minnow]

of f(x) or f'(x)?

[chilly widget]

f(x)

So only 1 root

[tropic minnow]

But how did u think of it?

[chilly widget]

Mind calculation

Like -5 will have a greater effect

value of x is just ±1/√3

[tropic minnow]

x³ - x = 5
(±1/√3)³ - (±1/√3) = 5

kkakssks too complicated for me to picture

but i think ur right

[chilly widget]

We r looking into the value of f(x) at the points where slope is 0

[chilly widget]

Do you have a rough idea how the graph of a cubic polynomial looks?

[tropic minnow]

+-

i guess its like / \ /

or

[chilly widget]

Yea

[tropic minnow]

\ /\

[chilly widget]

That's how ours look like

Since x^3 coeff is +ve

[tropic minnow]

hmmm

[chilly widget]

We just found out the y coordinate of the peak in the graph

[tropic minnow]

so this rule also works for cube roots

[chilly widget]

And found out it's -ve

Which rule?

[tropic minnow]

of x² being +ve affecting the graph

[chilly widget]

Mm

[tropic minnow]

in this case being x³

[chilly widget]

Yes

[tropic minnow]

i understand

[chilly widget]

U understood this?

[tropic minnow]

well.. i kinda got it

[tropic minnow]

this

[chilly widget]

So can u get that there will be only 1 root?

[tropic minnow]

I’m still not really used to this, so it’s a bit hard to fully get it, but Im kinda just accepting it for this exercise 💀

but thanks

im gonna do the other parts

do i have to end this chat? i don't remember

[chilly widget]

yes

Like .close

[tropic minnow]

.close

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[vast shale]

how can i solve this question?

[vast shale]

i tried squaring both sides of the equation but that doesnt help

[paper depot]

$ab^4aab^4a=b^{14}$

[twin meteor]

Ann

[vast shale]

i will share my solution wait

[vast shale]

what is this?

what i dont understand in my solution is in the line where we are considering b^n = ab^ma and b^k = ab^ma we consider that m has the same power which is m and then we solve the question and get n as 40, in my solution i have not proved what it is required to prove

[paper depot]

it's the third line of your work

or the first after givens

[paper depot]

this solution looks correct

[vast shale]

But in this solution I have assumed that b^n and b^k is equal but why have I assumed that??

[paper depot]

you didn't assume it, im reading it as more of an "i will set out to prove this instead"

[vast shale]

ouhhh

[vast shale]

thank you so much for helping me so many times

hi, can you help me with this question too? sorry to disturb you ann, but this chapter is very difficult for me and i dont understand whether the solution is correct or not even if i prove it

this is my solution

[paper depot]

,rccw

[paper depot]

abab is not known equal to a^2 b^2

we don't know that a and b commute

[vast shale]

so how can we do it?

can we write a = a^2 by cancellation of the left and right b's?

[paper depot]

ba^2b^-1 = a

square both sides here

careful not to commute

[vast shale]

okay

now we can square both sides after doing this?

[vast shale]

btw, can we do this?

[paper depot]

no

you can cancel if both b's are on the left or both on the right

but not the way you say generally

[vast shale]

then we can cancel b in this equation?

[paper depot]

no

they are not next to each other

[vast shale]

what does it mean? i dont understand sorry😭

oh okay i think i understand

if it was a^2bb^-1 we could have cancelled it then?

@paper depot is this correct? 🥲

[paper depot]

yes in this case they can be cancelled

[paper depot]

yes looks ok

[vast shale]

thank you very much 😄 I have my exam tomorrow

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

hi can anyone explain me that if we have a group G which is defined on multiplication then how is the subset of G which contains only identity element is also a subgroup of G?

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

This is the trivial subgroup right?

See it satisfies all the group axioms

[vocal sleet]

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[vast shale]

where do I even get started with this problem

[vast shale]

should I use quadratic formula?

[vital sable]

Yes

[vast shale]

we have the same first name

.close

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[vast shale]

Use quadratic formula

X and Y are integers

So it's easy

[peak matrix]

Why is the give up button more appealing than the input field, it almost looks like the problem wants you to give up

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[quartz zenith]

How do I solve 1000=8000/x^3

[quartz zenith]

I tried first by

Multiplying x^3 on the other side to get rid of it on the one side

But then I’m stuck with 1000x^3=8000

[scenic ravine]

yes, now divde across by 1000

[quartz zenith]

Ohhhhhhhhhhhhhhhhh

So

[glacial silo]

It’s easy

[quartz zenith]

X^3=8

And x is 2

[glacial silo]

Yeah

[quartz zenith]

Okayy tysm I was a bit slow 🙏

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[quartz zenith]

.reopen

[vocal sleet]

✅

[quartz zenith]

I feel dumb again

How do I get x by itself in this 5x-ax=10

Idek what to start dividing here

[cyan shadow]

"factor out" x here

5(10)+5(5) = 5(10+5)

as an example

[quartz zenith]

I don’t understand how to start

[cyan shadow]

$ab + ac = a(b+c)$

[cyan shadow]

you need to apply ^^^^ rule to get the x out of the two terms

[quartz zenith]

Ohhh so like

X(5-a)

[cyan shadow]

yes, that's one side

x(5-a) = 10

[quartz zenith]

Ohhhhhhh

[cyan shadow]

can you keep going from here?

[quartz zenith]

Ohhhh yeah now I divide it on both sides right

[cyan shadow]

depends what you mean by "it", but probably yes

[near pawn]

If you're struggling I'd suggest you check out khanacdemy. They have videos on pretty much all topics

[quartz zenith]

So it’s x=10/5-a

[quartz zenith]

What is that

[cyan shadow]

no!! be careful of parenthesis

[near pawn]

It's a website

[quartz zenith]

Ohhhh

[near pawn]

With videos and exercises for math

For all levels

[cyan shadow]

x = 10/5-a is $x = \frac{10}5-a$ which is not right

[near pawn]

The videos are really good

[cyan shadow]

you mean $x = \frac{10}{5-a}$

[quartz zenith]

Yessss

That’s what I mean

[cyan shadow]

which is written as "x= 10/(5-a)"

you need the parenthesis to show that they are both on the bottom

[quartz zenith]

Ohhhhh tysm I’ll remember it

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[vocal timber]

"Person started his journey in the morning. At 11:00 a.m. he covered 3 / 8 of his journey and on the same day at 4:30 p.m. he covered 5 / 6 of the journey. When he had started his journey?"

I am feeling so much dull that i am not able to solve this type of problem that is asked in 8th grade. 😰 What is wrong in my thinking?

[near pawn]

What is your thinking?

[vocal timber]

We have to tell the initial time taken before all the distance covered and time taken

[near pawn]

Also not beat yourself up if you can't solve a lower grade problem. It just means you haven't learnt those topics when standard curriculums teach them, it doesn't mean you're dumb or anything

[vocal timber]

The thing what lows down my confidence is i have learnt them when it was taught

[near pawn]

Well if you practice you'll get there dw

Some topics are harder than others for some people so don't stress

You'll get it eventually

Anyway

[vocal timber]

i have practiced and worked such problems in smaller grade and in higher grade i should be able to solve quickly ain't it?

[near pawn]

How much distance did he cover between 11 and 4.30?

[vocal timber]

5 and half hours

[near pawn]

Thats the time, yes

But how much of the journey does he complete

As a fraction

[vocal timber]

In that time he completed 3/8 and 5/6 that is 29/24

;(((

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[green kraken]

Can someone help me solve i with working

[quiet echo]

What can we note from the properties given?

And namely what can we do with AOC (the triangle)?

[charred moss]

math is a esoteric study

[cosmic arch]

An* LOL

[charred moss]

purely esoteric and dependant on feedback from others to reach a shread of value

i just realized how bizarre learning math works

[tranquil charm]

#discussion

[charred moss]

were solely dependent on the passions of others or there sheer perseverance alone to educate ua

OH

MB gng

[green kraken]

aoc is isoceles

meaning

angle aco = angle oac

wait

i think i got it

nvm

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[inland sail]

how do i determine the sign before the j and k

and i

in cross product

hi

[tall rain]

iM(i)-jM(j)+kM(k)

if you're expanding by the top row

M represents the minor

[inland sail]

its always i - j + k?

[viral glade]

yep

[tall rain]

yes, but it can be of the form iM(i)+jM(j)+kM(k) if the minor of j is negative

[inland sail]

yea i get it

k thanks

[quiet echo]

Oh, I had a really good example on this

I think there is a method, its called the "clock method"

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[frosty galleon]

I need to find the domain here.

[frosty galleon]

I've followed all the steps, and I understand all of them, but I can't understand how they got that answer

This is the graph I've been using:

(I got the same upper limit of 2/pi arc cos x, but my lower limits are different)

[vocal sleet]

@frosty galleon Has your question been resolved?

[frosty galleon]

<@&286206848099549185>

[vocal sleet]

@frosty galleon Has your question been resolved?

[white palm]

https://www.desmos.com/3d/vc7gnybacf

[white palm]

I think you have something wrong with the x^2+y^2-2x+2y+1 > 0 condition

i think it should impose (x,y) to be outside of a disk of radius 1 centered at (-1,1)

this would fit with the solution where they completed the square to make the circle more explicit

[frosty galleon]

I solved the equation to get (x-1)^2 - (y+1)^2 > 1, hence the hyperbola

I see where I made the mistake

Thank you!

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[whole dagger]

Why do I need a again to recreate the same graph from the factors?

[whole dagger]

I'm confused as to why it is 2(x-(1/2))(x+1) and not (x-(1/2))(x+1)

[glass oak]

well when you multiply it out and replace the variable x with p, do you get the original polynomial?

[whole dagger]

no

https://www.desmos.com/calculator/wmav8ujteb

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[latent bough]

can i get some help please, im always confused as to how to start these vector proofs when im not given magnitude or any actual vector

[latent bough]

im always confused how to start proofs

[quiet echo]

You should start with a claim or a well known statement.

[quiet aspen]

can someone help me, im new to server

[quiet echo]

In this case, what does perpendicularly imply with respect to the dot product?

[quiet echo]

!occupied

[vocal sleet]

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).

[quiet echo]

You can also just label a generic parallelogram ABCD and possibly use some vector addition (and dot product action) to show it satisfies the requirements of a rhombus (i.e., all lengths are the same).

If I recall correctly, the dot product is linear so the vector addition thing should work.

[latent bough]

uhh whats tha

[latent bough]

how tho? do i just use points a1a2a3

[quiet echo]

A fact that is obvious (possibly trivial).

[quiet echo]

Yeah.

However you like to label it.

[quiet echo]

^ This would be a nice starting statement.

The answer to the question, rather.

[latent bough]

Dot product =0

[quiet echo]

Yes.

[latent bough]

Ok so

Should I get 4 points

Abcd

As vertices of the parallelograms

btw

How do I prove that it’s a rhombus…..

[quiet echo]

^

[latent bough]

Sorry

😭

[quiet echo]

That the lengths of the vectors AB, BC, CD, and DA are all equal.

[latent bough]

how do I

First make sure that the diagonals are perpendicular using the dot product tho

they’re all variables

[quiet echo]

Read the question again. Is it asking you to verify that?

Or is it asking you to use that?

[latent bough]

How exactly do I use it…

😭

[quiet echo]

Trust me.

[quiet echo]

a•(b+c)=a•b+a•c

[empty bear]

i would label the width and height of the parallelogram as follows and try and find an expression for both diagonals using a and b, then use the given

[woeful stream]

probably add arrows in the diagram to make the direction of the vector obvious

[latent bough]

omg

I’m so lost

[empty bear]

with what

[latent bough]

Idek what im doing

[proven spire]

,rcw

[latent bough]

I thought it should be 0

[latent bough]

But nothing canceled out

[empty bear]

why is your vector 3 dimensional

[latent bough]

..

[proven spire]

I'm sure that's overkill

[latent bough]

……

nahhh

nahhhh

i was too lost in the third dimension

😭😭

Oml

sorry

[proven spire]

(after labelling these with arrows)
1 - Find expressions for the diagonals
2 - Deduce that their dot product is 0, and expand the brackets out (carefully)
3 - Determine whether |a| = |b|

(Is that the process you'd have gone with? @empty bear)

[empty bear]

yep

[proven spire]

(I tutor HS maths and I still hate vectors )

[latent bough]

honestly how do you even approach these proofs like

💀

[empty bear]

this is how lol

[latent bough]

Ya but im general

I

Un

In

[empty bear]

prob just practice

[latent bough]

[empty bear]

you're new to this so it'll seem out of left field but the more you do the more your pattern recog bank stores and it'll just become easier

[latent bough]

Ty

[latent bough]

ok ik I included an extra axis but is what im doing correct

I thought if yiu expand everything out it would = 0

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[vocal sleet]

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[warped osprey]

Hi I don't get what part d should look like

[gaunt finch]

it should look something like this

this is obviously for another graph/function

but at each grid point, there is a line segment of length 0.5cm which is angled in the direction of the slope given in the table

[warped osprey]

Oh this is part g of the same question, why does its slope field lines look different

[gaunt finch]

yeah that's what yours should look like
what do you mean by "they look different"

like the 3 lines they've drawn are just 3 "selected" paths

[warped osprey]

right

[gaunt finch]

they could've selected other paths and they would all look slightly different

[warped osprey]

okok got it thank you

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[glad cove]

question no. 29,30 and 31 is same as question number 24....

i wanna know which question or questions should i do out of these four

and save time

[flat whale]

Top to bottom

[glad cove]

i don't wanna do all the questions...

[flat whale]

Oh then don't do any of them

0 seconds

[glad cove]

😫

do u understand what i wanna say?

[flat whale]

Yea and it's not a math question

Go pick a problem, attempt it, and show your work here

If you get stuck

[paper depot]

then don't. who's forcing you?

[glad cove]

I am doing self study..
i thought of doing more questions in less time..
i thought...
nvm..

i think i have to stay restricted here

.close

[vocal sleet]

Channel closed

Closed by @glad cove

Use .reopen if this was a mistake.

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
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[brazen wharf]

The condition that x³-3px+2q may be divisible by a factor of the form x²+2ax+a² is ___________________________________________________________________________________________________________________.

[fervent wasp]

Wat hav u tri?

[brazen wharf]

x^3 - 3px + 2q = (x^2 + 2ax + a^2) (mx + b)

[fervent wasp]

good, can you factorize (x^2 + 2ax + a^2) ?

[brazen wharf]

(x+a)(x+a) = (x+a)^3 * (x+a)^(-1)

= (x+a)^2

[fervent wasp]

Do you know the value of m?

[brazen wharf]

1?

[fervent wasp]

yeah

[brazen wharf]

Ok

Let me update the system.
(x^3 - 3px + 2q) = (x+a)^2 (x+b)
(2q) = a^2 (b)
b = 2q/a^2

I put x = 0

[fervent wasp]

Cool, do you notice something is a bit off about the given cubic function?

[brazen wharf]

It's a depressed cubic

[fervent wasp]

It misses something

[brazen wharf]

It misses things

x^2 term?

[fervent wasp]

yes

Can you yield the solution from this information?

[brazen wharf]

x^2 b + 2ma x^2 = 0

b + 2ma = 0

m = 1

So b + 2a = 0

[fervent wasp]

Almost here. How about the x term?

[brazen wharf]

Oh

(x^2 + 2ax + a^2 ) ( x + b) = x^3 - 3px + 2q
a^2 x + 2ab x = -3px
a^2 + 2ab = -3p

I am really sorry that, I went somewhere else for, 2 minutes

[fervent wasp]

Same

Take your time

[brazen wharf]

a^2 b = 2q

[fervent wasp]

Now you have three equations at your hand, do you know how to describe them in English?

[brazen wharf]

system of equations?

[fervent wasp]

Remember, b is the root you set, which is not given in the problem

[fervent wasp]

You can simply explain the relation between a, b, and p.

[fervent wasp]

I assume the original question requires you to vividly explain it, doesn’t it?

[brazen wharf]

yes, it's a subjective question.

Uhm
We have p, q and a.

b = 2q/a^2 = -2a = (-3p - a^2)/(2a)

[fervent wasp]

a^2 + 2ab = -3p
2a + b = 0
a^2 b = 2q

Let’s make it simple, just explain the relative relation between a, b and p, b

Because b is the root we are looking for

[brazen wharf]

oh

a = -b/2
b = 2q/(-b/2)^2 = 8q/b^2
q = b^3/8
p = (a^2 + 2ab)/(-3) = (b^2/4 + 2 (-b/2) (b))/(-3)?

,w (b^2/4 + 2 (-b/2) (b))/(-3)

[twin meteor]

Results provided by WolframAlpha

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Requested by philosophizingphilosophy

[brazen wharf]

p = b^2 / 4

...

[fervent wasp]

Just leave the first and the third equation

It’s enough

[brazen wharf]

leave? discard?

[fervent wasp]

No, to leave it on the table

To preserve

[brazen wharf]

yes

okay

[fervent wasp]

Do you need a demonstration on how to explain a = -b/2 ?

[brazen wharf]

yes

[fervent wasp]

When the given cubic function is divisible by the given quadratic function, a = -b/2, where (x-b) (Q function) = (C function)

That’s all you need to explain. You can make it simpler by removing those trivial words

It’s up to you.

[vocal sleet]

@brazen wharf Has your question been resolved?

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[brazen wharf]

Thank you very much.

[vocal sleet]

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[naive vessel]

The question is factorise fully. My answer i got was the same as the answer sheet however instaed of 2sqrt3 i got 4 sqrt3

[steep crater]

@naive vessel

Step by step

[naive vessel]

ye

[steep crater]

First pull out the 4 term

[naive vessel]

?

[steep crater]

like take out the 4

[naive vessel]

soo
(x-4)^2 - 12

?

[steep crater]

Yes

Now treat x-4 as a variable a

[naive vessel]

yes

[steep crater]

And use difference of squares

what do you get?

[naive vessel]

hold up

ye its right but like

i didnt know thats how it worked

i thought

u leave the 4 til the end

and it js stays there

[steep crater]

ah

But you understand now?

[naive vessel]

yeuh i get it

i'll try another one hol up

[steep crater]

okay

[fair oak]

Urgent, who is good at math on homothety? It's for a multiple choice test.

[naive vessel]

oh ok

so i did it again

with this

and when i eliminated the 3 i got everything right except in the final answer there is meant to be a 3 on the outside, i didnt have that cuz i elimnated it

[steep crater]

!occupied

[vocal sleet]

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).

[steep crater]

also i might be able to help with geometry

[fair oak]

ok

[steep crater]

but you understand the correct method?

its correct you missed out the 3

[naive vessel]

but are yo umeant to add it again at the end?

[steep crater]

you took out a 3 term right

[naive vessel]

yu

[steep crater]

so everything should have been 3(...)

so the end is that

[naive vessel]

ah i think i get it

[steep crater]

yay

[naive vessel]

so the methoad is

like

taking the 3 out then adding it back in

to calculate it?

[steep crater]

the 3 is always supposed to be there

[naive vessel]

ohh i am acc blind

[steep crater]

its supposed to be on the left of all your equations

[steep crater]

is ok

[naive vessel]

i forgot bcz its 3()^2 - 21

the 21 is seperate

[steep crater]

m h m

[naive vessel]

so its not applied to the 3 yet

mhmmmhmmm

so itd be

3((x-1)^2 - 7)

is that right

what ur essentially doing is that

[steep crater]

yes

correct

[naive vessel]

then factorising w the other stuff

okayyy thanks

[steep crater]

Indeed

[steep crater]

Np

[naive vessel]

mhm

[steep crater]

!done

[vocal sleet]

If you are done with this channel, please mark your problem as solved by typing .close

@naive vessel Has your question been resolved?

[vocal sleet]

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[torpid sequoia]

Hey guys, I'm trying to prove the following:

I have a list X of numbers. I want to return the maximum value that a sublist S of X can sum, S of length k.

What I say is that the above is the same as sorting the list from bigger to smaller, and returning the sum of the first k elements

[torpid sequoia]

But I need to formally prove that.

on the other hand, is there a channel in this server for proofs specifically?

[river minnow]

By sublist you mean a subarray, as in the elements need to he contagious, right?

[heavy yoke]

the proofs channel is #proofs-and-logic but you can also ask here if you want

[torpid sequoia]

I believe S should just have elements in X

not contagious elements

[torpid sequoia]

Thanks 🙂

[river minnow]

You can use the fact that the sum should equal (the smallest element) + (smallest sum in the rest of the list) and do induction

[torpid sequoia]

Hmm why smallest?

I want to maximize the sum, you mean biggest (not smallest)?

[river minnow]

Oh, right, I meant to say biggest

[torpid sequoia]

Still don't get it 😄

[river minnow]

Didn’t get the suggestion or how to write the proof?

[torpid sequoia]

the suggestion

my friend recommended prooving <= and >=

[river minnow]

Well, say the >= direction is straightforward: Let m denote the largest element in X. Say the sum is f(S) = s1 + s2 + … + sk. We can let s1 = m, so largest sum of k elements in S >= m + s2 + … + sk. We can now let s2, …, sk be elements of S{s1} such that their sum is the largest and the inequality still holds

The <= direction, we can start by stating that the largest sum of k elements in S must contains m, can you see and show why?

Wait I think I messed up above

[river minnow]

s1 <= m is irrelevant here, we can simply let s1 = m and the inequality follows by the definition of the largest sum

[torpid sequoia]

I wanna prove the one with the arrow

The other one is easy

[river minnow]

Where O is the sorted list?

[torpid sequoia]

right

from bigger numbers to smaller ones

[river minnow]

Alright, you can rewrite the left hand side as O[0] + (same sum starting with i=1) and show that the sum on the right hand side must contain O[0] as well, meaning we can cancel them out and assert the resulting inequality holds by induction

[torpid sequoia]

I'll call this max, for simplicity