help-27

[warm hearth]

staticstic

my asnwer is right?

[devout snow]

@warm hearth Has your question been resolved?

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@warm hearth Has your question been resolved?

[warm hearth]

<@&286206848099549185>

[zinc veldt]

can you elaborate on what you're trying to do?

[warm hearth]

position measurements

quartiles

[devout snow]

@warm hearth Has your question been resolved?

[warm hearth]

NOOOOOOOOOOOOOOOOO

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[cosmic frost]

\

[stuck field]

What have you tried?

Do you know the criteria for congruency?

[cosmic frost]

i only know what it stands for

[stuck field]

Like what can you say about these two triangles?

[cosmic frost]

3 sides

[stuck field]

Wdym?

[cosmic frost]

the triangle is 3 sides

and

it has lines marked

[stuck field]

Do you understand what it means?

[cosmic frost]

not at all ngl

they just give us the homework

with a assigned video

lol

[stuck field]

Check that video then.

It's really essential to know the meaning of those lines and all.

[cosmic frost]

yeah im watching it still confused

[stuck field]

That's not randomly done.

[primal nebula]

can u understand this thing?

[cosmic frost]

oh crap

yeah i can lol

[stuck field]

Then compare this to the original problem.

[cosmic frost]

AAS

.close

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[tender lotus]

Discuss the sign of $1-x^2-e^x$ for all $x \in R$

[woven radish]

adam03

[patent tartan]

You can guess one root, then look at the derivative

[tender lotus]

Derivative? Why?

I need to find the sign of that function

[patent tartan]

Well the sign is not the same everywhere

Derivatives give you info about which direction a function grows

Combined with knowing the roots that can give you some hints

[tender lotus]

Yeah ik that it's not the same, i need to find in which intervals the function is positive and in wich intervals the function is negative

[patent tartan]

Yup

[tender lotus]

Ok how can i find the sign of the derivative then 🤦‍♂️

[patent tartan]

By thinking

[tender lotus]

$-2x-e^x$

[patent tartan]

and trying different methods

[woven radish]

adam03

[patent tartan]

That's the idea of such problems

[tender lotus]

I can't even find the roots of the derivative

I need to prove the roots and not by guessing

[woven radish]

adam03

[tender lotus]

<@&286206848099549185>

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[tender lotus]

.close

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[restive river]

suppose you surrounded everything in this image with parenthesis (image) and then did (image)^2. how would u simplify? x_sub_1,...x_sub_n are vectors and same applies to y variables

[restive river]

for more context, this was somehow simplified to this

[supple knot]

This is Cauchy Schwarz inequality

Look up the proof of it

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[frail lake]

did i get this right

[devout snow]

Please don't occupy multiple help channels.

[frail lake]

.close

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[frail lake]

is this right?

[orchid sierra]

yes.

good job.

it gets worse, I promise.

❤️

[supple knot]

Why did you say it's wrong

[orchid sierra]

if the program won't accept it, simplify the argument of the third root

(distribute that 3)

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[odd saddle]

Can someone please explain how these last 2 statements include lambda?

[odd saddle]

Like how does (P^-1)Ax just equal Lambda(P^-1)x

Where did we kick off the A to

[grim swallow]

What is the definition of an eigenvector along with its eigenvalue?

[devout snow]

@odd saddle Has your question been resolved?

[odd saddle]

Lambda A = Av is what ur alluding to?

[torn vessel]

that definition is not correct.

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[rare mantle]

Yes

[fallow wave]

Train harder

Do you try to become better or the parts that you dont know or do you focus on the things you already kinda know

Then you dont know everything

[stuck field]

No one knows everything.

[fallow wave]

Knowing everything isnt just about knowing the content, but also how to do it. Including knowing how to compute in the right way

You just said you dont

[stuck field]

Also knowing what kind of mistakes you're prone to do.

[fallow wave]

Do you do careless mistakes because of stress or because you are trying to save time?

[stuck field]

Make those mistakes in mock contests, so that you know what not to do in the original exam.

[fallow wave]

So because of stress then

I would say that you dont need to focus on getting better at math, but rather on gettting better at relaxing under pressure

have you tried meditation?

Well then there is you answer

Meditate, and like with everything else you will become better with time if you put your energy into it

Because if you cant meditate for more than 20 sec, then trying to meditate for 5 min will stress you out which is what you need to learn to handle

just like in the compitition

I did too once, until i learned to relax a bit and rewind my thoughts

Okey so there is another thing i did to become less sloppy

i only write with a ball point pen

i dont like messy notes, and so it forces me to think a bit more before i write

now i do everything with a ball point pen, even my exams

[patent pivot]

That is smart! But I also do all my math in pen and it’s still crazy

[fallow wave]

its a ball pen point :^)

it feels like you took that different

its not about the effectiveness of the pen, but rather the limitation it introducs to you just writing mindlessly

BUT

if you want a smooth pen

get a gel pen

waaay better than ink

pilot 0.5 gel pen are the best

I just wrote above

[patent pivot]

The brand is pilot

[fallow wave]

yes

i use size 0.5 for less smooshyness from the gel

They are perfect because you can just buy refil and dont need to spend money on a whole new pen

[patent pivot]

I feel like those always rip my paper tho

Or bleed

[fallow wave]

i use thicker paper so that it dosnt show throguh

[patent pivot]

The sharp tip on thin graph paper

Idk

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[merry stream]

posting problem one sec

[merry stream]

this is what I have and idk if I'm heading in the right direction

[supple knot]

do you know what $\frac{1}{\sec(\theta)}$ equals?

[woven radish]

riemann

[merry stream]

lmao

I hit send and was liek OH cosine

mind if I ping you if I reach another issue?

[supple knot]

just close and open a new ch

[merry stream]

.close

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[wet iron]

help

[devout snow]

@wet iron Has your question been resolved?

[devout snow]

@wet iron Has your question been resolved?

[wet iron]

Help

[devout snow]

@wet iron Has your question been resolved?

[wet iron]

<@&286206848099549185>

[vagrant mural]

what help do you want?

what's the problem?

[devout snow]

@wet iron Has your question been resolved?

[dark tundra]

If it’s a graphical method, that entails you to draw the vectors on a piece of paper such that the head of one vector is the tail of the next

After drawing everything, draw the vector from the tail of the first vector to the head of the last one

Measure that vector’s magnitude and direction

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[celest tide]

How do I find the range of $\frac{a - b\sqrt{1 - \cos^{2}{\theta}}}{\cos{\theta}}$

[woven radish]

kinglacto

[deep forge]

Do you have an idea of when this function could become arbitrarily large?

[celest tide]

um when cos theta is very small?

[deep forge]

Yes. Basically, 0

[celest tide]

yeah infinity

[deep forge]

And what are the possible values for sin²θ when Cosθ goes to 0

[celest tide]

1?

[deep forge]

Remember, Cos(90)=Cos(270)=0

And Sin (270)= - 1

[celest tide]

and y cord will be 1 and -1 at 90 and 270

so sin^2 will be 1??

[deep forge]

Yes

Now, since a>b, the value when cos goes to zero is +inf

[celest tide]

why is a > b?

[deep forge]

I remember reading that in your question earlier.
Am I hallucinating???

Weird

[celest tide]

umm yeah... lets work under the presumption that you never saw that

i.e the relationship between a and b isnt given

[deep forge]

Well, then if a>b then it goes to +inf. And it goes to -inf in the other case.

So, now you need the other range.

[celest tide]

and if we dont know that?

[deep forge]

You consider both cases separately and then see what happens

[celest tide]

Im not really interested in solving questions in hand.... just want to learn how to solve such questions ... ill apply them in a whole ton of exercise questions i have

[deep forge]

Oh

[celest tide]

alright

so first we take a > b?

[deep forge]

Yes

[celest tide]

upper limit is +inf

lower?

[deep forge]

The lower limit will be -inf. Lemme explain how

You see, the point 90° is the point where the function blows up to infinity. So, you see which way it blows up.
If you approach 90° from 0, then it blows up to +inf. But if you approach 90° from the other side, then it blows up to -inf

[celest tide]

wolfram says range is (-inf, -3] U [3, inf) though......

[deep forge]

Yes.

U just needed the max and minimum right?

[celest tide]

oh so just to the right of 90°

[celest tide]

no the range as i stated in the rendered image

[deep forge]

Yes, of you approach from the right, it's negative, because cos is negative.

[celest tide]

yeah

got it

[deep forge]

If you approach from left, ots +ve because cos is +ve

[celest tide]

ok

so we know min and max

but we cant assume that to be the complete range?

i mean it isnt this case, we know for a fact

[deep forge]

That is the complete range of the function

[celest tide]

then whats this?

[deep forge]

Tell me, what's the union of that

That's the range.

[celest tide]

its union is already simplified

i mean they are disjoint

[deep forge]

Oh wait. My bad. I see the fault now. Yeah, I'm wrong.

[celest tide]

np how do we arrive at 3 and -3?

[deep forge]

Well, kinda wrong. Cuz that disjoint/non disjoint will depend on the value of a

[celest tide]

oh im terribly sorry

i forgot to mention

[deep forge]

Idk how you got 3 or -3. Because that depends on the value of a. At least that's what I can see by plotting the function.

[celest tide]

i chose a = 5 and b = 4

as an example

yeah so that 3 will actually be an expression in a and b

which for a = 5 and b = 4, happens to be 3

well ik for a fact that, that expression is ||√(a^2 - b^2) for a > b btw|| cuz my TB says so

[deep forge]

OK. So, one way to find out the expressions is to find the derivative and look for the points where the derivative is zero. And those would give you your values.

[celest tide]

this is irrelevant ignore it

[celest tide]

ah... well i havent learnt calculus yet

[deep forge]

Oh

[celest tide]

any other way.....?

[deep forge]

I can't really think of any other way man. Sorry.
Maybe someone else can help.

[tiny minnow]

as far as i know, using differential math is the only way to solve this kind of thing

[celest tide]

aight np

[celest tide]

oh alright

now if I added the condition

that a > b

does that ...

erase the need for calc?

i think not....

[deep forge]

Nope

[celest tide]

ok

[tiny minnow]

^

[celest tide]

my tb solves it without tho

a very weird solution

[tiny minnow]

can u show it rq

[celest tide]

yeah sure lemme take a pic

i mean the actual question asks for minimum value

is that possible without calc?

[deep forge]

Here, a video to help you visualize your function

[celest tide]

and condition that a > b

[deep forge]

What's a tb?

[celest tide]

textbook

also the original func is a sin theta - b tan theta

so ig that makes it irrelevant

nvm

[celest tide]

oh thanks

in any case... here goes....

I dont like such solutions... they seem far fetched?

[celest tide]

ik this expression is not the same as the one in the textbook... just wanted to solve it for learning

thanks anyways ... finding ranges of such functions needs calc aight

.close

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[jade vector]

Need help computing this integral (dont have a good calculator)

[deep forge]

Well, break the long expression. It has a lot of terms that are added. Find their integrals independently and then add them.

For example, the first term would simply become πr²h/2 after integration.

[devout snow]

@jade vector Has your question been resolved?

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[junior lark]

help

[devout snow]

@junior lark Has your question been resolved?

[stable smelt]

first do them in order from smallest to highest, middle one is median

min / max is first and last element

quartiles definition vary by textbook

if n is even quartiles are acquired from the element on position (n+3) / 4, if uneven the position (n+2) / 4 (this is the definition in my textbook, but this can vary, so see what ur book sais)

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[copper flower]

to calculate the determinant we have $$\det A = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad-bc$$ but I'm really confused as to why this is.

$$A_n = c \cdot A_o$$
(new surface area = determinant times old surface area)

$$\implies c = \frac{A_n}{A_o}$$

the length (or whatever you call it) of a vector can be calculated with pythagoras:
$$L(\begin{vmatrix} a \ b \end{vmatrix}) = \sqrt{a^2+b^2}$$

just looking at the basis vectors, and calculating the area they're giving us (before applying the linear transformation):
$$A_o = L(\mathbf {\hat \imath}) \cdot L(\mathbf {\hat \jmath}) = 1 \cdot 1 = 1$$

after the linear transformation:
$$A_n = L(T_\mathbf {\hat \imath}) \cdot L(T_\mathbf {\hat \jmath}) = L(\begin{vmatrix} a \ b \end{vmatrix}) \cdot L(\begin{vmatrix} c \ d \end{vmatrix})$$

now let's say the linear transformation doesn't change anything (i.e. we're applying a linear transformation of $I_2$), then the new area would be the same as the old one:
$$A_n = L(\begin{vmatrix} 1 \ 0 \end{vmatrix}) \cdot L(\begin{vmatrix} 0 \ 1 \end{vmatrix}) = L(\mathbf {\hat \imath}) \cdot L(\mathbf {\hat \jmath}) = A_o$$

plugging this into the equation (our equation) of the determinant:
$$c = \frac{A_n}{A_o} = \frac{A_o}{A_o} = 1 \implies \det I_2 = 1$$

therefore the determinant must be 1

but plugging the identity matrix into the actual determinant equation gives us:
$$\det A = \begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} = 1 \cdot 0 - 0 \cdot 1 = 0 \implies \det I_2 = 0$$

what's wrong here?

[woven radish]

illuminator3

[pulsar dock]

Your notation is very difficult to follow and doesn't really make much sense to me. But at the end of the day you did not calculate the determinant of a 2x2 matrix correctly. Obviously an identity matrix has determinant 1

[copper flower]

oh...

I always calculated ac-bd in my head lol

thanks

.close

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[mighty acorn]

How do I calculate (27)^1/4

[neon folio]

you have to know this:

[devout snow]

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[civic oak]

I can bring all written notes and whatever books I want to my exam in May, do you guys recommend any books or techniques of writing down things?

[patent tartan]

depends on the exam/course

[civic oak]

the book is called Sinus Forkurs

[oak harness]

what is it on

[civic oak]

it stands for Sine Pre-course

it's a course in my university

[oak harness]

trigonometry?

[civic oak]

it's high school math tho basically

it's a course I have to take in university beause I didn't go all 3 years in high school cuz I had an electrician education where I dont need the last year in high school

it's a course I need to take in order to be eligible to start on a bachelor

there is trigonometry, integration, derivation, all sorts of graphs

[oak harness]

well, for trigonometry I would highly recommend taking a good reference table with various identities.

[civic oak]

equations, algebra

[oak harness]

something you can feel comfortable with

[civic oak]

trigonometry is easy for me tho

[oak harness]

an integral table if you can use that then

[civic oak]

I have a table for that ye

I can write down whatever and use whatever is on paper

and bring all the books I want

[oak harness]

https://tutorial.math.lamar.edu/ is a great resource too

It has pre-calc too

[celest tide]

just curious: what kind of an exam is this??

[civic oak]

nice thanks

[civic oak]

they decided to do it because they werent sure if it was gonna be a home exam or a real exam irl

[oak harness]

Sounds like some kind of revision exam

one of those general ones

[civic oak]

so now it wont be a home exam, but we still are allowed to use books n stuff

because they cant change it up now

[celest tide]

but arent you supposed to memorise any identity or formula you want to use?

[civic oak]

nah

[celest tide]

wow thats interesting lol

[civic oak]

if it was real exam we would have been allowed to bring a formula table book anyways

but now we can bring any book and notes we want

[celest tide]

I see

aight thanks sorry for taking your time

[oak harness]

trig identities can sometime be hard to memorise

That's why I recommend using a trig table

[celest tide]

we dont get to use for identities or values

lol

[civic oak]

my weakness are functions

[oak harness]

what specifically about them?

[civic oak]

like if they give a graph and I have to find the function to the graph

and all the other stuff

is there some "cheat sheet" for that kinda stuff

[oak harness]

I mean, I guess you could print out the graphs of common functions

🤷

[civic oak]

if I had a printer I would lol

I need to buy one before my exam

[oak harness]

then you can draw them by hand

[civic oak]

my test exam is tomorrow tho

yea I can

[oak harness]

You can also try playing around with https://www.desmos.com/calculator for a while

Get a feeling what transformation does what

[civic oak]

we also are allowed geogebra version 6

[oak harness]

well then

checking solutions to equations just became trivial

[civic oak]

yeah I can check solutions to things then

like even integrations

issue is I dont know how to use geogebra for that

but I can find out on youtube prob

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[vast cradle]

Anyone know anything about numerical analysis, because I don't understand how it wants us to answer this question, we use matlab for this subject, but I still don't know where to start.

[vast cradle]

?

Hello savior!

righttt

Welp, if anyone really wants to help, feel free

[rose girder]

Can you post your definition of backwards stability?

[vast cradle]

https://en.wikipedia.org/wiki/Numerical_stability

Hf, bye

In backwards stability, we take the small error(1+δ) the results of our calculation would normally have, and multiply it with the input instead, so the resulting output is without error.

x =transpose( [ξ1, ξ2])
xprog = [ξ1(1 + δ), ξ2(1 + δ)]

fprog(x) = f(xprog).
fprog(x) is the function done in our system that returns the result with the error, which we multiply by (1+δ) to remove the error, while f is our ideal function which takes the xprog=x*(1+δ) so the result is without errors

Not sure if I'm doing it justice tho

As you can see from my explanation, I'm also confused

[devout snow]

@vast cradle Has your question been resolved?

[rose girder]

Ok, so the definition I'm familiar with is the following: An algorithm for computing $y=\phi(x)$ is called backwards stable if for each $x$ there exists an $\tilde{x}$ such that $\tilde{y}=\phi(\tilde{x})$, where $\tilde{y}=(1+\varepsilon)y$ is the result with error $\varepsilon$ and $$\frac{|\tilde{x}-x|}{|x|}\dot\leq\sigma\varepsilon_0$$

[woven radish]

Zanarcane

[vast cradle]

Yeah, that sounds about right

[rose girder]

As inputs, we have the coefficients of the polynomial

Let's look at the coefficient $\alpha_n$ first, in the case $n=3$ if you want to

[woven radish]

Zanarcane

[vast cradle]

Aight

nvm....

kinda lost

[rose girder]

I'm on my phone rn, so it takes a while to write things down...

[vast cradle]

it's okay

[rose girder]

You know how the Horner algorithm work? Assume that you produce an error $\gamma_k$ and $\delta_k$ when multiplying and adding in the k-th step, respectively

[woven radish]

Zanarcane

[vast cradle]

good old horner

[rose girder]

How would the n-th coefficient of the result $\tilde{y}$ with error look like in terms of $\gamma_k$ and $\delta_k$?

[woven radish]

Zanarcane

[rose girder]

Well, this is not Horner

[vast cradle]

?

[rose girder]

This is Horner

The idea is to factor out the x terms

[vast cradle]

Huh, thought they were equivalent

[rose girder]

Mathematically, yes

Numerically, it's better if you use the second expression

[vast cradle]

Interesting

[rose girder]

So you start with $a_n$, multiply it with $x$ and add $a_{n-1}$ in the first step

[woven radish]

Zanarcane

[vast cradle]

Aight

[rose girder]

In the next step, you take the previous result, multiply it with $x$ and add $a_{n-1}$

[woven radish]

Zanarcane

[rose girder]

And so on until $a_0$

[woven radish]

Zanarcane

[vast cradle]

How do we show it's backwards stable tho?

[rose girder]

Assume that you produce an error $\gamma_k$ and $\delta_k$ when multiplying and adding in the k-th step, respectively. Determine the coefficients of the result in terms of the errors.

[woven radish]

Zanarcane

[rose girder]

That's where you get your $\tilde{x}$

[woven radish]

Zanarcane

[rose girder]

Then you have to show the inequality I provided above

[vast cradle]

Aight, thanks a bunch!!

[rose girder]

Np

[vast cradle]

.close

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[lucid zinc]

Sorry if this is a dumb question but given the conditional, "xy is an odd number, then x AND y are both ODD" (shortened to P->Q), if I wanted to negate it Q would the negation apply to the logical operator 'AND' in tandem with the the two variables being 'ODD' producing the negation, "only one of either x or y being even"?

[devout snow]

@lucid zinc Has your question been resolved?

[lucid zinc]

<@&286206848099549185>

[devout snow]

@lucid zinc Has your question been resolved?

[restive river]

Yes it would apply

and yes you're statement is correct

only one of either x or y is even

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[proper lark]

Alright. This question requires me to prove the AM >= GM inequality using weak induction. However the best part
"Let a1, a2, ..., an be an infinite list of positive real numbers."
"The AM of the first n numbers is defined as; An = (a1 + a2 + ... + an)/n"
"The GM of the first n numbers is defined as; Gn = (a1 * a2 * ... * an)^(1/n)"
"Prove AM >= GM for all n in the positive integers"

[proper lark]

When I started to do my work I decided to using gauses formula to make it 2 numbers over 2. Then use the factorial of (ak! * ak+1)^(1/2). Then try and use the classic proof, but I cant really combine a factorial to the left side. So I am stuck on how one goes about using the inductive hypothesis when proving said problem.

AM and GM being Arithmetic and Geometric means *

[devout snow]

@proper lark Has your question been resolved?

[proper lark]

<@&286206848099549185>

[devout snow]

@proper lark Has your question been resolved?

[devout snow]

@proper lark Has your question been resolved?

[proper lark]

unfortunate its all good bot man

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[young vortex]

Hi

[young vortex]

Can someone help me this?

Been trying to do it for 2 hours already

[patent pivot]

Is this a hw assignment?

[young vortex]

Yes

[patent pivot]

Do u have any work or thought processes?

Or guesses for any of them?

[young vortex]

None at all

I have no Idea

;((

[patent pivot]

I’m just going to put the suvat eq for reference in the chat. Are u familiar with them?

[young vortex]

Not at all

I'm sorry

[patent pivot]

I believe these are the eq needed to answer your questions

[young vortex]

Can you help me solve them?

[patent pivot]

I’m not an expert at all and it has been a bit since I’ve done projectile motion

[young vortex]

Thank you though

No worries

<@&286206848099549185>

[patent pivot]

So I was doing some researching as a refresher. Ik im not supposed to say the answer straight up but idk how to explain besides it just being googled the first one is zero.

[young vortex]

Ohhhhh Thanks

I'll try to do it but I can't promise to do it efficiently

Been cooking my brain for like 2 hours :c

[patent pivot]

I mean I can try them. But cannot promise they will all be right. I just want to disclaim that. Or you can of course wait for someone more knowledgeable and confident.

I feel that. School and physics can really do that

[young vortex]

I mean I can use your answers as a guide

If possible

[patent pivot]

So first, do u understand with projectile motion the vertical and horizontal component are calculated separately and use different info?

[young vortex]

Yeah

The given questions are blank for me

and I can't figure it out

[patent pivot]

So for the second one

We need to find the final velocity

Then we can find the time

Using the eq without t

Do u think u can start to plug in some values?

We have enough info to find the v

[young vortex]

Okay wait so Initial vertical velocity is 0m/s the displacement is 113m

Right?

[patent pivot]

Yes

[young vortex]

what about the acceleration (a)?

[patent pivot]

It is 9.81

Because it’s gravity

[young vortex]

its 5.0 for me (rounded up)

4.8 right?

[patent pivot]

How did u get that?

[young vortex]

I used this eq

[patent pivot]

Where is that from?

[young vortex]

From my notes

[patent pivot]

Doesn’t matter. It works

Great. I got the same thing

[young vortex]

Yey

:)))

Then what are we finding in the 3rd one again?

[patent pivot]

Better yet, it’s an answer in the multiple choice

The initial horizontal velocity

[young vortex]

is that U?

[patent pivot]

With horizontal component acceleration is zero

Yes

[young vortex]

our initial velocity is 113m right?

wait

now I'm confused

[patent pivot]

Since we are now dealing with the horizontal component, we don’t use that number anymore. Instead we use the number 248 m

Additionally, since in the question it says it lands, I am going to assume v is zero. Hopefully that thought process works and makes sense.

[young vortex]

Wait calculating it

[patent pivot]

So u can use any of the eq I think since u can use 4.8 from the past question

[young vortex]

Yeah

I did

s= 248m a=0m/s t=4.8s

[patent pivot]

Yes

[young vortex]

Thank you for your time I hope I'm not disturbing you

I'm sorrt

[patent pivot]

You’re fine. I’m choosing to help u!

[young vortex]

Thank you

[patent pivot]

It’s good review for me

[young vortex]

give me like 3 mins

[patent pivot]

Np

[young vortex]

I can't provide some pictures of my written eq cause my phone is dead

Haha 😅

[patent pivot]

No worries. What did u get as your answer?

[young vortex]

Wait almost done

its 51.7

[patent pivot]

Great. I got that too

[young vortex]

is it the same as yours?

ohhhhh

great :)))

What about the 4th and 5th one?

That's the most confusing for me

[patent pivot]

So u want to treat it like you’re finding v and you use the time 1.5

[young vortex]

What is V?

*v

[patent pivot]

Final velocity

[young vortex]

okay so our u=51.7 a=0 t=1.5s?

[patent pivot]

Yes

[young vortex]

Okay 2 mins

still

51.7 😅

[patent pivot]

Actually. Ig that makes sense if horizontal acceleration is constant

[young vortex]

Yeah

wow

I can't believe that was quite easy 😅

Last one though

What are we finding?

[patent pivot]

Displacement

Vertically

[young vortex]

ahmmm?

[patent pivot]

This may be too easy, but technically it gives you the height in the question

And all the other answers are smaller than it, which to me doesn’t make sense

But u could verify with a calc

[young vortex]

What was yours?

[patent pivot]

I haven’t done a calc

But just looking at the answers

[young vortex]

😅

HAHAHHAHAHAHAHAHAHA

now its getting all over

I don't get it why does it says 248.16

[patent pivot]

Alright I did it with math and my guess was right.

What says?

[young vortex]

248.16

[patent pivot]

As what?

How did u get that?

[young vortex]

u= 51.7 t=4.8 g=0

[patent pivot]

G?

[young vortex]

is acceleration due to gravity

[patent pivot]

It would be 9.81

Because the question is asking about vertical displacement

[young vortex]

I'm not getting it :c

[patent pivot]

Use the last eq

[young vortex]

what was our a again?

0?

[patent pivot]

9.81

Because it’s vertical

Vertically there is always acceleration

[young vortex]

our s?

248 or 113?

[patent pivot]

I don’t believe one of those is an option

[young vortex]

hmmmmmmmmm

I'm sorry im dumb dumb

[patent pivot]

You’re fine

One is correct

One isn’t a choice

[young vortex]

Ahmmm

I can't figure it out

[patent pivot]

Wdym

U said two numbers

One of them is the correct choice

One of them isn’t even a choice in your multiple choice

[young vortex]

so 113?

😅

[patent pivot]

Yup

[young vortex]

429.3888 m

🤦‍♂️

[patent pivot]

Wat

I’m so confused what u are doing

[young vortex]

Yeah same :(((

[patent pivot]

Read the question

What is it asking?

[young vortex]

rock's vertical displacement

ohhhhhhhhh

is it -113m?

[patent pivot]

Yes

[young vortex]

OH GOD

🤦‍♂️ 🤦‍♂️

[patent pivot]

You’re good you’re good.

[young vortex]

my brain was too fried up

im sorry

HAHAHAHHAHAHAHAHAHA

[patent pivot]

All good

Are those all of them?

[young vortex]

Yeah

Thank you so much

[patent pivot]

Wonderful. Please feel free to follow up and let me know how we did.

[young vortex]

you didn't just help me from this

you also helped me for my final exam

[patent pivot]

???

[young vortex]

You helped me alot

I can used what I've learned

on my exam

[patent pivot]

Glad to hear it

[young vortex]

Thank you so much Ellie

[patent pivot]

Np

[young vortex]

Do you mind if I add you on discord?

so that I can let you know how it did?

[patent pivot]

Sure

[young vortex]

.close

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[cyan tartan]

{(y,y) | y ∈ {j, k}} need some help finding the elements in this set, from my understanding i got (j,j) and (k,k) but im not sure if that's right cause i don't really know what im doing tbh

[rustic epoch]

what's j and k here

[cyan tartan]

elements of a set? and (y,y) is an n tuple?

[rustic epoch]

yea but unless it's given y is a whole number or natural number or real number wtv idt you can solve this

cuz depending on what y is, the set can have different number of elements innit

[tall knoll]

Innit

[cyan tartan]

that's all the problem says unfortunately

[rustic epoch]

I see

do u happen to have a pic of this question

[steel sage]

It’s correct

[cyan tartan]

it is a mc problem though, some choices are j, (j), (j,j), {j,j} with the other variations consisting of k

[cyan tartan]

are those all the possible elements? or are there more

like (j,k) or (k,j)? and () vs {} matters right?

[rustic epoch]

yea by {} u mean y can take values including j and k right?

i used to be taught [] instead of {} which means it includes values including j and k

just makin sure

[cyan tartan]

yea they do matter thanks for the help guys appreciate it

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[weary narwhal]

when dealing with the quadratic formula of y=x^2-5x how come the x intercepts at 5 and not -5? Isn't the formula to both factorize and find the intercepts ax^2+(m+n)x+mn=0?

[restive river]

forgetting formulas for a moment, what's an x-intercept?

[dapper heath]

the polynomial root

[weary narwhal]

the places in the curve that intercept the x axis, right?

[dapper heath]

the point where x = 0

[celest tide]

y = (-5)^2 + (-5)(-5) = 25 + 25 = 50 ≠ 0

also
x(x - 5) = 0
so
x = 0 OR x - 5 = 0 => x = 5

[weary narwhal]

ahhhh

I think I get it now, thank you.

.close

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[lethal mirage]

i’m not sure if the answer is 769.42 ft or 81

[hybrid snow]

Ag

Ah

Yeah so set up a triangle

[feral scarab]

769.42 should be the answer you should get, not 81

[lethal mirage]

okay thanks

.close

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[rough token]

needed help to figure out how to determine whether it is and what it's invertible linear transformation is

I'm aware that T=Ax T-1=A-1x

but idk how that applies here

[devout snow]

@rough token Has your question been resolved?

[pulsar dock]

You can get started by applying T to the standard basis on P_2(R)

[devout snow]

@rough token Has your question been resolved?

[steel sage]

Of course it is

The corresponding matrix is triangular with all elements on the diagonal being -1

[rough token]

That will get me my Range right?

what do I do with that?

{1, 2-x, 2+4x-x^2}

= R(T)

and how do I know why I am doing that?

[devout snow]

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[last umbra]

how would i know the derivative is negative or positive?

[deft pumice]

Because of the (6-ln(x)) term

ln(e^k) = k

So when it is greater than e^6 it becomes negative

[last umbra]

ahh yes

thank u

[deft pumice]

No problem

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[shy fossil]

what is a random variable?

[weak tundra]

Formally, it is a function from the sample space to the set of real numbers. It would be helpful to indicate your level of study to get an explanation that is useful to you

[shy fossil]

i am a ug student

my major is CS

and i need to learn Probability and statistics to pass the exams

[weak tundra]

Have you tried google? What level of understanding are you hoping to obtain? What is your background?
https://en.wikipedia.org/wiki/Random_variable https://www.investopedia.com/terms/r/random-variable.asp https://www.mathsisfun.com/data/random-variables.html http://www.stat.yale.edu/Courses/1997-98/101/ranvar.htm

[shy fossil]

i am reading these rn

[devout snow]

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[rough flume]

in the quadratic equation y=ax^2+bx+c what does the "a" stand for? for example: y=-4x^2

[rough flume]

I've watched videos read articles and seen all the resources my teacher has recommended and I cant seem to wrap my head around quadratic equations

[winter patrol]

a represents the coefficient of the x^2

(the constant being multiplied to x^2)

[rough flume]

so it just multiplies x^2?

what about graphing it? how would that work?

so if "c" is the y-intersept and "bx" is the offset

then "a" is the slope

but how do you calculate the slope on a curved line?

oh

https://www.google.com/search?client=opera-gx&q=calculate+the+slope+of+a+curved+line&sourceid=opera&ie=UTF-8&oe=UTF-8#kpvalbx=_KHpMYsGoDMSwqtsPiLa_gAo20

got it

.close

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[sharp pond]

need help explaining how to do paraterization

<@&286206848099549185>

[rose girder]

Given the x- and y-coordinates of a point on the surface, can you determine the z-coordinate?

[sharp pond]

i can sub in the values if that whaty ur askig yes

[rose girder]

And then solve for z

[sharp pond]

i need it in this form

i don tknow where r u v come from

[rose girder]

r(u,v) is the parametrization

[sharp pond]

ok

[rose girder]

You can choose u = x and v = y

[sharp pond]

ok

[rose girder]

What are the three components of r(x,y) then?

[sharp pond]

i don tknow

[rose girder]

r(x,y) is the point on the surface with x as x-coordinate and y as y-coordinate

[sharp pond]

ok

[rose girder]

And you already know how to determine the z-coordinate

[sharp pond]

if im given an x and y coordinate

[rose girder]

Yes

[sharp pond]

ok

am i supposed to sup in the 4 points i get from using 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and get 4 values of z

[rose girder]

Not only 4 points, but all points with 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

[sharp pond]

but theres only 0,0 0,1 1,0 1,1

[rose girder]

Wdym?

[sharp pond]

what do u mean?

i dont know what im doing

at all

i usually follow the example then just remeber that

but he skips changing it intto paramtrization

[rose girder]

You want to parameterize the surface, right?

[sharp pond]

and just has r(uv)

i think so

idk

i neeed to get this type of info

specifically this

[rose girder]

Yes

[sharp pond]

from x+y+z = 2

[rose girder]

Again, just choose u for the x-coordinate and v for the y-coordinate

[sharp pond]

ok so u + v+ z = 2

[rose girder]

The z-coordinate can be determined using the equation

[sharp pond]

ok

how

[rose girder]

Using this

[sharp pond]

hown

v= x

u = x *

[rose girder]

You know how to solve for z, I hope

[sharp pond]

y = v

theres no numbers

z = u + v -2

ik that

[rose girder]

Not quite

[sharp pond]

oh shoot

• u - v +2 = z

[rose girder]

r(u,v)=(x(u,v),y(u,v),z(u,v))

We did choose x(u,v)=u and y(u,v)=v

[rose girder]

And here, you have z as a function of u and v

[sharp pond]

u thin ku could gkve me the answer for parametrization and than explain it

[rose girder]

Well, I'm not here for giving answers

[sharp pond]

yea its to learn but i learn easier if ik what the outcome is

its not the answer to the full question

its just for getting started

casuse otherwise this is gonna take a while

r(u,v) = u, v , - u - v +2

is what i think ur trying to say

[rose girder]

Yes

[sharp pond]

but idk

ok \

how

i get the last part

the z

why is x = u y = v

[rose girder]

That's just a choice, you could have chosen something else, but then it's more complicated

[sharp pond]

ok

so where do i go from there

[rose girder]

The reason why I did choose x=u and y=v is because we want to also have easy limits for u and v since 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1

[sharp pond]

so limts of u and v is same as x and y right

[rose girder]

Yes

[rose girder]

You compute the partial derivatives of r

Then plug everything into the integral and compute

[sharp pond]

?

how do i take the partial derivative of r

its not a full equation

[rose girder]

You're taking vector calculus, I think you should already know how to compute partial derivatives...

[sharp pond]

i can do partial derivative of like x^2 +y^2

not of u, v , - u - v +2

[rose girder]

Just differentiate w.r.t. u and v

[sharp pond]

what is w.r.t

[rose girder]

With respect to

[sharp pond]

is it liek partial derivative of 3 expressions

[rose girder]

Yes

[sharp pond]

or so i need to figure out how to use commas in math

ok

[rose girder]

Consider each component one by one

[sharp pond]

so derivative of u

derivative of v

and derivative od -u-v+2

[rose girder]

Yes, first fix v and consider the derivative w.r.t. u

[sharp pond]

fix v?

[rose girder]

Assume that v is a constant

[sharp pond]

ok

[rose girder]

What you usually do if you take the partial derivative

[sharp pond]

so do i take a derivativbe of all of them twice once wrt u and once wrt v

i use symbolab

[rose girder]

Yes

You don't need symbolab for this...

[sharp pond]

i use it for derivatives and integrals

ok

ive got the parital wrt u

doing v now

ok ive got those

hello?

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[sharp acorn]

Hello. Please tell me why this is wrong...

 32 * 16^(x+1) = 8^(x+2) * 4^(x+4)

= log4(32) * log4(16) * (x+1) = (log4(8) * (x+2)) * (log4(4) * (x+4))

[hazy totem]

Remember, log(ab)=log(a)+log(b)

Not log(a)log(b)

Notice you turned everything into a multiplication

[sharp acorn]

Ok. Do you mean it should be right this:
= log4(32) + log4(16) * (x+1) = (log4(8) * (x+2)) + (log4(4) * (x+4)) ?

[hazy totem]

that looks right

[sharp acorn]

Ok. I will retry this equation now. I will tell you if I could solve it..
thank you so far

Thank you @hazy totem
It actually worked.
Made my day 🙂

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[neat cobalt]

Hi guys, please help me to solve this question.

Normal distribution related

[devout snow]

@neat cobalt Has your question been resolved?

[neat cobalt]

<@&286206848099549185>

[neat cobalt]

?

[polar bolt]

What are you struggling with?

[devout snow]

@neat cobalt Has your question been resolved?

[neat cobalt]

I cant solve this question, what does the ‘maximum’ mean?

[polar bolt]

Find the time, x, such that P(X<=x)=0.85

[devout snow]

@neat cobalt Has your question been resolved?

[devout snow]

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[wanton rapids]

for the second derivative f''(x) >0 in the region around x = a, why is f(x) in x=a convex?

[devout snow]

@wanton rapids Has your question been resolved?

[gentle estuary]

f'(a) = 0
So x = a is either a minimum or a maximum.

If f''(a) > 0, then the gradient of the curve around x = a is increasing. Which means it looks something like:

[wanton rapids]

One moment

for the second derivative f''(x) >0 in the region around x = a
So that means there's a positive slope after x=a, right? As you said
If f''(a) > 0, then the gradient of the curve around x = a is increasing.
But what does that tell us about f'(a) or f(a)?

This is difficult because we have to work backwards from f''(x) to f(x)

[gentle estuary]

1. Yes, it means that f'(x) < 0 for x < a, and f'(x) > 0 for x > a

And f'(x) = 0 at x = a

[wanton rapids]

How do we know that f'(a) = 0, if f''(a)>0?

[gentle estuary]

Imagine the curve in your head.

Convex means that the curve is like a $\bigcup$, while concave tells you it's like a $\bigcap$

[woven radish]

Vigne

[gentle estuary]

You can convince yourself that, for whatever rotation of a convex curve until +- pi/2 rad, that there exists a single point a such that f'(a) = 0

You should also be able to use the mean value theorem.

[weak tundra]

It's not the case that if f''(x) > 0 then f'(x) = 0. Consider f(x) = x^2.

For the convexity, basically, just imagine the gradient. If you have f''(x) > 0 in an interval, means that f'(x) is increasing in that interval. So it will appear to bend upwards.

[gentle estuary]

Actually, my argument is wrong, but I think you can still use the mean value theorem.

All you have to show is that the secant line is above the point on the curve which has the same gradient.

[wanton rapids]

If you have f''(x) > 0 in an interval, means that f'(x) is increasing in that interval
Is this always the case?

And so f(x) is always increasing in that interval?

[gentle estuary]

If the curve is convex, then there exists some point c such that for x < c, f'(x) < secant gradient. After the point c, f'(x) > secant gradient. So the gradient must have increased, and thus f''(x) in the interval has to be positive?

[weak tundra]

If you have f''(x) > 0 in an interval, means that f'(x) is increasing in that interval
Is this always the case?
Yes
And so f(x) is always increasing in that interval?
f'(x), not f(x). This doesn't say anything about f(x). f'(x) can be negative but increasing.

[weak tundra]

What level of study are you at?

[wanton rapids]

High school.

f'(x) can be negative but increasing.
I don't know what that means.

[weak tundra]

Negative means less than zero, increasing means its value is increasing. For example, f' going from -2 to -1.

[wanton rapids]

I don't know this ^

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@wanton rapids Has your question been resolved?

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[rain cradle]

i know this question sounds weird, but it was a real question for my exam

there is y every x. there are also five y that are also z . there is no x that is z. if the number of y is 15, and three of them are neither x nor z, then the number of y is?

[plucky stone]

is says "if the number of y is 15...", so the number of y is 15

[topaz axle]

makes sense

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[celest tide]

ive already asked this but i didnt quite understand

what exactly is an "identity"??

[winter patrol]

true (where defined)

[young plover]

it just means that the equation holds

[celest tide]

umm

what?

the answer is 0 btw

[winter patrol]

wdym

[celest tide]

there are 0 values of k

[young plover]

yeah that makes sense

[celest tide]

for which it is an identity

[young plover]

because it has to hold for all x

[celest tide]

how do we arrive at that?

[young plover]

i dont know off the top of my head

what kind of course is this for?

is it a proofs course?

[celest tide]

no just trig

grade 11 - 12

[young plover]

ok

lemme give it a bit of thought

i think a simple argument is as follows. If the equation were an identity then we have $1+k=(sin^2xcos^2x)(3+2k)$ but then we can see that for some values of x, say $2\pi$ the RHS will be zero, but the LHS will not be unless k=-1, but if k=-1 then the LHS is zero when the RHS doesn't have to be, take $x=\pi/4$ for example. So no matter the value of k the equation can't hold for ALL x

[woven radish]

llspacebarll

[young plover]

it could hold for some x keep that in mind, but identity in this context means it must hold for all x

[celest tide]

oh

so if, for some k, both (1 + k) and (3 + 2k) were 0

that k would make it an identity

[young plover]

then you could have an identity

yep

[celest tide]

but such a k doesnt exist?

[young plover]

yeah exactly

[celest tide]

ohh ok i think i understand

thanks

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[vernal lion]

I am stumped on how to find x, which is required to use cosine to find what that side of y equals. I am in geometry and we just started learning trig functions such as Sin, Cos, and Tan if thats any help on what past knowledge I have.

[vernal lion]

This was a practice test and I am reviewing for anyone wondering

also, I am confident this is 27

[silk zephyr]

Yup

[stuck field]

It is.

That does help.

[faint zinc]

@vernal lion sin = O/H, cos = A/H, tan = O/A, and there's two special triangles that you need to know, the 45,45,90 and the 30,60,90, which have side ratios 1:1:sqrt(2) and 1:sqrt(3):2 respectively.

You only need the special triangles knowledge for this problem.

[stuck field]

y is the sum of bases of two triangles, figure them out separately.

[vernal lion]

Oh i have those in my notes, and i took down the notes word for word but i'm not really able to see how its used on this problem

THe 30-60-90 and 45-45-90

[stuck field]

This is how.

[silk zephyr]

You have an angle and 1 side for the left triangle, you can find out the rest the same way you did for the right triangle

[vernal lion]

ok ill see if i can get it

So, from what I understand, x is 27*(sqrt)2, and the left triangle value of y is 27?

[stuck field]

Well that particular part of y.

Yes.

Correct.

[vernal lion]

Alright thank you, I'll figure out the right sides triangle really quick

[stuck field]

You're welcome.

[vernal lion]

Would the right triangles value of y be 46.8 if rounded to nearest tenth, leaving the whole value of y to be 73.8?

[stuck field]

Correct.

[vernal lion]

Ok, I see now, thank you so much for your help!

i use .close right?

.close

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[humble hollow]

Hey! This is probably not the most common question here, but i was wondering if anyone here could help me figure out how to solve this situation. My class isnt too good at mathematics so most of us are kind of slow, but i am pretty competent in it so i end up being bored every class cus everything is so easy. Ive tried to talk to my teacher about it but she says she has to focus on the class entirely and not just the "good people". Im scared i might lose my "flow" if i dont learn anything new, so the question is, Anyone know somewhere i can learn math(very broad question, i know)

[main gull]

Not a math related question, best to ask in #discussion but my advice, Khan Academy is a good way to learn new material

[humble hollow]

I appreciate it!