help-36

[tranquil pine]

They aren't talking about the 5th game, they're saying 5 games

[abstract imp]

oh i understand thanks

.close

[final saddle]

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[spare glen]

hello! I am stuck on this problem, how exactly do I find the derivative of this function using the chain rule?

[astral moss]

First define u and v

This is a combination of a product and chain rule

[spare glen]

u would be 2x and v would be (x^3-1)^2

[astral moss]

Yeah

Now to find their derivatives

[spare glen]

got it, so
derivative of u is 2 and v is 2(x^3-1)

[astral moss]

u is right

But for v, the chain rule states that you also need to then multiply that by the derivative of the function inside

[spare glen]

ah, got it. so it would be 6x^2(x^3-1)

[astral moss]

Yeah

[spare glen]

i do have a question though, why is the chain rule only applied to v and not u?

[astral moss]

Because chain rule is used when we have composition of two functions

[spare glen]

STRICTLY when there’s a composition of two function, right? i seem to be having trouble distinguishing that

[astral moss]

Yeah, strictly. Then the multiplication of two functions is your product rule which is seperate

[spare glen]

can you define composition of two functions in simple terms for me? I think I got it, but it would be great for some clarification

thank you so much in advance

[astral moss]

Sure, so

If we ignore the 2x for now in the function

[spare glen]

right

[astral moss]

Let’s define f(x)=(x^3-1)^2

Now define u(x) = x^2, and v(x)=x^3-1

f(x) can be written as u(v(x))

That is a composition

[spare glen]

I understand now, so it’s mostly recognizable from the exponent right?

[astral moss]

Yep, whenever there’s some function in an exponent, you should use the chain rule

[spare glen]

got it! thank you so much

okay so back to the problem, we have f’(x) and g’(x) figured out

[astral moss]

Yeah

[spare glen]

what exactly do I do from there?

[astral moss]

you just plug it into the product rule

[spare glen]

let me go ahead and give that a go

okay, so I have 2(x^3-1)^2 + (2x)6x^2(x^3-1)

[astral moss]

That looks good

[spare glen]

perfect, now how exactly do i simplify that?

the answer given from my teacher is (x^3-1)(2x^3+4x-2)

[astral moss]

Yep, so you can multiple the 2x and 6x^2 out

Or rather you could take out common factor out of the whole result

[spare glen]

how exactly can i do this? i can’t figure out what to do with the ^2 for 2(x^3-1)^2

[astral moss]

I don’t think the answer given by your teacher is correct

[spare glen]

here is what I was given from my teacher

[astral moss]

I got (x^3-1)(14x^3-2)

[spare glen]

perhaps you may be right

can you explain to me how you simplified what I had earlier?

[astral moss]

Sure

So a common factor of the result is (x^3-1)

So we get (x^3-1)(2(x^3-1)+2x*6x^2)

(x^3-1)(2x^3-2+12x^3)

(x^3-1)(14x^3-2)

You could also take 2 out of the second parentheses

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

[spare glen]

i see how you did it now

thank you so much

[astral moss]

Your welcome

[spare glen]

i will check in with my teacher tomorrow regarding the difference in answers

[astral moss]

I graphed it

It’s not correct unless I entered something incorrectly

[spare glen]

you definitely entered it correctly

I believe that you did it 100% correct

thank you again

ONE more question right now,
would you by chance be able to tell me what i’m supposed to do here just to get me started? I will try and attempt to do it myself though I am not 100% sure

so the quotient rule, right?

[astral moss]

Yep, combined with chain rule again

We know square root of x is the same as x^(1/2)

[astral moss]

Yeah, now just define u and v, find their derivatives, and plug into the quotient rule

[spare glen]

let me go ahead and give that a try!

thank you for the guidance once more

[spare glen]

okay, would this be right so far?

[astral moss]

Looks good

[spare glen]

GREAT that makes me happy thank you let me continue on

actually, no, so what exactly do I do from there now?

also, the answer given is -90x^2+48x-25 all divided by 2(2-5x)^3/2

[astral moss]

I would first do the simple things like multiplying -5 by -1/2 in the numerator, and simplifying the exponent on the denominator

[spare glen]

okay SO, would that turn out to be

[astral moss]

You wanna multiply the exponents on the denominator rather than adding

[rough crystal]

Which question do you need help with?

[spare glen]

oh shoot, I didn’t see that mistake, it would be 2-5x

[astral moss]

Yeah

[spare glen]

i am currently doing this one @rough crystal :)

[spare glen]

what should I do exactly after that particular step?

[astral moss]

Did you forget to multiply by f(x) actually

On the second term at the top

[spare glen]

Oh.. I see it now

thank you for pointing that out let me go ahead and fix that

here you go

[astral moss]

Sorry I gotta go now, you can probably ping helpers

I have an exam soon haha

[spare glen]

no worries! thank you so much for helping me out as much as you did. I might just call it a night for me as well

i really appreciates your dedicated help

[spare glen]

that is A LOT

[rough crystal]

I'm gonna double check for errors.

Yes its a lot.

[spare glen]

let me look at this, thank you so much

[rough crystal]

Your teacher really hates you man.

[spare glen]

it’s so tough I really am suffering but trying my best to survive

[rough crystal]

Just keep working hard, you'll do great. Be prepared my friend cus next unit is application of derivatives for you I presume?

[spare glen]

I’m not too sure, but I would assume so if you say so

I really struggle with all of these rules, and especially when they start combining

product rule, quotient rule, and especially chain rule

[rough crystal]

Hmmmm, you need understand the chain rule.

[spare glen]

also, here is the final answer, or the answer my teacher provided

[rough crystal]

Yep you could also state it like that.

If you don't understand the chain rule and are trying to find derivatives of functions like this it won't help you.

You need to first understand the chain rule and believe me its amazing once you understand the "why." I can teach you that if you want rq and it is gonna save you from a lot of trouble down the line.

[spare glen]

that would be so awesome, but I actually need to head to bed soon because it’s midnight where I am. would it be okay if I came back to you tomorrow?

[rough crystal]

Haha. Tmr I'm busy, so unfortunately no. If you want it'll only take like 5-10 minutes. No longer than that and ur troubles will be gone.

[spare glen]

that would be awesome!

i’m up for it

thank you so much for be willing to do so

[rough crystal]

Alright then lets get down to business rn.

Now what is your main confusion regarding chain rule?

Is there any specific area or r u confused in general?

[spare glen]

mostly in general, but I personally believe my troubles come in when there’s a mix between another rule and the chain rule

[rough crystal]

Alright no worries, well here is the thing the functions your differentiating our ULTIMATE goal is to have it with respect to x, not a composition. I'm going to show a simpler example so you understand.

Find dy/dx y = (x^2)^2

Its alright if you don't understand this I'm going to explain the why part and it should make much more sense.

[spare glen]

that would be 2(x^2) • x

[rough crystal]

f'(g(x)) x g'(x)

[rough crystal]

Yes. Now here is the interesting thing about the chain rule observe the notation

I just mentioned.

If we were asked to find the derivative dy/dx and only did f'(g(x))

[spare glen]

I THINK I got those simple ones down, but the ones where I have to use multiple rules and apply the chain rule to only one part of the function because I have trouble understanding why you only apply the chain rule to one part but not the other

[rough crystal]

I'm going to teach you that way.

First you must understand this and it will make other things clear. You know mechanically how to solve it, but don't understand why.

Alright so if we just did dy/dx = 2(x^2)

[spare glen]

you really are so much more understanding than my teacher is, i’m up for it thank you again

[rough crystal]

That would be incorrect because our ultimate goal is we want to find the slope

for any x value we plug in

We want to find the slope for the x value we plug into the composition.

But currently if we observe what we have we notice that the input is with respect to the output of a composition not our original number. I'm gonna use numbers to show.

[spare glen]

yes thank you so much, i am more of a visual learner

[rough crystal]

So here is the thing when we plug in a value into the function that is our INPUT the x value. If we currently have it in this form the problem is that when we plug in a value here the input should just be 2 inside, and so we expect that we will get the derivative for when x = 2

Nope that is wrong the actual x value we would be plugging in is 1.41

You see the problem with the composition is any value we plug in

That is the OUTPUT of the composition.

Its not the input

That is what we are essentially saying. Take a second to digest this and really think about it once you come across this flaw nothing will stop you.

And its okay if you don't digest this immediataly but observe the images really hard and think abt what im saying and respond. I have more to say.

[spare glen]

let me reread everything you have just said multiple times

give me ONE moment please, thank you

[rough crystal]

No worries take ur time.

[spare glen]

okay i THINK i got what you are explaining

[rough crystal]

Tell me then if you think so. Take a deep breath and tell me. Its okay if you don't get this immediataly

but here is the goal when we say dy/dx. We want to find the slope WITH RESPECT TO THE x. But the problem is with compositions is that its not the initial x we plug in, because what we are really doing is plugging in numbers that are the OUTPUT of the composition, and we then find the x value for that composition we would see that it would actually give us the derivative for a different number of x as shown in the image

For instance we plugged in 2 as our x value expecting a derivative for that, but upon closer inspection after plugging it in we see that in REALITY unbeknowst to us we actually plugged the sqrt2 = 1.41421

But that is an issue because that is giving the derivative for x = 1.41421 but we don't care about that as we want to know what is the derivative for when x = 2

[spare glen]

YES I believe I understand that psrt

is it okay if I reread it in the morning? I’m going to take a quick little power nap and reread everything you have just said when I wake up because it is 1am where I am and school is not too far away from now

[rough crystal]

Same time zone noice.

Do what u wanna do man. Just make sure to re read this in the morning promise its gonna plug in.

[spare glen]

I just wanted to thank you so much for your explanation and dedication to help me, I know that you are busy tomorrow but I will send in any questions I may have and it would be super awesome if you would answer whenever you’re free I don’t want to bother you at all but IF you can thag would be great thank you so much again

I really really appreciate it man

[rough crystal]

and if u r confused on applying it to scenarios where it involves multiple rules feel free to reach out tmr, if I am free at any point I'll see what I can do.

Np, Have a good man. Get some rest.

[spare glen]

sounds super great, thank you once again! have a goodnight man i really really am thankful for that

[final saddle]

@spare glen Has your question been resolved?

[final saddle]

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[vernal kernel]

Five digit odd numbers are to be made using the digits 1-7. determine how many such numbers exist where the number exceeds 50000 and no digit may be used more than once in a number

[vernal kernel]

sorry last channel got locked because i accidentally deleted original message

my working:

no. of odd numbers=6x5x4x3x4=1440

no. of numbers>50000=3x6x5x4x3=1080

no. of numbers total=7x6x5x4x3=2520

so why is there no intersection

[willow nacelle]

what do you mean no intersection

your working seems correct to me

[vernal kernel]

i mean that 1440+1080=2520

[willow nacelle]

oh

[vernal kernel]

this means that no numbers would be such that they are both odd and greater than 50000

[willow nacelle]

is this line supposed to be only using 1-7 or just

in general?

wait nvm

I see what you did

[vernal kernel]

answer is 600 if that helps

[willow nacelle]

I don't think the reasoning about no intersection is correct

The reason is because the number of odd numbers and the number of odd numbers more tahn 50000 are not two mutually exclusive sets

[limber sedge]

Wait, isn't it 3×6×5×4×3/2-1?

[willow nacelle]

They can have elements shared between the two of them

[vernal kernel]

yes i was hoping they werent and so i could get the intersection

[willow nacelle]

makes sense

[vernal kernel]

why tho?

[maiden widget]

hello can someone help me

[willow nacelle]

Ok, lets go over the problem more broadly. You already have the right approach but I think we can keep building on it

the first digit can be either
5,6,7. The last digit can be either 1,3,5,7. I think you realized this already

[vernal kernel]

yes

[willow nacelle]

If we pick 5 or 7 as the first digits, we lose cases

[limber sedge]

For the first digit: 3 values
2nd: 7-1
3rd: 7-2
4th: 7-3
5th: 7-4
Half of them are even, so /2
50000 is not included, so -1

[willow nacelle]

If we pick 5 as the first case, then we get:
1x5x4x3x3

[willow nacelle]

mb ill let u take it

[vernal kernel]

i was calculating all possible odd numbers

[limber sedge]

?

[willow nacelle]

nvm

Ok basically I think how you do it is you take

If we pick 6 as the first number + if we take 5 as the first number + if we take 7 as the first number

[vernal kernel]

yes thats the way to do it

[willow nacelle]

does it get 600?

[vernal kernel]

but the issue is about why my way doesnt work

[vernal kernel]

yes it does

[willow nacelle]

The reason why is because the first and second sets are not mutually exclusive

they each will have elements within each other

[vernal kernel]

yes they shouldnt be mutually exclusive

i am getting the number of elements in both sets

OH

ive got it

[willow nacelle]

sorry

[vernal kernel]

there is an overlap

[willow nacelle]

yeah

[vernal kernel]

but

we didnt take into account the even numbers

[willow nacelle]

yeah

sorry i explained that really poorly

[vernal kernel]

oh

we assumed that the numbers must either be odd, or greater than 50000 but there are even numbers smaller than 50000

so the method i am doing simply doesnt work unless we get those numbers as well

[willow nacelle]

that makes sense

[vernal kernel]

right thanks for helping

.close

[final saddle]

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[wet glen]

hello

[wet glen]

is this correct

[candid hull]

maybe

what are you trying to do here ?

[wet glen]

uhh i wrote the question

[final tangle]

is illegible

[wet glen]

what illegible

oh 8!

so is that right anyway

[final tangle]

and the word before it?

[wet glen]

wdym?

it's a geometry

[wet glen]

expression

unit

[final tangle]

what's that supposed to mean

[wet glen]

the 8th row

[candid hull]

how do you expect us to know your geometry problem like that ?

[final tangle]

what is
look for unit 8!
supposed to even mean and is that related to the stuff below it

[candid hull]

we're not gods you know

[wet glen]

look for the 8th row in the geometry row

idk the english language sry

[crimson flame]

what does that mean? is there any more context to the question?

[wet glen]

ughh I don't know the language of maths in english

[final tangle]

do you have a pic of the original question

[wet glen]

wait

its in indonesian (b)

[candid hull]

ah ok, geometric sequence

[final tangle]

google translation is

Determine the required ratios and terms for each of the following geometric sequences
8,7 th term

the value you're using is incorrect

[wet glen]

ohh its called sequences

[final tangle]

then $n$th term in a geometric sequence is
$$ar^{n-1}$$

[soft zealot]

ℝαμΩℕωⅤ

[wet glen]

wym value

yeah i did exactly that

[final tangle]

what's your n value for question b)?

[wet glen]

root2.root3⁷

8 but -1 so 7

[final tangle]

why 8?

[wet glen]

were looking for the 8th?

OH

[final tangle]

they wanted 8th for question a)
they want 7th for question b)

[wet glen]

okay so like this?

[final tangle]

yes

write $u_{\red{7}}$ or $t_{\red{7}}$ (or use variable of choice) to indicate the 7th term

[soft zealot]

ℝαμΩℕωⅤ

[wet glen]

okay thanks

wait one more

im stuck here

oh wait i did the formula wrong

wait did i, i think I didn't

NOO WHAT SHOULD I DO

[final tangle]

what's the original question

[wet glen]

the b one

i think its called summary

[final tangle]

these are geometric series

[wet glen]

yeah so uhh what should i do

[final tangle]

seems ok so far

apply the definition of division to simplify further

how would you simplify something like
$$\frac{3}{\br{\frac 15}}$$

[soft zealot]

ℝαμΩℕωⅤ

[wet glen]

3x5

[final tangle]

same idea applies here

[wet glen]

so 2²⁶ - 2/2²⁵?

[final tangle]

missing ()

but you didn't really need to expand the numerator like that

the initial step gives you
$$\frac{2(2^{25} - 1)}{2^{25}}$$

[soft zealot]

ℝαμΩℕωⅤ

[final tangle]

note that the numerator and denominator have a common factor of 2, and you can simplify accordingly

[wet glen]

so 2²⁵-1/2²⁴ ?

but im stuck there now

[final tangle]

missing ()

[wet glen]

sorry

[final tangle]

$$\frac{2^{26}-1}{2^{25}}$$

1 sec

[wet glen]

uhh i think the question kinda broken

[final tangle]

sry didn't catch this earlier but you used the wrong n value again

[wet glen]

oh

what n i should use

26

[final tangle]

yes

[wet glen]

duh

okay so like this

now what

[soft zealot]

ℝαμΩℕωⅤ

[final tangle]

and this would be a reasonable place to stop

[wet glen]

thx

[final saddle]

@wet glen Has your question been resolved?

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[feral plover]

My first idea was to get the intersection of G1 and G2, the intersection of G1 and G3 and then to compare them.
When I set G1 = G2 to get the intersection and rearranged it to 0 I got:
a1x1 + a2x2 - b - c1x1 - c2x2 + d = 0

I did the same to G1 = G3 with dissolved parentheses and got:
a1x1 + a2x2 - b - c1x1 - λa1x1 - c2x2 - λa2x2 + d + λb = 0

When setting both equations equal, cancelling several terms out and dividing by -λ I got:
a1x1 + a2x2 = b

which is exactly the first equation. That confused me although it did not seem neccessarily wrong,
because this indicates, that the equation must be independent of λ. Assuming that everything is correct, however, I do not know how to draw a conclusion.

It would be great, if someone could give me a full explanation.

[feral plover]

Btw, this is translated, so just ask, if something is unclear.

[final saddle]

@feral plover Has your question been resolved?

[feral plover]

<@&286206848099549185>

[weak saddle]

Tell us

[feral plover]

This is my Question

[tranquil pine]

plzz solve this

[feral plover]

@tranquil pine Please us another channel, this one is already occupied by my question...

Here my Question again:

My first idea was to get the intersection of G1 and G2, the intersection of G1 and G3 and then to compare them.
When I set G1 = G2 to get the intersection and rearranged it to 0 I got:
a1x1 + a2x2 - b - c1x1 - c2x2 + d = 0

I did the same to G1 = G3 with dissolved parentheses and got:
a1x1 + a2x2 - b - c1x1 - λa1x1 - c2x2 - λa2x2 + d + λb = 0

When setting both equations equal, cancelling several terms out and dividing by -λ I got:
a1x1 + a2x2 = b

which is exactly the first equation. That confused me although it did not seem neccessarily wrong,
because this indicates, that the equation must be independent of λ. Assuming that everything is correct, however, I do not know how to draw a conclusion.

It would be great, if someone could give me a full explanation.

[final saddle]

@feral plover Has your question been resolved?

[feral plover]

<@&286206848099549185>

[trim crown]

so you set the intersection of G1 and G2 = intersection of G1 and G3 and got the equation of G1 back

that means the intersection being equal is equivalent to the formula of G1 being true

since G1 is given, that means the intersections are equal

@feral plover

[feral plover]

Ok thank you, I think I overthought it! 🙂

.close

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[clever coral]

(x^5+x^4+x^3+x^2+x+1)×(x−1), i know this is extremely easy and stupid but i don't understand how this goes to x^6 - 1

[clever coral]

i tried just multiplying it using distributive property and im like 105% sure i did it wrong

[thorn grotto]

Try multiplying it out, yeah

What did you get?

[clever coral]

well it goes like -x^6 and then goes on and on like that ?

[thorn grotto]

How did you get -x^6?

[magic rampart]

why -x^6 though

ye

[thorn grotto]

And what do you mean with "and then goes on and on like that"?

[clever coral]

wait give me asec let me try again

oh so x ^ 5 * (x - 1)is just x ^ 6 - x ^ 5

ok yeah i did not need a whole forum for this shit

ty anyways

.close

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[rapid sky]

How many ounces are in a pound?

[hollow bane]

16

[rapid sky]

OK, so how many ounces is the loaf of bread?

[hollow bane]

8

[rapid sky]

Right, but you have 2 pounds plus 8 ounces in total.

So, how many ounces are there altogether?

[hollow bane]

40

[rapid sky]

Right.

So, they want dollars per ounce.

That means dollars divided by ounces.

So, what fraction would that be?

[wanton pine]

jesus christ what are these freedom units, there's 16 ounces in a pound?

[hollow bane]

?

[rapid sky]

Well, make a fraction with the dollars on top and the ounces on the bottom.

[hollow bane]

2.68/40

like that?

[rapid sky]

Right, now simplify that.

Or do the calculation.

[hollow bane]

.067

[rapid sky]

Right, so that's how many dollars it takes to buy an ounce of bread.

[hollow bane]

oh wow thank you

[rapid sky]

You're welcome.

[hollow bane]

.close

[final saddle]

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[keen nacelle]

Struggling with where to go next with task C

Am following these lecture notes from a few years ago that don't completely line up with what my textbook shows, just really confused about how to define the vector U here

Followed the textbook examples for making my matrix with some help from gpt with dubious value to help me understand how to find values for a, not sure if this is done correctly

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[keen nacelle]

Imma skip it and ask my TA next week I think

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[vital palm]

Need a sanity check about lagrangian mechanics/holonomic constraints:

considering the case of a planar pendulum, I know that the x^2 + y^2 = l^2 is the constraint, reducing the degrees of fredom from 2 to 1. My question is this:
if I separate that equation into two equations i.e.

x = l * cos(theta)
y = l * sin(theta)

where theta is angle from verticle, is the inclusion of theta what violates them being holonomic constraints? since it's I've used three coordinates for a system with, at most, two degrees of freedom?

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[tranquil pine]

A contractor wishes to build 8 houses, each distinct, in how many ways can he place those houses if 5 lots are on one side of the street and the other 3 are across?

[tranquil pine]

i am thinking [
\binom 8 5 \cdot 3! \cdot 5!
]
currently

[thorn grotto]

That's 8!

[tranquil pine]

you first choose 5 (or 3 of the houses, they are equivalent), and then permute them individually on each side

[tardy trellis]

well yeah

if u have 8 lots and u want to build 8 houses

then 8! is your answer

whether a house is on one or the other side of the street wouldnt matter

[tranquil pine]

oh i guess so lmao

[tardy trellis]

i suppose theres more to this qn given that they brought up such a specific scenario?

[tranquil pine]

i paraphrased but pretty much the same structure

[tardy trellis]

interesting

the 8! already does the job of considering all the lots

if im not mistaken

[barren hound]

this question is funny

[tardy trellis]

i would say phrased excessively

[tranquil pine]

i guess my answer wasnt incorrect but went overboard lmfao

[tardy trellis]

or maybe intentionally just to confuse you

which in that case has done its job

[tranquil pine]

okay thanks

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[tranquil pine]

,reopen

.reopen

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✅

[tranquil pine]

wait for c) are they also trying to fuck with me because isnt it literally just 654-330

[tardy trellis]

well

no

because each digit is used only once

so 001 002 etc is not an option

so if your answer to b was half of 654 then you would be wrong

[tranquil pine]

okay actually lets start from a) because i guess i misread "number" as any string of 3 digits

[tardy trellis]

well it is

just that each digit is used once

[tranquil pine]

but 012 isnt a number

[tardy trellis]

well

[tranquil pine]

so its not valid that way

[tardy trellis]

3 digit numbers meaning

xxx

where x is any integer

012 would be a number in this case

judging from this question i would assume they would like you to conisder 0xx to be numbers as well, as it is the case irl anyway

[tranquil pine]

7 * 6 * 5 isnt correct for a) i believe tho

[cosmic warren]

Seems ambiguous to me tbh, if it's for a class could ask teacher what they expect

[tranquil pine]

oops

[cosmic warren]

but I'd go with mesh for this help session lol

[tranquil pine]

edited

[tardy trellis]

depends on your syllabus i guess

[tranquil pine]

nah im just going through this book, not really given by anyone

[cosmic warren]

Oh bet just pick whichever assumption you want then

[tardy trellis]

where i am i would be accustomed to this

[tranquil pine]

lemme check the solution key for a)

[tardy trellis]

as it would generally be done like so during exams

[cosmic warren]

In US I could see something like a PIN number being 4 digits even with leading 0s

[tardy trellis]

^

[tardy trellis]

ah then ig go with that then

[tranquil pine]

but wait

question

does 000 not count as a number even under that principle

ok guess not

[tardy trellis]

yeah but you only use each number once

so

[tranquil pine]

fair

okay so for b)

uh how are u meant to do it

any nudges i guess?

[tardy trellis]

well

just count the cases

odd numbers are just numbers where the last digit is a odd number

so summate the numbers which last digitis 1,3,5

[tranquil pine]

oh thats smort

any other intuitive methods?

[tardy trellis]

not any i can come up with atm

imo thats the most straight forward

[tranquil pine]

yeah

so 3*5*5

okay cool

now for c

lets think

[tardy trellis]

for this you could use the conjugates

actually you dont nvm

[tranquil pine]

okay wait

lets do cases

like

if 4,5, or 6 are chosen in like

the hundred position

then you get like

[tardy trellis]

sound reasoning there, go on

[tranquil pine]

3*6*5 numbers

not including like the 300's yet

[tardy trellis]

mm hmm

[tranquil pine]

then for the 300 case

then like again assuming that the tenth digit is from 4 to 6

then 3*5?

wait no

[tardy trellis]

yeah

[tranquil pine]

okay

so we are left with the case where we have 33x?

so like

[tardy trellis]

nope

remember the requirement

[tranquil pine]

just like 6 lol

oh

yeah i guess 330 cant even happen

so what i got so far is enough

so totally

3*6*5 + 3*5

unless i messed up haha

,calc 365 + 3*5

[soft zealot]

Result:

105

[tardy trellis]

ig lol, im just following your logic here

sounds good, ill leave the calc to you

[tranquil pine]

yeah its good

thanks

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[tranquil pine]

can someone write down the work and the answers

[vital crag]

nope

try it yourself though and show your work here

[tranquil pine]

i have no idea how to docit

thats why im aksing

🧠

[vital crag]

what did you try

[tranquil pine]

i sont know how todo it

i didnt try

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[limber ice]

the first one equals 1

what about the second what can i do

[warm ether]

exact same thing

[limber ice]

it also equals 1?

[warm ether]

$w^a\cdot w^{-a}=w^0$

[soft zealot]

AℤØ

[warm ether]

the x isnt changing that

[limber ice]

alpha is a function of x

[candid hull]

Even if it's a function

[limber ice]

okay thank you

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[uneven light]

,rotate

i got the implicit differentiation althought not 100% sure if its correct

but idk how to get the slope after that

,rotate

[floral nova]

Plug in (1,pi)

[uneven light]

i did

my answer didnt match

any of the answer choices

[vital crag]

!show

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

[uneven light]

sec

[tranquil pine]

I think you fucked up here

you didn't apply chain rule and pull out a dy/dx out of the cosy

[uneven light]

you are probably right

let me redo it

i got it

thanks

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[halcyon glade]

Hey so i have a really simple qestion but im very confused

[halcyon glade]

(question)

they are asking if this is an even or odd function

now i would say it is odd

(answer sheet)

which it is

however i dont udnerstand the "odd about" part

[fallen vigil]

it means the function being odd

its that little blue thing, and there a diagram of an example odd function.

[halcyon glade]

iknow but what does the "odd about (2,0)" mean

and how do i find that point

[fallen vigil]

i mean, they defined it a f(2-x) = -f(2+x)

[halcyon glade]

that is in the answer sheet

[fallen vigil]

oh, hm I have not seen odd or even about other points before ><. my bad.

[halcyon glade]

no worries, thank you regardless

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@halcyon glade Has your question been resolved?

[halcyon glade]

@ me

[halcyon glade]

<@&286206848099549185>

[lusty sage]

an odd function about a point means if you rotate the graph 180° around that point the graph will line back up with it self. So if you rotate graph f(x) = 1/x-2 180° around point (2,0) you will have the same graph

[halcyon glade]

ok thank you

[unkempt saddle]

Sup

Lmfao

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[mental roost]

Hey

[mental roost]

I have a formula for the energy of something

It is in terms of delta x and delta p

Where those are the uncertainties in x and p

X being position and p the momentum

I want to differetiate this fornula

I think i can write
delta p=m d/dt delta x

If i now differentiate this

Does that even work?

With respect to delta x btw

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[mental roost]

$\int\frac{dk}{2\pi} (4\pi a)^{1/4} e^{\frac{a(k-k_0)^2}{2}} e^{ikx}=(\pi a)^{-\frac{1}{4}} e^{-\frac{x^2}{2a}+ik_0 x}$

[soft zealot]

Martin

[mental roost]

Any idea on how to do this?

I have to show that this is true

For theoretical physics

I think they want me to evaluate the integral from -oo to oo

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[hot patrol]

.close

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[solar pelican]

Prove that if x,y and z are nonzero rational numbers for which 1/x + 1/y + 1/z = 1 then the number A = (xy/z +1)*(yz/x+1)(zx/y+1) is nonnegative and sqrt(a) is a rational number

[solar pelican]

I've expanded A

Im just stuck now

Also this

<@&286206848099549185>

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[supple sierra]

How do you get the images for the standard basis matrix

H: R4 --> R2

So would I multiply by the standard basis for R4 or R2

[candid hull]

you can only input vectors from R4 in this function anyway

it's the only logical choice

@supple sierra

[supple sierra]

sorry H is a 4 * 2 matrix

so it can only work if i use the ones for R2

is that the way you're supposed to reason?

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[supple sierra]

@candid hull

because [1,0] isn't in R4

unless i make it [1 0 0 0 ; 0 0 0 0] in two rows

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[fresh relic]

you want the images of the standard basis of the domain, which is R^4...

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[rocky drum]

Hii

[final saddle]

Please don't occupy multiple help channels.

[rocky drum]

Sorry

Can anyone help there please?

[tranquil pine]

what are you tasked to find

[rocky drum]

@tranquil pine this is a test from university.
The first one is matrix.

[tranquil pine]

can you please close the other channel

[rocky drum]

There is actually no contex😕...

[tranquil pine]

yes and what are you looking for the determinant? eigenvectors? solution set?

[rocky drum]

@tranquil pine oh sure. I will..

[rocky drum]

I just need the solution way.. I have no contex here unfortunately.

[tranquil pine]

solution way? is this an augmented matrix

no 1

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[rocky drum]

🥲I'm sorry

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[rocky drum]

.close

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[little python]

,, \mathrm{\sum_{r=1}^{n} (r+1)(2)(2)^r}

[soft zealot]

!Yajat!

[little python]

how can i solve this using telescopic method

this is part of a question

@ when reply

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@little python Has your question been resolved?

[rustic wedge]

hmm well we know the 2 comes out

so $2\sum_{r=1}^n (r+1)2^r$

$\sum_{r=1}^n (2^r \cdot r) + 2^r$

[soft zealot]

ItzKraken

ItzKraken

[rustic wedge]

nowsplit

$\sum_{r=1}^n (2^r \cdot r) + \sum_{r=1}^n 2^r$

[soft zealot]

ItzKraken

[rustic wedge]

$(\sum_{r=1}^n (2^r \cdot r)) + 2(2^n - 1)$

(second simplifies using sum of finite GP formula )

[soft zealot]

ItzKraken

[little python]

how did you get this @rustic wedge

.

[soft zealot]

ItzKraken

[rustic wedge]

Ah I missed a 2

@little python sorry 😅

[rustic wedge]

just multiply this whole thing by 2

[little python]

youre helping me, why sorry, also i said to ping me lol

wait let me have look again

[rustic wedge]

kk

[little python]

wait

i understand nothing

could you just tell me what you did and i will try to understand it

[rustic wedge]

K

[rustic wedge]

,, \mathrm{\sum_{r=1}^{n} (r+1)(2)(2)^r}

[soft zealot]

ItzKraken

[little python]

yes we take 2 out

its constant okay

whats next

[rustic wedge]

lemme write

[little python]

lol sorry

[rustic wedge]

$2\sum_{r=1}^{n} (r+1)(2)^r$ \
$=2\sum_{r=1}^{n} (2^r \cdot r) + (2^r)$ \
$=2\sum_{r=1}^{n} (2^r \cdot r) + 2\sum_{r=1}^{n}(2^r)$ \

[soft zealot]

ItzKraken

[rustic wedge]

do u understand? @little python

[little python]

till now yes

[rustic wedge]

good

now do you know the formula for the sum of a finite GP?

[little python]

yes

[rustic wedge]

what is it?

[little python]

i got the same what you have in the last line, but idk how to solve further

[little python]

a(r^n-1)/r-1

[rustic wedge]

so u got till here right?

[little python]

till last equation of this tex

[rustic wedge]

ah Okay

[rustic wedge]

now apply this

what is our common ratio?

2,4,8,16...
(all are of the form 2^n)

what is common among them?

and what is the first term

[little python]

powers of 2, and also first term is 2?

[rustic wedge]

yes a=2 and r=?

[little python]

common ratio?, i dont get if you are asking this

[rustic wedge]

I am asking for 'r' in this with relation to the second sum

[little python]

sorry idk

2?

yea it should be 2

@rustic wedge dude you there?

[little python]

<@&286206848099549185>

.close

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[supple jolt]

If the question says: find the derivative of x^2 at x=3 computationally, am i forced to use the definition of the derivative or can i also use the power rule

[unborn sun]

depends on what your teacher wants

if you havent covered the power rule in your classes so far you probably are forced to use the definition

[supple jolt]

what exactly does compute mean

[unborn sun]

i would assume compute the derivative and plug it x=3

[amber holly]

Does it say "Find the derivative of x^2 at x = 3 computationally" or "Compute the derivative of x^2 at x = 3"?

[amber holly]

Yeah just differentiate f and plug in x = 3

[supple jolt]

Rechnerisch means computanionally

So i can use the power rule

[amber holly]

I would suppose so

[supple jolt]

Ok thanks

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[supple jolt]

,w compute

[soft zealot]

Results provided by WolframAlpha

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

Requested by masteroogway0599

[supple jolt]

Yeah compute just means calculate

He probably meant that i mustnt do it graphically

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[dreamy gull]

I need help doing (vi) and (vii)

[dreamy gull]

this is what I’ve done so far for (vi) but i dont know where else to go from here

and i have no idea how to start (vii)

[fallow wren]

vi is right

[dreamy gull]

would i not need to simplify it more though?

[fallow wren]

you have to find for -750pi a more naturale writing

[dreamy gull]

yeah thats where i got stuck

[fallow wren]

i think its only 2pi or 0

because -750 is even

[dreamy gull]

this is also all i have for (vii) i have not one clue

[dreamy gull]

is that for vii?

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[fallow wren]

yes

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[oblique kraken]

Hi, im really confused what to do here, just give me an example with 1.a.

[oblique kraken]

Its about drawing graphs of linear equations

[final saddle]

@oblique kraken Has your question been resolved?

[barren hound]

choose a value of x, doesn't matter what you use. let's say you chose 6. then solve the equation for y like:

$3y = -4\cdot\orange6 + 3$

[soft zealot]

Haylsune Miku

[barren hound]

3y = -24 + 3
3y = -21
y = -7

so for that equation, we know that the graph passes through the point (6, -7) so plot that on your graph. do this again for two more x values

and connect all three with a straight line

[oblique kraken]

Oh so if were looking for x that means y= a number?

And the other way around if were looking for y?

[barren hound]

if you fix a value of x, then that will force a certain value for y, and vice versa

[oblique kraken]

Ohhh alright alright

Thanks

[barren hound]

so yeah do that for both equations (3 points each then connect them)

then look where the lines cross

[oblique kraken]

Alright tysmm

[barren hound]

it might not be exactly at a grid line, that's okay just do your best

[oblique kraken]

Oki tyyyyy

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[rough pebble]

So what do I start with?

[stone wagon]

\begin{center}
\begin{tabular}{ c c c }
_ & $x$ & $+2$ \
$x$ & $x^2$ & $2x$ \
$+1$ & $x$ & $2$ \
\end{tabular}
\end{center}

[soft zealot]

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

[rough pebble]

Ohh

[rough pebble]

Okay I see

[stone wagon]

you multiply the first term of the first bracket (x) by every term in the second bracket. then do the same for the second term of the first bracket (2) and add those 2 results together

[rough pebble]

Rightt I see

Thanks! .close

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[alpine prairie]

I want to show that the arc length of f(x)=lnx for x \in [a,1] is greater than |ln(a)|, where 0<a<1.

[alpine prairie]

i solved this integral and got this result

now how can i show that this is greater than |ln(a)|

?

the first two terms are >0

and i can write the last term as +ln

but then it comes down to showing that the last term >= |ln(a)|

this is where im stuck

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[alpine prairie]

<@&286206848099549185>

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[ancient thistle]

Hello

[ancient thistle]

Can someone tell me where I am wrong?

Oh I solved it

nvm

.close

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[dull thicket]

my eigenvalues come out incorrect though, I think I might be doin my initial force equation incorrectly

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[rough crystal]

Hey there I was wondering what does the notation for f(x (asterik)) mean?

[tranquil pine]

to go from a exponential graph to an equation

[rough crystal]

Please go elsewhere Nav kindly.

[thick ledge]

i think it just means the x-coordinate of the midpoint of each rectangle

[rough crystal]

Thats what I was thinking as well.

But I can't seem to find any answers online.

[thick ledge]

yeah i doubt that its some standardized notation used by mathematicians

probably just something ur textbook/teacher is using

[rough crystal]

Yeah.

Ima still research it tho jus to be sure.

[thick ledge]

if you want extra clarity id suggest just directly asking your teacher, but this is my guess

[rough crystal]

Yeah.

Thanks.

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[nocturne stratus]

can someone help me understand absorption law?

(A and B) OR negated C

how would i apply it to that ^

[tranquil pine]

A AND (A OR B) = A

this is

A AND A OR A AND B

[nocturne stratus]

im confused tho

is (A and B) OR negated C elgible to use absorption law on it?

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[nova beacon]

how would you even use it there

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[vocal wharf]

The f(x) is 4x^3-2

[fallen estuary]

Alright, so we'll start with the more simple transformations you've been given

what do you know about reflections across an axis with polynomials?

[vocal wharf]

You need to subract either inside or outside depending on the axis

[fallen estuary]

Yes, now for b, you'll need to first stretch it and then reflect it across the x-axis

so, you know that at some point, you'll need to make something negative to reflect it across an axis

Now, if you want to stretch your function, $y = f(x)$, vertically (i.e in the y-direction), then what would you do?

[soft zealot]

Ammardian

[vocal wharf]

Would this be the answer for C?

[fallen estuary]

yep (that is correct)

seems like you're on the right track

[vocal wharf]

Would this be the answer to B.?

[fallen estuary]

Not quite, the reflection across the x-axis is different to reflection across the y-axis

what you've done is stretched it vertically by a factor of 5 and reflected it across the y-axis

to reflect a polynomial about the x-axis, what you need to do is:

$$f(x) \rightarrow f(-x)$$

[soft zealot]

Ammardian

[vocal wharf]

So the minus should be inside

Google says its on the outside

Would it just be -5x^3?

[fallen estuary]

what you're looking for is $-5(4x^3 + 2)$

[soft zealot]

Ammardian

[fallen estuary]

which is the same as

$5(4(-x)^3 - 2)$

[soft zealot]

Ammardian

[fallen estuary]

generally think of it like this

if you have $y = f(x)$

[soft zealot]

Ammardian

[fallen estuary]

to reflect it in the y-axis, you need to flip the sign of y

$y = f(x) \rightarrow y = -f(x)$

[soft zealot]

Ammardian

[fallen estuary]

to reflect a function in the x-axis, you need to flip the sign of x

$y = f(x) \rightarrow y = f(-x)$

[soft zealot]

Ammardian

[vocal wharf]

? Khan Academy says otherwise

[fallen estuary]

really? maybe I have gotten it backwards

ahhh yes, I flipped the y-values so that's reflection in the x-axis, yeah my bad lol

[fallen estuary]

consider yourself correct then

[vocal wharf]

Wait for which answer the first one or second one

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@vocal wharf Has your question been resolved?

[fallen estuary]

the second one

the first one was correct

[vocal wharf]

This one?

Or -5x^3

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[civic ether]

This might be a basic question, but what does this even mean?

Like i'm not sure what the question requires

If i have a 67.5 degree angle then i only need one bisector to 33.75

to get the 67.5 i'd need an angle bisector of 135

and to get a 135, i'd need the angle bisector of 270

270 is an "easy" angle so that's where i stop(?)

Not sure if this is what the question means or is it something else

[tranquil pine]

from the right angle they have there, maybe they want you to do it starting from 90?

[civic ether]

but then i can't get to 33.75?

[tranquil pine]

oh I see now

yeah you can

divide 90 by 2 what do you get

[civic ether]

45

[tranquil pine]

45 by 2

[civic ether]

22.5

[tranquil pine]

22.5 by 2

[civic ether]

11.25

[tranquil pine]

11.25 + 22.5 = ?

[civic ether]

o

wait

i'm thinking how all that connects lol

like okay i guess the accumulation of those two give the answer

but is what i did wrong?

[tranquil pine]

I mean they didn't really specify much other than having an angle. So like why not say

angle + angle = angle

[civic ether]

true

honestly i wouldn't think about adding angles

This is what i thought about doing

Not sure if this checks out

it would help a lot 😭 if the question was a bit more specific

[tranquil pine]

honestly the question is really vague and didn't really specify much. Can't say for certain that what I did or what you did will roll with their specific (lacking) wording

[civic ether]

hahaha yeah i guess

for all we know

130 could be easy enough to draw

[tranquil pine]

but yeah nothing wrong with this me think