help-13

[sand cradle]

Yeah

So f(x) = -1750x + 16000

Or rather, let's use t, to make it clearer that it is a time, in years.

f(t) = -1750t + 16000

Now you are asked to determine the volume, after this model, after 3 years (==> t = 3)

[tepid scarab]

So (-1750) *3 +16000 =f(3)

[sand cradle]

Yeah

[tepid scarab]

3f =10750

[sand cradle]

3f?

[tepid scarab]

Ah I thought f(3) meant I times it

[sand cradle]

No, it means that 3 is the argument of the function

[tepid scarab]

Ahh okay

[sand cradle]

f(t) = -1750t + 16000, if you plug in t = 3, then you write f(3) to show that.

3 * f(t) (or 3f(t), without the *) means 3 times the function

[tepid scarab]

Ohhhhh okay

The f button on my calculator

[sand cradle]

You don't need it

f(3) = -1750(3) + 16000, you already said it

You just need to calculate -1750(3) + 16000

[tepid scarab]

Oh yeah

10750

[sand cradle]

You don't need to tell your calculator that we made a function for it

Yeah

[sand cradle]

You technically can, you could input f(t) = -1750t + 16000 and then you could tell him to calculate it at t = 3.

But that's not necessary here

Maybe if you wanted to calculate a lot of values and not type it in all the time, it'd be good

[tepid scarab]

I get you yeah yeah

So now for the other questions I just do the exact same?

With model B

[sand cradle]

Well, first, (a), (ii)

[tepid scarab]

You could say something like the drills could stop working so it would cause the model to be less accurate

[sand cradle]

could stop working?
If they tell you they won't stop working, then, then it's great?

Look at the three assumptions

Are all of them taking into account?

(1) is, sure, the initial value is 16000

(2) also fine

What about (3)?

[tepid scarab]

The drills could become less effective over time and not extract as much oil in the time it used to

?

[sand cradle]

Ah, wait

This is the daily volume of oil per day, so (3) should also be fine, actually, since it's decreasing

[sand cradle]

Well, the problem I would say is that it never actually becomes negative, does it?

But the model does

You can't extract a negative amount of oil though

[tepid scarab]

You can’t no

[sand cradle]

So you can only use the model in a fixed interval, between 0 and the zero of f(t)

Now for (b), (i), you do almost the same as for (a), (i)

But keep in mind the different form of the equation

[tepid scarab]

Ah so this is the limitation

[sand cradle]

Yeah

[sand cradle]

If you look at B, it doesn't have that limitation

[tepid scarab]

No because it never reaches zero

I think

[sand cradle]

Yeah

[sand cradle]

And that's probably closer to reality, since you most likely will always get some oil

[tepid scarab]

Exactly yes

[sand cradle]

Well maybe in reality you can reach 0, but it's definitely better than something that goes into the negative

[tepid scarab]

Yeah yeah

So does this one use a different equation?

Because it is a curved

line

[sand cradle]

(b), (i) already tells you it's an exponential model

[tepid scarab]

Ae^-kt

Exponential decay

[sand cradle]

Yeah

[tepid scarab]

because it’s going down

[sand cradle]

You already know A

[tepid scarab]

16000

[sand cradle]

Yeah

Then plug in the second point and solve for k

[tepid scarab]

Time is going to be 3

now I’ll solve it

[sand cradle]

What do you mean?

The second point is P(4, 9000)

[tepid scarab]

Ah

I’m not sure how to put it in the equation

[sand cradle]

You have that $f(t) = 16000 \cdot e^{-kt}$.

[tepid scarab]

ohhhh okay

Yeah I thought so

I’ll do it now

[sand cradle]

$f(4) = 9000 \iff 16000 \cdot e^{-4k} = 9000$.

[tepid scarab]

0.144

Is k

[sand cradle]

Yeah

[sand cradle]

So put that into this, into your equation of f(t) as k

[tepid scarab]

16000e^-4•0.144

[sand cradle]

I meant the original $f(t)$ equation. \ $f(t) = 16000 \cdot e^{-0.144t}$.

Now you just need the value of f(3) and you're done

[tepid scarab]

Is that what we did before?

[sand cradle]

Yeah, when we plugged in 3 for t

[tepid scarab]

10750

So we put that in

[sand cradle]

No, we plug in 3 into t

To get the value of f at that point

Which is the volume of oil

[tepid scarab]

10392.3

[sand cradle]

,w 16000 * e^(-0.144 * 3)

Yeah, around that

[tepid scarab]

Yeah I put the exact value

[sand cradle]

10387.4 is what I get though

[sand cradle]

Ah, then yep

[tepid scarab]

So that’s the answer for b(ii)

10387.4

[sand cradle]

Yes

[tepid scarab]

Thankyou very much I appreciate it

[sand cradle]

as a sentence when answering an application problem, with correct units

[tepid scarab]

Yeah I get you

[sand cradle]

"After 3 years, the daily volume of oil that will be extracted will be around ... barrels."

[tepid scarab]

Ohh you told me to actually type the sentence sorry

I thought you meant that’s how you’d answer it

But yes I would have typed that anyway

Appreciate it man

[sand cradle]

np

[cedar kiln]

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[winged oasis]

Help

[idle tusk]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[winged oasis]

1

Just the activity 1 number 1 and 2

[frozen valve]

When using the principles of Mathematical Induction, you should first check for the basis stem (i.e. the base identity for N = the first element defined)
Then you should state an Inductive Hypothesis
For some integer k, the above statement is true
Finally, you should prove the statement to be true for the integer k+1

@winged oasis im not going to provide a solution mate

try to understand the concept

and work on it

if you are stuck tell me

[winged oasis]

Idk how does that sigma workk can u tell me how does it work

[frozen valve]

ok right

lets understand the elements of the sigma

we have
(1) Sigma -> thats the notation
(2) i = 1 -> we assign the variable i to 1
(3) n -> limit of i
usually the step for increase for i is 1
(4) (3i-1) -> this is just the function passing through i

this is just a for loop in coding

you find the sums of (3i-1) for i =1 then i =2 then ... then i = n

for n = 3 this would be

(3 times 1 -1) + (3 times 2 - 1) + (3 times 3 - 1)
specifically didnt use * to avoid discord font change

[winged oasis]

Oh okay thankss

[cedar kiln]

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[cedar kiln]

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[eager trail]

hi anyone can help me on how to find the derivative of (x-2)e^-x

[crimson sedge]

what have you tried

[eager trail]

i ve tried w the formule u'v + uv'

[sharp lotus]

$\frac{d}{dx}(x-2)e^{-x}$

[wraith dagger]

Soosh

[crimson sedge]

that's correct

[eager trail]

but i block

[blazing dune]

$\frac{d}{dx}[(x-2)e^{-x}]$

[wraith dagger]

Sherif Player

[crimson sedge]

$\dv{}{x}\left[(x-2)e^{-x}\right]$

[wraith dagger]

Pride

[blazing dune]

just use the chain rule

or i mean the product rule

[crimson sedge]

what is u? what is v?

[blazing dune]

$\dv{}{x}[f(x)\times g(x)] =f'(x)g(x) + f(x)g'(x)$

[wraith dagger]

Sherif Player

[eager trail]

i ve had 1 * e^-x + (x-2) e^-x * (-1) * e^-x

but then i found e^-x ( 1 + (x-2 ) (-1) )

and then i block

[sharp lotus]

i think youa re getting a bit confused differentiating e^-x part

[sharp lotus]

this is close, so you're doing this right:

[blazing dune]

so it is correct
$e^{-x} - (x - 2) e^{-x}$

[wraith dagger]

Sherif Player

[sharp lotus]

$=(1)(e^{-x})+(x-2)\frac{d}{dx}(e^{-x})$

[pallid cave]

Can somone help me with math

[wraith dagger]

Soosh

[eager trail]

i did derivative of e^-x

it s -1 * e^-x

right

[blazing dune]

look to an unoccupied channel like #help-32

[sharp lotus]

over here you are adding an extra e^ (-x) for no reason i think

all you chain rule the -1 and thats it you are done

[eager trail]

formule of derivative of e^ (-x) is u' * e ^ u tho

[sharp lotus]

since $\frac{d}{dx}(e^{-x}) = e^{-x}(-1)$

[wraith dagger]

Soosh

[eager trail]

yeah

i m supposed to find (-x-1) e^-x

[sharp lotus]

right you have it right here underlined already, but then you are adding another e^(-x) at the end

[eager trail]

but i can t find it

ohhh

okay

and then ?

[sharp lotus]

so i mean you are pretty much done

here for clarification, you have all the parts there

you could simplify a bit i suppose but also that answer is already valid

[eager trail]

but then i should find this

like i am at 1 * e^-x + (x-2)(-1)( e^-x)

and then i can factorise by e^-x i guess

[sharp lotus]

$e^{-x} + (x-2)e^{-x}(-1)=e^{-x}-(x-2)e^{-x}=e^{-x}-xe^{-x}+2e^{-x}$

[wraith dagger]

Soosh

[sharp lotus]

$=3e^{-x}-xe^{-x}$

[wraith dagger]

Soosh

[sharp lotus]

something like that unles si messed up

just adding like terms

[eager trail]

can u help me by doing it the factorised way

[sharp lotus]

oh so factor e^(-x) after that step:

$=e^{-x}(3-x)$

[wraith dagger]

Soosh

[eager trail]

i need it there tho 😭

to arrive at e^-x (-x-1)

[sharp lotus]

thats what i have right here, follow the 3 images for step by step

they all follow each other in order

i just didnt write the 1 * in the front because well you dont need to

[eager trail]

u don t get the result i said tho so there s an issue

the result is given by the xercice

they just want the steps and i can t find it

[sharp lotus]

what is it supposed to come out to?

[eager trail]

this

[sharp lotus]

sounds like the solution you were given is wrong

wolfram confirms this is the correct solution

[sharp lotus]

or written as what we wrote above

so probably typo by your teacher or book

its unfortunately rather common : )

@eager trail anything else?

[eager trail]

okay i will search thank you

[sharp lotus]

!done

[cedar kiln]

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

[sharp lotus]

you can check correct solution by basically typing it into google:

[eager trail]

the solution i gave is not uncorrect it sjust in a different systen

it s ok

got it dw

[sharp lotus]

its definitely incorrect, unless you typod the problem when typing it here

(-x-1) is not the same thing as (-x+3)

[eager trail]

nope it s just factorised it s how it is

dw

french system

[sharp lotus]

um no it isnt the same...sigh

maybe just how you are typing it, anyway

[eager trail]

dw thanks for the help tho

[cedar kiln]

@eager trail Has your question been resolved?

[cedar kiln]

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[earnest kettle]

help

[earnest kettle]

this is the question and solution

but like

how did he get 5/4???

this is how I solved it

I got 1 as my answer tho so ???

[green thorn]

im not following his last step; it looks like he goes from 4c-3=1 to c=(1+4)/4 but it seems like it should go to 4c=4 and then c=1

[earnest kettle]

exactly why I’m so confused

how did he get 5/4

And why is my answer incorrect?

[green thorn]

well i'd ask him about his last algebraic steps honestly

[earnest kettle]

okie

[green thorn]

for some reason he went from 4c-3=1 (the same thing as you have) to c=5/4 (not the same)

but that seems incorrect

so you might have spotted an error he missed?

[earnest kettle]

maybe he’s incorrect and I’m right?

ahh okie

that’s probably it

😭

[earnest kettle]

but I don’t get how he solved this qs

[cedar kiln]

@earnest kettle Has your question been resolved?

[cedar kiln]

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[mental trail]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wispy topaz]

prk tu fais ça

[mental trail]

déjà t'as de la chance que je sois fr, et ensuite c'est pour savoir où t'en es

[wispy topaz]

HJAJJA T’ES FRANÇAIS

COUCOU

[mental trail]

yo

[wispy topaz]

je suis vraiment nul en marh

marhmath

math frr

[mental trail]

donc 1. ?

Tu vois comment faire la première question?

[wispy topaz]

j’ai fait la 1 2 et 3

la 4

je galère

[mental trail]

la 4, tu as trouvé f'(x)?

[wispy topaz]

non

[mental trail]

tu as déjà fait les dérivées ou pas ?

Sinon on peut faire autrement

[wispy topaz]

je pleure j’y arrive pas je vais aller dormir

[mental trail]

:(

[cedar kiln]

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[cedar kiln]

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[fierce widget]

How would I even do this?

[cedar kiln]

@fierce widget Has your question been resolved?

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[steel ridge]

If I have a number line: 2 2 3 1 2
the table with answers will be:
5
4 4 6 2 4
4 4 6 2 4
6 6 9 3 6
2 2 3 1 2
4 4 6 2 4
(multiplied the numbers of each row and column and wrote them down in the form of a table)

What I get:
5
0 4 6 2 4
4 0 6 2 4
6 6 0 3 6
2 2 3 0 2
4 4 6 2 0

and I need to find: 2 2 3 1 2

My problem is that if I get:
5
0 3306 2432 2660 2888
3306 0 5568 6090 6612
2432 5568 0 4480 4864
2660 6090 4480 0 5320
2888 6612 4864 5320 0

I must get: 38 87 64 70 76

I don't know how to find this line (38 87 64 70 76)

[steel ridge]

visual example

[cedar kiln]

@steel ridge Has your question been resolved?

[cedar kiln]

@steel ridge Has your question been resolved?

[cedar kiln]

@steel ridge Has your question been resolved?

[cedar kiln]

@steel ridge Has your question been resolved?

[agile wyvern]

lets say the numbers are, (a,b,c,d,e)

using you example, 0 4 6 2 4
4 0 6 2 4
6 6 0 3 6
2 2 3 0 2
4 4 6 2 0

we know that ab=4, ac=6 and bc=6

multiplying the first 2 gets us a^2(bc)=24

substituting bc, we get a^2=4 or a=2

we can continue by saying, bc=6, bd=2, cd=3

using the same proccess, we get b=4

@steel ridge

[steel ridge]

Understood, thanks a lot!

[agile wyvern]

no problem

[cedar kiln]

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[blazing hedge]

stuck on how to do c

[patent hearth]

solve the equation

for c)

and find values of x

[nimble mountain]

https://tenor.com/view/substituir-kakashi-hatake-naruto-clássico-gif-18550808

you can try the substitution just. with 🐬=1/x

then solve the quadratic equation to find the value of 🐬 the rest is pretty straightforward

[steel venture]

Let $ u =1/x $ and $u^2=1/x^2$

[wraith dagger]

KTMath

[steel venture]

Then solve for u

[nimble mountain]

yes. solve for me.

[blazing hedge]

DEAD

tysm

.close

[cedar kiln]

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[nimble mountain]

🐬

[cedar kiln]

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[dawn junco]

hey

[ruby barn]

PLEASE PLEADE PLEASE

[dawn junco]

you tried something I guess?

[ruby barn]

Well uh yea

Should I tell you my working?

[dawn junco]

yea

[ruby barn]

Alright, basically I put n= 1 and saw if it gets divided by any of those options

[sharp lotus]

im not even sure i am reading the problem right, the last term is barely visible

$7^{2n}+(3^{n-1})(2^{3n-3})$

[ruby barn]

It's 7^2n+3^n-1× 2^3n-3

[hollow minnow]

Plugging in n=1 should give u the answer

[ruby barn]

It didn't

[hollow minnow]

I see

[ruby barn]

You get 49 when you use n=1 and it isn't divisible by any of those options

[dawn junco]

it's 51 not 49

3^0 = 1 same for 2^0

even then the question seems broken

51 = 3*17

[hollow minnow]

Should be 50

[ruby barn]

It's 3 not 4

My bad

[hollow minnow]

49 + 1 x 1

[dawn junco]

yea why is that so blurry

[ruby barn]

Oh yeah

[wraith dagger]

Soosh

[ruby barn]

My dumbass thought 3^0 =0

It is 50

Which means 25

TJANK YOU GUYS

[hollow minnow]

Np

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[frank nacelle]

yo tommorw is my aths test help mee

[frank nacelle]

maths i mean

[vague rapids]

well Good luck on your aths test

[nova snow]

Lmao so true

[frank nacelle]

aah i mean maths

not aths

whatever

[nova snow]

Well we can’t help but good luck!

I’m kidding you need to ask a question first

[frank nacelle]

is that so

thanks for the good wishes

[hollow minnow]

Do you have a question

[frank nacelle]

aa whole rd sharmaa

[nova snow]

Shawarma?

[hollow minnow]

Indeed

[nova snow]

So true my friend

[frank nacelle]

isnt it relatable

[nova snow]

Anyways good luck with trying to find a question

[vague rapids]

I'd recommend solving the NCERT first before it (assuming you're an Indian)

[frank nacelle]

thhanks buddy

[cedar kiln]

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[spiral bramble]

Guys what is my mistake here :
a^2 - 24a = b^2
(a+b)(a-b) = 24a
For this to be true a must divide b so b = az
(a+az)(a-az) =24a
a(1+z)(1-z) = 24
But here there are at most 8 possibilities for a and the ans for this is 10

[crimson delta]

did you forget the minus from the b^2-4ac ?

[spiral bramble]

Tf yeah

[crimson delta]

why only 8 options

a can be positive or negative

and does each of those have exactly one option z ?

[spiral bramble]

Check by plugging in

Either only a positive or a Negative works

Not both

Or z becomes complex

[crimson delta]

also you divided by a

so a=0 also works

[spiral bramble]

Yes...

I forgot

So the last?

[crimson delta]

wait why are you making the assumption that a itself has to be an integer

[spiral bramble]

The roots are integers

So a is not 0

[crimson delta]

well if a=0 then the polynomial is just x^2=0 which has only integer roots

[spiral bramble]

Yeah so the roots are 0

It doesn't say distinct

[crimson delta]

yes and?

[spiral bramble]

Idk

Any ideas?

[crimson delta]

so something is definitely weird

from a^2-24a=b^2 we can immediately conclude a >= 24

cause otherwise lhs wouldnt be positive

but your last equation suggests a <= 24

[spiral bramble]

a can be negative

[crimson delta]

right sry

[spiral bramble]

Something is Definitely lol

[crimson delta]

still, where are we leaving the other possible values of a

[spiral bramble]

Weird *

[spiral bramble]

Idk

It's AIME

Something needs to be tricky

[crimson delta]

For this to be true a must divide b so b = az
this doesnt have to hold I think

from a^2-24a=b^2 we conclude a divides lhs, so it divides rhs, aka b^2. but that does not mean it divides b

btw a is an integer because sum of roots = -a

and roots are integer, so a is aswell

product of roots = 6a. maybe thats something

[spiral bramble]

Wait I think I have something

\sqrt{b^2 - 4ac} must be natural, soo \sqrt{a^2 - 4 * 1 * 6a} must be natural... so a^2 - 24a must be a perfect square

a^2 - 24a = b^2
a^2 - b^2 = 24a

a - b^2/a = 24

So.. I can see here that b doesn't necessarily need to be a multiple of a, but instead can be a multiple of \sqrt{a} and the equation would still yield a natural number.

What do you think?

[crimson delta]

well if sqrt(a) isnt an integer then thats kinda shit I think

especially if a is negative

[spiral bramble]

That's the problem I guess

And there isn't such a factor of 24

Greater than 6

[crimson delta]

so, if a is positive, then from a^2-24a=b^2 we can conclude b<a

wlog b positive

but then 24a=(a+b)(a-b) <= 2a*a <= 2a^2

12 <= a

hmm

not what I hoped for

I'm not seeing it

and I have to go

good luck

[spiral bramble]

OK thanks I will try a bit longer before stack exchange

<@&286206848099549185>

[west silo]

i need help w smth

[cedar kiln]

@spiral bramble Has your question been resolved?

[cedar kiln]

@spiral bramble Has your question been resolved?

[cedar kiln]

@spiral bramble Has your question been resolved?

[spiral bramble]

.close

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[kindred storm]

What are you trying to understand about it?

[cloud saddle]

Just period is decreasing

as e^x grows

[kindred storm]

Oh, normally, you have cos(x), right?

So, the inside is a nice linear polynomial.

[cloud saddle]

hmm

[kindred storm]

It goes up at the same rate all the time, so the period of cosine will remain the same.

But e^x goes up at an increasing rate.

So, the period will shorten as you go along.

Well, the thing inside the cosine parentheses doesn't go up at the same rate all the time anymore.

Like cos(57x) would have a stable period.

But something like cos(x^2) wouldn't because the rate it increases at gets higher and higher.

It approaches 1 because e^x approaches 0.

cos(0) = 1.

It would also approach 1 on the left.

But that increases slower and slower, so the period on the right would get longer and longer.

,w plot cos(e^-x)

[wraith dagger]

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[kindred storm]

Oh, I was wrong.

It just gives you the mirror image.

Because the x value is negated.

Yes, like e^(-(-2)) = e^2.

Any time you change all the xs into -xs, it mirrors the graph across the y axis.

That works with any function.

Like 2x and -2x.

Or like x^2 + 3x and (-x)^2 - 3x.

Do you mean how do you find its domain and range?

Or do you mean the graph of it?

Well, sqrt increases slower and slower.

3^-t decreases slower and slower.

3^-t approaches 0 as you go higher and higher.

So, you'll approach sqrt(1 + 0).

[sharp lotus]

if you want to visualize a graph the good ol fashioned way, and it's not something you are familiar with (not just punching it into a graphing calculator), do stuff like: find the roots (zeroes) of the function, find the y intercept, find the domain of a function (where is defined \ undefined), then pick some random points (maybe integer values of x \ ones easy to calculate) near those other "interesting" points and plot a few points to see a trend, find asymptotes. being able to visualize any random algebraic function isn't going to come automatically right away but you will get better at it somewhat if you practice and start to see some trends

if you are in calculus you can also use first \ second derivative and see trends of increasing \ decreasing, concavity etc., not sure if that is the class you are taking or not

[kindred storm]

Well, if the function is like position, then the derivative is velocity and the second derivative is acceleration.

Velocity tells how fast the position is changing.

Acceleration tells how fast the velocity is changing.

I'm not sure what you mean.

It tells how fast something is changing.

Well, how fast is velocity.

Like you can be going 40 km/h down the road.

You're going 40 km/h, but you're not changing that velocity.

So, there's no acceleration.

Velocity is how fast you're moving. Acceleration is how fast the velocity is changing, not how fast you're moving.

It tells you how your position is changing.

Or how the graph of the original function is changing.

What do you mean?

It shows the slope, not the change in it.

Like each point on the graph of the derivative is a slope.

So, if (5, 3) is on that graph, then the slope at x = 5 is 3.

It does model the change.

The slope will tell you how the original function is changing.

Oh, OK.

[cedar kiln]

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[rancid grail]

Hi there, I am currently learning about SAT problems, implication graphs and resolution proofs.

I am having a difficult time understanding this SAT problem: {x, y, z} {-x, -y, -z}

From a quick glance, it's easy to find a simple solution to this problem from the following truth assignment:
x = true
y = false
z = true (or false)

From my understanding, this makes the problem satisfiable.

However, if I attempt to use a resolution proof for this problem as such:

1. {x, y, z} axiom
2. {-x, -y, -z} axiom
3. {} line 1,2

This indicates that the problem is un-satisfiable.

How can that be the case? Am I missing something?

[cedar kiln]

@rancid grail Has your question been resolved?

[signal cave]

Read section 6.3 from here: http://intrologic.stanford.edu/chapters/chapter_06.html

The catch is that you can only remove one complementary pair at a time.

[rancid grail]

Ohhhhhhhhhh I think I get it now

Thank you

this is something that my professor forgot to mention in the lecture....

[signal cave]

So the problem proceeds as:

1. {x,y,z}
2. {-x,-y,-z}
3. {y,-y,z,-z} lines 1,2
4. {x,-x,z,-z} lines 1,2
5. {x,-x,y,-y} lines 1,2

All three of which are tautologies.

Please close the channel with .close if your question has been resolved. @rancid grail

[rancid grail]

Thank you, I will!

.close

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[crisp rover]

Hi, can I get some help with this exercise? "solve for x xeR so that the sum of the 3rd and 8th terms of the development of (2x^3 - 1/x)^9 be equal to 0"

[crisp rover]

I can't find a way of the pharenthesis = 0. I think its the only way that all goes to 0

[cedar kiln]

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[trim surge]

I would really appreciate if you could help me with this question

@cedar kiln

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[trim surge]

<@&286206848099549185>

[cedar kiln]

@trim surge Has your question been resolved?

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@trim surge Has your question been resolved?

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[jagged wave]

Which is faster, gauss, gauss-jordan, LU or Cramer?

[crimson delta]

gauss, gauss-jordan and LU are basically the same thing

cramer is garbage

[jagged wave]

Okay thanks, but are they as fast as each other?

[umbral dew]

is this a math question?

[crimson delta]

I dont even know where you want to make the distinction between gauss and gauss jordan?

[crimson sedge]

there is a distinction?

[crimson delta]

for some people, yes

and LU is just gauss jordan but you write down the result a bit better so you can reuse it

[jagged wave]

That's how I was taught, blame Spain

[plucky owl]

Why?

[dawn junco]

for speed it is utter garbage yes

[crimson delta]

well you need to compute a lot of determinants

[jagged wave]

I guess it's because it's worse for matrix with order four or more

[crimson delta]

computing one determinant alone takes basically as long as gauss

[plucky owl]

I would use software for bigger matrixs tbh

[crimson delta]

well of course you wouldnt do it by hand

but that doesnt make it fast

[jagged wave]

I can't even use a calculator in the exam which is weird

But I guess that Cramer is not an option for me in that case

[crimson delta]

cramer is a theoretical tool

[jagged wave]

Thank you people 🙏

[crimson delta]

nothing more

[jagged wave]

Basically, a poop

[crimson delta]

well for theoretical considerations it can be nice

just not for actually solving stuff

[jagged wave]

Thanks 👍🙏👍

.close

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[static egret]

Solving for x : y^(x+4)-y^(x+2)=8/27 without log

[hollow minnow]

Solve for what

[static egret]

x

[upper garnet]

Try factory

[hollow minnow]

Does it say integer solutions

[upper garnet]

Oh, y I thought

[static egret]

What is integer solution

The answer is -5

But no idea how to get it

[hollow minnow]

That can’t be right…

What do you mean the answer is -5

There are x and y here

[static egret]

X is -5

Solve for x

We don’t need to solve for y

I just been pluggin in random numbers for x and -5 works but no idea how to get -5

Teacher didn’t teach log yet so I can’t use log but the calculators online are all using log so I have no idea how to get it

[hollow minnow]

Honestly no idea how especially when you haven’t even done log

[static egret]

Dam

Teacher gave an impossible test question

[hollow minnow]

Hm

I misread the question

[static egret]

oh uh

i did something wrong

……

y is 3

mybad

question should be….
3^(x+4)-3^(x+2)=8/27

[hollow minnow]

Lmao

That makes things 100x easier

[static egret]

[hollow minnow]

You know how to simplify that

[static egret]

no,,,,,,,,

[hollow minnow]

One sec

Okay so

This makes sense?

[static egret]

I thought when u multiply 3^3 on both sides 3^(x+4) and 3^(x+2) turns into 3^(3x+12) and 3^(3x+6)

[hollow minnow]

No a^n x a^m = a^(m+n)

That’s exponent rule

[static egret]

Oh

Well

Shit

Sorry but uh

Can you do the steps to get to -5

[hollow minnow]

Simplify my last step

What is 3^2 -1

[static egret]

8

[hollow minnow]

Great so now we have

3^(x+5) = 1

[static egret]

Yes

[hollow minnow]

This can only happen if the exponent is 0

[static egret]

Ohh

Ok!

Tysm I get it now

[hollow minnow]

Cool

Np

[static egret]

im never gonna miss a math class again

.close

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@tropic vale Has your question been resolved?

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[unborn night]

akak and bjvj instead?

Those are meant to be subscript sorry. And the sorry that the images are like that

are those the same?

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[viral wraith]

I don't understand what a telescoping product is and why this person has decided to minus a half from just the numerator

I understand everything before than, just how would you mathematically explain and represent taking half away from each numerator except the last?

[dire geode]

they just used $\frac{a}{b} > \frac{a - 1/2}{b}$

[wraith dagger]

riemann

[dire geode]

which is true for b > 0

[viral wraith]

Can there be a different approach other than taking 1/2 away?

[dire geode]

yes

stirling's formula

[viral wraith]

what is a telescoping product?

where terms cancel out

like a/b * b/c * ...

[dire geode]

correct

$b / b = 1, a/a = 1, etc.$

[wraith dagger]

riemann

[viral wraith]

And why do they use the greater than or equal to, why not just greater than?

[dire geode]

it doesn't matter

[viral wraith]

and how would you phrase taking half out of the numerator smth like: since n-1/2/n > n-1/n we can subtract half from the numerator expect the last term (as 1/2 -1/2 / 1 would be zero)

or is there a more formal way of stating it?

[cedar kiln]

@viral wraith Has your question been resolved?

[dire geode]

it is basically just this

no more formality is needed

[viral wraith]

ok ty

[cedar kiln]

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[stark stream]

I need help with a combinatorics problem involving the bars and stars method. The problem is listed in the image below. I know that when solving for bars and stars with constraints, whenever you have a variable for x >= n , you can simply subtract the n value from the top of a choose notation. For example, Lets say I had 3 bars and 8 stars with this constraint. My choose solution would be (8-n+3)C3. Further, I know that when given a constraint such as x <= m, I can use the negation of it and subtract it from the total. So that ~(x <= m) = x >= m+1. I can then find a solution with this by subtracting it from the total. This is my solution thus far,

[stark stream]

Is there anything wrong with my solution logically?

[cedar kiln]

@stark stream Has your question been resolved?

[stark stream]

<@&286206848099549185>

[cedar kiln]

@stark stream Has your question been resolved?

[stark stream]

<@&286206848099549185>

[prime panther]

Non negative, not a combinitronics guy but i have done some stuff with combinitronics in programming

prob add limit that [0 to existing limits

because it cant be negative if you havent already

x2 can only be 3 nums

x4 limit isnt defined so im assuming its also [0,inf)

the fact that non negative solution also means x3 limit is changed to [0,3]

so than you probably start off by making lists of valid ranges for variables you can

0,1,2,3 for x3

2,3,4,5 for x2

x1 >= 4
and x4 >= 0

sum of all must = 15

pulling 1 num from list x3 and x2 and adding will always be < 15

[stark stream]

hmm so a more brute force approach?

[prime panther]

That's what im going for : P i can programatically aproach this too

im just coming up with an algorithm to base this off of

[stark stream]

i see

using brute force i think the problem becomes muche asier

cause there are just set scenarios and you can use simple combinatorics concept sto solve it

but its just a lot longer aha

thank you!

[prime panther]

let me generate it real quick just to see what the answer is

[stark stream]

but i was wondering if my approach right now was wrong or not cause if it logically makes sense, im just gonna use it for my solution

tysm

[prime panther]

here is the code i have written so far

oscilating through valids for x4 and x1 shouldn't be hard

Heres a list of all possible sums for x3 and x2

the delta of 15 and sums *2 = possible combinations for that sum

[stark stream]

is there a total?

[prime panther]

not yet

[stark stream]

im actually also in compsci and this course is a requisite ahah

is this C++?

[prime panther]

yes

my favorite lang 😍

[stark stream]

i've kinda learn C, but id much rather use python

but pythosn slow af

[prime panther]

Manual memory management is painful but the fun kind of pain

[stark stream]

masochist behaviours ahahah

[prime panther]

that's how i introduced myself in my programming interview 💀 it worked

[stark stream]

LOL

maybe i should try that

learn C and just say that

[prime panther]

I did, i told them my love for c/assembly i taught myself both

[stark stream]

whats JHI?

[prime panther]

the company i work at

[stark stream]

what do they doi

[prime panther]

idk im just a programming grunt 💀

[stark stream]

LOL

accurate tbh

ur in high school?

[prime panther]

I was when i got hired, how u know?

im a senior now

erm no

[stark stream]

idk i just clicked ur pfp it says pre uni

[prime panther]

i graduated sry

in collage now

[stark stream]

same

[prime panther]

212 is what i got

[stark stream]

congrats on already landing a coding job thats crazy

[prime panther]

im very lucky but i also worked hard in the grind set

[stark stream]

good job

u shuold be proud of that

[stark stream]

also thank u

[stark stream]

tysm

ill work towards the answer LOL

[prime panther]

I reccomend looking at my code under microscope since im not actually combinitronics, just did some leetcode with it

// makgin it.cpp : This file contains the 'main' function. Program execution begins and ends there.
//

#include <iostream>

int main()
{
    std::cout << "Hello World!\n";
    int combinations = 0;

    for (int x3 = 0; x3 <= 3; x3++) {
        combinations += 1;
        for (int x2 = 2; x2 <= 5; x2++) {
            combinations += 1;

            int x4 = 0;
            int x1 = 4;
            if (x3 + x2 + x1 + x4 != 15) {
                int result = x3 + x2 + x1 + x4;
                std::cout << result<<'\n';
                int delta = (15 - result)*2;
                combinations += delta;
                //cycle through x1 and x4
            }
        }
    }
    std::cout << combinations << " is combinations" << '\n';

}

[stark stream]

alright

i will

thanks bro

[prime panther]

no prob

[stark stream]

.close

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[heady granite]

Could someone help me find out the notation for this? - IM pretty sure the denominator is just supposed to be sqrt k but the numerator I cant figure out

[tribal marsh]

To get the alternating sign remember that (-1)^k is 1 when k is even and -1 when k is odd

[surreal cave]

and as for the numerator would k-1, not work? :)

[heady granite]

yeah the -1^k I know of - so am I putting it as (-1)^k * (k-1)?

[tribal marsh]

Yep

[surreal cave]

[heady granite]

ok

got it

ty

[surreal cave]

yw!

[heady granite]

I knew I was getting close to it

.close

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[spark moat]

can someone help me with this question

its set relation

[cedar kiln]

@spark moat Has your question been resolved?

[cerulean sail]

You have a common element between the equivalence classes [3] and [5]

That should imply something, all pairs of equivalence classes are either...

[spark moat]

Hmm

I still don’t get it

[cerulean sail]

First reply to the bot with ❌ so the channel doesn't close

[cerulean sail]

Do you know any properties of equivalence classes?

[spark moat]

Reflexivity, transitivity, and symmetry

[cerulean sail]

Those are the properties of equivalence relations

[spark moat]

Ohh

[cerulean sail]

The equivalence classes are the sets of elements that are equivalent to each other, under that equivalence relation

So [3] is the set of all elements equivalent to 3, and [5] is the set of all elements equivalent to 5

[cerulean sail]

So is there anything you know about the equivalence classes under some given equivalence relation?

[spark moat]

I need to go back to my notes and review them

[cerulean sail]

Yep, a very good idea there should hopefully be some statement along the lines of "all equivalence classes are either..."

[spark moat]

They are either identical or disjoin

[cerulean sail]

Yep, and so, what can you say about [3] and [5], seeing they have the common element 6 in them?

[spark moat]

Let me think

So [3] are [5] are identical and they have the same element 6?

[cerulean sail]

Yep that's it

[spark moat]

That’s the answer ? They are identical ?

[cerulean sail]

Yep, they're identical/equal to each other

You'll have the same elements in both, cause they'd all be equivalent

[spark moat]

Ok okay thank you 🙂

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[cursive thicket]

A sample of size n = 100 produced the sample mean of 16. Assuming the population
standard deviation = 3, compute (calculate) a 95% confidence interval for the population mean.

[cursive thicket]

please i need help

this actually makes no sense

my teacher sucks

and chatgpt is making it hard to explain

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[nova snow]

Did I do anything wrong here

[nova snow]

So far

[patent hearth]

nope

[royal loom]

Yes

[patent hearth]

one thing

[royal loom]

You must adjust your bounds

After making a substitution

[patent hearth]

you need to change the integral ranges

[nova snow]

Ohh

I see

okay I’ll do that

[royal loom]

Or

Most of the times at the end you’re going to back substitute

The x’s back in right

[zealous compass]

Changing the bounds is easier in most cases however

[royal loom]

So really you don’t have to change the bounds, but you have to indicate that in the intermediary they are different

Like labeling u1 to u2

And then at the end just getting everything back in terms of x

Evaluating with normal bounds

If u prefer

[nova snow]

Like this?

Or should the order of the bounds be the same where g(2) is upper limit

And g(1) is lower

[royal loom]

The upper bound should be the same as g(2)

[nova snow]

okay I see

[royal loom]

Recall that you can flip the integral bounds around by adding a negative sign out front

[nova snow]

$-\frac{1}{2}\cdot-\int_{-1}^{-3}u^{-1}du$

yeah

[wraith dagger]

water beam

[nova snow]

so that the negatives cancel out

[royal loom]

If u add the negative out front, the bounds flip

Here you just added the negative but didn’t flip them

Which was the whole point

[nova snow]

ohhhhh

yeah

[royal loom]

This gets kind of tedious in my opinion right? And like a bit more steps means more of a chance of a mistake

I suggest just doing u1 to u2

That’s what I like

[nova snow]

$-\frac{1}{2}\cdot-\int_{-3}^{-1}u^{-1}du$

[royal loom]

But do it as u please

[wraith dagger]

water beam

[royal loom]

Maybe this works better for you

Yes that should be right

[nova snow]

i see

okay got -0.549

thanks for the help

.close

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[royal loom]

Np

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[west frigate]

how to check this diverges or converges by using Ratio or Root test

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[olive briar]

can sm1 help me here

this what i got so far not sure what to do

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[olive briar]

<@&286206848099549185>

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[still sentinel]

Someone help me solve this question pls

[safe ermine]

english?

[still sentinel]

yea

just ignore that sentence its nothing related to the question

is there a way to solve this question?

[safe ermine]

what am i solving

[still sentinel]

dis summation

[cedar kiln]

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

[slim pond]

=infinity

Lol

What's being asked of you

[dense locust]

I dont think this is the original question, did you simplify it? @still sentinel

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[still sentinel]

im confused now

forget this question man this is useless i wasted my time too XD

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[dense locust]

@still sentinel https://brainly.in/question/47974181

[cedar kiln]

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[crimson sedge]

In some sporting events you can see “hat” goals, that is, when a player makes the ball make a parabolic movement above the goalkeeper's reach.If the goal line is 7 m from the striker, and the goalkeeper, who has a reach of 2.4 m in height, is 2 m from it, is it possible to make a “hat” that describes a parabolic movement? ? If the answer is yes, determine the angle, initial speed and height at which the ball enters the goal. Accompany your answer with a graph (g = 10 m/s2).

I have never learned kinematics at school and they are asking me to solve this, and this is a math homework. I really like physics, but I will start learning more about this once I know calculus.

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[fervent pike]

quick question, conjunction is dealing with (\wedge) and disjunction is dealing with (\vee) in proofs?

[wraith dagger]

Aff07

[fading narwhal]

yeah

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[crimson sedge]

Hi everyone, I'm new to Laplace's transform and I need help with this problem. I need to find the inversed laplace transform of F(s)= e^(-3s)*(s^2+3s+6)/(s^3+3s+7s+5) I'll post my current work but there is clearly something wrong

[subtle wadi]

If you use latex, it'll be much easier to help 🙂

[crimson sedge]

yes can you help me im new to this discord

this is what i have done so far

[subtle wadi]

ok, let me check

[crimson sedge]

thanks!

I did type it in wolframalpha but it seems absurd

but as I staded earlier I'm new to laplace so it my be right idk

[crimson sedge]

@long arrow do you have any steps for this problem as i am learning it could help me understand

and it's 3s not 3s^2

[long arrow]

so it's 3s + 7s?

just 10s, nonsense

but ok

[crimson sedge]

yes I know

sorry

oh no sorry

it was afterall 3s^2

.close

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[raw kernel]

how do i show algebraically that 8/5x = (120/49)t^2 shows an inverse correlation between sqrt x and t

[wraith dagger]

! What the hell am I doing here?

[raw kernel]

ye

[dusty hazel]

Do you know what an inverse correlation means?

[raw kernel]

i would assume i would manipulate it to sqrt x = k/t

but idk how to get there

[dusty hazel]

Sweet.

[raw kernel]

or do i use f'x

to prove

[dusty hazel]

not necessarily at all.

Just sqrt both sides.

[dusty hazel]

Here.

[raw kernel]

sqrt both sides dont make them equal though

if i sqrt a side i would have to time the other side by a^(-1/2)

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[raw kernel]

<@&286206848099549185>

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[stoic gale]

what do you mean by this?

unless there are some conditions not written above, there is no issue with sqrt-ing both sides to show inverse correlation

you will get K/sqrt(x) = Lt, where K, L are constants, and so 1/sqrt(x) is proportional to t, i.e. sqrt(x) is inversely proportional to t

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[floral hinge]

What is wrong here?

[floral hinge]

I found delta x and xi
Put them in equation

I don't understand what the problem is here

Oops

Mistake located

.close

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[arctic lava]

Hiiii

[arctic lava]

I need some help please

How would you prove that √(2x^2+1) is continuous in all real numbers?

[astral steppe]

find the domain

[arctic lava]

By definition and without derivatives

[arctic lava]

it's all real numbers

[astral steppe]

i know

but can you prove it

[arctic lava]

because 2x^2+1 is always >0

so the square root is defined always

[astral steppe]

okay so there aren't any holes on the graph

[arctic lava]

yep

[astral steppe]

is it more complicated than that?

i'm confused

[arctic lava]

but how do you know there aren't any jumps?

[astral steppe]

because the domain is R

[arctic lava]

then floor(x) is continuous