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[fickle owl]

@dawn birch can you explain step 6->7?

for me i multiplied the sqrt by c^2u^2

and i have $\sqrt{1-c^2y^2}$

[ocean seal]

! matthewzz

[fickle owl]

@dawn birch

<@&286206848099549185>

also what is the integral of the left side

wouldnt it jsut be x* the integrand

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[forest turtle]

How do I get the stuff in re

D

Red

[wary stream]

From the first line

It solved for a, then plug a back into the first line

[forest turtle]

So I put 1 into (x-5)

@wary stream yo

So how do I solve exactly?

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[forest turtle]

.reopen

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✅

[tacit arch]

.close

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[forest turtle]

Bruh

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[white sphinx]

can somebody explain why log_8(49)-log_8(7)=log_8(7)

[vocal hawk]

this is one of the logarithm rules

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[alpine sable]

exponential growth. Though, in order to solve the formula i'd need the years no? How would I determine the years with

[tough fulcrum]

@alpine sable can I see the rest of the page?

[alpine sable]

Its a spreadsheet, but I have the full question if you want? it doesn't seem to give much info however

[tough fulcrum]

you can get expected revenue and expected expenses by multiplying the current revenue and expenses by 1+ the growth

and then get Total profit by subtracting expenses from revenue

[alpine sable]

oops wait plus the growth too

let me give it a try

Idk what to do for years but supposedly, the years are determined by the break even point. Resulting to some division of 2 values possibly?

Eh, honestly i'll figure it out eventually, thanks

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[earnest spruce]

Hello

[earnest spruce]

https://www.desmos.com/calculator/hguanwbkbu?lang=fr

I have found a link to this very cool desmos graph. I want to understand how it works

Its equation is $$-xcosa+ysinb=(xsina+ycosb-h)^2+k$$

[ocean seal]

Ze_Beeg_Almond

[earnest spruce]

I can see the vertex form of the parabola popping out here $$f(x)=a(x-h)^2+k$$

[ocean seal]

Ze_Beeg_Almond

[earnest spruce]

.close

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[soft needle]

Is this correct?

[wary stream]

What's the original equation?

[soft needle]

x² + 4x + 11 = 0

It's at the top

[wary stream]

I couldn't tell if there was a number next to the x^2

[soft needle]

Ah

I put a 1 in there so it's easier

[wary stream]

And to answer your question, no, that is not correct

[soft needle]

Oh

What did I do wrong?

[alpine sable]

Have you done imaginaries in quadratics yet?

[soft needle]

No

[wary stream]

Because you somehow made $\sqrt{-28} = 5$ which is wrong

[ocean seal]

dldh06

[alpine sable]

This equation will give you imaginary zeroes

[wary stream]

Then you wrote the question wrong

[soft needle]

Huh?

I'm confused now

[wary stream]

If you don't know anything about imaginary numbers, you wrote the question wrong

Because the roots for that equation are imaginary

Aka complex numbers

[soft needle]

I don't think I wrote the question wrong

[wary stream]

Then answer this, do you know anything about complex numbers?

[soft needle]

No

[wary stream]

Mainly $i = \sqrt{-1}$

[ocean seal]

dldh06

[wary stream]

If you answered no to that, then there is a high chance you wrote the question down wrong

[soft needle]

It was what the teacher gave me

It's #2 on this sheet

[wary stream]

Then I suggest you should ask your teacher

[tacit arch]

Is "no real solutions" a valid andwer

[wary stream]

That too

[soft needle]

Can you help me do it tho? Just in case

I don't wanna come off as a dick to the teacher

[tacit arch]

What is there to help. You already got the answer

[soft needle]

Oh

How did you get no real solutions?

[wary stream]

Because imaginary

Negative number under square root = imaginary solutions

Aka no real solution

[soft needle]

Ah ok

[soft needle]

So whenever I see that it's no real solutions

Everytime

[wary stream]

Yes, normally

[soft needle]

Ok

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[vagrant cobalt]

<@&268886789983436800>

[night geyser]

?

[sly mantle]

@karmic cosmos is this a test?

[karmic cosmos]

homework

[vagrant cobalt]

Oh

google forms is usually test

okay what have you tried

[karmic cosmos]

im having a hard time answering it

missed my class yesterday so idk how to answer it

[vagrant cobalt]

https://www.economicsdiscussion.net/dispersion/how-to-calculate-inter-quartile-range-explained/2523

you need to find $Q_1$

[ocean seal]

usernamephobic

[karmic cosmos]

thank u so much

[vagrant cobalt]

but like let me know if you get stuck anywhere

[karmic cosmos]

okay thnxxx ill try understanding it

okay i dont understand everything kekl maybe ill pass on this homework

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[cedar field]

how is the deffinition of a derivative this?

[tacit arch]

Do you have an alternative definition?

[cedar field]

no

[alpine sable]

it's more intuitive graphically

[tacit arch]

Then just accept it as the definition

[cedar field]

can you explain it too me?

[cedar field]

do i need to memorize it or does it not matter?

[tacit arch]

Read
https://tutorial.math.lamar.edu/classes/calci/DefnOfDerivative.aspx

[alpine sable]

If you understand it, then you won't need to memorize it. But it's important at the start of most calc courses

[tacit arch]

If you haven't seen limits yet
https://tutorial.math.lamar.edu/classes/calci/Tangents_Rates.aspx

[cedar field]

will this tell me how to understand it?

[cedar field]

never mind

[cedar field]

can you explain it too me?

this

[tacit arch]

Just read and ask questions about it

[cedar field]

ok

[vale wigeon]

@cedar field are you familiar at all with the concept of slope

[cedar field]

yes

[vale wigeon]

right

$\frac{f(x+h)-f(x)}{(x+h)-x}$ is the slope of the line through $(x, f(x))$ and $(x+h, f(x+h))$

[ocean seal]

Ann

[vale wigeon]

what you might call a secant line for the graph of f

does that make sense to you

[cedar field]

yes

[vale wigeon]

as the two points that define the secant line get closer together, the secant line looks more and more like the tangent line (assuming it looks like anything at all)

and the slope of the tangent line is what we call the derivative of our function

[cedar field]

Ok

[vale wigeon]

is this the explanation you were looking for?

[cedar field]

yes

does that explain the image?

[vale wigeon]

what image?

[cedar field]

.

[vale wigeon]

i mean, yes, that's what i was explaining... but it is kind of weird to hear YOU ask whether it does

surely if you're the one with the doubt then you are also the one who knows whether my explanation was sufficient for you

[cedar field]

i was double checking

anyway

thank you

.close

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[small quarry]

why is this not correct guys

[small quarry]

when the plug the x in it just equals 0

so 2 is the only thing left

so 2 should be C

[oak perch]

1, not 2

e^0=1

[small quarry]

oh oops I thought e never changes

thanks for the help!

[median oar]

e doesn’t change

The x changes

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[alpine sable]

i know that i can put pw of benefits = pw cost * n where n is B/C ratio. but it still leaves me with 2 unknowns

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[obsidian wagon]

How did they get x=3root2

[obsidian wagon]

This is the question btw

[blazing saffron]

It says x = 3sqrt(t)

[obsidian wagon]

Sorry I meant underneath that

When t=2

[blazing saffron]

They just plugged in t=2

To 3sqrt(t)

[obsidian wagon]

Guess I was just dumb in figuring it out 🫡

And because of that it also changes the bounds as well?

*domain

[blazing saffron]

The domain of the function is the interval of the input where x is the input variable in this instance

Since t goes from 0 to 2 means x goes from 0 to 3sqrt(2), yes that would be the domain when in terms of x

[obsidian wagon]

In this instance if I’m trying to find the range, what should I do? Because in part b I’ve already done the differentiation to show t=all root 2/3

Dy/dx = 3t^2-2 whic gives t= whole root 2/3

[blazing saffron]

Yeah so that should be the minimum output

[obsidian wagon]

Ahh okay second derivatives and whatnot

[blazing saffron]

You just need the first

The value of y(sqrt(2/3)) is the min

And the function clearly goes to infinity

[obsidian wagon]

Ahhh okay yeah I get what you mean

And do I sub the sqrt 2/3 back into y equation to get the range?

[blazing saffron]

So the range should be sqrt(2/3)^3 - 2sqrt(2/3) to infinity

[blazing saffron]

Well that's just the minimum but yeah

The range is the entire interval of possible outputs

[obsidian wagon]

Yup so I got the minimum as -4 root 6 all over 9

[blazing saffron]

So it would be,
\[-\frac{4\sqrt{6}}{9} \leq y < \infty\]

[ocean seal]

Skid

[obsidian wagon]

Ahh okay

Let me also cross check here in the answer booklet

So the minimum was right

I’m lost with the 4

[blazing saffron]

Oh we are bounded by t=2

I forgot

[obsidian wagon]

Ahhhh

So sub back into y equation I’m guessing

To get 4

[blazing saffron]

yeah

[obsidian wagon]

Phew… this makes a bit more sense for me at least 😅 thanks for your help @blazing saffron

[blazing saffron]

It's just the possible outputs of the function

[blazing saffron]

The graph here

Has y values between the range

Of -4sqrt(6)/9 the minimum and 4 at t=2

[obsidian wagon]

I guess 4 being where it’s marked by C

And the other where the curve bends

[blazing saffron]

Yeah the range is between the min and max of the y axis for the graph

In this instance

There could be discontinuous parts in other graphs

But here there is not

[obsidian wagon]

Gotcha!

I also have one more question if you don’t mind? It’s to do with cartesian equations…

[blazing saffron]

I can try to help

[obsidian wagon]

So what I’ve learnt is to make t as subject and sub into y

And I’ve always struggled in finding the best way to make t as subject for equation x

I’m curious as to how you would do it? Because the answer booklet sometimes just doesn’t help for me

[blazing saffron]

If you solve for t from y you'll get t= +-sqrt(4-y)

[obsidian wagon]

Since y has t^2 could I just leave it as 4-y?

Now I’m lost on what to do next after finding that

[blazing saffron]

To get it into the x^2 form I would square t^3-t

Hopefully, the t terms will all have multiples of 2 as exponents

Then

You can just use (4-y)

like if theres t^4 you could do t^4 = (t^2)^2 = (4-y)^2

[obsidian wagon]

Ah okay I get what you mean

[blazing saffron]

x^2=t^6 - 2 t^4 + t^2

Which is
(t^2)^3 - 2 (t^2)^2 + t^2

And we know t^2

Is 4-y

[obsidian wagon]

Yeah

[blazing saffron]

After simplifying I get x^2 = (4-y)(3-y)^2

Where a = 4 and b=3

[obsidian wagon]

Ahhh okay yeah makes sense for me now!

This is what they did

[blazing saffron]

Yeah its the same thing I just expanded x^2

[blazing saffron]

To this

Instead of leaving as parentheses

[obsidian wagon]

Yup, l kinda like the way you did it tho, sometimes the answer booklet feels like they just skip through some lines 😥

[blazing saffron]

I think the booklet did it a better way

Expanding it makes the algebra worse

But it's really a stylistic thing

[obsidian wagon]

I get what you mean

[blazing saffron]

Either way works

[obsidian wagon]

yeah true

[blazing saffron]

The main premise was to find t from y then plug into x^2

[obsidian wagon]

Yup, guess I gotta do a bit more of these before I move into using trig identities for this chapter 😳

Thanks for your help again @blazing saffron appreciate it

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[alpine sable]

In a set of data that has a normal distribution, the scores within one standard deviation from the mean ranges from 45 to 51. What is the mean score?

Would that just be (45+51)/2 or (45+46+47+48+49+50+51)/7

[median oar]

they both work

[alpine sable]

oh, so 48?

lol I got it now, thank you!

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[verbal elbow]

i assume this is an existing formula but i thought about it a couple of years ago and cant get it out of my head: x*x = (x-1)(x-1) + (2y + 1)

[median oar]

put \ before the *

[verbal elbow]

oops sorry discord took asterisks out of my formula

thx

[clear stump]

are you sure you wrote it right

[median oar]

do you mean (x-1)(x-1)

[verbal elbow]

yeah

[median oar]

so you'd get (x^2 - 2x + 1) + (2x + 1)

then x^2 - 2x + 2x + 1 + 1

x^2 + 2

so it should be (2x - 1) in your equation

x*x = (x-1)(x-1) + (2x -1)

like this

[verbal elbow]

no i mean 2x + 1

so like

[median oar]

then it's false

[verbal elbow]

oh frick mb

i mean y = x -1

hold on

y = x - 1

lemme repost down here

x*x = (x-1)(x-1) + (2y + 1)

is this correct if y = x-1?

[placid zinc]

,w simplify (x-1)^2 + (2x - 1)

[ocean seal]

Results provided by WolframAlpha

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Requested by Kaynex#7520

[placid zinc]

Yeah, it is

[verbal elbow]

cool

also my question was: is this formula already used for finding the square of a number 1 greater than another number

[placid zinc]

You could use it for that, sure

[verbal elbow]

cool

thanks

[median oar]

look at 41^2

it's 1600 + 80 + 1

[placid zinc]

13² = 12² + (2•13 - 1)
13² = 144 + 25

[median oar]

from (40 + 1)^2

yeah but that's looking backwards

[median oar]

this should say finding the square of a number 1 less than another number

[verbal elbow]

oh yeah but it can also be used in reverse

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[kindred anchor]

Is y=1/x function decreasing or increasing?

[median oar]

have u done calculus

[kindred anchor]

It's derivative or slope is -1/x².

[median oar]

ok

so for all values of x

what is the slope

[kindred anchor]

x² is always positive.

[median oar]

is it positive, negative, zero, undefined?

[kindred anchor]

it can't be 0

since the slope is negative

it must be decreasing

[median oar]

except at 0

so the function is always decreasing except at 0 where the slope is undefined

as a matter of fact not only is the slope undefined, the point is also undefined

[kindred anchor]

just wanted to confirm

there is discrepancy in the answer key of my assignment

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[turbid blaze]

how do you go from top to bottom

[bitter vault]

multiply top/bottom by e^(-1/5)

[turbid blaze]

ok thanks

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[alpine sable]

is that right for a function of s the area between a and b

[high rapids]

,rotate

[alpine sable]

n=infinity on top i know.. fhdh

[zenith lodge]

can you type this up in latex please?? i really doubt this would be comprehensible for most people here

[alpine sable]

ooo

um

idk how

[zenith lodge]

also, if i'm understanding the question correctly, you can get the area between functions a(x) and b(x) with just a simple integral

[alpine sable]

whut

between a and b of function s

[zenith lodge]

ohh

wait a sec lemme reread the latex documentation

[elder grail]

$\lim_{n\to\infty}\sum_{k=0}^{n}\frac{b-2}{n}\cdot S\left(\frac{(b-2)k}{n}\right)$

[zenith lodge]

$$ \int_{a}^{b} s(x) \dd x$$ should work

[ocean seal]

newbienoob

cis(kitten)

[elder grail]

correct the mistakes

@alpine sable

[elder grail]

is this ur question?

[alpine sable]

my handwriting dhdhdh

[elder grail]

so is the expression within the summation correct?

[alpine sable]

ig?

[elder grail]

i mean

u should be knowing right

like

ur q so

[alpine sable]

OOO

u mean what u did is the same what i did

[elder grail]

yes

[alpine sable]

yea

$\lim_{n\to\infty}\sum_{k=0}^{n}\frac{b-a}{n}\cdot S\left(\frac{(b-a)k}{n}\right)$

[elder grail]

so this is the same as what u've written?

[ocean seal]

Lenny

[elder grail]

ok

[zenith lodge]

ehh after staring blankly at the screen for like 15 sec i think you're probably writing out explicitly the riemann integral but the -2 seems out of pl..... oh nevermind, that looks correct.

[alpine sable]

AYYY

[elder grail]

yeah that looked like a 2 mb

[alpine sable]

So it’s correct?

[zenith lodge]

yes, at least to me

[alpine sable]

THANKS

.close

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[vestal pilot]

Hey, quick question:
Anyone knows how to get from this
https://cdn.discordapp.com/attachments/779022977425604649/982260335127961650/20220603_153134.jpg

[vestal pilot]

to this?
https://cdn.discordapp.com/attachments/779022977425604649/982260393370066974/unknown.png

The beginning of the question is at the top of the first image.

The subject is Ordinary Differential Equations

I'm not sure if the x next to the C is their mistake or if I'm missing something.

If it's their mistake, then I've reached the correct solution. I just can't find any way to reach a point where there's an X instead of the e 🤷‍♂️

[stone glen]

There is a mistake in this

Go through ur page again

U will recognise it

[vestal pilot]

Can you hint at which part it's at..?

Cause I've been looking at this question for a long time now.

[stone glen]

Right when u apply logarithm on both sides

[vestal pilot]

Are you talking about the ln(c)?

[stone glen]

No

Oh wait

Gme a min with this

[stone glen]

R u sure this isn't the wrong solution?

[vestal pilot]

question 69, answer 69

[stone glen]

Lmao

But i think there is a typo , it should be e instead of that x there

[vestal pilot]

that's what I said above

that I think that's what happened

and that they had a typo

[stone glen]

Yeah so it's a mistake

[vestal pilot]

cause I can't find any way of getting an x there

[stone glen]

There isn't one

Cuz it's written wrong

Close it

[vestal pilot]

Well, at least I'm not as dumb as I thought

.close

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[rough forum]

Can someone help me with this problem

[rough forum]

Im not sure if I did it right

The highlighted part

is what I had originally

then I added the other 2 curves

cause it looked weird

[merry depot]

the highlighted part seems fine.

[gray ingot]

This isnt even a function with the 4 parts

[rough forum]

okay

[median oar]

that is not a function

XD

[rough forum]

so I can remove the non highlighted part right?

[thick yoke]

yes

[rough forum]

okay ty!!! it was just weird looking at an asymptote with only 1 curve

[thick yoke]

you can put a random value for f(0) but I don't think its necessary

[rough forum]

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[median oar]

there are many solutions to this question btw

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[violet sphinx]

It seems that to solve the question, I need assume the origin like a midpoint. Why is that so even though it's not stated in the question?

[gray ingot]

diagonals of rectangles are bisectors

so it is stated indirectly

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@violet sphinx Has your question been resolved?

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[velvet hatch]

Any ideas on how to prove c? Below is what I have so far.

We argue by contradiction. Consider the sequence of partial sums $S_n(x)=\sum_{k=0}^n f'k(x)$. Assume that for all $x \in (0, \infty)$ $S_n$ converges uniformly to some $s$. Moreover, the differentiality of standard functions gives
$$f'k(x)=2k^2x\exp(-k^2x) - k^4x^2\exp(-k^2x)$$.
Since $S_n$ is defined on a metric space, by proposition 11.3.4, we have that $S_n$ is a Cauchy sequence on (0, $\infty$) with $\norm{\cdot}{\infty}$. Choose $\epsilon = 1$. Let $N \in \nat$ such that the definition of Cauchy sequence holds. Choose $n = N, m = N + 1$.
\begin{align*}
\norm{S{N+1} - S_{N}}{\infty}&=\norm{f'{N+1}}{\infty}\
&= \sup{x\in(0,\infty)}\abs{f'{N+1}(x)}\
&= \sup{x\in(0,\infty)}\abs{2(N+1)^2x\exp(-(N+1)^2x) - (N+1)^4x^2\exp(-(N+1)^2x)}\
\end{align*}

[ocean seal]

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

[alpine sable]

use proof by induction?

[velvet hatch]

Can you elaborate a bit more?

[violet sphinx]

What do you mean by that?

[median oar]

They are the radius of the circle the rectangle is inscribed within

The intersection of the bisectors is the origin and centre point of the circle

[violet sphinx]

I see

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[velvet hatch]

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[buoyant depot]

Hi, please could anyone give me a hi t for this qn, thx

[oak perch]

For i) just find number of non-negative integers (a,b,c) such that 6a+6b+3c=24
And =n

For ii) just find number of non-negative integers (a,b,c,d) such that 6a+6b+3c+d=24

Or=25/66 respectively

[buoyant depot]

oh thanks, but i havent figure out how to expand the polynomials

oh is this binomial expansion?

[oak perch]

Other cases are specific, let’s consider 6a+6b+3c=n
So non-zero when n=3m
2a+2b+c=m
2(a+b)=m-c
So when m is odd,c=1,3,…,m
Answer is 1+2+…+((m-1)/2+1)=(m+1)(m+3)/8
When m is even, c=2,4,…,m
Answer is 1+2+…+(m/2+1)=(m+2)(m+4)/8

I didn’t use any binomial coefficients

Anyway together answer is ([m/2]+1)([m/2]+2)/2
Actually we can consider number of non-negative solutions of 6a+6b+3c+d=n too
n=3m+r, then d=r,r+3,r+6,…,n, so Σ([k/2]+1)([k/2]+2)/2 where k is from 1 to m

Okay in summary
Coefficients of $x^{n}$ in expression in i) is 0 when n isn’t divisible by 3, is $\frac{([\frac{m}{2}]+1)([\frac{m}{2}]+2)}{2}$ when $n=3m$
Coefficients of $x^{n}$ In expression in ii) is $\sum_{k=1}^{m}\frac{([\frac{k}{2}]+1)([\frac{k}{2}]+2)}{2}$ when $n=3m+r$

[ocean seal]

Cogwheels of the mind

[oak perch]

Then you plug in n=24 or 25 or 66 whatever you need

[buoyant depot]

thank you

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[valid thunder]

m = 27, does that mean i plug that into y =mx + b?

y=27x + b

then i need to find y

x^2 + 9x + 20, x = 9
9^2 + 9*9 + 20 = 182

182=27x + b

plug in x then

[placid zinc]

What's the question?

[kindred oyster]

What's the correct answer for a?

[valid thunder]

why is a wrong, and is my newer solution correct

[placid zinc]

I agree with your answer. The slope of the tangent is 2x + 9

[valid thunder]

182 = 27(9) + b, solve for b

[valid thunder]

but it says its wrong

so it wants it in y = mx + b form

[kindred oyster]

^

[merry depot]

probably need x_0

not just x

[valid thunder]

im not sure, but im figuring that out right now, i assume i need it in y = mx + b form

[placid zinc]

I would be very concerned if, when asking for the slope of the tangent, it instead accepted an equation for a tangent line

[valid thunder]

couldnt i just plug in 9 with f(x)

i get the y value then

[merry depot]

for part a? no, it's asking for the general formula for the slope when x = x_0
your formula is correct, but it's a function of x when you know x = x_0

[valid thunder]

i kinda get what youre asking for

[valid thunder]

b = 182/243

y = 27x + 182/243

would this be the equation?

i think i can reduce the fraction

[valid thunder]

im not sure what you mean

wait i think i got it

y-y0 = m(x-x0)

y - 182 = 27(x - 9)
y = 27x -243 + 182

y = 27x - 61

[kindred oyster]

Oh yeah that makes sense

[valid thunder]

the answer was wrong

wtf

i have 1 attempt left

[lone heart]

@valid thunder Has your question been resolved?

[valid thunder]

<@&286206848099549185>

[nimble fern]

Did you try this? DeltaF(x)

[valid thunder]

i already have x_0, its 9

[nimble fern]

I think for (a), I means for general x=x_0, and x_0 is still not 9 yet in question (a)

[nimble fern]

Since it say general point, that's why I guess this

[placid zinc]

You are overthinking to the extreme

If you can input this, I think this is what they are expecting:

$2x_0 + 9$

[ocean seal]

Kaynex

[placid zinc]

If neither are correct, talk to the teacher, the question was borked

[valid thunder]

yea i got it

[valid thunder]

yup this is correct

i plugged that in a while ago

tyty @placid zinc

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[alpine sable]

would this be right?

[tacit arch]

Naaahhhhh

The $\times$ symbol means cross product.

[ocean seal]

riemann

[alpine sable]

ah but dont i need an angle for that?

[tacit arch]

What definition of cross product uses angle?

[alpine sable]

o i get it

its using the a2b3... stuff

this

[tacit arch]

Looks right

Now memorize it

[alpine sable]

ok but first i gotta multiply but 2 right?

[alpine sable]

like using this 2u value?

[tacit arch]

Right

a=2u and b=v

[alpine sable]

n did i do the 2u right?

in that image above

[alpine sable]

does this look right?

so the ans would be none of the above

.close

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[alpine sable]

I don’t understand how they got there, (the cos, I understand the a^2 thing

[high rapids]

do you know the cosine rule?

law of cosines

[alpine sable]

no

[mortal trellis]

double angle/half angle formula for cosine if you were asking about the last simplifications

[alpine sable]

oo

But, then why was the 2aa term removed

[mortal trellis]

it's factored out

[alpine sable]

oh

i need to study trig

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@alpine sable Has your question been resolved?

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[willow bough]

Is this correct?

[high rapids]

Where did the l go?

Oh nvmd that's 1

Umm how is b3 - a3 = (b-a)(b2+ab+a2)

[abstract fractal]

Is that not the difference of 2 cubes formula?

[high rapids]

Oh nvmd I didn't know that

[tacit arch]

$\sqrt{z^2} = |z|$ and not just $z$

[ocean seal]

riemann

[tall hearth]

@willow bough add ± to consider that

[willow bough]

Oh true so I add a ± after the sqrroot

but are there any other mistakes?

[tacit arch]

no

[willow bough]

okay thanks guys

[tacit arch]

.close to free up the channel

.close

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[wanton nova]

how did you close his channel?

[tacit arch]

helpers have the ability to close just by typing .close

i'll open a channel and you can test

[wanton nova]

really?

thx

[lone heart]

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[shy elk]

can someone do this partial fraction

[shy elk]

ive tried but dont know how to

[wanton nova]

sure

[wary stream]

Post your work

[abstract fractal]

Can you do partial fractions? The denominator doesn't factor outside the complex numbers

[shy elk]

i have no work idk where to begin

[shy elk]

its on a question paper im not sure but id assume so

[wanton nova]

on a test?

[shy elk]

a past paper yea

i dont have it tho my friend just showed me that question

[wanton nova]

you cant factor $x^2+9$

[ocean seal]

Mike Oxbig

[wary stream]

I think the only option is long division, not PFD

[abstract fractal]

It's possible if you're willing to use complex numbers, no?

[wanton nova]

sure

[shy elk]

Idk what that is

I'm in UK year 13 not uni level

[wanton nova]

ur friend messed with you?

maybe...

[abstract fractal]

Complex numbers are just a + bi, where i² = -1

[wanton nova]

he didnt learn them]

[wary stream]

,w apart (2x^2+x-27)/(x^2+9)

[ocean seal]

Results provided by WolframAlpha

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

Requested by dldh06#2659

[wanton nova]

yea

thats basically all you can do

[shy elk]

could be idk

[abstract fractal]

Is it possible the denominator was actually x² - 9 and you misread it?

[shy elk]

Maybe actually

he's in work now so i cant even triple check with him

would that just be a/(x+3) + b/(x-3)

if it was x^2 - 9

[abstract fractal]

Yes

[shy elk]

ok

thanks everyone

[wanton nova]

btw

the numerator cant be factored

just saying

so...

[shy elk]

u dont need to tho?

[wanton nova]

idk

how

[shy elk]

2x^2 + x - 27 = a(x-3) + b(x+3)

Then u make x = to 3 to work out a

And x = -3 to work out b

[abstract fractal]

Wait, can you do partial fractions when the denominator has a smaller degree? Because I'm imagining combining the fractions back after partial fractions and it doesn't seem to work

[wary stream]

You have to do long division first

If n > m or n = m, you have to do long division

When n < m, that's when you can do PFD

[shy elk]

havent done long division

I think its just

2x^2 + x - 27 = a(x-3) + b(x+3)

where x = 3
2(3)^2 + 3 - 27 = a(3-3) + b(3+3)
-6 = 6b
b = -1

where x = -3
2(-3)^2 - 3 - 27 = a(-3-3) + b(3-3)
48 = -6a
a = -8

didnt write a lot of steps but isnt that right?

[wary stream]

No

As I mentioned
If n > m or n = m, you have to do long division
When n < m, that's when you can do PFD

[shy elk]

what n and m

numerator?

[wary stream]

The power

The highest power in the numerator = n

Highest power in denominator = m

[shy elk]

oh

[wary stream]

Now that is possible without long division

[shy elk]

cuz the bottom is x^3?

[wary stream]

Because n = 2, and m = 3

[shy elk]

ooooh ok

i think for us we will only get n<m cuz we havent done long division

[wary stream]

Yes

[shy elk]

Okay 👌

Thanks very much everyone

[worthy lichen]

The solution would be x²+x+13= A/(x+2) + B / ( x+2)² + C /(x-3)

[lone heart]

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[visual gyro]

so i got this formula, it contains a really complicated quadratic equation and i would like to get help in separating the a,b and c component of it (to be able to use the quadratic solving formula)

[visual gyro]

sorry this channel is occupied

[copper thorn]

Ok

[alpine sable]

Yes

We know that b is equal to -delta x *v

A is equal to 1/2g delta x^2/ v^2- delta y

C is equal to 1/2g delta x^2/v^2

[visual gyro]

not to argue but 1/2g delta x^2/v^2 is in the bottom part of the formula that should be 2a in the solving fomula?

[lone heart]

@visual gyro Has your question been resolved?

[alpine sable]

Whoops yeah you are right

[visual gyro]

in the meantime i implemented the equation in my code...and turns out it gives inaccurate results...i gotta look (or have someone look into) the formula itself becuase there might be a chance that my formula is wrong

.close

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[alpine sable]

have to use integration by parts, but if we can use anything simplier we can but I'd rather do it IBP to learn how to do it completley lol

what have you tried?

[alpine sable]

nothing yet, I should have specified but I get confused when I see "e"

integration by parts I don't neccesarily have a problem with its how to go about all these variables

[weary wyvern]

just do ibp

[alpine sable]

Well try it first and see, ask when and where you're confused

do you think it would be good to use the tabular method here?

it has different methods

names***

but thats what I call it lol

Pick a u and dv that you think would be appropriate and try it.

[mortal trellis]

use the method of IBP that you can use correctly. which one you use is not important

[dusky blaze]

x/e^x ok let u =e^x x=lnu

then go from there

either find

dx/du=1/u

or

du/dx=e^x

[alpine sable]

I got 5e^5 -e^-1

💀

[dusky blaze]

show work?

[alpine sable]

if you insist

I did the table method btw

here ya go

[dusky blaze]

hm

[alpine sable]

the answer is completley wrong btw lol

[dusky blaze]

ik

bro

u can use ibp

x/e^x is xe^-x

[grim sedge]

Hey

I have a question

May i send it?

[alpine sable]

not here silly

[dusky blaze]

let u=x u’=1
v’=e^-x v=-e^-x

u can

dw

[grim sedge]

Sorry where?

[alpine sable]

in the help section

[grim sedge]

This is help-

[wanton nova]

#help-9

here

[grim sedge]

Oh fuci

Nvm

[wanton nova]

send it here

[alpine sable]

@dusky blaze was my U and DV right before?

Okay so

U=1

DU= dM

DV= e^M

V= ???

thats what has been confsing me lol

the V i guess

V= e^-M

right?

[dusky blaze]

mbb

i was doing something

du/dx=1

dv/dx=e^-x

[alpine sable]

so V=e^-M , right? lol

[dusky blaze]

yeahh

not positive e

[alpine sable]

so its -e^-M

[dusky blaze]

e^-m integrates to (e^-m)/-1

yh

then apply the formula

∫uv’=uv-∫vu’

or use the method u were using

[alpine sable]

so how do I subtract this, -5e^-5 -1e^-1

the e's confuse me always lol

.close

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dense mural]

i am trying to form a linear programming formulation for this problem
two types of plants must be planted
no more than 1000 plants must be built per km^2 and there are 6km^2 in total
and the ratio of the 2 types of plants are S:F = 10:1 how would i represent this, i’ve attempted and it doesn’t seem right

[dense sleet]

I dont get whats the question?@dense mural

[wary stream]

Dude

Read #❓how-to-get-help

[dense sleet]

What are we trying to solve for / get

[wary stream]

Like seriously

[dense mural]

it’s linear programming,formulations , i’m trying to optimise a certain value and i have to represent those conditions in a linear form

[iron osprey]

the problem is that i don´t know how represent the conditions

my restrictions

[wary stream]

No one is talking to you. Please read #❓how-to-get-help for the 5th time, and understand how to properly get a help channel

[dense sleet]

Yes but what are you trying to do? What is that value

[iron osprey]

I don´t understand you rules

sorry

[dense sleet]

Everyone has their own help channel

Want help ask in a available one

[iron osprey]

I still thank you I will look for another channe

this my firt time in discord

[dense sleet]

Just read what #❓how-to-get-help says

[dense mural]

i’m trying to represent it in these linear forms but i don’t know whether they are correct

[high rapids]

,cw

,rotate

[lone heart]

@dense mural Has your question been resolved?

[lone heart]

@dense mural Has your question been resolved?

Channel closed

Closed by @dense mural

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[south sun]

need a lil help on this

[south sun]

done part a

on part b i get sin2theta=0

Like this

But then i cant seem to get the right solutions

[tacit arch]

,rotate

[tacit arch]

that is the right answer

,calc sqrt(2)/2

[ocean seal]

Result:

0.70710678118655

[tacit arch]

,calc 1/sqrt(2)

[ocean seal]

Result:

0.70710678118655

[tacit arch]

oh you said part b

do you know all solutions for theta? here

[south sun]

this is part 1

right

and then the second bit

i got 0, pi then 2pi

as solutions for theta

but since its 2 theta

i expanded the range to 4pi

[tacit arch]

did you plug those 3 into the original equation?

[south sun]

plug what 3

[tacit arch]

.

[south sun]

no

[south sun]

ohhh

[tacit arch]

verify your answer by checking

[south sun]

lemme do that rq

0 works

for pi i get -1/root2

so pi doesnt work

[tacit arch]

and that doesn't work

[south sun]

hmm

[tacit arch]

that's because you squared both sides

(-1) = 1 is wrong

[south sun]

2pi works

waitttt

[tacit arch]

but when you square both sides it works

[south sun]

huh

[south sun]

?

[tacit arch]

square both sides here

is it true?

[south sun]

yes

[tacit arch]

you did the same thing

[south sun]

if u square both sides that is true

right

[tacit arch]

no

ok, more clearly

(-1) is not equal to 1

[south sun]

true

[tacit arch]

$-1 \neq 1$

[ocean seal]

riemann

[tacit arch]

But if you square both sides like you did

[south sun]

right

u get 1=1

[tacit arch]

$(-1)^2 = 1^2$

[ocean seal]

riemann

[south sun]

ooooooh

is this a problem?

[tacit arch]

you did that here

[south sun]

yeah

i see

so this is not right

how can i avoid doing this though?

[tacit arch]

alternatively, you could also write $\sin(\theta) + \cos(\theta) = B\sin(\alpha)$

[ocean seal]

riemann

[tacit arch]

and solve for $B$ and angle $\alpha$

[ocean seal]

riemann

[south sun]

u would get sin theta plus cos theta =1

How do i solve this

💀

[tacit arch]

$\cos(\theta) = \sqrt{1-\sin^2(\theta)}$

[ocean seal]

riemann

[tacit arch]

then let $x = \sin(\theta)$ and quadratic formula this

[ocean seal]

riemann

[south sun]

God i forgot about that identity

Gimme a sec

Is this correvt?

[south sun]

Or is this squared just 1-sin squared theta

No it can't be

I would just get 1=1

Nooo

Root 3 times root 3 is just 3 not o

9

What im doing here is wrong

@tacit arch

How would u turn this into a quadratic

I subbed in costheta

But it turns into 1=1

btw im confused cuz

the ms didnt do any working

it just shits out values for theta with meaningless limits leaving me with no idea as to how they did anything

[north adder]

which step

[south sun]

there is no step

idek what they did

they just took what we figured out in part a

then replaced lhs

with the question form

sin(theta + pi/4)

made it equal to root2/2

[north adder]

ok so whats the problem

you proved the identiyt

so you could use it for part b

[south sun]

OHHHHH

i see

but

how did they get the theta values

so i see they did arcsin of

root2/2

which gives pi over 4

[north adder]

mhm

[south sun]

but can u explain what they did with the limits

OOOOH

i got it

i got it

[north adder]

great :)

[south sun]

pi over 4

so u get theta plus pi over 4

the inside of the bracket

=pi over 4

add pi to the right side until u hit within range of 2 pi

then subtract pi/4 to get just theta right?

is that what theyre doing?

[north adder]

yes

[south sun]

neat gnna try it now

[south sun]

sorry i have a question

[tacit arch]

Oh yea I said that here

[south sun]

when i try it

i don't get 3pi/4

i only get pi/4

and 9pi/4

because sin's periodicity is pi right?

[north adder]

wdym

[south sun]

never mind my rambling

[north adder]

yes sin has periodicity

[south sun]

im an idiot

i drew cast diagram

[north adder]

ok

[south sun]

just now

and got the 3pi/4

[north adder]

great

[south sun]

Like this

[north adder]

yes the unit circle is a handy tool

[south sun]

thats what its called?

we're maths babies, we just call it the cast diagra,

m

cuz a s t c

[north adder]

ahhh

i figured

[south sun]

yeahh

Is this sufficient for part c?

[south sun]

Part c btw

[north adder]

your solutions are incomplete

[south sun]

whats missing

[north adder]

take for example pi

nvm

im balding

[south sun]

hold on hold on

[north adder]

youre right

[south sun]

same interval

[north adder]

gg

[south sun]

kekw

lemme check ms

[north adder]

no

youre right

dude

[south sun]

no i get that

but

i wanted to see how ms says it

and wtf is wrong with them