help-13

[carmine pagoda]

yes

that is how

coefficients work

if its below 1 it will bloom more

if its above 1 it will narrow

[cedar kiln]

@tender valley Has your question been resolved?

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[cold bridge]

how would i solve this, answer is already given but i am still confused

[dim finch]

Rationalise denominator

[muted bear]

more like common denominator

pretty much the same thing tho

in this case

[dim finch]

Ya it's a

Particularly symmetric case

[clear berry]

You can directly evaluate it like subtraction of a fraction

[cold bridge]

how do i do that in this case

[clear berry]

forget the i for a moment how would you have done it if it were

$\frac{1}{a-b} - \frac{1}{c-d}$

[wraith dagger]

numbpy

[cedar kiln]

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[vital burrow]

hi

[vital burrow]

hello, I would like you to help me with a problem, basically I can't find the difference to be able to solve the problem

[crimson sedge]

what is the proble

[vital burrow]

I am confused and cannot find the difference to solve the problem

[crimson sedge]

i dont speak

spanish

[vital burrow]

The problem says: Calculate the sum of the first 100 terms of the following sequence:

1;3;5;-7;9;13;-15;17;19;21;-23

I have alternatives but I don't get a single one

[cedar kiln]

@vital burrow Has your question been resolved?

[fallen moat]

did you miss a 11?

[vital burrow]

nop

but the alternative is 4950, 3750, 2950, 3850, 4850

[fallen moat]

then i dont know 😦

[cedar kiln]

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[jagged solstice]

i just don't understand what ab, st, xyz is

[vital burrow]

and as it would be as 11

.reopen

[cedar kiln]

@jagged solstice Has your question been resolved?

[small dust]

context?

also... why..?

[cedar kiln]

@jagged solstice Has your question been resolved?

[dire geode]

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[dire geode]

looks like they left

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[radiant cove]

a bit puzzled here, i was able to solve part a (not much of a difference when sample size is increased but im puzzled by b, c & d. perhaps someone could shed some light on the problem for me? i've tried googling a bunch of references for prediction intervals but not really provide answers/clues to any of the questions below

[radiant cove]

here is the equation im using

i would want to say that for b there is no change in the width
for c the width increases slightly (?)
for d i'm pretty lost altogether

[small dust]

standard deviation is sqrt(variance)
and standard error is std dev / sqrt(n)

so std error is sqrt(variance/n)

so yes, there is a change

[radiant cove]

so if the variability in predictor values increase the prediction interval will widen?

i was assuming x bar (mean of x values) was the predictor variable so if we were to increase the variability in the values of all x's it wouldn't have much change because that wouldn't drastically increase x bar

[cedar kiln]

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@radiant cove Has your question been resolved?

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[dense gate]

How do I differentiate A’

[cedar kiln]

@dense gate Has your question been resolved?

[pale fern]

@dense gate if you multiply the (-1) into the 3, you simplify the first term to be -3(64-y)^2. From there, the process is no different than the process you used for differentiating A

[dense gate]

Yeah what will the answer be

Im confused

[pale fern]

Let me see if the latex bot works here, I'm new to the server. Otherwise give me a moment and I will send a picture

[pale fern]

$x = 4$

[wraith dagger]

Kappa Mikey

[dense gate]

Ahh?

[pale fern]

That's correct

[dense gate]

You sure?

Cool

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[dire tundra]

.close

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[crimson sedge]

Where does 1/3 come from in "For f(x) = 4/1-3x, that means 1/3 is excluded from the domain, since 1-3(1/3) = 0"

[valid tendon]

Because when x=1/3 the function is undefined

[crimson sedge]

why is it specifically 1/3 and not 1/4 or 1/2

[buoyant latch]

Because dividing by 0 is undefined

[crimson sedge]

wut

yea isn't that just for the domain of gof

[buoyant latch]

You can’t divide by 0

Ok um

[crimson sedge]

im just confused where they pulled the number 1/3 from

[buoyant latch]

You can’t pass an undefined into a function

[crimson sedge]

yssir

[valid tendon]

Because 1-3x=0

[buoyant latch]

If a function is undefined for the parent function

It’s also undefined for the inner function

[crimson sedge]

OHHHH

[crimson sedge]

w8 this it

[buoyant latch]

So if it’s undefined for g(a) then doing f(g(a)) is like

[crimson sedge]

1-3x=0 = -1/-3 = 1/3 ?

[buoyant latch]

What is f(undefined)

Which is also undefined

[valid tendon]

frost

too complex

[crimson sedge]

still appreciate frfr

got wut i wanted

tyty

[buoyant latch]

:/

[crimson sedge]

.close

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[valid tendon]

sorry frostie

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[indigo crag]

Lil clue

[cedar kiln]

@heady osprey Has your question been resolved?

[heady osprey]

i know but i dunno how to continue from there

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[velvet minnow]

A little question what's the zero of x - 5 ( basically p(x) = x - 5)

[quartz frost]

you mean, when is p(x)=0?

p(x)=0
p(x)=x-5
therefore
x-5=0
add 5 to both sides
x-5+5=5
x=5

[velvet minnow]

Thank you

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[earnest saffron]

how would i know if ill do subtitution or table of values?

[earnest saffron]

on ex 7 if get the its left side

[jaunty mural]

if substitution fails, then table

[earnest saffron]

f(3.001) = 29.02

it's not going to the 4 or approaching 4

[obsidian coral]

Why are you plugging in 3.001?

[earnest saffron]

because on table of values u need to get the right and left side before the actual c or x

[obsidian coral]

The limit is when x approaches 5, not 3

[earnest saffron]

oh riight, then i should plug in 4 lel akhsjdha

[livid hound]

no

[earnest saffron]

ahdasdhajh

i just forgot the log XDDD

nvermind iz f(4.001) = 3.59

[livid hound]

there's no real point in plugging stuff that far way from the location of interest

[earnest saffron]

but it is said to pplug in number that's closer to the c or x ?

[livid hound]

very close to the c value

[earnest saffron]

yeesh

[livid hound]

despite what's shown in the table, you don't really care about:

[earnest saffron]

i see i see

so is it only right to plug in 4.001?

[livid hound]

no

4.001 isn't close to 5

[earnest saffron]

4.999 ? > . >

[livid hound]

yeh

[earnest saffron]

shagdhasg got it

thank uuuu!

btwww howd u know if subtitution failed?

[livid hound]

you'd use a few values that get progressively closer to 5 to deduce the limit as x→5

substitution won't work if the function isn't continuous/defined there

e.g. for something like sin(x)/x

[earnest saffron]

the function is undefined?

anyway augshd i think i understand it now

thnk uuu

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[remote zodiac]

.close

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[languid vector]

How do I integrate the left side

Stuck

I just realized I wrote dx/dy lol

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[sullen spire]

how do u solve this?

like

set it up

[heady osprey]

theres 1 main ste

step*

try to realize that first

[sullen spire]

move all to one side

and divide to get dy to dx?

dy/dx

?

@heady osprey

[cedar kiln]

@sullen spire Has your question been resolved?

[sullen spire]

<@&286206848099549185>

[long swan]

so we have an equation of the form:

$P(x,y) dx + Q(x,y) dy = 0$

[wraith dagger]

Mr. Gamer

[long swan]

let's write this as:

$P dx + Q dy = 0$

[wraith dagger]

Mr. Gamer

[long swan]

now

i want you to find:
$$\frac{\partial P}{\partial y}$$
and:
$$\frac{\partial Q}{\partial x}$$

[wraith dagger]

Mr. Gamer

[long swan]

@sullen spire

because if this is true, then there exists a potential function $\Phi(x,y)$ such that $\frac{\partial \Phi}{\partial x} = P$ and $\frac{\partial \Phi}{\partial y} = Q$

[wraith dagger]

Mr. Gamer

[long swan]

and if there does exist such a potential function, we know that this must hold:
$$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$$
and we know that because this must hold:
$$\frac{\partial}{\partial y} \left(\frac{\partial \Phi}{\partial x}\right) = \frac{\partial}{\partial x} \left(\frac{\partial \Phi}{\partial y}\right)$$

[wraith dagger]

Mr. Gamer

[sullen spire]

oh right

[long swan]

so find dP/dy, and dQ/dx, and check to see if they are equal.

now.

[sullen spire]

yes they r

-sin(x+y)

now i can find the f function

easy peasy

[long swan]

yes

how do you propose we do that

[sullen spire]

sin(x+y)+y^3+2y^2

u integrate w respect to x

for P

then derive with y

and compare

to find the g(y) function

then put it into the first integral

to get this

[sullen spire]

^

[long swan]

yep, all good

you have found the potential function.

how does this let you solve the differential equation though?

[sullen spire]

uhh

is that not the solution to the ode/

C=sin(x+y)+y^3+2y^2

then u could sub in any initial conditions

[long swan]

yes. but why do we let $\Phi(x,y) = C$?

[wraith dagger]

Mr. Gamer

[sullen spire]

uhm

ngl

thats just what i remember learning

[long swan]

if we let phi(x,y) take on a constant value C, that gives us the level surface of the function phi(x,y) at C. this accounts for the fact that the derivative of a constant is 0 and that there are infinitely many solutions to any given DE

nice work!

[sullen spire]

ah ok that makes sense

also

@long swan do u know how to parameterise intersections

[long swan]

show problem

[sullen spire]

ive got the curl done

i just need help param the curve

i got

rcosx + rsinx + sqrt(5-r^2)

with i,j,k

respectively

[long swan]

,w curl (x^3exp(x)z, y^3ze^z, xy)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Mr. Gamer#6658

[sullen spire]

wellp i did that by hand

but yes

what about the curve?

[long swan]

it's actually easier to avoid using stoke's theorem here, given that we can easily parameterize the curve

we have two equations:
x^2+y^2 = 1
x^2+y^2+z^2 = 5, from this one we get that 1 + z^2 = 5, so z^2 = 4, and because z is positive this gives us z = 2

so this gives us the unit circle in x and y on the plane z = 2

parameterize

[sullen spire]

i think the lecturer wanted us to do it using stokes

unfortunately

so

the param

is

rcosx

rsinx

z=2

?

[long swan]

parameterize the CURVE

1 dimensional.

[sullen spire]

OHHH

rip

i see

wait

but when doing the Sr x Stheta

whats the z component?

[long swan]

would you agree that if we have a unit circle in x and y on the plane z = 2, that would give us the 1 dimensional curve:
r(t)= (cos(t), sin(t), 2)

just one variable, it's a curve.

[sullen spire]

yes

i agree

so for my dot product of n

id do the cross of Sr x St

and the k component

is 0 both tims

times

right?

[long swan]

find F(r(t))

then take the dot product with r'(t), which will just give you a line integral from t = 0 to t = 2pi

[sullen spire]

thats avoiding stokes tho right?

cuz i remember thats how i used to do line integrals

[long swan]

yeah it is, if you insist on using stokes for this, you will have to find a surface

[sullen spire]

before learning stokes

[long swan]

whose boundary curve is r(t)

[sullen spire]

in that case

would my surface be

S(r,x) = rcosx + rsinx + sqrt(5-r^2)

[sullen spire]

anyways ty for the help ahha

at least i can do it as a line integral

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[high kernel]

@haughty flare is it like this

[haughty flare]

The top image is correct, it wouldn't hurt to have an exaggerated slope for infinity approaching infinity, but you got the point across in my opinion

[high kernel]

Tysm

[haughty flare]

That's better

Just more aggressive of a slope ig

[high kernel]

I really do appreciate you

[haughty flare]

You're welcome

[high kernel]

.close

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[jaunty pumice]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[jaunty pumice]

4

I know this is a maths server but please if anyone know help me

[jaunty pumice]

Is the ans of 5 b -1000cm³

[crimson sedge]

,rotate

[crimson sedge]

Use $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

[wraith dagger]

Ryuzaki

[jaunty pumice]

Yeah i know that

I just want my answer checked

[crimson sedge]

$T_1 = T_2$, and $P_2 = 2.5P_1$

[wraith dagger]

Ryuzaki

[jaunty pumice]

?

[crimson sedge]

i can't be bothered to do the algebra for you

just use a calculator

[jaunty pumice]

2500/2.5=1000?

[jaunty pumice]

Ok

I have one more ques

[twilit aurora]

1+1=10

[jaunty pumice]

18 - a

,rotate

[jaunty pumice]

@crimson sedge is this correct?

[cedar kiln]

@jaunty pumice Has your question been resolved?

[trail bridge]

How many moles of hydrogen can this hold? (Convert mass to moles)

[cedar kiln]

@jaunty pumice Has your question been resolved?

[jaunty pumice]

1000 moles

[trail bridge]

Remember PV=nRT, in the same environment, the amount of mole of hydrogen gas that the container can hold is same as the amount of carbon dioxide

[jaunty pumice]

Avogadros law now

So 1000 moles of CO2?

[trail bridge]

So basically 1000 moles of cardon dioxide molecules, yes

[jaunty pumice]

Mass of CO2= 1000×44g=44kg?

[trail bridge]

Yes (if molecular mass of carbon dioxide is 44g, I don't have the table next to me)

[jaunty pumice]

CO2-> 12+2(16)

[jaunty pumice]

But in ans it is given 22 kg

[jaunty pumice]

And if we do like this ans is different

[trail bridge]

Wait...

[jaunty pumice]

V.D=mass of gas 'X' / mass of H2 gas under similar conditions

We get diff. Answer

Oh wait

I understood now

Moles of H2 gas = 1000/2 = 500 moles

[trail bridge]

Oh, right

I just realised that as well after realising the answer is halve of what I have up there lol

[jaunty pumice]

Thanks

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[random flare]

Can someone please help me with this

The question that I circled

[timber violet]

Ya so basically what u want is the hypotenus of the right triangle formed by one edge of the cuboid and one of the diagnols

[fallen moat]

,rotate

[timber violet]

The diagonal is root(l^2 + b^2)

The edge is h = 30cm

Now u can find hypotenus

[random flare]

How does h = 30 and what is h

oh wait

[timber violet]

I meant 25

My bad

[random flare]

I got it

[timber violet]

h is the height

[random flare]

in class today our teacher explained it way differently

You just add the sides together and then change them to root form

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[tired inlet]

how would I do this question

[tired inlet]

$$\frac{1}{a}\left(a-\frac{1}{a}\right)$$

[wraith dagger]

Lex1729

[tired inlet]

I know I can write this as:

$$a^{-1}\left(a-a^{-1}\right)$$

[wraith dagger]

Lex1729

[short blade]

distribute

[tired inlet]

so a^-1 * a^1?

@short blade

[short blade]

youre missing half the expression

[frosty ocean]

Why don't you just directly distribute instead of writing in terms of exponents?

[tired inlet]

I really don't know how

could you walk me through it?

@frosty ocean

[mellow loom]

a(b+c)=ab+ac ?

[tired inlet]

so a * a = a^2

what about the a^-1?

[tropic oxide]

where do you see any a * a?

we're telling you specifically to distribute and do nothing else

i.e. $a^{-1}(a - a^{-1}) = a^{-1} \cdot a - a^{-1} \cdot a^{-1}$

[wraith dagger]

Ann

[tropic oxide]

and only AFTER you've done that should you worry about what happens to either term

[tired inlet]

but do I add or multiply the exponents?

[short blade]

review your exponent laws

then try to answer the question on your own

[tropic oxide]

$\frac{1}{a} \cdot a = ; ? \ \frac{1}{a} \cdot \frac{1}{a} = ; ?$

[wraith dagger]

Ann

[tired inlet]

could you explain the steps?

1/a * a = 1

[tired inlet]

just not sure what to do with the exponents

@tropic oxide

[rough rapids]

a^-1(a-a^-1)
1/a(a-1/a)
1/a((a-1)/a)
(a-1)/a²

[tropic oxide]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[tired inlet]

Honestly, it's more helpful than asking you. No offence.

[tropic oxide]

that answer isnt even correct by the way

a - 1/a is (a^2 - 1)/a, not (a - 1)/a

anyway

i strongly suspect you're overthinking things

[tired inlet]

Well I may be, although I just don't know how to approach the question.

[tropic oxide]

well can you tell me this

what does 1/a * 1/a simplify to

[tired inlet]

it simplifies to a^-2

right?

[tropic oxide]

yeah, sure.

or 1/a^2

so you see that you are able to correctly simplify 1/a * a to 1 and 1/a * 1/a to 1/a^2

so please explain what else is troubling you

[tired inlet]

so a^-1 * -a^-1 = -a^-2?

[tropic oxide]

yes, of course it is

[tired inlet]

so then how would I do a^-1 * 1?

[tropic oxide]

where do you see a^-1 * 1?

did you mean a^-1 * a?

or do you want me to answer your question exactly as asked, with zero regard to its relationship with the problem at hand?

[tired inlet]

damn I wrote the question wrong

my bad

$$\frac{1}{a}\left(1-\frac{1}{a}\right)$$

[wraith dagger]

Lex1729

[tired inlet]

this is it

[tropic oxide]

ok in that case there are two possible interpretations of your question

a) "How do I multiply a number by 1?"
b) "How can the product a^-1 * 1 be viewed formally through the lens of exponent laws?"

which of these sounds more like the question you want to ask?

[cedar kiln]

@tired inlet Has your question been resolved?

[tropic oxide]

well it sounds like c) "Fuck you for even suggesting that my question could reduce to either of such idiocies. Fuck you and never talk to me again"

[cedar kiln]

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[fallen ravine]

solve the Diophantine equation and explain like I'm 12

[cedar kiln]

Please don't occupy multiple help channels.

[fallen ravine]

.close

[cedar kiln]

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[graceful pasture]

Signal processing:
Filter: y(𝑛) = 𝑥(𝑛) + 𝛼 ⋅ 𝑥(𝑛 − 𝑑)
Autocorrelation: 𝜙_xx(𝑘)= sum(𝑥(𝑛)𝑥(𝑛 − 𝑘)) - n=...-2,-1,0,1,2...
𝜙_yy(𝑘)= (1+𝛼^2)𝜙_xx(𝑘) + 𝛼(𝜙_xx(𝑘+d) + 𝜙_xx(𝑘-d))
what is the right way to design algorithm that finds both alpha and d. I'm quite stuck, would love to get any directions 🙂
Thanks!

[royal loom]

I can't help with this, but if you know a specific channel that it might be better in you might have better luck there

maybe this could be dynamical-systems or modeling? I am sure you know better than me

check them out it might help

[cedar kiln]

@graceful pasture Has your question been resolved?

[cedar kiln]

@graceful pasture Has your question been resolved?

[cedar kiln]

@graceful pasture Has your question been resolved?

[cedar kiln]

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[lean jolt]

How can i solve this limit and it should be equal to 0

[crimson sedge]

,rccw

[crimson sedge]

uh

[
\lim_{x \to -\infty} \f{x^2\times e^{\f 1x}}{x-1} -x-2
]
correct?

[crimson sedge]

@lean jolt

[lean jolt]

Yes

[crimson sedge]

[lean jolt]

Istg if my teacher gave it wrong...

[crimson sedge]

nah

it is right

[lean jolt]

Okay

[crimson sedge]

i think l'hopital is the easy way out

but i am going to guess you can't do that

[lean jolt]

Yep i cant

[unborn pike]

How about taking limit x->0 and replacing x with -1/x?

[crimson sedge]

yeah i was thinking of a substitution too

seems like the most ideal approach

[lean jolt]

Imma try it

Still didnt figure it out

<@&286206848099549185>

[leaden fractal]

hello @lean jolt

[lean jolt]

Hii

[leaden fractal]

is your problem solved?

[lean jolt]

No it is not

[leaden fractal]

okay

what procedures you have tried?

[lean jolt]

Factorization with x then i tried the same with x^2 then i itried to factorize with exp(1/x)

None worked

@leaden fractal

[leaden fractal]

try substitution method once

lt h->0 where h= 1/x

replace x terms by 1/h

and in limit h->0

[lean jolt]

Alright imma try it again

[leaden fractal]

yeah

is the ansr 1 ?

[lean jolt]

Its supposed to be 0

[leaden fractal]

okay

yeah its zero only.. i missed a factor..its zero

[lean jolt]

I couldnt figure it out

[leaden fractal]

you tried Lhopitals?

[lean jolt]

We didnt learn that its not in our programme

[leaden fractal]

oh okay

no worries

after substitution what is the reduced form you get?

[lean jolt]

I got this

Tried to simplify it but led to nothing useful

t=1/x here

,rotate

[leaden fractal]

the reduced form is wrong try to do it once

[lean jolt]

Can you please point out where i am wrong

,rotate

[lean jolt]

Forgot -1 in the end

[leaden fractal]

yeah

[cedar kiln]

@lean jolt Has your question been resolved?

[cedar kiln]

@lean jolt Has your question been resolved?

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[lean jolt]

.reopen

[cedar kiln]

✅

[cedar kiln]

@lean jolt Has your question been resolved?

[cedar kiln]

@lean jolt Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[topaz ginkgo]

can i just assume that the function is a linear application given the fact that it only contains sums and multiplications with scalars?

[topaz ginkgo]

"data la funzione" means "given the function"
"così definita" means "defined as"

[cedar kiln]

@topaz ginkgo Has your question been resolved?

[cedar kiln]

@topaz ginkgo Has your question been resolved?

[cedar kiln]

@topaz ginkgo Has your question been resolved?

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• Be polite and have a nice day!

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[crimson sedge]

where k is a constant.
The graph of g has either zero, one or two horizontal tangents depending on the constant k. b) Determine k so that the graph of g has exactly one horizontal tangent.

[scenic crest]

Find derivative of g(x)

And than equate it to 0

g'(x) = 0

[latent bloom]

Essentially make it a binomial

After you find the derivative ofc

[crimson sedge]

like this

[latent bloom]

Nope

What's the derivative of kx?

[vagrant elbow]

$\dv[2]{t}x + kx = 0$

[wraith dagger]

NEONPerseus

[vagrant elbow]

don't you love physics

[crimson sedge]

1?

[latent bloom]

Well

Let's see another example

What's the derivative of 6x?

[crimson sedge]

Yes

[crimson sedge]

6

[latent bloom]

Okay

That means that whatever's being multiplied by x, stays while x reduces its power

Can we apply the same concept on kx?

k is a constant btw

[crimson sedge]

k

[latent bloom]

Exactly

So what's your derivative now as a whole?

[crimson sedge]

but the calculater said 1 😦

[crimson sedge]

umm

[latent bloom]

Hmm

Forget the calculator for now

Sometimes it's hard-coded

[crimson sedge]

let me think

[crimson sedge]

$x^(2)+4x+k$

[wraith dagger]

EmmyChan🌸

[crimson sedge]

latex

$x^2+4x+k$

[wraith dagger]

EmmyChan🌸

[latent bloom]

Yep

[crimson sedge]

looks good 🙂

btw TYSM for your help

[latent bloom]

Happy to help

[crimson sedge]

x^2+4x+k=0

[crimson sedge]

i like that emoji

[latent bloom]

Aww thanks. I edited into what it is now :)

Btw you try to make the roots of this equation 1 root

And in quadratics, you do that when your have 1 bracket

Which is essentially a binomial

[crimson sedge]

talk english 🙂

[latent bloom]

Lol alright

x^2+4x+k=0

This equation

[crimson sedge]

yes

[latent bloom]

When you try to find the solutions to it, you will get 2 solutions

That's if you use quadratic formula

[crimson sedge]

ahh yes

[latent bloom]

However

You don't want it to give you 2 solutions

You want it to give you 1 solution only

[crimson sedge]

√(3)+2

and

√(3)-2

[latent bloom]

Wait

[crimson sedge]

√(3)±2

[latent bloom]

Hmm

No

The question doesn't want those

Let's revisit quadratic equations

$(x-a)^2 = 0$

[wraith dagger]

VulcanOne

[latent bloom]

How many solutions does this have?

[crimson sedge]

2?

[latent bloom]

What are they?

[crimson sedge]

idk

[latent bloom]

Hmm

[crimson sedge]

tell me

[latent bloom]

What about

$x^2 = 0$

[wraith dagger]

VulcanOne

[latent bloom]

How many solutions does it have?

[crimson sedge]

2

no

1

[latent bloom]

Ooo

How did you know?

[crimson sedge]

i am euler

[native jungle]

Hj

[latent bloom]

[crimson sedge]

x=0

[latent bloom]

Yep

Now let's go back to this

https://cdn.discordapp.com/attachments/903430334681608232/1073305749230460928/1000057838007226480.png

[crimson sedge]

2

[latent bloom]

2 solutions?

[crimson sedge]

no 1

[latent bloom]

Yepp

What's the solution?

[crimson sedge]

hmm

x=a

[cyan wren]

🥳

[crimson sedge]

https://tenor.com/view/smart-lady-sarah-sherman-saturday-night-live-clever-girl-you-go-girl-gif-23380054

[cyan wren]

So what they want is for this equation to also have 1 solution

[latent bloom]

Exactly

[cyan wren]

Which means it must be written as (x-a)^2 as well

[crimson sedge]

not agin

[cyan wren]

You need to figure out what k achieves that

again?

[crimson sedge]

agin

[latent bloom]

Actually I wanted to expand (x-a)^2 first

[crimson sedge]

ez pz

a^2 - 2 a x + x^2

[cyan wren]

Isn't that the next step? But yeah feel free. Just popped in cause you were busy in Sonny's channel

[latent bloom]

Exacccctttly

[crimson sedge]

https://tenor.com/view/wolfram-alpha-my-beloved-wolfram-alpha-my-beloved-gif-21254252

[cyan wren]

Hey!

No WA

Bad

[crimson sedge]

https://tenor.com/view/sorry-gif-18453366

[cyan wren]

But what's done is done.
So now we compare x^2 + 4x + k and x^2 - 2ax + a^2.
Note how we know the coefficient for x in both cases. Can you figure out what a, and by extension, k is?

[crimson sedge]

idk

[cyan wren]

Well...try again ig

It's been a minute, surely you can put a bit more effort into it

[crimson sedge]

okay

let me think

https://tenor.com/view/think-pooh-winnie-the-pooh-thinking-brain-gif-17755939

[latent bloom]

Hint: make +4x = -2ax

[cyan wren]

Nooooo Vulcan

I guarantee 50% of the thinking time was spent on finding that gif

[latent bloom]

[crimson sedge]

-2

[cyan wren]

Correct! Better not have been WA

[latent bloom]

What's -2?

[crimson sedge]

a?

[latent bloom]

Okay

[crimson sedge]

lol

x^2 + 4x + k = x^2 - 2ax + a^2.

• 4x + k = - 2ax + a^2.

4x + k = - 2ax + a^2.

4x + k = - 2*-2x + -2^2.

k = -2^2.

k=4

[cyan wren]

📈

[crimson sedge]

time for a gif

[cyan wren]

Yes

[crimson sedge]

https://tenor.com/view/book-bookworm-bowtie-glasses-dino-gif-26046453

[latent bloom]

[crimson sedge]

can i ask somethig

[cyan wren]

Sure

[crimson sedge]

i shouldnt have use WA

because i font understand how i got to the Expanded form

[cyan wren]

Yes

(x-a)^2 = (x-a)(x-a)

[crimson sedge]

yess

and then if you (x-a)(x-a) =x^2 - 2ax + a^2.

or something?

[cyan wren]

Yeah, it's called distribution

In general:
a(b+c) = ab + ac

So (x-a)(x-a) = x(x-a) - a(x-a)

surprised your school hasn't tauhgt you anything like that

[crimson sedge]

thank you master Ternimus and vulcan

https://tenor.com/view/thank-you-stitch-gif-25595739

[cyan wren]

@latent bloom You got fucked bro

Sorry

[crimson sedge]

no

vulcan also helped a lot

[crimson sedge]

i ment this for both

[cyan wren]

But I only see my name 🙂

Get wrecked Vulcan

[latent bloom]

Haha it's fine lol

[crimson sedge]

THANK YOUUU VULCAN

[cyan wren]

kekw

[crimson sedge]

look agin

[cyan wren]

Now .close before we get bonked by mods

[cyan wren]

still my name. Now it's fixed

[crimson sedge]

agin

look

👀

.close

[cedar kiln]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[crimson sedge]

i am sorry but i have another question

[crimson sedge]

or the same

[latent bloom]

Sure ask away

[crimson sedge]

how do i get to k=4 without a

[vagrant elbow]

Imagine scrolling all the way up to this

[crimson sedge]

$g'(x):=x^2+4x+k$

[wraith dagger]

EmmyChan🌸

[crimson sedge]

better?

[cyan wren]

Search -> mentions: user -> mentions:NEONPerseus. YW

[vagrant elbow]

Oh nah I was talking about Vulcans reply not your question dw

Whats the problem btw I can help if you give it to me in full

Hopefully

[vagrant elbow]

😔

They always have a solution don't they

[crimson sedge]

how to get to k=4

without a

[vagrant elbow]

I need to read the previous question I guess

[tired portal]

complete the square

[latent bloom]

Well

[crimson sedge]

2*√(4-k)=0

[latent bloom]

$x^2 -2ax + k = (x-a)^2$

[wraith dagger]

VulcanOne

[latent bloom]

So k = a^2

Or a = sqrt(k)

So

[crimson sedge]

but then i have a agin

[latent bloom]

Substitute sqrt(k) into a

It will be

$x^2 -2\sqrt k + k = (x-\sqrt k)^2$

[wraith dagger]

VulcanOne

[latent bloom]

I can't ignore replies lol

[crimson sedge]

so it is not possible to do it without a?

[latent bloom]

It is

[crimson sedge]

hmmm

[latent bloom]

$(x+\sqrt k)^2 = x^2 + 2\sqrt k x + k$

That's another valid formula to use

[crimson sedge]

g'(x)=x^2+4x+k

[wraith dagger]

VulcanOne

[latent bloom]

So you do

$4x = 2\sqrt k x$

[wraith dagger]

VulcanOne

[latent bloom]

Or in other words: $4 = 2 \sqrt k \implies 2 = \sqrt k \implies k = 4$

[wraith dagger]

VulcanOne

[crimson sedge]

i am still stuck at g'(x)=x^(2)+4x+k

[latent bloom]

Hmm

[crimson sedge]

https://tenor.com/view/i-might-be-an-idiot-liz-lemon-30rock-idiot-stupid-gif-19745990

[latent bloom]

No

Factually incorrect

Anyways

You want 1 solution

[crimson sedge]

yes

[cyan wren]

Vulcan is super nice

And super patient

[latent bloom]

And your g'(x) = x^2 + 4x + k

And we notice that this is a quadratic

[crimson sedge]

indeed

[crimson sedge]

yes

[latent bloom]

Quadratics have 3 situations

Either they have 2 solutions, 1 solution, or no solutions

[crimson sedge]

and quadratic with n can have n-> solotion

[latent bloom]

We want to have 1 solution

[crimson sedge]

yess

TYSM for being so patient

[latent bloom]

That's called a polynomial

[crimson sedge]

foget what i said

[vagrant elbow]

n possible solutions*

It's entirety possible for a degree three polynomial to have one root

(real)

[latent bloom]

[crimson sedge]

how do we get 1

[latent bloom]

If your quadratic is a binomial bracket.

Binomial bracket = $(a+b)^2$

[wraith dagger]

VulcanOne

[crimson sedge]

hmm

[latent bloom]

g'(x) = x^2 + 4x + k let's remember

[crimson sedge]

yes

[latent bloom]

And we wanna transform it to (x-something)^2

[crimson sedge]

x^2 + 4x + k
(4x + k)^2

[latent bloom]

We know $(x+a)^2 = x^2 +2ax + a^2$ \
Or $(x+\sqrt k)^2 = x^2 + 2\sqrt k x + k$

[wraith dagger]

VulcanOne

[crimson sedge]

i dont understand that part

[latent bloom]

Those are both valid formulas we can use

[crimson sedge]

hmm

i am so stupid

dont understand

[latent bloom]

[crimson sedge]

sorry

[latent bloom]

Don't say that about yourself though

[crimson sedge]

and tysm for helping me

[latent bloom]

It's perfectly fine to not understand

[crimson sedge]

you are a heor

[latent bloom]

I can recommend you go back to quadratics and complete the square

[crimson sedge]

soo kind

the square?

[latent bloom]

https://youtu.be/C206SNAXDGE

Here's a really good video

[crimson sedge]

ty

[latent bloom]

[hard belfry]

$2π r h + 2π r²$

[latent bloom]

Actually line 2 is sufficient

$x^2 + 4x = k

Wait no

[crimson sedge]

i did it wrong

[crimson sedge]

i dont understand

[latent bloom]

Hmm you could this

$x^2 + 4x = 0 \
x^2 + 4 x + 4 = 4 \
(x+2)^2 = 4$

[wraith dagger]

VulcanOne

[latent bloom]

This also tells you that the missing part should be = 4 to make the square bracket = 0

[crimson sedge]

k

[latent bloom]

You could have made it like this too

$x^2 + 4x + k = 0 \
x^2 + 4x = -k \
x^2 + 4x + 4 = -k + 4 \
(x+2)^2 = -k + 4$

[wraith dagger]

VulcanOne

[latent bloom]

And we know for sure that -k + 4 = 0

So k = 4

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[crimson sedge]

Quick question, but what does this notation mean?
[
\dv{\vb{v}}\bracks{\div (\vb{v} \otimes \vb{v})}
]

[crimson sedge]

what is \otimes supposed to mean?

[dull oxide]

divergence?

I'm not too familiar with tensors though

[tame wraith]

Oh that

[dull oxide]

Oh. otimes

[tame wraith]

I think it's the outer product

[dull oxide]

yes

[crimson sedge]

outer product huh

[tame wraith]

Yh i learnt of it recently aswell @crimson sedge

Naviar stokes right?

[crimson sedge]

yeah it is all tensors

[dull oxide]

$\vb{v} \otimes \vb{v} = \vb{v} \vb{v}^T$

[tame wraith]

@crimson sedge a (×) b = a b^T

[wraith dagger]

SWR

[tame wraith]

I think

[crimson sedge]

hm thank you

i will research this further

.close

[cedar kiln]

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Use .reopen if this was a mistake.

[dull oxide]

It's a neat way to represent rotations.

That's the only time I've ever used it myself

[tame wraith]

It looks cursed though

The symbol itself

The :: one

Is even wierder

I acc hate tensors so much

[crimson sedge]

[cedar kiln]

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[unreal tiger]

Is this correct

[rustic siren]

no

[obsidian coral]

First, you can expand that to check

Two, is that a poorly written 7?

[unreal tiger]

Huh

[unreal tiger]

Where

[crimson sedge]

how do we know this

[rustic siren]

what are your factors?

[unreal tiger]

-28x2 -4x -7x -3x

[rustic siren]

not that

did you write (x-4)(x-1)?

or (x-4)(x-7)

[unreal tiger]

X-1

How y’all get 7

[rustic siren]

expand this

[unreal tiger]

Wym?

I never heard that term in my life

Not gonna lie

Well I have

But we have t learned expanding yet

We just have to do that

If that brakes sense

That’s what we’re learning next

[rustic siren]

do the multiplications

[unreal tiger]

Writhed I’m not getting what you’re saying or you’re not getting what I’m saying idk 😭

Or I’m just dumb

[rustic siren]

just do them

what you get if you multiply (x-4) by (x-1)

[unreal tiger]

My final answer is x-4 x-1

Ion gotta do alat

[rustic siren]

its wrong tho

and u wont even check if its right, even when i advised you to

[unreal tiger]

Well the reason I created this was to ask if it was right bc I didn’t know

[rustic siren]

and we told you, that you can check if its right, by doing the multiplication of the brackets

(x-4)(x-1) equals x^2-5x+4

[cedar kiln]

@unreal tiger Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[swift anchor]

@sudden stratus

[sudden stratus]

Yeah well if you add all that it comes to 100% so that's a strange question

oh wait

[swift anchor]

random question

are u indonesian

[sudden stratus]

No I'm from NZ

lol

[swift anchor]

new zealand

[sudden stratus]

yup

[swift anchor]

isnt it like really early there

[sudden stratus]

its 11:02 AM

[swift anchor]

oh

ok back to question

no its like 30% of the 50%

[sudden stratus]

yeah that's hard

[sudden stratus]

that would be 15%

damn my brain isn't braining

[grim elm]

Rather than thinking how much you're saving. Think about how much you still have to pay. If I saved 70% then 40% That's as simple as saying I have to pay 60% out of 30% which is as simple as 0.6*0.3=0.18, so 18% of original

[swift anchor]

isnt it 5/10 +(3/10*5/10)+(3/10 times 5/10 times 2/10)

[swift anchor]

oh yeah

18% is not an answer

these are the answers

possible answers

[grim elm]

Yeah yeah I was giving example with two different numbers so you'll have to do yours on your own. Don't want to just give number

[swift anchor]

ohh

alr

so its basically

.5*.3 times .2

oh

no

its

.5*.7 times .8

equals .28

[grim elm]

And that's how much you have to PAY, so how much have you SAVED? Cause that's what the question is asking

[swift anchor]

ohh so its 72%

is there a way to do it by multiplying how much i save

[grim elm]

(.5+.5*.3)+((.5+.5*.3))*.2

[swift anchor]

ty

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[cedar kiln]

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[undone star]

is there any convenient way of proving this

[terse bolt]

Is it for calculus?

[undone star]

yeah

[terse bolt]

Bc then I'd say a volume integral

[undone star]

im guessing u just different in respect to r

[terse bolt]

Then the derivative

[undone star]

*differentiate

i havent done integrals

most ive done is primitives

only differentiation as of now

forgot to add, it has a fixed surface area

it has a fixed surface area with the top open

so its just SA= 2pi*rh +(pi)r^2

[terse bolt]

Argh, you're asking me to get far into my memory, weren't there a story of slicing? My professor would explain it as a dishwasher

[undone star]

idk

its only highschool math

and havent done integration yet

[terse bolt]

Yeah maybe I've done that in calc 2 you're right.

Sorry, from there I don't know sorry

[undone star]

yeah allg