help-23

[dense wadi]

i see your changed your pfp again haha

[wraith locust]

substitute and multiply

[dense wadi]

ah okay tyy

[sweet saddle]

Hi there can someone please help me understand how the domain works within the intervals

[solar hazel]

[exotic snow]

use a different channel

#help-26

[dense wadi]

i would but i genuinely forgot everything about cal 1

[sweet saddle]

Alr thank you

[dense wadi]

are you in graduate studies? i forgot if its you or p norm

[sweet saddle]

No I am in Grade 9 doing a hard level of Kumon

[solar hazel]

i was

[dense wadi]

not kumon 💀

[dense wadi]

oh you left?

O_O

[sweet saddle]

Yeah the math hell

[solar hazel]

yes

[dense wadi]

bro is bringing back my PTSD

[exotic snow]

what's kumon

[dense wadi]

:O

[sweet saddle]

My bad

[dense wadi]

its hell but without the fire

[dense wadi]

still hope its going well for you

[sweet saddle]

Yeah thank you

[safe radish]

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[dense wadi]

i dont understand why it has infinitely many solutions

hii chartbit

[junior smelt]

Hiii

And tl;dr "because there's that free variable", you can change that to be any real number you want, and that gets you a different solution

[stone coyote]

Can I ask a quick question off topic

[dense wadi]

ah okay so like

even though its like 5 or 9 it can be 47628593

[dense wadi]

yes?

[stone coyote]

Why are there so many math channels

I'm so confused

Where do I ask

[dense wadi]

because its a lot of people asking questions

[dense wadi]

here

under math help (Available)

see how im in the occupied section

[stone coyote]

Oo wait wait wait so, in this channel, you're the only helper?

[dense wadi]

well im the one being helped lol i opened this channel

basically you can open a channel and veryone can help you

[stone coyote]

Huh how

[junior smelt]

,w rref [[1, -3, 7, 2, 5], [0, 1, 2, -4, 1], [0, 0, 1, 6, 9]]

[flat frigate]

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[cedar void]

https://discord.com/channels/268882317391429632/488120190538743810

[dense wadi]

well go in math help (available) and send your question

[stone coyote]

Ohh

[dense wadi]

oh whats htat

[stone coyote]

So max 16 channels?

[junior smelt]

Basically it's doing the algebra they talk of (I'm sure you'll get to reducing matrices soon )

[dense wadi]

is this what you mean btw?

or am i still confused

what i mean is like 3x can be 2 or 29373

[junior smelt]

But, like, what it basically says is $x_1 - 88x_4 = -109, x_2 - 16x_4 = -17, x_3 + 6x_4 = 9$, so then you can say that $x_1 = 88x_4 - 109, x_2 = 16x_4 - 17, x_3 = -6x_4 + 9$

[flat frigate]

@junior smelt

[dense wadi]

and therefore a free variable can take on any value

[junior smelt]

So like you can choose x_4 to be whatever, that then gets you something for x_1, x_2 and x_3

[dense wadi]

ahh okay

sooo is what i said right? just to make sure

like above

[junior smelt]

Yea free variables can take on whatever value you want: there's no restriction, here, on x_4

[dense wadi]

ahh okay tysm!

.close

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[dense wadi]

whats the use

[dense wadi]

of knowing that the pivot position A had -10 and -5 before being a echelon form

because my textbook just wrote this and ended the chapter like a cliffhanger

[marsh walrus]

my guess would be they want you to notice the matched 2's 4's and 6's

or otherwise sort of re-point your brain at the original matrix

which you can kind of stare at, after the fact, and see that what ended up happened might have been anticipated from the beginning

just a guess though

[junior smelt]

What comes before the example here?

(also you're already doing matrix reducing )

[dense wadi]

nothing thats the thing...nothing at all

[dense wadi]

oh yeah probably

anyway ty!

.close

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[gusty canyon]

Question 4 <@&286206848099549185>

[plucky elk]

!15m

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[gusty canyon]

X1=-3 and I plugged it in and got Y1=-324.11

It doesn’t look correct tho

[gusty canyon]

I did in other channel and don’t get help so I closed it and opend this

[plucky elk]

yea don't do that either

[gusty canyon]

Got it

[junior smelt]

And how did you do this?

[gusty canyon]

I plugged in -3 into the Xs

Then I got 4(-81)+3 over -27

Which is -324+0.11

[junior smelt]

-81

,calc (-3)^4

[flat frigate]

Result:

81

[gusty canyon]

What the heck

So a - to the power

Is always a positive

,calc (-3)^3

[flat frigate]

Result:

-27

[gusty canyon]

I see

Odd numbers are negative

Positive are positive

2,4,6,8

1,3,5,7

So in that case

324+0.11

324.11?

@junior smelt

[junior smelt]

Well you should have $4(81) + \frac3{-27}$, remember, the odd powers still are negative, only the even ones turn positive

[flat frigate]

@junior smelt

[gusty canyon]

Yes i got there

And then I got 324for the left

And 3 over -27 is -0.11

[junior smelt]

Yep, negative

[junior smelt]

[gusty canyon]

So 324+(-0.11)

,calc 324+(-0.111111)

Is that the answer?

[flat frigate]

Result:

323.888889

[gusty canyon]

How much 1 should I put

[junior smelt]

Better

,w 4x^4 + 3/x^3 when x = -3

[flat frigate]

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[junior smelt]

,calc 2915/9

[flat frigate]

Result:

323.88888888889

[gusty canyon]

So 323.8

Would work

[junior smelt]

(it's best to keep things exact for as long as you can though!)

[gusty canyon]

So don’t?

So Y1=323.8

@junior smelt

@junior smelt

[junior smelt]

Best to use the fraction/exact one honestly

But you could round it if you wanted to, I guess

[gusty canyon]

I’m still going to minus it tho?

Is this a better way @junior smelt

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[safe radish]

@unkempt hawk Has your question been resolved?

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[unkempt hawk]

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[rocky isle]

Solving this and checking for extraneous solution

[rich stratus]

okay first of all -1 could be -(x+1)/(x+1)
So right side would be [2x-(x+1)]/x+1=(x-1)/x+1

Then you multiply both sides by x+1 and x+6
Left side = x^2+x
Right side = x^2+5x-6

5x-6=x
4x=6
x=1,5

[rocky isle]

I’m gonna be honest I’m so cooked in math

Wait it’s wrong one sec

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[faint seal]

What is your question about this?

Are you asking how to derive this?

[short finch]

solve for the ii

[obsidian oracle]

the ii?

[short finch]

yes

[obsidian oracle]

you mean solve for this

so you know how is "defined" i factorial?

[short finch]

yes

[safe radish]

@short finch Has your question been resolved?

[proud grail]

This is an interesting and non-trivial mathematical identity,but just that

u cant solve nothing

like Euler Identity : exp(iπ)+1=0

[safe radish]

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[manic radish]

we're not a "do my homework for me" discord

what have you tried?

[proud grail]

yeah with the gamma function

[soft matrix]

<@&268886789983436800> he's trollinh

[wet wigeon]

He is fr

[proud grail]

<@&268886789983436800> true

hahahaha

[fathom adder]

They can see deleted messages iirc

[wild cape]

.close

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[wild cape]

yes we can

[fathom adder]

Perfect

Ty

[safe radish]

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[twilit moat]

can somebody help me with this question?

[twilit moat]

working it out I got dx/dy = 2y-2, dy/dx = 1/(2y-2)

given it should be parallel I assumed that dy/dx = 1 so 1 = 1/(2y-2) giving y = 3/2

finding the coordinate I substituted for x = (3/2)^2-2(3/2)=9/4-3=-3/4

but the answer (-3/4,3/2) isn't an option

here are the options

wondering if I made some kind of mistake in the process

[timber karma]

im not great with calculus, but can you flip from dx/dy to dy/dx the same way with a fraction?

[twilit moat]

that's what I did at the start of my working out

[twilit moat]

.

[terse lichen]

you can rearrange y = x + 2

[timber karma]

i was asking since ive heard dy/dx isnt actually a fraction or something like that haha

[terse lichen]

honestly what they've done here is just changed the coordinate systems a bit

[twilit moat]

wdym sorry?

im a little confused

[twilit moat]

yeah it isn't a fraction but it can be used similar to one if I'm not mistaken

[terse lichen]

ok here, in the usual 2d coordinate system, usually x is the horizontal axis and y is the vertical correct?

so you write functions like y = something x

but here, they changed the horizontal axis to y and vertical to x

so they write functions like x = something y

tl;dr: change all x to y's and all y's to x and its the same thing

[twilit moat]

hmm I see. What I'm confused with is what is wrong with my solution though

because in some cases you can't rewrite the functions and in that case the dx/dy = 1/dy/dx property becomes useful

[terse lichen]

well again like i said about the coordinate system

the value of y you calculated is actually the first coordinate

and x is second

(y, x)

[twilit moat]

yes I get that

but I'm saying for this question I can't apply the property of rewriting

and need to use the intgration technique

dx/dy=1/dy/dx

[twilit moat]

I was wondering how to get to the answer considering my working was slightly different

[terse lichen]

integration?

[twilit moat]

yes

[terse lichen]

i dont see how integration applies here

[twilit moat]

sorry I meant differentiation

scatterbrained revising for both

[terse lichen]

your answer is correct except the coordinates are flipped

also idk why theres no -3/4

[twilit moat]

what I was wondering was if I made a mistake in the working

in terms of finding that it equals -3/4

and also I'm a bit confused why you would need to flip the coordinates in this question considering I thought it would still be x and y regardless of whether it is rotated

[terse lichen]

nope

[twilit moat]

alright

.close

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[wraith swift]

how to draw 3^x/2?

[wraith swift]

i get some of it

but would appreciate some details

[thin bridge]

if in doubt, make a table of values

[wraith swift]

yeah

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@wraith swift Has your question been resolved?

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[gloomy scaffold]

I got a understanding on how the 1st formula works but I can't find out a formula for three terms.

[exotic cypress]

so it would start off with 1/1x2x3 + 1/2x3x4 + ...

[gloomy scaffold]

no

[exotic cypress]

no?

what does it ask for then idk i must have interpreted wrong

[gloomy scaffold]

I think it would start off with the last term of the first fraction

1/1.2.3 then 1/3.4.5 and so on

[exotic cypress]

hm

im not sure i remember getting some qu with it just increasing by 1 reach time

[idle frigate]

it's not very complicated

[gloomy scaffold]

whenever I decompose the fraction I get a little complicated fractions

[idle frigate]

tell me what 1/(n(n+1)) - 1/((n+1)(n+2)) is, @gloomy scaffold

[gloomy scaffold]

for the the first formula the fraction can be broken into 1/n -1/(n+1)

[idle frigate]

yes

[gloomy scaffold]

I did same thing for three terms and got rather complicated fractions which do not cancel out each like the first one

[idle frigate]

show what you did

you decomposed 1/(X(X+1)(X+2)) in partial fractions I suppose?

[gloomy scaffold]

I rewrite the last fraction of three as 1/2(n-1) - 1/n + 1/2(n+1)

[idle frigate]

= 1/2· 1/X - 1/(X+1) + 1/2 · 1/(X+2)

yes

it does cancel out

it's just doubly telescopic

[gloomy scaffold]

I considered the form as 1/n(n-1)(n+1)

[idle frigate]

you get 1/2 ( [1/n - 1/(n+1)] - [1/(n+1) - 1/(n+2)] )

yeah same thing

so if you call f(n) = 1/n - 1/(n+1), then 1/(n(n+1)(n+2)) = 1/2 ( f(n) - f(n+1) )

so the sum is 1/2 f(1) - 1/2 f(n+1)

from 1 to n

[gloomy scaffold]

last term: 1/n(n-1)(n+1)
after decompose: 1/2(n-1) - 1/n + 1/2(n+1)
second last term: 1/(n-3)(n-2)(n-1)
after decompose: 1/2(n-3) - 1/(n-2) + 1/2(n-1)

[idle frigate]

sure

factor the 1/2 out

[gloomy scaffold]

I don't see any terms canceling out

[idle frigate]

you get 1/2 ( (1/(n-1) - 1/(n-2)) - (1/(n-2) - 1/(n-3)) )

[gloomy scaffold]

just one term gets doubled

[idle frigate]

well except you got the expansion wrong

just do it with 1/(n(n+1)(n+2)) it's easier

$\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2}\right)$

[flat frigate]

suut

[idle frigate]

that's the partial fraction decomposition

now, group one of the 1/(n+1) with the 1/n, and the other with the 1/(n+2)

[gloomy scaffold]

ok so I did something wrong in my decompostion?

[idle frigate]

$\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\left(\frac{1}{n} - \frac{1}{n+1}\right) - \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \right]$

[flat frigate]

suut

[idle frigate]

you get this after the grouping

it's clearly telescopic

[idle frigate]

yes but it's not a really important part, just do it again later until you get it right

just focus on the two equations I gave

the same pattern repeats, if you do it with 4 factors you get a difference of differences of differences

[gloomy scaffold]

I see

[idle frigate]

if you use 4 variables you would get:

$\frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{6}\left(\frac{1}{n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)$

[flat frigate]

suut

[idle frigate]

which again can be put in a telescopic form

as $\frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{6}\left[\left(\frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2}\right) - \left(\frac{1}{n+1} - \frac{2}{n+2} + \frac{1}{n+3}\right) \right]$

[flat frigate]

suut

[idle frigate]

and so on

[gloomy scaffold]

now I understand

[idle frigate]

we're just grouping terms semi-cleverly

you see the pattern from one stage is repeated in the next stage

and the coefficients are not random, it's simply binomial coefficients

for 3 variables: 1 2 1
for 4 variables: 1 3 3 1
for 5 variables: 1 4 6 4 1
etc.

you read it directly from Pascal's triangle

[exotic cypress]

does it go -,+,-,+ between the numbers?

[idle frigate]

and the fraction in front is simply:
for 3 variables: 1/2!
for 4 variables: 1/3!
etc

[idle frigate]

yes

[exotic cypress]

nice and shrimple

[idle frigate]

yeah

[gloomy scaffold]

do all these stuffs relate to combinatorics?

[idle frigate]

you can rewrite these equations with the ∇ operator, where ∇f(n) = f(n) - f(n+1)

[idle frigate]

not exactly, I guess you can lookup "umbral calculus" or "discrete calculus"

[gloomy scaffold]

I just had a problem on my mind where I need to find the genarel formula of sums of terms like
1/n +1/(n+1) +1/(n+2)+...

[idle frigate]

there's no general form of these

there's only an approximation using a logarithm

[gloomy scaffold]

up to 1/(n+x)

[idle frigate]

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

[gloomy scaffold]

I watched a video about it . It had something with the gamma function and the gamma constant

[idle frigate]

the Euler-Mascheroni constant γ yes

approximately 1/1 + 1/2 + … + 1/n ~= ln(n) + γ

[gloomy scaffold]

yeah that one

do u know how to find sums of these kind of complicated problems. As an example sum of terms like:- n(n+1)+(n+1)(n+2)+...

these series are a combination of geometric and arithmetic series

[idle frigate]

that's not too hard to find out

again just lookup discrete calculus or umbral calculus

[gloomy scaffold]

what are the prerequisites for the topic?

I know school level math and I just graduated from school

[idle frigate]

we have the relatively easy formula $\sum_{n=0}^N n(n-1)\ldots(n-k+1) = \frac{(N+1)N\ldots(N-k+1)}{k+1}$

which you can use to compute all the stuff of the form Σn^k

at least for the first few terms

for instance for the sum of squares Σn² you write n² = n(n-1) + n and apply my formula

you get $\sum_{n=0}^N n^2 = \sum_{n=0}^N n(n-1) + \sum_{n=0}^N n = \frac{(N+1)N(N-1)}{3} + \frac{(N+1)N}{2}$

[flat frigate]

suut

[idle frigate]

with some factoring you get the well-known formula N(N+1)(2N+1)/6

[flat frigate]

suut

[gloomy scaffold]

so these are the formulas of discrete calculus?

[idle frigate]

yes

or umbral calculus

[gloomy scaffold]

I see

[idle frigate]

more or less the same thing

[idle frigate]

this one is fairly easy to prove by induction

or using the forward difference operator and the falling exponent as you can read about on the umbral calculus wikipedia page

[safe radish]

@gloomy scaffold Has your question been resolved?

[safe radish]

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[wary wolf]

Could you explain this question in simple terms please

[violet stratus]

Can you show your work

[safe radish]

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[wary wolf]

.reopen

[safe radish]

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[final radish]

im approaching this by taking an example fucntion like y = x though i dont know if itll work

also given that x1<y<x2

<@&286206848099549185>

[safe radish]

@final radish Has your question been resolved?

[safe radish]

@final radish Has your question been resolved?

[hasty wagon]

I think this question is about Mean Value Theorem , have you tried using the theorem?

[final radish]

uh well not yet

lemme see if i can apply it

[hasty wagon]

first use MVT, then use IVT (or just takin mean? not sure for 2nd step yet)

[final radish]

so what im doing is
f(x1) = F'(x1)
f(x2) = F'(x2)

and MVT
f(x2) - f(x1)/x2 - x1 = f'(y)

or its something else 💀

[hasty wagon]

(F(x1)-F(0))/(x1-0) = f(a1) for some a1 within [0,x1]
similarly for x2
so we have
F(x1)/x1 + F(x2)/x2
= f(a1) + f(a2)
with a1 a2 constructed above

[final radish]

aahhh this makes sense

so now i just have to prove that f(a1) + f(a2) lie within the interval given

right?

[hasty wagon]

well, actually it's ( f(a1) + f(a2) )/2 lies within

[final radish]

i forgot why this is the result, could you explain pls 🙏

[hasty wagon]

ohhhh it's IVT

intermediate value theorem

do you rememebr that?

[final radish]

f(a) < f(c) < f(b)
then
a<c<b

this one?

[hasty wagon]

yaaaa

with f is continuous

[final radish]

right

[lean sparrow]

this only works when function is increasing

[final radish]

yeah

[final radish]

so how do i get this from here
do i start with
0<a1<x1?

and so and so?

[hasty wagon]

I thought it's an Multiple choice question only? 🤔

[final radish]

yes but im the one asking
cuz idk💀

also the 1/2 factor

hows that coming into play

[hasty wagon]

well, we can split cases:
a1<a2
a1≥a2

oh

about the 1/2

I'm just taking the mean

[final radish]

why

[hasty wagon]

so, that (f(a1)+f(a2))/2 is within the interval between a1 and a2,
we can also take weighted mean like
(2f(a1)+f(a2))/3
or even
( r f(a1) + s f(x) )/(r+s)

for positve r,s

[lean sparrow]

is the ans b and c?

[final radish]

ohhh i get it
but if it was just f(a1) + f(a2) theres a possibility that it exceeds or preceeds the interval right

[final radish]

i dont know
this question was put up a while back

[hasty wagon]

true

[final radish]

and according to the question f(a1) + f(a2)/2 is already within the interval?

[hasty wagon]

but we might have to investigate more if y needed to be in such condition

[final radish]

wdym

[hasty wagon]

by IVT, there exists a y within the interval between a1 and a2 with f(y)=(f(a1)+f(a2))/2

[final radish]

right thats what i was thinking

so then the ans is b

[final radish]

hows c coming then

oh by the weighted mean

as you said

right?

[lean sparrow]

i just used f(x)=1 since it satisfies

[final radish]

right i got it now

[final radish]

i see

makes sense

thank you so much

[hasty wagon]

cheers!

[final radish]

.close

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[edgy venture]

Hey guys

[edgy venture]

Can you help me and explain it?

[safe radish]

@edgy venture Has your question been resolved?

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@edgy venture Has your question been resolved?

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@edgy venture Has your question been resolved?

[safe radish]

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[idle birch]

How does a left-hand limit not exist? Stuck on the wording for this

[idle birch]

The question I'm working for a homework assignment on says: "Describe or draw a way that a left-hand limit does not exist"

I'm just confused by its meaning since if it could just exist from the right, you'd just have to make left go to +/- infinity

♾️

[lean sparrow]

root x

[idle birch]

wdym

@lean sparrow

[lean sparrow]

is there something on the left?

[idle birch]

no

[lean sparrow]

but its there on the right, ryt?

[idle birch]

ya

but couldnt you just say that if it approached infinity

[lean sparrow]

so lhl DNE

[idle birch]

ohhhhh

so it just has to be a number that cannot be less than 0

Would it not just be the same for something like this?

[lean sparrow]

yes for this too

[idle birch]

okay

it just cant be approaching a definitive number on the left?

[lean sparrow]

yes

[idle birch]

dope thanks

.close

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[rain bay]

Hey guys sorry but this is my first introduction to Eigen values and vectors. As you know I have to multiply the identity matrix first with the lamda value before the subtraction although this is just 1 and inconsequential but I wanna know if the identity matrix would be affected by the negative sign.. so will the new matrix be -1 all through diagonal before the subtraction.. cox I saw an example somewhere that got me confused

[safe radish]

Please don't occupy multiple help channels.

[rain bay]

<@&286206848099549185>

Anyone here to help?

[devout shale]

Are you trying to find the eigenvectors associated with lambda=1?

[rain bay]

I know how to do that I just want to understand if the negative signs affects the identity matrix in the case of multiplication

[devout shale]

It doesn't look like you're multiplying them it looks like you're subtracting

Actually it doesn't make sense what you've written because I*(x1,x2,x3) will be a column vector

which you can't subtract from a 3x3 matrix

[rain bay]

Aye.. here is the thing if the lamda value were to be 3 and I have to multiply the identity matrix by 3 first

[devout shale]

yea

You forgot your parenthesis in the above picture

it should be like this

And then you're asking what, if you should multiply the -1 through the matrix or just subtract them?

Either way it will be the same

[rain bay]

Wouldn’t the diagonal be -1 all through if I multiply?

That’s: what I’m trying to know

[devout shale]

yes, but then you'd be adding the matrices together instead of subtracting, so it will be the same either way

[rain bay]

So you mean the negative sign should only be applied when subtracting?

So that new matrix would be

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[iron bolt]

could i get help with this?

[delicate bobcat]

So Energy = Power * Time

Power is watts, time is hours on this graph

Graphically, the energy is the "area under the power curve"

You have a power x time graph, energy will be the total area under the graph

Units will be Watts*Hours

Or Watt Hours

[iron bolt]

i think reading/understanding the graph is also giving me issues

i see that from 8am-4pm it's 200w?

[delicate bobcat]

Imagine it's able to be broken into several rectangles

[iron bolt]

how do i read what it is from 12am-5am?

[delicate bobcat]

12-6am is same height as 8am - 6pm

[iron bolt]

but isn't 200w total from that time or for every hour?

[delicate bobcat]

We assume that from the diagram cause otherwise we couldn't answer

[delicate bobcat]

200w is the current power use at that moment

Energy is after a certain amount of time at 200W how much energy did you use

You use the watts to find energy

E.g. from 12am to 6am

There was a constant 200 W of power

The energy consumption in that period is

200 W * 6 Hours

Or in other words the area under the graph from 12am to 6am

[iron bolt]

800 W * 2 hours

[delicate bobcat]

For 6am to 8am right?

[iron bolt]

yep

[delicate bobcat]

Yes exactly

[iron bolt]

200 W * 10

[delicate bobcat]

Because the graph is simple, we can answer this question by cutting it into rectangles

[iron bolt]

1200 W * 4

[delicate bobcat]

Yess

[iron bolt]

200 W * 2

[delicate bobcat]

All of those areas added together will be the total energy used

[iron bolt]

so the W displayed is per hour

and we assume it's the case for the other sections where it's not displayed bc it's the same height as where it shows 200W for 8am-6pm

[delicate bobcat]

More like

Watts in use at a particular time rather than per time

[iron bolt]

i see

[delicate bobcat]

Right now at this moment we are using 200W

This went on for 6 hours

So energy used is

200W * 6 hr = 1200 Watt Hours

The last thing to keep in mind..

They dont want watt hours

They want

kWh

Kilowatt Hours

So you have to convert the watts to kilowatts

[iron bolt]

do we divide by 1000?

[delicate bobcat]

Yea that should do it correct

So 800 W becomes 0.8 kW

Then the area will be kW*Hour

Which is what they want

[iron bolt]

so:

a) 10kWH
b) 10kWh/24h = ?

[delicate bobcat]

Just leave it as kWH

That's the final unit

kW * Hours

[iron bolt]

i see

how would i calculate b?

the conversion confuses me a bit

it would be easier if i divide using W?

[delicate bobcat]

So b is a bit different

Average power per hour

Is actually average y value of the graph

This is an average value of a function question

[iron bolt]

ohh

[delicate bobcat]

The graph you have is power

[iron bolt]

so 10000W total displayed in the graph

/24

would give us the average per hour

[delicate bobcat]

So average power is gonna be average value of the power function

Not quite like that

[iron bolt]

hm

[delicate bobcat]

We use a formula

One sec

[iron bolt]

no worries

[delicate bobcat]

Ok so for a situation like this you can think of it this way

Average value of function in an interval = area under curve/length of interval

So for us we have

Average power in 24 hours (12am - 12am)

Area under curve = Energy

Length of interval = 24 hrs

You're gonna do

Total Energy/total Time = average power used

[iron bolt]

10000W/24h?

[delicate bobcat]

Total energy is the kWH number

The area under the power curve

Maybe that was 10000

I didnt check xD

[iron bolt]

it was

[delicate bobcat]

Oh ok yes then

[iron bolt]

ok ty!! that makes sense

ty for the help

[delicate bobcat]

Great! Yw

Good luck

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[fierce egret]

is there a hero that can explain how to solve a linear diferential equation rn? i cant understand the procedure for solving y' = x + y - 1 ... voice if possile

[plucky elk]

equation 9 and 10
https://tutorial.math.lamar.edu/classes/de/Linear.aspx

[safe radish]

@fierce egret Has your question been resolved?

[plucky elk]

plug your y into your DE to check

[fierce egret]

Not sure what to do here, can you help?

[plucky elk]

what are you doing?

[fierce egret]

I dont know dude y' = dy/dx right? Since x is not dependant?

[fierce egret]

omg, i forgot to multiply by C as well.

solved it now

thanks for help 🙂

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[proud tide]

Could someone explain these to me, step by step?

[soft prism]

Have you tried them? Read the solutions? Read some sample questions?

[old chasm]

Hint: when you want to find the inverse, you want to reflect it with the line y = x (which is essentially saying that you should switch y and x and solve for new y = f(x))

[proud tide]

I was able to answer number 1, but I got really confused in number 2

[lean sparrow]

how did u solve 1?

[proud tide]

I'll send a pic, one moment

[soft prism]

Yes, you're supposed to. Question 2 is literally non invertible.

[proud tide]

what do you mean by that?

[soft prism]

Are you aware of objective functions?

Bijective

[proud tide]

My first time hearing about it

[soft prism]

Actually there's a theorem that a function has a unique inverse of and only if it is bijective

Now let me briefly explain bijective

Bijective functions are defined as functions which are both surjective and injective.

Surjective: also called onto function. It means that the function should "cover all codomain" for a given domain. For example, Say 1/x. Can you ever get some value for x where 1/x = 0? No right? So 1/x is not surjective in all real codomain but it is surjective in R - {0} codomain.

[delicate sphinx]

You're probably going into too much detail if he hasn't heard of bijective

The goal is to find the inverse of that each function

Nothing about proving it

[proud tide]

oh yeah, also graph it, forgot to mention TT

[delicate sphinx]

Where does it say in the instructions to graph?

[proud tide]

It's not shown in the picture, but it was required as said by our teacher TT

[delicate sphinx]

The process for question 2 is the same as what you did in 1

You swap x and y, then solve for y

[proud tide]

I'll try to solve it again

[delicate sphinx]

So given g(x) = 1/x^2, if you swap x and y, what would it look like?

[proud tide]

x = 1 / y^2

[delicate sphinx]

How would you start solving for y?

[proud tide]

by removing the exponent?

I'm not sure TT

[safe radish]

@proud tide Has your question been resolved?

[proud tide]

<@&286206848099549185>

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@proud tide Has your question been resolved?

[safe radish]

@proud tide Has your question been resolved?

[proud tide]

.close

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[bleak epoch]

This is my first time asking for help on a math discord server.

[bleak epoch]

I'm having trouble integrating this particular problem

[severe pond]

congratulations

[bleak epoch]

I will send an image, give me one moment

Whenever I am solving for the coefficients, I get two different results for E

[desert pasture]

start by subtracting and adding 1

to get less of a headache

[bleak epoch]

That's funny, I literally have a headache right now

I will give it a try

[split fulcrum]

@junior smelt is a master at algebra they can totally check it working out for you

[bleak epoch]

Okay :D

I am liking adding by zero a lot more I forgot I could do that

[junior smelt]

https://tenor.com/view/look-at-your-life-look-at-you-foolish-nonsense-rubbish-gif-14191134

[bleak epoch]

I think the way I was trying to do it first should be possible, no?

[desert pasture]

did you even try my suggestion

[bleak epoch]

Yes, it's working out better

[desert pasture]

what did you get

I think I have an idea

[bleak epoch]

I'll share the solution when I'm done

I'm eating while also doing work, so working slowly

My apologies

[desert pasture]

$\int \frac{dx}{(x+1)^2+1)^2} +\frac{dx}{(x+1)((x+1)^2+1)^2}$

no>

oops

[bleak epoch]

lmao

[desert pasture]

notice that subbing in (x+1)=t helps even more

[flat frigate]

Veni, vidi, perii

[bleak epoch]

I think my brain is so fried right now

I think I'm gonna stop here, as this is my last homework question before going to bed

I just need to complete it by tomorrow

But there was more progress made by adding by zero

Thank you for the help, @desert pasture

.close

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[molten finch]

Can someone help me with this problem My first step to graphing these quadratic surfaces is to plug in 0 for each variable and then graping while an axis is 0 and then cobining with the end Is that the right approeach?

[molten finch]

i am not sure how the elipsoid was formed at the end cause we are not using the y=0 one

[safe radish]

@molten finch Has your question been resolved?

[hasty wagon]

to be honest, I haven't done such questions before, but I'd like to try.

i think your approach is good and also you might wanna mark the extrema in each x,y,z=0 pieces so that it will be easier to plot with the 3D one

by then we'll have 6 points (all on the axes) then the ellipsoid would look nicer

[molten finch]

i see do you know how to graph z=4x^2+y^2 if we make x=0 thats z=y^2 and if y is zero z=4x^2 what about if z is zero 0 is 4x^2+y^2?

[hasty wagon]

oh, i thought we are talking about a, haha, i was drawing

for b and c, we can take different values of z, e.g. z=1,4

[molten finch]

ok thats looks good with what i got

i am just having trouble graping b

[hasty wagon]

lemme try for B

[molten finch]

wait so y and z axis are interchangable

[hasty wagon]

nah

maybe i didn't draw narrow enough, hehe

the point is
when you are drawing graph with the fixed values, I'd use the "same scale" with all those if possible

[hasty wagon]

so that, e.g.
we can see x=0 parabola is wider than y=0 parabola

[molten finch]

that makes so much sence ty so much

[hasty wagon]

Cheers!

[molten finch]

same thing with other one just aplly 0 and use same scale correct

[hasty wagon]

correct

[molten finch]

alright ty

!done

[safe radish]

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

[hasty wagon]

z=1 and 4 too

[molten finch]

we need 4 aswell

or is 1 sufficent

[hasty wagon]

it will look better

[molten finch]

true and negative parobloa just faces downward right

[hasty wagon]

not sure, i forgot about hyperbolae lol

[molten finch]

sorry for spelling mistake

[hasty wagon]

me too😆

[molten finch]

when graping -x^2 +y^2 =1

thats an elipsoid

right

how shall we graph that

[hasty wagon]

nah, it's a hyperbola

[molten finch]

yea for the third equation i plugged in 1 for z

[hasty wagon]

you can have a hint with graph calculator online, like desmos
or you can draw point by point to see what's happening

i prefer the former one if you have already done before about these conics

[molten finch]

so for that the horizontal is alwaus half of z

[hasty wagon]

-x²+y²=1
e.g. points
(-1,√2) (0,1) (1,√2)
(-1,-√2) (0,-1) (1,-√2)
making two curves

[hasty wagon]

I'm not sure what that means, i will consider the non squared term and plug in ±square numbers to test usually
so you might also need z=-1, -4

[molten finch]

sorry for the elipsoid how far up down right left is gooes depends on coefficents of x and y?

or z?

[hasty wagon]

for left and right, it's along y-axis, so its -3 to 3

[molten finch]

i think i figured it out

ty

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[fathom jewel]

What have you tried so far?

[flat frigate]

Austin

[safe radish]

@willow fossil Has your question been resolved?

[willow fossil]

Ive tried getting dx, du

But none of them seem to fit

[burnt notch]

Wdym?

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@willow fossil Has your question been resolved?

[fathom jewel]

Show us how calculated u'

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[cobalt agate]

help I don't know how to find the answer

[cobalt agate]

the need is displacement

[silk valve]

@cobalt agate Could you please share the question? Is it asking how far point A is from point E?

[cobalt agate]

ahh yes yess

sorry, I forgot to add that part

.close

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[hardy cedar]

hey guys wanna join a minecraft server pls dm me

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[floral cave]

how do u draw this in 5 lines, the lines cant overlap

[floral cave]

<@&286206848099549185>

[verbal falcon]

but its been 2 minutes

[floral cave]

cus i only got like 5 left yo solce kt 😭

srry

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[limber edge]

pretty sure you can't do that

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[visual pond]

Why does this graph just stop at ~710?

[visual pond]

$\frac{3-e^x}{3+2e^x}$

[flat frigate]

smeagol

[visual pond]

I was solving for horizontal asymtope when x -> +infinity, when you graph it looks like DNE because no numbers appear, but if you multiply by 1 as e^x/e^x then you get -1/2 🤔

[finite yacht]

Because computers and calculators have a limit on how large numbers can be

When x = 710 is probably when 2e^x + 3 becomes that limit

[visual pond]

ah so the computer just gives up because e^1000ish is really big

[finite yacht]

Yes

[visual pond]

gotcha thank you

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[umbral solar]

guys when looking for homogenuity of this function

why we had to (ln lambda * y - ln lambda * x)

why then make it into fraction?

i dont understand

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[sinful wyvern]

guys, how to do (ii)?

[finite yacht]

The perpendicular bisectors of (i) are diameters of the circle, do you agree?

[sinful wyvern]

how do you know this?

[finite yacht]

It is a relatively well-known property

[sinful wyvern]

oh…

OK!

ohhh

[finite yacht]

If you have two points in a circumference, then the perpendicular bisector of the line between them is a diameter

[sinful wyvern]

ohhh okk im going to remember this

okay i think i have some plans now

i’ll try solving it

[sinful wyvern]

yes i agree

wait what do i do next tho

what should i do with the equation of the diameter

[finite yacht]

So from question (i) you have two different diameters of the circumference

And if you have two diameters then you have the centre

[sinful wyvern]

ohhh

[sinful wyvern]

but one of the chords have undefined gradient in (i)

can i still plug the gradient to the equation y-y1=m(x-x1)?

[finite yacht]

No, you need a different type of equation

Because it is a vertical line

[dense pulsar]

Use the general equation of circle If you have found the radius

[finite yacht]

We don't know the radius yet

[dense pulsar]

Oh

Found the diameter?

[finite yacht]

No

[sinful wyvern]

ohhh

if the gradient in chord AB is undefined, then am i allowed to make the gradient be 0?

for the y-y1 = m(x-x1)

[finite yacht]

No

That would be wrong

If the gradient is undefined then you simply can't use that formula

Use ax+by=c instead

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[hot urchin]

Explain why numbers are moved past the equal sign and what it implies.

[peak estuary]

wdym "implies"

[hot urchin]

Explain it and the context that needs it.

[peak estuary]

it will be clear later why the -c/a was moved over

[thin bridge]

is that what the question is asking you
or what you're thinking about

[hot urchin]

I need to know. Are you here to help or not?

[peak estuary]

why are you quoting everything

thats not how that feature is supposed to be used

[thin bridge]

we're trying to help,
the wording of the question is a little strange
so additional context helps

[hot urchin]

I'm asking for the context behind the operation in question where a number is removed and moved past the = sign as seen in the image.

[thin bridge]

ok.

[hot urchin]

He is not.

You are.

[soft matrix]

How do we do that look kool

[hot urchin]

Use a >

[thin bridge]

the main focus of completing the square is
considering the x^2 and x terms
and identifying the value that when added, will "complete the square"
when you separate the c/a term is up to personal preference
(here that was achieved by subtracting c/a from both sides of the equation)

[soft matrix]

Regarding this completion of square relied on the identity a² + 2ab + b² = (a+b) ²
Here is an example of completing the square
x² + 6x + 18
x² + 6x + 9 + 9 = (x+3) ² + 9 = (x+3)² + 3²

[soft matrix]

a ok >

a how >

[hot urchin]

Before it with a space

m

[soft matrix]

a >m

[hot urchin]

Remove the a and add a space before m

[soft matrix]

m

Nice

[hot urchin]

I honestly don't care at all.

The identity part was decent but I don't need the rest.

[soft matrix]

ok

[hot urchin]

Actually not too bad but it could've been alot better.

[hot urchin]

please answer the original question

[thin bridge]

read what i typed up earlier

[hot urchin]

The way you explained this makes me want to throw a chair at the screen.

I'm not saying I'm mad because I don't understand.

[soft matrix]

x² + b/ax + c/a
What should be added or subtracted to bring this to form a² + 2ab + b²
|| it can be rearranged as x² + (2b/2a) x + (b/2a)² + c/a -( b/2a)² = 0 ||

[hot urchin]

I'm mad because I do understand and can see how flawed of an explanation that really is.

But shrug you tried so thank you.

[thin bridge]

what's flawed about it

[hot urchin]

Here

[soft matrix]

U know it's flawed so u must know what the actual theory is
Ahh so u dont really have a doubt do you?

[hot urchin]

1. "Considering" makes zero sense to someone learning this even if you did understand you still wouldn't know that. Consider learning to use brackets () they are probably my favorite thing in all of English.

2. Completing the square is a stupid term that doesn't explain anything about the concept itself (I know what that is)

3. My entire question was how subtracting from both sides of the equation worked.

[thin bridge]

ok then
its simply subraction property of equality

[hot urchin]

4. Your explanations operated under the idea that I already would've understood making it inoperable.

Please consider such points I say this to be purely critical.

[hot urchin]

I don't know this either.

[soft matrix]

points " , " I say

You forgot a comma

[hot urchin]

Alright.

[thin bridge]

I'd recommend you look that up.
Along with stuff like addition/multiplication/transitive

[hot urchin]

Alr

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[real turtle]

The numbers x,y,z satisfy the equation
x + y + z - 2(xy + yz + zx) + 4xyz = 1/2
Prove that at least one of them equals 1/2

[real turtle]

I know I'm meant to use a third order polynomial as that's what the hint says

but I can't figure out exactly how

I wanna factorise the 3rd order polynomial using FTA but still can't figure it out

[hazy elbow]

put p=1/2 in the cubic so formed (assuming the cubic to be p(x))

@real turtle

[real turtle]

wait lemme wrap my head around that

what's the cubic formed

is it x^3 - x^2 - 2x + 4

[hazy elbow]

p^3+ap^2+bp+c

[real turtle]

oh just a general one ok

then what u get 1/8 + 1/4a + 1/2b + c

[hazy elbow]

assuming x,y,z are roots of p(x), we can rewrite x + y + z - 2(xy + yz + zx) + 4xyz = 1/2 as a+2b+4c=-1/2

[real turtle]

wait wldnt it be -1/2

[hazy elbow]

why do you think so

[real turtle]

cos a is equal to -(x+y+z) by vieta's no?

yh i see what to do

u take that eq divide it by 4 and u can plug it back in and see it holds consistent

[hazy elbow]

ah yes you are right

[real turtle]

thx a lot

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[nimble stirrup]

I need some help on limits, so our teacher gave us a assignment pretty much everything in this was discussed but the problem are the number 6 and 7 i have no idea on how to solve it since it isnt discussed i just need some help on steps to solve it.

[junior smelt]

For 7, think about what happens if you common denominator those

[nimble stirrup]

oooohhh

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[polar knot]

i figured out its a periodic sequence every 6 terms

idk what next

[eternal carbon]

consider, for example, what a1 + a4 and etc would be

[polar knot]

huh

i did mod 6 on 1492 and 1985

[eternal carbon]

and?

[polar knot]

so that everythiing would boil down to

the first 6 terms since its periodic

then i got

a1 + a2 +... + a6 + sum from n = 1 to 1488 = 1985

oh i get it

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[narrow chasm]

How do I read the n# notation

Like how do I get an answer from it

[median vigil]

it stands for any integer. so replacing n# with 1,2,3, etc would give you a different but valid solution

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[real silo]

I need help with counting a perimeter of two triangles , using rule of the same triangles ( CCC ( corner, corner, corner ).
From the book answers I know, that the first triangle has a perimeter of 37.5, and the second one 14,4. But I don’t know how to come to them
Task 2 A and B

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[latent tendon]

is my answer wrong?

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[lean otter]

I have a question about this one it says to solve each rational inequality and express solution set in interval notation. In order to do that I have to find the zeros of numerator and denominator, but there is no zeros.

[lean otter]

So is the answer all real numbers and how should I graph in the number line?

Does this seem right?

[junior smelt]

(Yep that'd do also for your inequality you could notice that both the numerator and the denominator are always strictly positive too, hence it's all real numbers )

[lean otter]

Ok thank you!

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[rigid birch]

Hiii i need help with j) and isolating θ

[safe radish]

@rigid birch Has your question been resolved?

[rigid birch]

<@&286206848099549185>

I keep trying to work it out but i dont even know where to start here

[glossy cypress]

for j?

[rigid birch]

Yup

[glossy cypress]

i might have forgotten these entirely, but cant you just use arcsin

[rigid birch]

How would i do that? Is it something on the calculator?

Sorry for asking so many questions lol

[glossy cypress]

there's no real other way to solve this

it's

[rigid birch]

Yea i have to isolate theta

Not solve

[glossy cypress]

long story short it takes the output for sin(x), and gets you the input needed to get that value

so for example: sin(90) degrees = 1, arcsin(1) = 90

there's a weird thing where it only includes angle values between 90 and -90 because of how sine works but

that's

the only way i know how it solves

[rigid birch]

But i dont need to solve it

I need to isolate theta

The funky 0

[glossy cypress]

yeah i know

[rigid birch]

Oh my bad sorry

[glossy cypress]

you just apply arcsin to both sides

im

not sure what notation your teacher wants you to use

some people just do something like

$arcsin(x)$ or $sin^{-1}x$

[flat frigate]

Serphic

[rigid birch]

Oh is that the inverse sine ?

[glossy cypress]

yeah

[rigid birch]

ur such a life saver man tysm

[glossy cypress]

np lmfao

[glossy cypress]

[rigid birch]

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