help-27

[earnest palm]

What abt this one

Is it b

For 6h = 750
Additional charges = 100 so 750+100
=850
And 850+20+20=890?
And more additional charges for distance over 100km

Hell.

O

Anyone help

[copper mango]

be patient.

[earnest palm]

Ok but a question aren’t u too online n not solving : (

[solid trail]

consider the base charge, then add to it the additional charge

[earnest palm]

Yes

[solid trail]

then what dont you understand

[earnest palm]

Yeah I meant solving as in helping @restive river

Ahhh

Look the total additional charges are

100+20+20= 140

Base charge is 500 for 4 hours but he was riding the taxi for 6 is for one hour it is 1w5 so for 6 it will be 125x6=750

750+140=890 which isn’t even a option : (.

[solid trail]

ok this car travelled for 6 hours

so base charge 500

what about the remaining 2 hours

it travelled for 2 more hours, so additional charge + 100

also it’s 20 km over the 100 km limit, so additional charge + 400

option C is Rs 1000

consider the base options, then the conditional options

[earnest palm]

Yeah thx a lot

Thx

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[restive river]

for the cauchy mean value theorem's proof, everyone uses rolles theorem ; why not use mean value theorem over two functions

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

[polar bolt]

Wdym?

[sweet zinc]

what do you mean?

[restive river]

like in every book (and in class) our teacher used rolle's theorem to prove cauchy's mean value theorem

even tho we already proved mean value theorem

so why cant we just use mean value theorem on f and g then divide the two equations

<@&286206848099549185>

[copper mango]

Cauchy's generalizes Lagrange's, so you'd be more or less proving a corollary then the thm

you prove theorems, then derive corollaries

(Also Rolle's thm is Lagrange's MVT, it's just the case when f(a)=f(b))

[restive river]

oh alright

thanks

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[strange basin]

Aye

[strange basin]

So we got hw about quadratic equations but our teacher is sick and just told us to see the books explanation but didn't get it

How to work out

7(x+3)(x-1)=0

[solid trail]

if x times y = 0, either x or y MUST be 0

do you get this principle

[strange basin]

Yes

[solid trail]

ok you have 7 (x + 3) (x - 1) = 0 right

[strange basin]

Yep

[solid trail]

7 is apparently not 0

x + 3 can be 0

x - 1 can also be 0

so two solutions: -3 and 1

[strange basin]

Ah I wrote 3 and -1, what I don't understand is how x+3 can be 0

If x is already 0

[solid trail]

x is not 0

for example

(-1) (0) and (0) (1) both lead to 0

right

[strange basin]

Yes

[solid trail]

uhh

so essentially

if two (or more) things are multipled together

one of them must be 0

and one is enough

[strange basin]

Mhm

[solid trail]

if i have x (x - 1) = 0, x can be 1 or 0

[strange basin]

So I understand all but the 7 you just ignore it?

[solid trail]

because 0 (-1) and 1 (0) are the same

[solid trail]

yes

[strange basin]

Alright lemme try one rq

[solid trail]

7x = 0 is the same as x = 0

7 is constant so it does not matter

[strange basin]

5x (x+4)=0

[solid trail]

yea

[strange basin]

Would be just x is '3

-4

[solid trail]

3???

[strange basin]

Misspel

[solid trail]

-4 is one solution

there is another

[strange basin]

And the other one is x=-5

[solid trail]

no

do you know what linear expressions are

[strange basin]

Yes

[solid trail]

what are they

[strange basin]

Basically when you have a sum like

(3x+2)(3x-2)

[solid trail]

im guessing you understand the notion of constants

[strange basin]

I'm not good with definitions but basically if I have the sum I can work it

[solid trail]

if you have something like

x (x + 1) (x - 1) = 0

you see x actually has 3 possible values

do you see

[strange basin]

Yes

[solid trail]

what are thise

those

[strange basin]

0 1 -1

[solid trail]

yep

so it’s when x = 0, x + 1 = 0, and x - 1 = 0

basically

(expression1) (expression2) = 0

then expression1 = 0 or expression2 = 0

that’s why we omit the 7 because it can never be 0 lol

[strange basin]

Ohh

[solid trail]

do you get it

[strange basin]

Yep understand it

[strange basin]

So to clarify, would this be x-4 and x=20, not sure of this answer

[solid trail]

no

it wouldn’t

we omit the 5 right

so it’s just x (x + 4) = 0

[strange basin]

So what would the other be if there's always w

2

[solid trail]

it’s apparently -4 or 0

[strange basin]

Oh so it's just 0

[solid trail]

yea

[strange basin]

X(x+6) = 0

X= -6 or 0

[solid trail]

correct

[strange basin]

How are you so good in all these and know all the terms like notions of constants etc

Thanks tho

[solid trail]

i deal with them a lot

[strange basin]

Alright thank you I'll try to work more of these

Have a good one

Appreciate it

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[solid trail]

np

[devout snow]

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[bleak bluff]

Need help

[bleak bluff]

why does 1/sin²(x) * sin²(x)/sin²(x)-1 = -1/cos²(x) ?

[frozen elm]

The sin^2s cancel out

Then you can use the identity

cos^2 x + sin^2 x = 1

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[shy wing]

stop

[severe summit]

2

[devout snow]

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[shy wing]

<@&268886789983436800> this guy keeps trolling

please do something about them

this person

[copper mango]

Can confirm.

[wicked rover]

@raven basin yes

[shy wing]

not just that

[surreal mason]

B&

[shy wing]

you’re posting nonsense in other help channels

nice

[surreal mason]

There was spam that he deleted roketto

[copper mango]

and a poor comment in another help channel

[wicked rover]

ic

[copper mango]

(in lecture so couldn't ping)

[devout snow]

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[shy juniper]

What do they mean by regression summary here

<@&286206848099549185>

[shy wing]

gotta wait 15 minutes to ping helpers

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[shy juniper]

<@&286206848099549185>

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[devout snow]

@meager rose Has your question been resolved?

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@meager rose Has your question been resolved?

[umbral cargo]

Wouldn't it just be defect rate times amount?

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[scenic cloak]

my examination is tomorrow and I need to study like 74 more pages, im gonna sleep now and wake up after 4 hours as I have been studying for 12 hours straight

the examination begins 26 hours from now
what's the best strategy to ace it?

[restive river]

well for the future, study earlier, ig review core concepts and know the application of formulas and what not

not much info to really answer this on

[winter patrol]

74 more pages of what...

[umbral cargo]

Get good rest for sure, you don't want to pass out during your exam, that's a definate bad idea.

[devout snow]

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[restive river]

$7^{961} \equiv x (mod 1716)$

[woven radish]

Cüneyt

[restive river]

I found that 7^6*6 is 1 in mod 1716

That's all I could do

[pseudo basin]

,w 7^36 mod 1716

[woven radish]

Results provided by WolframAlpha

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

Requested by Ann#0413

[pseudo basin]

@restive river looks like you messed that up

but i think repeated squaring should let you do this

[restive river]

No

Its not

$7^6 times 6$

[woven radish]

Cüneyt

[restive river]

If I knew that 7^36 equals 1 then question is done

[pseudo basin]

$7^6 \cdot 6$?

[woven radish]

Ann

[restive river]

Yea yea

[pseudo basin]

that sounds like it cannot equal 1 mod 1716 tho

[restive river]

But it doesn't help much

[pseudo basin]

it's going to be something even mod 1716

and besides

repeated squaring

[restive river]

Ahh you're right

[pseudo basin]

find $7^{2^n} \mod 1716$ for $n=0, 1, \dots, 9$ and then multiply the necessary ones together to get the exponent to 961

[woven radish]

Ann

[restive river]

Okay just give me 5 mins ILl back

I found this

$7^{{\left(7^3-7^2\right)} \cdot 4} \equiv 1 (mod 1716)$

[woven radish]

Cüneyt

[pseudo basin]

i think you're making this more complicated than necessary

but ok

,calc (7^3 - 7^2) * 4

[woven radish]

Result:

1176

[pseudo basin]

oh lol that's bigger than 961 so this is definitely useless for your problem

[restive river]

This is just the simpliest form of 7 to the totient 1716

Looks like it

[pseudo basin]

again

repeated squaring is right there

why do anything else lol

[restive river]

I have no idea what else to do

If I square root of them would it be equal to 1 again?

[pseudo basin]

why are you refusing to do repeated squaring

it sounds like youre just looking and looking for something else

[restive river]

Tell me then what to do lol

[pseudo basin]

.

[restive river]

That's why I am asking

[pseudo basin]

find 7^1 mod 1716 then 7^2 mod 1716 then 7^4 mod 1716 then 7^8 mod 1716 etc.

[restive river]

Good method if you are crazy to calculate them all by hand

[pseudo basin]

each one of them is the square of the previous

like

7^1 mod 1716 = 7

7^2 mod 1716 = 49

7^4 mod 1716 = 49^2 = 2401 = 685

[restive river]

Yeah continue

7 to the 8

[pseudo basin]

7^8 mod 1716 = 685^2 = 469225 = 757

[restive river]

Hahha lol

We are not allowed to use calculators

[pseudo basin]

this is much easier than raising 7 to the 961st power

[restive river]

During exam

It's still too hard

[pseudo basin]

oh sorry

are you supposed to do this in 10 milliseconds

[restive river]

Anyways I'm looking forward to see those crazy guys that spend 50 mins for one question

Ty

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[restive river]

How do i find m and n

pls explain cus im so lost

@restive river

[austere rock]

hey

do u know trignometric identities?

[austere rock]

?

[restive river]

Nope

its special right

oh do u mean

sin cosine

tan

?

[austere rock]

there is a simple way to do it

yes

tan 30

[restive river]

yeah ik them

[austere rock]

what is tan

?

[restive river]

tantagent

idfk how to spell it

[austere rock]

i know it

but like what is it

p/b

[restive river]

ohh

its o/a

opposite

over adjacent

[austere rock]

in this question tan 30=n/3

[restive river]

oh

[austere rock]

yes

[restive river]

so i just do that

for both?

except switch the identity

cus its different!

?*

[austere rock]

no first u will use tan 30 then u

use sin 60

[restive river]

Okay

[austere rock]

which is m/3

[restive river]

Ty daddy

[austere rock]

daddy?

[restive river]

i mean bro

[austere rock]

😂

[restive river]

ty bro

[austere rock]

np

[restive river]

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[real patrol]

Name: Brayden C
Topic: Geometry, Graph Parallel Lines
Grade / Level : 9th Grade (Highschool Freshman)

Description :
I need help with how to solve it because I don’t understand what the problem wants me to do

I’ve tried several times before asking, I even searched it up on YouTube for help but it didn’t help at all because the explanations are too complicated for me
Thank you.

[real patrol]

^
<@&286206848099549185>
Please Help Me, Thank You!

[placid rover]

what does 'parallel' mean

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[digital egret]

i do this right?

[graceful cosmos]

Assuming a z-score of 2.33 gives a probability of 99.01%, then yes

[digital egret]

okie tyty

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[worthy quest]

hi

[worthy quest]

was wondering if anyone could look at this and tell me how what i'm not seeing

why does cows x hay = "the total yearly amount of hay produced on sophias farm"

what do the cows have to do with it

oh nvm

i see it now.

[frozen elm]

🐄

[worthy quest]

F(t) is the amount per cow not the amount produce total per year

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[austere dust]

Well let me try opening a help channel for what I hope is just a definition question

On Wikipedia the Laurent series and Z-transforms are defined with sums from -infty to infty, but I see one source (which covers the topic I'm interested in, applied math) with an index starting from 0. What gives?

Also, on Wikipedia itself also, the Laurent series has coefficients z^n but the Z-transform is z^-n. Given the indexing sum over all integers I suppose they are the same, but is there any difference actually?

[devout snow]

@austere dust Has your question been resolved?

[devout snow]

@austere dust Has your question been resolved?

[leaden aurora]

if you can't figure it out what makes you think we can? i think stack exchange would serve you better

[austere dust]

Uhh I think this is a definition question

But anyway I'll read more into the Z-transforms and Laurent series

I have a feeling some 0s are creeping into applied math without them saying it outright

Another issue is I'm supposed to assume analyticity presumably since these series don't make sense (?) without analyticity

[leaden aurora]

If nobody helps, maybe the stack exchange chat room can help you out. A bunch of phd and professors hang out there https://chat.stackexchange.com/rooms/36/mathematics

[austere dust]

Hmm I'll try that tomorrow

Thanks

I'll also ask my peers tbh

I'll close for now since I also need to sleep soon

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[lusty gulch]

Can Someone Help Me..??

[lusty gulch]

I'm Not very Good at math, it's only certain subjects i am good at, I am in 8th Grade.

<@&286206848099549185>

[cursive smelt]

Do you remember how to find the gradients of lines?

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@lusty gulch Has your question been resolved?

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[idle zinc]

Can someone help me solve this?

[stiff ruin]

Do they not give you an equation?

[idle zinc]

No they don’t

[stiff ruin]

Well you will have to find that out in part b. Part a is when cosine graph starts, and cosine always starts at top of interval.

[idle zinc]

So how would I go on about solving part b?

[stiff ruin]

apply transformations using this formula

[idle zinc]

After that?

[stiff ruin]

then that's it

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[burnt falcon]

im not sure if there is a way to simplify it passed (2x+b+2xe^x+be^x)/e^x for part a

and for b i got -(2x+b/e^x)-2x-3=b

[dim lotus]

for b can you not just sub in x = infinity into the equation

or first try to divide all by x or something

[burnt falcon]

see i wouldnt know because i have no clue what this shit is

is that what that means sub in infinity?

[dim lotus]

it means as x approaches infinity what would the equation be like

[burnt falcon]

instead of infinity would i sub in a really high number?

cuz how are you gonna do infinity minus 3

<@&286206848099549185>

[dim lotus]

thats why maybe divide all by x so you would have the eqautions as something/x so when u sub in infinity it would just equal 0 for all terms that have /x

[burnt falcon]

if i plug in infinity rn il have 2x+b

=-3

for part b

.close

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[green wagon]

$f(x) =
\begin{cases}
x^2 & \text{for all positive x such that } \frac{1}{x} \in \mathbb{Z} \
0 & \text{otherwise}
\end{cases}$

[woven radish]

Sacha

[green wagon]

if i'm trying to find $\lim_{x \to 0^{+}} f(x)$, can I take the limit of each case separately?

[woven radish]

Sacha

[spice flare]

you can. If the limits exist and are the same then it will work out

here a comparison may be easier like $0 \leq f(x) \leq x^2$

[woven radish]

Tom

[green wagon]

ah right thank you!

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[restive river]

this = 0 is putting me off

and im a bit confused

i think it wouldve been (x+6)(x+2)

[dreamy holly]

forget about the quadratic for a moment

what would it mean if i said 8x + 12 = 0

@restive river

[restive river]

would I simplify it first?

but it would been it = 0

[dreamy holly]

well what do you know and not know from that equation

[exotic stump]

this is factored correctly, set equal to zero and use zero product property

[dreamy holly]

think about what you are trying to find

[restive river]

x

[dreamy holly]

yes thats right

[restive river]

whats zero product property?

[dreamy holly]

so for a linear example like 8x + 12 you would rearrange for x

[restive river]

yeah

do i divide 8

then -12

[dreamy holly]

this is a quadratic meaning there are 0-2 results

[restive river]

oh

[dreamy holly]

thats because when graphing it, when x = 0 it is actually the points that the line crosses the x axis

[exotic stump]

if $a\cdot b=0$, a or b is zero

[woven radish]

a disappointing son

[restive river]

oh yeah

[dreamy holly]

i meant when y = 0

[restive river]

ok

[dreamy holly]

see how the x axes is the same as the line y = 0?

[restive river]

wait what do you mean

[exotic stump]

this is all you need to do lmao

[dreamy holly]

what an equation like x^2 = 0 is asking you, is "for which values of x, does x^2 = 0?"

[dreamy holly]

ok but im trying to explain it so it makes sense

[restive river]

idk what to put tho

[exotic stump]

what is a and what is b?

[exotic stump]

if given this

[restive river]

6 and 2

oh

0

and something

[exotic stump]

not quite

you have (x+6)(x+2)=0

that's in the form (a)(b)=0

so what is a and what is b

[restive river]

is a everything in the brackets

[exotic stump]

yes

everything in one of the pairs

you're just multiplying to quantities together

cause if the product is zero, if you multiply something by zero, you get zero

therefore if one of either (x+6) or (x+2) is equal to zero, the entire thing is equal to zero

[restive river]

understood

where would i put it though?

[exotic stump]

put what

[restive river]

0

like

im not sure how to put it into the brackets

[exotic stump]

you're not putting 0 in any brackets

you have (x+6)(x+2)=0, which is true if x+2=0 or x+6=0

solve for x in those two equations and you have your zeroes

[dreamy holly]

for example with (x+6)(x+2)=0 you can divide both sides by (x+2 to get (x+6)=0 meaning that x=-6 as one solution

[placid rover]

$$ab = 0 \implies a = 0,or, b = 0$$

[woven radish]

Shuri2060

[exotic stump]

can't do that, you're automatically assuming x != -6 and x != -2

[dreamy holly]

then what are you doing to go from ab = 0 to a = 0

[placid rover]

its literally a theorem

[exotic stump]

zero product property lmao

[placid rover]

if 2 things multiply to make 0. One must be 0

[dreamy holly]

yes so you are dividing by the non zero one

both times

[exotic stump]

??

[restive river]

ok sorry guys im back

[dreamy holly]

if 2 either a or b have to equal 0 and both equal 0 depending on which solution you are finding, you are dividing by a non zero value each time

[restive river]

do i just do opposite

or is there some opposite way

[dreamy holly]

?

first factorise it all

[restive river]

ok

ye

(x+6)(x+2)=0

[exotic stump]

you're assuming the roots if you do this lol

[restive river]

wait

[placid rover]

The way to prove the zero product property is what you said.

But there is no need to do that.

[dreamy holly]

should be 6 instead of 8 there

[restive river]

sorry my bad

accident

[placid rover]

Assume a, b not 0. Then divide by a for a contradiction. Yes.

[dreamy holly]

i dont like memorising things unless i understand them 🤷‍♂️

[placid rover]

Ofc we do

but you are NOT going to write that in your working

[restive river]

what do i do now

[placid rover]

its just a basic fact

[dreamy holly]

true but id want other people to have the same understanding

[dreamy holly]

now use the rule we have just been talking about

[placid rover]

You are going to create a misunderstanding

by dividing through by factors

without being clear

assume it is non-zero. divide. Case 2, it is 0.

[dreamy holly]

not rlly what i was saying but sure i wasnt being clear

[placid rover]

@restive river 1st. Do you recognise this fact

$$ab = 0 \implies a = 0,\textnormal{or}, b = 0$$

[restive river]

yes

[woven radish]

Shuri2060

[placid rover]

That is what you need to use then.

[restive river]

im unsure

[restive river]

ok

what would i write though???

[analog swan]

are we solving quadratic?

$x^2+8x+12=0$

[woven radish]

abe

[analog swan]

this right?

@restive river

bruh they're gone

[nocturne willow]

he could just use the formula right?

[analog swan]

no

that is a waste of time

[exotic stump]

he already factored it lol

[analog swan]

when there are obvious factorisations

[exotic stump]

he could use quadratic formula but yeah lot easier to factor

[nocturne willow]

oh of course

[analog swan]

¯_(ツ)_/¯

[restive river]

sorry i didnt notice

[analog swan]

yeh

[restive river]

yes

so

[analog swan]

so u managd to factor it into

(x+6)(x+2)=0

[restive river]

ues

yes

[analog swan]

yeh, so the property you will use when solving quadratics

is tht when u multiply 2 things together and they equal 0

one of theme has to be 0

so u end up with 2 solutions

[exotic stump]

don't just give people the answers

[dreamy holly]

maybe a diagram explanation would help

[analog swan]

i explained

[exotic stump]

you're still not supposed to give answers

also this is purely algebraic, he just needs to use zero product property as i've said before

[analog swan]

yeh

i think that's their problem atm

they don't understand it

[dreamy holly]

but whats the point for them to memorise some algebra if they dont understand

[analog swan]

its not memorisation

its just common sense

[restive river]

ok so what are the exact steps

[exotic stump]

.

[analog swan]

$ab = 0 \implies a=0$ or $b=0$

[woven radish]

abe

[analog swan]

this is common sense

no memorisation here

[restive river]

yes

so one of the brackets?

[analog swan]

yeh so u had

(x+6)(x+2)=0

so then u have to solve 2 equations

to find the 2 solutions

either the first bracket is 0

or the 2nd one is 0

[restive river]

wouldnt that give me the same thing?

[analog swan]

no

(x+6)=0

is the first equation

(x+2)=0 is the 2nd

[analog swan]

why would u think this

[restive river]

OHHH

i thought it was

(x+6)(x+2)=0

which is what the same thing is factorised

SO WHERE DID U GET

sos caps

where did u get it lke tha

bruh

i just watched a mathwatch video

cuz

i wanted to see if it included it

and now i get it

was so easy

didnt even realise

,close

.close

[devout snow]

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[civic lion]

Hi, i just have 2 questions

[civic lion]

for this, i did all the work for the other parts, but im not really sure what the 2 functions are when separating the integral

<@&286206848099549185>

would f1(x) be x^3 - 9x or something?

[placid rover]

draw a picture.

$$|x^3-9x|$$

[woven radish]

Shuri2060

[civic lion]

I tried graphing it and i thought it was g(x) - f(x) and f(x) - g(x)

[placid rover]

.......

graph it by hand

dont need a computer

factorise that cubic.

[civic lion]

so just x^2 * x?

[placid rover]

wut.

???

[civic lion]

by cubic, youre referring to x^3?

[placid rover]

thats the cubic term

in that expression

the entire expression inside modulus is a 'cubic'

[civic lion]

oh ok, i see

so x(x+3)(x-3)?

[placid rover]

yes.

now you can sketch.

without desmos.

[placid rover]

sketch without modulus sign first. then think about it after.

[civic lion]

ok, give me a second

now i have a sketch of the graph

[placid rover]

with the modulus?

then consider how to do the integral.

[civic lion]

would f1(x) be 9x - x^3 from -3 to 0 and f2(x) be x^3 - 9x from 0 to 3?

[placid rover]

idk - if u show me your pic i can immediately tell 😄

[civic lion]

uhh so is my answer correct?

[placid rover]

idk

let me see

yh ok

so in the original thing without the modulus

the graph between -3 and 0

is positive

so that is unchanged

between 0 and 3 is negative

I think your answer is wrong if I'm not misunderstanding you

[civic lion]

welp i just tried this and it was wrong

[placid rover]

Think carefully about what your graph means

[civic lion]

so from -3 to 0, would the function be f1(x) = x^3 - 9x

and from 0 to 3, f2(x) = 9x - x^3

[placid rover]

agreed.

just look at your graph to be sure

[civic lion]

ok, ill try that then

[placid rover]

if you sketch the mod and no mod version on same graph

its fairly obvious

[civic lion]

ok, sweet

i have one more question if thats ok

[celest quail]

So first you need to find where it intersects, its also helpful to sketch it so you can tell what bounds you integrate between

[civic lion]

so i need the points where y = 3x interesects with y = 2/x + 1 and where y=6x intersects with y = 2/x + 1

right?

[placid rover]

the points of intersection are probably important.

[celest quail]

yes

as those points, and the origin will form a three sided polygon

[civic lion]

ok...

so i have the pois

[celest quail]

If you've got the points, it should be clear enough what 3 areas you'll be intergrating to get?

[civic lion]

i think so?

[celest quail]

theres 2 parts to this, a few are negative and a few are positive, so your gonna have a sum of integrals

[civic lion]

right

but i have 4 pois

[celest quail]

and if you have just algebra instead of nasty words, you should be able to solve?

do you mind showing me your working out so far?

[civic lion]

sure one sec

Would it be something like this?

[celest quail]

oh i seeeeeeee

you forgot a part of your question, only in the first quadrant

so you can just yoink 2 of your points and then write out your equatios

[civic lion]

thats the top right, right?

[celest quail]

yeah, so both x and y have to be positive

remember that O (the origin (0,0) ) is one of your points as y=3x and y=6x intersect at it

[civic lion]

oh ok so i only need (1,3) and (0.667,4)

[celest quail]

yes, leave it as a fraction though

[civic lion]

ok, so just 2/3

[celest quail]

yes

and you'll have 1 integral for each equation

[civic lion]

so i integrate both 6x and 3x from 1 to 2/3 and add them together?

[celest quail]

no

have you sketched it?

[civic lion]

yeah, i just used desmos

[celest quail]

Well my similar sketch shows this:

remember that you need to know how to sketch them in the future

[civic lion]

oh really? my teacher never said that

[celest quail]

so, the top line (y=6x) you need to integrate between the points it intersects

[civic lion]

but yeah mine looks like that

[celest quail]

What course are you studing for?

same with the other 2 curves

[civic lion]

so i integrate 2/3 to 1 for y = 6x

?

[celest quail]

0 to 2/3

2/3 to 1 is for the green line

[civic lion]

ohh mb

[celest quail]

and 0 to 1 is for red

[civic lion]

ok so 0 to 2/3 for y = 6x and 2/3 to 1 for y = 3x

and add them together?

[celest quail]

no, 0 to 1 for y = 3x

and 2/3 to 1 for the y = 2/x + 1

and you add the ones you want area underneat, and subtract above, so add the y=6x and y=2/x+1 and subtract y=3x

[civic lion]

Ok so just making sure

0 to 2/3 for 6x

0 to 1 for 3x

2/3 to 1 for 2/x + 1

[celest quail]

yes

[civic lion]

and add the first and last and minus the middle

[celest quail]

correct

[civic lion]

ok, lol sorry for the stupid questions

i got 0.97759

gonna see if its right

[celest quail]

Uhh good luck!

[civic lion]

ay, its correct!

[celest quail]

Woop woop

[civic lion]

im gonna have to double check the bounds to better understand it but thanks alot for the help

[celest quail]

No poblem, you can !close if you are done to let it open up the channel for someone else

[civic lion]

.close

[devout snow]

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[languid glen]

What does it mean if the solution is one dimensional flat in R^4

[languid glen]

I can’t tell wether this means there’s 1 or 3 free variables

[spice flare]

idk but one dimensional makes it sounds like 1

[placid rover]

needs context.

sounds vague.

Perhaps original language will make it clear?

[languid glen]

What do u mean?

English? Lol

[placid rover]

'one dimensional flat' just sounds off

[languid glen]

Honestly I have no idea that’s what it says lol

[placid rover]

show

original thing

[languid glen]

Ugh it’s not letting me attach a picture

But it’s just a matrix with unknowns a and b and it says to find the values of a and b where the set of solutions of the system is a) one dimensional flat in R^4 and b) two dimensional flat in R^4

[placid rover]

god.

im guessing one dimensional flat

means a line

[languid glen]

Been tryna do this shit for a couple hours now

[languid glen]

Yeah I think so aswell

Otherwise i can’t find a solution

[placid rover]

flat just refers to the line not being able to wiggle

maybe

[languid glen]

Do u know how to find a Cartesian description of a hyper plane

Literally last question of homework and I’m stuck

[placid rover]

u seen plane equation in 3d right

sure its not just an extension of that?

hyper plane being 2d or 3d? idk which u mean

[languid glen]

R^6

[placid rover]

ok so ur in r^6

and u want the thing of dimension 5?

or?

[languid glen]

I think a hyper plane is n-1 ?

So 5 yeah

[placid rover]

concept of dot product

for plane equation in 3d

should still apply

[languid glen]

I get that

But it also needs to contain $S\subset of R^6$ where $R^6$ is defined as x1+x6=2 and x2=8

[woven radish]

AK78778888

[languid glen]

We can get a position vector (2,8,0,0,0,0) but then what do we do

[placid rover]

🤔

[languid glen]

a hyper plane in R^n is described by the Cartesian equation of the form
<x-p,n>=0

Literally only thing it says in the teacher notes

[placid rover]

yh its the same idea as 3d plane

r.n = r_0.n

[placid rover]

what is this

[placid rover]

also do you mean 'where S is defined as ...'

[languid glen]

Yes

[languid glen]

Honestly idk what I’m even saying anymore

Brain is fried

[placid rover]

i mean you solve the system of equations

you can do that right for those 2 equations?

[languid glen]

I mean yeah?

I mean yeah?

[placid rover]

$$\begin{pmatrix}1&0&0&0&0&1\0&1&0&0&0&0\end{pmatrix}\begin{pmatrix}x_1\x_2\x_3\x_4\x_5\x_6\end{pmatrix}=\begin{pmatrix}2\8\end{pmatrix}$$

[languid glen]

X1=2-x6

[woven radish]

Shuri2060

[languid glen]

X2=8

And we’d have x3 x4 x5 and x6 as free parameters?

[placid rover]

yh

been too long since ive done this

[languid glen]

Ok then what do we do

Would we need to get it in vector form then change it to Cartesian?

[placid rover]

no, keep things in vector form

thats what i always do

[languid glen]

Question asks for Cartesian description tho

[placid rover]

u can do that at the very end

[languid glen]

Ok what do we do before it

[placid rover]

$$\begin{pmatrix}x_1\x_2\x_3\x_4\x_5\x_6\end{pmatrix}=\begin{pmatrix}2\8\0\0\0\0\end{pmatrix}+a\begin{pmatrix}1\0\0\0\0\-1\end{pmatrix}+b\begin{pmatrix}0\0\1\0\0\0\end{pmatrix}+c\begin{pmatrix}0\0\0\1\0\0\end{pmatrix}+d\begin{pmatrix}0\0\0\0\1\0\end{pmatrix}$$

x1 + x6 = 2

x2 = 8

[languid glen]

What’s with the question marks

[placid rover]

me filling it in

[languid glen]

Should they just all be different variables?

Wait are you trying to write it in vector form?

[placid rover]

yh

just a pain with latex

still filling it in

[languid glen]

I get how to do that

What would we do next

[woven radish]

Shuri2060

[placid rover]

Hope I got this right???

[languid glen]

Shouldn’t the -1 and 1 be switched

[placid rover]

a -> -a

doesnt make a dif?

[languid glen]

Idk

[placid rover]

you can set

a = -e

as a new param and gives the same thing

Anyways, this represents a 4 dimensional thing

that is offset by 280000

And we wanted a 5d thing

ok, that sounds unique

[placid rover]

I can make this 5d by putting a new free variable

in front of the offset

can you see why?

It satisfies the original equation and goes thru the origin

I can't rlly think unless things are in vector form

[languid glen]

What do u mean infront if the offset

As in infront if 280000

[placid rover]

yes. that is like

y = mx + c

the c

it offsets from the origin

shifting the graph up c

same concept.

$$\begin{pmatrix}x_1\x_2\x_3\x_4\x_5\x_6\end{pmatrix}=a_1\begin{pmatrix}2\8\0\0\0\0\end{pmatrix}+a_2\begin{pmatrix}1\0\0\0\0\-1\end{pmatrix}+a_3\begin{pmatrix}0\0\1\0\0\0\end{pmatrix}+a_4\begin{pmatrix}0\0\0\1\0\0\end{pmatrix}+a_5\begin{pmatrix}0\0\0\0\1\0\end{pmatrix}$$

$$\begin{pmatrix}x_1\x_2\x_3\x_4\x_5\x_6\end{pmatrix}=\begin{pmatrix}2\8\0\0\0\0\end{pmatrix}+a\begin{pmatrix}1\0\0\0\0\-1\end{pmatrix}+b\begin{pmatrix}0\0\1\0\0\0\end{pmatrix}+c\begin{pmatrix}0\0\0\1\0\0\end{pmatrix}+d\begin{pmatrix}0\0\0\0\1\0\end{pmatrix}$$

[woven radish]

Shuri2060

Shuri2060

[placid rover]

Can you see how this new thing

[placid rover]

contains S

and goes thru the origin

Hence thats the vector equation of our new 5d thing we need

[languid glen]

Ok

And now to get into Cartesian form

[placid rover]

theres a few ways

find a new vector perpendicular to all 5 of those direction vectors

and converting is easy

or just write out the system

[languid glen]

Write it out as in

[placid rover]

u have 6 equations

this is parametric form

[languid glen]

X1=

X2=

[placid rover]

yh

[languid glen]

Etc

[placid rover]

this is parametric form

uhhh how do u go to cartesian... think u solve or something

[languid glen]

So in this case x1=2a1+a2

I think leaving it like this is fine idk

[placid rover]

actually

cartesian form aint hard

[placid rover]

notice x3, x4, x5 are free

they won't appear in the final equation, there's no restriction on these coords

its like me saying y = 3 in 2d. x does not appear

youll only related x1 x2 x6

x1 = 2a + b
x2 = 8a
x6= -2b

2x1 = 4a + 2b

2x1 = 4a - x6

4x1 = x2 - 2x6

pretty sure its this but yh

[languid glen]

Shouldn’t x1 = 2a-b if x6 is =b

[placid rover]

wheres the error

reply to the line

[languid glen]

This?

[placid rover]

wut

[languid glen]

Originally we had x1+x6=2

So x1=2-x6

[placid rover]

this original equation

applies to S

anyways theres a typo there

x6 = -b

x1 = 2a + b
x2 = 8a
x6= -b

[languid glen]

Ok that makes more sense

Thanks

Time to sleep

.close

[devout snow]

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[merry arch]

please give hint what to do in the left side

[merry arch]

i did some manipulation in the right side

[languid glen]

Are you tryna find the value of theta

[placid rover]

if u assume this thing is true

[languid glen]

Or are you showing the 2 equations are equal

[placid rover]

and get rid of the denominators

i think that shows you how to do the working (then reverse the process)

[merry arch]

this yeah

[placid rover]

cus bottom left times top right is diff 2 squares

and that looks nice

[merry arch]

should i only get rid of the bottom of the left or both?

[placid rover]

just try my hint and i think ull see it

doesnt matter.

[merry arch]

ok thanks

.close

[devout snow]

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[high gulch]

is the range of this [0, infinity)U[3]U(2, infinity)

[winter patrol]

that can be heavily simplified

[languid ocean]

the function is also defined for the negatives....

while your interval (simplified) is just [0,infinity)

[high gulch]

how do you simplify if @languid ocean

[languid ocean]

[3] is in [0,infinity)
same with (2,infinity). it's a subset of [0,infinity).

[high gulch]

ohhhhh

ok thanks

wait so the domain

(-infinity, 1] U (1, 2] U (2, infinity)

so that turns into

(-infinity, infinity)

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[normal zephyr]

can someone explain this to me? I don't really understand

[normal zephyr]

this is the explanation the textbook gives but idk what is missing from it

[placid rover]

urgh

I prefer explanations that involve matrices

The row operations can be expressed a matrices multiplied by the original system iirc

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

Multiplying by elementary matrices results in row operations

[placid rover]

What I'd say is missing from this is maybe a representation?

to show how exactly it is true

[normal zephyr]

hmm maybe

but it also says that it omits some details

[placid rover]

How solutions for old system = solutions for new system

[normal zephyr]

true

[placid rover]

yh the words are a bit vague

[normal zephyr]

2 linalg lectures and I'm already mental boomed

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[radiant bronze]

not sure how to do this ive tried subracting the 8 and 4, and ive tried distributing the -4 and -x etc. stuck

[crystal summit]

oooooh i know what you've done here

[radiant bronze]

.-.

godsend

[placid rover]

$8-4(-x+5)=8+(-4)(-x+5)$

[woven radish]

Shuri2060

[placid rover]

$$=8+(-4)(-x)+(-4)(5)$$

[radiant bronze]

then wouldnt the last one be -20?

[woven radish]

Shuri2060

[copper mango]

yes, -4*5 is in fact -20

[placid rover]

Just make sure you know how to distribute

[radiant bronze]

then wheres the 20

in the answers

[placid rover]

whats 8 - 20

[placid rover]

u know theres both an 8 and -20 here. We have 3 terms

[radiant bronze]

uh oh

forgot to combine

have test do today

well thanks though

xD

kk

.cole

.close

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[lucid nymph]

We are given that ab = ba where a and b commute. We are assuming that they commute

[placid rover]

whats the issue

iff

is

$$\iff$$

[woven radish]

Shuri2060

[placid rover]

if and only if

[lucid nymph]

Yes

[copper mango]

Yeah, tbh you've not really said much of anything meaningful. You repeated the fact a and b commute twice

[placid rover]

What are you stuck on

note that in basic group theory stuff, like showing commutativity, you should be justifying every single manipulation

Otherwise all of these can be solved in 1 line of working

[lucid nymph]

I was just confirming that we are supposed to assume that it's commute

[copper mango]

you assume that for the forward direction

[lucid nymph]

I didn't get much from the problem, like what's asking

[copper mango]

you prove that for the converse

[placid rover]

lets start with

$$ab = ba$$

[woven radish]

Shuri2060

[placid rover]

Whats the first thing you do ?

[lucid nymph]

I'm not sure

[placid rover]

The first thing to do is right multiply by (a^-1)(b^-1)

ok?

$$ab = ba$$
$\implies$
$$(ab)(a^{-1}b^{-1}) = (ba)(a^{-1}b^{-1})$$

[lucid nymph]

Like (ab)(a^-1b^-1)?

[placid rover]

That's the 1st step in the proof

[lucid nymph]

Ah

[woven radish]

Shuri2060

[placid rover]

then?

if u can see what they need u to do then 👌

[lucid nymph]

So when I get beb^-1, it's technically saying bb^-1?

[placid rover]

well

i would use brackets

and show the steps involving associativity

eb = b by the identity axiom