help-39

[signal atlas]

The quadratic formula tells you what b and c are, but nothing about a

[brisk lichen]

ahh thats what you meant by correct scaling

^

[signal atlas]

Yeah

[pearl pond]

@brisk lichen Has your question been resolved?

[pearl pond]

@brisk lichen Has your question been resolved?

[pearl pond]

@brisk lichen Has your question been resolved?

[pearl pond]

@brisk lichen Has your question been resolved?

[brisk lichen]

I got it! Thank you so much for the insight. Also appreciated the fact you didnt just gimme straight answer(otherwise id just be like " oh cool" and move on without knowing fully how it works.) , instead patiently guiding me.

After finally finding the time and motive to the tedious algebra, i solved it without a hatch and matched the original equation too. Funny earlier i actually solved it already, i got "12". But i wasnt thinking clearly, when inputed to desmos it didnt match the og equation, so i gave up and and started to procrastinate without knowing the answer was just its reciprocal LOL. So glad the scale is just simple number. I was fearing that the expression is gonna become nasty and require me to do long division with the polynomials just to find the scale factor.

But yeah i got the answer, its scale factor is 1/12, though only for this form. Turns out if i factor out the 6 from each of i "adjusted" factor form, and put together with 1/12:

$\frac{6^2}{12}=\frac{36}{12}=3$

[jolly parrot]

jj

[brisk lichen]

So for this form, the scale factor is actually 3 exactly. Same as what the original simplified equation, also 3. I have another question, is it really required doing all this works just to find the scale factor? Couldnt I just take the og equation 3, and call it a day? Or was it just happens to be a 3 and doesnt work all the time? Or there is therem/ method to find the scale factor quickly?

[signal atlas]

The quickest way is to do what you did at first: find the roots, call them a and b, and write it as (x-a)(x-b). Then just multiply this by the leading coefficient of the original standard form polynomial

So basically you can find the scale factor by just looking at the leading coefficient

[brisk lichen]

so its just *Couldnt I just take the og equation 3, and call it a day? *

the lead coeffi of this which is my simplified one

[signal atlas]

Yeah

[brisk lichen]

i see

[brisk lichen]

thank you so much once again, ig its still a good algebra practice :D

[brisk lichen]

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[pearl pond]

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[topaz mulch]

Why is it 10x0.1x9.8 instead of 8x0.1x9.8?

[uneven smelt]

u are talking about moment of A on X?

what is the distance from A to X?

[topaz mulch]

A to X

[uneven smelt]

r u fr

[topaz mulch]

I just dont understand where 10(0.1g) comes from

[slate pond]

A to B + B to X = 8 + 2 = 10

[pearl pond]

@topaz mulch Has your question been resolved?

[pearl pond]

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[mighty basalt]

I wrote something random in my notes but I didnt label it. Do we use integration when acceleration is not constant basically?

[mighty basalt]

like here?https://cdn.discordapp.com/attachments/903430333888884736/1286800862605349027/Integration_Example.png?ex=66ef3a31&is=66ede8b1&hm=7349927f634a207424967e43ee9c01a3e159a384179a4a6e991676473c20fcfb&

[spiral pivot]

@mighty basalt you can use integration even when acceleration is constant

Integration always works

[mighty basalt]

Hm, alright then. Thought I wrote it after my teacher said it during this.

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[cyan bronze]

can anyone please verify whether delta = min{1,epsilon/7} is a valid delta as a function of epsilon to say that yes the limit exists

[pearl pond]

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[tough viper]

I'm not sure how to do 9b

[merry carbon]

Expecting you’re e.g. to write that as something like Rsin(x + a)?

[tough viper]

Apparently I'm supposed to solve for x

I did try converting to r formula but I'm still kinda stuck

[merry carbon]

Did you manage to convert it, or did you struggle with doing so?

[tough viper]

i did but well i belive i did it wrongly

the prob well a problm i stuffle with is do i convert to R sin(a+b) or Rcos(a-b)

since its + in the center it shld only be thes 2 scenarios right

[merry carbon]

Yep those are the ones you’ll probably want

[tough viper]

ohh

thing is tho

[steep oar]

https://discord.com/channels/268882317391429632/1287031114740928512

[tough viper]

what do i do with the 1.2 when converting

[steep oar]

help please

[merry carbon]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[merry carbon]

And just leave it until you converted that left hand side

[tough viper]

hmm okay

lemme try it

[merry carbon]

Then you’d have e.g. Rsin(x + a) = 1.2 and you can solve that one

[tough viper]

Pardon my handwriting haha

But here I can't use 3cosx to simul eqn and solve cause it's cos alpha

Wait erm wat.

or do i use cos(x+alpha here)

[merry carbon]

Well, the idea is that you’d find alpha, which you can do by e.g. squaring coefficients of sin(x) and cos(x) with the original, say sqrt{13} * cos(a) = 2 and sqrt{13} * sin(a) = 3 for coefficients of sin(x) and cos(x), respectively

[tough viper]

Ermmm not suree I rlly understand here I tried using cos instead and I cld change it to r formula but when I use cos-1(1.2) it's math error forgot cos graph doesn't go above one Soo ermm not sure how to continue

,rotate

[tough viper]

Ignore the sin part yes

[merry carbon]

If you’ve found alpha, you’re solving sqrt{13} * cos(x - a) = 1.2, or sqrt{13} * sin(x + a) = 1.2 (for the a you found)

Both of those should be easier to solve

[merry carbon]

E.g. that + should be an = as those are the same expression

[tough viper]

OHHHH i forogot abt the r mb...

[tough viper]

ahhh i seeee

ohh done im dont tks a lot!

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[jolly parrot]

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

[pearl pond]

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[midnight haven]

rip

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[silver sonnet]

Could anyone look over my work and tell me what I'm doing wrong? Here is my work;

[silver sonnet]

and the question;

nvm I was able to figure it out!

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[chrome niche]

how to solve the geodesic equation numerically?

[pearl pond]

@chrome niche Has your question been resolved?

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[civic ledge]

How do you solve this

[rough forge]

Do you know chain rule?

Say you have f(g(x)), how would calculate the derivative of f(g(x))?

[jolly parrot]

bacc

[rough forge]

(f(g(x))' = f'(g(x)) * g'(x)

[civic ledge]

f`(g(x)) * g'(x)

[rough forge]

yea

So you need

(f(g(4))' = f'(g(4)) * g'(4)

what is g'(4)

and g(4)

[civic ledge]

2 and 3

[rough forge]

yes

[civic ledge]

G prime of 4 is 3 g of 4 is 2

[rough forge]

(f(g(4))' = f'(2) * 3

what is f'(2)?

[civic ledge]

5

[rough forge]

(f(g(4))' = 5 * 3

what is result

[civic ledge]

15

[rough forge]

yes

[civic ledge]

That's weird wait

Oh basically

Just get the value of that

And get it again

Wait does this even make sense on a real function

[rough forge]

You wanna try it?

[civic ledge]

Yeah

[rough forge]

Say f(x) = x² and g(x) = ln(x)

f(g(x)) = ln²(x)

The derivative of that is

2ln(x) * 1/x

Say we want the derivative at x = 1

2ln(1) * 1/1 = 0

[f(g(1))]' = f'(g(1)) * g'(1)

g(1) = 0
g'(1) = 1

[f(g(1))]' = f'(0) * g'(1)

f'(0) = 2(0) = 0
and g'(1) = 1

[f(g(1))]' = 0 * 1 = 0

[civic ledge]

soo

[rough forge]

chain rule works

[civic ledge]

Sec lemme process

OK thanks for the help I see

.close

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[upbeat lily]

hi 🫠

[violet geyser]

hi how are you

[upbeat lily]

hi!! all good on my end

you?

[upbeat lily]

I thought I would just get the derivatives and subtract but that just gives me a constant so I can't do any derivative tests

as in,
P(x) = 440-0.7x - 15000+90x
P'(x) = 89.3

[snow sail]

Are you sure it should be constant @upbeat lily

I mean I would think we'd have some kind of calculation like

Profit = Revenue - Cost

and Revenue is going to be the number of units sold, times the cost of each unit to the consumer

[jolly parrot]

jan Niku

jan Niku

jan Niku

[snow sail]

so we must have to multiply U by Units, if we wish to subtract C from it

[upbeat lily]

and to get U units I have to find the maximum of that specific function no?

[snow sail]

We don't need to do any problem solving here, this is just the initial step

[upbeat lily]

ah okay

[snow sail]

The problem is asking you to maximize a profit function

so we're trying to find that function

[upbeat lily]

mhm

[snow sail]

I think you've proposed that profit is U - C

[upbeat lily]

yes

[snow sail]

but I'm arguing that it should be xU - C instead

[upbeat lily]

yeah ur logic checks out

cuz there's no real revenue yet

[snow sail]

well, its not that

U only tells us how much a single unit costs a consumer to buy

so it doesnt directly tell us how much money we made from selling these

[upbeat lily]

ahhh okay

gotcha

so should I solve for x?

[snow sail]

no

@upbeat lily before you do anything you need a profit function

that means you need to figure out how much money you made, and how much money it cost you

[upbeat lily]

hmmm

so

$p(x)=x(400-0.7x)-15000+90x$

?

[jolly parrot]

Matz

[snow sail]

yea this seems right

[upbeat lily]

$-0.7x^2+490x-15000$

[jolly parrot]

Matz

[snow sail]

so this tells us how much money we make, net, having sold x units

[upbeat lily]

mhm

[snow sail]

whats the problem ask you for?

[upbeat lily]

so I find the derivative of this

[snow sail]

okay

[upbeat lily]

maximum

[snow sail]

maximum profit, right

[upbeat lily]

and then do the first derivative test

[snow sail]

alright, what do you get?

[upbeat lily]

brb 💀

[upbeat lily]

is 70,750 correct?

[snow sail]

,w maximum -0.7x^2+490x-15000

[jolly parrot]

Results provided by WolframAlpha

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

Requested by janniku

[snow sail]

,calc -0.7 * (350)^2 + 490*350 - 15000

[jolly parrot]

Result:

70750

[snow sail]

[upbeat lily]

yay!!!

awesome

thank u so much :)

[snow sail]

np

[upbeat lily]

.close

hello?

why idditn it

lol?

,close

??

weird

[snow sail]

thats a good question

wonder if the bots down

.close

[pearl pond]

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[snow sail]

[upbeat lily]

😭 okay then it just hates me

[pearl pond]

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[rapid wagon]

I'm given matrices A, B, C that belong to R^nxn such that C is invertible and
AB=C-A

[rapid wagon]

How do I know which one of these is correct:

1. 1 is an eigenvalue for A but not for B
2. 0 is not an eigenvalue for A and -1 is not an eigenvalue for B
3. 1 is not an eigenvalue for A and 0 is an eigenvalue for B
4. 0 is not an eigenvalue for A and B
5. -1 is an eigenvalue for A and B

I have no idea how to start with this i just need a clue or a thing to work with

[plush bramble]

Add A to both sides and factor

[rapid wagon]

A(B-I)=C

C is invertible so, A and (B-I) has to be invertible

Oh so 0 can't be eigenvalue for A

For B..?

I don't think (4) is the answer so it only leaves (2) but that means -1 is not an eigenvalue of B but by what I thought it's 1 that's not an eigenvalue

Why is it -1?

[pearl pond]

@rapid wagon Has your question been resolved?

[pearl pond]

@rapid wagon Has your question been resolved?

[pearl pond]

@rapid wagon Has your question been resolved?

[pearl pond]

@rapid wagon Has your question been resolved?

[pearl pond]

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[cedar leaf]

Having some hard time here, can some explain me how to solve this?

[rough forge]

All deserve a nice u-substitution treatment

[cedar leaf]

m gonna try, im going insane with this one

[rough forge]

They are all in a way similar, especially the 1st and 2nd

[cedar leaf]

what abt the 3rd one, makes me feel fear

[rough forge]

Substitute u = sqrt(x)

[cedar leaf]

still having troubles with the first one, 5^ 2 = 10, what i have to do with that value, im new on this so is still confussing 4 me

😔

[rough forge]

5^2=10?

what is going on

[cedar leaf]

wait

5x ^2 = 10 isn't it?

[rough forge]

no?

[cedar leaf]

dang it

idk what i am doing then

[rough forge]

guess judging by that statement you will have a hard time to solve anything

[cedar leaf]

yes

😶

[woven garnet]

Which one are you working on right now

[cedar leaf]

first one

[woven garnet]

What part are you struggling with

[cedar leaf]

everything, ig, i derived 5x^2, that gives me 10x, it's that right?

if it is, then i don't know what to do next

[woven garnet]

So you let 5x^2 = u?

[cedar leaf]

yes

[woven garnet]

and found du=10xdx

[cedar leaf]

i think that i have to do smth with the 3x, but i'm not sure what

[woven garnet]

Next step is find what dx is

[cedar leaf]

yeah

[woven garnet]

And it is?

[cedar leaf]

yes

[woven garnet]

I'm asking you what dx is

[cedar leaf]

😶 i don't really know how to get dx

[woven garnet]

Well you know du=10xdx

So solve for dx

Don't overcomplicate it

[cedar leaf]

5x^2 ?

[woven garnet]

That's u

[woven garnet]

Solve for dx from this

[cedar leaf]

du/10 = dx

?

[woven garnet]

Yes

Now substitute that into the problem for dx

Dont forget the x

du/10x

[cedar leaf]

like this?

[woven garnet]

Yes but make sure you have u for 5x^2

[cedar leaf]

like this then?

[woven garnet]

Yes

[cedar leaf]

i understand, what's next?

[woven garnet]

You see an x and an x

[cedar leaf]

yes

[woven garnet]

Do you see what to do with them?

[cedar leaf]

delete them?

[woven garnet]

Yes they cancel out

$\frac{3}{10}\int e^udu$

[cedar leaf]

like this?

[woven garnet]

Just integrate from the first one

Keep it in terms of u

[cedar leaf]

wdym

[woven garnet]

Not sure how else to say it

[cedar leaf]

ok lemme try

[woven garnet]

Integrate from here

Wait

[jolly parrot]

mari

[cedar leaf]

have to be like this?

[woven garnet]

Where did the 10x come from

[cedar leaf]

from 5x^2, not sure how to proceed

[woven garnet]

It's not right

The integral of e^u is e^u + C

So your final answer is?

[cedar leaf]

this

[woven garnet]

Yes, good job

[cedar leaf]

what with the 5x^3 on the second one? i understand the process itself, but not sure what to do with that

[woven garnet]

It's the same process

Let u = the exponent of e

[cedar leaf]

so du= 48x^3, how do i get dx? i think that the process change here bc there is an exponent

[pearl pond]

@cedar leaf Has your question been resolved?

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[steep pier]

is it side angle side? and if it is, how would i explain it

[north talon]

they have the same angles, since you dont know the side lengths

[steep pier]

so its angle angle

.close

[pearl pond]

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[mild socket]

why dont channels instantly reavailablize

im assuming its bot powered right

[sharp vigil]

discord has rate limits which limit the speed of bot activity, + the server hosting the bot has limited resources

[pearl pond]

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[midnight haven]

So I haven't seen a problem like this. I see that the squares would be a rotation of the x-axis, and the rectangles about the y-axis. I'm a little lost in how I would find the area between the equations

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

.close

[pearl pond]

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[spice condor]

How is the norm defined in C^n?

[hallow cedar]

first define an inner product on C^n

[spice condor]

This may be the wrong channel (if it is please inform me I'll take it down)

[spice condor]

Then?

[hallow cedar]

the norm is defined as sqrt(<v,v>), where <,> is the inner product function

[spice condor]

I forgot about this

.close

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[finite stump]

no idea

[pearl pond]

@finite stump Has your question been resolved?

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[vernal cobalt]

if the second and seventh term of a geometric sequence multiplied= 6, what is the sum and product of the 1st, 3rd, and 4th term

[vernal cobalt]

i don even know where to start w this question

[verbal whale]

It seems to me you don't have enough info

[vernal cobalt]

the only info given that i remember is a2a7 = 6

and it was looking for the sum of a few other terms

[finite stump]

wdym remember ?

take a pic of question

[vernal cobalt]

it was a test question

[finite stump]

oh

[vernal cobalt]

dont have the test paper rn sorry

[finite stump]

oh ur just checking answer

[vernal cobalt]

yea

[finite stump]

bro just dont worry about it focus on other things

im sure u have other tests

[vernal cobalt]

how would u go about solving it

[vernal cobalt]

id skip it normally but a girl id asking me

[finite stump]

oh

not too sure sorry

try asking her what she remembers from teh question

[vernal cobalt]

ight

[pearl pond]

@vernal cobalt Has your question been resolved?

[vernal cobalt]

<@&286206848099549185>

[odd surge]

you don't have the full question

we can't solve it without all the given information

[vernal cobalt]

huh

alr

.close

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[pearl pond]

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[golden crane]

how do u do 251?

[north talon]

1/e^3 = e^-3

[naive orbit]

it is ln(e^-3)

isn't it

[north talon]

what does the question want you to show?

[golden crane]

this was the asnwer

how did u get t ?

[north talon]

1/a^b = a^-b

[modest crater]

raise e by both sides

e^(ln(1/e^3)) = e^-3

[golden crane]

why are r we doping ?

why r we doing that?

[odd surge]

1/e^3 = e^(-3)

[naive orbit]

$ln(\frac{a}{b}) = ln(a)-ln(b)$ right?

[jolly parrot]

MetuMortis

[golden crane]

yes

[odd surge]

just e^ both sides, so the ln cancels

[naive orbit]

$ln(\frac{1}{e^3}) = ln(1)-ln(e^3) = 0 - 3$

[jolly parrot]

MetuMortis

[naive orbit]

is it clear?

[golden crane]

Like this

[naive orbit]

that is also a way

[golden crane]

what would you do after that step?

[sinful nebula]

ln is natural log

of base e

e^(ln (x)) = x

[naive orbit]

$a^{log_b(c)} = c^{log_b(a)}$

[jolly parrot]

MetuMortis

[cosmic garnet]

does this rule have a name?

[sinful nebula]

I don't think it does

[pearl pond]

@golden crane Has your question been resolved?

[pearl pond]

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[pearl pond]

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[shy ginkgo]

(8, 600) and (16, 200) are points on a circle. I want to find the point (12, y) but idk how

[plush bramble]

Two points aren't enough to determine a circle

[shy ginkgo]

oh rip

ok then

sads

.close

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[shut island]

<@&286206848099549185>

[next dove]

!15m

[pearl pond]

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

[midnight haven]

Which question ?

Oh

Lol

Nvm

[shut island]

this

[midnight haven]

Telangana ?

[shut island]

nah

[midnight haven]

Ouch

Do it this way

A can be written as x I

Agreed ?

[shut island]

no

how

[midnight haven]

Why🤨

[shut island]

in I the elements are 0101

right?

[midnight haven]

I is identity matrix

Oh

F

A can be written as x P

P is 1111

[shut island]

ok

[midnight haven]

Agreed now?

[shut island]

then

[midnight haven]

Sorry about that

Add and subtract P

Now Take P common from P+xP+x^2P/2.....

[shut island]

what

[midnight haven]

Let me write it down

[shut island]

alr

[midnight haven]

My method is wrong
F

Sorry

[shut island]

what did u try?

[midnight haven]

Thought I could factor out P
So we can have P*e^x

[shut island]

oh

[shut island]

u from here?

[midnight haven]

Oh ysss

Got it

Want the solution or should I help you by giving hintz

[shut island]

hint

[midnight haven]

Okay

So

2min

I made a mistake again😭

[next dove]

you need some help?

[shut island]

yes

[next dove]

have you got the g(x)?

[shut island]

no

[next dove]

erm okok

well you can factor out the x from A

then try to find A^2, A^3 and notice the pattern they follow (if any)

[midnight haven]

There is

[next dove]

By that I mean keep the x aside for it

[midnight haven]

A^2=2A
A^3=4A

So on

[next dove]

erm, it was supposed to be OP who should answer it

[midnight haven]

A^n=2^n-1 A

Oops

Orry

Sorry

[shut island]

so f(x)= e^x?

[midnight haven]

Ohhhhhh

[shut island]

i got it thx!

[rose robin]

not sure if that's not totally correct
because you have all the 2^(n-1) stuff

[pearl pond]

@shut island Has your question been resolved?

[pearl pond]

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[sharp ridge]

I have a bunch of questions but the first is that in the equation I need to sub f(t) for 2 or do I sub in any number between 0 < t ≤ 2

[ivory plaza]

Is this calc 3?😭

[sharp ridge]

this is review for a pde course lol

[merry carbon]

How do you mean? You would want to split the LT integral at 2, and between 0 and 2, replace f(t) as 2 (and f(t) as 0 when you’re greater than 2)

[hybrid basin]

im about to do these! i just started my differential equations course

Does this help?

[sharp ridge]

no

[sharp ridge]

can you explain in depth please?

are you saying that I need to find the LT of 2 and 0 first?

so like

y"+y=2 and y"+y=0

[merry carbon]

Well, more that you have the LT integral as
[
\int_0^\infty f(t) e^{-st} \dd t
]
where
[
f(t) =
\begin{cases}
2 & 0 < t \leq {\color{green} 2} \
0 & t > {\color{green} 2}
\end{cases}
]
and you can split up any integral like
[
\int_a^c g(x) \dd x = \int_a^{{\color{green} b} } g(x) \dd x + \int_{{\color{green} b}}^c g(x) \dd x
]

[jolly parrot]

@merry carbon

[sharp ridge]

so my first step is?

[merry carbon]

...separate the LT integral, $\int_0^2 f(t) e^{-st} \dd t + \int_2^\infty f(t) e^{-st} \dd t$

[jolly parrot]

@merry carbon

[merry carbon]

for which you know what f is in each of those parts

[sharp ridge]

...

[merry carbon]

...you're telling me you don't realise you replace f(t) with 2 for the first integral, and with 0 for the second one...(?)

[sharp ridge]

ngl I did not realize until told

[merry carbon]

awww

[sharp ridge]

now I integrate as in let me get a calculator to speed up the process

damn all I got was that it diverges lol

[merry carbon]

Did you replace them correctly? The second integral is just zero, the first being just an integral of a continuous function on a closed interval

[sharp ridge]

I did it again and got this

[merry carbon]

Yep, cool, then you just need to LT that y'' + y, using what you were given

[sharp ridge]

so the equation looks like

$s^2Y\left(s\right)-sy\left(0\right)-y'\left(0\right)=-\frac{2\left(e^{-2s}-1\right)}{s}$

[jolly parrot]

arrow891

[sharp ridge]

which means that it really is
$s^2Y\left(s\right)+1+Y(s)=-\frac{2\left(e^{-2s}-1\right)}{s}$

[jolly parrot]

arrow891

[sharp ridge]

right?

[merry carbon]

Don't forget the transform of y itself, you're missing a Y(s) on that left hand side

[sharp ridge]

$s^2Y\left(s\right)+1+Y(s)=-\frac{2\left(e^{-2s}-1\right)}{s}$

[jolly parrot]

arrow891

[sharp ridge]

after I do that do I just solve for Y(s) and then I'm finished?

[pearl pond]

@sharp ridge Has your question been resolved?

[sharp ridge]

.close

[pearl pond]

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[ivory plaza]

Oo ncie! Im taking an ode course rn super fun

[pearl pond]

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[limpid tiger]

what other input could I have used instead of 1/a

oh were they trying to get rid of a

[sacred crest]

Use a different input with what intention?

[limpid tiger]

like to find the interval

cause they used intervals [1/a,0]

i js wanted to know if there couldve bee nanother interval

instead of 1/a

[sacred crest]

I think you can’t choose a narrower interval, as that would mean that h(i)<0 (i being the lower bound of the interval) wouldn’t hold for all a<0

As for choosing a wider interval, I don’t see why that wouldn’t be possible

[limpid tiger]

so 1/a would be the smallest it can go ?

[plucky python]

it should be possible to get a smaller interval

lemme actually try tho

well ok u can certain get better bounds for specific values of a

[plucky python]

well i mean the solution to e^x - ax = 0 is

-W(-1/a)

where W is the lambert W function

so ur basically asking for optimal bounds on the lambert W function

[limpid tiger]

yea i aint got to that part 😭😭😭

my bad

[plucky python]

which exist

[plucky python]

lambert W isn't normally taught cus it's kinda niche

but it's well studied

[plucky python]

but yeah so to answer ur question, ppl have found better bounds

[limpid tiger]

oh maybe il look into it in the future

thanks tho

.close

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[round epoch]

i need help with a) and b)

[round epoch]

I know for b) i put in 1 into the first equatuon and second, and 7 into the second equation and third

bht when i checked the answer i got it wrong

and i have no idea for a)

[wooden merlin]

You got three lines of the function, which line has its domain 7 included?

[west bluff]

A piecewise function is defined along different sets of x values

[round epoch]

the last one

[round epoch]

Yeah but for b) what would i do?

[round epoch]

would i do what i said i should or

idk if im doing it wrong or right

[west bluff]

How's continuity defined?

Like, do you recall the definition of continuity?

[round epoch]

when theres like breaks?

no breaks*

[west bluff]

That's a discontinuity

[round epoch]

yep my bad

[west bluff]

The actual definition

That's what happens when q function's continuous

But not the definition

[round epoch]

something that never breaks and continues forever? or

i dont really remember

[west bluff]

Nope

I'll give ya the definition, if you don't understand it, ask me though

[round epoch]

okok

we havent done lim yet though thats the thing

[west bluff]

Well, that's what's required to actually justify the continuity

Though I'll give ya some clues

[round epoch]

okok

[west bluff]

A function can only be discontinuous when it:
a) Shifts part(as in, goes from one function to another in a piecewise function)
b) it “breaks“, that is, stuff like a division by 0, a negative square root in reals, and stuff like that

[pearl pond]

@round epoch Has your question been resolved?

[pearl pond]

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[distant patrol]

hey, can anyone help with this problem:
Suppose | |vVector| | = 5, | |wVector| | = 9. What are the largest and smallest values possible for | | vVector - wVector | | ?

[distant patrol]

so far

with this formula

i have the upperbound

= 14

i'm trying to find the lower bound / smallest value

[rough forge]

Did you use the reverse triangle inequality?

[distant patrol]

no

i'm going to try it now

[rough forge]

well use it

[distant patrol]

ok

i got 0 >= 14 as final answer

its the same answer again

so is the lower bound and upper boun for this problem both 14?

[distant patrol]

.

[rough forge]

5-9 = -4

abs makes it 4

[distant patrol]

i see

i did try -4 as one of the values and it was wrong

i guess i forgot to absolute it

[rough forge]

yea

[distant patrol]

yup i tried it, its correct

thanks alot for the help

.close

[pearl pond]

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[pearl pond]

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[elder patrol]

how do i solve this?

[elder patrol]

I'm not sure how to solve it and my teacher hasn't taught me a lot about this, especially reading a histogram. please help.

This is the exact directions from my teacher:

For each graph identify the scale of the axes, the clsss width of the histogram, and the SHAPE, CENTER, and SPREAD that we would report. Also, identify the minimum, maximum, range, mode, and # of observations whenever possible. Assume that any unlabeled axes have a scale of 5.

[pearl pond]

@elder patrol Has your question been resolved?

[elder patrol]

<@&286206848099549185>

[pearl pond]

@elder patrol Has your question been resolved?

[foggy parrot]

do you still need help?

[elder patrol]

yes i do

i tried to do the first one , not sure if i did it right

[foggy parrot]

Ok give me like 2 minutes.

Everything checks out. You seem to have it down.

[pearl pond]

@elder patrol Has your question been resolved?

[elder patrol]

thanks!

[pearl pond]

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[slender belfry]

b

[slender belfry]

hello

Linear algebra:

no solution
0 free variables
1 free variable
2 free variables

Someone please just tell me the right answers so i can sleep

[sharp vigil]

no solution: all 3 planes do not intersect in the same place
0 free variables: all 3 planes intersect at a point
1 free variable: all 3 planes intersect in a line
2 free variables: all 3 planes are identical, so they intersect as a plane

[slender belfry]

thank you so much

saved me

[pearl pond]

@slender belfry Has your question been resolved?

[pearl pond]

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[lone comet]

im looking at the explanation the website is giving and I cant make any sense of it

[heady finch]

ok so what's confusing

which part are you struggling with?

[lone comet]

what am I supposed to be applying the general power rule to

[heady finch]

where are you applying the general power rule?

you're using implicit differentiation here

[lone comet]

nowhere right now

is that not what the second step says to do

[heady finch]

sorry I'm not sure where it says that

nvm I see it

not a term im familiar with

it's basically just applying implicit differentiation rules to the equation

[lone comet]

alright thank you

I couldnt find a video explaining it exactly but it lines up after the differentiation

the text book is very bad

.close

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[inner agate]

Anyone know what I did wrong?

Would answer be d?

[pearl pond]

@inner agate Has your question been resolved?

[pearl pond]

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[royal gull]

I haven’t been at school for the whole week and don’t know how to do these for an upcoming quiz so was wondering if somebody could teach or help me

[pseudo oxide]

which ones, exactly?

[royal gull]

I think I know the simplify 1 and 2 but the rest I dont know

[pseudo oxide]

for 3, it's the typical formula for the volume of a cuboid (rectangular prism)

length times width times height

so you just multiply all those together

similar strats for 4

i'll send you a vid really quickly for factorization

[royal gull]

Thank you

[pseudo oxide]

one sec

https://www.youtube.com/watch?v=9MUDAr1PnlE&t=6s&ab_channel=TheOrganicChemistryTutor

[royal gull]

So for 3 x+2 is length and x-4 is width and 2x +5 is height?

[pseudo oxide]

doesn't matter

multiplication is both commutative and associative

so you can multiply them in any order

[royal gull]

Ok

[pseudo oxide]

(commutative means ab = ba, associative means a(bc) = (ab)c)

[royal gull]

X times X is x^2 or 2x

?

[pseudo oxide]

x^2

x + x = 1x + 1x = (1+1)x = 2x

x^2 is x times x

[jolly parrot]

ren

[royal gull]

So the total ok and x^2 times 2x should be x^4

[pseudo oxide]

no...

think again

[royal gull]

Sorry typo

x times x is x^2, x+x is 2x but how do I multiply x^2 and 2x

[jolly parrot]

ren

[royal gull]

So 2 times x^2 makes it 2x^2 and the other x multiplied into that will make it go up one power

[pseudo oxide]

yes

[royal gull]

Thank you

So 3. is 2x^4 -40

[pseudo oxide]

one sec

no

try again

[royal gull]

Oh yeha i see what i did incorrect

x + 2 times x - 4 would be x^2 -2x -8

and x^2 -2x -8 times 2x + 5 is
2x^3 + 5x^2 -4x^2 -10x -16x -40

[pseudo oxide]

hang on lemme check

[royal gull]

Highest I know how to simplify
is 2x^3 + x^2 -26x -40

[pseudo oxide]

yeah

that's correct

[royal gull]

Can i simplify any more?

[pseudo oxide]

always simplify it as much as you can

[pseudo oxide]

no

[royal gull]

Ok so thats 3

the perimeter would be 2(3x^2 + 2x -1)

[pseudo oxide]

you mean

okay no nvm

yeah, that's right gj

[royal gull]

Ok and when I multiply a squared by 2 its cubed

Right

so Perimeter would be (6x^3 + 4x^2 -2)

Wait

Nvm

It only goes up if I multiply by x

So it would be 6x^2 + 4x - 2

and the area would be
4x^3 -18x^2 +8x -30

Ok for factor

This is grouping factoring right

Or

Nevermind its trinomials

5. (x+2)(x+7)

6. (x-3)(2x+5)

Does it matter which order I put the factored?

<@&286206848099549185>

[frank plinth]

no

[royal gull]

Ok thanks

I dont understand the last question though

[pearl pond]

@royal gull Has your question been resolved?

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[pearl pond]

Please don't occupy multiple help channels.

[midnight haven]

OH

[sterile turtle]

oh wait

[midnight haven]

I was doing sin

[sterile turtle]

sine rule

[midnight haven]

yes

so the equation is

1/2 x 5 x 13 x sin(49) ?

[sterile turtle]

yeye

[midnight haven]

theres more wait

so

60=1/2 x 15 x ? x sin(30degrees)

Its 16

[sterile turtle]

i feel like you got this

[midnight haven]

Ok ready

area = 1/2 x 16 x 11 x sin(14.4)

[sterile turtle]

ye

[midnight haven]

84.98

Oh

what

its wronnng

nnvm i tgyped it wrong

its 21.88

[pearl pond]

@midnight haven Has your question been resolved?

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[pearl pond]

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[pulsar stump]

here (e1, e2) is a standard base of R^2

why are these the coordinates of Te1?

[pulsar stump]

i get for the top (x axis) that it's tan alpha

oh it's not tan alpha

[plush bramble]

What are the coordinates of the unit circle when the angle is alpha?

[pulsar stump]

but i dont see how he got there

[pulsar stump]

idk

that was Te2

he was lazy to draw the vector

[plush bramble]

No

T(e1) rotates e1 an angle of alpha counterclockwise

[shell socket]

who want voice

?

[plush bramble]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[pulsar stump]

yes

that doesnt give me the actual formula he used

[plush bramble]

It should if you've learned the unit circle

https://math.libretexts.org/Bookshelves/Algebra/Algebra_and_Trigonometry_1e_(OpenStax)/07%3A_The_Unit_Circle_-_Sine_and_Cosine_Functions/7.03%3A_Unit_Circle

[pulsar stump]

ty

.close

[pearl pond]

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[pearl pond]

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[visual path]

Here, it states two ponds are related~
Does that imply that the relation consists of two distinct ponds in which case there would be no occurrence of (a,a) in it?

[pearl pond]

@visual path Has your question been resolved?

[pearl pond]

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[visual path]

.reopen

[pearl pond]

✅

@visual path Has your question been resolved?

[pearl pond]

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[pulsar stump]

if we are asked to find this, does it mean that T is a linear transformation neccsearily?

[pulsar stump]

or could it be just a "transformation"

[spiral pivot]

A linear transformation is a transformation that is a linear.

Linearity is the property that the object in question is transparent to scalar multiplication and addition. In other words, if we have a linear function f(x) then f(ax + by) = af(x) + bf(y) (assuming that a and b are scalars, and x and y are the correct type for the function. For a linear transformation, this means x and y are vectors.)

@pulsar stump

[pulsar stump]

ik what it is, im asking if here T is a LT or not necessarily

[spiral pivot]

I have no idea

$[T]_E^E$

[jolly parrot]

OmnipotentEntity

[spiral pivot]

I'm assuming this is the transformation written in the basis given by the basis set E on some space

but I don't have nearly enough information or context to decide anything about T whatsoever.

I will say that if T has a matrix representation, which this seems to imply, then it's a linear transformation

But I don't know that from a fact based on what is written.

@pulsar stump

[pearl pond]

@pulsar stump Has your question been resolved?

[pulsar stump]

yeah that's the definition of $[T]_E^E$

[jolly parrot]

Ayanokoji (ALWAYS PING ME)

[spiral pivot]

So I guessed correctly?

[pulsar stump]

i was almost sure it is a LT

[spiral pivot]

In that case, yes it's a linear transformation

[pulsar stump]

yeah

[pulsar stump]

alright great, ty mate

.close

[pearl pond]

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[stiff falcon]

im trying to simplify this and dont know how to get the a to the top of the fraction

[rough forge]

You can get a to the top by bringing the denominator together as one fraction

[fast geyser]

you mean that you want to cancel the fraction?

[stiff falcon]

yea

[stiff falcon]

i think this helps so like the right term on the demonimator(the 1) becomes a^2/a^2 which then the bottom term are both /a^2 so it can be brought to the top?

[rough forge]

yes

[jolly parrot]

bacc

[stiff falcon]

okay, thanks! that helps a ton

.close

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[woven hill]

Can someone help me with this please

[pearl pond]

@woven hill Has your question been resolved?

[wooden folio]

Can you translate the language please?

@woven hill

Also what question specifically are we looking at?

Or is it a grouped amount

Are you still stuck on this problem?

Sorry for the nagging

[woven hill]

oh thank u for help but i already solved it

but thank you

[wooden folio]

Sounds good! Just use .close to free up this channel

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@woven hill Has your question been resolved?

[pearl pond]

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[rose dagger]

with part c
ik how to get AB but how do I get CB

[wet swallow]

C is the midpoint

So use the ratio formula(in this case midpoint)

[rose dagger]

.close

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[boreal dawn]

Hi, i've been trying to solve this problem for a while.
I dont know what to do, im stuck.
I need to calculate the area of the black part of the square.

I know the square has an Area of a^2
The longitude of the oval-looking part of the black figure is a/2

I just wanna know how to calculate the black area, I have no idea how and I need to study this 😭

Ive tried calculating a part of it using pithagorean theorem but it took me nonewhere.

[wicked mortar]

There's two ways you can do this
Either calculate the areas of all the components or take a shortcut
(the shortcut would involve finding what other area has an equivalent area to the 'oval-looking part')

[boreal dawn]

Yes

I was trying to go by the 1st way but I get stuck on calculating the area

Of the highlighted part

[pearl pond]

@boreal dawn Has your question been resolved?

[boreal dawn]

I can wait

: )

<@&286206848099549185>

[pearl pond]

@boreal dawn Has your question been resolved?

[boreal dawn]

😭😭😭

<@&286206848099549185>

[boreal dawn]

Its been almost 2 hours.. : (

[pearl pond]

@boreal dawn Has your question been resolved?

[sharp sail]

Hey

I’ll help

Step 1: Area of the Square

The area of the entire square is given by:
area of a square = a^2

Step 2: Area of Circular Sectors

Looking at the diagram, it appears that two circular sectors are formed inside the square. Each sector has a radius of a/2 (since the square’s side length is divided in half).

The area of one circular sector (which is a quarter-circle) is given by the formula for the area of a sector: area of one sector 1/4 pi r^2
Here, r = a/2, so:
area of one sector = 1/4 pi (a/2)^2 = pi a^2/16
Since there are two such sectors, the total area of the sectors is:
2x pi a ^2/16 = pi a^2/8

Step 3: Area of the Shaded Region

The area of the shaded region is the area of the square minus the total area of the circular sectors: area of shaded region : a^2 - pi a ^2/ 8
Step 4: Factoring the Expression

We can factor out a^2 from the equation to simplify: area of shaded region: a^2 (1-pi/8)

Step 5: Approximation

We know that \pi \approx 3.14, so we can substitute this value into the equation to get a numerical approximation:

Step 5: Approximation

We know that pi is approx 3.14, so we can substitute this value into the equation to get a numerical approximation:

[boreal dawn]

Ohhh now i got it

Tysmmm

.close

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[gusty prism]

sry

[gusty prism]

what is ma

[gusty prism]

and why does ma just equal ma here lol

[finite jolt]

m is mass

[gusty prism]

what determines ma equal to ma here

oh

[finite jolt]

a is acceleration

[gusty prism]

because mass is on the y?

mg is weight?

or idek

oh mass x acceleration got it

[finite jolt]

Mass is just the mass of the object

[gusty prism]

Oh

[finite jolt]

the force of gravity aka weight is equal to mg

[gusty prism]

Ok perfect

thnx

[pearl pond]

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[kindred geyser]

Hello

[kindred geyser]

Idk how the answer is b)

[pearl pond]

@kindred geyser Has your question been resolved?

[kindred geyser]

<@&286206848099549185>

[pearl pond]

@kindred geyser Has your question been resolved?

[acoustic path]

use column operations to subtract columns form the 3rd column in both matrices

[kindred geyser]

it will become a 2 x 2 ?

[pearl pond]

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[pearl pond]

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[sharp smelt]

Only interested in the lcm part

[north talon]

maybe express a and b as a product of primes, p_1 p_2 and so on each to some power

[sharp smelt]

No prime factorisation allowed

[north talon]

oh

[tardy reef]

are you able to do the gcd part?

[sharp smelt]

We haven't "done" gcd in calss yet, and I'm revising for a quiz,so not yet

[tardy reef]

Ohh ok

[sharp smelt]

so to start

[glass meadow]

What are the multiples of ka, and what are the multiples of kb?

[sharp smelt]

wait

[tardy reef]

you can start by checking if ka | k * lcm(a,b)

[sharp smelt]

yeah, I know

$a \mid lcm(a,b) \implies ka \mid k lcm(a,b)$

[tardy reef]

only \gcd is there no \lcm

[jolly parrot]

Veni, vidi, perii

[sharp smelt]

and similarly $b \mid lcm(a,b) \implies kb \mid k lcm(a,b)$

[jolly parrot]

Veni, vidi, perii

[sharp smelt]

and we also know that $ka \mid lcm (ka,kb) $ and $kb \mid lcm (ka, kb)$

[jolly parrot]

Veni, vidi, perii

[sharp smelt]

now what

[rustic gate]

what's your definition of lcm

[sharp smelt]

The smallest natural number that's both a multiple of a and b

[sharp smelt]

so from this we can conclude $lcm(ka,kb) \geq k lcm(a,b)$

[rustic gate]

no?

[sharp smelt]

oops

one minute

[jolly parrot]

Veni, vidi, perii

[rustic gate]

still no

[sharp smelt]

[rustic gate]

The smallest natural number that's both a multiple of a and b
lcm(ka, kb) is the smallest number both ka and kb divide

you've shown that ka and kb both divide klcm(a, b)

[sharp smelt]

yes

so $klcm(a,b) \geq lcm(ka,kb)$

[jolly parrot]

Veni, vidi, perii

[rustic gate]

that's the right direction

you still need to show the other direction

[sharp smelt]

yeah

[rustic gate]

you had it the wrong way round