help-33

[tacit elm]

tysm

.close

[marsh citrus]

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[humble river]

Hi

[humble river]

I have the work sending but i got it wrong

I felt like it went smoothly but 😭

[humble river]

apparently the answer is wavelength is equal to

122nm

[desert dirge]

E=hc/l l=hc/E

,calc [(6.626 * 10^(-34)) (2.998 * 10^8)]/[ 1.633 * 10 ^-18 ]

[elfin berry]

Result:

1.21645731782e-7

[humble river]

uh

[desert dirge]

,calc (1.633 * 10^-18)/(6.626 * 10 ^ -34)

[elfin berry]

Result:

2.4645336552973e+15

[desert dirge]

wait how did you get ^23

also the way you write stuff is confusing

[humble river]

idk i use the formula sheet and it seems different

gave me 1.216 x 10^23

[desert dirge]

why do you write if a=bc then a/c = b c/c, you can just go straight to a/c = b you know, will make things a lot simpler

[humble river]

maybe i wrote it bad let me see

o

[desert dirge]

,calc (2.998 * 10^8)/( 2.465 * 10^15)

[elfin berry]

Result:

1.2162271805274e-7

[desert dirge]

you must have just made a calculation error

[humble river]

what the

[desert dirge]

10^8/10^15 in no way is going to give you 10^23

[humble river]

do i have to do it like a fraction dviison

[desert dirge]

you need brackets

[humble river]

O

[desert dirge]

you wrote a * b / c * d
its seeing it as a * (b/c) * d

[humble river]

Ok perfect

1.216 x 10^-7

multiply by 10^9 and i got 121.6 nm

Wooho

[lone heart]

Fyi, ti-84 has a button that does x 10^ so you can use that instead

If you press 2nd, "," (button above 7), it's the blue EE

That's the notation for x10^

[marsh citrus]

@humble river Has your question been resolved?

[humble river]

O

thnx ok

[marsh citrus]

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[glass quiver]

I just had a question about derivatives and critical points.
If I have a graph of f'(x) then I know when f'(x) = 0, so where the graph crosses the x-axis are critical points for f(x), but when the graph has a vertical tangent or there is a sharp corner, would that also be considered where a critical point is for f(x)?

[vernal mantle]

No

[split elm]

it really depends on what you define as "critical point". most of the time in calc class they will refer to critical points as when the slope is equal to 0, but you can just as easily argue that when you have a vertical tangent that its also a critical point.

however for the sharp corner, the derivative would not be continuous, which means its not differentiable at that point

[vernal mantle]

Isn't this a graph of f'

And not f

[glass quiver]

its f'

[split elm]

yeah this is a graph of f'

oh my bad

[glass quiver]

and i know like a crtical point is when f'= 0 or f' is undefined

[vernal mantle]

Just clarify ig

[vernal mantle]

But in this case f' is defined

it's only some parts where f'', the second derivative is not defined

[glass quiver]

so since its defined, then it isn't a critical point?

[vernal mantle]

Yes

[glass quiver]

it would be a possibly inflection point instead of second deriv?

but that would be for a corner, but the vertical tangent would be considered a critical point since its still undefined?

[vernal mantle]

No the derivative is undefined when there's a sharp corner in f not f'

[glass quiver]

what about the vertical tangent?

[vernal mantle]

Tbh it might depend

If there's a vertical tangent at a certain point in f'(x) then f'(x) might not be defined at that point

Because like there might be a discontinutity

[glass quiver]

for this since its a graph of f'(x) and theres like a gap at x = 1, then would be that mean theres a critical point at x=1 on f(x)

[split elm]

just looked it up and did an example, for the vertical tangent in f', f would have a cusp/sharp bend, so f is not continuous at that point

[vernal mantle]

Yea I guess so

[glass quiver]

so would that mean it would be a critical point for f(x) since f'(x) is not defined there?

[split elm]

uhhh technically yes.. i guess so

but it is a sharp bend

when f'(x) is vertical

[glass quiver]

so its a corner?

[split elm]

yes its a corner

[glass quiver]

then doesn't that mean its a critical point if going off the definition of critical points are when f' = 0 or undefined

[split elm]

this graph of f' represents a corner in f also right?

[glass quiver]

yeah since its undefined

[split elm]

think of the vertical tangent like the gap, only its infinitesimally small

so in that case, yeah its a critical point

[split elm]

its the length of however long the graph is vertical*

[glass quiver]

so if given a graph of f' then critical points of f(x) would be when it touches xaxis, vertical tangents, and discontinuity in the graph?

[split elm]

yeah

[glass quiver]

okay thank you

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[buoyant jetty]

For what base-$6$ digit $d$ is $2dd5_6$ divisible by the base $10$ number $11$? (Here $2dd5_6$ represents a base-$6$ number whose first digit is $2$, whose last digit is $5$, and whose middle two digits are both equal to $d$).

[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[marsh citrus]

@buoyant jetty Has your question been resolved?

[marsh citrus]

@buoyant jetty Has your question been resolved?

[devout mauve]

its 6 options

you could have gone through them in way less time than waiting here

[marsh citrus]

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[still temple]

how would i go about doing this?

[late geode]

partial fractions

[still temple]

does the coefficient of x has to be zero to apply partial fractions?

why am i asking this here when i can google this smh

thanks for your help

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[stoic slate]

Factor numerator and denominator in left side

[lime inlet]

how do i factor the numerator

i did put cos2

[lapis atlas]

take tan common in the denominator

[stoic slate]

$a^2-b^2$

[lapis atlas]

youll get sin -1

[elfin berry]

Samuel

[lapis atlas]

and on the top (sin + 1)(sin-1)

[lime inlet]

ohhh

[lapis atlas]

cancel sin -1

you get the rhs

[lime inlet]

thank u i got it now

i was replacing sin - 1 with cos 😭

[lapis atlas]

oh haha

[lime inlet]

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[grand smelt]

"One of the angles in a triangle is a right angle. Show that the other two angles are complementary."

how am i supposed to show that the other two angles are complementary

[grand smelt]

i asked chatgpt and he said 90 + A + B = 180

but isnt complementary limited to 90

i thought that was supplementary

[sharp ivy]

sum of all 3 angles of a triangle is constant.

well, you move the 90 to right side, to find A+B.

[grand smelt]

well yes but im getting a little confused with complementary in this context

[sharp ivy]

complementary=sum of them is 90.

You have to show A+ B add up to 90, when C is 90.

This one time, I can say chatgpt did a pretty good job.

[grand smelt]

if a question asks "One of the angles in a triangle is 120, show the other two angles are complementary" is this question asking me the right thing or is it asking for a supplementary

[sharp ivy]

The question is asking you to use whatever properties you can, to show that the other 2 angles are complementary (= their meausures add up to 90).

It only works if 1 angle is 90 though.

If it said 120, like in your last comment, it would be impossible to show.

[lone heart]

That's not valid because complementary means they add to 90 so if (a + b) = 90 but you have 120, then the sum is 210, which is more then 180

[grand smelt]

so complementary is 90 + 90, supplementary is 180 + 180?

[sharp ivy]

no...

[grand smelt]

??

[sharp ivy]

complementary is x+y=90.

2 angles add up to be 90. no requirements on them.

[sharp ivy]

the property we use to show they're complementary is a different 1.

[grand smelt]

oh okay

[sharp ivy]

are you familiar with property of the 3 angles in a triangle that they add up to 180?

[grand smelt]

the existing right angle confused me

[grand smelt]

yes ofc

[sharp ivy]

well...you just use that.

[grand smelt]

so i thought its asking for a supplementary

cuz 90 + A + B = 180

[lone heart]

Because it's a triangle, you have 3 angles in a triangle, and it gave you one of them

[sharp ivy]

It's asking you only about THE OTHER 2.

so, you substract the 90 from both sides...and get a+b=90. definition of complementary.

[lone heart]

You're using that info to prove that (a + b) = 90

[grand smelt]

ok .close

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[novel juniper]

Why are there no solutions for a, b

[small stream]

By calculations

Wait just noticed a typo, it's I not E

[rotund violet]

this is the same as showing that $3$ isn't an eigenvalue of $A^2-2A$ right?

[elfin berry]

Arnavutköy

[small stream]

Hmm wdym?

Why is this related to eigenvalues

[rotund violet]

An eigenvalue of matrix $M$ is any value $\lambda$ such that $\det(M-\lambda I)=0$.

[elfin berry]

Arnavutköy

[rotund violet]

Have you seen this definition of an eigenvalue?

[small stream]

I know this, but i still don't understand why is this q related to ew

I just want to know why the method I used is invalid

[rotund violet]

well im confused in your solution what $E$ is exactly, and why any inverse would be of the form $aA+bE$.

[elfin berry]

Arnavutköy

[small stream]

.

[rotund violet]

moreover, if you do define $E$, then how do you prove your statement?

[elfin berry]

Arnavutköy

[rotund violet]

oh

how do we know that this matrix has an inverse which isn't of the form $aA+bI$?

[elfin berry]

Arnavutköy

[small stream]

Idk

That's my question

[rotund violet]

well your proof assumes this question's answer

it assumes that if you just show that there is no inverse of the form $aA+bI$, then there is no inverse at all

[elfin berry]

Arnavutköy

[rotund violet]

and thus is not invertible

[small stream]

Ye so what form it should be

[rotund violet]

yeah wait also if there are no solutions this is showing that it is NOT invertible

[rotund violet]

it can be any arbitrary matrix the same size as A

[small stream]

Uhm

...

Ok thx

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[novel juniper]

Let $x,y,z,w$ be elements of a group $G$
\
solve for $y$
(a) $xyz^{-1}w=1$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

Would it simply be $y=x^{-1}w^{-1}z$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

I can show my working

[rotund violet]

Yes

[novel juniper]

Yes show my working?

[rotund violet]

as in yes that is correct

[novel juniper]

Thanks!

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[novel juniper]

I suppose I should have written this as $y=(wx)^{-1}z$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[rotund violet]

That also works but isn't neccessary

[upbeat sable]

both are the same

yeah

[slate yarrow]

both are the same lol

[novel juniper]

I know

Thanks everyone!

Suppose $xyz=1$ does it follow that $yzx=1$. I follows that $yz=x^{-1}$ and subsequently $yzx=1$ so yes

[elfin berry]

ƒ(Why am. I here)=I don't Know

[slate yarrow]

xyz=1 does not mean yzx=1

[novel juniper]

it does

.reopen

.reopeon

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[still temple]

Lol

[marsh citrus]

✅

[slate yarrow]

it does if its commutative ig

[novel juniper]

$xyz =1 \implies yz=x^{-1}$
\
I now post multiply both sides by $x$ to get
\
$yzx=x^{-1}x=1$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[rotund violet]

oh yes it does. $xyz=1$ does imply that $yzx=-1$

[elfin berry]

Arnavutköy

[rotund violet]

sorry $1$ not $-1

[slate yarrow]

ah alright

[elfin berry]

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

[novel juniper]

One more follow up

what about $yxz$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[novel juniper]

I suspect not

oh nvm

it does

$xyz =1 \implies y = z^{-1}x^{-1} \implies yx=z^{-1} \implies yxz = 1$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[keen mist]

shouldnt this second bit be the other way round

[novel juniper]

why

[keen mist]

you multiply both sides on the left by x^-1 then on the right by z^-1

[novel juniper]

yes

oh

riggt

right

so they aren't the same

[keen mist]

probably not

[novel juniper]

cool

thanks

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[wise oxide]

I do integral test for all of these?

[odd orchid]

integral test is for series

use a limit

evalute the integral normally

[wise oxide]

So if it equal any number that’s not inf its convergent ?

[odd orchid]

not really

if the integrand is undefined at one of the bounds

its improper as well

[wise oxide]

So how do I know if it converges or not

[odd orchid]

let the problematic bound be t

set a limit as t approaches the bound

evaluate the integral

evaluate the limit

[wise oxide]

Yeah so if the answer is any number it converge?

[odd orchid]

yes

[wise oxide]

How do I integrate this

[odd orchid]

partial fractions probably

[wise oxide]

So do I keep my denominator like that

[odd orchid]

no

hang on

i dont think you can factor that denominator

[wise oxide]

So how am I suppose to do partial fraction

[marsh citrus]

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[vernal herald]

I don't get why the alternating series test doesn't work. The limit to infinity of the non-alternating term is zero, and it's decreasing...? So why doesn't it converge absolutely?

[rugged spindle]

it converges absolutely if $\sum_{n=1}^{\infty}\left \lvert\frac{(-1)^n}{2n} \right \lvert$ converges

[elfin berry]

fish

[vernal herald]

huh... I though it would be absolute for some reason as the root and ratio tests depict absolute convergence

[rugged spindle]

it converges, but to converge conditionally means that it only converges because it's an alternating series and would otherwise not converge if it isn't alternating

[vernal herald]

are you an alternating series because you converge or do you converge because you're an alternating series?

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[fierce kite]

Need some help w complex numbers

[fierce kite]

is there a faster way to solve this then to rewrite it as (2+i)(2+i)(2+i)(2+i) ?

[tight furnace]

binomial theorem

or square twice

[desert dirge]

or write stuff in exponential/polar form, speedy

[tight furnace]

god no don't do that

[desert dirge]

[marsh citrus]

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[last hound]

How to solve the second question pls

[last hound]

Like show that A³ = A?

[void elm]

that would suffice

though it might be easier (less computationally painful) to simply let take a vector v in R^4, and compute f^3(v)

which you can do by computing A^3v

[last hound]

Okay and then the next question, should I calculate the eigenvalue or like how I can deduce it?

[void elm]

characteristic polynomial of A

compute it, and find the roots

those will be the eigenvalues of f

[last hound]

Alright Roger

[void elm]

actually, there's a better way to do it

if $f^3(v) = f(v)$, then $f^3(v) - f(v) = 0$

[elfin berry]

higher!

[void elm]

then you can factor

that'll instantly get you the eigenvalues when you split it into linear factors

much less painful than the char poly

[last hound]

Oh yeah
So
We get lambda(lambda-1)(lambda+1) = 0
And then get
Lambda= 0, 1, -1??

[void elm]

well, not lambda

[last hound]

The eigenvalues

[void elm]

,, 0 = f^3(v) - f(v) = f(v)(f^2(v) - I(v)) = f(v)(f(v) - I(v))(f(v) + I(v))

[elfin berry]

higher!

[void elm]

so either $f(v) = 0, f(v) = I(v) = v$, or $f(v) = -I(v) = -v$

[elfin berry]

higher!

[void elm]

these correspond to eigenvalues 0, 1, and -1, respectively

[last hound]

Yeah that exactly I mean 😅

[void elm]

does that answer your question?

[last hound]

Yes sir 🫡 thanks you so much

[void elm]

no worries, happy to help

if you're done here, you may close the channel

[last hound]

Yes yes, I forget about that

!close

close

[void elm]

it's .

[last hound]

.close

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[still current]

I'm stuck on this and I don't remember when I learned to so I don't really know where to start

[desert dirge]

i see potential to use the product rule

in reverse

with FTC

[still current]

the first part of ftc?

[desert dirge]

yeah

consider adding those two integrals

[still current]

so 5 + second integral?

[desert dirge]

keep it as an integral, lets call the value of the integral we want 'Z' or something
then int [f'g] + int[fg']=5+Z

see what can be done on the left

[still current]

we would take the derivative?

im very confused sorry

[desert dirge]

sorry i meant ftc 2 not part 1

[still current]

ok

[desert dirge]

remember i said the product rule is relevant, in reverse

on the left we have the integral of [ f'g + fg' ]

which is what

[still current]

5+z?

the product rule in reverse thing is throwing me off

youre not talking about by parts right

[desert dirge]

what does f'g+fg' look like

no

[still current]

ohhhhhh

i see it now

f(x)g(x) = f'g + fg'?

[desert dirge]

derivative of, yeah

[still current]

oh yeah

[desert dirge]

so you can apply the ftc now and find Z

[still current]

wait but if i integrated f'g + fg' it would be f(x)g(x) so it would be f(x)g(x) from 0 to 1 = 5+ Z?

[desert dirge]

yeah

[still current]

alr thank you

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[rough ingot]

This probably isn't the clearest diagram, but how would I find what Theta 1 and 2 are in this diagram?

[rough ingot]

this is all the information i was given

[odd orchid]

probably a derivation of the sin addition formula

[rough ingot]

And what is that?

[odd orchid]

sin a+b = sin a cos b + sinb cos a

[rough ingot]

Alright thank you

.close

[marsh citrus]

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[sweet quarry]

I dont understand what im doing wrong, it said that the answer is partially correct

[sweet quarry]

(-3,-1) and (5,-1) are the local maximums

[shadow flame]

it says local max

max means highest

u gave the local minimum

[sweet quarry]

but theres 2 highest

[shadow flame]

theres one

[sweet quarry]

its for the csc graph

not sin

[shadow flame]

oh

lmao lemme take a look at that first img

[shadow flame]

no?

ok do you have the graph?

like do you know how to get the graph

cause after that, you have to look at the domain

it says from [0,2pi]

so theres only one local max

[sweet quarry]

thats the graphed part

ohhh i see

i didnt do the 0,2pi thing

[shadow flame]

hold on, whats ur question, for cos or for csc?

[sweet quarry]

csc

[shadow flame]

yea so the graph should look like this

[sweet quarry]

so yea there is only one cause thge other isnt in the 0, 2pi range

[shadow flame]

then with the domain

should just be 5,-1

[shadow flame]

yup you got it!

[sweet quarry]

alr thanks

.close

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[marsh citrus]

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[soft oar]

3-7x4

[soft oar]

Is it possible

[brave marsh]

What?

[shadow flame]

yes?

theres no equation here

[wary kite]

no, it’s way too hard

[soft oar]

Uh

[odd orchid]

,calc 3-7*4

[elfin berry]

Result:

-25

[soft oar]

Uh

[odd orchid]

👏

[soft oar]

I think is (7X4)-3

[wary kite]

sir were you born after 2011

[soft oar]

Uh

Why

[wary kite]

just curious

[soft oar]

I can choose to not answer this question

.close

[marsh citrus]

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[shadow flame]

hm

im not surprised

[marsh citrus]

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[still temple]

.smart

[wary kite]

oh lol

.close

[marsh citrus]

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[shadow flame]

whyd u proceed to dm him about this....

[wary kite]

i didn’t dm him

[shadow flame]

that sure looks like a dm

[wary kite]

well aren’t you sharp

[shadow flame]

thats quite the random first message

4 days ago

[wary kite]

what

[wary kite]

hence why i responded with "what"

[shadow flame]

eyy ur 18

congratz

[shadow flame]

EY AYO NO

I

I DIDNT MEAN IT LIKE THAT

[marsh citrus]

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[marsh citrus]

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[novel juniper]

Determine if $G=GL_n(\C)$ is a subgroup of $H=GL_n(\R)$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[static quarry]

do you mean the other way around?

G isn't even a subset of H

[novel juniper]

oops, yes

[static quarry]

what do you need to check in general to see if something is a subgroup of another group?

[novel juniper]

I suspect it is

[novel juniper]

Closure, identity and inverse must be present

[static quarry]

yea

so any thoughts on those?

[novel juniper]

Identity (I) is present and the inverse of a real matrix is real, so both are present

The product of two real matrices is real, so it's closed under compsition too

[static quarry]

yep

could even just say, GLn(R) is a group and it's a a subset of GLn(C) with the same identity and group operation

so it's a subgroup

[novel juniper]

Oh

yeah that's much better

thanks

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[vapid oar]

can someone please help me with iii)

[vapid oar]

The most work I could get up to is just showing that RT is just OT-OR, and then SU is just OU - OS

but after that i'm unsure what to do

<@&286206848099549185>

also, ive found that the length of rt or |RT| is just |OT-OR| and likewise |SU| is just |OU - OS|

<@&286206848099549185>

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[vapid oar]

help with iii) please

[vapid oar]

ive only managed to identify that RT = OT - OR and vice versa with SU = OU - OS

[marsh citrus]

@vapid oar Has your question been resolved?

[vapid oar]

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[undone surge]

Help with this

[marsh citrus]

@undone surge Has your question been resolved?

[vale jacinth]

use sin, cos

[undone surge]

Is BD: 5.0 and CD: 8.1?

[raw hawk]

Yes 5 and 8

[undone surge]

@raw hawk what about this?

[marsh citrus]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[undone surge]

1

I cant solve it

[raw hawk]

Well if you have x=√3
What will x² be

-# no calculator

[undone surge]

3

[raw hawk]

So to find BC use pythagoras's theorem

[undone surge]

So ill end up with 5.9? for BC?

Wait

@raw hawk 33.4?

Its 33^2 + 5^2 right?

[raw hawk]

No √(33+5)

@undone surge

[undone surge]

So its 6.2?

or 38?

@raw hawk

[raw hawk]

√38

[undone surge]

6.2

I got 21.3 for angle C, is that right?

[raw hawk]

Yes

[undone surge]

So the other one is 68.7 right

180 rule

[raw hawk]

Yes

[undone surge]

Thanks david, really helped out here.

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[marsh citrus]

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[marsh citrus]

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[sand wind]

Hi

[sand wind]

I need help with this

I am failing to understand wwhere did the 2x came from

[normal moss]

Do you know how to calculate the slope of a line?

[sand wind]

No

I am like

Really bad at math

Like really bad at it

But I am trying to re learn it

[normal moss]

Okay, so if you have two points on a line the slope can be calculated via a formula.

[sand wind]

Formula is the fancy F thing right?

[normal moss]

It's just an expression you can plug in numbers in this context to get out another number.

[sand wind]

I see

[normal moss]

Gimme a sec

https://images.squarespace-cdn.com/content/v1/54905286e4b050812345644c/f3d20328-86c8-4c0d-ac45-aae258d2c1c1/Formula-for-Slope-01.jpg

Here is a formula for calculating the slope of a line if you know two points on the line.

[sand wind]

What's a subscript? is that like the first location?

[sand wind]

And then you divide them both?

[normal moss]

A subscript is a little symbol you put below another symbol

A superscript is a little symbol you put above another symbol

[sand wind]

Is it like

[normal moss]

Like when you write $x^2$ the 2 is a subscript

[elfin berry]

DootDooter

[sand wind]

A power?

[normal moss]

Oops

[sand wind]

Aaaaah

Ok ok

[normal moss]

Sorry the 2 is a superscript

[buoyant pond]

the subscripts denote the two points

[normal moss]

Subscripts are the same idea but you put them at the bottom right of the symbol rather than the top

Yes the subscripts are used here because there are two points

[sand wind]

So what significance do they bring?

[normal moss]

So they are using the subscripts to distinguish between the two points

[sand wind]

Ah ok

I was about to say

[normal moss]

The point here is if you have two points you can plug their x,y values into this formula to get back the slope m

[sand wind]

Can you elaborate more on that or give an example?

[normal moss]

Here you have the graph of the line. So you can use this picture to find two points on your line. For example (1,1) is on the line and (0,-1) is on the line.

So you can plug these values into the formula for slope I posted and it should give you 2.

[sand wind]

I see

But must the 2 be in the x? Can't it be on the y?

[normal moss]

Say $(x_1,y_1)=(0,-1)$ and $(x_2,y_2)=(1,1)$. Then $m=\frac{y_2-y_1}{x_2-x_1}=\frac{1-(-1)}{1-0}=2$.

[elfin berry]

DootDooter

[sand wind]

So if I divide these numbers I should get 2

We have to divide them correct?

I'm understanding this weirdly

[normal moss]

Yeah there is a division in the formula.

People sometimes call this the rise over the run.

[sand wind]

Can we swear on this server?

[normal moss]

Just don't say slurs and stuff

[sand wind]

I see

Ok what the fuck I put it on the calc and it showed me 2

This is so trippy bruh

Math so trippy

[normal moss]

Wdym?

[sand wind]

I mean it works

[normal moss]

It should give you 2 that is the correct slope lmao

[sand wind]

Yeah it gave me 2

[normal moss]

So like the reason they call it the rise over the run is that you divide the amount the curve rises by the amount the curve runs by

[sand wind]

I see

But I am confused why x tho

Why can't it be y?

[normal moss]

In the m formula?

[sand wind]

Oooooo

I forgot about that

[normal moss]

There are x's and y's in the slope formula

[sand wind]

Let's say the formula changes instead of x it's y, the same rule still applies but it's y that changes correct?

[sand wind]

Yeah I know

Ok wow

This helped

[normal moss]

Mmm no I wouldn't think of it this way

The y's go on top and the x's go on the bottom.

[sand wind]

So it's always the x then?

[normal moss]

The point is the slope of the line IS the rise over the run. The rise is a measurement of how much the line changes along the y axis.

The run is a measurement of how much the curve changes along the x axis.

[sand wind]

I see

[normal moss]

The m is like the rste of change the line makes per unit.

[sand wind]

rste?

[normal moss]

Like if we measured the y axis in miles and the x axis in hours m would be an amount in miles per hour for example.

[sand wind]

I see

ok

interesting

[normal moss]

Probably for now it is best not to think too hard on that.

The slope formula is just a formula you can use when you know two points on a line. You plug in the values from the points into the formula correctly and you get out the slope of the line.

[sand wind]

I am like a 1 GPA student and I have to study math for an entrance exam to go to the #2 university in my country, I can get in if I pass the entrance exam the only issue there is math 😭

I might ask more help in this server tho

[sand wind]

Got it thanks man

[marsh citrus]

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[modest mantle]

see my method is simple alright its to find max and min value of complex no
if i have been give an expression suppose |z-1| =2 and an ecpression whose max and min you tryna find suppose |z-(3+4i)| now we should try creating z-1 in that other expression so you will get |(z-1) - (2+4i)| and now we get max min by, (z-1)^2 + whole root(x^2+y^2) xand y being 2 and 4 now with that method solve this question

[marsh citrus]

@modest mantle Has your question been resolved?

[modest mantle]

<@&286206848099549185>

[marsh citrus]

@modest mantle Has your question been resolved?

[unkempt orbit]

Well, first you need to understand what the first statement was trying to say

[marsh citrus]

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[rocky lark]

Hi

[rocky lark]

How do you apply gauss to a cylinder?

Hi

How do I request help?

!help

[marsh citrus]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[rocky lark]

I did the procedure

[unkempt orbit]

What exactly did you try?

[rocky lark]

I used a gauss surface

But still wrong

To find the potential

,w cylinder

[elfin berry]

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[rocky lark]

It's like the first one

[wide citrus]

Hi

[rocky lark]

Hi

[rocky lark]

I used a gauss surface

But still wrong

What

Its gauss not gas

I had a vision

I'll try this one

.close

[marsh citrus]

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[marsh citrus]

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[wise oxide]

Is this correct to find integral of (1+4x) times cos(2x)

[marsh citrus]

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[marsh citrus]

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[fleet fossil]

My teacher gave me this problem in 10th grade trig. She said we havent learnt how to solve this but I really want to figure out how

[silk sable]

,rotate ccw

[fleet fossil]

ok

nvm

[mossy turret]

isnt BE just 45*

for both B and E

[fleet fossil]

couldnt it be different values?

like 46 and 44

[mossy turret]

yeah mb

[fleet fossil]

also i forgot to mention, 60 degrees was the only angle that was defined except for the right angles

[mossy turret]

there are rules tho

you can look up. icompletly forgot

[mossy turret]

isosoles right angle triangle i was referring to but itt dont seem like it

side lengths equal mean they are 45 each

if one is 90

[fleet fossil]

Exact question

Ignore the writing on there I didn’t make it

[mossy turret]

oh

[mossy turret]

what i see is

there are two triangles

maybe scale it?

[fleet fossil]

3 triangles

?

2 right triangles one is not

[mossy turret]

im talking about

proportional

ones

[fleet fossil]

ahh

what do you mean by scale it?

[mossy turret]

see if you look there is a small one and a big one

[fleet fossil]

oh i know now

[mossy turret]

B to E and B to A

seem to be the same

ish¨

but different size

they should have the same

angles then

[fleet fossil]

but how do i prove that both are the same?

or is it just assumed

[mossy turret]

They are if you look at it

[mossy turret]

nah

i did this exact question before

and you solved it using that method

[fleet fossil]

okay

[mossy turret]

i just dont remember anything about angles

it was years ago

E = A

[fleet fossil]

i was thinking

could i use the perpendicular bisector theorem?

[mossy turret]

is the answerfor B

70 degrees

[mossy turret]

idk tbh

dont think it would be useful here

[fleet fossil]

because since theres a perpendicular line

BC = CD

or am i mistaken?

[mossy turret]

thats right

do you have an answer sheet btw

[fleet fossil]

no 😭

i wish i did so i could reverse it

[mossy turret]

you can use trig here too

[fleet fossil]

it is a trig class

so

but she said we havent learnt about this, so im clueless

[mossy turret]

maybe try it

[fleet fossil]

i cant use sine law, cosine law, nor soh cah toa i dont think

[mossy turret]

B should be 60 degrees

CEA are the same side lengths so using one of the laws we know that CE which is 30 is equal to EA which should also be thrity

180 - 60 is 120

E is 120

180-120 gives EC

180 - 120 gives B?

thats how i would solve it

[fleet fossil]

which law did you use?

[mossy turret]

the law where when the side lengths are equal

then both angles are equal

hold up

lemme find it

[fleet fossil]

isocoles triangle theorem?

[mossy turret]

"Isosceles triangle theorem states that, if two sides of an isosceles triangle are equal then the angles opposite to the equal sides will also have the same measure."

yeah

CEA

EA = EC

are equal in length

[fleet fossil]

i see

[mossy turret]

tho

wait

idk if its correct

im just giving my solution

since we dont have the answer sheet we wont know

[fleet fossil]

ill ask my teacher tomorrow if this is right

im unsure since this had no trig and she said that we learn about this in grade 11 even tho we learnt about ITT like last year

but the solution does make snese

[mossy turret]

yeah those are laws you gotta know

when dealing with this¨

there are more

"u v + = 180°
Side angles
w v =
Vertical angles
L1 intersects two parallel lines
L2 and L3
v w =
Corresponding angles
u w =
Alternate angles
Angle sum
in an n-sided polygon:
S n = (n - 2) * 180°"

[fleet fossil]

Ive done some of these before

ifnot most just in different wordings

[mossy turret]

yeah they teach u that in ur grade

you prob have it on ur

or that classes

formula sheet

years

[marsh citrus]

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[wise oxide]

How u do this

[main idol]

see example 3
https://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx

[wise oxide]

X^2(x^2+1+ 3/x^2)?

[main idol]

i don't know what you're asking

[wise oxide]

That’s the denominator

I factor it?

[wary kite]

quadratic in terms of x^2

factor u^2 + u + 3

where u = x^2

[wise oxide]

I can’t factor it

They did this in the answer sheet but I still have no idea what’s going on

[wary kite]

bruh

you should’ve gave more context

evaluating this integral is probably really ugly actually

,w \int_0^{\infty} \frac{2x^2+9x-2}{x^4+x^2+3}

[elfin berry]

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[regal coral]

it looks like they didn't evaluate (a), they just showed it converges

[wary kite]

yea i know but i was suggesting you find the zeros

but honestly that’s just ugly

yep

[wise oxide]

Ok well I have no idea how this works

[regal coral]

i think they used the comparison test

[main idol]

giving context about what the question is requires 0 math knowledge

[regal coral]

actually it says "direct comparison test" idk if that's different than "standard comparison test"

[wise oxide]

Yeah but I don’t get the split integrals and then idk what they’re even comparing it to

[regal coral]

i think they split it because it goes negative near 0

the comparison test requires it to be non-negative

[wise oxide]

What

For the third question they didn’t split it

And it’s also 0

Although I have no idea how they got any of these

[regal coral]

it's positive on (0,1)

red is finite

blue converges because green+blue converges

therefore red+blue converges

[wise oxide]

I don’t get split integrals thing tho

Like how they split it and why they just ignore some

Or any of it actually

[regal coral]

how they split it?

do you mean how they chose the bounds?

how they chose to split at x=1?

[wise oxide]

They changed it to 2 equations with different bounds

[regal coral]

that's allowed

$\int_a^{c}f(x)dx=\int_a^{b}f(x)dx+\int_b^{c}f(x)dx$ with $a<b<c$

[elfin berry]

Axe

[wise oxide]

So why didn’t they also for the third question

[regal coral]

the integrals are positive on (0,1)

[wise oxide]

So when r u suppose to split the integrals

To 2 different bounds

[regal coral]

you can split it in order to apply the comparison test on just part of the interval in which the functions are positive

[wise oxide]

But the third one they also did comparison but didn’t split it

Bruh is there any easier way

[regal coral]

uh

idk

[wise oxide]

😭

<@&286206848099549185>

[marsh citrus]

@wise oxide Has your question been resolved?

[regal coral]

the third one is positive on (0,1) like i said

but besides that

$\int_0^{a}\frac{2x^2+9x}{x^4}$ diverges

[elfin berry]

Axe

[regal coral]

we want to use this function for its behavior at infinity, not at 0

[marsh citrus]

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[fiery root]

did i do this right

[spark mason]

you know you can check your work yourself

[fiery root]

how

[upbeat sable]

At x=0 the value of the function is incorrect

[spark mason]

select a point on the graph like (0, 2) and put y=2 and x=0 to original equation

to see if it checks

do it with another point and if both check, then you’re good to go

[upbeat sable]

Well it doesnt check at (0,2) so it isnt correct just by that

[honest sage]

no this is not right. Start with the point 4, -1

and then use the slope which is rise over run

to graph the next point

[marsh citrus]

@fiery root Has your question been resolved?

[marsh citrus]

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[limpid bison]

FOr part c

I wasn't 100 percent sure if I could just integrate the curve c equation with the bounds of 0 to 5

or I needed to equate hte line l with the curve equation

[still temple]

no need to equate mate

[limpid bison]

thats what the mark scheme does tho

[still temple]

the formula for the area enclosed between two fnctions is just

area enclosed by line on top - area enclosed by line on bottom

[limpid bison]

theyre subtracting

the line equation

with the curve

[still temple]

That is what I am trying to say

They aren't equating the lines

[limpid bison]

oh

[still temple]

They are subtracting them

Lemme give you a nice graphic for this

I have two functions f(x) = x^2 and g(x) = x^3

when you subtract the area of the curve on top (f(x) = x^2)

from the area of the curve on the bottom

g(x) = x^3

you are left with the area enclosed between them

[limpid bison]

i think i get that

but in that question

they are not subtracting an area

but the equation of the line

oh yh

ok ill delete that

thx

[novel juniper]

Cool

[limpid bison]

will j resend this pic

[limpid bison]

can u do one with a straight line equation

rather than a curve pls

nvm sry i think i got it

thanks

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[vast tide]

Так делают, потому что площадь, в определении определенного интеграла, можно разбить на множество прямоугольников. Смысл в том, что эта разность даёт высоту столбиков, на которые разбивается эта фигура

[vast tide]

.close

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[trail gulch]

How do you setup part a, i keep getting the wrong answer (answer is 0.89%)

[vital oracle]

They’re asking for the rate of growth per year
367000000/33900000 is the growth over all 9 years, to split that into one year you do ^(1/9)
so you have to calculate (36700000/33900000)^(1/9) instead

[trail gulch]

Omg thank you

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[fair gate]

Yo

[fair gate]

How do u do this?

[amber birch]

The denominator must have x - 1

There must also be (x - 1/3)/(x - 1/3)

Multiply this function by some constant k and sub in x = 1/3 to the cancelled equation, and y = -2

You can thus find k

[fair gate]

I understand the first part

I need help with b)

[proud ice]

How I would do it is choose $p(x)=Ax+B$ and $q(x)=Cx+D$, and find the $A,B,C,D$ that satisfy the conditions

[elfin berry]

SWR

[fair gate]

Huh?

P means the top

And q means bottom

So we plug in 2 for both the top and bottom?

Im not sure

U think u can write it down?

[proud ice]

correct

[proud ice]

correct

[fair gate]

Hmm

Ok

[proud ice]

You want $p(2)=0$, so you can solve $0=2A+B$. You're still left with an unknown, but you get just a little bit closer, imo

[elfin berry]

SWR

[fair gate]

I see

Ok but the first equation is: f(x) = 4(3x-1)/3(3x-1)(x-1)

So...

What do we do now?

[fair gate]

Part a btw

[proud ice]

I think the hole gives us some good info

[fair gate]

Idk man

Can u write it down pls?

If u want lmao

[proud ice]

We want $f(x)$ to satisfy $\lim_{x\to\frac15}f(x)=\frac56$. Let's say there was no hole. Then we want $\frac{p(x)}{q(x)}=\frac56$ when $x=\frac15$. You can use the linear equations $p(x)=Ax+B$ and $q(x)=Cx+D$ to solve for another variable

[elfin berry]

SWR

[proud ice]

This is one of those problems where I know exactly what to do, but explaining it is really hard without having to ramble on for a very long time

[fair gate]

So do it 😭

[fair gate]

This doesn't make sense

Cx + D???

[proud ice]

We do not hand out answers here

[fair gate]

Damn 💀

[proud ice]

Here's what we know so far

$f(x)$ must be a rational function

[elfin berry]

SWR

[fair gate]

yea

[proud ice]

which means it is the ratio of two polynomial functions, which we are calling $p(x)$ and $q(x)$.

[elfin berry]

SWR

[proud ice]

What am I doing first is starting at the simplest possible case: $p$ and $q$ are two linear functions.

[elfin berry]

SWR

[proud ice]

Although, I suppose, the simplest case would be that they are each constant functions.

It might be easier to do this a different way though.

What's the simples rational function you can think of that has a vertical asymptote of $x=2$?

[elfin berry]

SWR

[fair gate]

No idea

1/x-2?

[proud ice]

yup

So let's start with $\frac{1}{x-2}$. So, for now, we would have $p(x)=1$ and $q(x)=x-2$. Note that $q(2)=0$. However, $p(2)\ne 0$. Any idea what we could do here?

[elfin berry]

SWR

[fair gate]

No idea what we can do here

[proud ice]

Recall that $p(x)$ must be a polynomial function. So, if $p(2)=0$, that is saying that $p(x)$ has some root $x=2$, and therefore has a factor $(x-2)$. We can write $p(x)$ in terms of some other polynomial function $r(x)$, $p(x)=(x-2)r(x)$

[elfin berry]

SWR

[proud ice]

Now, regardless of what $r(2)$ is, we would get $p(2)=0$, as we want

[elfin berry]

SWR

[fair gate]

This isn't helping tbh 💀