help-26

[cobalt vine]

Just differentiate the options
The answer is coming out to be D

[restive steeple]

what if in a parallel universe there were no options?

[ivory sorrel]

let -x be u

maybe that will help

[restive steeple]

it worked, thanks a lot

thank you to kyanite too for poor man's strats

[cobalt vine]

np

[coarse tusk]

yeah differentiating the options can be a good strat

in like exams and such

[cobalt vine]

yeah agreed

[coarse tusk]

wouldn't recommend doing that for exercises and all though, since there's no point

[restive steeple]

yeah but what about the ⭐ learning⭐

[coarse tusk]

learning doesn't matter in exams

just do the integration properly when you get home

but yeah learning the integration is important i fully agree

[cobalt vine]

Kya yaar shikhar

[restive steeple]

jee nikal lenge aur kya

[coarse tusk]

real

jee has wild integrals though

[cobalt vine]

gupta ji aapne disappoint kardiya

[coarse tusk]

usually doing the integration is faster

[restive steeple]

samajhdari khoon mai chalti hai bhai inlog ke

[coarse tusk]

nah bro

practice se bas

[cobalt vine]

College mein ho kya?

[restive steeple]

yeah plus they give definite integrals mostly to prevent differentiators

[coarse tusk]

exactly

[coarse tusk]

no 11th

[restive steeple]

coaching?

[cobalt vine]

im in 9th class

[coarse tusk]

fiitjee

u?

[restive steeple]

in 9th foundation coaching

[coarse tusk]

ohh

then why the hell are you doing integrals

lmfao

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[violet hornet]

How do I solve an equation that doesn't have a fixed point or accuracey ?

[crystal goblet]

Are you aware of the rational roots theorem?

[violet hornet]

no

[crystal goblet]

You can try guessing a root

[violet hornet]

Ok

I know this one

[crystal goblet]

If you know this theorem you will know what guesses to make

[violet hornet]

but how does it end ?

[crystal goblet]

First try guessing a root

Try 1,-1,2,-2

[violet hornet]

I did

[crystal goblet]

Did you get any?

[violet hornet]

ok lemme see

Oh

wait

I required to use the iteration method

[crystal goblet]

Oh okay

Which method, Newton's method?

[violet hornet]

I have the equation as x = 7/x - 1/x

no x = f(x)

sorry

7/2 - 1/x

[crystal goblet]

2x^3=7x-2

x=7/2x-1/x^2

[violet hornet]

wait

Yh that's right

my bad

didn't pay attention

then what ?

And it's 2 not 1

So it's x = 7/2x - 1/x^2 ok

[coarse tusk]

there's no need for any of this

x=-2 is a root

just factor it out

[violet hornet]

I'm required to do it this way

[coarse tusk]

oh

[crystal goblet]

We know

[violet hornet]

Do I just assume a root ?

[crystal goblet]

Yeah make a guess

[violet hornet]

ok

[crystal goblet]

Unfortunately someone already spoiled it

[violet hornet]

it's ok

I assumed x = 0.5

and got X2 = 3

[crystal goblet]

,calc (0.5+(3.5/0.5-1/(0.25)))/2

[thorny flame]

Result:

1.75

[crystal goblet]

Oops sorry wrong calculation

[violet hornet]

yh

[crystal goblet]

,calc 3.5/0.5-1/(0.25)

[thorny flame]

Result:

3

[crystal goblet]

Okay yes

[violet hornet]

now what

[crystal goblet]

Keep applying it

,calc 3.5/3-1/(9)

[thorny flame]

Result:

1.0555555555556

[crystal goblet]

Find x3 and x4, x5,... till you get |x(n)-x(n-1)|<0.001 like they asked here

[violet hornet]

So this is not different from the equations that have a starting point ?

[crystal goblet]

What does that mean?

[crystal goblet]

You chose the starting point to be 1

They chose the starting point to be 0.5

The starting point(x1) is a guess that you make

You guessed 1

[violet hornet]

Ok got it

thank you

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[noble laurel]

I am having a doubt about this example from my class

T(p)=d/dx((x-1)p) is the linear map from P2(C)->P2(C)

and we're trying to find its eigenvalues/eigenvectors

I am confused why we are somewhat randomly considering the vector v=x^2

Is there a reason?

[shut spindle]

Is your phone 20 years old ?

or is your classroom the size of a stadium

[noble laurel]

It is a screenshot from the panapto camera, which is a very bad camera

[shut spindle]

ahh

[noble laurel]

Apologies

[shut spindle]

makes sense

[worthy storm]

so hard to read

exccept for the red scribble

[noble laurel]

I can translate

let v=x^2

Tv=...

T^2v=..

T^3v=..

do the row reduction to get the linear combination of T, T^2, T^3 and I that equals 0

it turns out it is (T-3I)(T-2I)(T-I)v=0

once factored

and now my sir is claiming that, we just substitute v one by one to figure out what the eigenvalue is

if (T-I)v=0 then it is 1

if not substitute into the next and so on

but my doubt is about why does v=x^2?

[shut spindle]

my sir?

[noble laurel]

my ma'am*

[shut spindle]

more confused lol

[worthy storm]

(T-3I)(T-2I)(T-I)v=0 tells you that at least one of T-3I or T-2I or T-I fails to be injective

so 1 or 2 or 3 is an eigenvalue

[noble laurel]

right

right

for v=x^2 being the eigenvector

but why did we randomly have v=x^2 to begin with

[worthy storm]

if (T-I)v = 0 then boom 1 is an eigenvalue and v is an eigenvector associated with 1

[noble laurel]

I understand that part of it

[worthy storm]

i think there's nothing magical about working with x^2

for any nonzero v, you could consider
v, Tv, T^2 v, T^3 v

a list of four vectors in a 3 dimensional space

[noble laurel]

yes

[worthy storm]

hence linearly dependent

[noble laurel]

I agree

[worthy storm]

so some linear combination gives you zero

then a similar argument will apply

[noble laurel]

Will that not imply any arbitrary v is an eigenvector though? Which surely can't be true

[worthy storm]

no it won't

refer back to this: (T-3I)(T-2I)(T-I)v=0

if (T-I)v is not zero, then we proceed:

either (T-2I)(T-I)v = 0

in which case 2 is an eigenvalue with (T-I)v an associated eigenvector

or

3 is an eigenvalue with (T-2I)(T-I)v an associated eigenvector

[noble laurel]

ah and (T-I) like "adjusts" v into being the eigenvector we're guaranteed?

[worthy storm]

well it adjusts you to something that might be an eigenvector

[noble laurel]

right

assuming it goes to 0

[worthy storm]

one of the following has to be an eigenvector:
v
(T-I)v
(T-2I)v

[noble laurel]

yes

So there's no reason we're using v=x^2 except for the hell of it?

[worthy storm]

i think that's true

[noble laurel]

amazing

[worthy storm]

anything nonzero would work i think

don't quote me on that, check it for yourself to be sure

[noble laurel]

gosh I fucking wish my maths sir would teach me matrix algebra for hours

[noble laurel]

I'll check it out

ty

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[noble laurel]

seems right, it does seem like this process will only ever give us one eigenvalue/eigenvector though, is that right?

What about the other ones

[knotty ledge]

honestly i would just find the matrix of the map wrt the standard basis

[worthy storm]

yea, it's not immediately obvious how to coax this process into revealing other eigenvalues/eigenvectors

[noble laurel]

My only issue with this is then I am not allowed to use a determinant to solve T-lambda I = 0

[noble laurel]

I think if we try v=x, v=C

[knotty ledge]

why not lmao thats stupid

[noble laurel]

it should give us the others

but that seems particular to this map

[noble laurel]

because we are using Axler and my prof wont let us use determinants for anything because we haven't introduced nor proven any determinant facts

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[knotty ledge]

.well then yeah i would just do the same thing with x and 1

[noble laurel]

. it just seems particular to only this map, and just because 1, x, x^2 form a basis for P_2(R) and maybe it works out here, we could've had other less obvious basis, but maybe the key is like, so long as you're plugging in a vector not in the span of other ones you've already tried you will get a new eigenvector?

[knotty ledge]

oh or perhaps alternatively since all these maps commute you can just change the order and probably get new evectors

[knotty ledge]

fuck

whatever

bc the evalues are actually 1,2,3

[noble laurel]

how do we know they commute

[worthy storm]

T commutes with itself and with any scalar multiple of the identity

use that and distributivity

[knotty ledge]

i think if you did go ahead and do two other LI vectors you might just get the same thing but in a different order

or maybe the poly is lower degree idk

[noble laurel]

hm this is really dumb no offense to axler but lemme just use a little bit of determinants

or don't make me compute eigenvalues

[knotty ledge]

well yes do v =1 lol

[worthy storm]

yea it seems like just rearranging the product you can re-use the same v each time

[noble laurel]

idk why my ma'am would never give any of these details

[knotty ledge]

did she not finish the question?

like she just found one evector and called it quits?

[worthy storm]

i'm not sure if with this method you're guaranteed to get a polynomial whose factors cover all the eigenvalues

like this might just divide the minimal polynomial as opposed to being the minimal polynomial

[noble laurel]

yes actually

[knotty ledge]

yeah thats definitely true bungo bc of v=1

[worthy storm]

yea

so you may need to repeat it, possibly n times (n the dimension of the space), with v equal to different basis vectors

[noble laurel]

each time with L.I vectors?

[knotty ledge]

its just interesting that you do get all the evalues in this case by doing it with x^2

[noble laurel]

neccessarily the basis vectors or just L.I?

[worthy storm]

yea is that just a quirk of this particular linear map?

[worthy storm]

amounts to the same thing really, any n LI vectors is a basis

[noble laurel]

My ma'am has a thing for maps with eigenvalues 1,2,3,4,5,....

[rapid meteor]

Hi ! Can I have some help here

[knotty ledge]

!occupied

[topaz sinew]

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).

[worthy storm]

ocupado

[noble laurel]

Alright well he can have my doubt channel anyways

I have to go to office hours

[worthy storm]

what was your ma'am's intent with this example, just showing how to find an eigenvalue/eigenvector pair?

[noble laurel]

Unfortunately not for this class

[noble laurel]

I believe so yes

She is not my most organized or clear maam of all time

[rapid meteor]

I want some recommendation books

[worthy storm]

#book-recommendations

[noble laurel]

I want you to leave my doubt channel

Anyways thanks bungo and ZAC I’ll get back to my linear algebra in about an hour after these office hours

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[north river]

can somebody explain the solution to me? Why did they put both expressions in the integral?

[topaz sinew]

@north river Has your question been resolved?

[north river]

<@&286206848099549185>

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@north river Has your question been resolved?

[north river]

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[neon iron]

will you please help me understand how the order of the compression and shift transformation on f(bx - h) works and why it is important? i can understand that (x+h)/b but when i change it to the expression (x/b + h/b) im not sure if the order of transformation matters?

[neon iron]

will you please help me understand how the order of the compression and shift transformation on f(bx - h) works and why it is important? i can understand that (x+h)/b but when i change it to the expression (x/b + h/b) im not sure if the order of transformation matters?

[radiant marlin]

it comes from what happens when you solve for x =

so for f(bx-h) you go from bx-h=[ ] to x=([ ]+h)/b

[neon iron]

yes

(x+h)/b

the pre transformation function

[radiant marlin]

how you represent that last part doesn't matter

[neon iron]

(f(x)+h)/b that better?

[radiant marlin]

yea I'll work with that

it's better written as 1/b*f(x) + h/b

like fully simplified

[neon iron]

ok

well

my question is this

having f(x)/b + h/b

does it matter wether or not the compression or shift is performed first?

im concerned its different for verticsal versus horizontal

because one is domain and one is range

im thinking hard

[radiant marlin]

it's always compression then shift

just because of how it's "standard" to write things as y=ax+b

you multiply then add

[neon iron]

ok, check out this photo i took from openstax, it is saying to do a shift first... any clues as to maybe how they are representing the transformations or maybe anything::::

[radiant marlin]

it just works better visually

ok

[neon iron]

f(bx-h)

ty for ur time and help btw

[radiant marlin]

it's all confusing because sometimes it's going backwards haha

[neon iron]

lol

ive been trying to litteraly figure this out for a day or two i really think this is challenging my thinking hahaha

[radiant marlin]

like usually I'd just rewrite f(bx-h) into f(b(x-h)) form when I'm thinking about the graph, but this image is sort of telling you to just memorize both cases

[neon iron]

b(x-h/b) ?

[radiant marlin]

right

that gives the same transformations as the image

just a different way of looking

[neon iron]

oh ok i see what ur saying about memorize both ways

one each way of using order of the transformations?

just to learn one way of compressing then shift, then a way to shift then compress?

[radiant marlin]

yea that's one way

for me I just always try to make it f(b(x-h))

[neon iron]

yeah that seems like the gold standard.. hey thanks captain! btw i saw u on here earlier u put a lot of time in this? go karma! lol cya thanks !

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[naive zinc]

how do i do question d

[topaz sinew]

Please don't occupy multiple help channels.

[noble laurel]

.close

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[naive zinc]

i closed my old one

[noble laurel]

Do not close and reopen a channel to post the same question

[naive zinc]

bruh

[misty gust]

l' hopital would be ashamed of u

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[cyan lantern]

I'm a little rusty with my high school math, my son had a test and he left this question blank. The work written down is by the teacher and when I put it in my calculator I get 15 for width... which makes length zero. That makes no sense, kids can't play in zero space haha. Where did I go wrong?

[radiant marlin]

not sure what you're plugging into but to finish the work shown, you take the derivative of that bottom equation and set it equal to 0 to find the maximum

[cyan lantern]

ohhh, i found the x-intercept, why the maximum?

[radiant marlin]

like here's the graph of A=30w-2w^2

and the thing about maximums/minimums is that the derivative there is 0, so that's what we use to find them

[cyan lantern]

gotcha, makes sense, thanks

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[ashen rose]

i need calc help bro

[ashen rose]

so for this question, I found the first deriviative then i set it to zero

and i got the wrong answer

how would i do it

[radiant sparrow]

Maximum velocity, not the distance

So you need to put zero for the double derivative

[ashen rose]

wait

so i get the second dervitiative

[radiant sparrow]

Yea

[ashen rose]

ok i did that and i got zero

but then 2 in the interval is higher than 0

[radiant sparrow]

[ashen rose]

and 1 idk

i didnt calculate 1

@radiant sparrow

even for this one, i tried getting the first dervitive then setting it to zero and it didnt work

[radiant sparrow]

that doesnt look correct

,w second derivative 1+2t-8/(t^2+1)

[thorny flame]

Results provided by WolframAlpha

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Requested by ankitwa

[radiant sparrow]

put that =0

[ashen rose]

ill just do that one later

[ashen rose]

how would you do the second qusestion

this one

i have the derviative of cost then i basically set itto 0 and solve

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[twin heath]

Is there a way to prove that the limit of x^y as x and y both approach zero equals 1? not sure if i'd be wrong here since i dont have a lot of background on multivariable calculus, but would it work to just approximate it with the limit as x approaches 0 of x^x?

If this limit is considered true, then why is 0^0 sometimes considered undefined?

[wary tulip]

there is usually no reason to leave it undefined

[topaz sinew]

@twin heath Has your question been resolved?

[twin heath]

but why would it be 1? the way ive seen x^0 defined is by dividing x^1 by x to get x^0. But 0/0 is clearly indeterminate, so 0^0 has to be proven by something else, right?

[wary tulip]

"indeterminate form" is just a limit computation phrase and how limits work often motivative how you define expressions but they don't have to

the value of $$\lim_{(x,y)\to(0,0)}x^y$$ is its own thing that can be computed. $0^0$ is its own thing that you choose how you define

[thorny flame]

chmonkey #1 simp

[wary tulip]

there isn't in any choice in what the limit is once you define limits and a^b with nonzero a,b

but 0^0 is something you define

just as you choose how a^b for nonzero a,b is defined

and what a limit is

you should not look at like the value of 0^0 is "discovered" by computing that limit

[twin heath]

i know 0^0 is something that you can just define, but the value that it's given still has to be a reasonable value. i just want to know why we can just define it as 1, but cant define 0/0

[wary tulip]

you can define 0/0 but it's pointless

it won't behave like division

no matter what you try to define it as

you can't define it if you require that it has the characteristic property that division by nonzero numbers has

which, morally, you should

[twin heath]

right but i could say that 0^x is always equal to 0 or is undefined and that any x^0 is equal to 1

[wary tulip]

back to this

how limits work don't need to determine how you define expressions

[twin heath]

but 0/0 is considered indeterminate because of limits :(

[wary tulip]

who

cares

[twin heath]

because it's weird to have one considered indeterminate and the other to have a given value when they're literally the same thing

are 0^0 and 0/0 not the same expression

[wary tulip]

??????

[twin heath]

i think

[wary tulip]

no, one has a ^ and the other has a /

[twin heath]

my brain is probably doing smth wrong

this might might be the wrong way to think about it but

if raising an exponent by one is multiplying by that number

then decreasing it by one is dividing it

then x^0 is equal to x^1/x^1

or x/x

[wary tulip]

as a reason for why 0^0 and 0/0 are the same?

[twin heath]

i meannnnnn
Yeah??? maybe??

[wary tulip]

you're applying "rules" where you shouldn't be. the fact you're thinking of is not

x^m/x^n = x^(m-n)
because that doesn't make sense as a mathematical statement. rather, it's
x^m/x^n = x^(m-n) for any x > 0, and m and n any real numbers

this is not your fault really, lack of quantifiers is a devastating issue in math education

[twin heath]

okay i think i kind of see why but

still a bit weird that one's defined and one isnt when they can both equal the same values depending on how you want to define it

but it works ig

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[native vale]

This question is not exactly what I need at the moment as i have completely missed the 'exact form' and pi section. What simpler/earlier areas of math do you recommend I study first to understand questions of this type?

[lucid junco]

whats the question asking you?

[native vale]

Use a trapezium to find each area under the curve.

[lucid junco]

look up videos on the trapezoidal rule

[native vale]

i understand the trapezoidal rule, but i mean alot simpler, like this type of stuff

like pi and roots

[topaz sinew]

@native vale Has your question been resolved?

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@native vale Has your question been resolved?

[native vale]

.close

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[fallow chasm]

Where does your -40 come from?
Maybe I'm slow but I don't think -1/0.026 gives -40 :/

[topaz sinew]

@dark pond Has your question been resolved?

[fallow chasm]

No worries, best of luck 😊

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[lunar merlin]

hey im up to the first question I am confused how to do this

[lunar merlin]

I understand rational root theorum im just not sure how to use it to help factorise these

[winged cove]

do you want to find 3 factors or just a single factor

I can help with this

[shut belfry]

Try finding a single factor by applying the rational root theorem

[winged cove]

but Im not sure what the rational root theorem is,

it seems way too complicated for something as easy as this

[lunar merlin]

im not sure just whatever the quesstion wants

[winged cove]

[lofty salmon]

compute the roots via RRT and then represent in the simplified form

[lunar merlin]

ah

so

[winged cove]

seems too complicated

[lunar merlin]

solve for p or q

and just use either

wait no

we arent find a factor we have to factorise it

nvm im confused again

[lofty salmon]

p is either positive or negative factor of the constant term a_0

[lunar merlin]

yeah

so q = +- 1, +-2

sorry no

looked wrong

[lofty salmon]

same for q but it is the factor of the leading coefficient a_n

[lunar merlin]

okay so how do I figure out which is which

[lofty salmon]

you gotta then test the possible p/q values in respective polynomial p(x) if it evaluates to zero then you've found a root!

[lunar merlin]

how do i sub in p and q

wait

omg thats so much trial and error

so try all values of p till we get 0

[lofty salmon]

not p and q the fraction p/q

[lunar merlin]

oh

so

idk

[lofty salmon]

its indeed a tedious task but very useful in high degree polynomials

[shut belfry]

for ex part 1 has possibilities ±(3/2) , ±(5/2)

[lunar merlin]

wow

so

we have to try

[shut belfry]

yeah

[lofty salmon]

you have to try

[lunar merlin]

+-1/+-1

then +-3/+-1

then etc

i just dont know which ones to choose

[lofty salmon]

test out everything

[lunar merlin]

lmao

[lofty salmon]

altho you could analyse for some p/q

[lunar merlin]

im good

this is fine

[lofty salmon]

go ahead

now this is all computation

[lunar merlin]

my brain is gonna get too jumbled if I do that

[lofty salmon]

if you need further help ping me!

[shut belfry]

Be a smartass check for the signs (+,-). in p(x) ex if all are positive then no negative possibility is viable root. Hope it helps

[lunar merlin]

okay thank you ill leave this open and lyk if I get it

okay

@lofty salmon it took a bit

but i found f(3/2) = 0

what do i do next to factorise the polynomial

[lofty salmon]

you need to find 2 more

[lunar merlin]

wdym

two more what?

[lofty salmon]

a cubic equation has 3 roots

[lunar merlin]

lets say i find the roots

what do i do next

[lofty salmon]

,tex ```tex
\begin{align*}
\text{Given cubic polynomial: } & ax^3 + bx^2 + cx + d \
\text{Factorized form: } & (px + q)(rx^2 + sx + t)
\end{align*}

[thorny flame]

ℕaive

[lunar merlin]

where is r s and t coming from

[lofty salmon]

a cubic polynomial is factorized into 2 simplified polynomials that are linear and quadratic.

[lunar merlin]

yep

and how does the rational root help find that

[lofty salmon]

those are coefficients that you need to determine based on the specific cubic polynomial you are dealing with

[lunar merlin]

yup

[lunar merlin]

^

.close

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[lofty salmon]

You find these values through methods like synthetic division or hit and trial

if you find any 1 root

[lunar merlin]

its okay dont worry

im gonna ask my teacher

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[wild brook]

Ysaac is standing on the origin of a 2-dimensional cartesian plane. Suddenly, Ysaac is running away from his problems with a rate of 2 units/s in a direction determined by y=2x. Exactly 2 seconds after Ysaac started running, Some random guy also starts running away from Ysaac’s problems at the same speed as Ysaac but in the direction determined by y=x. If Ysaac’s problems stay at the origin point of the cartesian plane, calculate the equation for the rate of change of the distance between Ysaac and Some random guy. Assume Some random guy stands and moves on the same plane and quadrant as Ysaac.

I have made a question I cannot solve

[wild brook]

how does one solve a rate of change question

[fair thorn]

OH HEY NAUT

"ysaac is running away from his problems" 💀

[wild brook]

hey there

[fair thorn]

"some random guy also starts running away from ysaac's problems" 💀

[wild brook]

I'm back with another great problem I made and I cannot solve

[wild brook]

ikr, my name is weird

[fair thorn]

OH i see

hmmm

ok i think i got it

[wild brook]

do I have to view y=2x as the x-axis

that'd be tedious but like it'd work I think

[fair thorn]

wait wait wait

hang on

im... struggling

OHKAY i see

[wild brook]

ain't no way....

[fair thorn]

$\frac{\dd\left(\sqrt{\left(x_{Y}-x_{R}\right)^{2}+\left(y_{Y}-y_{R}\right)^{2}}\right)}{\dd x}$

[thorny flame]

ren

[wild brook]

bro wifi is making the timing super peccable

[fair thorn]

"peccable"

[wild brook]

yes

[wild brook]

impeccable is a word meaning perfect, therefor peccable

[fair thorn]

ik

[wild brook]

how do I do this 😭

[fair thorn]

well

y_Y = 2x_Y and y_R = x_R

[wild brook]

yr=xr, yy=2xy
(xy-xr)^2+(2xy-xr)^2
5(xy)^2-5xyxr+2(xr)^2

[fair thorn]

tf??

[wild brook]

idk I just placed it in ¯_(ツ)_/¯

[fair thorn]

ok hang on

y_Y = y of ysaac = 2x_y (the x of ysaac)

y_r = y of rando = x_r (the x of rando)

now, since we know that x_r = x_y - 2

just sub it in

im 70% sure this is the way to go

[wild brook]

but they're moving 2 units per second

in the line, not in the x-axis sense '<'

[fair thorn]

well

just assume x = time parameter

and btw Ysaac is moving at 2u/s, since the derivative of 2x is 2

[wild brook]

ohhhhh I've been thinking from a vector sense

[fair thorn]

xD

so, how's it going?

solved it?/

[wild brook]

nah I'm dying

[fair thorn]

show ur work

sorry, DC crashed smh

@wild brook ?

@wild brook ??

[wild brook]

yo

lag

[fair thorn]

fax

[wild brook]

so how do I solve this '<'

[fair thorn]

i just told you

$\frac{\dd\left(\sqrt{\left(x_{Y}-\left(x_{Y}-2\right)\right)^{2}+\left(2x_{Y}-\left(x_{Y}-2\right)\right)^{2}}\right)}{\dd x_Y}$ fairly certain this is what you need to do

[thorny flame]

ren

[fair thorn]

75% sure

[wild brook]

mm hmm, now I have 0 knowledge on doing rate of change questions, so now how do I do this '<'

[fair thorn]

...

this is a derivative-

y = that big thing

[wild brook]

I know

I don't know how it works though

[fair thorn]

sorry, powercut

am back

naut for now just simplify and give me the expression

[wild brook]

uh

UHHHHHHH

[wild brook]

(5(xy)^2-5xyxr+2(xr)^2)^(1/2)?

[fair thorn]

simplify, x_r = x_y - 2

and no idea what that is, use latex

[wild brook]

(xy^2+4xy+8)^(1/2)

I'm stupid

do I just differentiate that?

there's no way it's that simple

💀

I'VE BEEN QUESTIONING MY KNOWLEDGE IN MATH FOR 3 HOURS

(xy+2)/(sqrt(xy^2+4xy+8)

[wild brook]

it doesn't work

[fair thorn]

ok naut

i have no idea what ur saying

USE LATEX

take smth from desmos and copy the mess here

[fair thorn]

give me the simplified version of this; remember that x_r = x_y - 2

@wild brook ??

@wild brook bro ded?

[wild brook]

yes

[fair thorn]

$\frac{d\left(\sqrt{\left(2\right)^{2}+\left(x_{Y}+2\right)^{2}}\right)}{dx_{Y}}$

[thorny flame]

ren

[fair thorn]

simplifies to this

[wild brook]

this is the simplified verson

[wild brook]

yeah

[fair thorn]

$\frac{d\left(\sqrt{8+x_{Y}^{2}+4x_{Y}}\right)}{dx_{Y}}$

[thorny flame]

ren

[fair thorn]

,w differentiate sqrt (x^2 + 4x + 8)

[thorny flame]

Results provided by WolframAlpha

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

Requested by justarenegade

[fair thorn]

there

that's your answer

i think

[wild brook]

supposedly

but if you place a value it doesn't really do the distance '<'

[fair thorn]

wait

you wanted DISTANCE

I THOUGHT YOU WANTED ROC OF DISTANCE

[thorny flame]

ren

[fair thorn]

@wild brook

[wild brook]

wait

I mean it doesn't find the distance

[fair thorn]

what

explain what you WANT

naut

don't disappear periodically please @wild brook

[wild brook]

formula for the distance

or preferably a more solvable rate of change problem

[fair thorn]

i mean

[fair thorn]

this is distance b/w ysaac and rando

[fair thorn]

can't do that for ya

[wild brook]

Welp

I gotta disappear for now I will discuss this tomorrow

.close

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[fair thorn]

alr, DM me

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[frail bay]

How do I calculate cosα when sinα = –0.6 and α is a fourth quadrant angle.

[woven creek]

do u know sin37degrees

[frail bay]

nope

[woven creek]

oh nvm

u dontt need it

relation b/w sinx and cosx?

[frail bay]

uhhhh

i honestly have no clue

[woven creek]

(sinx)^2 + (cosx)^2 = 1

wait actually there is another way

do u know what sinx stands for

[frail bay]

nope

[woven creek]

sin theta = opp./hypotenuse

that thing?

[frail bay]

oh yeah

[woven creek]

u dont know?

[frail bay]

i know now

[woven creek]

0.6 =6/10 = 3/5
so draw a right triangle with hypotenuse 5 and one sidee 3 and fnd other side ussing pythagoras theorem

and then mark theta

and find cos theta using similar definition

[bitter hemlock]

do you know SOH CAH TOA

[woven creek]

and since ur given angle is in 4th quadrant, cos wud be -ve

[frail bay]

i feel like i should know that but english really isnt my native language

ive been studying math in a foreign language all my life

[bitter hemlock]

oh ok then i guess you must have whatever language memory device

[gloomy pilot]

Hi can someone teach me 2.3 polynomial identities

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[patent dome]

Need help with e) 2

[topaz sinew]

@patent dome Has your question been resolved?

[patent dome]

<@&286206848099549185>

[hearty raptor]

looks similar with 1)

[patent dome]

With one I got 23

[patent dome]

As an answer

[hearty raptor]

Did you try solving using the same ideea?

Imagine that you would amplify with a (-1) and then just add and remove what you have extra

if it makes any sense

from what I believe I can t tell you the exact solution

[patent dome]

What is the solution

[patent dome]

<@&286206848099549185>

[hearty raptor]

it seems so simple

like 7th grade

but I ve been busting my head for a little more than 10 minutes

[patent dome]

Bro, I feel you

[hearty raptor]

what grade is this in?

[patent dome]

11

[hearty raptor]

how did you solve a?

trying to do a sistem

system

[patent dome]

I don't think you really solve for a. I think you are trying to find out what e) 2 would equal to

[hearty raptor]

what s the result for 1?

I mean

sorry

[patent dome]

The result is 23

[cursive patrol]

to do part E you kind of have to

the answer is very easy to find once you do that

[hearty raptor]

BRO KNOWING 1 you can solve it

GOT DAMN

[patent dome]

When I do what?

[hearty raptor]

try to amplify that thing

and you can get a

[cursive patrol]

a + 1/a = 23
a - 1/a = the thing you're trying to find

(what happens when you add these equations together?)

so to answer the question, just solve for a

[hearty raptor]

I was right with the system

hell yeah, 11th grade didn t beat me

[hearty raptor]

This guy is right

[patent dome]

I do not know

[cursive patrol]

i'm not asking u if u know

i'm asking u to do it

[patent dome]

How do I solve for a

[cursive patrol]

sqrt(a) + 1/sqrt(a) = 5

multiply both sides by sqrt(a)

[patent dome]

Isn't this the answer for e) 1

[cursive patrol]

no..

that equation i wrote is the equation you are given at the start

[patent dome]

Ok

[patent dome]

But why 5?

[cursive patrol]

??

[cursive patrol]

this is the given information

you see they use 5

[patent dome]

OK, so the answer is 25?

[cursive patrol]

no

do you know what it means to "multiply both sides by sqrt(a)"

@patent dome yes or no

[patent dome]

Yes

[cursive patrol]

try to do it

[cursive patrol]

no

[cursive patrol]

this equation

[patent dome]

It is that equation

[cursive patrol]

the equation i give you is the one the problem gives you

that is where you solve for a

[cursive patrol]

not this

[patent dome]

Can please just show me how it should be done

[hearty raptor]

note sqrt(a) with a t

and substitute

then you can solve for a

rewrite the equation

as followed

[patent dome]

Quadratic formula?

[hearty raptor]

t will be the value of sqr of a

[hearty raptor]

ofc

[patent dome]

So I assume the left one is probably the answer right?

[hearty raptor]

a1 should be (23-5sqrt(21))/2

a2 should be (23+5sqrt(21))/2

check which ones respects the condition

sqrt(a)-1/sqrt(a)>0

[patent dome]

How did you get this?

[hearty raptor]

Using the quadrulic formula

delta=b^2-4ac

x1,2=(-b(+/-)sqrt(delta))/2

[patent dome]

How did you get 5sqrt(21)

Because it's the answer for e) 2

[hearty raptor]

told you alr

after you calculate that is the answer

[patent dome]

FUCK I GETT IT Now!!! Sorry about being so stupid and thank you

[hearty raptor]

now that you get it this should explain

happy that I was able to help, keep on grinding

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[stone verge]

i need clarification pls

[stone verge]

for example suppose f(x) = x [1,2]. Does the derivative at x = 2 exist??? because according to the derivative formula we need (2+h)-2, but 2+h would be greater than 2??

[keen venture]

f(x) = x/x?

[stone verge]

oops

[nocturne sparrow]

no i think y = x from the domain

[1,2]

and think of it this way

derivative is your instantaneous rate of change

[keen venture]

So what's stopping you from carrying out (2+h) - 2?

It's not really a big deal that one is larger

[nocturne sparrow]

what would the instantaneous rate of change be at x = 2

[shrewd horizon]

f(2+h) isn't in the domain of f

if h > 0

[keen venture]

Ah yeah, I see. Then indeed that will stop you entirely

[stone verge]

there would be no right limit for h --> 0+

then no limit then no derivative

[nocturne sparrow]

yes

.close

@stone verge

[stone verge]

ok

.close

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[rough girder]

Pwease help 🥺

[rough girder]

Guys am i cooked

<@&286206848099549185>

[neon iron]

y=sqrt(x), as a parent function. Looks something like this

For A, it's reflected across the y axis, so we would negate what's inside the square root.

So A would be y=sqrt(-x)

B has a translation left 3, so we'd have to do the opposite and add 3 to x inside of the square root. So it would be y=sqrt(x+3)

C has a reflection across the x axis, so we have to negative the entire square root. It also has a translation up 4 units, so we need to add outside the square root. So will both transformations in mind, we'd write y=-sqrt(x) + 4

I hope that helps, I'm not the best teacher

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[glacial pumice]

Hey, I have $||u||=3$ and $||v||=2$ and $\theta=\frac{\pi}{3}$. What is $u+v$?

[thorny flame]

Totalani

[glacial pumice]

I know you can use $||u+v||^{2}=(u+v)\cdot(u+v)$ but I think im missing something

[thorny flame]

Totalani

[glacial pumice]

book says sqrt(19) but im not getting that

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[glacial pumice]

.close

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[vocal token]

how do i go about solving this

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[vocal token]

.close

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[visual dome]

Hi I need help if I solved this correctly

[bold nebula]

How you can say that cos(x) = x?

[pearl shuttle]

1.415/4 is not 0.285 🤔

[visual dome]

mb thats 1.14159

[pearl shuttle]

He's just doing substitution which might cause problems but was fine

[bold nebula]

What's more you solve for cos(x) not for x

[visual dome]

our course we can let cosx be temporarily x

[maiden cipher]

ick... use "u" instead

[pearl shuttle]

Better not for harder questions.. Lol

[gleaming reef]

huh

you wanna use diff variables bud

[visual dome]

its what the course instructed

[gleaming reef]

fine let x = cos(\theta)

[pearl shuttle]

Weird but whatever

[maiden cipher]

I mean.. you can... but it's too easy to get confused on which "x" you're working with

[gleaming reef]

then the course wanted them to use small angle approximation

[maiden cipher]

Global/local variable conflict

[gleaming reef]

that's what I'm getting with cos(x) = x

[pearl shuttle]

I c..

[visual dome]

is the result correct

[pearl shuttle]

x is not 0.29

[visual dome]

what should i do then

[pearl shuttle]

U found the range for cos x but not real x

What u need to do is cos inverse to find x

[visual dome]

right i move cos to right then would be arccos 0.28540

[pearl shuttle]

Yes

[visual dome]

its very close

but i think its slighty greater than pi

[pearl shuttle]

Which is?

[visual dome]

subbing it back to orignal

after arccos

[pearl shuttle]

But there are 2 answers

And question is asking in radians mode

[visual dome]

right 1st and 4th quads

we use the cast rule in this course

[pearl shuttle]

No

Oh

Yes

[visual dome]

oki i solve rq

i got x=1.28 and x= 5.00

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[lament fractal]

Is this correct?

[topaz sinew]

@lament fractal Has your question been resolved?

[lament fractal]

<@&286206848099549185>

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@lament fractal Has your question been resolved?

[lament fractal]

.close

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[crisp raptor]

well what was your answer?

maybe you put 2 ^ after 6^x ?

it looks like that a little bit

but the math is correct

$\ln\left(6\right)\left(2x+8\right){\cdot}6^{x^2+8x}$

[thorny flame]

tobi

[crisp raptor]

ln(6)*(2*x+8)*6^(x^2+8*x)

your answer is correct

[bright musk]

i think its a problem with the answer key

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[hasty surge]

Could someone help w this?

Would it be 3(sqrt)2 - 2

Instead of plus/minus

<@&286206848099549185>

[radiant marlin]

that's right

none of the other options are even on [1,4]

[hasty surge]

Ty !

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[icy iris]

can someone help me with the above please

[sleek gate]

dont ping 9000 times now ok?

also no one is obligated to help you

[icy iris]

i understand that but no one has helped me since the first time i pinged

[sleek gate]

.

[icy iris]

i came here looking for help and my work should not take to lo long

if you could help me wrap it up that would be awesome

[sleek gate]

im just here to tell you this, cuz your pings annoyed me so im going now and wish you for someone to help you!

[icy iris]

can you tell someone to come in this chat and help me

i really need help

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[lofty hearth]

Hey could someone help me solve the inverse of this

[plain nest]

Make f(t) y

Then switch x and y

And solve for y

I might be able to do this in my head rq

[lofty hearth]

What’s the rule to isolate the exponential again?

[plain nest]

Log

Log is inverse of exponents

Logbase.92

[lofty hearth]

So this?

Sorry it was 35, not 26

[plain nest]

Yeah

But make y inverse notation

[lofty hearth]

What do you mean?

[plain nest]

Instead of x it is t for you

[lofty hearth]

So this is what i write

[plain nest]

👍

[lofty hearth]

Also what do they mean by this?

[maiden cipher]

Bacteria are reproducing...

How many exist after time

[lofty hearth]

Aim i isolating t?

Or just finding the fct

[plain nest]

I think you just write the formula

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[neon iron]

im just here to check if I did my work right, was gone for a week and im struggling to grasp the concept

[neon iron]

equation d is cut off

should be 35,500(.88) to the power of 9 (since its from 2015)

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[neon iron]

no

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[cursive mango]

guys im convinced this is impossible

[loud oasis]

what have you tried?

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@cursive mango Has your question been resolved?

[cursive mango]

briefly tried integrating by y first, and then i was gonna try to maybe substitute u for 1+xy

gave up and tried an online calculator and it couldnt do it

[loud oasis]

that seems like a fine approach

[cursive mango]

gimme a sec i guess

got it thanks man

.close

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[grim urchin]

If ACO wasn't iscoseles and ADO was iscoseles would C be 50°, D be 20°, O1 be 140° and O2 be 60°
If so what would the reasons be for each one

[grim urchin]

Someone please just answer my question

<@&286206848099549185>

<@&286206848099549185>

[devout socket]

O2 should be 40 degrees not 60

If ADO is isoceles then angle D must be the same as angle A making it 20 degrees

sum of all angles in a triangle must be 180 therefore angle ADO is 140

Again for triangle ACO, we have a 90 degrees at A and 20 degrees at D, so angle C will be 70

O1 is same as angle AOD making it 140

O2 is 40 since O1 + 02 = 180

@grim urchin

[grim urchin]

Thank you

If they asked to give a reason for the statement for C, D, O1 and O2 what would I say for each of them then

[devout socket]

for D you can reason that the two equal angles must be acute therefore D = A

the value of C is a result of angle sum property

O1 and O2 are results of vertically opposite angles

[grim urchin]

Thanks so much

[devout socket]

You’re welcome

[grim urchin]

.close

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[formal edge]

I want to discuss m. So basically the answer I'm supposed to get is 0,

My first step is to just simplify what is actually standing here, so I did so.

My first step:

((Sin A)/(-cos A)) + ((-sin A)/(cos A))

[formal edge]

High school math, sorry.

[quiet carbon]

Which question

[formal edge]

M

[quiet carbon]

The denominator for the second one

It is cos (360+A)

[formal edge]

Yes.

[quiet carbon]

It can be written as cos (180+(180+A))

[formal edge]

I'm supposed to simplify it as cosA

[fading zephyr]

yes

do you something called the unit circle?

[formal edge]

The letters represent which one is positive

[fading zephyr]

yess

[formal edge]

Unit circle?

I guess that

[fading zephyr]

this one

yes so lets say angle a is somewhere between 0 to 90 degrees

[versed cairn]

nvm

[tulip monolith]

Can you explain your first step?

[fading zephyr]

and let us simplify sin (A - 540)

[formal edge]

Ok so we're at 360 degrees. So adding a sharp angle(A) will put us in the 1st quarter which is where cos is positive

[fading zephyr]

everything is positive in the first quadrant

[formal edge]

I just simplified the equations sorry guys English isn't my home language

[formal edge]

Exactly hence the A for all

S for sin

T for tan