help-38

[obtuse brook]

@past river

[past river]

yes sorry, am playing a game at the same time. One moment

Yeah, I get that too!

now you could solve for b if you wanted but the question only wants m so, you're pretty much done :)

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[frosty glacier]

How do I evaluate this limit without L'hopital's rule (We haven't learnt it yet)

$$\lim_{x\to a} \frac{\sqrt{2x+3}-\sqrt{2a+3}}{x-a} $$

[solid kiln]

Sb

[frosty glacier]

I tried rationalising but it didn't work out

[prisma drift]

Sqrt(2x+3)+sqrt(2a+3)/Sqrt(2x+3)+sqrt(2a+3)
Multiply

[solid kiln]

Sb

[frosty glacier]

Which is still 0/0

[zinc ginkgo]

Simplify that numerator

Things should cancel after you rationalize

[frosty glacier]

OHHHH

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[wraith hinge]

im struggling with this question as it seems to not be a GP or an AP

[wraith hinge]

<@&286206848099549185>

hey cant u use infinite sums?

i dont know how to do that

we use a notation to find the sum of a sequence of terms

this sequence is unbound

it does not diverge to a specefic value like in the answer 64

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[dim widget]

Completely lost on this

1. Let f(x, y) = x cos(𝜋y) − y sin(𝜋x).
   Find the second-order Taylor approximation for f at the point (3, 4).
   g(x, y) =

[dim widget]

i didnt learn second order taylor approx ;-;

[ripe valley]

then what form of taylor approx did you learn

[dim widget]

none

😭

[ripe valley]

....

do u atleast know single variable taylor approximation?

[dim widget]

my professor did not finish the course, but its on the homework so i must complete it

we didnt do taylor series at all

[ripe valley]

Huh? They're asking for multivariable taylor series when u dont even know single variable..?

[dim widget]

supposed to learn it on my own

ahh idk

[ripe valley]

well

Iemme see if i can find a video for u

u first should understand how it works

then we can help u if u had any problem finding it

[dim widget]

okay thank you0

[ripe valley]

https://www.youtube.com/watch?v=3d6DsjIBzJ4

I suppose I'll have to teach u taylor series in 2 variables

https://www.youtube.com/watch?v=a58otkmkHtk&pp=ygUNdGF5bG9yIHNlcmllcw%3D%3D

[dim widget]

oh

[ripe valley]

https://www.youtube.com/watch?v=Iub16Y1ZXRw&pp=ygUNdGF5bG9yIHNlcmllcw%3D%3D

[dim widget]

i am so sorry i actually do know single variable taylor

[ripe valley]

bruh

its okay then

well

[dim widget]

i thought u said first order taylor approx

wait is that the same thing

[ripe valley]

No no ofc not

[dim widget]

okay

[ripe valley]

a n order taylor series means that u use the first n derivatives

[dim widget]

yes

[ripe valley]

now

if n =1

i.e. a linear approx

then

if the taylor polynomial for some function f(x) around a

is

$f(a) + f'(a)(x-a)$

[solid kiln]

ItzKraken

[ripe valley]

then would would it be for f(x,y)

[ripe valley]

this is some constant + the equation of the tangent * another constant

because f(x) is a curve

but if f(x) is a surface

i.e. a function with 2 inputs

then

what do u think this becomes

I am pretty sure u wont get the intuition first try and its tough to explain this on chat, so I do have a few links read up on those

[dim widget]

thank you

im assuming its related to vectors, but im honestly lost

[ripe valley]

nah

well u can if u want to

but not necessary

[dim widget]

okk

[ripe valley]

n order taylor series for f(x) is $f(x) = \sum_{i=0}^{n} \frac{f^i(a)}{n!} (x-a)^i$

[solid kiln]

ItzKraken

[ripe valley]

correct?

where f^i(a) is the ith derivative of f(x)

@dim widget

[dim widget]

uhhh

yes

[ripe valley]

nice

[dim widget]

i do recall this

[ripe valley]

now if f(x) had two inputs

and u had to take the derivative

what would u do

[dim widget]

partial

derivative

?

[ripe valley]

yes

[dim widget]

ah

[ripe valley]

well but if u take only 1 partial derivative

u would be able to approximate on one input right?

i.e. the 'x part' of the range

[dim widget]

yes

[ripe valley]

so u need both partial derivatives

agreed?

[dim widget]

yes

as in

x and y

right

[ripe valley]

yes

now

similarly

u would also need the (y-b)^i

because else the partial derivative w.r.t y evaluated at b is just a constant right?

[dim widget]

(y-b)^n is replacing which part in the single variable taylor?

[ripe valley]

its not replacing anything, u would also need it correct?

[dim widget]

oh

[ripe valley]

also not ^n ^i i accidently wrote n

[dim widget]

ohh i see, so its like the x-a^i part but for y as well since we also take it into account now

okay

[ripe valley]

now another thing

if ur gonna approximate on a surface

do u think 1 summation is enough?

i mean think about it what would a single summation represent

lemme write down what have rn

[dim widget]

thank you

[ripe valley]

$f(x,y) = \sum_{i=0}^{n} \frac{f^{i(x)}(a) f^{i(y)}(b)}{n!} (x-a)^i(y-b)^i$

[solid kiln]

ItzKraken

[ripe valley]

i(x) means the ith derivative w.r.t. x and i(y) means that but w.r.t y instead of x

[dim widget]

oh okay

why are the two multiplied here?

[ripe valley]

huh wait 1 sec

oh shit

[dim widget]

hm?

[ripe valley]

okay I think at this point I should just the taylor series in 2 variables

this is it

[dim widget]

a summation of summations?

[ripe valley]

yes

even I dont have the full intuition for this

[dim widget]

this is crazy to me lmao

ok, but i can see how its like adapted to fit 2 variables

i guess can we move onto how to utilize it

[ripe valley]

n=2

[dim widget]

whats f(a,b)?

is it the gradient

wait no thats fx(a,b)

uhhhhhh

oh are a and b the point like (3,4) in my case

[ripe valley]

no the function

[ripe valley]

yes

[dim widget]

oh

[dim widget]

so this is what it means by second order

should i find $f_x f_(xx) and. f_(yy)$ first

lmao

[solid kiln]

risa★

[dim widget]

damn i cant format this right but iu know what i mean

[ripe valley]

yes

[dim widget]

okayy

[dim widget]

oh nooOOO

oh y not x

still wrong :(

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[delicate jasper]

For each of the following pairs of hyperplanes, determine the affine subspace D defined by their intersection. Represent the affine subspace as v + span(b1, . . . , bk), where v is a vector and {b1, . . . , bk} is a basis for the underlying vector space of D.

[delicate jasper]

How do I do it? I can't find any examples

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[wraith hinge]

Where $i$ is the usual imaginary unit ($i^2 = -1$) and $\vb q$ is a constant vector, write
[
F(\vb q) = \int_\Omega e^{i\vb q \vdot \vb r} \dd V, \textqq{where} \Omega = B^3(\vb 0, R) \subsete \R^3.
]
By introducing appropriate polar coordinates, evaluate $F(\vb q)$.

[wraith hinge]

So I think I already solved this by orthogonal transformation and spherical coordinates. I'm looking to see other people's approaches on how to attempt doing this though

[zinc ginkgo]

Try stack exchange

[ripe topaz]

is this math ? (i'm genuinely asking)

[wraith hinge]

I'm probably going to get told off, in all honesty

[zinc ginkgo]

Who cares?

Or try #advanced-analysis

Your question is very niche.

[edgy cedar]

Does this get you to some bessel function stuff

[wraith hinge]

no

well not how I did it at least

[wraith hinge]

sure

.close

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[edgy cedar]

I’ll look at it when I get home maybe

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[sinful flume]

ello

[sinful flume]

im solving for amount of integer solutions. i will upload a picture.

everything scribbled is a failed attempt. where ive circled 1 and 2 is two equations that account for the equality on x3

im not sure if i include the <= 18 correctly

<@&286206848099549185>

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[terse parrot]

I am having trouble figuring out how these two intersect, as I believe that the depreciation curve decreases linearly and the savings rate increases linearly. Can someone show me how to graph this information so that the two points intersect?
Suppose that in the economy where technology 1 is used, the capital
depreciation rate is 10% and the savings rate 18.4%. Sketch a graph of the
depreciation and savings curves, showing the steady state equilibrium.

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<@&286206848099549185>

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[tribal fractal]

i have to find the solution to dy/dx=f(x)/y^2

[tribal fractal]

im doing this using seperation of variables

this is what i have so far

$\frac{\mathrm{d}y}{\frac1{y^2}}=f(x)\mathrm{d}x$

[solid kiln]

talk_less

[tribal fractal]

$y^2\mathrm{d}y=f(x)\mathrm{d}x$

[solid kiln]

talk_less

[tribal fractal]

$\int{y^2\mathrm{d}y}=\int{f(x)\mathrm{d}x$

[solid kiln]

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

[tribal fractal]

$\frac{y^3}3=\bdots ?$

[solid kiln]

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

[tribal fractal]

not sure what to do next

[rugged latch]

can you show the problem as it appears in ur homework

like take a pic

is f defined as the derivative of F?

@tribal fractal

[tribal fractal]

part b

[rugged latch]

ah ok

[tribal fractal]

am i doing the right method?

[rugged latch]

uh lemme see

hmm

[tribal fractal]

also im not sure why they give me f(x)=F'(x)

oh wait

that means F is the antiderivative

[rugged latch]

yes

[tribal fractal]

so the right hand side of this should be F(x)+C

[rugged latch]

yea

[tribal fractal]

so do i just plug in x=7 now

y^3/3=F(x)+C
y^3=3F(x)+C1

y^3=3*F(7)+C1=519+C1

what do i do with this?

ohhh

i have to use the point (3,2) to find C

ok i see how to do this

thanks!

.close

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[rugged latch]

lol no need to thank me, u did it all urself, nice job

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[tender steeple]

hi how to do part b

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[winter tendon]

Y=1/2(x-1)^2-1 I don’t know what to do

[winter tendon]

I’m stuck on something’s

<@&286206848099549185>

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[blissful bison]

what don't you know to do? you want me to set a question for you ?

[winter tendon]

Like how do I start the question

[blissful bison]

the potential question would be: Please sketch the graph of the given function using appropriate transformation of the parent function

[winter tendon]

I have to find the vertex though

[blissful bison]

so first formulate nicely question, since, the given function is not a quadratic function

and only quadratic function has a vertex

correct the function plz

if

$y=2\left( x-1 \right)^{2}-1$

[solid kiln]

Joanna Angel

[blissful bison]

then the vertex is (1,, -1)

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[winter tendon]

.reopen

[trim jolt]

✅

[blissful bison]

if

$y=\frac{1}{2}\left( x-1 \right)^{2}-1$

[solid kiln]

Joanna Angel

[blissful bison]

vertex is also (1, -1)

[winter tendon]

Ok

How did you get the half 1/2 to go away

Like how did you solve the problem

[blissful bison]

there is a general formula for the vertex form of any quadratic funcion

$y=a\left( x-p \right)^{2}+q$

[solid kiln]

Joanna Angel

[blissful bison]

then the coordinates of the vertex are: (p, q)

so coefficient a does not play a role above

unless, you have a general form of the given quadratic function, but it is not your case now

[winter tendon]

So I inform the 1/2

Ignore

Ok

@blissful bison

[blissful bison]

right 🙂

[winter tendon]

Sorry if that sounds dumb

.close

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[winter tendon]

Reopen

Wait how would I find the domain and range

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[winter tendon]

.close

[safe flower]

question
is $$\infty + \infty = indefinite$$

[solid kiln]

Maladroit

[blissful bison]

indefinite is only:

$+\infty -\infty $

[safe flower]

but

what if

$$\frac{\infty + \infty}{\infty - \infty}$$

[solid kiln]

Maladroit

[safe flower]

is that considered as indefinite

because of the denominator

[blissful bison]

then you need to cope with denominator

[safe flower]

wdym by that

[blissful bison]

cope with denomiantor = calculate it

[safe flower]

oh wait so only denominator is

the indefinite? not the whole fraction?

[blissful bison]

correct, only the denominator, but if later denominator = + or - infty, then

all fractin becomes

indefinmite

but first you must deal with denomiantor

do you want to see such example ?

[safe flower]

if it involves limits

i have an example here actually

[blissful bison]

great

[safe flower]

$$\lim_{x \rightarrow \infty} \frac{5e^x + e^{2x}}{e^{3x} - e^x}$$

[solid kiln]

Maladroit

[safe flower]

e is 2.72

[blissful bison]

lol

i did not know it hahahah

🙂

[safe flower]

what 😭

[blissful bison]

i am joking

ok look at it : and pay attention carefully

[safe flower]

oh

[blissful bison]

:

[safe flower]

??

[wooden mural]

2.718281828459045........... 🙂

[blissful bison]

$\lim_{x \to \infty } \frac{5e^{x}+e^{2x}}{e^{3x}-e^{x}}=\lim_{x \to \infty }\frac{5+e^{x}}{e^{2x}-1}=\\=_{\cdots }$

[solid kiln]

Joanna Angel

[blissful bison]

can you continue that ?

[safe flower]

wait

why did u use e^x?

not 5e^x?

[blissful bison]

please thikn why i did that

[safe flower]

our prof taught us

[safe flower]

because its the common of everything?

all terms have e^x

[blissful bison]

yes

[blissful bison]

i want you to write such a form of the fraction I offeerd, to make it clear, that the expression reaches 0

[wooden mural]

I reasoned that it would be 0.

[blissful bison]

yes )

[wooden mural]

And to explain why

Basically when a limit approaches infinity often times what we care about analyzing is the horizontal asymptote, if it has one to speak

here are the 3 cases

you must know of

should know*

[safe flower]

1/e^x = 0??

[wooden mural]

you plug infinity in

to the x

[blissful bison]

it is not = but limit

[wooden mural]

and you get e^infinity

[safe flower]

right right

[wooden mural]

1/infinity

[safe flower]

that is 0

[wooden mural]

yep

[blissful bison]

but no need to tal about horizontal asymptotes here

[wooden mural]

nah its good to know

[safe flower]

our teacher proff doesnt care if the horizontal asymptote exist

he wants to teach us the behavior of the graph

[wooden mural]

x^n/x^b if b > n then horizontal asymptote is 0

if ax^n/bx^c if n = c then horizontal asymptote = a/b

[safe flower]

yeah

thats the shortcut, but our proff wants to see solutions

[wooden mural]

if x^n/x^o where n > 0 then there is no horizontal asymptote

[safe flower]

so usually when x apporaches infinity, you use the term variable with the HIGHEST degree

[blissful bison]

again, to evalue the limit, this theory is not necessary, unless, the exerice is about identifyign asymptoesw

i fi give you a limit to find i do not care, whether you know or not asymptote defintion

[wooden mural]

You could use L'Hopitals rule as well if you know it.

[safe flower]

were not there uet

[wooden mural]

Note the following rule is applicable when we evaluate limits that are interdeterminate where when evaluated we see 0/0 or infinity/infinity. If f(c) = g(c) = 0 or f(c) = g(c) = infinity, you can use the rule.

example

lim x --> 0 sin(x) / x

Up until now you may have been memorizing that its 1

lim x--> 0 sin(x) / x = sin(0) /0 = 0/0 (Indeterminate form 0/0 there L'Hopitals Rule is applicable where we differentiate the top and then the bottom. Note we are not using quotient rule common mistake
lim x--> 0 cos(x)/1 = cos(0) = 1

[safe flower]

what quotient rule common mistake?

[safe flower]

so how do i like

or whats the process of Lhopitals?

[wooden mural]

A lot of ppl when they see f(x)/g(x) they think they have to use the quotient rule for derivatives, but no.

Here is the process to identify whether L'Hopitals rule is applicable or not

[safe flower]

im not in derivatives yet 😭

[wooden mural]

ohhhhhhhhhhh.

[safe flower]

so Lhopital is used for indeterminate

[safe flower]

were in the basics basics

[wooden mural]

I see. Limits laws, etc.

Epsilon delta definition of limits.

nice.

[safe flower]

rn were discussing limits in exponential and logarithim function

[safe flower]

idk what that means but

our proff told us that limits help indiciate what is happening in a very small region of a graph

[wooden mural]

No worries, your going to learn abt the sandwich theorem soon.

Take a peak into proofs

they r rlly enjoyable dw this ain't apart of ur course

[safe flower]

https://tenor.com/view/little-big-planet-what-da-hell-little-big-planet-what-ayo-what-gif-13768843195184919620

[wooden mural]

but if u wanna know the intuition behind a theorem learn it.

[safe flower]

ofc

[wooden mural]

Gl grinding brother.

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[heady ermine]

I was trying to solve for the volume using washing machine method

and I get 168 pi which is not the right answer

was wondering where I messed up

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[untold lichen]

Does the a(a+1)(a+2)-1=b²+2=10c+3 have solutions over integers?

[untold lichen]

It is a question from the math olympiad I was participating in that happened like a month ago. The provided solution turned out to be wrong. It isn't the question itself, but the solution basically comes to it:
If it doesn't have any solutions, the answer should be 8.
If it has 5 or more, then it should be 5.
If it's something between 1 and 4 then it'd be pretty weird ngl.

The answer considered correct was 8 but mine was 5 and it got marks as if it was correct

Which is weird af

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[surreal eagle]

using the idea of the fundamental theorem of arithmetic, and p is a prime number and n is a positive natural number
5p+9=n^2
5p=n^2-9
5p=(n-3)(n+3)
then we are split into 3 cases,
case 1:
5=n-3
p=n+3
thus n=8 and p =11
case 2:
5=n+3
p=n-3
thus n=2 and p=-1
and the 3rd case n-3=1?

[surreal eagle]

i get splitting into case 1 and 2 but why do we have the third case where n=4?

[marble wharf]

well why shouldnt we

you have 2 prime factors, 5 and p

which means that the product can either be 1*(5p) or 5*p or p*5 or (5p)*1

[surreal eagle]

oh

and the third case covers when its 1*(5p)

[marble wharf]

\* for *

yes

[surreal eagle]

okay that clears it up, thank you. i didnt see it like that

[marble wharf]

and for completeness you should also do the fourth case with n+3=1

(but that quickly leads to a contradiction)

[surreal eagle]

yea, clearly false

thanks for explaining it so clearly 👍

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[chilly dune]

(12x - 1)(6x - 1)(4x - 1)(3x - 1) = 5

[chilly dune]

Where should i begin with this problem?

<@&286206848099549185>

[zinc ginkgo]

,w solve (12x - 1)(6x - 1)(4x - 1)(3x - 1) = 5

[solid kiln]

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[zinc ginkgo]

Rational root theorem?

[tepid cobalt]

Eeh, where did this problem come from? Is it an olympiad thing, or otherwise consider the rational root theorem?

[chilly dune]

IDK, the math teacher gave a couple of problems and this is just one of them

[tepid cobalt]

It might be interesting that 2(6x-1) = 12x-2, 3(4x-1) = 12x-3 and 4(3x-1) = 12x-4

[chilly dune]

I have no idea about that theorem

[tepid cobalt]

What level of math are you at?

[chilly dune]

grade 10 (high school)

so the roots are expressed in the rational form, p/q

but to apply it you will have to painfully multiply everything to find the coefficients

[tepid cobalt]

Yeah, but only the "top" coefficients and the "bottom" coefficients.

Like 12*6*4*3 is the coefficient of x^4

[chilly dune]

right so it is like the coefficient of x^4 and the constant term

[tepid cobalt]

(-1)(-1)(-1)(-1) - 5 is the coefficient of x^0.

Yeah

If you want to give the rational roots theorem a go, yes.

[chilly dune]

just read about it

ill try

but this only applies for the rational roots of the polynomial, not for the imaginary roots ?

[tepid cobalt]

Ok, otherwise it might be interesting to perform a trick and multiply both sides by 2*3*4 to get
(12x - 1)(12x - 2)(12x - 3)(12x - 4) = 5*4*3*2

[tepid cobalt]

Yeah, only rational roots. But when you find the two rational roots, you can use them to get a quadratic polynomial (x-r1)(x-r2) that you can factor out of your 4th degree polynomial.

The other factor will itself be a quadratic, and you can show that this quadratic has two imaginary roots.

[chilly dune]

not really sure how to apply it but

the theorem states that p and q are the factors of the corresponding coefficients

so the real root will surely have a one on the numerator, but how do i calculate the denominator of these roots?

[tepid cobalt]

Yeah, if p/q is a rational root, then p divides the constant term -4, and q divides the top coefficient 12*6*4*3 = 864

[chilly dune]

wait, it is not 1

[tepid cobalt]

The idea of the theorem is that this restricts what p and q can be. For instance, there are very few numbers that divide -4 (they are -4, -2, -1, 1, 2, 4, so these are all options for p).

[chilly dune]

right, but still the q value is pretty large.

[tepid cobalt]

Yes, it is pretty inconvenient for q, since 864 has a lot of factors.

One would usually start with p=1 and try different values of q.

[chilly dune]

it simplifies out to -1/216. But still, 216 has a lot of factors

[tepid cobalt]

Ah, of course. So for sure p = 1 or p = -1.

Eh no

The polynomial has to have integer coefficients.

before you can apply the rational root theorem

[chilly dune]

it does have int coefficients?

[tepid cobalt]

Yeah, but I thought you got 216 from 864/4, i.e. by dividing all coefficients of the polynomial by 4.

[chilly dune]

no. the last term by the coefficient of x^4 yields -4/864, which simplifies to -1/216

[tepid cobalt]

,w expand (12x - 1)(6x - 1)(4x - 1)(3x - 1) = 5

[solid kiln]

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[chilly dune]

yeah so by the theorem you mentioned, the zero would be a factor of -1/216

216 is 2^3 * 3^3

so i think it is possible

[tepid cobalt]

I don't follow.
p can be any of the numbers -4, -2, -1, 1, 2, and 4 - maybe it is 1.
q is a factor of 864. It can be 864.

In this case you would have that p/q = 1/864, which I don't think fits into "being a factor of -1/216".

(also, when dealing with numbers, being a factor of something else is a phrase usually reserved for integers)

[chilly dune]

no, not a factor of -1/864. A factor of -4/864 which on simplification yields -1/216

Oh, right

i see your point

p can be any factors of -4 and q can be any factor of 864 so simplification here is not possible?

[tepid cobalt]

Right. Like we could also have p = 4 (a factor of 4) and q = 1 (a factor of 864), but 4 = p/q is not "a factor of -1/216".

[chilly dune]

yeah. I seemed to have consider it a fraction and accidentally simplified it

But even then, there are lots of factors of 864

864 = 2^5 * 3*3

So i think the rational root theorem is maybe not applicable here. Although it restricts the limit of zeroes, it still leaves us with a lot of possibilities.

[tepid cobalt]

Yeah, it will be a tiresome approach.

If you start from the low numbers, i.e. first put p = 1 and try - then you will first see that q = 1 doesn't work, but luckily q = 2 does work.

Next solution is p = 1 and q = -12, which takes some cases to come to.

[chilly dune]

i mean if you could find 2 roots then it will be easy

but anyways, i will maybe find solutions to this tomorrow

Thank you very much for you help. Learned something about the rational root theorem.

✌️

[tepid cobalt]

Happy to help and np

[chilly dune]

thank you

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[velvet parrot]

Prove that given that CD || EA. They have 1 side in common they still need to to have at least 2 equal angles in order to prove that. Im having trouble to find those.

[velvet parrot]

Someone says that <BCE = <CDE. I dont know why

[dapper flicker]

remember the stuff with transversals

its only true with two parallel lines

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[woven nova]

hey^^

[woven nova]

im doing stochastics right now

we are talking about probabilityspaces

$(\Omega ,\mathfrak{A}, P)$

[solid kiln]

Martin

[woven nova]

then, we talk about events, which we call A

so we write

$A\in\mathfrak{A}$

[solid kiln]

Martin

[woven nova]

the question is, how do i say this? A in A?

i mean, everyone i talk to know what is meant, but it sounds so weird to me

is there another consensus i am missing?

i guess i could say "A in fracture A"

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[loud wasp]

for part c where did the 1/2 come from?

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[dire matrix]

how to get aec?

[reef plaza]

You have 3 angles of a 4 sided shape

What do angles in a 4 sided polygon add up too

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[thorn pawn]

been having difficulty understanding this

[zinc ginkgo]

What did you try

[thorn pawn]

substituting k+3 and k-3 in the equation

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[thorn pawn]

would youhave a clue as to how i could go about solving it

[zinc ginkgo]

That should work

Use the definition of root

Alternatively you can use the definition of root and factor the quadratic as a product of terms that look like x-root

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[slim mountain]

I don’t understand what they wanted me to do in here?

[sour field]

You did the first step right

Now you just have to continue you it

Do you know power rules with logs? Like how you can move the power of the argument to the front of the log

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[slim mountain]

Like the exponents to the front?

[sour field]

Yea

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[tender thistle]

can someone help verify these?

[tender thistle]

I am pretty sure the first one id right but I dont think for the second one J made use of the definution right and I am kinda struggling

[crimson rover]

which part of the second one are you unsure of?

[tender thistle]

I don’t know if I utilized the definition provided in the question properly

[crimson rover]

whats definition of antisymmetric?

@tender thistle

that its simply not symmetric right?

[tender thistle]

mhm

[crimson rover]

so why dont you provide an example instead?

unless your teacher wants you to provide a general case?

[tender thistle]

yeah think so

[crimson rover]

one second

let me read it

[tender thistle]

okie thanks!

[crimson rover]

would the symmetric case be

$b\leq a $ if $k\in N$ s.t $b+k=a$ ?

[solid kiln]

Joshii

[crimson rover]

or would it be

$b\leq a $ if $k\in N$ s.t $a+k=b$ ?

[solid kiln]

Joshii

[crimson rover]

If its this one, then you can provide a counter example

[tender thistle]

do you mean antisymmetric?

[crimson rover]

well you would have to look at the symmetric case

and see if its true or false no?

and if its false then its antisymmetric

true then its symmetric

[tender thistle]

ohhh I see what you are saying!

[crimson rover]

this would be symmetric

because its clear in this case that a will always be greater or equal to b

this is becase $\mathbb{N}={0,1,2,3,4, . . .}$

[solid kiln]

Joshii

[tender thistle]

mhm!!

[crimson rover]

if its this then its antisymmetric

hm

if we have a=2, b=3 then it should be the case that $a\leq b$ but if we have a=3 and b=2 then its not the case that $b\leq a$

[solid kiln]

Joshii

[crimson rover]

and i think thats enough because a+k can still equal b in this case

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[wise pecan]

how do i find asymptote of (20+10x)/(2x) ?

[flint sluice]

vertical or horizontal?

[wise pecan]

horizontal

[flint sluice]

okay

are you familiar with limits?

[wise pecan]

we are in alg 2

so no

[flint sluice]

okay so

you can look at the highest degree term for the nominator and denominator

and simplify them

so here that would be 10x/2x

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[wise pecan]

which is 5

[flint sluice]

yea

so there's an horizontal asymptote at y = 5

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[maiden parrot]

Is the first step I did correct

And how to do last question

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[silent canyon]

Seems incorrect

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[crimson rover]

how to integrate using u substitution?

[crimson rover]

i already split it up

so

$\int(x\sin^2x)-\int(x^2\sin(x)\cos(x))$

[solid kiln]

Joshii

[crimson rover]

then idk

i havent done u substitution for trig functions yet

only power functions/rational functions

[whole coral]

do you even need to do it by substitution? You could do it by spotting it by reversing product rule (not even needing IBP)

[crimson rover]

im not that smart to see that 😔

[whole coral]

Think about what $x^2$ differentiates to, and what $\sin^2(x)$ differentiates to

[solid kiln]

@whole coral

[crimson rover]

sin^2(x) differentiates to 2sin(x)cos(x)

x^2 =2x

[whole coral]

Yep and yep

[crimson rover]

o

[whole coral]

Do you notice it yet?

[crimson rover]

should i be looking at this

[crimson rover]

or this

[whole coral]

The original think about the fact that x^2 differentiates to 2x, and that sin^2(x) differentiates to 2sin(x)cos(x)

And how all of those terms (almost) appear in that original - almost as if you differentiated something(!)

[crimson rover]

$\int x\sin^2(x)-x^2\sin (x)\cos (x)$ $ dx$

[solid kiln]

Joshii

[whole coral]

Oh wait that minus sign makes it not as nice

For some reason I read it as +

[crimson rover]

oh

[whole coral]

Wonder what they want you to substitute then

[crimson rover]

i havent really done anything like this before, so sorry if it takes me a bit

but my intuition tells me to substitute sin(x) for u since thats whats common

[whole coral]

seems like it might be a bit painful might be it though!

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[rich stag]

I think it's B but

It's so hard to decipher

What it's asking

😭

[sweet hare]

yes B looks correct

[rich stag]

Yay

Ty

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[errant forge]

Hi

[errant forge]

How do i know if im ready for calculus

[half oracle]

e

[zinc ginkgo]

Don't use math help channels for non math problems

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[half oracle]

Be able to do pre-calc first I guess

[half oracle]

Ohh okay sorry~

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[woven bone]

Help I don’t know how to set up the equation!

[flint sluice]

don't need to set up any equation here

[woven bone]

Oh how do I solve then?

[flint sluice]

do you know how composite functions work

[woven bone]

💀

No

[flint sluice]

ok so (g o f) is like

g(f(x))

[woven bone]

Oh thatt

Yeah

[flint sluice]

so the x first goes through the f function, and then the output goes through the g function

[woven bone]

Yes

[flint sluice]

now the output is = 7

so you gotta work your way back

to the first input

[woven bone]

Ohhh

So x=7?

[flint sluice]

the output of g is 7

so find for which x value is y = 7

in the g table

[woven bone]

Mh

Is this what you mean

[flint sluice]

yeahh

so you see that the output of the inside function (f(x)) is -4

[woven bone]

Bc x of g is 4, then f(x) = 4 is x=3

[flint sluice]

so the input gotta be 3

according to the table

[woven bone]

Yeahhh

Tysm

[flint sluice]

np!

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[lament stone]

how do i remove the fraction

it’s exponents laws btw

[lament stone]

like i know u have to turn it into negative exponents

but i don’t understand how

[south brook]

I can write 1/c as c^(-1)

[lament stone]

yes ik

[south brook]

1/23 as 23^(-1)

So I can write 4/x² as?

[lament stone]

uh

4*x^(-2)

[cedar flower]

hint: ||a/b = a * 1/b||

[lament stone]

idk 😭

[south brook]

I can write 1/4 as 1/2² which is 2^(-2)

[lament stone]

oh

so it’s like 4*1/x^(-2)

[cedar flower]

yeah, because you can write 4 as 4/1, and then it's basic multiplying of fractions

but also be careful with doing this 1/x^-2

you're mixing fraction notation (1/x) with negative exponents (x^-2) which is like doing 1/(1/x)

[lament stone]

isn’t 1/x^2

[cedar flower]

or doing x^-(-2)

[lament stone]

x^(-2)

[south brook]

Yes

[lament stone]

okk

[cedar flower]

yeah you've done it twice by accident

[south brook]

and 1/x^(-2) is x²

[lament stone]

is it like this

[south brook]

Times 1/y³

Then eliminate the fractional part

[lament stone]

so then it’s 4*x^(-2)*y^(-3)

[south brook]

Yes

[lament stone]

okk

like this

[lament stone]

can u double check

[south brook]

Yes

Or

If you wanna get fancy

You could do

4(xy)^(-2)y^(-1)

[lament stone]

wait do u need to multiple the 4 with the y as well?

like what we did with x

[south brook]

I used brackets

[lament stone]

wait so the last line is correct right

[south brook]

Yes correct

[lament stone]

okkk

tysm

[south brook]

Cheers

[lament stone]

could u help me with another similar one?

[south brook]

Which one

[lament stone]

12

[south brook]

12?

[lament stone]

yea question 12

[south brook]

Try it yourself

[lament stone]

does this make sense?

[cedar flower]

(I think you have once again tried to use both 1/x and x^-1, rather than choosing one or the other)

[lament stone]

oh

so how do i change it to only one

[cedar flower]

1/r^-1 = 1/(1/r) = r

or the other way to show it, 1/r^-1 = r^(-1)*(-1) = r

[lament stone]

what if it was r^(-2)

[cedar flower]

then that's 1/(r^2)

[lament stone]

ok wait so for the s

part

is it just s then?

[lament stone]

is this correct

[lament stone]

could u check for me

[south brook]

Answer to 12 ) is 2^(-1)r^(-2)s^(-1)p³q⁵

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[wispy karma]

radius of convergence

i dont get the final step

nvm

.close

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[solid robin]

<@&286206848099549185>

[prime badger]

yes

[solid robin]

so we're given the above equation

we need to proof that the second one is also true

[prime badger]

i cant unnderstand it

i am sorry

[solid robin]

?

[cosmic meadow]

He means your writing is not good

[solid robin]

ah ok, imma rewrite it give me a second

[prime badger]

yeah

[solid robin]

if it's still not understandable:
We're given:
ln(1 - U/E) = - t/tau

Proof that:
ln(1 - U/E) = t/tau

[prime badger]

hmm give me a minute

have you ttied taking

the

exponential sides of both eq

e^(ln(1 - U/E)) = e^(-t/tau)

[solid robin]

yeah I tried that

[prime badger]

then simplify the left side

1 - U/E = e^(-t/tau)

[solid robin]

but I got that: - ln(1 - U/E) = t/tau

so it didn't work

[prime badger]

we can rearrange the equation to isolate ln(1 - U/E)

ln(1 - U/E) = ln(e^(-t/tau)

@solid robin

[solid robin]

can you show me how plz

[prime badger]

just rearrnte it

reaggrange

[solid robin]

wait wdym

[prime badger]

rearrange

[solid robin]

i'm still getting - ln(1 - U/E) = t/tau

[prime badger]

show me your solving

[wooden plover]

you're sure you have the question written correctly anyway ?

[solid robin]

if we take the ln of both sides, we get this: - ln(1 - U/E) = t/tau

[solid robin]

yep

[wooden plover]

cause unless everything is 0 that won't work

that's why I'm suspicious

[cosmic meadow]

!original

[trim jolt]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[prime badger]

no no hea has the rgith question but there is something wrong with his solving ig

[wooden plover]

I mean the question is literally
"assume B = -C
show B = C"

unless I'm misreading

[prime badger]

yeah ig that makes sense

[wooden plover]

so this is about RC or RLC circuits I suppose

it's been a while I looked at these lol

[solid robin]

yep

[cosmic meadow]

start by negating the original equation

[solid robin]

what do you mean

[cosmic meadow]

If you multiply by -1 you’ll prove that is not the same using logarithmic identity

[wooden plover]

that second equation makes no sense physically anyway

[cosmic meadow]

That’s why I wanted t9 see the original question

[wooden plover]

the tension would go to -infty

[solid robin]

i don't know if we have to use some physical formula tbh

[cosmic meadow]

Do you have a picture of the original question?

[solid robin]

that is the original question

[wooden plover]

I'm just trying to argue why the question as written right now is BS, I'm not trying to solve it

[cosmic meadow]

That is not a picture

[solid robin]

this is what the teacher wrote on the board

[cosmic meadow]

Then you copied wrong or he made a typo

[solid robin]

r u sure sure?

like there's no way to solve this?

[cosmic meadow]

That’s my take

[wooden plover]

also my take, go ask your classmates or something

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[solid robin]

.close

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[trim jolt]

@wraith hinge Has your question been resolved?

[sweet hare]

I think you'll have to take the log

logarithm

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[proven steppe]

can someone help please?

.close

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[wraith hinge]

What output does a compound interest formula give you in terms of units?

[wraith hinge]

say for example this

$1500=500(1+\frac{0.12}{56})^{56t}$

[solid kiln]

mj

[wraith hinge]

since we're using weekly, shouldn't the answer be in weeks?

[brave iron]

why is it 56? arent there 52 weeks in a year?

[brave iron]

in this case, with your current formula, no, you multiplied t by 56 (52?) to account for years into weeks, so the answer is in years

[brave iron]

it depends, you need to pay attention to what the 't' in your formula means

[wraith hinge]

,w 1500=500(1+(0.12/52))^(52t)

[solid kiln]

Results provided by WolframAlpha

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

Requested by 857mnj

[wraith hinge]

okay

makes sense

thanks

[brave iron]

good luck with the material!

[wraith hinge]

🙂

.close

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[short trellis]

with some money people are trying to buy some sugar but then the sugars price went up by 25% and with the same amount of money they bought 1 kg less sugar how many kgs did they buy

[short trellis]

100x equals 125(1-x) would this be correct?

[warped plaza]

x=1,25(x-1kg)

:/

Isn't it

Why you wrote 1-x

[short trellis]

yea x-1

you here?

@warped plaza

[warped plaza]

no

[short trellis]

.close

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[merry perch]

there is another answer of 81 how do i get it?

[merry perch]

thanks but why didnt calcuator get it?

.close

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[stone heron]

Hi, quick question about logarithm laws and order of operations.
How would you solve the example: -ln (4) -( ln (2) - ln (3) )

[red loom]

This is $-\ln(4)-(\ln(2)-\ln(3))$, yes?

[solid kiln]

PajamaMamaLlama

[warped plaza]

do -ln4-ln2+ln3 or -ln4-(ln2-ln3) as you wish

[stone heron]

Yess

[red loom]

then I'd suggest by starting with distributing the negative to the terms

[stone heron]

Yepp

[red loom]

so then what would the equation become?

[stone heron]

Ln 3/8

What about - ln 2 - ln 4

Would it be ln 1/8

[red loom]

ln(3/8) is right :)

however -ln2-ln4 becomes -ln(2*4)=-ln(8)=ln(1/8)

[stone heron]

The last two equalities, can you expand on those? ln(8)=ln(1/8)

[red loom]

ln(8) =/= ln(1/8) because that means 8=1/8

however

-ln(8)=ln(1/8)

[stone heron]

Thought so, sorry I missread the earlier sentence

Tired 😅

AWesome thanks

I know all I need then, thanks yall

[red loom]

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[red loom]

do you know the formula for arc length?

[proven steppe]

no

is it theta r actually?

[red loom]

Arclength from a to b is $\int_{a}^{b}\sqrt{1+\frac{dy}{dx}^2}\dd x$

[solid kiln]

PajamaMamaLlama

[proven steppe]

oh

Y

answer is wrong

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[slow hawk]

Hello! If I have circle x^2 + y^2 = 8 and point P = (4, 0), how do I find the tangent line (the touching lines) that touch the circle and that go through point P?

[slow hawk]

x^{2\ }+\ y^{2\ }=\ 8

[solid kiln]

A brief description and guide on how to use me was sent to your DMs!
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