help-38

[trail pike]

wouldnt it be different

on the graph

[compact sigil]

do you want me to visualize the graph?

[trail pike]

i mean right off the bat

like 4x+1

and (4x + 1) ^ (1/2)

its different in terms of its graph

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[hollow yarrow]

How does this work, how do i isolate w here?

[hollow yarrow]

rn i am solving a differential equation

[zinc ginkgo]

integrate both sides

[hollow yarrow]

so i have 1/w = ln(x)-3x ?

i think the left side is wrong, i am unsure how to integrate w'/w

[marble wharf]

u-sub

[zinc ginkgo]

$\int \frac{dx}{x} = \log(x) + C$

[solid kiln]

riemann

[marble wharf]

in general w'/w is a common integral tho, good idea to remember it

[hollow yarrow]

this is for integrating the right side?If so, I am clear on that that. I am unsure how to integrate a function and its derrivative at the same time

[zinc ginkgo]

do u sub like dena said above

[hollow yarrow]

yeah ok i got it thanks guys

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[broken steppe]

Hi

[broken steppe]

!close

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[toxic herald]

is this right

idk what a positive or negative correlation is

ik what a causal is

[silent canyon]

Positive correlation means if x increases/decreases y also increases/decreases.

Negative means if x increase, y decrease and if x decrease y increase.

[toxic herald]

so for the paycheck part

it wouldnt be either correlation or causation because days doesnt matter for ur paycheck

and it stays same

wait

by the days part increases

the amount doesnt change at all

what would that be

<@&286206848099549185>

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[wispy dew]

1 —> 2 , 2 —> 1, 3—>4

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[toxic herald]

like that

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[toxic herald]

.reopen

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✅

[warm citrus]

first one goes too bottom

2nd one goes too top

3rd goes to middle

[toxic herald]

whats the causal on number 1

i feel like the amount of days increasing wouldnt cause the amount to do anything

[warm citrus]

it has no casual, thus the 'no casual relationship'

[toxic herald]

u said 2nd one goes to the top

[warm citrus]

and what does the top one say

[toxic herald]

positive with causal

hows this

i think

[warm citrus]

???

look at it

read it

[toxic herald]

positive with no causal

im so confused

[toxic herald]

would this be right

[warm citrus]

no

my

my answers r right

why r u confusing ur self

u just told yourself 2 different things

[toxic herald]

whattt

im so confused

so the number of days you received your paycheck and amount of your paycheck remaining is positive with no causal as you said?

[warm citrus]

bro ur trolling

[toxic herald]

a

[warm citrus]

?

ur not mAking any sense

[toxic herald]

i have no clue

[warm citrus]

<@&268886789983436800> no way this guy isnt trolling

[toxic herald]

this is what u instructed

[night zodiac]

Or maybe if you didn't give out answers and better explained how to approach the question, that would result in a better outcome

[warm citrus]

nope that is not what i said

[toxic herald]

wtf

[warm citrus]

look at your orignal u sent, i said 3rd one (alligator) goes to middle one (positive casual)

[toxic herald]

u coulda said

days-> this
students->this
alligator->this

positive is x and y increase
negative is x increase y decrease

[warm citrus]

u know how to do it now?

[toxic herald]

i thought u were referring to the right

makes more sense i was confused because i thought u said move the right part

but for the first one

[warm citrus]

didnt say nything about that

[toxic herald]

wym

[warm citrus]

the paycheck?

[toxic herald]

yea

[warm citrus]

what bout it

[toxic herald]

the days increases but how does the amount decrease

since its a negative correlation like you said

you dont lose money for each day passed or smth

[warm citrus]

Yea so basically its saying as the days go on, hes buying more food ect, thus the money from his paychck is going down

[toxic herald]

hm

[warm citrus]

thats how i interpreted itanyway

[toxic herald]

cause if there was no correlation and causation then no days pass

prob the most sensible interpretation

aight ty

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[wraith hinge]

is this what im suppose to do to find x

[sturdy leaf]

yes

you combined two steps in one, but it looks goos

goos

good

[wraith hinge]

this is what i did

do i expand 3x into 2 ln's now

@brisk solar

[sturdy leaf]

8=2×2×2=2^3

thus
log(8)=log(2^3)=3log(2)

log4=2log2

that makes it easier to simplify

[wraith hinge]

ok ty

thank you for helping me

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[solemn wigeon]

hello, how can I solve this?

[modest lion]

Do you know the section formula?

[solemn wigeon]

no haven't heard of it

[modest lion]

Hmm i can teach you how to solve the problem but for more intuition for where you get the answer from you will need to see a video or two on section formula

[solemn wigeon]

the thing I am supposed to do is called dividing line segments

[modest lion]

Yup the process of dividing a line segment gives you the section formula which is used to find the coordinates of a point which cuts the given line segment in a particular ratio

[solemn wigeon]

ok so I should take the change in x and y and divide them by something?

or how do I go about solving this

[slender shard]

Perhaps an explanation of how m1 and m2 relate to the ratios and why you divide by their sum. 😉

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[solemn wigeon]

could you explain? I don't know how I can apply that section formula to my problem

[modest lion]

Take the x coordinate of point C as x2 ,the x coordinate of point A as x1 , and m1= first number in the ratio, m2= second number in the ratio. By this you get x coordinate of the required point

Similarly use the formula for y coordinate and you'll get y coordinate

By the way this formula only works for internal divison ( the point you are trying to find lies on the the given line)

Sorry could try to explain to you why this works but im kind of bad at explain geometry and you'd get even more confused

[slender shard]

C is the Section Formula.

D finds the unit length according to the ratio and adds it to A. If you simplify the equation in D, you get the same value as in C.

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[solemn wigeon]

.reopen

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✅

[slender shard]

Still confused about how this is solved?

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@solemn wigeon Has your question been resolved?

[solemn wigeon]

yes could you explain please

[slender shard]

Ok, I'm going to explain this rather thoroughly that way you intuitively understand the process.

[solemn wigeon]

would be appreciated

[slender shard]

So first I want to explain the concept of a unit.

A unit has the same meaning as one. What one actually equals is a matter of perspective.

The line, AC, is composed of 6 units.

AB is one unit.
BC is five units.

[solemn wigeon]

makes sense

[slender shard]

The ratio of AB to BC is 1:5.

Similar to your question.

[solemn wigeon]

yes

bc 5 times greater

[slender shard]

Your question has this line which you are asked to find the point B that is 1 unit away from A and 5 units away from C.

Does that make sense?

[solemn wigeon]

yes I tried taking the change in x, then divide that into 6 so I got one unit

[slender shard]

Which is the correct step.

[solemn wigeon]

then I could take that times 5

but I was wrong

incorrect in fact they told me

[slender shard]

So you can simplify this problem in terms of the x-component and the y-component.

You solved for 1 unit with respect to the x-component.

[solemn wigeon]

yeah

4/6

[slender shard]

$\frac{2 - (-2)}{6} = \frac{4}{6} = \frac{2}{3}$

Texit bot is down again. 😦

[solid kiln]

Kookiemon

[solemn wigeon]

where did you get that from

[slender shard]

The x-value of C minus the x-value of A divided by 6; 6 being the number of units in the segment.

The point C has an x-value associated with it which is 2.

[solemn wigeon]

I don't understand where you got 2 - (-2)

[slender shard]

The point A has an x-value associated with it which is -2.

A point is defined by an x value and a y value and we represent that point as (x,y).

So when I refer to the x-component, I mean the value of x in a point (x,y).

The point A is defined as (-2,5) in your question.

The x-component is -2 and the y-component is 5.

Do you understand?

So if I want to find the length of the orange line in the image above, I need to subtract the x-component of point A from the x-component of point C which is 2 - (-2) = 4.

[solemn wigeon]

yes i think so

[slender shard]

Once you know that length, you can find the length of 1 unit by dividing 4 by 6.

Each of those units are represented above by the vertical dashed lines.

If you count them, you will see that the orange line is divided by the dashed lines into six segments.

Do you understand?

[solemn wigeon]

yes but that was what I did

4/6

and it said I was wrong

[slender shard]

So you found the length of 1 unit along the x-axis but you also need to find the length of 1 unit with respect to the y-axis.

One unit with respect to the y-axis is equal to (-5 - (-5))/6 = -10/6 = -5/3.

Which gives you a unit point equal to (2/3, -5/3).

Each of the points on the line AC will be n*(2/3, -5/3) units away from point A.

Notice where the dashed lines intersect on the line.

@solemn wigeon Still with me?

[solemn wigeon]

apologies I was getting some water

so 10/6 aswell

?

[slender shard]

It will be -10/6 because you need to calculate the value in the same manner that you calculated the x-component in which we subtracted C-A.

[solemn wigeon]

but the change was 10

[slender shard]

Yes.

[solemn wigeon]

in y

[slender shard]

So you have a unit point which is U = (4/6, -10/6). The formal name is a vector but you don't need to worry about that, I'm just going to call it a unit point.

If you add that unit point to point A, you will get each segment on the line AC.

U1 = A + 1*U
U2 = A + 2*U
U3 = A + 3*U
U4 = A + 4*U
U5 = A + 5*U

[solemn wigeon]

this seems very complicated

[slender shard]

It's not once you understand it.

What I'm trying to do is teach you the process of what is actually going on. There are methods to calculate these values more directly as hello showed you earlier with the Section Formula.

This formula is derived from the method I am teaching you.

You are given the point A which equals (-2,5).

The unit point, U, equals (4/6, -10/6).

All you are doing is adding U to A.

[solemn wigeon]

but how can I apply the section formula on my problem

[slender shard]

$A = (-2,5) = (x_{2}, y_{2})\
C = (2,-5) = (x_{1}, y_{1})\
\therefore\
x_{1} = 2\
y_{1} = -5\
x_{2} = -2\
y_{2} = 5$

[solid kiln]

Kookiemon

[slender shard]

m1 and m2 are the ratios 1 and 5.

$A + U = (-2,5) + (\frac{4}{6}, -\frac{10}{6}) = (-2 + \frac{4}{6}, 5 - \frac{10}{6}) = (-\frac{8}{6}, \frac{20}{6})$

[solid kiln]

Kookiemon

[solemn wigeon]

I don't quite understand the formula could you tell me step by step possibly?

of how I would go about solving this

@slender shard

[solemn wigeon]

<@&286206848099549185> could someone help me with this

[lyric phoenix]

what is the question exactly?

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[solemn wigeon]

I need to solve this

but I don't know how

[lyric phoenix]

If you know coordinates of A and C

[lyric phoenix]

Use this formula

https://youtu.be/EMG76A6OlOA

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[mint peak]

is class of regular languages closed on this operation?

L_1 and L_2 are regular

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[wraith hinge]

If I reach 8’10, how high would I need to jump to dunk.

[mint peak]

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[compact sigil]

.close

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[gaunt axle]

Have to find the derivative through logarithmic differentiation

[gaunt axle]

Ugh it's a file, give me a sec

There we go

[onyx fulcrum]

There's a process on how to do these

[gaunt axle]

Yeah

I just like I'm not too sure how to apply the process with this one

[onyx fulcrum]

use quotation rule

[gaunt axle]

ok so funyn story, no idea what that is

[whole coral]

Don't you start by taking [natural] logs of both sides tho? (assuming you mean "quotient rule"(?), but the OG question asks for logarithmic differentiation?)

[gaunt axle]

ohh quotient rule

yeah i was planning on doing just quotient but I realized it wants me using log differentiation so that part is the icky part

[whole coral]

Log differentiation is actually pretty nice imo 😂

[gaunt axle]

i agree, I really like it

just not with this problem

so like do the ln of both sides first

[whole coral]

Yea

[gaunt axle]

so like the right side will be ln of everything

then I apply the derivative to both sides

i think

[whole coral]

Yep, but first, try to make use of log rules...

[gaunt axle]

ah gotchu

yk making them their own little lns

[whole coral]

Yea, separate ln's

[gaunt axle]

right

ok and then I find the derivative of those things

[whole coral]

Yeee that's it

[gaunt axle]

ahhh alr

I got it from here

tyyy

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[daring karma]

in part b, that means that x is an n x m zero matrix, right?

[daring karma]

that's what the trivial solution means, right? Please @ me if anyone respodns

[fleet yarrow]

@daring karma I think x here is just a vector

[daring karma]

it's not, the book doesn't talk about vectors until chapter 3

at least, i dont think it's a vector lol

oh wait

it would have to be an n x 1 matrix then

[fleet yarrow]

yeah

[daring karma]

so yeah, a vector with n variables, right

[fleet yarrow]

exactly

[daring karma]

but they're all 0, right

[fleet yarrow]

thats what trivial means, yeah

[daring karma]

okay just wanted to make sure i understand

thank you!

[fleet yarrow]

no prob

[daring karma]

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[elfin escarp]

The cosine of an angle can be used to calculate the length of the side of a triangle opposíte the angle. Differebt angles will give different lengths, so it does matter which angle is used. In this case, the cosine of 15.8 is used to calculate the length of the side opposite the 15.8 degree angle.

The angle used in this calculation was likely determined based on the given lengths of the other side and/or angles of the triangle. If the lengths of the two sides and one angle of a triangle are known, the length of the remaining side can be found by using the cosine law.

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[cinder oar]

I am studying laplacian filter, which is a second derivative filter. The gradient in x direction is given by: L_x = f (x, y +1) − 2 f (x, y) + f (x, y −1)
My question how can I demonstrate this?

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[mossy summit]

Hello

[mossy summit]

can someone explain why

converges to 3/8

and not 3/4

I use the formula $\frac{a}{1-1/2}$

[rugged latch]

Ok so what does the denominator become if u break up the exponent

[solid kiln]

heka gigantes

[mossy summit]

it becomes 2^n * 2

so I take out the 3/2

from the series no?

[rugged latch]

No

(2^n) * (2^2)

[mossy summit]

OOOH

OHAOSHOASHF

WEFWERGV

S RGÑONJLKYN SETPKIN SR

GBAET

VBSRGNKI SETOLN SEGF

thank you

[rugged latch]

Lol

[mossy summit]

AHAHhaha

[rugged latch]

Np

[mossy summit]

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[wraith hinge]

Ive done a, and b(i) am not sure how to do the rest.

[wraith hinge]

I got ln y = ln a + bln x for 7 (a)

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[brisk solar]

yeah thats right

you should be able to plot 7bi

[wraith hinge]

i plotted it, its the linear regression bit i cant do

<@&286206848099549185>

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[viscid wagon]

need help with the chinese remainder thm

[tough snow]

The Chinese Remainder Theorem is a mathematical result that states that if a1, a2, ..., an are integers which are pairwise relatively prime (i.e., no two of them have any common factors except 1), and if b1, b2, ..., bn are any integers, then there exists a unique integer x satisfying the following conditions:

x is congruent to a1 modulo b1
x is congruent to a2 modulo b2
...
x is congruent to an modulo bn
In other words, the Chinese Remainder Theorem provides a way to find a single integer x that is "compatible" with multiple modular equations, meaning that it satisfies all of the given equations simultaneously.

To use the Chinese Remainder Theorem, you first need to find the least positive integers x1, x2, ..., xn satisfying the above conditions. Then, the solution x can be found using the following formula:

x = x1 * b2 * b3 * ... * bn + x2 * b1 * b3 * ... * bn + ... + xn * b1 * b2 * ... * bn-1 modulo b1 * b2 * ... * bn

In this formula, the xi's are the least positive integers satisfying the given modular equations, and the bi's are the moduli from the equations.

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[red kite]

Would I use a half angle formula here?

[red kite]

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[wraith hinge]

What multiplier can be placed before the parentheses in the expression x(x-2)+3(x-2)?

[wraith hinge]

(x+3)(x-2)

Hiw

How @wraith hinge

factor out (x-2)

and you will see

Factor what means

Not my first language @wraith hinge

its ok

Google translator does shlt job

i will use math

to explain

$x(x-2) + 3(x-2) = (x-2)(x+3)$

[solid kiln]

45

[pure hatch]

Or you can do this

Let u = (x-2)

[wraith hinge]

U?

[pure hatch]

Any letter you want yeah

[wraith hinge]

Answer

Is (x-2)

Or

(x+3)(x-2)

[pure hatch]

No

[pure hatch]

This is correct

I'm trynna explain how

You can let m = x-2

So x(x-2) +3(x-2) = mx + 3m

Since we have multiple things added as multipliers of m

We can take m as a common factor like this

mx + 3m = m(x+3)

[wraith hinge]

Right

And m =

X-2

Cool

[pure hatch]

Yep

[wraith hinge]

Thx

[pure hatch]

[wraith hinge]

.cloze

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[versed hill]

how can i solve this question?

[versed hill]

<@&286206848099549185>

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[novel fern]

Why did we only differentiate x^k here and not (-1)^k as well?

[novel fern]

s

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[novel fern]

.reopen

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✅

[novel fern]

^

[marble wharf]

(-1)^k is constant with respect to x

[novel fern]

Aha its because its a constant, ty

Also why did we cross out that +2 here?

[zinc ginkgo]

$(-1)^{j+2} = (-1)^{j}(-1)^{2} = ?$

[solid kiln]

riemann

[novel fern]

(-1)^j, aha

thank you

When doing the extra tests for the interval, why is it that we use whats in the black box in particular?

Like why can't we use this or that to test instead for example?

.

[zinc ginkgo]

you're testing the endpoints because the test excludes them

between the endpoints

testing separately

[novel fern]

Yes but why are we picking the series in the black box in particular to test on?

[novel fern]

^

[zinc ginkgo]

oh. the earlier formulas are inconclusive

but you have to try it yourself to see that

the latter one could equivalently work

the real answer is it doesn't which form you take. just pick one that's most convenient and that works

try this one. it's almost identical

[novel fern]

aha okay I see, thanks

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[astral mirage]

Need some help on a quick clarification on some notes in the subject Nonlinear Programming:
Does this definition mean that if there exists any point in the set S that outputs a negative vector, P is superconsistent? Or would it be any critical point with the notation vec(x^*)

[zinc ginkgo]

this? https://en.wikipedia.org/wiki/Slater's_condition

[astral mirage]

yep

[zinc ginkgo]

how does P relate to S and $\vec{x}^*$ in your def

[solid kiln]

riemann

[astral mirage]

ah mb

P is a program

[zinc ginkgo]

looks like you just need one vector

https://classes.engineering.wustl.edu/2010/spring/ese415/ch5-2.pdf

[astral mirage]

oh i see

doesn't have to be a feasible solution?

[zinc ginkgo]

Slater implies feasible

[astral mirage]

hmm

so if we have an question like

how do we check for superconsistency before we apply KKT if we don't know the feasible region yet?

or do we have to map it out first?

[zinc ginkgo]

you can find the region from your conditions

[astral mirage]

got it

guess and check, right?

[long basin]

I fucking hate KKT holy shit I never wanna see that

Wouldn't it just be easier to just test whether any local extrema exist within the region

[zinc ginkgo]

yea

[astral mirage]

alright tyvm!

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[loud dragon]

Hi, if I have to find how many equivalence classes can be defined in $X={x \in \mathbb{N} :\ x \leq 18}$ that satisfy this conditions:\
1- $[1] \cap [3] = \emptyset$\
2- $#[x]=6\ \forall x\in X$\
3- $[2] \cap [x] = \emptyset$ if x is odd

[loud dragon]

How can I get the result?

[solid kiln]

Neo

[loud dragon]

I have a resolution by another person but I dont really understand why they reach that result

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[wraith hinge]

Idk where to go but I need help with algebra

[wraith hinge]

function notation and evaluating functions if f (x) = -x - 7, and f (5)

show as many steps please and thank you

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[split chasm]

substitute all instances of x in
f(x) = -x - 7
with 5 to get the equation explicitly telling you what f(5) is equal to

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[lilac wing]

Need (a)

[kindred pier]

in terms of what

c?

[lilac wing]

i just need to find the variable im stuggling wit hit

[uncut yacht]

I think the c was just there to identify where the hypotenuse was

[kindred pier]

not enough information here to solve for a then

[lilac wing]

i also have the area if that helps

its 1200+6a

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[surreal eagle]

why does the limits change here? or how

[trim jolt]

Please don't occupy multiple help channels.

[fathom ivy]

you have an even function, i.e. your function is symmetric about the y-axis

so it's equal to twice the volume of one side of the integral

[surreal eagle]

so like, how do i recognize this or what do i do to get this

do i always put it as 0 to 2 if its symmetric about the y axis

[fathom ivy]

anything that's the sum/difference of x^{even number} is an even function and satisfies this property

as well as a few other functions, but in general you just recognize them

in general$\int_{-a}^{a} [\text{even function}] dx = 2 \int_{0}^{a} [\text{even function}] dx$

[solid kiln]

IV

[surreal eagle]

neat!

and the even function is if it has an x^a where a is any even number?

[fathom ivy]

rigorously speaking a function $f(x)$ is even if $f(x) = f(-x)$ for all $x$

and you can check pretty quickly that this holds for all $x^a$ where $a$ is even

[solid kiln]

IV

[surreal eagle]

awesome

so in recap, if I have an even function, and the limits go from -a to a, then i can just multiply it by to and change the limits to 0 to a

perfect, thanks!

[open idol]

@fathom ivy can u come to my channel

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[bronze echo]

Hey would anyone be able to help with this sum problem

[bronze echo]

It seems really easy, so I can't tell what I am doing wrong

I was planning on index shifting and then using geometric series

so I index shift to k=0, and get 2^(k+6)/5^(k-1)

but I seem to be getting a wrong answer from there

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[sonic vector]

sup, it kinda looks like you can telescope this

or you can just do a ratio test

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[trim jolt]

@lime wigeon Has your question been resolved?

[lime wigeon]

<@&286206848099549185>

its been 15 mins

dude

its been 20 mins'

[cerulean star]

still need help?

[lime wigeon]

bo

no

i finished

the problem

skipped like 8 people over me

nicee

[inland orbit]

Everyone here is just volunteering dude. It's not first come first served.

[lime wigeon]

its not?

[inland orbit]

No, people help if they are able to and want to

I guess nobody likes two column triangle proofs 🤷 sorry

[lime wigeon]

bruh

[inland orbit]

truth be told, they're just not very easy to explain quickly over text

and they're kind of boring tbh

[lime wigeon]

can u just tell me the asnwer pls

[inland orbit]

but yeah sometimes it takes a minute for someone to respond

[lime wigeon]

took 25 mins

[inland orbit]

I meant

a figurative minute

[lime wigeon]

can u help me pls

[inland orbit]

Oh I thought you finished it

[lime wigeon]

no like for a different problem

[inland orbit]

yeah let me see

[inland orbit]

First of all, look at what you're trying to prove

and make sure you even believe that it's true in the first place

Because otherwise you're just gonna be wondering aimlessly

[lime wigeon]

oyeah i do

[inland orbit]

How do you know?

[lime wigeon]

because its juist a linear pair

idk how to say it with like geometric terms though

[inland orbit]

is this checked by your teacher or by the computer?

jw

[lime wigeon]

computer

why

[inland orbit]

ugh ew ok

because there's not necessarily just one way to phrase it

what are your choices in that dropdown menu?

[lime wigeon]

theres like 15

[inland orbit]

ok well basically the gist is

you said they're both linear pairs

so that means for one,
m∠ABD + m∠CBD = 180 degrees

and likewise for the other

[lime wigeon]

ik

[inland orbit]

m∠ABD + m∠CBD = 180
m∠EDB + m∠CDB = 180

And m∠ABD = m∠EDB

I'm not sure exactly what your choices are there so it's hard to tell you exactly what the flow of logic will look like

[lime wigeon]

ik

[inland orbit]

But if you treat those angle measures like variables, do you see how you could substitute ABD for EBD

and then end up with CBD = CDB

[lime wigeon]

ik

but i need to know how to write it out

like vernbally

[inland orbit]

Can I see what all the choices are?

can you screenshot or something

[lime wigeon]

ok

i got it

[inland orbit]

oh

you finished the proof?

[lime wigeon]

yea

[inland orbit]

Oh

okay cool

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[wraith hinge]

Find the complex form of the Fourier series of
[ f(x) = e^{-x} \quad \text{in } -1 < x \leq 1
]

[solid kiln]

♡LexQa♡

[left oriole]

i suppose you have a formula for the fourier series coefficients, did you try computing it?

[wraith hinge]

yeah one sec i am just latexing this out

[left oriole]

something like
$$\frac{1}{2}\int_{-1}^{1}f(x)e^{-i\pi n x}\ dx$$

[solid kiln]

Bungo

[left oriole]

although i always screw up these formulas

[wraith hinge]

yeah it is like -l though

like

[left oriole]

i have to go do some dishes alas, will check back in case the question is still open after that

[wraith hinge]

[
\frac{1}{2l} \int_{-l}^{l}f(x)e^{-i\frac{n\pi x}{l}dx
]

[left oriole]

yea, here $l = 1$ right?

[solid kiln]

Bungo

[wraith hinge]

yep

[left oriole]

hey i got it right! mine equals yours when l = 1

amazing, i'm usually off by a scale factor somewhere

[solid kiln]

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

[wraith hinge]

okay u can keep bugging latex

[wraith hinge]

ahah its okay i just want to check my calculations because it is trippy

[left oriole]

it actually is displaying correctly, who knows what the tex error is

[wraith hinge]

oh i have the dx

in the exponent

[left oriole]

oh right

ah well who cares about the dx, we all know how it goes

feel free to post your calcs and i'll check them in a bit if no one else did

[wraith hinge]

or do we

[wraith hinge]

aight! time to MEGA LATEX

[amber python]

[wraith hinge]

SNOOOOOW MY SAVIOUR

[amber python]

unfortunately i'm on mobile so i won't be typing out any latex

pure lurking

[edgy cedar]

layla

[amber python]

[wraith hinge]

[
f(x) = \sum_{n = -\infty}^{\infty}C_n e^{i\frac{n\pi x}{l}}
]
[
\therefore C_n = \frac{1}{2}\int_{-1}^{1}e^{-x}e^{-in\pi x}dx
]
[C_n = \frac{1}{2}\eval{(\frac{e^{-(1+in\pi)x}}{-(1+in\pi})}_{-1}^{1}
]
[C_n = \frac{1}{2(1+in\pi}(e^{-1}e^{-in\pi}-e^{-1}e^{in\pi)}
]

[amber python]

i wonder if you can taylor expand this and get something

[solid kiln]

♡LexQa♡

[edgy cedar]

LGTM

[wraith hinge]

okay the brackets are fucked but idc

[amber python]

lgmt

[edgy cedar]

[wraith hinge]

lgtm?

[edgy cedar]

Calm

[wraith hinge]

serene

[amber python]

c a l m

[wraith hinge]

anyways from this, i do a bit of calculations mumbo jumbo which icba to write out to:
[
C_n = \frac{(-1)^n}{1+in\pi}(\sinh(1))
]
[
\to f(x) = \sinh(1) \sum_{n= \infty}^{\infty}(\frac{(-1)^n(1-in\pi)}{1+n^{2}\pi^2})e^{in\pi x}
]

[solid kiln]

♡LexQa♡

[wraith hinge]

pls send help idk anymore

[amber python]

kinda sus

[wraith hinge]

[amber python]

why is the second term e^-1

well you got sinh(1) anyway so

i guess it's fixed

[edgy cedar]

Yeah

Should be e^1

[wraith hinge]

oh wot

[amber python]

you can probably pair up the n and -n terms in this and get something nicer

also not sure why you rationalised the denom lol

[wraith hinge]

i dont know i have been writing this for like an hour now

[amber python]

[wraith hinge]

but i guess i did it correct even though i probs messed up the signs

soooooo its a win in my book 😎

[amber python]

you can check it on desmos

well

you can check it after you pair up the n and -n terms lol

[wraith hinge]

aightyy sounds good tyy snow

.close

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[umbral maple]

ight turns out i knew how to do it and i just somehow mistyped the answer 4 times

.close

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[atomic portal]

help

on this

[zinc ginkgo]

• Show your work, and if possible, explain where you are stuck.

[atomic portal]

i dont know how to start

[cedar maple]

$\rho^2 \propto \frac{1}{A^3}$ can be written as $\rho^2 = \frac{k}{A^3}$ where $k$ is some constant

[solid kiln]

Mixnik

[cedar maple]

see where you can go from there

[atomic portal]

i dont understand

[zinc ginkgo]

what part

k is a number you're solving for

[cedar maple]

k will always be the same, no matter the scenario. So in the first scenario, you are given both the density and the surface area, but not k, what could you do here?
Could you use your solution to the first scenario to find the density when the surface area has increased in the second scenario?

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[loud dragon]

Hi, if I need to find for what k's [ 3 1 3 ] can be generated by the vectors in that set, is it ok to reduce the matrix like

1  2   k | 3
1 1-k -1 | 1
1  1   1 | 3
```?

[loud dragon]

and solve that ofc

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[sweet fable]

Hey, is 1, 2, & 3 correct?

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[carmine parcel]

Hi

[fallen plank]

find the slope

[carmine parcel]

Bro can you answer it please

Just one

Then I will submit it

I just need those 2

Then i will sumbiy

<@&286206848099549185>

[tawdry bison]

use substitution elimination

[carmine parcel]

Where you eliminate answers? Bro Thats risky

[tawdry bison]

after you found the size for 1 pounds, multiply the meal size with 30 pounds

[carmine parcel]

1 meal would be 24 pounds

Hm

[fallen plank]

so if 1 = 24 what does 30 = x

[carmine parcel]

1.25

BRO

[tawdry bison]

wait you cant do that

[carmine parcel]

ITS B

[tawdry bison]

10 pounds is 0.75

[fallen plank]

yeah i just saw that

[carmine parcel]

Yea

[fallen plank]

so 10*3 = 30

[carmine parcel]

Oh

[tawdry bison]

but 40 pounds is 1.50

that is half of the size

but the weight has 30 differences

[carmine parcel]

Dude this is confusing

[tawdry bison]

take a look at 50 and 80 pounds

[carmine parcel]

For real

What do i do then

[tawdry bison]

80 pounds are half of the size of 50 pounds

so

lets use

every 30 pounds difference has 2x the meal size

oh frick i underestimate this question

[fallen plank]

lol i would think it would be linear

you can eliminate D

[carmine parcel]

Hm

[fallen plank]

wait

cant you just fine the difference from 80 pounds and 50 pounds @tawdry bison

[carmine parcel]

Wait

[fallen plank]

maybe it won't work like that

[carmine parcel]

How about i just minus 10 pounds meal size to 50 pounds

BROO

I THINK I FIGURED IT OUT

IS A

@fallen plank is that right?

I got 55

BRUUHH

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[carmine parcel]

.close

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[left quartz]

Could someone help me with this

“What type of function is f(t) = 14(1/4)^5”

[left quartz]

1. Linear growth
2. Exponential growth
3. Exponential decay
4. Logarithmic growth
5. Natural growth
6. None of the above

I did this for my exam and just want to know if I got it right

[severe wave]

you are missing the variable if there is any

[left quartz]

Sorry

I meant t

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[left quartz]

<@&286206848099549185>

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[wraith hinge]

i know its easy but helppppppp

[vagrant marsh]

8(t)=5h20m

convert to minutes and youre golden

[trim lichen]

1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

which stage are you on?

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[wraith hinge]

2

[wraith hinge]

Ann seems to be busy, so I will continue. What did you try? @wraith hinge

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[trim lichen]

ok, i'm back from the things that were occupying me

@wraith hinge show your progress so far

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[feral herald]

Simplify: 2tgα x cos^2(α/2) -senα

[sinful elbow]

tga?

[feral herald]

Tg= tan

Sen =sin

[rugged latch]

So

2tan(a) * cos^2(a/2) - sin(a)

Where do u think we should start first

[feral herald]

Idk

[rugged latch]

Let’s start with the cos^2 (a/2)

[sinful elbow]

Is that cos to the power of 2(a/2)?

[feral herald]

Cos(a/2) is a formula

[rugged latch]

Half angle

[feral herald]

Ye

[rugged latch]

Ok so, do u know the cos (a/2) formula

[feral herald]

Yes

[rugged latch]

Now square it since we have cos^2 (a/2)

What do u end up with

[feral herald]

This is the formula

[rugged latch]

Yep

What do we get upon squaring it

[feral herald]

I have done that but it's wrong

[rugged latch]

U messed up at the cos^2 part

So we have cos^2 (a/2)

= cos(a/2) * cos(a/2)

[feral herald]

Eliminate ^2?

Oh yes

[rugged latch]

So knowing this, what do we end up with now

[feral herald]

,rotate

[rugged latch]

[Cos (a/2) ]^2 = [1 + cos(a)] /2

U keep adding another cosine

[rugged latch]

It’s just this

[feral herald]

And what i do With it

[rugged latch]

Look at ur problem and Refer to this

[feral herald]

I don't understand, i have same not?

@rugged latch

[rugged latch]

No

Look at ur original problem

Cos^2 (a/2)

What is that equal to

@feral herald

[feral herald]

Is ^2 i copied wrong

[rugged latch]

Wdym

What’s the actual problem then

[feral herald]

I need to simplify the all equation

Thats the problem

[rugged latch]

Can u please just state the actual problem again

Without anything copied wrong

[feral herald]

Simplify: 2tgα x cos^2(α/2) -senα

[rugged latch]

What did u copy wrong?

That’s the same one from before

[feral herald]

I copied wrong in the last photo

[rugged latch]

Oh ok

[rugged latch]

So what do we have now after knowing this

[feral herald]

Tan?

[rugged latch]

What’s the full expression we have after simplifying the cosine term

[rugged latch]

,rotate

[rugged latch]

Ur still messing up the cos^2 (a/2)

[Cos (a/2) ]^2 = [1 + cos(a)] /2

U keep putting an extra cosine before it

Do u speak Spanish

[feral herald]

Yes

U?

[rugged latch]

Deja de poner un cos(a) antes del (1+cos / 2)

[rugged latch]

Poco

[feral herald]

Pero entonces para que sirve el ^2 en el cos

[rugged latch]

Sip, pero ve esto cos^2 (a/2) = cos (a/2) * cos (a/2)

Así qué sucede cuando multiplicamos cos (a/2) con cos (a/2) ?

[feral herald]

^2

[rugged latch]

Perdemos la raíz cuadrada

Y es todo

Es todo q hacemos con esto

[rugged latch]

Si

[feral herald]

Aaaa

[rugged latch]

Entiendes?

[feral herald]

Si

[rugged latch]

Ok

[feral herald]

Y ahora que multiplicó el /2 en los otros términos?

[rugged latch]

Espera, primero, muéstrame lo q tenemos ahora en tu página

[rugged latch]

Siiiiiiip

[feral herald]

🤓🤓🥳🥳

[rugged latch]

Ahora el 2 con el tg cancela con el con el cos, tas de acuerdo?

[feral herald]

Que

[rugged latch]

El 2 cancela

[feral herald]

El que está dividiendo?

[rugged latch]

Sip

Con el que está multiplicando con el tg

[feral herald]

Pero porque va a cancelar

Sería multiplicar no?

Para quitar denominadores

[rugged latch]

2 * tg * (1+cos) * 1/2

Si?

[feral herald]

Sería así no?

[rugged latch]

Es esto

No necesitas hacerlo

[feral herald]

Y porque tú solo divides en -sen

[rugged latch]

Espera

2 tg * (1+cos) /2

Hazlo

[feral herald]

Esto?

[rugged latch]

Sí hazlo, y luego trabajaremos con el sen

[feral herald]

Que hago

[rugged latch]

2tg(1+cos) primero

[feral herald]

Paso sen/cos=tg

[feral herald]

Y porque quitas el /2?

[rugged latch]

Porque el primer paso es multiplicar

Usaremos el /2 para el sen luego

[feral herald]

Y entonces q multiplicas

2tg+tg cos?

[rugged latch]

2tg(1+cos) primero

Si

No

Poner el 2 afuera

2tg(1+cos)

[feral herald]

2(tg+tgcos)?

[rugged latch]

Si

Así tenemos

2(tg + tgcos)/2 - sen

Así ahora puedes cancelar el 2

Q tenemos ahora

[rugged latch]

No

Por que tienes 2s para todos los términos

[feral herald]

Así?

[rugged latch]

Por que 2sen

[feral herald]

Para quitar el denominador no?

[rugged latch]

No

cambia los tg a sin/cos

[feral herald]

Así?

[rugged latch]

Casi

Con el (sin/cos) * cos

Ves esto?

Simplificaste incorrecto

[feral herald]

Y que tengo que poner entonces?

[rugged latch]

Que es: 5 * x/5

[feral herald]

X/25

[rugged latch]

No

Simplemente es x

Los 5 cancelan

[feral herald]

Aaaa

[rugged latch]

Comprendes?

[feral herald]

Me he equivocado con la regla de la división

[rugged latch]

A

[feral herald]

Lo he hecho en cruz

[rugged latch]

Así q es cos * sin/cos