help-39

[daring wedge]

OML

tf is happening

[crude echo]

First line iss

[daring wedge]

What is this then dude

But..

I don't get it dude

Someone on YouTube

Told me that this is

The parts thingy

[crude echo]

Do you know all the values in first eq.

[crude echo]

Probably misunderstood them

[daring wedge]

Is this the corect one

[crude echo]

Yeah

[daring wedge]

Can you help me understand it

[crude echo]

Sure

[daring wedge]

What is happening in the last term

After the negative sign

[crude echo]

So we basically know what U And V are

Rightt

[daring wedge]

Yup

[crude echo]

Then what is integration V ?

[daring wedge]

Ummm

Integration of e^2x is e^2x/2

Right??

[crude echo]

Yeah

And du/dx ?

[daring wedge]

Nah but, In the second term, what we do is integrate the derivative of u and multiply it with the integrated form of v?

[crude echo]

Yeah

Right

So ?

[daring wedge]

That's it?

[crude echo]

Yeah

[daring wedge]

Ok I'ma solve the part one now

I HOPE I CAN THIS TIME

it's the 5th attempt

I was using the wrong rule all the time lmao

[crude echo]

Happens

Dw

[daring wedge]

😢

Integ is hard

I loved finding derivatives

But integ is much harder duse

Dude

[crude echo]

It'll be easy

After sum practice

[daring wedge]

Check the first linee

Did I writ it correctly

The parts rule

@crude echo

[crude echo]

Compare with this

Is it same?

[daring wedge]

no

ok ive corrected it

thx

[pearl pond]

@daring wedge Has your question been resolved?

[pearl pond]

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[sterile turtle]

hi, how would I do c)?

[sterile turtle]

i thought it would be (10C1 x 2)/10C2

[grim fractal]

Well once the first person sits down, there are 9 seats remaining. How many of those seats are next to the first person?

[sterile turtle]

2

[grim fractal]

so what is the answer

[sterile turtle]

im having trouble understanding

[grim fractal]

2/9?

[sterile turtle]

ah

but why isnt it that we do 10C1 for the first person

[grim fractal]

Can you explain why you think it's 10C1

[sterile turtle]

so we fix the first person, but whats saying the first person is part of the comittee

[grim fractal]

wdym

[sterile turtle]

hmm like

from the ten we chose 2 to be art of the committee

so i thought that was 10C2 as the major probability

and then from the 10 we choose 1 to be the frist person

[grim fractal]

I think it's asking to just consider two specific people: what is the probability they sit by each other

[sterile turtle]

and then x2 for the people next to them

ahhh

that makes more sense

thanks

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[rancid jay]

Can anyone help with this. its precalculus 2 on the circles topic:

A small radio transmitter broadcasts in a 46 mile radius. If you drive along a straight line from a city 54 miles north of the transmitter to a second city 56 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

[elfin walrus]

its easier if you try thinking of the picture

now how can you get the line segment thats the path of the car

if you know it goes through the points (0,54) and (56,0)

[rancid jay]

is it using soh cah toa?

[elfin walrus]

no

[rancid jay]

hmm

it would go up 54 on the x axis, then right 56 on the y

[elfin walrus]

how could you get the slope of the line

because remember a line is slope * x + b

[rancid jay]

ohh would the slope be 28/27

[elfin walrus]

think about it

you go from a higher point to a lower one

so should the slope be positive or negative

remember slope is rise/run

[rancid jay]

ohh so then it would be negative right

my bad if im wrong havent took a math class in a minute 😭

[elfin walrus]

the run is 56

the rise is -54

[rancid jay]

ahh, so -54/56, which is negative correct

[elfin walrus]

yeah

[rancid jay]

-0.96

[elfin walrus]

so

which points do you think are important here

[rancid jay]

te ones intercepting withthe circle

[elfin walrus]

yeah

and how can you find those

[rancid jay]

not sure ..

the one on the top is 9.5, somethign

[elfin walrus]

so

you know the equation of the line is -27/28x + 54

which means y = -27/28x + 54 for the line

[rancid jay]

okay

[elfin walrus]

what do you know about the y values for the circle and the line at the two intersecting points

[rancid jay]

that the distance between them will be the miles

[elfin walrus]

no i mean the

here just focus on 1

what do you know about the y values of the circle and the line at this point

[rancid jay]

yes i think its 45.5

[elfin walrus]

is your teacher chill if you use desmos

cuz you should also know the proper way to do it if you have to on pen and paper

[rancid jay]

Yeah he doesnt care

[rancid jay]

got it

[elfin walrus]

ok but

what i was tryna say is

the y values are equal

at the points of intersection

[rancid jay]

ohhh

[elfin walrus]

so for the circle we have x^2 + y^2 = 46^2

which means y = sqrt(46^2-x^2)

and for the line we have y = -27/28x + 54

[rancid jay]

Okay got it

[elfin walrus]

we can combine these to get sqrt(46^2-x^2) = -27/28x + 54

[rancid jay]

okay

[elfin walrus]

and then you would have to solve that

[rancid jay]

ohh that makes sense

okay ima solve it real qwuick

I got x = 44.688

[elfin walrus]

there should be two solutions

[rancid jay]

also x = 9.286

9.276*

[elfin walrus]

yes

do you know the quadratic formula?

[rancid jay]

I did but i forgot it bruh

[elfin walrus]

cuz thats how you could get the exact answer

not just the decimal

[rancid jay]

ohh i see

so with these 2 equations, only one of them is right

[elfin walrus]

wdym

[rancid jay]

Like how would i get the final answer of miles

[elfin walrus]

remember the car is driving on this line

so you want the distance between the two intersection points

[rancid jay]

okay

[elfin walrus]

do you remember how to get distance between two points

[rancid jay]

slighltly but not really

[elfin walrus]

its sqrt((x1-x2)^2 + (y1-y2)^2)

[rancid jay]

ohh okay so i would plug in those points for x and y right

[elfin walrus]

yeah

you got the two x coordinates

you can plug them into the equation of the line to get the two y coordinates

[rancid jay]

its sqrt((44.688(1)-(9.276)2)^2 + (y1-y2)^2)

[elfin walrus]

when i say x1 and x2 i dont mean x * 1 and x * 2

its just to show that they are different x's

[rancid jay]

Ohh my fault

[elfin walrus]

its like

x_1

and x_2

imagine if they were a subscript

[rancid jay]

so its sqrt((44.688-9.276))^2 + (y1-y2)^2)

[elfin walrus]

yes

[rancid jay]

and i solved it and got sqrt -41.293376 + (y1-y2)^2

[elfin walrus]

im confused

this is what i mean

[rancid jay]

ohh

[elfin walrus]

which goes to this

then you just need to find y1 and y2

[rancid jay]

Ohhh i tihnk i wrote it wrong

okay

yea im lost bruh

[elfin walrus]

so

the equation of the line is -27/28x + 54

what does y equal when x equals 44.688

[rancid jay]

10.93?

[elfin walrus]

yeah

and what does y equal when x equals 9.276

[rancid jay]

45.055

[elfin walrus]

yeah

so now just plug those in for y1 and y2

[rancid jay]

i did it and got 57.175

wait actually 66.244

[elfin walrus]

as the final answer?

[rancid jay]

Yeah after plugging both the y's into the euation and solving

but it wasnt correct as the final ansewr

[elfin walrus]

this is what i got

[rancid jay]

ohh, i think i probably messed up soimehwere

i just did it again and got the same answer

and it worked

thank you so much i really appreciate it

[elfin walrus]

np

[pearl pond]

@rancid jay Has your question been resolved?

[pearl pond]

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[hazy radish]

can someone explain to me what the difference between standard matrix multiplcation modulo 2 and boolean matrix multiplication?

[uneven smelt]

it should be the same

modular arithmetic is a homomorphism

[hazy radish]

thats what i thought but in certain situations they produce different results

Like here

row1 column 4

in boolean its 1

but in standard its 0

[uneven smelt]

oh nvm sorry

i read wrong sorry

• 0 1
  0 0 1
  1 1 0

• 0 1
  0 0 0
  1 0 1

and
^ 0 1
0 0 0
1 0 1

or 0 1
0 0 1
1 1 1

they act differently addition multiplication vs and, or

[hazy radish]

im not sure i follow

[uneven smelt]

they do different things so u get different results

[hazy radish]

so what does each one mean?

i know how to do them

but the results what do they mean

[pearl pond]

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[stuck widget]

how would I show the reasoning it is closed under addition and multiplication with a polynomial like this?

[modern crest]

let p and q be two polynomials. closure under vector addition is asking if p''(9) + 9p'(9) = 0 and q''(9) + 9q'(9) = 0, then can we guarantee (p+q)''(9) + 9(p+q)'(9) = 0?

we can use the property that the derivative of the sum of two functions is the sum of their derivatives, so (p+q)''(9) = p''(9) + q''(9)

you can do a similar operation on the 9(p+q)'(9) term and then rearrange to show we get zero

scalar multiplication is pretty much the same process, just using the fact that for a scalar k, (kp)' = kp'

[stuck widget]

sorrry for late reply

how would you rearrange to make it get zero?

I'm not the greatest when it comes to whole polynomials and their properties

[modern crest]

okay so if we split each derivative up separately using the fact i said earlier we get
(p+q)''(9) + 9(p+q)'(9) = p''(9) + q''(9) + 9p'(9) + 9q'(9)

does this make sense?

[stuck widget]

ok I get that

OHH

[modern crest]

alright, then we can rearrange and add some parenthesis and get
(p''(9) + 9p'(9)) + (q''(9) + 9q'(9))

now what do we know about p''(9) + 9p'(9) and q''(9) + 9q'(9)?

[stuck widget]

oh we rearrange and since we know they equal to 0 (given)

therefore its closed

and its the same with scalar isn't it

[modern crest]

yep, exactly!

yeah, it's pretty much the same thing

[stuck widget]

I NEVER THOUGHT OF THAT OHH

I thought we would have to expand to like ax^4+bx^3.....

[modern crest]

oh yeah that would be painful LOL

it would work... eventually...

[stuck widget]

thanks isabelle

just one quick one, how would you do the zero vector?

I already did it but I am wondering if I could've made it smaller

[modern crest]

so the zero function is just 0(x) = 0. so as long as 0''(9) + 90'(9) = 0

and ofc 0' and 0'' are just zero everywhere as well, so it works out

[stuck widget]

huh

I expanded it to the ax^4 and replaced all the constants with 0's

hopefully thats the same :V

[modern crest]

yeah it is, that works too

[stuck widget]

oh perfect

thanks isabelle, your a life saver 🙏 🙏 🙏

[modern crest]

i just usually get in the habit of using more general facts instead of the specific facts of that space (when possible), since you'll often work with vector spaces where you can't plug in constants

[stuck widget]

true, really good advice

I just never knew I could maniuplate a polynomial like that

thanks for the help!

[modern crest]

np!!

[pearl pond]

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[stuck widget]

.close

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[sick breach]

how do i find the area of the shaded region

[stoic pecan]

oh god

i remember this question

you gotta draw some triangles

[stoic pecan]

wait nvm this is easy

thats the idea

[pearl pond]

@sick breach Has your question been resolved?

[sick breach]

what do you do from here?

[stoic pecan]

okay think about those 2 sections

[sick breach]

its 14

[stoic pecan]

what happens if you subtract it

what do you get

[sick breach]

um

[stoic pecan]

lemme bust out ms paint

let me cook

[sick breach]

🙏

[stoic pecan]

okay its shit but like

big part - small

[sick breach]

so do i find the sector area?

and the triangle?

[stoic pecan]

the arc area yes

yes

good

[sick breach]

ohh

ok

[stoic pecan]

do you see it yet?

[sick breach]

i think i do

[stoic pecan]

i hope you do

[sick breach]

so its 2 times (sectorarea- triangle area)

[stoic pecan]

yep

[sick breach]

ohhh

[stoic pecan]

or 4x if you use the mini area

easy once you see it right

[sick breach]

yeah

[stoic pecan]

now go solve it

[sick breach]

ight

one sec

[sick breach]

is it 123u^2

[stoic pecan]

i got no idea i didnt do it

uhh gimme a sec

[sick breach]

okok

[stoic pecan]

what is u?

[sick breach]

i got 123 cm^2

or units

[stoic pecan]

8.17505543965

i got 33

send working?

[sick breach]

💀

wait what was the internal angle u found

for the sector

[stoic pecan]

36.8698976458

i did half the sector

so double if you did the full thing

[sick breach]

wait i got 106 wtf

[stoic pecan]

how

thats like

[sick breach]

i just wrote tan 8/6

[sick breach]

tan inverse

8/6

then get 53

[stoic pecan]

6/8

its 6/8

[sick breach]

and times by two

oh

😭

[stoic pecan]

at least u got the rest of it right

maybe

[sick breach]

yh lemme just plug it all in

what area did you get

i got 33

thank you

.close

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[prime bane]

k is supposed to be 18 but I got 1.8

[prime bane]

Can someone help

[light helm]

you need to pay attention to units

[prime bane]

Whats wrong

CM and seconds match up right?

Ohh

It's supposed to be m and s

[light helm]

proportion is for length in metres,
you're asked about when l is a certain number of cm

[prime bane]

So if I convert the cm to m before starting I'm good right??

[light helm]

yeh

[prime bane]

I'm still getting the wrong answerr

[light helm]

show work

[prime bane]

Nvm

It's correct

Tysm

[prime bane]

Now i am getting the second sub question wrong

[light helm]

what's the intended answer

[prime bane]

12.7

[light helm]

that seems to be an error on their end

,calc 18sqrt(4.5)

[jolly parrot]

Result:

38.183766184074

[prime bane]

You sure

Cuz I am getting the 3'rd one wrong too

[light helm]

show your work and their answer

[prime bane]

Here you go

[light helm]

,W (7/18)^2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by ramonov

[prime bane]

That's what even I got

[light helm]

results the same

[prime bane]

But it does not match the textbook answer

[light helm]

again pay attention to units

[prime bane]

I did

I checked it a 100 times

[light helm]

can you convert
0.151m to cm?

[prime bane]

15.1cm

[light helm]

no

[prime bane]

Wait whatt

[light helm]

it's correct now

[prime bane]

What are you talking about

[light helm]

so yeh, it's the same result as what they have

you said
15.1m earlier

[prime bane]

But why did you convert

There was no need to

[light helm]

you don't have to

it's just that you thought that what you and they had was completely different

[prime bane]

Explainm

[light helm]

when in fact they're just different ways to express the same result

[prime bane]

The answer is different

Ohh the mentioned cm

I get it

[light helm]

yes...

[prime bane]

What about b

[light helm]

like I said, pay attention to units

[prime bane]

B has no units

[light helm]

b as mentioned earlier is just a mistake on their end

[prime bane]

Oh okk

Tysm

.close

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[tired saddle]

How do I go about solving this?

[tired saddle]

I dont really know where to start tbh

[modest badge]

Well logarithms are the inverse of exponentiation

[modest badge]

Try to think about the powers b is raised to inside the logarithm

[tired saddle]

Oh would it just be -3 and 1/2?

[midnight haven]

Yes

[tired saddle]

Alright thanks

[pearl pond]

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[warm osprey]

I find this difficult to accept. My reasoning is this:
consider a = b+e for all e > 0
This theroem tells us that a = b now
but if a = b then we have a = a + e => e = 0
But we started with e > 0 thus this is a contradiction???
What have I done wrong?

[vestal tapir]

this says "a = b+e" is never true

what you considered can't exist, and that's what you proved, and that's also what they were proving in the first place

[warm osprey]

oh lol thanks

.close

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[vestal tapir]

actually i'm confused idk what i'm saying

[warm osprey]

.reopen

[pearl pond]

✅

[vestal tapir]

like, "a = b+e for all e > 0" is already absurd

there can't be multiple e that satisfy it

[warm osprey]

Oh yeah fair

[vestal tapir]

but that's maybe not the same as what we were trying to prove

that's the part idk

what they say after "it means" is clearly true, if a > b, then a is not less than any number larger than b, cuz a = a

[pearl pond]

@warm osprey Has your question been resolved?

[pine jay]

Also I think it’s worth mentioning that the premise does not say that there must be an equality, it’s an or statement and say if there was an equality for all such epsilon’s it would just be vacuously true.

On the other hand it’s also good to note in general that we can have an inequality for all epsilon’s which may have an equality for some specific epsilon; I don’t think this is the case here(?) but it’s good to remember what the quantifier means here, it’s essentially a bunch of different inequalities

[pearl pond]

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[languid heart]

what does it mean when someone says parallel lines meet at infinity

[languid heart]

parallel lines have the same gradient

so isnt it impossible

for them to ever

intersect

or touch

?

[glacial sequoia]

you never reach infinity so it's all good

[glacial sequoia]

tbf you really shouldn't say that

If you take two lines that are almost parallel, you'll see they intersect very "far away"

So as you keep getting closer and closer to true parallel, you'll find the point of intersection gets away further and further, the tangent function, tan(x), encompasses this in a way

You could try this for yourself on desmos

[languid heart]

yeah but in that case the are almost parallel

but if the lines are

completely parallel

they cant intersect at all?

[glacial sequoia]

They don't

You can't show me infinity on a number line anyways

[languid heart]

yes

[glacial sequoia]

Do you understand what a limit is?

[languid heart]

no-

[glacial sequoia]

No worries

I expected that

[languid heart]

but i've seen shit like 0 --> infinity

💀

[glacial sequoia]

Right, don't worry about it for now

The point is to think of lines that are almost parallel, maybe their slopes differ by 0.0000001, the point of interesection is at such a large distance

This distance is increasingly large as the lines tend towards being trult parallel

[languid heart]

yes

[glacial sequoia]

When they are, the distance is immeasurable, and some people pass this off as them intersecting at infinity

[languid heart]

ohkk

[glacial sequoia]

Infinity does not work the same way as a large number

It's not 100

It's not 1000

It's nothing you can think of

So "intersecting at infinity" is somewhat nonsensical

[languid heart]

we have been taught to make infinity=30000

as we cany

cant

imagine

it

[languid heart]

right

[glacial sequoia]

Yeah, the point of that is to make it large enough

[languid heart]

ohkk

[glacial sequoia]

So that it doesn't mess shit up for smaller inputs

If your input itself was 30,000 then you would set infinity to be 30,000,000,000

[languid heart]

ohk so basically they mean that the point of intersection is like really really far?

and thats basically intersecting at infinity

[glacial sequoia]

They don't intersect, like you say

[glacial sequoia]

But sure

[autumn fossil]

(It might be in context of projective geometry, where lines indeed intersect at infinity btw. But in standard geometry, they dont intersect at all as was already said.)

[autumn fossil]

The problem with this is that in standard euclidean geometry, there is nothing like "infinity"

"infinity" has no meaning there

[languid heart]

💀

[glacial sequoia]

Infinity hasn't been sold off to students in the right way

All you need to know is that let the infinitiy people keep their yap up, saying that they don't intersect won't cause you harm

[languid heart]

ohk

thanks

.close

[pearl pond]

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[languid heart]

i had another question, why is circle just a line

.reopen

[pearl pond]

✅

[glacial sequoia]

Same shit

As your radius gets very very large

The circle begins to look like a line

So some people say a line is just a circle with radius infinity

[languid heart]

but if we're increasing the radius

the diameter increases

as well

[glacial sequoia]

yes

[languid heart]

so its increasing both upwards downwards left and right

that just makes a bigger circle

i guess my imagination is not up to the level

[glacial sequoia]

The first image is a section of a circle of radius 1

notice how I'm only looking at 4 boxes

The second image is the same section but the radius is 100

Notice how it's much flatter?

[glacial sequoia]

But we stay the same size

It's like how you know the earth is curved but if you look around, the horizon looks like a straight line

[languid heart]

hmmm

but the only reason its a straight line is cuz the circle is too large

[glacial sequoia]

Yes

[languid heart]

and one last thing , if we take half a disc, like a semicircle

why does it have

infinite

edges

[glacial sequoia]

Well it doesn't again

Damn the infinity peeps really told you some stuff

I can't use desmos for this

Imagine you have a triangle

[languid heart]

yes

[glacial sequoia]

Like that

Now we cut off the top

[languid heart]

ohk

[vital estuary]

who told you all this

[glacial sequoia]

we get a quadrilateral

Now imagine we cut off the top two corners

[languid heart]

where did the quadrilateral come from

💀

[glacial sequoia]

We cut the top off the triangle

[languid heart]

ohkk

[languid heart]

OHH

[glacial sequoia]

Now we have a hexagon

So you keep cutting and cutting and cutting

So some imbecile would say it has infinite edges

But it doesn't

[languid heart]

so its has many edges?

?

[vital estuary]

i feel like whoever taught you these infinity concepts was trying to confuse you

[languid heart]

we have been told that infinity is a really large number

and that u cant imagine it

[vital estuary]

infinity is a concept

its not a number

[glacial sequoia]

This is for practical purposes

Not in math here

[languid heart]

can u suggest

a video

so that i can understand what infinity actually is

[glacial sequoia]

One curved side and one straight side

[languid heart]

but the edges are a lot

if we keep

cutting

[glacial sequoia]

Yes but it's also never truly a semicircle

It just looks like one because our eyes are trash

[languid heart]

ohk

thanks a lot

.close

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[midnight haven]

Help

[midnight haven]

.close

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[junior pawn]

about power series, since there're polynom in x are they C infinite?

[tropic saddle]

they arent polynomials. polynomials are finite sums

power series might not even converge

[pearl pond]

@junior pawn Has your question been resolved?

[junior pawn]

in the interval where it converge

[pearl pond]

@junior pawn Has your question been resolved?

[pearl pond]

@junior pawn Has your question been resolved?

[pearl pond]

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[errant wasp]

I got a (for me impossible to even comprehend) math question soo i hope someone can help me

I need to calculate the max speed and acceleration to said speed of an object with only 2 measurements

Measurement 1: 0 to 20 meters in 3 seconds from standstill

Measurement 2: 0 to 80 meters in 6 seconds from standstill

These are both with the same object and same accelleration, i just have no clue what it is

[errant wasp]

So basically 0 - 20m in 3 seconds and 20 - 80m in 3 seconds

[calm wing]

pretty certain that is not constant acceleration

or wait hold up

[calm wing]

use suvat

since the initial velocity is 0, u=0

so you are left with $s = \frac{1}{2}a t^2$

[jolly parrot]

artemetra

[calm wing]

solve for a

in both cases it ends up being ||40/9 ms^-2 which is around 4.444||

[errant wasp]

I should mention i'm horrible at maths

And what would the maximum speed end up being with a accelleration of that?

[calm wing]

in which experiment?

the final speed is different for them

[errant wasp]

The 80m one

[calm wing]

i know the answer but let's find it together

[errant wasp]

Okie

[calm wing]

have you seen any of these before?

like any formula here

[errant wasp]

Uhh no

[calm wing]

okay

these are called suvat formulas and they are super handy for solving kinematics-related problems

s is for displacement
u is for initial speed
v is for final speed
a is for acceleration
t is for time

[errant wasp]

So u is 0, v is ?, t is 6 seconds and accelleration is 4.44?

[calm wing]

precisely

and s is 80

so if you want to find the final speed you use the very first one

v=u+at

since u is 0, you can rewrite it as v=at

[errant wasp]

At being accelleration / time?

[calm wing]

no, acceleration times time

a*t

[errant wasp]

Ah right

So 4.44 x 6?

Being 26,64

[calm wing]

yep

that's the answer

[errant wasp]

And thatd be the final speed in 80 meters in kmh?

[calm wing]

yes but not in km/h

it will be in meters per second

[errant wasp]

oh

Damn

96kmh

That is pretty damn fast

Thank you for the help

[calm wing]

yep

[calm wing]

it is indeed

always welcome to help

if you don't have any more questions or doubts type ".close"

[errant wasp]

Alrighty

.close

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[pearl pond]

Please don't occupy multiple help channels.

[midnight haven]

what is this formula

[somber needle]

Just seeing it makes my brain crazy

[midnight haven]

P is the hypothetical economic parameter (e.g., price level, profit, etc.).
• N is the number of consumer segments.
• C_i is the consumption function for the i -th segment.
• \mu_i and \sigma_i are the mean and standard deviation of the i -th segment’s income distribution.
• \zeta(j) is the Riemann zeta function.
• \Gamma\left(\frac{j}{2}\right) is the Gamma function.
• \lambda is a fictitious decay constant.
• \theta is an arbitrary angle.
• \pi is the mathematical constant pi.
• i is the imaginary unit.
• \hbar is the reduced Planck constant.
• c is the speed of light.
• k_B is the Boltzmann constant.
• T_i is the temperature-like variable for the i -th segment.
• GDP is the Gross Domestic Product.
• I(t) is the investment over time.
• r is the discount rate.
• \alpha is a scaling exponent.
• K is the number of production factors.
• \beta_k is a coefficient for the k -th production factor.
• Q_k is the quantity of the k -th production factor.
• Q_0 is a baseline quantity.

[pearl pond]

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[charred niche]

how do i do this

?

[silver nest]

the 2 is multyplying the whole 5log2(4)?

[naive orbit]

i would say it is 5log16(2)

[silver nest]

maybe 10 log2(4) => log2(4^10) => log2(2^2^10)

log2(2^20)

so 20log2(2)

so20(1)

so the answer is 20

[charred niche]

its 5logbase16 2

or

logbase16 2^5

but i simplified it

[charred niche]

yes

[charred niche]

the answer says its 1/4

[silver nest]

oops, I did something wrong

[naive orbit]

are you sure 5 is not a question number?

[charred niche]

yes

[charred niche]

i can give you the original question

if u want

[naive orbit]

please

[charred niche]

its evaluate logbase16 52 - logbase16 7 + logbase16 4

[charred niche]

the base is 2^4

the little number

wait no the answer is 5/4

woops

[naive orbit]

yep

[last moth]

you're reading the question wrong

[charred niche]

idk how to get to it tho

[last moth]

so, the question is what power you need to raise 2^4 to in order to get 2

[naive orbit]

you are aware of this rule?

[charred niche]

yes

okay

before i simplified it

logbase16 32

and then i got stuck

yeah

[charred niche]

wait no

i didnt knwow that

what the

oh

i got it

thanks

.close

[pearl pond]

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[naive orbit]

no problem

[pearl pond]

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[brittle onyx]

I’m confused

[brittle onyx]

Can any1 tell me where I made the made the mistake in this

<@&286206848099549185>

[warm current]

You are given |AC|=|BD|

That'll help you finish up

[terse scaffold]

did I mog him chat?!

ignore the plus at the top

[pearl pond]

@brittle onyx Has your question been resolved?

[brittle onyx]

.close

[pearl pond]

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[visual moon]

dumb question but im not understanding how the lcd is x^3 and not x^3x^2?

[tulip ore]

whats the LCD of 8 and 4?

[visual moon]

1?

[tulip ore]

sorry, LCM of 8 and 4

(or LCD of 1/8 and 1/4)

[visual moon]

8

[tulip ore]

not 8 * 4, right

(because 4 is a factor of 8)

[visual moon]

yea

sorry i dont understand how that works with the variables

[tulip ore]

say you have x * x * x and x * x

compare this to 2 * 2 * 2 and 2 * 2 which we just did

LCM of x^3 and x^2 is x^3, not x^5

because x^2 is a factor of x^3

does that make sense

[visual moon]

not really ;-; sorry

[tulip ore]

do you know what LCM means

[visual moon]

lowest common multiple, like if im adding/subtracting fractions with different demoninators, i find a multiple that fits into both

[tulip ore]

alr

now say you have 1/x^2

and you multiply this fraction by x/x

what would you get

[visual moon]

i cant use the x method where i multiple the demons by the top and bottom of the other fraction and vice versa?

[tulip ore]

no idea what that means

lets just do things simple here

what is 1/x^2 * x/x

[tulip ore]

(you shouldnt consider methods if you dont know whats happening with variables)

[visual moon]

x/x^3?

[tulip ore]

yep

[visual moon]

fractions just make my brain turn off

[tulip ore]

what about fractions with numbers

we'll need to start somewhere where youre faster

[sonic patrol]

hm im curious, what do you find difficult about fractions?

[visual moon]

adding and subtracting expressions with like multiple variables in the demonintor

like i do this basic mathod for previous questions

[sonic patrol]

I see I see

[tulip ore]

thats not how you calculate LCD for regular fractions

so it wont be the LCD for fractions with variables

say you have $\frac18+\frac14$

[visual moon]

i see

[jolly parrot]

mtt

[tulip ore]

you recognized that the LCD here is 8

[sonic patrol]

It is true that in general $\frac ab + \frac cd = \frac{ad + bc}{bd}$

[jolly parrot]

Pseudonium

[sonic patrol]

But this isn’t always the most efficient way to add two fractions

[tulip ore]

see here that you dont need to x the fractions to have a denominator of 32

[sonic patrol]

Maybe fractions are hard cause it’s the first time you meet an equivalence relation

[tulip ore]

now part of what you did to recognize the LCD is to factor 8 and 4

8 factors into 2, 2, 2
4 factors into 2, 2

[sonic patrol]

Does it make sense that the lowest common multiple of $x^3$ and $x^2$ is $x^3$?

[jolly parrot]

Pseudonium

[visual moon]

i dont see why the lowest common multiple for x^2 is x^3

[sonic patrol]

I see I see

[visual moon]

should i break down x^3 and x^2

[tulip ore]

that works

[sonic patrol]

I guess there’s kind of 2 related but different statements here

[tulip ore]

you can do that to break down x^3 and x^2 to find the LCM of x^3 and x^2

[tulip ore]

try doing that just like with 8 and 4

[sonic patrol]

It’s true that $\forall a \in \mathbb{N}, \text{lcm}(a^3, a^2) = a^3$

[jolly parrot]

Pseudonium

[sonic patrol]

It’s also true that as polynomials, $\text{lcm}(x^3, x^2) = x^3$

[jolly parrot]

Pseudonium

[sonic patrol]

these are technically different

and I guess really you’re using something like the second result..

Like, it’s very possible that you’ve only seen lcm discussed in the context of numbers, right?

You know what it means to find the lcm of 5 and 10

Of 27 and 34

Or any other pair of numbers

But maybe it’s unclear what the lcm of x^2 and x^3 even means

[visual moon]

thank you both of you

[visual moon]

broke it down with the varibles

and watched a video

i got x^3 yea

[tulip ore]

nice

[visual moon]

x^2 *x

[tulip ore]

thats correct

if you look in the problem you posted,

theres a typo so maybe not

,,\lim_{x\to0}\qty(\frac{\sin(2x)}{x^3}+\frac b{x^2})
\=\lim_{x\to0}\frac{\sin(2x)+bx}{x^3}

[jolly parrot]

mtt

[tulip ore]

this is what was done

youll see there that * x being done, since afterwards b has an x next to it

[visual moon]

hmmm

[sonic patrol]

Yeah the point is that $\frac{b}{x^2} = \frac{b x}{x^3}$

[jolly parrot]

Pseudonium

[sonic patrol]

Whenever x is nonzero so that both sides make sense

[visual moon]

oh

[sonic patrol]

This is part of what makes fractions so hard

$\frac 12 = \frac 24$

[jolly parrot]

Pseudonium

[sonic patrol]

They’re not, like, literally the same latex symbols or anything

But they represent the same thing

Even though they look different

It’s a very analogous situation to $\frac{b}{x^2} = \frac{ b x}{x^3}$

[jolly parrot]

Pseudonium

[sonic patrol]

The two sides aren’t literally the same latex symbols

But they represent the same thing

And maybe it just takes getting used to..

[visual moon]

thank you, yea i gotta do more practice

[sonic patrol]

In more advanced mathematics, there’s a name for this kind of concept

An “equivalence relation”

It’s probably beyond what you need to know at the moment

But it’s this idea of, even if things don’t literally look the same, there might be a sense in which they’re the same

[visual moon]

interesting, cool to know

[pearl pond]

@visual moon Has your question been resolved?

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[strange quiver]

6i no clue where to start

[west sapphire]

do you have the characterization of contininuity in terms of inverse images of functions?

[viscid dew]

As someone who's never actually done any metric spaces, my intuition is to consider the continuous function f(x)-x
but hey, I might be completely insane

[strange quiver]

Do you mean the fact that f is continuous if f^-1(U) is open for all open U?

[west sapphire]

yep

try applying that to f(x)-x as suggested by nameless individual

[viscid dew]

do you mean iff?

[strange quiver]

Yea

[pearl pond]

@strange quiver Has your question been resolved?

[strange quiver]

Is my approach correct?
let g(x)=f(x)-x, we have g^-1(0)=A then since g(x) is continuous and {0} is closed A is also closed

[pearl pond]

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[sour wasp]

the maximum distance will be bound by the sum of the diameters and the distance between the sets
d(x,y)≤diamA+diamB+d(A,B)
therefore d(A∪B)≤diamA+diamB+d(A,B)

[strange quiver]

I'm actually doing 6i but thanks for helping

[warm current]

That's perfectly correct

@strange quiver

[strange quiver]

That's great to know thanks

[warm current]

Need anymore help?

[strange quiver]

Nope

[warm current]

.close

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[daring wedge]

What do I do next

[daring wedge]

How do I integrate the x/√1-x²

[quiet sentinel]

x is basically almost the derivative of (1-x^2) already

[daring wedge]

U sub?

[quiet sentinel]

ya

[daring wedge]

Love uuuuuu

Mr helpful stranger

Hoorah Ig

[quiet sentinel]

yay

[pearl pond]

@daring wedge Has your question been resolved?

[pearl pond]

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[ruby trail]

What's the next step after getting the general solution?

[pearl pond]

@ruby trail Has your question been resolved?

[pearl pond]

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[pulsar stump]

how to find all partial limits for the sequence?

[cursive wraith]

force factorisation inside the sqrt by 2^n

[pulsar stump]

wdym?

[cursive wraith]

like

a+b = a(1 + b/a)

i forced factorization by a

[pulsar stump]

okay but why? i mean also the partial limit we have here is just the singleton of 2 right?

[cursive wraith]

well you want to prove the only partial limit is 2

[pulsar stump]

well how then?

[cursive wraith]

.

[pulsar stump]

i know how to use squeeze theorem for 2 id say

but how do i prove there arent any other limits?

like partial ones

[cursive wraith]

...

[pulsar stump]

alright

[pulsar stump]

and for something like this?

[cursive wraith]

here the most annoying thing is the sin

[pulsar stump]

fr

[cursive wraith]

show that the left stuff has limit 1

and find all partial limits of the sin

then do limit * partial limit theorem

[pulsar stump]

how do i find for the sin functions the partial limits?

[cursive wraith]

oh it's gonna be easy

n = 1, n=2, n=3,...

[pulsar stump]

doubt it

[cursive wraith]

you'll see it pretty quick

[inland ivy]

What are partial limits?

[cursive wraith]

limits of subsequences

[inland ivy]

And what’s this theorem you speak of

[inland ivy]

Ah, I see

[cursive wraith]

if an has limit ... and bn has partial limits ...,...... then an*bn has partial limits...

[pulsar stump]

$lim a*b = lim a * lim b$

[jolly parrot]

Ayanokoji

[inland ivy]

Oh okay

Makes sense

[pulsar stump]

i see infinite partial limits

.close

[pearl pond]

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[sharp smelt]

to prove sin(x)<x is the taylor series overkill

[sharp smelt]

like am I overthinking this

sin(x)= x-x^2/!...

[limber oasis]

it is

||MVT|| suffices

[sharp smelt]

that $f'(c)=f(b)-f(a)/(b-a)$?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

hmm

I guess I could prove that the mean is less than 0?

[limber oasis]

more like sin(x) = sin(0) + sin'(c)x

conclude

technically two steps because of a detail

[sharp smelt]

sin(0)+sin(c)x?

[limber oasis]

I insist

[cursive wraith]

no sin'(c)

[sharp smelt]

that would be sin(x)=cos'(c)x

[limber oasis]

no

[cursive wraith]

no

[limber oasis]

apply the theorem properly

[sharp smelt]

that would give me sin(x)/x =sin'(c)

[limber oasis]

yes

[sharp smelt]

that means sin(x)/x is always between or equal to -1 and 1

from which I can conclude that $sin(x) \leq x$?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[cursive wraith]

there's a case step missing

[sharp smelt]

hmm?

[cursive wraith]

like

1/(-2) is between -1 and 1

is 1 <= -2?

[sharp smelt]

no

[cursive wraith]

ok

[cursive wraith]

so

[sharp smelt]

$-1\leq sin(x)/x \leq 1$

[cursive wraith]

?

[sharp smelt]

oops

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

$\implies sin(x) \leq x$

[cursive wraith]

sure, why?

[cursive wraith]

how is it any different from this?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

not sure of how I'd formally justify this

[cursive wraith]

again, what's the problem with my method

where's the flaw and why is it a flaw

[sharp smelt]

I dont know what your method is, sorry

[cursive wraith]

1/(-2) is between -1 and 1
thus 1 <= -2

[cursive wraith]

where's the flaw, why is it a flaw

[limber oasis]

he's evaluating your inequalities at x=-2

[sharp smelt]

oh , right

this is only true for $x \geq 0$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[cursive wraith]

at least this is an improvement

[sharp smelt]

0

my bad

[cursive wraith]

yeah better

so x > 0 there I agree

because you take sin(x)/x <= 1

and you multiply by x

all good

now, x = 0? x < 0?

[sharp smelt]

this sin't true any more

sin(x)>x for x <0

[cursive wraith]

yep

[sharp smelt]

thanks!

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[pearl pond]

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Remember:
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[daring wedge]

How to change into partial fractions for the top most part

Do I have to divide

Or is already a proper frac

[oblique river]

$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}$

[daring wedge]

Shouldn't the degree of the numerator be one less than the degree of the denominator?

[oblique river]

numerator's degree is 3, denominator's is 4

[daring wedge]

Taking ABT the C and the D

It should be Dx +E

[light helm]

you can split the fractions like that when you have repeated factors

[daring wedge]

Cuz the denominator has degree w

[oblique river]

oh thats for unfactorable denominators

[daring wedge]

2

[daring wedge]

Oh

What

Guess I'll have to revise partial fractions