help-13

[olive folio]

I dont know

[violet flume]

well whats 1^3 and 1^2

[olive folio]

1 and 1

[violet flume]

okay

so try to simplify

[olive folio]

A+B+C

[violet flume]

okay

so f(1) = A + B + C = 4

you need two more equations

so you need to repeat this process twice more

[olive folio]

what do you mean repeat

[violet flume]

with a diffferent x from the table

[olive folio]

like do the same thing with a different x value?

[violet flume]

yea

[olive folio]

okay

so I would do f(2) now right

[violet flume]

sure

[olive folio]

what do I do after I get F(2) = a(8) + b(4) + c(2) = 20

[violet flume]

that 8a + 4b + 2c = 20

[olive folio]

would that be the second equation?

[violet flume]

yea

[olive folio]

so I go onto the next x value?

[violet flume]

yup you need 3 total

[shut delta]

nice username 🤔

[olive folio]

do I use f(3)?

even though f(x) is unknown

[violet flume]

nah

use 4

[olive folio]

ok

so what do I do now that I have all 3 equations

[violet flume]

you have a system of equations

how have you solved them before?

[olive folio]

i dont remember

[violet flume]

https://mathbitsnotebook.com/Algebra2/Polynomials/POSystems.html

theres some suggestions here

substitution is a pretty common method

[olive folio]

I got a = 2 c=2 and b = 0

does that look right?

@violet flume

[violet flume]

hmm the graph doesnt look right

[olive folio]

x=3 I got 60

idk if its right

[violet flume]

what dyou get like first a+b+c=1, then 8a+4b+2c=20

[olive folio]

what do you mean

[violet flume]

64a+16b+4c=136

[olive folio]

like what did I solve for first?

[violet flume]

its not correct

have you ever worked with a matrix?

[olive folio]

no

when I check my a b and c values im getting 4=4 20=20

does that still mean im wrong

whats a matrix

[violet flume]

well when you get an a b and c

you can just graph your answer

and see if it works out or not

these system of equation things are pretty error prone

[olive folio]

alright

[violet flume]

lemme see

i mean you could solve the first one for ... lets say c

c = 1 - a - b

the nyou can substitue this into the next equation

[olive folio]

yea I did something like that

[violet flume]

so 8a + 4b +2(1-a-b) = 20

or 8a + 4b + 2 - 2a -2b = 20

or 6a + 2b = 18, i think?

say 3a+b=9

so b = 9 - 3a

[olive folio]

shouldnt it be c=4-a-b

[olive folio]

here

[violet flume]

i really gotta go to bed

im gonna make stupid mistakes

[olive folio]

sorry

[violet flume]

lol youre good its not your fault

so c = 4-a-b

and 8a+4b+2(4-a-b)=20

so 8a + 4b + 8 -2a-2b =20

then 6a + 2b = 12

or 3a+b=6

so b=6-3a

then we had c=4-a-b=4-a-(6-3a)

or c = 4-a-6+3a = -2+2a

lol this stuff takes so long

last one

64a + 16(6-3a) + 4(-2+2a)=136

,w simplify 64a+16(6-3a)+4(-2+2a) = 136

[wraith dagger]

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[violet flume]

a = 2

[olive folio]

I got that

[violet flume]

ive made a mistake somewhere im sure of it

[olive folio]

idk you solved it the same as I did

i got a = 2

b = 0

c = 2

[violet flume]

oh

OH

its right

you were right

i was wrong

sorry man im just so god damn tired

[olive folio]

so f(x)=60?

[violet flume]

i was reading the graph wrong

yes

[olive folio]

thank you for the help

[violet flume]

no problem thanks for bein patient with me

[cedar kiln]

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[umbral mirage]

Hi, can anyone help me with this

[tropic oxide]

!occupied

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[umbral mirage]

i tried finding the limits and they gave me 0

[tropic oxide]

weird defn of "differentiable" they got there.

[tropic oxide]

the limit of f'(x) should not have given you 0. show work?

[umbral mirage]

ok

Thats the first part

im confused about the whole limit to 0- and 0+ thing

because both of these just equal 0

to prove its continuous

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[cedar kiln]

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[bronze phoenix]

can someone help me with this prblm

Idk where to start

[valid yacht]

you know the formula to calculate arc length of a function?

[daring rune]

We use the formula
$L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} , dx$

[wraith dagger]

Jane

[bronze phoenix]

yea this makes sense

but then why do have to find the antiderivative of that

isnt the antiderivative the exact same numbers we get from the beginning

?

[royal loom]

'the numbers we get from the beginning' being what?

if you're talking about the distance between x=1 and x=9 then no

because if our function curves, we cover more distance than just 8

the shortest distance would be a straight line, which our function clearly is not

[bronze phoenix]

Like this I mean

[cedar kiln]

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[floral terrace]

Anyone good at Taylor series?
Translation is at the bottom of the image. I was struggling to approach this, because I sort of assumed that the answer would be 0 due to the fact that x0 is 0, and that would be substituted into the x inside the function.... right?

[floral terrace]

In this type of question, I would be using this:

[dawn junco]

do you really want to compute 17 derivatives of that thing

[floral terrace]

because we are looking for an index of 17, I started by approaching (x-a)^n first. I replaced a with 0 so it's just x, and I replaced x with the contents inside the ln(x), so --

[floral terrace]

exactly, I did not want to suffer

[dawn junco]

yeah just use the series for ln(1-x) or whatever

[floral terrace]

because I assumed that the derivative would just continue to equal 0

ohe

this thing?

[dawn junco]

yeah

and instead of x you have -x^2 here

& don't forget multiplying by x the whole thing afterwards

[floral terrace]

and the n... I would be able to find n through the use of the (1-x^2)^n * x from earlier.

becomign -x^{2n+1}, and 2n+1=17, n = 8...

[dawn junco]

yeah n=8

[floral terrace]

Is the answer (-)x^16 / 8? :3

[dawn junco]

that should be it

well - x^17 /8

but the constant in front is correct

@floral terrace

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[floral terrace]

Yeah I realized that later, thank you anyway :3

[cedar kiln]

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[misty venture]

how can I find the domain of this equation?

[undone epoch]

Is it x^2 + 1 or (x^2 + 1)?

[misty venture]

log(x^2) +1

[vague rapids]

well, what can you conclude about the domain of x when you have log_10(x^2)

[misty venture]

R ?

cuz x^2 > 0 for all real numbers

[vague rapids]

well can x be zero?

[misty venture]

no

so x!=0 is the domain?

because when I write it in geogebra it doesn't show the answers -1 and -10

only 10 and 1

<@&286206848099549185>

[spice kraken]

for $x = -1$ you get $(-1)^{\log_{10}(2)}$ though

[wraith dagger]

WhereWolf(ping if needed)

[misty venture]

@spice kraken

[spice kraken]

oh

yes -1 and -10 are indeed solutions

geogebra probably missed it as the function is not continuous for x<0

[cedar kiln]

@misty venture Has your question been resolved?

[misty venture]

ok ty

.close

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[spring tide]

Need a bit of help on this question

I’ve taken the area of A_1 to be a^2

And since the progression is geometric, the area of A_3 = (ar)^2 (where r is the common ratio)

So far all I’ve got is the equation (ar)^2 = a^2 + 1000

I have no clue what to do with it or how to solve for a and r

I’ve also got r = sqrt(1 + 1000/a^2) but that is derived from the above eqn so idk what to do

Any help will be greatly appreciated

[cedar kiln]

@spring tide Has your question been resolved?

[spring tide]

<@&286206848099549185>

[median grove]

?

[spring tide]

Need some help on how to solve this question

[median grove]

which

[spring tide]

Here

Not for school, it is for a competition but we’re allowed to ask for help as long as we can show our work

[median grove]

ALRIGHT

[cedar kiln]

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[spring tide]

.close

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[supple flume]

Can’t understand how to fully solve this

[supple flume]

Ima try a bit more before getting help.

.close

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[halcyon zephyr]

hello, i am absolutely lost on laplace transforms

[halcyon zephyr]

this is a question from a practice test

i've tried looking at videos for laplace transforms and information from the internet

but everything looks different

I'm also provided with 4 multiple choice answers but i dont understand how to get to those

my teacher has provided 0 learning material online so im kinda clueless rn

[cedar kiln]

@halcyon zephyr Has your question been resolved?

[halcyon zephyr]

i totally understand people avoid this stuff, I tried to do that too

[halcyon zephyr]

[vernal shell]

Hello, do you know how to get the laplace transform of a derivative?

[halcyon zephyr]

im using a transfer table

[vernal shell]

A laplace transform table I guess (?

[halcyon zephyr]

yes

[vernal shell]

It should have information about Laplace transform of df/dt and d²f/dt²

[halcyon zephyr]

yes df/dt = 𝑠𝑌(𝑠) − 𝑦(0)
and d²f/dt² = 𝑠²𝑌(𝑠) − 𝑠𝑦(0) − 𝑦′(0)

am i supposed to transform both sides of the differential equation?

[vernal shell]

Yes

Laplace transforms are a useful tool for solving differential equations

They "convert" those equations into "polynomials"

So you can solve for Y(s), or R(s)

In this case, they want you to isolate Y(s)/R(s)

Given y is the physical system's output, and r is the input, the transfer function is defined as Y(s)/R(s)

[halcyon zephyr]

yeah that makes sense

[vernal shell]

Remember that initial conditions are 0, that simplifies it a lot as y(0)=0 and y'(0)=0

[halcyon zephyr]

currently i have the left side of the equation, but im not sure how to continue writing

how does the text to math work, lemme give it a shot

$R(s) = M(s^2y(s)-sy(0)-y'(0))$

[wraith dagger]

♡ Mustard ♡

[halcyon zephyr]

amazing

but then for the right side it would be $R(s) = -b(sy(s)-y(0)+k(r(t)-y(t))$?

oopsie

$R(s) = -b(sy(s)-y(0)+k(r(t)-y(t))$

[wraith dagger]

♡ Mustard ♡

[halcyon zephyr]

i feel like im forgetting something

[vernal shell]

Why are both equal to R(s)?

[halcyon zephyr]

I dont know, I thought I'd get Y(s) by solving R(s)

im assuming R(s) is coming from the laplace transform

[vernal shell]

Start from this equation. Take the laplace transform of both sides of the eq

It's like when you have an eq. and take the derivative on both sides

[halcyon zephyr]

okay

[vernal shell]

$\mathcal L\left{M\dv[2]{y}{t}\right} = \mathcal L\left{-b\dv{y}{t}+k(r(t)-y(t))\right}$

[wraith dagger]

ELeonardo

[halcyon zephyr]

oo

$M(s^2Y(s)-sy(0)-y'(0)) = -b(sY(s)-y(0))+k(rt-yt)$

[wraith dagger]

♡ Mustard ♡

[halcyon zephyr]

[vernal shell]

What's rt and yt?

The laplace transform of some function f is usually written as (capital) F: F(s)

[halcyon zephyr]

well r(t) should be the input and y(t) the output
im not sure what the outcome should be after the transform

if all initial conditions are 0 then should rt and yt also be 0

at least im assuming that it means 0(t) so 0

i feel like younger me learning math for the first time like 12 years ago

[vernal shell]

XD

It happens from time to time

Well depending on what you're studying

[vernal shell]

But you're taking the laplace transform, not evaluating them at 0

I mean rn you're doing

$\mathcal L\left{M\dv[2]{y}{t}\right} = \mathcal L\left{-b\dv{y}{t}\right}+k(r(t)-y(t))$

[wraith dagger]

ELeonardo

[vernal shell]

Instead of

$\mathcal L\left{M\dv[2]{y}{t}\right} = \mathcal L\left{-b\dv{y}{t}+k(r(t)-y(t))\right}$

[wraith dagger]

ELeonardo

[halcyon zephyr]

oh yes

so k(rt-yt) is not the correct transform of k(r(t)-y(t))

theres nothing that looks similar in my table

[vernal shell]

they are just functions

for example, laplace transform of r(t) is R(s)

[vernal shell]

.

[halcyon zephyr]

is that something like this? or do i understand wrong

[vernal shell]

No, that's the laplace transform of an impulse function

you don't know what r(t) or y(t) are, so just write their transforms as R(s) and Y(s)

[halcyon zephyr]

ohhhh

so ...+k * (R(s) - Y (s))

[vernal shell]

yes

[halcyon zephyr]

so that means I have R(s) and Y(s) and now I have to solve it so i get Y(s)/R(s) on the left side of the equation

[vernal shell]

yes

[halcyon zephyr]

also, the transform table writes this

is the Y different than the y

[vernal shell]

Y(s) is the laplace transform of y(t)

$Y(s)=\mathcal L{y(t)}$

[halcyon zephyr]

oh that makes sense now yeah

[wraith dagger]

ELeonardo

[halcyon zephyr]

amazing, i'm gonna simplify and solve if i can

[wraith dagger]

♡ Mustard ♡

[vernal shell]

Ok

For future problems, if the statement says that initial conditions are 0, then you can use this as the remaining terms cancel

$\mathcal L\left{\dv[n]{f}{t}\right} = s^n F(s)$

[wraith dagger]

ELeonardo

[halcyon zephyr]

I think this the right answer

i am left with 2ks and one b

and s^2

I was wrong

oh well

[plucky owl]

Where did that come from?

Wouldn't it be s^2 + bs/M + k/M?

[halcyon zephyr]

Probably came from me doing something wrong. Ill check after dinner :)

[plucky owl]

Send your work if you can

[cedar kiln]

@halcyon zephyr Has your question been resolved?

[halcyon zephyr]

working on it :)

[plucky owl]

What is the answer?

[halcyon zephyr]

it didnt tell me

but we can retry

[plucky owl]

I can help in a bit, I need to do something

[wraith dagger]

♡ Mustard ♡

[halcyon zephyr]

.close

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[crystal raptor]

You might wanna try #foundations

[cedar kiln]

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[dusk remnant]

hiya, I'm currently working on part B but I don't really understand why the answer for x from cosine has 2 values, 39.2 and 140.8. I got 39.2 from cos^-1(1/5) but I can't find 140.8. Thank you so much.

[dusk remnant]

this is the graph i drew

[cedar kiln]

@dusk remnant Has your question been resolved?

[vernal shell]

Hello, try drawing the angle cos^-1(1/5) in the unit circle

That angle is 2x, also remember that 0≤2x≤360 according to the statement. Is there another angle in the unit circle that has the same cosine?

[cedar kiln]

@dusk remnant Has your question been resolved?

[cedar kiln]

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[crimson sedge]

i need to find the standard deviation in a dataset of 17, 22, 22, 30 (answer rounded 2 decimal places)

[crimson sedge]

<@&286206848099549185>

[dawn junco]

hey

[crimson sedge]

hey

i need help with the question above

[dawn junco]

so yeah what's up with this question ?

you did anything already or no?

[crimson sedge]

i need to find the standard deviation

[dawn junco]

yeah ok

you know what it is or no ?

[crimson sedge]

no

[dawn junco]

alright

you know what the mean/average is ?

we'll use that to get the standard deviation

[crimson sedge]

the mean is 22.75

[dawn junco]

,calc (17 + 22 + 22 + 30)/4

[wraith dagger]

Result:

22.75

[dawn junco]

okey

do you know the sum/sigma notation or no ?

if not I won't bother with it

[crimson sedge]

no i dont know

[dawn junco]

okey

so the idea of standard deviation is that it's number that measures deviation from the average

it takes a few steps to compute

[crimson sedge]

i got 5.75 and -7.25

[dawn junco]

for ?

[crimson sedge]

22.75 minus the extremes

[dawn junco]

could you compute value - 22.75 for each value of the dataset, not just the extremes ?

[crimson sedge]

-5.75, -0.75, -0.75, 7.25

[dawn junco]

okey

now could you square each one of them, then take the average of the values you get ?

(the square helps us count the positive and the negative deviations equally)

[crimson sedge]

for the squares that are negatives i got error

im using a TI-84 Plus CE

[dawn junco]

nah not square root

square as in (-5.75)^2

not √-5.75

[crimson sedge]

the squares are -33.0625, -0.5625, -0.5625 and 52.5625

[dawn junco]

you're sure for the negatives

you have to square the whole thing

[crimson sedge]

i'm putting what the calculator told me

[dawn junco]

you should only have positive numbers

that's the whole point of squaring here

[crimson sedge]

is it the same answer as before just without the negative sign?

[dawn junco]

yeah pretty much

[crimson sedge]

33.0625, 0.5625, 0.5625 and 52.5625

[dawn junco]

ok yeah

now take the average of those

we're averaging all the deviations in the dataset now

[crimson sedge]

21.6875

[dawn junco]

ok

now that's a number called the variance of the dataset

take the square root of this, and you'll have the standard deviation

[crimson sedge]

i got 4.656984003 abbreviated to 4.66

[dawn junco]

ah crap they wanted the sample standard deviation

[dawn junco]

sum all those and divide by 3 instead of 4 then

and then square root again

[crimson sedge]

is it because there's a mode that i divide only by 3?

[dawn junco]

nah it's because we're working with a small sample, this formula works better

(but it's quite theoretical to know where the n-1 comes from instead of n)

[crimson sedge]

i got 5.38

[dawn junco]

aight did it work ?

[crimson sedge]

yeah

i have the same type of question with the dataset 10, 10, 10, 12, 12, 12

[dawn junco]

I'll let you try first

and then we can check

you can reread the convo no problem

[crimson sedge]

mean: 11

-1, -1, -1, 1, 1, 1

[dawn junco]

yeah you typo'd there

the last three should be 1 1 1

[crimson sedge]

for some reason i read it as 22

[dawn junco]

aight

[crimson sedge]

1, 1, 1, 1, 1, 1

average of that is obv 1

[dawn junco]

yeah but remember, we want to divide by 6-1 = 5, not 6

I just wasn't sure which formula you used at the beginning

[crimson sedge]

1.2

is the square root of that my final answer

[dawn junco]

yes

[crimson sedge]

i have 1 more which is a table

the mean is given in this one

[dawn junco]

yeah you're just doing each step individually now

well we already computed deviations from the mean twice in this convo

I guess you can fill it out yourself

[crimson sedge]

is it each data item minus 108?

[dawn junco]

yeah

[crimson sedge]

.close

[cedar kiln]

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[formal pasture]

,rotate 270

[formal pasture]

i dont get the third and forth bullet points

for #5

like what r they saying

[dusky peak]

f(x)<=0 when x>3 means:
f is negative or 0 when x is greater than 3

[formal pasture]

thats for first bullet point

yea i understand how to read then

but

like

[dusky peak]

which means that visually speaking, f(x) will be in the gray area here for x>3

[formal pasture]

i duno how u apply

[formal pasture]

yea

but like thise two bullet points

dont restrict

anything?

like

i dont get it

[dusky peak]

they do restrict it

bullet 1: gives us the blue points
bullet 2: gives us the green point
bullet 3: gives us the orange region
bullet 4: gives us the purple region

so

because of the orange region, f(-oo)>=0

and since f will be a polynomial, we will get f(-oo)=oo

and because of the purple region, f(oo)<=0
and since f will be a polynomial, we will get f(oo)=-oo

does that make sense?

and since you just have to draw it, it should be doable

we can actually also calculate the easiest possible polynomial that matches the problem

if you are interested in that

[cedar kiln]

@formal pasture Has your question been resolved?

[formal pasture]

can we try

[dusky peak]

ok

so

we search a polynomial

[formal pasture]

ouhh

i think

i got it

[dusky peak]

let's first talk about the degree of the polynomial

[formal pasture]

its rly bad

but line

like

smthn like this?

[dusky peak]

yes, good 👍

[formal pasture]

third

degree

[dusky peak]

correct

good

[formal pasture]

how can u explain

the degree tho

like lets say

the teacher asks

how u got it

how can i show it

do i say like three roots

no

[dusky peak]

we expect f(-oo)=oo and f(oo)=-oo

if we look at a polynomial in the infinities, then only the term with the highest power matters
so in our case if we have a degree of n, then f(oo)=a*x^n

if n is even, then f will be positive infinity on both ends

which is not the case here

[formal pasture]

its odd

[dusky peak]

so we must have an odd power

correct

[formal pasture]

put odd can be

3

5

7

[dusky peak]

yes

[formal pasture]

etc

[dusky peak]

then we notice that we have at least 2 roots

[formal pasture]

yes

[dusky peak]

there is a theorem that states that a polynomial of degree n can have at most n roots

[formal pasture]

yes

[dusky peak]

so our polynomial with at least 2 roots must be at least of degree 2

[formal pasture]

and n-1 turning points

[dusky peak]

but since it has to be odd, it must be at least 3

[formal pasture]

yesokk

[dusky peak]

then we just look at the easiest possibility

[formal pasture]

okaoaky

do i say its third degree

or

just odd

[dusky peak]

we say the degree must be at least 3

what we now try to argue is that a polynomial of degree 3 can actually be valid here

that being said, polynomials of degree 5, 7, etc. will also work

[formal pasture]

yea

[dusky peak]

but they make it more difficult

[formal pasture]

but

^5

can look

like that

too

[dusky peak]

yes

[formal pasture]

and odd

[dusky peak]

x^5+... looks like that

but we could use -x^5+...

[formal pasture]

yea

[dusky peak]

the thing is that we only need 1 turning point here

[formal pasture]

ouesyes

[dusky peak]

you can see that in the graph you sketched

[formal pasture]

yea

[dusky peak]

wait

turning points

[formal pasture]

is n-1

[dusky peak]

are those where the derivative changes right?

or was it where the second derivative changes

[formal pasture]

yea form increasing

to decreasing

[dusky peak]

ahhh ok

[formal pasture]

and vise versa

[dusky peak]

i confused them

ok

then we need 2 of those

[formal pasture]

yup

[formal pasture]

yea i think i got it

[formal pasture]

this was ely helpful

[dusky peak]

and a polynomial of degree 3 satisfies that

now for the actual equation:

we use the fundamental theorem of algebra

we assume that we do have 3 roots

[formal pasture]

yes

[dusky peak]

this is similar to the graph we have

but on the left we have 2 roots

so we move them closer to arrive at the graph you had sketched

what does this mean?

it means that the root on the left is a double root

[formal pasture]

no

the left

is single

it goes theough the axis

it skims through it

double bounces off axis

in my sketch

its double

[dusky peak]

yes

[formal pasture]

yes

[formal pasture]

mb

[dusky peak]

like this

[formal pasture]

i thiught we r talking abt urs

[dusky peak]

in this sketch (like in yours) the graph bounces

that is what we call a double root

[formal pasture]

yess

[dusky peak]

that root on the left is x=-1

the root on the right is x=3

assuming that a,b,c are our roots, then:

f(x)=d*(x-a)(x-b)(x-c)

that is a theorem we can use

which i can't prove for you here

[formal pasture]

yea

[dusky peak]

if you are interested in why that works, look up fundamental theorem of linear algebra

we have the roots x=-1, x=-1, x=3

so we get:

f(x)=d*(x+1)^2(x-3)

[formal pasture]

eyew

yes

[dusky peak]

now we can use the other points to get d

f(-3)=d(-2)^2(-6)=-24a=16=>a=-16/24=-2/3

with the other point:
f(0)=d(1)^2(-3)=-3d=2=>d=-2/3

one of those would have been enough

but good that t hey match

[formal pasture]

yes

[dusky peak]

so now we can put it together

f(x)=-2/3 (x+1)^2(x-3)

and if we visualize it, it actually works

🪄

[formal pasture]

yessss

[dusky peak]

if we were to solve for degree of 5 or more, we would have more unknowns which we could not solve for

we would instead get a set of solutions

[formal pasture]

exactly

tyyy

[dusky peak]

you're welcome^^

have a good day

[formal pasture]

u too tysm :DD

.close

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[barren karma]

help

[dire geode]

,tex .exp rules

[wraith dagger]

riemann

[barren karma]

so it would be (p/r)^c?

[dire geode]

right

[barren karma]

thank you so much

appriciate it

.close

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[mossy pivot]

could someone explain this to me

[cedar kiln]

@mossy pivot Has your question been resolved?

[wraith dagger]

Mathematics (268882317391429632)

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[twin heron]

,wr give me unit circle

,wolfram x^2

[wraith dagger]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[twin heron]

@mossy pivot

Sin is positive, Cos is negative
sin (y) cos (x)

It's in Quadrant II (where degrees 150, 135, 120 are located)

[mossy pivot]

got it

and cos is negative in that quadrant

[twin heron]

Yes. I'd suggest remembering the unit circle (not hard memory, figure it out): I'll give you my method of figuring it out in a moment here.

[mossy pivot]

i think i kinda got it down its kinda the same but flips in quadrant 3 and 4

[twin heron]

Okey dokes! Remember, in Quadrant 4 the numbers are 5, 7, 11

goes out of the usual pattern.

[mossy pivot]

yeah

i have a question

[twin heron]

sure

[mossy pivot]

how do i turn the degrees into radians do i multiply by 180/pi

[twin heron]

Degrees > Radian Degree * (pi/180)

Radian > Degree Radian * (180/pi)

[mossy pivot]

ohh okay

also when it asks cos x = how do i get the 63/65

[twin heron]

putting this here for reference

It says sin^2x

[twin heron]

16/65^2 = 256/4225

[mossy pivot]

do i replace 16/65 in the equation?

[twin heron]

thenn you subtract the value from both sides adding 1. Square root the value and then you get your answer.

[twin heron]

Into Sinx, yes.

But it shows that you must square its value

Look at this:

(uploading work here, one second)

[mossy pivot]

like if i were to write it out is this wrong

[twin heron]

sin^2(x) can also be stated as [sin(x)]^2, thus sin(x) = 16/65 which (16/65)^2

[mossy pivot]

write it like this instead?

[twin heron]

No. Sin(x) = 16/65 in other words, sin(x) is 16/65

[mossy pivot]

got it

and do i do 16^2 and 65^2

[twin heron]

Yup. Square the numerator and denominator.

[mossy pivot]

256/4225

uh does this seem right

[twin heron]

Yup.

Just don't forget the x in cos^2

[mossy pivot]

how do i get rid of the 1

[twin heron]

1-(255/4225) = 4225/4225 - (255/4225)

when you add or subtract, you must get like bases so multiple the denominator and numerator of 1 by 4225

[mossy pivot]

im a lil confused now

[twin heron]

how come?

[mossy pivot]

how did u get 4225/4225

[twin heron]

oh ok. Look at this:

same thing for subtracting

We want like bases, in other words, same denomintaor.

[mossy pivot]

damn my ass cant add fractions lol

[twin heron]

Don't worry about it.

What math level is this?

[mossy pivot]

college 💀

Precalculus Algebra and Trigonometry

[twin heron]

Ok. Understood. You may not have done it in awhile. Don't worry about it.

[mossy pivot]

so i multiply 4225 by 4225

cuz im doing that to the 1

?

[twin heron]

1 is 1/1 so you multiply both the denominator and the numerator by 4225

you want to get rid of it, so when it's 4225/4225 you do not simplify it to 1, just subtracting 256/4225

[mossy pivot]

so

does this look right...

[twin heron]

1/1 * 4225/4225 = (4225/4225) - (256/4225)

I'll write it in latex so it's easier to read if I could remember how to work it

$\frac{1}{1} * \frac{4225}{4225} = \frac{4225}{4225} - \frac{256}{4225}$

there we go

[wraith dagger]

Tank_Driver011

[mossy pivot]

oh ok

🥲

and then 4225-256 sqrt that get 63

its sum decimal but

nearest whole number ig

and sqrt 4225

65

and cuz cos is negative in quadrant 2

thats why its -63/65

[twin heron]

Yup!

[mossy pivot]

alright thanks i have something else im struggling on but i wanna watch the video first

[twin heron]

Great! I have to leave, as it's getting late on my end but if you need more help, feel free to ping helper

[mossy pivot]

ty

.close

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[cedar kiln]

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[vast drum]

Whats the power set for A = { {}, {a,b}}?
Is it P(A) = { {}, {{a}}, {{b}}, {{a,b}}}

[dire thorn]

You only want the subsets of A. {a} is not in A, so {{a}} should not appear in P(A).

might help to substitute the current elements of A with some letter, then take the powerset, and substitue back

[cedar kiln]

@vast drum Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Prove that $||u+v||^2 = (u+v) \cdot (u+v)$

[wraith dagger]

Derivative

[crimson sedge]

no idea how to do this

like i could say that it equals to $||u+v|| \cdot ||u+v||$

but how do i interchange magnitude and direction

to assume that it is the dot product, it needs to be same direction

[wraith dagger]

Derivative

[crimson sedge]

for me to remove to make u + v a vector and not a norm, I need to make sure that the norm is in the same direction as the vector right

i know that the norm or length of v is $\sqrt{v_1^2 + v_2^2 + ....}$

[wraith dagger]

Derivative

[crimson sedge]

depending on R

so the norm of u+v vector is

$\sqrt{(u_1 + v_1)^2 + (u_2 + v_2)^2 + .....}$

[wraith dagger]

Derivative

[crimson sedge]

but now that is squared

so it becomes $(u_1 + v_1)^2 + (u_2 + v_2)^2 + .....$

[wraith dagger]

Derivative

[crimson sedge]

now what is that

thats where i am stuck

<@&286206848099549185>

[shut river]

Try expanding the right side using the properties of the dot product

Or you can try expanding these squared terms you've generated from the left side

[crimson sedge]

ok so here is what i see

$u_1^2 + 2u_1v_1 + v_1^2 + u_2^2 + 2u_2v_2 + v_2^2 + ....$

[wraith dagger]

Derivative

[crimson sedge]

now if i combine the u_1 ^2 together and v_1^2 together

i get

$||u||^2 + ||v|||^2 + 2(u\cdot v)$

[wraith dagger]

Derivative

[shut river]

There, almost done

[crimson sedge]

ok. but now what do i do

$||u||^2 + 2(u\cdot v) + ||v||^2$

[wraith dagger]

Derivative

[crimson sedge]

seems like its factorable

[shut river]

yep

[crimson sedge]

hmm

like i know it factors to $(u+v) \cdot (u+v)$

[wraith dagger]

Derivative

[crimson sedge]

but i don't know why

[shut river]

The dot product distributes

so $(u+v \cdot (u+v) = u \cdot (u+v) + v\cdot (u+v)$

[wraith dagger]

Xenophon

[shut river]

and so on

[crimson sedge]

but there you get

$u \cdot u + u \cdot v + v\cdot u + v\cdot v$

[wraith dagger]

Derivative

[crimson sedge]

ahhhhhh

[shut river]

Yeah

dot product is associative

and mag squared is dot product of a vector with itself

[crimson sedge]

but $u\cdot u= ||u||^2$ !!!!!

[wraith dagger]

Derivative

[crimson sedge]

ahhhhh

now i understand

[shut river]

great

[crimson sedge]

I just saw it in my textbook haha

that $v \cdot v = ||v||^2$

[wraith dagger]

Derivative

[crimson sedge]

well thank you very, very much @shut river for your help

[shut river]

np

don't forget to close the channel if you're done

[crimson sedge]

ok i will thanks

.close

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[heady granite]

Could someone help me graph this with the requirements - I know I messed up somewhere but idk where

[cedar kiln]

@heady granite Has your question been resolved?

[heady granite]

<@&286206848099549185>

[patent badger]

Is it an odd function? It doesnt go through the origin

[patent badger]

Odd functions have rotational symmetry around the origin, not reflective symmetry across y=x.

This is an odd function for example

[cedar kiln]

@heady granite Has your question been resolved?

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[fathom raft]

I'm a bit lost on the step where it says (product rule), what acctually happens here?

why can the term $3x2ex**3y be ignored?

[hollow trail]

try going from that step to the previous step

by differentiating

[fathom raft]

okey I will do

ah and these first order ODE's are always like a product of eachother?

.close

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[wise hedge]

Hi need some help with some discrete math: Prove the following statement:
Let A, B, C and D be sets. If B ̸= ∅ and C ⊆ B and A × B ⊆ C × D, then A ⊆ D.

[wise hedge]

Currently have something like this: suppose B ̸= ∅ and C ⊆ B and A × B ⊆ C × D, then A ⊆ D.
firstly, since B ̸= ∅, there exists an arbitrary element b such that b ∈ B. Let a be an arbitrary element such that a ∈ A. Since A × B ⊆ C × D, this must mean that (a,b) ∈ CxD. If (a,b) ∈ CxD, there must exist some arbitrary element c such that c ∈ C and an arbitrary element d such that d ∈ D where (a,b)=(c,d) because (a,b) ∈ CxD. This implies that a=c and b=d.

[upper abyss]

You've gone off the deep end. There exists some element c and d, such that a = c and b = d,

but if c and d are arbitrary, then it's not necessary that a = c, b = d

Or, okay I think that's what you're saying

[wise hedge]

My brain has been getting cooked I need help walking through the problem if that makes sense 😅

[upper abyss]

Don't use the word "arbitrary" since c and d are actually specific

The big find is to notice that A is a subset of C, and B is a subset of D

Both are implied from A×B being a subset of C×D

[wise hedge]

due to the cartesean product right

where (a,b) ∈ CxD and a and b are arbitrary

[upper abyss]

So you have that
A ⊆ C ⊆ B ⊆ D

Or, A ⊆ D

[wise hedge]

How do I get to this assumption

[upper abyss]

The Cartesian product one?

[wise hedge]

this:

A ⊆ C ⊆ B ⊆ D
Or, A ⊆ D

Ah wait

[upper abyss]

⊆ is transitive

Which, bonus points if you mention that

[wise hedge]

C ⊆ B is given...

I just want to make sure that a ⊆ C and b ⊆ D because of the (a,b) ∈ CxD

[upper abyss]

Let a ∈ A, b ∈ B. So (a,b) ∈ A×B.

That implies (a,b) ∈ C×D, or that a ∈ C, b ∈ D.

In other words, A ⊆ C, B ⊆ D.

[wise hedge]

And because we're given C ⊆ B, we get this: A ⊆ C ⊆ B ⊆ D which is A ⊆ D

Correct?

I'm assuming that is the case

<@&286206848099549185>

@upper abyss Just want to confirm, will close after 😛

[upper abyss]

yes

[wise hedge]

appreciate it ❤️

.close

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[rotund wind]

Hello, I have a simple problem. It states "Tony has 830 bottles. A small bottle is worth 10 cents and a bigger one is worth 40. If Tony got $159.50, how many of small and bigger bottles did he have?" I were able to solve it using two equations:

x + y = 830

0,1x + 0,4y = 195,50

However the question implies that it should be done in one equation and we haven't necessarily been teached ways with multiple equations. I can't think of a one equation answer. Does it exist? The answer was 575 small and 255 big.

[fair geyser]

you can skip straight to one variable

small bottles is x, and you use (830−x) for large bottles

[rotund wind]

Sorry, how could I combine the two equations with this info?

[fair geyser]

i don't understand what you mean

[rotund wind]

The question was that is it possible to solve this using only one equation instead of two

[fair geyser]

the first equation doesn't appear, you only have the second equation, without y

it's exactly like you just don't show any initial steps

and write the combined equation from the start

but that's what they must have meant

[rotund wind]

I still don't get how I could combine them

[fair geyser]

okay ping helpers

[rotund wind]

I don't know k ow if this is old enough

[fair geyser]

i guess not

[rotund wind]

Ig this is not that critical, I don't even have ro solve this, just wondering if I could with one equation

[fair geyser]

if you solved it, you must have had one equation at some point

[rotund wind]

Wdym

[fair geyser]

i don;t understand how you solved it otherwise could you show it?

[rotund wind]

It's in the post, using two equations

[rotund wind]

.

[fair geyser]

there's no solution

[rotund wind]

In the end there is

[fair geyser]

what

[rotund wind]

Look at the very end

[fair geyser]

that's the answer

[rotund wind]

Yes

[fair geyser]

you;re saying you gave the equations to the computer or something?

[rotund wind]

No

I solved it using them two

[fair geyser]

so at some point you had one equation, that's how it works as far as i know

[rotund wind]

Uhh

Lemme check if I have the whole workaround somewhwere

[fair geyser]

it would be natural to straight up write that equation as the first step, then you have one equation

it's like, the first equation only appears because you made 2 variables, if you start with only x, that stands for number of small bottles, or number of large bottles, you still can get an equation

[cedar kiln]

@rotund wind Has your question been resolved?

[cedar kiln]

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[vernal kite]

For Question 2 B, I am unsure if I overcomplicated it or if i’m doing it correctly

[untold torrent]

You got it right

[vernal kite]

so what do i do about the plus minus

do i form two equations and solve both?

[untold torrent]

They just asked you to complete the square, right?

[vernal kite]

yep

[untold torrent]

Are you supposed to solve for the roots?

Like, 3 - x^2 - 8x = 0?

[vernal kite]

i dont think so

[untold torrent]

Seems like that's what you did, and you got it correct. So what's the problem?

[vernal kite]

but what would be the maximum/minimum value then

wouldn't you need to solve for y

[untold torrent]

Oh I see what they're asking

y = 3 - x^2 - 8x

Complete the square from here

Don't plug in y = 0

[vernal kite]

if i factor out the negative sign, does y also become negative

so it becomes x^2 + 8x -3 = -y

[untold torrent]

Don't move the negative sign

Just factor it

[vernal kite]

alrlr

[untold torrent]

y = -(x^2 + 8x - 3)

You want to get y to be of the form:

a(x - h)^2 + k

[vernal kite]

what did i do wrong

[cedar kiln]

@vernal kite Has your question been resolved?

[stray drift]

@vernal kite

next time distribute the -

[vernal kite]

so how would you find the maximum/minimum vaule of that

[stray drift]

max min value of x?

[vernal kite]

y i think

is the working out above the max min value for x?

[stray drift]

none

you found the function in terms of x

if you want to find max or min you have to find the turning point because it a parabola

[vernal kite]

so that means i have to solve for y then?

[stray drift]

you have to solve for the turning point

do you know how to do that?

[vernal kite]

u find x, sub the x values in and then find y

[stray drift]

what do you mean find x?

find the x value of the turning point?

[vernal kite]

ye

[stray drift]

ok

how do you do that?

[vernal kite]

use the axis of symmetry formula thing

-b/2a

[stray drift]

ok

so do that

[vernal kite]

but isn’t it just question 2 a?

[stray drift]

ok

so you need help with completing the square right?

2b

[vernal kite]

yep

[stray drift]

you had it right up untill you sqrted everything

you dont need to solve for x,

you need to put it in the form a(x-h)^2+k

[vernal kite]

ohhhhh

[stray drift]

so you just had to minus 19 from both sides

[vernal kite]

so once it's in the form, what do i do then

[stray drift]

if its in the form a(x-h)^2+k then the turning point is at (h,k)

so the equation 3(x-2)+5 would have the turning point (2,5)

[vernal kite]

ohhhhh

wait i remember this

it's been a long time since i've learn this so i forgot everything

thank u :)

[stray drift]

np

[vernal kite]

wait but if the form has -h and +k and this has +4 and -19, then what

do i just switch the signs around?

[stray drift]

3(x-2)+5 would have the turning point (2,5)

so flip h and keep k

[vernal kite]

so its -4 and -19?

[stray drift]

yea

[vernal kite]

the answers says it's -4 and 19

max value at 19, x=-4

[stray drift]

but that is wrong

its becayse you let y = zero

[vernal kite]

wait so are u correct or is the answer correct

[stray drift]

the answer is correct

[vernal kite]

okay thank you.

.close

[cedar kiln]

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[dreamy zenith]

how do i convert this to m/sec

[dreamy zenith]

from m^2/sec

[opal basin]

why would you?

[dreamy zenith]

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[tawny drum]

need help on question 1, 2, and 5

[opal basin]

For 1 you can use difference of squares

[cedar kiln]

@tawny drum Has your question been resolved?

[opal basin]

For 2, you can combine the last 2 restrictions to form a relation between the number of red and white balls, then use number theory

and some bounding

[tawny drum]

can you explain what bounding is??

[opal basin]

What is this for btw?

these questions

[tawny drum]

its like monthly questions