help-19

[sand knot]

I dont see any roots

sorry I kinda messed up this quadratic semi-unit so uh I might sound like a crackhead

dont be

surprised

[fallow tapir]

when you solved the two curves

you formed a quadratic equation

and the roots of the quadratic equation (the value of x that satisfies it) are the x coordinates of point of intersection of the curves.

[sand knot]

oh roots mean the value of x that satisfies it?

[fallow tapir]

but here we have a single point (a single root)

[sand knot]

ohhhh

yeah

ok so now we know it has one unique solution

yipeee

what now though-

[fallow tapir]

do you know when a quadratic has both roots equal?

hint : ||discriminant||

[sand knot]

oh uh

the discriminant is

equal to 0

[fallow tapir]

correct

now just find the discriminant of the quadratic and put it = 0

[sand knot]

you lost me there

so I know the discriminant is uh

b^2-4ac

and the quadratic equation we have here is

2x^2-kx+k

[fallow tapir]

yeah and?

[sand knot]

okay I get that but I dont get something else

how do I know 2x^2-kx+k has both the roots equal

im new to the term roots in this way of using it so im slightly unsure

[fallow tapir]

i just explained the whole thing above

give it a read again

[sand knot]

oh is it because

the parabola is tangential so you can only have the same roots?

nvm

I checked my notes

yeah I get what you mean now

holy jesus

yeah

I get it now

thank you

/close

.close

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[fallow tapir]

lol

np

[odd edge]

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[cosmic grail]

Hi, I’m trying to figure out K on my simplifying radicals worksheet but I got 4 * the square root of 11 when the answer is 2 * the square root of 22
What did I do wrong?

[tropic hazel]

sqrt 88 right?

step 1: separate it into factors

[wooden python]

sqrt(8) is not 4, btw. @cosmic grail

[tropic hazel]

sqrt(88)=sqrt(2x2x2x11)

then pair up the numbers that get multiplied more than once

sqrt(2^3 x 11)

then like sqrt(2^2 x2x 11)

remove until power is even

[wooden python]

don't use the letter x for multiplication

[cosmic grail]

Oh my god I’m dumb

Thanks

Frick how did I screw that up

[tropic hazel]

well in this case there's not really variables involved

It's convenient cuz I don't wanna add spaces to *

[tropic hazel]

anyways if ur finished don't forget to close the cannel

channel

[cosmic grail]

.close

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[sand knot]

lmao wth is this

[sand knot]

im lost

[low locust]

the discriminant says something about the roots of f

what does a root of f mean for 2^f(x)

[sand knot]

well

I mean

f(x) is the exponent of 2

[low locust]

yes. what happens if the exponent is 0

[sand knot]

oh shi I forgot

uhhh

wait

yeah

equal to one

no matter the number, if the exponent is 0, it will equal to one

[low locust]

and now compare with the graph

[sand knot]

uhhh

when y is 1 x is at

-1.7 or 0.6 somewhere in that ballpark ig

im a bit confused

[low locust]

what does the discriminant say

[sand knot]

discriminant in math usually takes in the form of b^2-4ac but in this case uh

I cant rlly

find any

it says comment on the value of the discriminant of the function f(x) but idk

wait

f(x) can be negative because

the minimum value of y is below 1

[low locust]

the discriminant says something about the number of roots of f

[sand knot]

two

because f(x) = 0 means y=1 which has two solutions

[low locust]

f has two roots, yes

which means that the discriminant must be...?

[sand knot]

meaning the criminal is a positive number

discriminant*

[low locust]

yes

[sand knot]

ok thx

yay

.close

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[wooden python]

sadly none of this can be salvaged

[earnest dagger]

omg...

i want to kms w all this man

[wooden python]

are you actually suicidal right now?

[earnest dagger]

no haha

okay imma come back w another proof soon

.close

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[surreal sparrow]

This is my question for a quiz can someone help me out? it’s for Differential Equations

[surreal nymph]

what are you confused on @surreal sparrow

[surreal sparrow]

i thought this questions answer was that it is a PDE and a degree of 2 equation but i didn’t get full marks and i don’t understand

i j started this class differential equations

so i just need help with this question

[surreal nymph]

what is a PDE?

[fallow tapir]

partial differential equation

[surreal sparrow]

^

[surreal nymph]

I know what it is, i was asking you

[surreal sparrow]

oh

[surreal nymph]

why do you think its a PDE?

whats the difference between a PDE and an ODE?

[surreal sparrow]

pde depend on several variables and ode is like a derivative w only 1

[surreal nymph]

okay so when a function depends on multiple variables what changes in notation when we write derivatives

[surreal sparrow]

the partial derivative sign thing?

gamma?

whatever it’s called

[surreal nymph]

uhh not gamma

its just a curly d

[surreal sparrow]

yea

that

[surreal nymph]

does that appear here

[surreal sparrow]

i forgot the name of that

no

damn i’m stupid

okay so it’s not a pde

but it’s a degree of 2

right

because of the dt^2

[surreal nymph]

yes

and because?

[signal crown]

I wouldnt call that degree but order

🤔

[surreal sparrow]

i meant the top

d2x

d^2x

[surreal nymph]

yah i guess theres terminology here

its asking secifically for "degree"

did your class cover the difference between the degree and the order

[surreal sparrow]

we’ve only had 1 lecture and then we got a quiz

the first lecture was j over the syllabus of the class

[surreal nymph]

they should have given a list of definitions then

[surreal sparrow]

i’m j trying out the quiz cux it has 6 attempts

but this the the only question i was stuck on

there’s 3

and i’m assuming that this equation is not a first order differential equation?

[odd edge]

@surreal sparrow Has your question been resolved?

[surreal sparrow]

i j need explanation on why or why not this equation is a first order differential equation, and a linear differential equation

[signal crown]

its not a first order differential equation but a second order differential equation

[surreal sparrow]

i see

[signal crown]

for the linearity, you can take x = a*x1 + b*x2 and see for yourself

[surreal sparrow]

aii i got it

thank you

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[mystic saffron]

please dont try to cheat lol

[odd edge]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[mystic saffron]

ok

[odd edge]

Please stick to your channel.

[mystic saffron]

!onechannel

[odd edge]

Please stick to your channel.

[quasi sparrow]

then what do you want?

[tepid yacht]

If you're not asking for help, then what are you doing?

[mystic saffron]

@earnest gyro what grade r u in

ok then

[quasi sparrow]

asking for help on a test?

[mystic saffron]

yea it says test

so it could be cheating idk

[tepid yacht]

I read that as "I'm not asking for help" 🤦‍♂️

[mystic saffron]

💀

[tepid yacht]

But is that a test? Are you supposed to get outside help on it?

It does say test, and hand in to the teacher for a grade

[mystic saffron]

ok then

[tepid yacht]

Yes and cheating on tests is academic dishonesty and is bannable

[odd edge]

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[mystic saffron]

ok so for number one in the first image, you want to find the angle of FCB

oh nvm

[tepid yacht]

Can't ask in this server if you're trying to cheat on a test

[mystic saffron]

Its due today you say??

[tepid yacht]

Asking for help on a test is cheating

[mystic saffron]

@hallow pelican ?? Its due todayyy???

Yeah no worries we will answer tomorrow

Come back in 24 hrs

[tepid yacht]

If you aren't allowed to be using resources, then you using discord is cheating

Then you should refresh on what cheating means

I'll leave it to the mods <@&268886789983436800>

[mystic saffron]

because cheating = bad

cheating is a big no-no

ok i can help

.reopen

[placid lynx]

no this is cheating lmfao

[tepid yacht]

If they aren't supposed to get outside help, I don't suggest this until mods check

[mystic saffron]

for the 3rd image you want to create a coordinate plane with all the stuff that was specified

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[tardy yacht]

could someone explain surds and multiplication/division and adddition/multiplication regarding it?

[mystic saffron]

When the base is same

And you multiply , you add the power

You divide, you subtract the power of denominator from that off the numerator

Of*

[tardy yacht]

what if the base isn't the same?

[viscid flint]

then you can't do much

like $\sqrt2+\sqrt3$ is terminal

[clever fjord]

hayley!

[viscid flint]

oh for multiplying

[tardy yacht]

ohh alr alr

[viscid flint]

if you have $\sqrt2\cdot\sqrt5$ then that's the same as $\sqrt{2\cdot5}$

[clever fjord]

hayley!

[tardy yacht]

what if it was divide?

n there was another number on the other side of the root?

[viscid flint]

dividing works just like multiplying

if you had like

$\f{\sqrt{15}}{2\sqrt3}$

[clever fjord]

hayley!

[viscid flint]

it would be safest to rewrite that as

$\f12\cdot\f{\sqrt{15}}{\sqrt3}$

[clever fjord]

hayley!

[viscid flint]

and then do the surd stuff and then put it back together

[tardy yacht]

ohhh ok

so if it was a 3 instead of 2 u would the same

nvm i think i got it. thank you for the help!

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[honest robin]

Hi, how do I solve the limit x->0 of (a^x-1)/x³ I know I need to split it into a>1 and 0<a<1 and use l'hopitals rule but idk how

[quasi sparrow]

for the numerator, use $\frac{d}{dx} a^x = \log(a) a^x$ for $a > 0$

[clever fjord]

riemann

[honest robin]

so I calculate the derivative for both possibilities which comes down to lna/3 * lim x->0 of a^x/x² but how do I get an answer from this

[quasi sparrow]

Keep differentiating

[honest robin]

but if I fill in 0 it comes down to 1/0 so how

[odd edge]

@honest robin Has your question been resolved?

[quasi sparrow]

...

i said keep differentiating

[honest robin]

oh okay

.close

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[frozen bear]

Is this correct??

[odd edge]

@frozen bear Has your question been resolved?

[frozen bear]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

Maan, why is no one here

<@&286206848099549185>

[odd edge]

@frozen bear Has your question been resolved?

[quasi sparrow]

small font or bad picture, hard to read

[quasi sparrow]

picture of a screen, also hard to read

[frozen bear]

Yo

I actually have another question

Where did I go wrong?

@mystic saffron I used the Pythagoreon theorem

(I sent the problem in case you missed my message)

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[tidal harbor]

If we take the derivative of an antiderivative there is no constant C

[tidal harbor]

But what about the other way around?

If we take the antiderivative of a derivative

Does the C stay?

[icy kindle]

yes, when you find the antiderivative you should add a constant C

[tidal harbor]

Ok then if we apply that to the steps of integrating factor method of a differential equation

After multiplying the integrating factor, by product rule we are left with the derivative of some term

And we take the antiderivative of that (usually d/dx of some term)

Then we should be left with a C in this process don't we?

And that should be subtracted to the C we get from antiderivative we take on the other side

[ivory raven]

they're all arbitrary constants

if you subtract two arbitrary constants, you just get another constant

so you only write +C once

[tidal harbor]

Hmm..

That only apply to subtraction and addition right

When multiplied we have to specify the difference like C1 and C2?

[ivory raven]

if you're multiplying terms with x in them, yes

[tidal harbor]

Ah thanks

But one more question

e^(x-y^2) / 2y

=dy/dx

We want to separate variables here

Like we can multiply 2y to both sides

But what about the y^2 in e^?

[ivory raven]

use index laws

e^(a+b) = (e^a)(e^b)

[tidal harbor]

So we just multiply

2y/e^-y^2

To both sides?

[ivory raven]

that works

[tidal harbor]

Then we are only left with e^x on the side with dx right

Oh ok

Thank you..

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[restive canopy]

How do I calculate that area?

[odd edge]

@restive canopy Has your question been resolved?

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@restive canopy Has your question been resolved?

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@restive canopy Has your question been resolved?

[next wagon]

There is a couple ways to find that

We already know one side of the triangle you want to find the area, which is 8.

The remaining you can find by using the provided figues

figures

Take a look at

1. Pythagorean theorem
2. Triangle area using sin
3. Law of Sins

With those you can solve it

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[bold saddle]

just to be sure it would go y=-2^2? or y=-2x^2

[soft sage]

$y=(-2)^2$

[clever fjord]

FirstRuns

[soft sage]

You’re plugging the individual values into x

[bold saddle]

how would that be graphable tho its just a straight line

[soft sage]

It’s asking for the graph of $y = x^2$ not $y = (-2)^2$.

[clever fjord]

FirstRuns

[soft sage]

It will be easier if you complete the table because it gives you a bunch of points to plot

[bold saddle]

oh so like when x=-2 of the parent fucntion it would be 4?

[soft sage]

yeah

[bold saddle]

ah I get it now

because I was thinking we were changing the function

[soft sage]

the goal of that homework problem is finding what the function looks like by feeding values into it

you’re creating a table of points of the form (x, f(x))

aka (x,y)

[bold saddle]

wait im confused again

so im not plugging in x into y=x^2

wait nevermind i get it noe

now*

[soft sage]

you are lol

[bold saddle]

yeah

how do you solve |x| again

[soft sage]

$| -x | = | x |$

[clever fjord]

FirstRuns

[soft sage]

I’m not sure what you mean by solve

could mean a lot of things

But in general absolute values are always positive and the graph of | x | is mirrored because the negative and positive x outputs equal eachother in their y values

[bold saddle]

I mean this

thanks

.close

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[mystic saffron]

Find the parametric eqns for the tangent line to the curve x=t^3 y=t^2 z=t at the point (-1, 1, -1)

[mystic saffron]

so the point (-1, 1, -1) occurs at t=-1 (not sure if that helps

and r'(t) = <3t^2, 2t, 1>

and now idk where to go

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[mystic saffron]

okay well nevermind 😦

.close

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[near zealot]

z is in N

sorry I didn't study math in english

[viscid flint]

uh try $\lim_{x\to0} \f{1-\cos x}{x^2}$ first

[clever fjord]

hayley!

[viscid flint]

and then $\lim_{x\to0} \f{1-\cos x\cos2x}{x^2}$

[clever fjord]

hayley!

[near zealot]

$1-ABC=(1-A)BC+(1-B)C+(1-C)$

[clever fjord]

Ditril

[near zealot]

I used that

[viscid flint]

oh that's a neat identity

[near zealot]

i discovered it while trying to solve this limite

I already did but I'm wondering how I can implement sigma/phi

I'll go now but ping me whenever you find it

[viscid flint]

oh i'm going to sleep lol

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[mystic saffron]

It is a little bit physics but mostly about vectors. I was examining and studying orbits but i realised that i learned something wrong. Because according to newton, magnitude of velocities must be constant while their directions are not. So there is an acceleration which is affecting the velocity vectors. I thought we can do simple vector calculations but i realised that we actually can't (at least in my example) because no perpendicular triangle with same lengths for side length and hypotenuse exist. What am i missing can someone explain please thanks

[mystic saffron]

As you can see at the right top of image, vector calculation should look like that but it can not be because V2 has same magnitude with initial velocity and also degree between them must be 90

which is resulting in very unlogical vector calculation

[viscid flint]

ohhh ok i think i understand what you're asking

you're saying from v1 to v2 the particle experienced some kind of acceleration

which changed the direction of its velocity but not its magnitude

[mystic saffron]

yes

[viscid flint]

and there are no right triangles that would make that make sense

[mystic saffron]

exactly

[viscid flint]

yeah this is because the acceleration isn't constant either

the acceleration is always perpendicular to velocity

[mystic saffron]

ohh that make sense

[mystic saffron]

is there no way to result with finale velocity in delta time interval using also changing acceleration vector stuff?

[viscid flint]

no, you need to use calculus

[mystic saffron]

do you mean vector calculus?

[viscid flint]

the idea is that you construct an infinitely small right triangle

[viscid flint]

"vector calculus" usually refers to stuff with multiple independent variables; in this case we just have time so this is the sort of thing you'd be able to do (with some thought!) after a second-semester calculus class

or second-year if high school

[mystic saffron]

Because the acceleration must be perpendicular to the velocity at every infinitesimal interval of time?

[viscid flint]

yep!

and all of those things are constantly changing

[mystic saffron]

So is it possible to find the exact vector operation or calculation which result in vector of finale velocity in specific time interval if we use calculus?

I was wondering if this could be visualized as a vector calculation

[viscid flint]

yep! you'll end up with an equation involving sine and cosine

which makes sense because those are inherently tied to circles

[mystic saffron]

what should I need to follow to do this?

Because I don't even know how to approach this problem, but I am very curious about the solution to the problem.

[viscid flint]

uh hmm
i guess it is vector calculus actually

you'd say that like $\dv{t}\vec{v}(t) = \begin{pmatrix} 0 & 1 \ -1 & 0\end{pmatrix}\vec{v}(t)$ which looks scary but isn't

[clever fjord]

hayley!

[mystic saffron]

What does this matrix do?

[viscid flint]

it's a rotation matrix

[mystic saffron]

why do we need that

[viscid flint]

it takes a 2-vector and rotates it 90º counter clockwise

this is saying "the derivative of velocity is a 90º rotation of the velocity itself"

[mystic saffron]

It sounded me like super abstract stuff. Maybe I should learn more about the mathematical methods you suggest using before trying to solve this. Is what you are talking about called "vector calculus" in the literature or something else?

[viscid flint]

it would be, yeah; i'm trying to find the derivation i'm thinking of

but you definitely want to start with single-variable calculus and understand that first

[mystic saffron]

I have some basics such as fundamental theorem of calculus etc. but I probably can't solve complex calculus problems.

[viscid flint]

here's something https://physics.stackexchange.com/questions/332186/proof-that-a-constant-magnitude-of-velocity-and-a-constant-net-force-implies-tha/332374

[mystic saffron]

Do I need to learn every method in single variable calculus before learning vector calculus?

[viscid flint]

no

[mystic saffron]

thanks for the source i will check that

[viscid flint]

there's a lot of integration tricks that are neat but not necessary at all

[mystic saffron]

ah okay

[viscid flint]

but, you should be very comfortable with the mechanical process of taking a derivative, for instance

because what originally takes 10 minutes when you're starting, will become a tiny subpart of a more complicated problem later

just like you (probably) don't think about times tables anymore

[mystic saffron]

oh got it thanks for your help then it gave me new perspective

[odd edge]

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[quick needle]

Wait

[quick needle]

can you have negative induction

i.e. in the backwards direction

I'm trying to make a claim for all Z

so splitting into ≥0

and then <0

[quartz oracle]

you can claim “for all n, P(-n)”, and then proceed

so you just need to tweak normal induction a bit to do the negative numbers

[quick needle]

then what would the induction step look like?

P(-n-1)?

or P(-n+1)

[quartz oracle]

let Q(n) = P(-n)

then you would prove that Q(n) implies Q(n+1)

like normal induction

[quick needle]

so this?

[quartz oracle]

no, -n-1, because -n+1 is the one larger than it

you’re trying to go to negative infinity

[quartz oracle]

but to be safe, do this

then you’re just doing normal induction

[quick needle]

I am confused as how this would be applied in context lol

can you give me an example?

[quartz oracle]

Ok

[quartz oracle]

Do you have a proposition you want to prove for negative numbers

let’s use that

[quartz oracle]

or I guess I’ll just make one myself

“prove that the sum of all the negative numbers before 0, that is -1 + (-2) + …, is -n(n+1)/2”

then the proof would go like this

Let Q(n) be the statement that “(-1) + … + (-n) = -n(n+1)/2”

@quick needle and then you would do normal induction on n

[odd edge]

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[zenith jasper]

can someone please tell me what went wrong?

[zenith jasper]

,rotate

[zenith jasper]

the answer is this

,rotate

[zenith jasper]

thanks

[quartz oracle]

you didn’t apply the product rule correctly

check it again

[zenith jasper]

should be cos^nx - (n-1)cos^(n-2)xsin^2x?

oh

im not sure where went wrong except for the missing ^n in the cosx

[fallow tapir]

write sin^2 as (1-cos^2) and simplify.

[zenith jasper]

wait where is sin^2

oh

@fallow tapir do i multiply the (1-cos^2x) into the cos^(n-2)x?

[fallow tapir]

multiply everything

lol

[zenith jasper]

oh

alright i got it

thank you

.close

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[stable bobcat]

hi so i was asked in a trial exam for school relating to this function, but it revolved primarily around it and its inverse. i was wondering how i would approach finding the inverse for this as my t-inspire wasnt giving me any answers. i did take the domain into consideration.

[north crow]

can you inverse cubic functions?

[stable bobcat]

id assume for a certain domain it can? is that wrong

[north crow]

what do you mean?

[outer ivy]

,w plot x^3

[clever fjord]

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[outer ivy]

This function has exactly one y value for every x, so the inverse exist over R completely

Being x^(1/3)

[stable bobcat]

right, the vertical line test

for my function, it has an inverse from 0 to 4/3

but idk how to get it

[outer ivy]

We can maybe plot it to get an idea

,w plot sqrt(x)*(4-x)

[clever fjord]

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[outer ivy]

Here we have a couple of double y values, so that's why the domain of the inverse is a small part

[stable bobcat]

stationary point is at 4/3 so i assume past there is where it cant have an inverse

[tall veldt]

you want the horizontal line test to see where the function can have an inverse

you want no two x values to map to the same y value, which in this case is x > 4

[stable bobcat]

oops

wait x >4

?

ah

but how do we find the inverse function

swapping y and x variables didnt work for my calculator

[tall veldt]

do it by hand then

[stable bobcat]

i did do it by hand

which i got the root of a cubic

and

[tall veldt]

yeah this function doesn't really have a very nice inverse

[stable bobcat]

i cant tell if

it should be reflectable in y=x no?

[tall veldt]

and the blue line is

[stable bobcat]

the original function

it overlaps the black function

[tall veldt]

i meant

the blue line can be reflected

and give a function

[stable bobcat]

but for 0 to 4/3 it cant be?

[tall veldt]

if you reflect it, the resulting line wouldnt be a function (vertical line test)

[stable bobcat]

so itd be a relation

im assuming thats where the y^2 comes in

are these inverses?

[tall veldt]

are what inverses?

[stable bobcat]

the black graph i found

it was what i got by hand

its just a reflection right?

[tall veldt]

it is not a relfection of the blue line in the line y=x

[stable bobcat]

is the only way to find the inverse thru swapping the variables

[tall veldt]

pretty much

[stable bobcat]

so is the inverse for this not able to be found

[odd edge]

@stable bobcat Has your question been resolved?

[odd edge]

@stable bobcat Has your question been resolved?

[stable bobcat]

<@&286206848099549185>

[quasi sparrow]

You answered this already

[odd edge]

@stable bobcat Has your question been resolved?

[odd edge]

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[gusty blaze]

How do you take the derivative of 9sin4x

[gusty blaze]

i know the derivative of sin(x) is cos(x)

[fallow tapir]

chain rule.

[gusty blaze]

how do you know to use that

[fallow tapir]

its a composition of functions.

sin(t)

and 4x

[clever fjord]

Dyssrupt

[fallow tapir]

second last ^

[gusty blaze]

damn i was unable to recognize it was a composition of functions

so it would be 9cos(4x) * 4 = 36cos(4x) ?

or does the 9 go away when we take the derivative

to give 4cos(4x)

[fallow tapir]

correct

[fallow tapir]

no, its a constant in multiplication

not addition or subtraction

[gusty blaze]

ohh ok, thank you!

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[nova cosmos]

isnt this just 1/32^0.8? im honestly confused can someone explain

[mystic saffron]

i mean yes, but it isn't really the simplest form you can get

well you are forgetting a minus for the 32

how would u express 0.8 as a fraction

[nova cosmos]

1/-0.8

[odd edge]

@nova cosmos Has your question been resolved?

[ocean bramble]

how would you express -0.8 as a fraction using only integers

also, 1/(-0.8) is -1.25, not -0.8

[sand portal]

1\-32^0.8

i think so

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[silent jetty]

Can someone help me with part b? I tried using mathway and geogebra but both cannot solve this.

[uneven lodge]

,w integral of sqrt(1 + 36874x^4) from x = 0 to x = 3/4

[clever fjord]

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Stephen

[royal herald]

Some algorithm they used ig?

[uneven lodge]

No idea lol

[silent jetty]

Awesome, thank you!

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[cursive jackal]

Any short method?

[odd edge]

@cursive jackal Has your question been resolved?

[void yew]

Lhopitals rule

Take ln of that

And find the limit of that with l'hôpitals

oh wait hm

[serene mango]

Does this limit exist?

[void yew]

Yeah I don't think it should

[serene mango]

I am pretty sure it doesn't exist not only because of oscillation but also because it has roots of negative numbers

[cursive jackal]

Infinity ^0

Can you giys explain more how to do it i didn't understand even similiar solution available on youtube indian sites@serene mango @void yew

[void yew]

but it oscillates

For the limit as x goes to infinity to tend to infinity

For all N, there has to be an M such that for aaaaaaaall x such that x> M, f(x) >N

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[void jacinth]

is this a glitch or am i just stupid

[void jacinth]

oops

didn't mean to reply

¯_(ツ)_/¯

[neat robin]

They asked you to graph the intervals where the polynomial is positive

You graphed the interval where the polynomial is negative!

[odd edge]

@void jacinth Has your question been resolved?

[void jacinth]

it says that's the answer

i forgot i was in here ngl

i was waiting for somebody to tag me and say 'this and that'

maybe i shoulnd

shouldn't have gone here lol

but whatever

should've donea. whole screenshot then

i'm past that anyway

.close

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[void jacinth]

i put it where it was positive anyway

and it said iw as wrong and that was right

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[brisk aspen]

Hello

[brisk aspen]

How to demonstrate this assuming that

.tex a,c \in \mathbb{C}

[sharp oak]

What are a and b?

Something's not quite right here

[odd edge]

@brisk aspen Has your question been resolved?

[brisk aspen]

unknown values

that may be imaingary

so their conjugation should be done individually

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[fathom aurora]

i have a quick question: how did they get from the first equation to the bottom one by simplifying

[fathom aurora]

my mind is not understanding

[fathom aurora]

wow thats way simpler than i thought

thank you so much

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[thorn shore]

Hello I need help with this

Idk how else you can do this unless you graph

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[mystic saffron]

In how many different ways can five couples of husband and wife arrange themselves around a bonfire, if the men and women alternate?

[mystic saffron]

Numbers = Women
Letters = Men

If we take them as fixed groups, we can calculate how many** Group configurations** are for a certain fixed pairs
Which is (5-1)! = 24
Then if we take one group configuration, we can turn 🔁 5 times, so there are 5 arrangements for each group configuration.

So we have 24 x 5 = 120

[static totem]

turn what

like turn the women while keeping men fixed?

it's too correlated you don't get the other 2760 configurations

if 1 and 3 are in these spots, then you had C immediately after A, you can't turn it into A after C

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

the women or the men

[mystic saffron]

yes

[mystic saffron]

but I am multiplying the 5 possible ways for each group configuration (24)

[static totem]

and that counts 120 configurations

and you would never get something on my picture

if this is fine, and you just want to count 120, then you did everything right

[mystic saffron]

yes, this is why there are 5 ways for each group configuration and not 10

[mystic saffron]

yes, because there are 5 ways for each group configuration, and 24 group configurations (5-1)!

[mystic saffron]

I would, because the group configurations (24) would group them in different ways including the one of your picture when you turn them 🔄

[static totem]

no, i'm showing why it can;t happen

[mystic saffron]

it includes your picture

[mystic saffron]

I'll do it like that w8

[mystic saffron]

You are right, that couldn't happen, because the order would be fixed for either women or men

But how can you work this out using this strategy

[static totem]

what do you mean

if you do something else, then it's not this strategy anymore

[mystic saffron]

mm how can you represent it

[static totem]

5! × 4! would solve it

[mystic saffron]

can you please explain it

[static totem]

or i mean 4! × 5!

[mystic saffron]

I know that (5-1)! is bc of circular permutation

[static totem]

you start the same way, and then instead of rotating you mix the women, or the men freely

[mystic saffron]

and turning is the same

mm why there is only one 5!

(I am not saying you are wrong)

just questioning

like why not 5! x 5! or 5! x 2

[static totem]

i'll think

[mystic saffron]

mm okay

[static totem]

it's like, suppose you arrange them in rows first
5! for men, then 5! for women, then you have 2 ways to merge that so it alternates, either wmwmwm ot mwmwmwmw

then you divide by 10

so 5! × 5! × 2 / 10 = 2880

ok i can't explain it

[mystic saffron]

mm would it be divided by 10?

[static totem]

why not, there's 10 people

it's just circular arrangement

i seriously give up, good luck

[mystic saffron]

don't worry, I'll ask again, thanks a lot

👍

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[odd edge]

Please don't occupy multiple help channels.

[mystic saffron]

i do that right?

[mystic saffron]

i feel as if

the absolute value may change something

like p can equal -12?

because

-12+4 = -8

then abs value

is 8

so it may be -12,2

as the number line interval;

perhaps

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

i think answer for question two is wrong

it must be 4>= p >= -12

p can be -12, -5, 3... and many other numbers but they are not included in your answer

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[tidal harbor]

We are given this graph, and we have to solve the domain of h(2x+2)-1
I can think of finding out the equation of h(x) based on the graph and doing a transformation on that graph to get the domain
But that seems maybe too complicated
Is there a way to figure out the domain purely from this graph?
I forgot transformation of graphs a long time ago and according to what I found on the internet something like h(x) -> h(2x) is just a horizontal shrink so there shouldn't be a domain change
And h(2x) -> h(2x+2) we are doing a horizontal shift to the left direction 1 unit, which means the asymptote at x=4 should move to x=3
Then the -1 at h(2x+2)-1 should indicate a vertical transformation downwards so shouldn't affect the domain
That's why I'm thinking the domain might be -inf<x<inf, x=/=3
But is this wrong?
This is just my thought process so would appreciate if anyone could correct this

[quartz oracle]

notice that the domain is everywhere but x = 4

so the domain of h(2x+2) - 1 is everywhere but 2x+2 = 4

[tidal harbor]

But how do we account for the -1 at the end

[quartz oracle]

It doesn’t do anything to the domain, it just shifts the graph downwards by 1

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[mint coral]

Find the slope of the tangent to the curve y= 1/sqrtx at the point where x=a

[mint coral]

cant use derivitives so i need to show my work using lim h-> 0 f(x+h) +f(x) /h

[fallen remnant]

It's f(x+h)-f(x)/h

[mint coral]

ya sry, so would you multiply by the conjugate?

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[quartz oracle]

[quartz oracle]

Yes

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[fast ivy]

MY question is

[fast ivy]

how did it go from that to that 😭

like im at the part 2x(x+1)^3 (-x-2) and thought i was done and checked the answer key and saw that

[silk bramble]

He took out the negative from (-x-2)

if you're just referencing the last step

[fast ivy]

Oh

but why

😭

[silk bramble]

He just wanted to simplify it further, either way it's still the correct answer

[fast ivy]

Ohhhhh

Ok

i guess directions dont say factor it fully so

no need

[silk bramble]

The answer and the step before are the same thing, just written differently so he would most likely still give you credit if you put that

[fast ivy]

Ok

[silk bramble]

if not then well try to keep your x's positive since now oyu know your professor or teacher is like that

👍

[fast ivy]

Yeah

Thnx

[silk bramble]

be sure to close!

[unkempt violet]

Why do i feel like i never seen this

Bro just casually removed the negatives and put it in -2x

Math has too many scenarios

[odd edge]

@fast ivy Has your question been resolved?

[fast ivy]

lol

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[ashen ingot]

how many zeroes are there that can be determined

[odd edge]

@leaden kayak Has your question been resolved?

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[scenic spire]

If wanted to get more invested to into cryptography should I study abstract algebra and number theory or something else?

[signal crown]

you need both tho

[scenic spire]

Yeah thats what I was asking?
I said and

Then I was asking if not those would there be something else, as well?

[signal crown]

combinatory

linear algebra

sometimes, google is your friend

[scenic spire]

Yeah im sure but I wanted to know someone who actually was in the field

And what they actually used on a daily, not an ambiguous google search but thanks

Ig I shoulda rephrased my question

[graceful chasm]

I don't think abstract algebra will be useful, except for giving some alternate perspectives on combinatorics

[scenic spire]

Hmm but what about topics such as group in AES
Wouldn't that be abstract algebra

Imma check it out
Thanks, there's probably overlap

[low locust]

definitely abstract algebra

for example one topic in crypto is studying apn functions over finite fields

[scenic spire]

Ok ok makes sense
Exactly
The thing is took cryptography class last year and Ive been the tutor for it the last year and I understand the mathematics but I don't know which topic of the mathematical applications are from.
I only assumed abstract algebra bc I did reseaech on finding the inverse in Galois Fields

[signal crown]

||yes||

#❓how-to-get-help

[low locust]

the answer is an <@&268886789983436800> ping

[scenic spire]

For the record Google didn't mention abstract algebra.

But thanks

[low locust]

well you can work in crypto without understanding the math behind it. sadly

[signal crown]

well I remember hearing my computer science teacher using abstract algebra, but I dont think thats something you really need for it

its more like a theory

not a direct application

Im not talking to u bozo

[scenic spire]

I mean I understand it, what I need . I just would like to understand more of it.

[low locust]

understanding stuff like AES really benefits from knowing some abstract algebra

[scenic spire]

Yeah, the professor taught it the basis for it (but she didn't say "this is all part of abstract algebra")
But i figured the better I understand the better I can explain.

[low locust]

well dont scare students by calling stuff such scary words

but even stuff like the construction of F_256 is definitely abstract algebra

[scenic spire]

Ah yes thats what I thought
Yeah for sure
I always tell people what they need to know/ what they don't need to know.
But I don't want to be stumped when people ask more complex questions.

For example this
I want to know how I arrive at the inverse without guessing one of the other questions as an answer.
I gather it would take long manually (im not asking you to solve)

[low locust]

extended euclidean algorithm

[scenic spire]

Yea exactly that ik but I would definitely have trouble finding the factors for each of these

[low locust]

wdym finding the factors

eea is basically repeated polynomial division

[scenic spire]

Oh haha thats true
I think if I actually wrote it would make sense. Ive only use eea with actually numbers so

[low locust]

yes actually doing stuff also helps tremendously

[scenic spire]

LOL yeah but thats out of the scope of our course
(Not eea but using eea to find the inverse of polynomial in a GF)

But ill do it for fun ig
Thanks tho
I apperciate the input

[low locust]

youre welcome

[scenic spire]

.close

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[wary venture]

Is it correct to handle the equation sin(πcosx) = 1 like this?

[tacit wasp]

Yes, it's totally correct 👍

But those are actually the same equations (you can just take one of them)

[odd edge]

@wary venture Has your question been resolved?

[quasi meteor]

The other case would be pi + pi/2

Not pi - pi/2

[tacit wasp]

No, pi - pi/2 is correct

But for this particular case it ends up being the same as the first equation

[quasi meteor]

Yeah

So it’s just another way to write the first case

[tacit wasp]

Yes, but no😅

I mean, only here they are the same but in general they are not

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[wary venture]

Ty

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[tepid quarry]

Systems of linear congruences, if solvable have only one solution x. This must mean that there is only one, unique y1, only one unique y2 and so on. But then how does the following happen?:

[tepid quarry]

8y1 ≡ 1 (mod 7) means that there are infinite values that could be y1. But we need y1 to be unique so that x can be unique

[odd edge]

@tepid quarry Has your question been resolved?

[tepid quarry]

<@&286206848099549185>

[spice goblet]

The solution is only unique up to mod M

It can be found using the Chinese remainder theorem

I'm not sure how you get 8y1 =1 mod 7

[tepid quarry]

Oh ignore that line, idk why I wrote it, it’s useless

[tepid quarry]

Yeah but out of all the infinite y1-s, how do we know which one to use?

The one that I used and was listed as correct was y1=1

But there are many other y1-s that satisfy 8y1 ≡ 1 (mod 7)

[spice goblet]

The solution is unique up to mod 56, so there is exactly one solution between 0 and 55. That's the one you use.

[tepid quarry]

So we need to seek for a solution in the interval [0;55]

[spice goblet]

Yes

The Chinese Remainder Theorem gives an algorithm to find it.

[tepid quarry]

If the remainder was 2, would it be [0; 54]?

[spice goblet]

No, it's the least common multiple of the modulos, so 7 and 8.

Since the lcm of 7 and 8 is 56, the interval is 0 to 55.

[tepid quarry]

Why do we exactly need to find y that’s between [0;M-1]?

[spice goblet]

Because the solution is only unique up to that point. If the solution is x0, then x0+nM is also a solution for any integer n.

You are correct in that there are infinitely many solutions. We simply choose the least positive one by convention.

[tepid quarry]

Oh, so it’s conventional

But how come it’s just that if by picking another congruent y1, we get a completely different result?

[spice goblet]

I'm sorry, I don't understand the question. I also don't know what y1 is in this context.

Is your question resolved?

[tepid quarry]

As I said, one of the solutions is y1=1. But the equation would also be correct if, for example, y1=8

If we used y1=8 in the x=a1m1y1 +…., we’d get a different solution than if we used y1=1

[spice goblet]

Neither 1 nor 8 is a solution.

[tepid quarry]

I think they are?

[spice goblet]

Ah. In this case, you use the one in the interval 0 to 6

[spice goblet]

I misunderstood you, never mind

[tepid quarry]

Oh yeah ok

[tepid quarry]

But why this interval in particular

[spice goblet]

Because it's mod 7

[tepid quarry]

I see that it’s [0; 7-1]

But how does the upper boundary being the mod -1 help?

Why would it be wrong if we used a solution outside of that interval?

[spice goblet]

You would get another valid solution outside of the interval where the solution is unique.

There is one unique solution in the interval [0,55] and that's the one we go for because why choose anything else?

You are welcome to write out all solutions: x=1+n*56, where n is any integer.

[tepid quarry]

When we solve a system of linear congruences, we search for a solution mod M because only then the solution x is unique?

[spice goblet]

We search for any solution x and know that x+n*M gives us all solutions. By convention, we use the smallest numbers we can validly use because it saves work.

[tepid quarry]

Oh so systems have an infinite amount of solutions but finding just one is enough?

[spice goblet]

Yes. Exactly.

[tepid quarry]

And the easiest way to find such a solution is one that’s mod M?

[spice goblet]

One that's positive and below M, yes.

[tepid quarry]

Ohhh I understand now

So there’s multiple solutions mod M? Or was it just one unique one?

[spice goblet]

The solution is unique mod M, meaning there is exactly one positive solution below M and all others you get by adding or subtracting multiples of M.

[tepid quarry]

Ohhh I see

But I have another small question: when we solve singular linear congruences, we find the generalised solution formula as opposed to finding just one random solution and calling it a day. Why don’t we do the same with systems of linear congruences?

[spice goblet]

Because in real life, the smallest solution is usually the only relevant one. You can also very easily construct the general solution from any specific solution.

[tepid quarry]

Ohhh that makes sense

Thanks a lot

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[delicate rover]

I need help with this

[delicate rover]

Hence, write down the smallest possible value of "y = x² + ax + b" using the values of a = -10 And b = 25.

I have no clue on what to do

[nimble blaze]

this is a "hence" question

what was the part before this

[delicate rover]

Oh I did that part and got the a = -10 and b = 25

[nimble blaze]

can you show the full thing

[delicate rover]

Alright I scribbled over it quite a bit though

[nimble blaze]

your b is incorrect

[delicate rover]

Oh what is it supposed to be?

[nimble blaze]

expand/equate components/coefficient/constant

what's the result of expanding
(x-5)^2 + a

[delicate rover]

X² -10x -25 + a

[nimble blaze]

no

where's -25 coming from

[delicate rover]

Oh +25

[nimble blaze]

yes

$x^2 + \red{a}x + \blue{b} = x^2 \red{- 10}x + \blue{ 25 + a}$

[clever fjord]

ℝam()n()v

[nimble blaze]

from there you'll have
a = -10
and b = 25 + a

knowing the value of a
allows you to determine the numerical value of b

[delicate rover]

So b = 15?

[nimble blaze]

yes

as for the hence question,
note that you have an expression in vertex form
$$(x-5)^2 + a$$
what's the minimum value of this?

[clever fjord]

ℝam()n()v

[delicate rover]

I don't know

[nimble blaze]

$$\blue{(x-5)^2} + a$$
what do you want that blue component to be for the minimum value

[clever fjord]

ℝam()n()v

[nimble blaze]

as small as possible right?

[delicate rover]

Yeah but how do you make it smaller?

[nimble blaze]

well note that its a squared value

what's the smallest possible number for such a value

[delicate rover]

0?

[nimble blaze]

yes

[delicate rover]

So then what do I do?

[nimble blaze]

well you'll have the min amount when that blue part is 0

what do you have when that blue part is 0

[delicate rover]

0 + - 10?

[nimble blaze]

simpliy that

[delicate rover]

-10

[nimble blaze]

yes

that's it

[delicate rover]

Yo thanks

.close

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[manic niche]

hii! can i please have some help with this?

[manic niche]

its extremely simple but i keep getting it wrong haha

[mystic saffron]

Îs this correct

[manic niche]

im not sure...?

i'd just like some help

[delicate rover]

It's 17/24 I'm pretty sure

[grizzled tide]

this is right but you need to explain the method

[manic niche]

this is how ive tried doing it:

5/12 & 7/24 having the same denominator
5/24 + 7/24 = 12/24 - 6/12 = 3/6 = 1/2

[grizzled tide]

giving answers isn't useful

[manic niche]

so confused on how you got that haha

[delicate rover]

You gotta make the bottom one the same on both