help-27

[restive river]

I'm not going to bore you by making you find me the root for this polynomial

because the answer is in fact x = 1

and f(1) is the minimum

global minimum, in fact

[restive river]

unless you want me to

[solid osprey]

ok

thx

.close

[devout snow]

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[restive river]

f(1) = 31

so the minimum is 31

[solid osprey]

cya

.reopen

wait where did i gi wrong with this

[devout snow]

✅

[twilit plank]

If you time the left calculation by x^3 does't the left calculation become the right calculation? Therefore if the functions are the same the left function does constitute the right function

[solid osprey]

what

[twilit plank]

left function * x^3

or am I crazy?

[solid osprey]

ye

what are you saying?

[solid osprey]

yes we already got that, but whats that got to do with thiz

[solid osprey]

this is me attempting to divide by (x-1) btw

[devout snow]

@solid osprey Has your question been resolved?

[devout snow]

@solid osprey Has your question been resolved?

[devout snow]

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[brittle wharf]

that question with markers, i think the answer is c despite the video saying a, can you pls evaluate the abstract pattern?

[brittle wharf]

its sending

i think its c because of the alternating horizontal and vertical lines in regards with the slanted lines

[solid comet]

it is a

if you see every one of them has the vertical black side touching the central line

instead c would have it touching the external one

[devout snow]

@brittle wharf Has your question been resolved?

[brittle wharf]

i think there is no central line on the second one? it is a slant line and a horizontal line, or is it as if its imaginary?

[solid comet]

imaginary

you get the point

[devout snow]

@brittle wharf Has your question been resolved?

[brittle wharf]

ok thanks

[devout snow]

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[toxic kettle]

How do I do this?

[toxic kettle]

Anyone??

!15m

[devout snow]

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.

[cinder slate]

there is no hack for (2^2+2^3+2^4+2^5) ?
or im forced to calculate each of them and then add them up (edited)

[toxic kettle]

?

[eager nova]

Open your own channel

[cinder slate]

my bad

[eager nova]

#❓how-to-get-help

[toxic kettle]

its ok

#help-19

is free rn

<@&286206848099549185>

[toxic kettle]

this

[toxic kettle]

and this

<@&286206848099549185> !!!!!!!

[runic trellis]

for familiarity's sake let amount food items he buys be $x$ and amount of drink items be $y$. Establish an equation describing the situation using these 2 variables. Then, can you graph this somehow in your coordinate system?

[woven radish]

Crystopher

[toxic kettle]

how?

nvm

no bodys helping

.close

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[devout snow]

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[next nacelle]

Can someone explain to my why this inequality actually has NO solution? I did the whole problem out but at the end it said no solution as the answer.

[urban hornet]

|3x-7|+7≠ 3x this step is wrong

[heady tulip]

the absolute value of anything is always greater than or equal to 0

so |3x-7| cannot be less than 0 (in this case -5)

[next nacelle]

ahh @heady tulip

so i coulda saw the problem

and known it was NO solution

because it was an absolute value EQUAL to a negative

is that correct?

[heady tulip]

yes

anyways ur work is incomplete

when u split the abs value u have the red condition too

[next nacelle]

ah

wait circling back to before

what if the equation was the same

but it was the opposite sign

like

|3x-7| > / = -5

would it still be no solution

[heady tulip]

ok its better if u do it on ur own so u get how to break the inequality

an example here: |5x+1| > 8

case 1) 5x+1 > 8 whenever 5x+1 >= 0

case 2) -(5x + 1) > 8 whenever 5x+1 < 0

[next nacelle]

ok i see

thank you

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[winter torrent]

have you seen dy/dx before?

yes, dy/dx means the derivative of y with respect to x

i'm sorry, i have to go to work; I don't have time to teach a whole calculus class right now

[next nacelle]

LOLL sorry i cant help u because im in precalc rn but that was funny @marble otter

[devout snow]

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[rose pumice]

differentiate e^(-x)

[rose pumice]

is it just e^(-x)?

cuz you dont bring x's down

[lusty mantle]

-x is actually (-1)•x

[rose pumice]

so -e^(-x)

[lusty mantle]

Yes

[rose pumice]

thank you

if it was like e^(-2x) would it be -2e^(-2x)?

[lusty mantle]

Ye

[rose pumice]

if it was like e^(-2x^2) would it be -2xe^(-2x^2)?

[dusky drift]

Use Chain rule

where u = -x

[lusty mantle]

Almost right

[dusky drift]

my chat is glitching

I sent these messages long ago

[rose pumice]

would it be -2e^(-2x^2)

[stark anvil]

Yet the fact remains true

[lusty mantle]

No

[rose pumice]

-4xe^(-2x^2)?

[lusty mantle]

Yes

[rose pumice]

ah ok thank you for your help

🫡

.close

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[iron sun]

mfw when someone manages to help someone with just multiple Yes and No

[lusty sapphire]

yes

[rose pumice]

no

[iron sun]

https://media.discordapp.net/attachments/584485136414801936/1134238857605423134/IMG_5374.gif

[rose pumice]

thats not in my dictonary 😭

[devout snow]

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[toxic kettle]

what are the 2 types of formates for sloples?

[toxic kettle]

other than Y = mx +b (Slope intercept form)

I forgot

[tame matrix]

There are many types

[toxic kettle]

the one

in

like:

[tame matrix]

(Y-y1) =m(x-x1)

[toxic kettle]

4x+4y=4

X+y = a number

[tame matrix]

You want for slope or line

[toxic kettle]

line

formula

format

[tame matrix]

Yes this is line

[toxic kettle]

what is it called?

[tame matrix]

Y=mx +c

And

Y-y1=m(x-x1)

What do you exactly want

[tame matrix]

This is also a format to represent 2d lines

[toxic kettle]

I think I found

it

Ax + By = C

thx

@tame matrix

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[devout snow]

@pulsar edge Has your question been resolved?

[pulsar edge]

.close

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[lone stirrup]

Quick help with linear equations with 2 unknown values

[lone stirrup]

2x - y = 8
3x + 4y = 1

I did -y = 8 -2x

[lost laurel]

try making either the x or y coefficents the same as the other equation for one equation of your choice

[lone stirrup]

But in the answer in the book it recommends doing -2x + 8, wouldn't this be basically the same outcome?

I ended up with

-1(-y = 8 - 2x) -> y = -8 + 2x

[vapid pelican]

$-y =8-2x\iff y=-8+2x$

[woven radish]

Miliski

[vapid pelican]

its the same thing, now that you have $y=-8+2x$ you can simply replace y in the second linear equation with $-8+2x$ and find x

[woven radish]

Miliski

[lone stirrup]

i think i know my mistake

i think my mistake was doing the equation = 8 which made 32 + 8 = 40, but instead i shouldve done 1 + 32 = 33

.close

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[native thistle]

are these two methods the same thing or is the guide correct?

[frozen aurora]

something wrong happened here

there should be a + between the parentheses

(3x-1) + (-2x-2)

then they will be the same

[frozen aurora]

otherwise this looks like multiplication

[native thistle]

why does it have to have a +?

[frozen aurora]

because you wrote (3x-1)(-2x-2), which is the same as (3x-1)*****(-2x-2)

and you are clearly not multiplying them, but only adding/subtracting

[native thistle]

oh wait

oh yeah its that thing

i should’ve not study math at 2am damn 😭

[frozen aurora]

go sleep lol

[native thistle]

i cant

well thank you anyways

[novel locust]

Im too stupid too read how to get help

So i will ask

I have a very hard problem anyone wanna help

[native thistle]

im gonna close this you can reask on the same chat

.close

[devout snow]

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[novel locust]

How

[frozen aurora]

go ask in #help-35

[devout snow]

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[vague obsidian]

$$\left(\frac{16}{25}\right)^{\left{x^2+x\right}}<1$$

[woven radish]

Shinutsi

[vague obsidian]

How can I make the bases equal?

raise to 0?

but then $$0(x^2+x) < 1$$ $$ 0 < 1$$

[woven radish]

Shinutsi

[coral heron]

$a^0 =1$ for all $a\in \mathbb{R}$, so just make 1 $\left(\frac{16}{25}\right)^0$.

[woven radish]

fish

[vague obsidian]

OHHHHHHHHH

i was doing the opposite thing

got it now

thanks

.close

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[glacial agate]

can someone help me solve this

[glacial agate]

i drew the diagram but im stuck on what to do next?

i thought it would just be basic trig but guess not

i messed up the drawing, one of the friends are behind him

[outer jolt]

i kinda forgot how to do it but doesnt the angle begin at the dotted lines?

actually leemme calcualte

[glacial agate]

well as long as it connects to the floor from where the guy up top i think thats what matters

[outer jolt]

oh shoot i realized what u meant

[glacial agate]

i did the calculation and got 176 memmeters

dang typo

but

none of the answers match what i got

[outer jolt]

its prob none

[glacial agate]

so maybe its "none of the epxressions"

wait

bro

[outer jolt]

ill recalc tho cuz i messed up 💀

[glacial agate]

would you have your calculator set in radians or degrees

[outer jolt]

u could just calculate it as decimals

[glacial agate]

i dont get it

cause were using tangenet to calculate the distance

you know what i think it is just none of them

Can someone else look at this

.close

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[smoky sentinel]

How would i factor (x^6+x^-6+2)

[restive river]

write x^-6 as 1/x^6

[smoky sentinel]

ye

[restive river]

then it will be in the form of (a+b)^2

[smoky sentinel]

(x^6 + 1/x^6 +2)

i dont see the form

[fierce heath]

x^6 = (x^3)^2

[restive river]

yup

and 2*(x^3)*1/(x^3) = 2

[fierce heath]

2 * (x^3) * (x^-3)

you mean

2ab

[restive river]

yes

[smoky sentinel]

does this method have like a topic so i can search up more examples?

@fierce heath

[fierce heath]

search up square of binomial

or (a+b)^2

or factoring etc

[smoky sentinel]

ok thx

[restive river]

"Factoring Expressions Involving Powers and Reciprocals"

[devout snow]

@smoky sentinel Has your question been resolved?

[devout snow]

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[fading lion]

how would i prove this?

[devout snow]

@fading lion Has your question been resolved?

[lyric yarrow]

Let me see if i can help

[crystal obsidian]

can u use vectors to prove them?

[lyric yarrow]

I myself got stuck at the parity bit

so I had to phone a friend (ChatGPT) to help me out

hopefully this helps @fading lion

[fading lion]

yes i see now

thanks

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[woeful whale]

I need to find all zeros and their multiplicities of f(X)=(x-5)^2 (x^2-3)

[devout snow]

Please don't occupy multiple help channels.

[weak cove]

stick to #help-3 don't open multiple channels

.close

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[woeful whale]

This can be closed

[devout snow]

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[woeful whale]

Determine the largest open interval of the domain for which the function is increasing. f(x)= -5/6x^5

[woeful whale]

I have already graphed it

but I am not sure what they mean by the largest open interval ?

I know we are talking about ordered pairs and x cause they talk about the domain?

but what is the largest open interval ?

[iron sun]

i guess largest here means you can find an interval which covers the previous (smaller) one, so for example x<1 is "larger" than x<0 if i interpret that correctly

[woeful whale]

ok so I am looking for X<1 ?

[iron sun]

well that was just an example

[woeful whale]

oh

[iron sun]

In your case, can you atleast find some interval where the function is increasing?

as a start

[woeful whale]

i know the intervals I dont know how to tell increase?

[iron sun]

Do you recall the defintion of an increasing function?

[woeful whale]

Nope :/

I can look up real quick

A function f is increasing on an interval if for any two input numbers and in the interval, x 1 < x 2 implies that f ( x 1 ) < f ( x 2 )

but that doesn't really make sense to me

[iron sun]

So basically, given two points x1 and x2 in your domain, such that x1 < x2, if f(x1) <= f(x2) for all x1,x2 in your domain (or some subset) then that function is increasing on that domain (or subset)

[iron sun]

that is strictly increasing

So just graphically, what would you imagine they mean by the name?

Say youre going along the graph (in the right)

[woeful whale]

that if I plug in a real number for x and x1 =x2 its increasing ?

[iron sun]

what would you interpret as being increasing?

[woeful whale]

the line goes to infinity

that

[iron sun]

Ignore the symbols for a moment here and just try to picture the graph of an increasing function

[woeful whale]

is a terrible way to explain it but

it will drasticly climb or drop

OMG

I get

[iron sun]

Do you see it?

[woeful whale]

the domain doesn't increase

because it

only crosses the x once

while the y

increases to - infinity and infinity

@<@ THANK YOU

so much

holly crap

[iron sun]

Uh im not sure im following

[woeful whale]

I was staring at this for ever

[iron sun]

but im glad you think i helped!

[woeful whale]

XD

if you graph it you can see

[iron sun]

So what's your answer?

[woeful whale]

it doesnt have an increase in interval

for the domain

[iron sun]

Because?

[woeful whale]

cause it only crosses the x axis once

there is only 1 x intercept

oh shit

wait

[iron sun]

Well you can have an increasing function which does that too

[woeful whale]

I might be wrong

yeah

now I think I am dumb

XD

[iron sun]

No! just take it slow

you happen to actually get the right answer

but maybe in the wrong way

So lets start slow

Say you had the function g with the rule g(x) = x

would this be an increasing function? (on say any interval)

[woeful whale]

I think so because it will be multilyed by its self so it will grow larger infinitly

or no

becuase the who equation is just x

[iron sun]

I see, im not sure what you mean by muliplied by itself; but yes

[woeful whale]

like

[iron sun]

Look at the graph instead

,w graph y = x

[woven radish]

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[iron sun]

say we're traveling from left to right

say this is a mountain

Would we go up or down?

[woeful whale]

yeah it has an increase

[iron sun]

Yeah!

That is the point of the name "increasing"

it's in some sense increasing

[woeful whale]

correct

[iron sun]

in respect to going from left to right

So let's try a different function

say -x

,w graph y=-x

[woven radish]

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[iron sun]

With the very same premise as before, going from left to right

would we go up or down as we do so? (in any interval)

[woeful whale]

down

[iron sun]

Correct!

In this case, we call this function decreasing (on any interval)

[woeful whale]

oh

[iron sun]

Okay, so if this was your function in your question

[woeful whale]

so the reason

[iron sun]

what would be the answer?

[woeful whale]

So basically because the graph wasn't rising on the x-axis and and more on the y-axis

is why it didnt really have an increase

on a specific domain

it was just increasing steadly

while y was increasing expenentialy

[iron sun]

Or put more formally, imagine you have two x values, x1 and x2. One is to the right of the other, like this: x1 < x2

So if h(x) = -x

then we have that h(x1) > h(x2)

So by the definition of (stricly) decreasing h must be a decreasing function

[woeful whale]

h is y ? correct

[iron sun]

you can think of h(x) here as the corresponding y-coord for a given x yes

[woeful whale]

ok

then yes I follow your explination

[iron sun]

Great!

[woeful whale]

and it makes sense

[iron sun]

Now let's tackle your question from the start

[iron sun]

You said you graphed this?

[woeful whale]

yes

[iron sun]

Do you have a drawing maybe?

[woeful whale]

one sec

[iron sun]

Okay, what would you intuitively say about this graph? regarding "increasiness"

[woeful whale]

decreasing

[iron sun]

Right! that would be very reasonable

Let's put this feeling to test and try to maybe see it via the definition aswell if youre comfortable with that?

[woeful whale]

f(x)=x1>x2

?

[iron sun]

What do you mean by that?

[woeful whale]

never mind

I thought that what you

meant

like hx1>hx2

[iron sun]

Hm well not quite, im not letting f(x) be equal to that

Yeah right!

[woeful whale]

in how its decreasing

[iron sun]

yup!

So since we have f already defined

let's try to use it

[woeful whale]

awesome!!!

im sorry I have to go

[iron sun]

alright it's fine!

[woeful whale]

I have alot more stuff to study and

[iron sun]

Goodluck!

[woeful whale]

I already feel good on this

thank you so much

you really helped explain this to

me

[iron sun]

Youre welcome :D

[woeful whale]

I will try it with the inputs as well

YOU ARE THE BEST!!!

.close

[devout snow]

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[kindred lintel]

no idea how to do

45

Like from here

This da best i got

[frail bear]

this is factored

[kindred lintel]

Is it correct?

[frail bear]

ok I figured it out

[kindred lintel]

Ok

[frail bear]

you have one parameter a

wait

no two parameters my bad

Anyways if you don't know it you can use Lagrange's polynomial interpolation formula

but it may be a little bit overkill

[kindred lintel]

Wadafak is that

[frail bear]

$P = \sum_{i=0}^n y_i L_i$ with $L_i = \prod_{j=0; j \neq i}^n \frac{X - x_i}{x_j - x_i}$

[woven radish]

Valentin

[kindred lintel]

Yeah idk a single thing in that john

[frail bear]

ok forget that

[kindred lintel]

They got fucking E's in math now i guess

[frail bear]

Be confident

[kindred lintel]

yes sir

[frail bear]

You need to get a polynomial with a degree of 4

[kindred lintel]

3

[frail bear]

3 my bad

P = a (x + 1)(x - (2 + sqrt(5)i))(x - b)

is it ok for you ?

[kindred lintel]

Yeah

Whered u get x - b

[frail bear]

now plug in the solution point

[frail bear]

you need a third factor for the polynomial to be of degree 3

[kindred lintel]

also what happened to the x - (2 - sqrt(5i))

[frail bear]

but you don't know b

[frail bear]

you know that 2 + sqrt(5)i is a root of the polynomial, so X - (2 + sqrt(5)i) is a factor

I prefer to write X instead of x (to distinguish formal polynomials from polynomial functions)

[kindred lintel]

mmmmmmmmm but wasnt there already 3

They gavve us 3 zeroes no?

[frail bear]

-2 is not a zero

or I read it wrong

[kindred lintel]

no like 2 + sqrt(5)i and 2 - sqrt(5)i

And then the -1

Cuz it has the i

So its like plus or minus

[frail bear]

sorry I don't follow you

why would 2 - sqrt(5)i be a zero ?

[kindred lintel]

dont u need both?

[frail bear]

why ?

Let me break it down

P is a polynomial and its degree is 3.

[kindred lintel]

ok

[frail bear]

Therefore there exists 4 complex numbers a, b, c, d such that P = a(X - b)(X - c)(X - d).

This is the factored form. And you know that such a writing is unique (except for the order of the coefficients)

[kindred lintel]

how do i find the P, a, and d

[frail bear]

you need to plug in f(-2) = 42

[kindred lintel]

42 for y and -2 for x?

[frail bear]

exactly!

[kindred lintel]

and then how do i get d

[frail bear]

good question

I think there is a degree of freedom remaining

not sure

let me grab a piece of paper

Did you do some linear algebra ?

[kindred lintel]

wdym

Also are zeroes and roots the same thing

X int values

[frail bear]

I was going to be talking about vector spaces because they can help in that situation

help to understand*

Ok can you solve if you assume a=1 ?

[kindred lintel]

Uhhhh

no idea

[frail bear]

what are a and b ?

in the prompt

[kindred lintel]

a is completely factor form and b is expanded form

[frail bear]

ok

so (a) -> (b) is trivial (just computations)

[kindred lintel]

Computation

Whats that

[frail bear]

calculations

my scientific english is broken

[kindred lintel]

Oh yeah

But its so dam hard with the sqrt 5 i

[frail bear]

ok let's simplify for you to understand better

If I ask you to find a polynomial function f with a degree of 2 such that 1 is a root and f(2) = 4, what would you do ?

If you can do this you can do the whole exercise

And what is your math level ?

[kindred lintel]

Im taking pre calc rn

so (x-1)

and thennnnnn

4 = (2-1)

Yeah no idea

[frail bear]

ok

in this simplified exercise, f(x) = a(x - b)(x - c) right ?

but 1 is a root, so we can write f(x) = a(x - 1)(x - c)

[kindred lintel]

yes

[frail bear]

But f(2) = 4, so 4 = a(2 - c)

Do you see that we have an infinity of solutions ?

[kindred lintel]

O shit

so f(2) = a(2-1) as well

[frail bear]

no

[kindred lintel]

I mean 4 = a(2-1)

[frail bear]

no

[kindred lintel]

O

[frail bear]

c does not depend on x

[kindred lintel]

no i mean thats how i find a value riht

Cuz we know (x-1) is a factor

So i can plug in 4 and 2

[frail bear]

the exercise is unclear : do you want a solution or all the solutions ?

[kindred lintel]

A solution

[frail bear]

ok so just assume that the dominant coefficient a is equal to 1 (the polynomial is unitary)

and we are assured that there will be a unique solution

Alright

[kindred lintel]

Yes

[frail bear]

So if a=1 then 4 = 2-c then c = -2

[kindred lintel]

in this simplified exercise, f(x) = a(x - b)(x - c) right ?

but 1 is a root, so we can write f(x) = a(x - 1)(x - c)

wait

Need to see eq again

[frail bear]

yes !

but we will assume a = 1

as I say, we want a solution, not all of them

[kindred lintel]

Ye

[frail bear]

And we have « enough » solutions to drop out those with a ≠ 1

[kindred lintel]

ah

[frail bear]

When you will grow older you will learn the Lagrange interpolation formula and know about the Vandermonde matrix

(that's if you want to go further)

[kindred lintel]

😭

Yeah nawww

[frail bear]

ok did you progress ?

[kindred lintel]

ok so i got

X^3 - 3x^2 + 5x + 9

[frail bear]

let me just calculate it

(I don't use a calculator)

Did you calculate f(2 + sqrt(5)i) and f(-2) to verify ?

I don't have 42 for f(-2)

[kindred lintel]

oh shit i forgot to put that in

It should be -2x^3 + 6x^2 - 10x - 18

[devout snow]

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[kindred lintel]

hi

[kindred lintel]

can there be a vertical asymptote and a hole

on the same point

lets say here

[trail eagle]

I mean there can be a vertical asymptote on one side and a hole on the other.

[kindred lintel]

can the asymptote vertical

still be x = 2

there is also x=-3/2

[trail eagle]

Vertical asymptotes blow up to infinity or -infinity.

Your graph does not show this behaviour for x=2.

[kindred lintel]

on the left

mmmm

so in this case it cant?

[trail eagle]

Well it's not an asymptote

x=2

[kindred lintel]

mmm

oh its just a hole

aight thanks

[trail eagle]

And around x=-3/2, we can see this unbounded behaviour

[kindred lintel]

yeah

.close

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[gloomy loom]

this is done by setting up an integral with y correct?

[devout snow]

@gloomy loom Has your question been resolved?

[gloomy loom]

<@&286206848099549185>

[gloomy loom]

someone help me please

<@&286206848099549185>

[stone palm]

yeah

so first we can plot these graphs that will help us a lot

[gloomy loom]

hey!

should I use Desmos and post it here?

[stone palm]

yupp you can

[gloomy loom]

https://www.desmos.com/calculator/wag58jjdrf

idk how to properly use the graph for this problem

[stone palm]

that should be right

[gloomy loom]

cool cool

[stone palm]

now we see that the intersection point is around 9 right

[gloomy loom]

yes

[stone palm]

that means the function x is below the x axis till x=9

so we can ignore the area of function x till x=9

[gloomy loom]

ok ok

[stone palm]

hence we can integrate the function y from 0 to 9 to find the are of the curve from x=0 to 9

now the major things comes after x=9

we know that e^-x will never ever reach 0

or become negative

[gloomy loom]

mhm

[stone palm]

and the function x becomes positive only after x=9

seeing the graph you see function x covers majority area after x=9
and function y covers minority area
so we can integrate function x from 9 to infinity and then subtract integral of function y from 9 to infinity from that

later on add the 2 areas we got
integral of y from 0 to 9 and the other answer we got

[gloomy loom]

hmm

let me right it out and send it to see if i'm understanding properly

[stone palm]

yup sure

[gloomy loom]

like this?

[stone palm]

yup and you gotta add the other area integral 0 to 9 of e^-t

this is the area under curve after x=9
you forgot the area befor x=9

[gloomy loom]

like this?

[stone palm]

um no see function x is below x axis from 0 to 9

so we dont care about its area

the first thing the question says is that it needs the area above x axis

[gloomy loom]

oh

wait

i'm confused now

[stone palm]

solution would be
integral of e^-t fom 0 to 9 + (integral of t^2+8t+9 from 9 to infinity - integral of e^-t from 9 to infinity)

[stone palm]

ignore the graph below the x axis

[gloomy loom]

right

[stone palm]

now till x=9

we only have the function e^-t

right?

[gloomy loom]

OHHHH

area under the curve!

that's what you meant

mb mb

[stone palm]

yeah

lol its fine

[gloomy loom]

gimme one sec

lemme plug this into desmos

LUCKILY, I don't need to show steps of integration for this problem

that would've been TEDIOUS

[stone palm]

ok awit

wait

[gloomy loom]

mhm?

[stone palm]

other thing i think is we should take integral from -ve infinity to x=9 for e^-t

as they havent mentioned any bounds that it should be on the right side of y axis

[gloomy loom]

-ve?

[stone palm]

negative

[gloomy loom]

oh

[stone palm]

to calculate this area we should keep -ve infinity in our solution

[gloomy loom]

oh ok ok

[gloomy loom]

so then that would be adding to this right here

[stone palm]

yup

[gloomy loom]

so it'd be the new integral plus the other shennaningans

[stone palm]

yup an integral from -ve infinity to 0 should be added

[gloomy loom]

to 0 or 9?

[stone palm]

or you can modify the other integral instead of 0 to 9 to -ve infinity to 9

[gloomy loom]

yeah

this is better lol

[stone palm]

yup

so i will close this?

[gloomy loom]

oh, you mind if I check the answer rq while you're still here?

if you're busy, then you can close it

[stone palm]

upto you

[gloomy loom]

ok, let me put the answer rq then

when I put this integral, it just says undefined

[stone palm]

then ig just limit fom 0 to 9

[gloomy loom]

ok

this one is also undefined

[crisp zealot]

I have bad news

But this is not what this question is

[gloomy loom]

oh

[crisp zealot]

It's a parametric equation for a curve

[gloomy loom]

oh snap

that shenangigans

[crisp zealot]

There is one curve

And you are calculating the area between that and the x-axis

[gloomy loom]

ok

don't you have to use your as part of your inegral

and then differentiate x so that it becomes your dx in it?

like this?

@crisp zealot

[crisp zealot]

Yes

[gloomy loom]

and then just solve the integral correct?

[crisp zealot]

Yes

[gloomy loom]

got it!

@crisp zealot @stone palm

thank you both for you help!

https://tenor.com/view/leonardo-dicaprio-gif-21978827

[crisp zealot]

Nice np

[gloomy loom]

.close

[devout snow]

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[wooden veldt]

You might first try to simplify the second expression on the left hand side

[solid osprey]

ye

[mystic scarab]

Well, what do you have after that cancellation?

Yep

[devout snow]

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[fiery drift]

I found an old quiz we had to do when to do 10 years ago but apperently i cant get the answer to this simple quistion:

[soft umbra]

What have you tried so far?

[fiery drift]

5^2=20

x=2sqrt(5)

8.94427191

8,9442

and many more

but i think it is with the metes thing where the problem is

more of a formatting thing

[soft umbra]

Just apply a^2 + b^2 = c^2

I guess that’s everything

[fiery drift]

(2*x)^2+x^2=10^2

x^2=20

[soft umbra]

Correct

That’s the only solution

[wooden veldt]

Note what value the question actually wants

[fiery drift]

4,4721 meter?

[wooden veldt]

Oh your answer here is correct

[fiery drift]

what about this:

"Please enter the correct password."

not the right answer still

[wooden veldt]

What exsctly are you entering

[fiery drift]

8,9173m
8,9173meter
8,9173 m
8,9173 meter

[wooden veldt]

Why aren't you using this and rounding it

Something is off with your graph

[fiery drift]

Got it

the website is stupid

8.9442 was correct

[wooden veldt]

Oh it wanted it truncated

That is pretty stupid

[fiery drift]

ty

.close

[devout snow]

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[wooden veldt]

What is the quiz btw?

[devout snow]

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[novel grotto]

Could someone please explain what this graph is and how you get it?

Sorry for not being clear I’m just confused

I’m mainly confused with the cubic in the centre and where you’d get it from

[acoustic totem]

latom

[acoustic totem]

do you not see it

its so obvious im starting to believe you might be joking

[trail void]

you're confused on how to draw this graph?

just making sure

[novel grotto]

Yea but mainly the middle part of the graph

[trail void]

well the middle part is not a "cubic" like u said

it's just part of this graph

[acoustic totem]

exactly

i though he was messing with me

[trail void]

oh lmao well maybe we need to go over how to draw this graph w him

@novel grotto how would you start drawing the function? any ideas?

[novel grotto]

Finding horizontal and vertical asymptote

[trail void]

yeah good

[novel grotto]

By set denominator to 0

[trail void]

okay and what do you get when you equal it to 0

[novel grotto]

2 and -2

[trail void]

yeah

where does the function cross the x axis?

[novel grotto]

At 2 and - 2

[trail void]

no

[acoustic totem]

..

[trail void]

so the function crosses the x axis when you equal the numerator to 0

so x = 0

right?

[acoustic totem]

first he told us 0 was a cube and now this

he's messing with us

[trail void]

are you guys friends or not?

[acoustic totem]

i know him irl

he's a little dumb

[trail void]

oh okay

well hey you dont gotta say taht

lemme just explain to him how this works

[novel grotto]

I’m literally asking for help cos I don’t understand

If you don’t want to help that’s fine but please leave then

[trail void]

okay just calm down now, I'll help ok?

want me to explain or do we need to get some mods in here?

[novel grotto]

Thank you

[trail void]

okay

so far we know

that the function isn't defined at -2 and 2 right?

[novel grotto]

Yep

[trail void]

so we have this

right?

the little dotted lines meaning its not defined there

you agree so far?

[novel grotto]

Yep

[trail void]

okay so we have f(x) = x/4-x^2

lets say we put a really huge negative number as x

what will the output be.. negative or positive?

so if we say

f(-1000000)

it'll be

-1000000/4-(-1000000)^2

is that positive or negative?

[novel grotto]

positive

[trail void]

good

so we know that graph that goes from -infinity

will be above the x axis

right?

[novel grotto]

Yep

[trail void]

okay so we draw this

you agree?

[novel grotto]

Yep

[trail void]

and now where do I continue drawing

do you know?

[novel grotto]

On the other side?

[trail void]

yes but where on the top or bottom?

[novel grotto]

Top

[trail void]

liket his?

[novel grotto]

That’s what is confusing me. How did u get that?

[trail void]

that's wrong

so these vertical asymptotes

[novel grotto]

I meant on the other asymptote sorry

Like on the 2

[trail void]

oh you wanted to go from +infinity?

[novel grotto]

Yea

[trail void]

nono we'll just go from left to right

[novel grotto]

Ah okay sorry

[trail void]

so are you familiar with degrees of vertical asymptotes?

lets say we have

(x-2)^2

[novel grotto]

Like dividing with the highest power

Of denominator

[trail void]

what is the x here

if you equal it to 0

[novel grotto]

4

[trail void]

not 4

(x-2)^2 = 0

x has to be 2 right?

[novel grotto]

Yep

[trail void]

but

its of degree 2

see that because it's squared

[novel grotto]

Yes

[trail void]

so just keep that in mind

whenever you have vertical asymptotes of even degrees

you would draw like this

[trail void]

.

if it was even

but if its odd

you do this

[novel grotto]

Okay yes

[trail void]

okay so that is correct

becasue we have x^2 - 4 = 0

we only have one 2 and one -2

so its odd

[novel grotto]

Yep

[trail void]

but now we know it crosses the x axis at x = 0 right?

[novel grotto]

Yep

[trail void]

so we just continue drawing through 0

agree?

and now because the vertical asymptote x = 2

we only have one 2

so its odd

we do the same thing as before

[novel grotto]

Yep

[trail void]

that's it

so that middle "cubic" graph as you mentioned it

that is just a part of this whole graph

[novel grotto]

Ah okay. Thank you so much! 🙏🙏 I really appreciate your help

[trail void]

do you understand?

everything?

[novel grotto]

Thank you for clarifying

[novel grotto]

Yep

[trail void]

okay and by the way

the "cubic" graph in the middle, later if you'll ever decide to study math you will be able to approximate that graph but for now

just do things this way

good?

[novel grotto]

Yes thank you

[trail void]

okay

no problem

[acoustic totem]

.

[trail void]

its . yeah

instead of !

[novel grotto]

.close

[devout snow]

Channel closed

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[acoustic totem]

yw for the help

[novel grotto]

Sorry thank you for the help

[devout snow]

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[earnest canyon]

do i assume c is 2?

[sonic smelt]

Depends on what you mean by c

[earnest canyon]

like what you explained before

x= logc(c^x)

[sonic smelt]

Letting c be 2 will be helpful, yes

[earnest canyon]

ok perf

[devout snow]

@earnest canyon Has your question been resolved?

[devout snow]

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[woven wing]

Completely stuck, no idea what to do. All I did was labelling the radii of the three circles as a, b and c and created an expression of the area of the shaded region as the area of the large circle - the area of the two enclosed circles

<@&286206848099549185>

Is no one able to help?
<@&286206848099549185>

[slim lava]

what help do u need?

[woven wing]

This question

[restive river]

hey mate

don't give the answer just a hint

[slim lava]

im stuk too

ill try to solve

[slim lava]

kk

[woven wing]

How is is going? Have you figured it out?

[slim lava]

lmao no

i think E (its my prediction dont follow)

try math apps

[woven wing]

alr

[slim lava]

they give full solution

[restive river]

bro like i said

don't give answers

[slim lava]

its not a answer

its a joke

no worries

[crisp zealot]

My recommendation is

Try 2 different position of the line pq

That are easy to compute

If they both give the same answer it is very suggestive that that option is correct

If they give 2 different answers it must be E

So pq half way down the big circle and 3/4 down the big circle should be fine to compute

[woven wing]

ok, I'll try

[devout snow]

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[agile garnet]

help

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[lost laurel]

if gcd(a,b)=1,prove gcd(a^2,b^2)=1

[lost laurel]

I would like to use Bézout's lemma here

or is there a better way to solve this

[stone stump]

how would you want to use bezout

[lost laurel]

xa+yb=1

ta^2+ub^2=d

but that doesn't seem to work ver much

[faint zinc]

If gcd(a, b) = 1 then a, b are coprime and share no prime factors. a^2 has only the same unique prime factors as a, similarly for b^2 and b. Therefore, gcd(a^2, b^2) = 1.

[lost laurel]

ah, right

wait, is that valid for.a proof

[faint zinc]

Dunno, what are you allowed to know and use?

[lost laurel]

I technically haven't learnt prime factorisation in NT yet

[faint zinc]

Then you probably cannot use it

[lost laurel]

why is why Bezout's lemma came to mind

can I just have a hint please

[faint zinc]

I'm working through the problem

I don't know the answer yet

[lost laurel]

my book does have a hint

[faint zinc]

What is the hint?

[lost laurel]

first show that gcd(a,b^2)=1 and thatgcd(a^2,b)=1