help-13

[south tundra]

Yeah ik but they want to replace 1 with sin^2 + cos^2 so I went along

[jaunty mural]

that's the first thing you should clear up for yourself

[crimson sedge]

because 1+2 makes 3 and not 2

[jaunty mural]

yes and just "crossing out the 2's" does not get you the correct answer

,,\frac{1+2}{2}=\frac12 + \frac22

[jaunty mural]

Are you aware of this?

[crimson sedge]

yeah

[jaunty mural]

You can distribute division over addition

That's what you should be doing in the original

and eventually end up with this

[crimson sedge]

so like

1-sin^2x/cos^2x?

[jaunty mural]

ngl i lost track of whats going on

$\frac{\sin^2{x} + \cos^2{x}}{\cos^2{x}}$

[jaunty mural]

I thought u were trying to simplify this

at a certain step

[south tundra]

You mean (1 - sin^2)/cos^2 ?

[crimson sedge]

yes

[south tundra]

Yeah and just replace 1 - sin^2 with cos^2

[crimson sedge]

cos^2/cos^2?

[south tundra]

Yes

Which is?

[crimson sedge]

1

but how does that happen?

[south tundra]

a/a = 1 for nonzero a in general

[crimson sedge]

I know

but how do you get to that point

[south tundra]

How does what happen exactly?

[crimson sedge]

I meant to ask how do you get to that point

[jaunty mural]

1/cos^2 happens to be 1 + (sin^2/cos^2)

due to identities

pythagoras

[crimson sedge]

ok ok I'm getting it now

it's like finding b^2 through c^2-a^2

I get it

[cedar kiln]

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[hollow pewter]

please factorise x^3 pluse 6x^2 pluse 12x pluse 16

[sinful ocean]

Use photomath

[hollow pewter]

how can you tell

[solemn torrent]

x^3+6x^2+12x+16

[hollow pewter]

it is 12 x

[solemn torrent]

Rational root theorem?

[crude gale]

he means use a calculator to solve this kind of question. or you get try (x+a)(x+b)(x+c) and slowly guess out what are a,b and c.

[solemn torrent]

nah don't use no calculators you won't learn or you could

but RRT it

or equate (x+a)(x+b)(x+c) to your expanded cubic

[sinful ocean]

Youre right i forgot rrt existed lol

[dusk finch]

How would this help?

two of those constant will be complex btw

so it's really uneasy

and unhelpful

[hollow pewter]

please help

[dusk finch]

To speeden up the RRT process, notice that every term of x^3+6x^2+12x+16 is positive for positive x, hence all the roots will be negative.

Do you know what's rational root theorem?

[hollow pewter]

you can solve this by identities but which identity i dont know .yes i know

[dusk finch]

by identities? What kind of identities

I dont know any identity that could be applied to this

[hollow pewter]

i mean algebric identities like a pluse b the whole square

[wraith dagger]

methisalwaysright

[dusk finch]

It's quite non-trivial way to solve cubics though

[hollow pewter]

yes brother thank you.

actually yes but there is factor theorem which can also help

[solemn torrent]

in order to apply the factor theorem you'd need to have a root

[cedar kiln]

@hollow pewter Has your question been resolved?

[uneven garden]

SORRY

TO WHOEVER SAW THAT

THAT WAS MY BROTHER

[cedar kiln]

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[hearty oak]

https://gyazo.com/8ee9399e52f2cddfd421317e0efbc63c

[hearty oak]

how does that get b

[crystal raptor]

Well how did you get E

[hearty oak]

i had like 2 seconds left on timer i guessed randomly

[eager crypt]

wait i was trying to get one leg of CFB

i think bc it's 45-45-90

wait no that's not it

[hearty oak]

i just assumed since cb is 8 that b had to be 8 above since the distance was that long but i guess tahts the hyptonuse

[eager crypt]

actually idk i'm going to go again

yeah i think it's because 8 is root 2 * one of the legs

so you find the sides and follow it to where it would go on the plane

[hearty oak]

so they split cfb in two triangles?

if so where would they split it

[eager crypt]

no cfb is a triangle itself

[hearty oak]

i meant cfeb

[eager crypt]

yeah

tht's what i did

[hearty oak]

do they split in fb?

[eager crypt]

i did

and then since CFB is 45-45-90 the hypotenuse is root 2 times the other sides

[hearty oak]

i get that when we do that fb becomes 4sqrt 2

[eager crypt]

yeah

[hearty oak]

so the x axis is 4sqrt 2

but how is the y axis same

[eager crypt]

because c is on the origin

you go up one of the sides to get to fb

[hearty oak]

ohh it has to be same if its origin

[eager crypt]

and then you go right on the other side to that point

yes

[hearty oak]

when we split triangles does it have to be one point to another what if we split it from f to 8

[eager crypt]

then it would just be more steps to find fb

since we still would need to split fb into a line i think

[hearty oak]

we did it using feb

how we do it using cfb

[eager crypt]

oh well i was doing the exact same steps just on cfb i didn't know you wre on the other one

[hearty oak]

nvm i just figured out we just divide 8 by sqrt 2 then ti gives side

[eager crypt]

yes that was exactly what i was going to say

[hearty oak]

and its same as 4 sqrt 2

[eager crypt]

yes

[hearty oak]

thanks for help

[eager crypt]

alr now i have to do my own work

[hearty oak]

.close

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[hollow pewter]

can i help you i have found theanswer

[cedar kiln]

@hollow pewter Has your question been resolved?

[cedar kiln]

@hollow pewter Has your question been resolved?

[dire geode]

don't open a channel to help other people who aren't available

.close

[cedar kiln]

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[hollow pewter]

ok

.close

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[haughty python]

how would you do this

[cedar kiln]

@haughty python Has your question been resolved?

[dire geode]

!show

[cedar kiln]

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

[haughty python]

ummmmm thing is i dont even Know how to start this

[dire geode]

use definitions of consistent estimator

[haughty python]

like i know u gotta get plim y = y

but like

i dont know how to like approach it

[haughty python]

should i start by expanding the summation or

[dire geode]

use the weak law to find the limit of this

https://www.probabilitycourse.com/chapter7/7_1_1_law_of_large_numbers.php

[haughty python]

ok so then itd be limn-> infinity P(|yn-y| >e) = 0

(where lambda = y cuz idk how to type it on the keyboard)

[dire geode]

the second definition of consistent estimator here is the more useful one

[haughty python]

Ok Wait but then what 😭

[dire geode]

plug your definition of gamma_n into this

use all the information you're given from the problem

[haughty python]

ok one sec

[haughty python]

but like even when you do that like what are u supposed to like simplify it

[dire geode]

Simplifying is generally a good thing to do in math

[haughty python]

oops i meant how*

lmao

so like expand and then sub in gamma into wherever theres a exi^3

wait wait

hang on

Nvm 😦

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[eager crypt]

alr so the way i think of it is like

its 3:1 which is 4 total

20 is divisible by 4 easily

with the model they do that times 20/4 bc that's how many times you can fit it

that's how the model works basically if taht's what you were asking

pretty sure

[cedar kiln]

@leaden shore Has your question been resolved?

[eager crypt]

alr cool

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[native shuttle]

im working on d)

i cant seem to get the induction step right

i need to show the last part of the inequality to then link it back to the start to show that its valid for K+1

[inland fossil]

okay so you know what you have to show right

[native shuttle]

yes exactly

so i expanded the left

and got this

but i dont know how to proced from here

maybe squaring both sides?

[inland fossil]

yeah thats worth a try

[native shuttle]

can i cancel out k+1 from the left?

acc no i cant

[inland fossil]

okay yeah this problem is a bit annoying

[native shuttle]

my professor did scribble something down for me

here it is

[inland fossil]

ur prof is using products?

[native shuttle]

he just use sigma notation cause thats what we were working on at the momment

so he used it as a example of how it can be summed up as a summation

[inland fossil]

yeah it looks like ur prof just rewrote the problem

didnt really provide anything useful past that

[native shuttle]

look at the bottom line

thats where he started to sorta solve it

[native shuttle]

everything from before is restating this

[slate lintel]

okay i don't want to give you too much but this can be done with relatively simple algebra

note that everything here is positive

[native shuttle]

yeah thats what i was thinking

is just state that $2 \sqrt(k+1)$ is greater then $\sqrt(k)$

omfg how do u use this

close enough

but the issue is, is that you need to show that the statement is true for the 1/sqrt(k+1) case

,help

[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!

[inland fossil]

i found a way thats horrible but works so rn im looking for a nice elegant way

[native shuttle]

@inland fossil

just so we are on the same page

this is all we need to show right

[inland fossil]

yea

its a ridiculously tight bound

i cant find any nice inequality approximation tricks that work

[native shuttle]

ill email my prof

[inland fossil]

if you want i can show you my horrible way

but i doubt thats what ur prof is looking for

[native shuttle]

yeah he really wants elegance

[native shuttle]

ill tag u in #discrete-math once i get a solution if u want lol

[inland fossil]

sure lmao

my solution is just repeated squaring

starts from 1/(n+1)^2 >= 0

[native shuttle]

hmmm yeah idk if thts exactly what we are looking for

he did it in those pictures and it was super clean when he did

ofcourse he scribbled it and skipped steps in his writing lol

[inland fossil]

¯_(ツ)_/¯

[native shuttle]

ill find out soon, thanks for the help anyway!

.close

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[eager grove]

hello i just need someone to verify my solution

[eager grove]

one second

this is the question

this is the solution for a

this is the solution for b

the answer for B was 1.51 for anyone who is confused on the hand wriring

[slate lintel]

hmm why do you have a different formula for P in part b than you do in part a?

[eager grove]

because it says to begin at 12

for B

oh shoot

i jus realized

i think B is incorrect

but what about A

[slate lintel]

A is right

,calc 2.9^0.65

[wraith dagger]

Result:

1.9978307742067

[slate lintel]

but you have $P = 100(2.9)^{t/0.65}$ in part b which isn't what you had before

[wraith dagger]

Hayley

[eager grove]

got it

wait so

it says the life is 12

so that would mean a half life of 6

[slate lintel]

wat

do you know what half life means?

[eager grove]

sorry sorry

it would be 1 =

[slate lintel]

you start with 12g, so after one half-life there would be 6g left

[eager grove]

the equation

[slate lintel]

yeah

you can either do it by percent or by grams

[eager grove]

so it would be

1 = 100(2.9)^t/6

and i just solve

[slate lintel]

what

no, do it by percent because itll be easier

if you have 1g left over out of 12 starting grams, what percent do you have left?

[eager grove]

8.3 percent

[slate lintel]

okay, so forget about problem (a), look at the question and set up your eqn

[eager grove]

1 = 8.3(2.9)^-T

?

[slate lintel]

there's a negative sign but yes

wait no

P is percent left over after t days
why did you replace 100 with 8.3?

$P = 100(2.9)^{-t}$

[wraith dagger]

Hayley

[slate lintel]

and you said there would be 8.3% left over right?

[eager grove]

so its this but P = 8.3

[slate lintel]

yeeeeee

[eager grove]

thanks a lot 🙏

[eager grove]

so the answer would be 2.3 right

2.3 days

[slate lintel]

,calc 100*2.9^-2.3

[wraith dagger]

Result:

8.6394203453389

[slate lintel]

,calc 100*2.9^-2.4

[wraith dagger]

Result:

7.7668479839676

[slate lintel]

yea

[cedar kiln]

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[jovial snow]

Its on non-homogeneous poisson process. This is an assignment Question. I did part (a). I would like a hint or blue print to get started on (b). I am completely lost.

[cedar kiln]

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@jovial snow Has your question been resolved?

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@quartz salmon Has your question been resolved?

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[quartz salmon]

I'm currently working on a problem involving a water tank that contains 30,000 liters of pure water. The tank is being supplied with water containing 0.008 g/l of a contaminant at a rate of 600 liters per minute. Additionally, 800 liters per minute of the well-mixed contaminated water is being removed from the tank.

I'm seeking assistance in graphically representing the variation of the contaminant's quantity in the tank over time. Specifically, I would like to create a graph that illustrates how the contaminant's concentration changes within the tank as a function of time.

[cedar kiln]

@quartz salmon Has your question been resolved?

[cedar kiln]

@quartz salmon Has your question been resolved?

[cedar kiln]

@quartz salmon Has your question been resolved?

[cedar kiln]

@quartz salmon Has your question been resolved?

[edgy oyster]

err

[fallen moat]

what have you tried?

hmmm how to add spoiler to picture?

[cedar kiln]

@quartz salmon Has your question been resolved?

[austere hull]

when you send an image you click to the eye button then it makes it as spoiler

[fallen moat]

oh thanks!

👁️

[grizzled geyser]

.close

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[fluid anchor]

Shivangi deposits Rs. 500 every month in a recurring deposit scheme and receives Rs. 16550 at the end of 5/2 years. The rate of interest given by the bank is ?

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[fluid anchor]

<@&286206848099549185>

[somber valve]

"Stop that."

[fluid anchor]

?

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[fallen moat]

is it compounded monthly?

[cedar kiln]

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[cedar kiln]

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[wicked brook]

I'm trying to prove the forwards direction of Wilson's theorem: if p is prime then (p-1)! = -1 mod p.

My argument is that if I take all the elements of $\mathbb{I}_p^x$ and multiply them, I'll get the class [-1] since [-1] is the only order two element in the group. The result follows immediately.

[wraith dagger]

zander_a

[wicked brook]

Does this work?

[crimson delta]

why does order two have anything to do with that?

[wicked brook]

Cause I'm multiplying everything together, a = a^-1 iff <a>=2. Thus, the other elements will be cancelled by their inverses

[crimson delta]

ok and why is -1 the only order two element?

[wicked brook]

Cause If I'm squaring to 1 mod p I'm either 1 or -1, and the identity is special

[crimson delta]

why are those the only two?

[wicked brook]

Hmmm.
My first thought is to look at it as a class like if [a]^2 = 1 then [a^2]=1, so like a^2 = 1 mod p. Or (a-1)(a+1)=0 mod p. As far I see, this implies that it's just 1 and -1 cause division mod p is nice

[crimson delta]

we have no zero divisors mod p

and this now is the complete argument

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

Would this be correct?

[rare vault]

i think they want you to use numbers, not a, b, c

because right now i could say (a=-2) and this wouldnt be right

[crimson sedge]

it says sketch tho-

Why wouldn't -2x be right?

As long as its the ax^2+bx+c formula, wont it be all good?

[slate lintel]

if a = -2 in your eqn then it would point up not down

[rare vault]

^^

[slate lintel]

just pick some numbers

[rare vault]

or bound your variables but thats harder

[crimson sedge]

x^2 + 2x + 3?

would that work?

[rare vault]

does that go up or down?

[crimson sedge]

oh-, it goes up

-x^2 + 2x + 3?

[rare vault]

does that have any x intercepts?

[crimson sedge]

nope

i dont think so?

[slate lintel]

I think it does

[rare vault]

it absolutely does

[crimson sedge]

oh-

[rare vault]

its upside down and you shifted it up

[crimson sedge]

it shouldn't

hmm, how about -2x^2 + 4x - 7?

I think that works

Cause its no x-int

[rare vault]

nope

wait

yep

its far enough down

[crimson sedge]

But idk if it satisfies the second condition

ohhh

[slate lintel]

it's quadratic so it's a parabola

it's not centered at the axis like your drawing but whatever

[crimson sedge]

y=x+3?

Would that have an x-int?

[rare vault]

Do you know how to check?

you want to see when its equal to 0

so does 0 = x+3 have any solutions?

[crimson sedge]

yep

-3+3=0

[rare vault]

then it does have an x int

[crimson sedge]

wont there always be a solution for it to equal 0 tho?

[cunning forum]

no

there might be imaginary solutions

but no real solutions

[rare vault]

whats the solution to

0 = -2x^2 + 4x - 7

(this is your original one)

[slate lintel]

lines will always have both a y and an x intercept unless they're horizontal or vertical

[cunning forum]

the coefficient of x^2 should be negative
b=0 for it to be centered at the y-axis
discriminant must be negative

[crimson sedge]

It says no x-int tho?

[crimson sedge]

there would be 2 solutions

[cunning forum]

these three combined will give a solution

[rare vault]

okay this is true but they are both imaginary so they are not x intercepts

[slate lintel]

right this is a line though (but I think you just forgot the ² symbol maybe)

[crimson sedge]

-2x^2 + (-3)?

[cunning forum]

yes this works

so a=2, b=0, c=-3

[crimson sedge]

yep a=-2

[cunning forum]

(since the original thing was -ax^2+bx+c)

[cunning forum]

no

[crimson sedge]

oh

[cunning forum]

you put -ax^2+bx+c for some reason?

[crimson sedge]

i thought x^2 was negative?

[cunning forum]

ah either way put it such that the coefficient of x^2 is -2

[cunning forum]

yeah so if a=-2 (-ax^2) works out to 2x^2

[crimson sedge]

OHHH

[cunning forum]

ok first are you going with y=ax^2+bx+c or y=-ax^2+bx+c

[crimson sedge]

So -2x^2 + (-3) is the equation that would work?

Does it satisfy the other condition tho-

[cunning forum]

what other condition

[cunning forum]

any quadratic with negative coefficient of -x^2 satisfies the x->inf, y-> -inf thing

dont worry about it

[crimson sedge]

as x approches negative infinity, y will too

[cunning forum]

yeah that too

[crimson sedge]

does -2x^2 + (-3) satisfy that?

[cunning forum]

as x approaches plus minus inf, y will approach -inf

it will approach inf instead of coefficient of x^2 is positive

[crimson sedge]

but x should only aprroch -inf

[cunning forum]

yes since the coefficient of x^2 is engative

[cunning forum]

yeah im saying IF the coefficient was positive

its like additional info

[crimson sedge]

OHH, that makes sense

Ty! :>

Its just a downwards parabola right?

Because its supposed to be a function

And a parabola wont count as that

[cunning forum]

yeaah

[cunning forum]

wait what why not

its a quadratic function

[crimson sedge]

it says on google that a parabola isnt a function-

[cunning forum]

what

wrong kind of parabola

those that open left or right are not

@crimson sedge dont be confused

[crimson sedge]

OHHHH

[cunning forum]

it says those arent functions

[crimson sedge]

Okayyy! Thanks!

[cunning forum]

since one x value has two y values

yeah

[crimson sedge]

Yea- got that

thankyuu! :>

[cedar kiln]

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[crimson sedge]

im supposed to explain why this is false but do not know how: Let n ∈ N. If n is not prime, then it has a divisor d with 1 < d < n

[rare vault]

If n is not prime, then its composite

or

||its 1||

[golden sundial]

contradiction

assume that there exists a natural number n that is not prime, but it does not have any divisors d with 1 < d < n.
because n is not prime, it must be composite (product of integers). consider n the product of two primes, p and q.
since n = p * q and we implied to not have divisors then n must be also prime and not composite but this goes against our initial assumption that n is not prome

[crimson sedge]

I see thank you

.close

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[hard pawn]

in this example, we did we add pi instead of subtracting pi like what we do with sine and cosine?

[rare vault]

Is this asking why you added pi?

[hard pawn]

yes

because for sine we subtracted pi

and for cos we subtracted 2pi

[rare vault]

it has nothing to do with the trig function, you just need all the answer between [0,2pi)

and since tan has a periodicity of pi, you can add and subtract pi to get more valid answers

so in this case we need to add it to get another valid answer in the domain

[hard pawn]

and in what case do we need to subtract pi?

[rare vault]

When you do part b, you will find that there are infinite solutions

you need to only find the ones between 0 and 2pi

1.3... + pi is less than 2pi

but 1.3 - pi is less than 0 so its not valid for this

and 1.3 + 2pi is greater than 2pi so its not valid answer

[cedar kiln]

@hard pawn Has your question been resolved?

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[vestal bear]

!status

[cedar kiln]

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

[fallow oracle]

1

[bitter obsidian]

i think u should start by simplifiing the fractions

[livid hound]

first step would be to sub in your values

[fallow oracle]

to decimals?

[livid hound]

no

keep fractions as they are

don't feel the urge to convert everything to decimals

[fallow oracle]

so 2 x (4/5) x (2/7) + root (4/5)(2/7^2)

[fallow oracle]

okay

[livid hound]

missing ()

also avoid using x for multiplication

[fallow oracle]

where?

[fallow oracle]

so 2(4/5) (2/7) + root (4/5)(2/7^2)

[livid hound]

$2\br{\frac 45}\br{\frac 27} + \sqrt{\br{\frac 45}\br{\frac 27}^2}$

[wraith dagger]

ℝamonov

[fallow oracle]

yes

[livid hound]

2(4/5) (2/7) + root**((4/5)((2/7)^2))**

missing those bolded () to clearly reprsent that in text

[fallow oracle]

oh yes

that's my mistake

\frac{16}{35}+\frac{4}{7\sqrt{5}}

is that the answer

[bitter obsidian]

u must put ur answer in a single fraction

[livid hound]

$\frac{16}{35} + \frac{4}{7\zqrt{5}}$

[wraith dagger]

ℝamonov

[fallow oracle]

\frac{4\left(4+\sqrt{5}\right)}{35}

so is that right

[vestal bear]

use $

[livid hound]

yes, whether you factored out that 4 is optional

[vestal bear]

$\frac{4\left(4+\sqrt{5}\right)}{35}$

[fallow oracle]

$\frac{4\left(4+\sqrt{5}\right)}{35}

[wraith dagger]

Arctic

[fallow oracle]

thanks

.close

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[short crater]

I have absolutely no idea where I am going wrong I keep getting 3.66460663 but since i round to two decimals i get 3.37 which is still wrong

[royal loom]

are you sure your answer is also in radians

[spare trout]

3.664 rounds to 3.66

[short crater]

Apologies i get 3.3664

[short crater]

Im actually so dumb

thank you

[royal loom]

np

[short crater]

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[cedar kiln]

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[short crater]

.reopen

[cedar kiln]

✅

[short crater]

So I got 0.00102548 and I dont know if i just risk it and put 0

and this is my last try on the question

[royal loom]

no don't do that

the angle in between them is very small

so it could be correct

but I'll check first

[short crater]

Thank you so much

[royal loom]

you can also just visualize the angle to give yourself a sanity check

something like this could never hurt

[royal loom]

this is wrong

[short crater]

so what would it be then?

[royal loom]

What formula are you using

$\cos{(\theta)}=\frac{uv}{|u||v|}$?

[wraith dagger]

austinu

[short crater]

That one

[royal loom]

ok

plug in

and then take arccos

the result is not 0.001

tell me what you got for u dotted with v

and for the magnitudes

[short crater]

the magnitudes I got u=sqrt29 and v=sqrt10

for the dotted I got 17

[royal loom]

okay so

$\cos{(\theta)}=\frac{17}{\sqrt{29}\sqrt{10}}$

[wraith dagger]

austinu

[royal loom]

now take arccos of the right hand side to find theta

[short crater]

so arccos=17/sqrt29 times sqrt10?

[royal loom]

what

[short crater]

wdym right hand side

[royal loom]

^

[short crater]

Oh wait so you want me to put all that into an arccos?

[royal loom]

if $$\cos{(\theta)}=\frac{17}{\sqrt{29}\sqrt{10}}$$ Then $$\arccos{\cos{(\theta)}}=\arccos{\frac{17}{\sqrt{29}\sqrt{10}}}$$ Then $$\theta=\arccos{\frac{17}{\sqrt{29}\sqrt{10}}}$$

[wraith dagger]

austinu

[royal loom]

note that in all cases arccos(cos(x)) is not necessarily equal to x, but in this case it doesn't matter

[short crater]

0.05875582?

Okay and then that becomes my radians?

[royal loom]

Do you understand the difference between degrees and radians?

[short crater]

not really to be completely honest I always thought it was just another way to put degrees in

or vice versa

[royal loom]

well, it is another way of representing an angle, yes

but

[royal loom]

if you put arccos(whatever) into a calculator in radians mode

then the result will tell you the degree in radians

it doesn't "become" the radians

like that is just the angle

if you wanted to see if your result makes sense

you could confirm that 0.0587... radians is about 3 degrees

and then you can look back to your picture

and be like

yah that's about 3 degrees

[short crater]

Ohhhhhh alrighty that makes alot more sense actually thank you

[royal loom]

mhm

[short crater]

.close

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[night forum]

I have tried to study the functions my points made, like when they intersected and etc., to maybe combine some expressions to get the value of the parameters (I checked those with Wolfram Alpha again and again, of course).
The problem is, I got an expression for the x value of the inflection point of f(x), and an expression for the intersection point of f(x) and f'(x).
I got a domain of both of them and when I combine them, there is no solution. I am right in both expression, I can see with my eyes that these 2 conditions can co-exist, but somehow the equation says they can't. Where am I wrong?
f(x) = a/(1+be^(-cx+d))
An expression for the x value of the inflection point of f(x): 0.15 ≤ ln(be^d)/c < 0.2
An expression for the x value of the intersection point of f(x) and f'(x): 0.85 < ln(b * e^d * (c-1))/c < 0.9
f(x): https://imgur.com/a/opSSyLM
integral of f(x): https://imgur.com/a/5Ik7HxB
f'(x): https://imgur.com/a/fSm7ov9

[cedar kiln]

@night forum Has your question been resolved?

[night forum]

<@&286206848099549185>

[cedar kiln]

@night forum Has your question been resolved?

[flat oracle]

can you explain what the problem is again? i cant really understand

you got f(x)=a/(b+e^(-cx+d)) and you have found (somehow) an expression for the inflection and intersection point

[night forum]

Indeed

I will try to explain it

f(x)=a/(b+e^(-cx+d))
The x value of it's inflection point is when f''(x)=0

So I found the second derivative of f(x), made it equal to zero, and got the value of x (which is ln(b*e^d)/c)

Now from the points I have, I can see that the inflection point is between 0.15 (or equal to) and 0.2

Same for the intersection, calcualted the first derivative of f(x), made it equal to f(x), got the value of x, and from the points I can see it's between 0.85 and 0.9

ln(b * e^d * (c-1))/c

Now I can get a range of values for d from both equation:

ln(e^0.15c/b) ≤ d < ln(e^0.2c/b)

ln(e^0.85c/b(c-1)) < d < ln(e^0.9c/b(c-1))

But there is no value for d that can satisfy both ranges of values

And I can't understand how

[flat oracle]

from this you conclude that the inflection point is between 0.15 and 0.2?

[night forum]

Not from this

Inflection point of f(x) = Extrema point of f'(x)

This is f'(x)

You can see pretty clearly that there is a minimum point

Maybe it's on 0.15, and it can be after that, but it's before 0.2 for sure

[flat oracle]

i dont see any 0.15 or 0.2 on the numberline...you do that just by looking at it?

[night forum]

Right because it's really zoomed out, but I have the exact coordinates of every point, so I can check

[flat oracle]

alright

[flat oracle]

i dont think this is completely true

if the minimum you are talking about (which i cant see) is at the edge of the interval (or whatever than means for a function that consists only of points) then f'' isnt necessarily defined at that point

and also

[night forum]

Wait I will draw it

Ignore my drawing skills

The extrema can be on this exact point

But it's also can be a little bit after that

Like that

[flat oracle]

yeah that is an inflation point ok

[night forum]

You mean inflection?

[flat oracle]

yeah lmao

[night forum]

It's the extrema of f'(x)

And inflection of f(x) yes

[night forum]

and also?

[flat oracle]

what i was going to say is that if f'(x) has a minimum or maximum at the edge of the interval then it isnt an inflection point

but thats not the case so

[night forum]

Oh right

But this is not the case

Do you have any idea why I get this weird situtaion? Of no solutions for d?

I must be wrong somewhere, because there is clearly a solution

[flat oracle]

so this is 0

considering a,b,c,d arent 0

a=0 and c=0 dont make sense anyway

so as you said x=ln(be^d/c)

so e^0.15<=be^d/c<=e^0.2

[night forum]

They aren't

I can tell you more, all of them are positive except of a which is negative

I can tell by how they change the shape of the function

[night forum]

How you got this?

[flat oracle]

0.15<=x<=0.2

[flat oracle]

i dont see how you got this for the intersection point though

[night forum]

I asked wolfram alpha

He said so

I trust him

Try it yourself

[flat oracle]

i have been

i get this

oh it s the same alright

[night forum]

Is it?

[flat oracle]

how do you get the expression for d from this?

yeah it is

[night forum]

I am sorry I have a feeling I am really weak at log rules

[night forum]

Again, wolfram alpha

I wrote

0.85 < expression < 0.9

And it gave it to me

[flat oracle]

this is =ln(b/(c-1)*e^d)/c so ln(ab)=lna+lnb

[night forum]

Wtf

I don't get it

How

I will trust you that you wrote the same but in different form

But how you got a solution and I didn't?

[flat oracle]

no idea

but u got 3 variables it s kind of weird if you couldnt just choose some for it to work

let me try to find some values to make it work ig to make sure we got it

[night forum]

Imma do the same

[night forum]

Hold upppppppppppppp

You didn't wrote the same function

Oh wait

[flat oracle]

it s the same

[night forum]

Or maybe I am dumb?

[flat oracle]

i checked

[night forum]

Oh fuck

[night forum]

I wrote the original function I was trying to work with

[flat oracle]

i was confused too u just multiply up and down with e^2cx

[night forum]

I changed it a little but I didn't write it here

So the function and the range of d don't match in what I wrote

Ok fine

[night forum]

Wait where the range of c came from?

Something during the calculation?

I mean c must be in these intervals or it's just some of the solutions?

[flat oracle]

it just gives values for when the 2 inequalities are true at the same time

it says we got 3 cases

b>0 1.47<c<1.49 and the restriction of d

[night forum]

So they are true at the same time just in those values of c?

[flat oracle]

and then the other case for c=1.49...

well yes it seems so

if c is below 1.49759 or above 1.52205 we dont have any solutions

[night forum]

Understandable

[flat oracle]

for c=1.48 and b=1 there are d so that both are true

[night forum]

Why you wrote b=1 and c=1.48? just as an example?

[flat oracle]

yes

[night forum]

Ohhh ok

[flat oracle]

so yeah everything makes sense

[night forum]

Yea

[flat oracle]

weird problem

[night forum]

You just did some kind of witchcraft and wolfram decided to work

[flat oracle]

you gotta check if wolfram is doing what you are telling it to do...most times it s pretty easy to realise something is wrong

but for example wolfram may decide that a is dependant on x or something crazy

and then you gotta realise that it s doing that and type the question in an other way

[night forum]

I will try

Thank you so much

Em

How do I close it?

[flat oracle]

np

.close

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[ember geyser]

can someone explain to me why 8 isnt correct?

[dreamy sleet]

if t is 8

then t + 1 is 9

try drawing a triangle with sides 8, 9, and 17

because that’s what triangle D would look like if t were 8

[ember geyser]

okay

now what?

[dreamy sleet]

have you done this?

[ember geyser]

wait nvm i get why

bc i made it equal not more

[dreamy sleet]

Exactly

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[ornate sequoia]

oh i'm sorry

didn't know that was a rule here

[night forum]

Good luck in that, a triangle like that doesn't exist

[dreamy sleet]

That was the point of my message

[night forum]

Opssie

Sorry for no understanding

Oh right he asked why t=8 is wrong

[cedar kiln]

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[steady fossil]

😭

[steady fossil]

Could anyone tell me how +infinity works in integrals?

[runic garnet]

Well

U rewrite it as

$\lim_{a \to \infty} \int_c^a {\frac 2{x^3}} ,dx$

[wraith dagger]

northsteve

[runic garnet]

Then u go from there homey

Then solve accordingly

[steady fossil]

whats with the c?

NVM

[runic garnet]

lol

[steady fossil]

these integrals fucking me up in statistics, why they gotta be so hard

[runic garnet]

U using speed of light in statistics?

[steady fossil]

c is just a parameter, but wolfram thinks its that

[silent finch]

that’s why there isn’t a solution

if you use a different variable it should show

[cedar kiln]

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[hard pawn]

is it really necessary to do all this working just to prove this identity?

[hard pawn]

since tan is sin/cos, could i just do

(sin/cos)^2 + 1 = (1/cos)^2
therefore tan^2 + 1 = sec^2?

i dont see how the first and second steps are necessary

[south tundra]

Sure but you then need to show why (sin/cos)^2 + 1 = (1/cos)^2

[hard pawn]

oh

so thats why i have to use the first property?

[south tundra]

Yes you start with sin^2 + cos^2 = 1

[hard pawn]

the one that says cos^2 + sin^2 = 1

[hard pawn]

so for the 3rd property, will we also have to use this property to prove it?

why choose the first property to prove this tan property(the one in the ss)?

also, sorry if im asking too many questions

i need to get a solid understanding

[south tundra]

Yeah

[south tundra]

In a proof you usually start with what is known/assumed to be true

And the fastest way here would be to start from the Pythagorean identity

[hard pawn]

i see

[south tundra]

Cuz you just divide by cos^2 and simplify

[hard pawn]

wait

[sand cradle]

https://math.stackexchange.com/a/2688641

[hard pawn]

in the first step, whyd we only use cos^2 + sin ^2 for the left side, and not the entire first property? (cos^2 + sin ^2 = 1)

is it because the cos and sin are relevant to the equation, and not the 1 from the first property?

[south tundra]

Not sure what you mean by that

[hard pawn]

ok so like

[sand cradle]

If sin^2(x) + cos^2(x) = 1 can be used

[hard pawn]

in the first property, the cos and sine squared is equal to 1. so im wondering why we didnt bring back that 1 into the left side. as in, the '1' from the first property isnt used to prove the 2nd property

[hard pawn]

ill take a look at this

[south tundra]

The 1 is still there though, it's just divided by cos^2

a = b -> a/c = b/c (for nonzero c)
sin^2 + cos^2 = 1 -> (sin^2 + cos^2)/cos^2 = (1)/cos^2

[hard pawn]

wait

[hard pawn]

so is the 1 in the 3rd step the same '1' from the first property?

[south tundra]

If you are talking about the second '1' then yes

The first '1' is just a result of cos^2 cancelling itself out in cos^2/cos^2

[hard pawn]

OHHH

I get it now

the 1 has been there all along

thank you sir

or ma'am

.close

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[versed fjord]

help

[sinful ocean]

x^3(x-7)(x+7)

[crimson sedge]

A polynomial of the form [
\map f x = a_nx^n + a_{n-1}x^{n-1} + \hdots + a_1 x + a_0 ]
Is said to have degree $n$

[cedar kiln]

@versed fjord Has your question been resolved?

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[crimson sedge]

https://media.discordapp.net/attachments/1127334904229789727/1129752609482354791/image.png?width=343&height=37

[crimson sedge]

So i calculated the formulas and also resulted in x -5 v x 3 and i also understand the <= sign but i do not understand where the 0 <=x comes from

if the formula is needed to explain do let me know but it's kinda hard time typing it over and i've seen this coming back literally every time

[ornate sequoia]

@crimson sedge wait uh do we know f(x) and g(x) or

or what exactly do you need to do

[crimson sedge]

yea

ill send them i guess thats easier for sure

but it's not really related to this question

i'm more needing to understand as to how 0<=x<=3 came to be

i understand the x<=3

but not the 0<=x

[ornate sequoia]

well i suppose the 0<=x can only be explained if we have the functions right

[crimson sedge]

like i dont know how to calculate it or whatever how it works i dont knwo what it is at all

ive only worked with x<=3 v x<=-5 kind of format

but one sec

ill send u the formulas

[ornate sequoia]

okay

(do ping me 🙂)

[crimson sedge]

i alreayd got them

[ornate sequoia]

great

[crimson sedge]

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

g(x)=7x

i understand i put them like

x^3+2x^2-8x=7x

and then sqrt

x^2+2x-8=7

x^2+2x-15=0

i put them in the abc formula, got -5 and 3

i reckon i see the mistake rn

sqrt so - or +

it's 7 or -7

so -1 or -15

but i still don't really get where the 0 comes from

[ornate sequoia]

sqrt?

[crimson sedge]

yea i used that i think

i am told i can just make x disappear everywhere

[ornate sequoia]

sqrt (x³) is not x²

[crimson sedge]

1 at a time

if everything has an x

or is it dividing by x?

[ornate sequoia]

i get it but then ur doing x(x²+2x-15)=0

so ur saying
x=0 OR x²+2x-15=0

[crimson sedge]

Yes

wait i understand

[ornate sequoia]

I suppose in this case it's <= right? instead of =

[crimson sedge]

yes

so that's where the 0<=x from?

because 2 solutions

so solution 1 <= x <= solution 2

[ornate sequoia]

uh i'm not entirely sure, i'm doubting myself

[crimson sedge]

Hm okay yeah i've never got teached anything so self taught but this is kinda the problem of it

cause these tiny must know things i dont know

[ornate sequoia]

self taught? gl

[crimson sedge]

yea

it's not great idea

but it is what it is

no other options

i do have an exercise book with answers

they do list the steps

but the 0 is nowhere to be found in there either

[ornate sequoia]

urgh i wish i had a piece of paper i hate doing stuff in my head

[ornate sequoia]

if you have f and g, you can calculate where f is above (so >=) g or vice versa

i'm belgian i can read it dw

[crimson sedge]

oh that's great

uh yea so the 0 <= x <= 3 is the problem

i know now the 0 where it comes from okay

[ornate sequoia]

oh you can use the graph?

[crimson sedge]

but i dont understand how the heck it ended up in the answer at that place

[crimson sedge]

yea but i never do that

i dont know how to 😭

[ornate sequoia]

u don't know what website to use?

[crimson sedge]

Wdym what website to use

[ornate sequoia]

for graphs

[crimson sedge]

i dont know how to read graphs and get this as result

i got the graph right here

[ornate sequoia]

oh

[crimson sedge]

like i see the graph visually

but i don't know how it is useful

[ornate sequoia]

can you show me the graph if possible?

[crimson sedge]

ill need to take a pic with phone idk how good that is

i only got the answers online

the book is on paper

[ornate sequoia]

dw just try your best it'll be more than fine

[crimson sedge]

one sec

i took the pic

need to find this hcnanle on phone

[ornate sequoia]

the x³+2x²-15x = 0 stuff just calculates the places where f=g equals 0

[crimson sedge]

I know how to calculate all the numbers

like the 0 is just a stupid mistake

i couldve known

like finding the numbers is fine

[ornate sequoia]

f=g means that on the graph, they cross

[crimson sedge]

the answering format with ...<=x<=... i dont understand

[ornate sequoia]

you see? at x=5 the functions intersect

oh okay

[crimson sedge]

i do

and 3 crosses aswell

and 0 aswell

the exact same answers

[ornate sequoia]

so f<=g means that the y value for g is bigger than the y value for f

look at the functions between x = 0 and x = 3

[crimson sedge]

but if they intersect they are at the same y right

and x

[ornate sequoia]

yea

[crimson sedge]

Yea

it's below

and after 3 it's up

is that how th ey answer the format

0 <= x <= 3

[ornate sequoia]

you see that the blue line is under the red, so f<=g

[crimson sedge]

Yes

[ornate sequoia]

yeah this means
f(x) is less than or equal to g(x) when x is between 0 and 3

i could write it out better in dutch if it helps for u

[crimson sedge]

Yes

[ornate sequoia]

idt it's allowed here

[crimson sedge]

that's much better

[ornate sequoia]

okay

[crimson sedge]

it doesnt really matter i think

if it helps me better they wont care

[ornate sequoia]

sure

[ornate sequoia]

dit betekent dat de functie f(x) op de grafiek onder de functie g(x) ligt wanneer x tussen de 0 en 3 ligt

[crimson sedge]

Oke dat snap ik dus maar om op deze manier te beantwoorden heb je de grafiek ook echt nodig

visueel

anders zou je het niet kunnen beantwoorden?

[ornate sequoia]

ik denk dat je het zelf kan berekenen ook

[crimson sedge]

of moet je hem anders gewoon tekenen

[ornate sequoia]

nee je kan een verschilfunctie opstellen

verschilfunctie is f(x) - g(x)

dus per toeval is dat die x³+2x²-15x

snapje?

[crimson sedge]

Ja

[ornate sequoia]

en dan bereken je de nulwaarden

dus die -5, 0 en 3

[crimson sedge]

Dus vul je 0 in bij y en x?

[ornate sequoia]

nene wacht

nadat je die nulwaarden hebt

maak je een tekenverloop

ken je dat

[crimson sedge]

nee]

iig bij naam

[ornate sequoia]

wacht benje nederlander of belg want het heeft misschien een andere naam

[crimson sedge]

nederlands

[ornate sequoia]

je kijkt of de functie negatief is of positief in een tabel

[crimson sedge]

daar heb ik ook nooit iets van gehoord

[ornate sequoia]

uhhh ik kan proberen het te tekenen in m'n notes alsje wilt

[crimson sedge]

ja

misschien ken ik het wel

[ornate sequoia]

brb

[crimson sedge]

ik heb wel 5 jaar les gehad in wiskunde

havo diploma gehaald

maar moet nu certificaten vwo wiskunde b halen

[ornate sequoia]

heb geen idee wat dit betekent sorry hahaha

[ornate sequoia]

dit ding dus

[crimson sedge]

geen probleem

dus je vult de formule op je rekenmachine in bij de table functie

en dan krijg je dit

en met die nummers kan je zien wanneer het boven of onder is

[ornate sequoia]

ja dus de verschilfunctie v(x)

is dus f(x) - g(x)

[crimson sedge]

ja

dus g naar de andere kant gebracht basically

[ornate sequoia]

en als die positief is, dan is f(x) > g(x)

en als die negatief is dan is f(x) < g(x)

dus jij had f(x) < g(x) nodig

[crimson sedge]

Ja

[ornate sequoia]

dus je kan in die tabel zien dat dat is tussen 0 en 3, en alles onder 5

[crimson sedge]

aha

ik snap hem

[ornate sequoia]

hoe je dit doet:

1. nulwaarden berekenen en die erbij zetten ( de -5, 0 en 3)
2. een testwaarde (dus iets dat er tussen ligt, bij 0-3 bijvoorbeeld 1) invullen in de functie

[crimson sedge]

het antwoord geeft aan wanneer f(x)<=g(x) is

[ornate sequoia]

ja de <= of de < is gewoon of je de nulwaarden erbij neemt of niet

[crimson sedge]

en dit is wanneer x is tussen 0 en 3 of kleiner is dan -5

[ornate sequoia]

ja