help-19

[clever fjord]

SaltnPepper

[south plume]

go on..

[wispy geode]

given a recursive function, f, is there a way such that it can be rewritten to become recurive without requiring feeding the result of f back into itself?

$f(x) = 1.08x+100000 \implies f(f(f(\cdots f\relax(x)} \text{ where} f(x,n) = \overbrace{f(f(f(\cdots f\relax(x)}^{\text{n}}$

[clever fjord]

SaltnPepper
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wispy geode]

nvm i give up

.close

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[gilded bane]

How do i draw diagram for this

[wooden ibex]

!show

[odd edge]

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

[wooden ibex]

70.7m ?

@gilded bane

what is answer

[gilded bane]

50 root 2

Is ans

Ye urs is correct

[wooden ibex]

yeah same 50√2

[gilded bane]

Could you send the solution

[fair summit]

!noans

[odd edge]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[wooden ibex]

base level for amit and Deepak is same, but you have to calculate the vertical height of bird from base to compare how high is bird from deepak

bird height is 100m, and deepak is on 50m high so bird is 50m high from deepak. and you can use this 50m as perpendicular for triangle

did you understand @gilded bane

@gilded bane you there? stop copying answer and understand

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@gilded bane Has your question been resolved?

[odd edge]

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[gilded bane]

alright gott it thx

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[tiny granite]

.close

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[restive night]

What is the 1- poisson dist?

how does that work

It says at least 10 so more than 10 and above to infinity, I always thought it was 10 11 12, since it says average, but I guess not

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[weary compass]

What the fart does this even mean 😭

[vale vapor]

you get x=0

which is just a vertical line

[mystic saffron]

Through 0 there is a line

[weary compass]

Answer key and it still ain't make no sense

[vale vapor]

lol what

ohhh

[weary compass]

I know that

But lik

[vale vapor]

i see what they did

thats fkin stupid

[weary compass]

Theres 2 lines?? 😭

[vale vapor]

the left hand side is an equation

and the right hand side is an equation

we get y=3x+6 and y=6

you need to graph both those lines

[weary compass]

⁉️

[cold sage]

who does this

[prime falcon]

@vale vapor bro i need answer of my question in my room why do you leave me😭

[weary compass]

How do I know where to put the Y's

[amber schooner]

@vale vapor when did you get green

[vale vapor]

a few hours ago

[weary compass]

He's mine pal 😼

[vale vapor]

told you i would get it soon 🙂

[amber schooner]

you were right

[vale vapor]

ngl, i have no clue what youre asking, so i was hoping someone else would answer 🙂

[vale vapor]

the LHS is a line, the RHS is a line

but for some reason they put them equal to each other already

[weary compass]

So replace the size with a Y?

[vale vapor]

thats hwoy ou find their intersection

[weary compass]

Six

[vale vapor]

no

[weary compass]

😔

[prime falcon]

And you call yourself helpful 😔

[vale vapor]

\begin{align*}3x+6=6\3x+6\quad\quad\quad 6\end{align*}

[vale vapor]

!volunteers

[odd edge]

Helpers are just people volunteering their time to help you. Be polite and patient.

[clever fjord]

Bonk

[vale vapor]

that left one is an equation

so y=3x+6

and the right is an equation

y=6

[weary compass]

Okay where do you see y=6

[vale vapor]

right here

[weary compass]

Ohhh

OHHH

Okay so basically

Y= 3x + 6 AND 6?

[vale vapor]

if i give you \begin{align*}y&=3x+6\y&=6\end{align*} how would you find their intersection?

[weary compass]

Graph it 🤑🤑

[clever fjord]

Bonk

[vale vapor]

exactly

and thats what they want you to do

but instead of giving the two lines separately

they already write it in one equation

which makes absolutely no sense

[weary compass]

Why my teacher do that to me 😓😥

[vale vapor]

but like, look at the second question

[weary compass]

How do I show -17

It's only 10x10 graph

[vale vapor]

one line is 2x-17

and the other is x-10

[weary compass]

Yeah

[vale vapor]

why would you need to start at -17?

[weary compass]

The graph only goes to -10

[vale vapor]

you can just plug in x=5 for example

[weary compass]

Y=2x-17

woah

You're like a genius

[echo loom]

is there somone who understand frensh

[vale vapor]

it does go off the plot

but that counts for almost every function right?

[weary compass]

Okay thank you 🤗

.close 🤑

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[umbral epoch]

Couldn't this be written as (ln(-5x)/9) + C ?

[umbral epoch]

From log rules

[wanton bison]

which log rule does that

[worn sandal]

no because the square is on the outside of the log

$\ln(a^2) = 2\ln(a)$ but $(\ln(a))^2 \neq 2\ln(a)$

[clever fjord]

Ari

[umbral epoch]

ohh

okay i see

thank you :)

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[umbral epoch]

I got marked incorrect for this

[umbral epoch]

and the answer is -1/12 ln|cos(12x)| + C

[sweet needle]

u wrote sec 2x

[umbral epoch]

are these not the same thing ???

oh

damn me

thank you 😭

[sweet needle]

lmao

yw

[umbral epoch]

but just to make sure these are the same?

[sweet needle]

yes

[umbral epoch]

alrighty

thank you

!

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[oblique heart]

I need help with this exercise, it wants to calculate the determinant.

[odd edge]

@oblique heart Has your question been resolved?

[oak crown]

use the “box” method

here, i’ll see if i can draw it out

(basically the first part is that diagonal you have

that would be your first determinant

so $a_1a_2a_3...a_n$

[clever fjord]

;(

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[dry sparrow]

https://www.youtube.com/watch?v=ePWlt72K9zM&list=PLPyAO0bbCyXc5XT6NhSgQjCnrpdYV1Dma&index=8

[dry sparrow]

funny?

or no?

[sand horizon]

?

Whats your math question ?

[quasi sparrow]

don't open help channels to spread videos

.close

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[uncut plaza]

hi, not sure what to do here ;-;

[strange aspen]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[uncut plaza]

1

[tepid pelican]

If you compute f^-1(33), you will have 2 points of g

(which will allow you to find the line passing through them -> g(x))

[limber glade]

g(3) = 20 give you one linear eq in a and b

and after calculating f^-1(33) and making it equal to g(1), you get another one

solve those 2 equations to get a and b

[uncut plaza]

i'm confused

[limber glade]

ok if g(x) is ax+b

then what is g(3)

[uncut plaza]

a(3)+b so 3a +b

[limber glade]

yeah

and set that equal to 20

so 3a + b=20

[uncut plaza]

why 20?

[limber glade]

cuz g(3)=20

that's given

gotcha?

[uncut plaza]

oh

[limber glade]

now

can you find the inverse fuction of f(x)

[uncut plaza]

f^-1(x) = (x-3)/5

[limber glade]

yeah

now in the same way calculate f^-1(33)

[uncut plaza]

is it equal to a+b?

[limber glade]

yeah

so a+b=30/5

now you got a pair of linear equations in a and b

so you can solve for them

[uncut plaza]

wait where did the '30/5' come from?

[limber glade]

f^-1(33)

[uncut plaza]

can u be a bit more specific? :,)

[timid sky]

plug in x = 33 here

[uncut plaza]

ohhh

[limber glade]

just like how you did with g(3)

[uncut plaza]

so a= 7 and b= -1 ?

[limber glade]

yeah

[uncut plaza]

ohh

ty

.close

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[limber glade]

open your own channel

[limber glade]

welcome and good luck for your mocks

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[amber plank]

hello guys.. i have another question

[amber plank]

?????

could someone please explain why?

[nimble blaze]

unit circle

[amber plank]

what?

doing the same operation but using 3.14 instead of pi gives -1

(-1.99..)

[nimble blaze]

2 * -1 = -2

[amber plank]

ooh

wait

you telling me that firstly is doing the cos(pi) and then multiplying it by 2?

i want to think it actually does it..

[nimble blaze]

yes

[amber plank]

alright thank you

i guess it does the same for the sen?

[nimble blaze]

yes

[amber plank]

thanks

that's everything for now

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[deep orbit]

Difference between bar chart and histograph?

[vale vapor]

a histogram has frequency on its y-axis, that doesnt have to be the case for a bar chart

[deep orbit]

?????

show example

[vale vapor]

also the x-axis is divided into bins, the order of the bars has meaing

histogram

its kinda like all histograms are bar charts

but not all bar charts are histograms

[shut ember]

can you*

[deep orbit]

no

they dont deserve anything

can u make two charts with the same theme but in histograph and bar?

[shut ember]

..

thats rude

?

theyre helping u

<@&268886789983436800>

We arent working for u

we r helping u

[deep orbit]

its my ticket..

[shut ember]

bc u dont know whats the diff between barnchart and histograph

ok bye

[oak crown]

no need to be rude tho

[deep orbit]

.close

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[oak crown]

it’s not just you coming here for your time

[shut ember]

fr

[vale vapor]

!volunteers

[odd edge]

Helpers are just people volunteering their time to help you. Be polite and patient.

[oak crown]

we could be doing actual things

[lone elbow]

https://www.cuemath.com/questions/what-is-the-difference-between-a-bar-graph-and-a-histogram/

[oak crown]

but we decide to do this out of our generosity

so don’t be rude when we are already “wasting” our time

tl;dr: you asked a question, someone attempted to help you, but you didn’t receive it, wasting both you and the helpers time

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[stiff sluice]

I cant do number 85. Where do I go from here?

[vale vapor]

did you also put the 1/h on the 1/x^2?

[mellow tundra]

u got the sign wrong

[stiff sluice]

care to elaborate?

[stiff sluice]

the whole thing is over h so surely when u subtract the fraction by multiplying yada yada yada then u have a combined denominator u can multiply by h ?

[wooden gorge]

:p Bonk mentioned another error in your step 1 ._> plz be sure to correct that as well

[stiff sluice]

ahh i see... brb

[stiff sluice]

how do I correct that? do I just wrap the whole thing in braqckets again to signify the entire thing is multiplied by 1/h

[wooden gorge]

Now it fien

[stiff sluice]

so where do I go from here? do I just expand the denominator then take a h out?

[wooden gorge]

Are you maybe ignoring this?

[mellow tundra]

u missed the h in the deno again lol

[stiff sluice]

how

oh yeah

true ngl

my bad

[wooden gorge]

._>

End result should be -(2x + h)/[x²(x² + 2xh + h²)]

[mellow tundra]

[stiff sluice]

ohhhhhhhh

I see it now

I see the light

so annoying how such simple mistakes can rlly stump u lol

thanks guys

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[ionic raft]

Hi everyone!

Q: [Water in a 50 meter high waterfall is moving 11 m/s at the top and 5 m/s after it has dropped. By how many degrees does the water temperature rise?]

I only know of a formula for positional energy which is:
Q = mass * gravity * height

There is no mass given for the question, and there is also information about the water flow speed which im wondering about the importance of.

[strange aspen]

i guess the info of the water speed could relate to kinetic energy but i'm not sure how to use it for this question

[odd edge]

@ionic raft Has your question been resolved?

[ionic raft]

Youre right, most likely. Not sure how to connect kinetic energy with positional energy though

[strange aspen]

you can try energy conservation law

[odd edge]

@ionic raft Has your question been resolved?

[ionic raft]

Sorry, im still uncertain how to approach this. This is a math discord afterall, ill be patient and try to get more specific help on a physics discord.

Thank you for trying to help me though! I really appreciate it :)

[odd edge]

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[obtuse saddle]

can some one please explain the pattern here I feel like my teacher didn't explain it well

[obtuse saddle]

also an example would be apperaciated

like tell what i^145 would be

[quasi sparrow]

do you understand these 4

[obtuse saddle]

I understand why i^2 is -1 but that's it

[quasi sparrow]

anything^1 = anything

applies to i as well

and anything^0 = 1

did either of those 2 confuse you

[obtuse saddle]

uh no cause anything^1 = anything is the same thing as saying 1^1 is just 1 and anything to the power of 0 is just anything cause your not modifying it

[quasi sparrow]

no

i recommend ignoring i and just learning what exponents are

[obtuse saddle]

I understand i^1 but 0 is confusing me

I should of just said that

idk I tried to put it into my own words lol

[steel spire]

for starters, you know x^a * x^b = x^{a+b} right?

[obtuse saddle]

yes

[steel spire]

so we can sorta figure out what x^0 is this way
x^a * x^0 = x^{a+0}

[obtuse saddle]

yeah

[steel spire]

x^a * x^0 = x^a so multiplying by x^0 is the same as multiplying by 1, does that make sense?

[obtuse saddle]

yes it does

[steel spire]

cool

[obtuse saddle]

I still don't understand the pattern though

[steel spire]

where are you stuck on the list

[obtuse saddle]

I just don't understand like why would i^16 = 1 or why i^19 would = -i

[steel spire]

gotcha, do you know what i^4 is

[obtuse saddle]

without looking no

can you try to explain to me what like i^27 would be and what did you do to figure that out

[steel spire]

there are multiple ways to figure it out, one way is to factor the exponent or write it out completely

as i * i * i * i or (i^2)^2

[quasi sparrow]

figure out i^4, i^5, i^6, i^7 first

[obtuse saddle]

if i is 1 then i x i would be i^2 which is -1 and -1 x 1 is -1 and that is where I am stuck

I am trying to solve i^4 here btw

[steel spire]

"if i is 1" no i is never 1 it's a separate thing altogether

[obtuse saddle]

but on the chart it says i^0 is 1 and you said that i^0 is just i

[steel spire]

no i^0 = 1 I never said i^0 is i

i^1 is i though maybe that's the confusion?

[obtuse saddle]

sorry thats what I meant

[steel spire]

you're getting on the right track I think

I'll help show how to work out i^3

i^3
expand it out:
i * i * i
you can group i*i to make -1
-1 * i
so now you have
-i

[obtuse saddle]

is there a way to get this instantly cause my teacher called out a girl and said what is i^303 and she answer within like 2 seconds

[steel spire]

there is but we have to walk before we can run

[obtuse saddle]

ok

[obtuse saddle]

this makes sense ill try i^4

[steel spire]

cool, yeah once we know what i^4 is it'll help us understand the trick

I have to step away for a bit but I'll be back but maybe riemann can keep helping too in the mean time or anyone else

[obtuse saddle]

ok i*i is i^2 which is -1 and -1 x i is -i and this is where I am stuck I don't know what just i is but the next step is -i x i

[quasi sparrow]

just keep doing what merosity showed you and answer the rest of these powers

specifically the part where he showed you i^3 = -i

[obtuse saddle]

I looked up what i is and it says the square root of -1 so then -i must be -(-1) right ?

[quasi sparrow]

yea that's equivalent to this part after you take square roots of both sides

why do you think $-i = - (-1)$ ?

[clever fjord]

riemann

[obtuse saddle]

cause if i is the square root of -1 then -i must be -(-1)

[quasi sparrow]

$i = \sqrt{-1}$ yes

[clever fjord]

riemann

[quasi sparrow]

how does that imply $-i = - (-1)$

[clever fjord]

riemann

[quasi sparrow]

what algebraic operations did you do

[obtuse saddle]

I'm gonna be honest I don't know but what I do know is that if I have i and I know that i is -1 and then I have -i then -i must be the same exact thing as i but with just a - sign

[quasi sparrow]

you need to stop saying $i = -1$

[clever fjord]

riemann

[quasi sparrow]

do you know what this symbol is? $\sqrt{}$

[clever fjord]

riemann

[obtuse saddle]

yes square root

[quasi sparrow]

e.g. $\sqrt{4} = 2$

[clever fjord]

riemann

[obtuse saddle]

yep

[quasi sparrow]

so why do you keep leaving out square root

[obtuse saddle]

cause I don't know how to type it in

[quasi sparrow]

omitting is worse

[obtuse saddle]

huh ?

[quasi sparrow]

you can write this as sqrt(4) = 2

[obtuse saddle]

oh ok

[quasi sparrow]

full words is also fine

[obtuse saddle]

got it

[obtuse saddle]

so this is wrong then if i is the sqrt(-1) then -i must be -sqrt(-1)

[quasi sparrow]

right

$i = \sqrt{-1}$ multiplying by -1 to both sides gives $i (-1) = -i = (-1) \sqrt{-1} = -\sqrt{-1}$

[clever fjord]

riemann

[quasi sparrow]

and $-1 = i^2$ so both sides also equals $i (i^2) = i^1 (i^2) = i^{1+2}$

[clever fjord]

riemann

[quasi sparrow]

$i^3 = -i$ should follow after that

[clever fjord]

riemann

[odd edge]

@obtuse saddle Has your question been resolved?

[odd edge]

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[magic quail]

what do i search up to find a video to help me do this

[magic quail]

Elementary Row Operations ?

[worn sandal]

yes

or you don't really need to here

You should be able to tell just from looking at the matrices

[magic quail]

how would i show my work

just write them out?

like hold up

1x1 + _ 8x3 + (-5x4) = 6
_ + x2 + 4x3 + (-9x4) = 3
_ + _ + x3+x4 = 2

[worn sandal]

Well I think it's easiest to start with C

can you explain which one C is

[magic quail]

can a linear equation = 0

[worn sandal]

wdym

[magic quail]

like is it a linear equation if it equals to 0

the equation

[worn sandal]

yes

But the main thing I'm looking at in the matrix in (c) is the 3rd row

0 0 0 1

what does this tell you

[magic quail]

theres no solution..?

or

idk i think i see

like

its just =1

[worn sandal]

Right

Because $0x_1 + 0x_2 + 0x_3 = 1$ makes no sense

[clever fjord]

Ari

[worn sandal]

So (c) has no solution

[magic quail]

so because the last row makes no sense the whole matrices has no solution ?

how would i show my wokr for that

do i write out

all the equations

and like show that the last row isnt a linear equation

[worn sandal]

Yes

We say the matrix is "inconsistent"

you don't need to explain this that much, just point out the 3rd row

[magic quail]

im not really understanding

[worn sandal]

Any matrix with a row
0 0 0 c
where c is nonzero
can't have a solution

[worn sandal]

yes this is correct

So it doesn't matter what the other rows are

0 0 0 1 forces no solutions

[magic quail]

how would i solve b since it doesnt have an obvious row like c

[worn sandal]

have you learned what an "upper triangular matrix" is

[magic quail]

no i dont think so

unless thats what my professor is explaining in the video and im just not understanding]

[worn sandal]

ok that's fine

Rewrite each row in the matrix as an equation

i.e convert the matrix to a system of equations

[magic quail]

like this?

[worn sandal]

right

And you can see that this gives you a unique solution

[magic quail]

what does unique solution mean

OH

wait

so

because x3=5 i plug that back into the 2nd equation

and solve

like

wait hold up i think its piecing together in my brain

[worn sandal]

right

[magic quail]

actually

idk

i thought i was onto something

[worn sandal]

I think you're right

have you learned the term "pivot"

in linear algebra context

[magic quail]

probably but i dont recall

[worn sandal]

A pivot is the first non-zero entry in a row of a matrix

You could also argue that in this matrix, you have 3 rows and 3 pivots

each row has a pivot means that there's a unique solution

You can also use this logic for the matrix in (a) (different answer but you can use pivot argument)

[magic quail]

im not picking up what ur putting down

[worn sandal]

if "pivot" doesn't really ring a bell you don't need to worry about it for now

but it'll probably come up later

[magic quail]

so does b have many solutions

[worn sandal]

No

It has only one

[magic quail]

oh

[worn sandal]

which is the one you found here

Have you heard of linear independence

[magic quail]

no

[worn sandal]

o

nvm then

Let's do (a)\

[magic quail]

wait

ik how i solved b

but

is it one solution because

the like

first equation

when plugging everything in

it just equals back to 7

7=7

-37-3(-8)+4(5)=7

[worn sandal]

That means it's a solution

you know you have a unique solution because Row 3 fixes the value of x_3

Row 2 is an equation in terms of x_2 and x_3. Since x_3 is fixed, x_2 is now fixed as well

And Row 1 fixes x_1 with similar logic

[magic quail]

this one seems tricky

how do i go about this

[worn sandal]

So here, we have 4 variables, but only 3 equations

this means there isn't a unique solution

[magic quail]

so is this solveable

or would it be infinitely many solutions

and youd know that just by knowning that is has 4 variabls and only 3 equations

[magic quail]

this because i can plug in different numbers and get the answer to the equation?

[edgy junco]

since number of columns greater than number of elements per column, it's automatically linearly dependent

I'm pretty sure this is something in the invertible matrix thereom

[odd edge]

@magic quail Has your question been resolved?

[odd edge]

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[magic quail]

what does it being linear dependent mean

[steady tide]

if a set of vectors whose length is higher than the dimension of the space it lives in, it is guaranteed to be linearly dependent

there are three rows, so the dimension of column space of the matrix is maximally 3, but there are 5 columns vectors

[magic quail]

so with that info how do i know how many solutions

[steady tide]

since the matrix's column vectors are linearly dependent, you know that the matrix does not have unique solution; that is, it can only have no solution, or infinite solutions

by the glance of it you can already find one solution, thus it's the latter

[steady tide]

formally, a set of vectors v_1, ..., v_n are linearly dependent if and only if at least one of v_i lives in the span of v1, ..., v_i-1

another definition is that for real a_1, ..., a_n, the set of vectors v_1, ..., v_n is linearly independent if and only if

a_1v_1 + ... + a_nv_n = 0

has no solution other than a_1 = ... = a_n = 0

[odd edge]

@magic quail Has your question been resolved?

[magic quail]

i am not understanding

[odd edge]

@magic quail Has your question been resolved?

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[worthy falcon]

What am I doing wrong?

[steady tide]

,rccw

[steady tide]

e^(x + y) is not e^x + e^y

[worthy falcon]

Ok let me try again

[steady tide]

likewise, ln(x + y) is not ln(x) + ln(y)

[worthy falcon]

Wait rlly?

[steady tide]

yes lol

[viscid flint]

yes, e^2 is not equal to 2e

[steady tide]

e^(x+y) = e^x * e^y

[worthy falcon]

Oh it’s ln(x)*ln(y) right?

[steady tide]

no

[viscid flint]

,tex .exp rules

[steady tide]

it's not the same for ln

[viscid flint]

,tex .log rules

[clever fjord]

hayley, who shakes the world

hayley, who shakes the world

[worthy falcon]

Ok

what do I do next from here

[steady tide]

sub y(1) in

[worthy falcon]

Got to this step

I can’t ln from here though so what do I do

[steady tide]

ok, now solve for c

[worthy falcon]

how? I can’t ln because other side is negative

[steady tide]

therefore no solution

[worthy falcon]

uhh I think there’s supposed to be an answer to this question

let me check

C = -3/e^2

then we plug that back in

How do we get to that step? @steady tide

[steady tide]

can you show me the original question

[worthy falcon]

Number 8

,rccw

[steady tide]

oh i see the problem

[steady tide]

what's the integral of 1/y?

[worthy falcon]

lny

[steady tide]

wrong

[worthy falcon]

ln(y)?

[steady tide]

wrong

[worthy falcon]

ln(|y|)?

[steady tide]

the answer is ln|y|

yes

[worthy falcon]

what’s the difference

[steady tide]

the difference is that this guarantees the integral of 1/y is defined for all nonzero y

now do it again

[worthy falcon]

Ok

I think this is still wrong

Because it’s supposed to be C = -3/e^2

What am I doing wrong? @steady tide

[steady tide]

,rccw

[steady tide]

uh, what are you doing?

how did you arrive to the second step?

[worthy falcon]

plugged in y(1) = -3

[worthy falcon]

From this

@steady tide uhh r u there

[steady tide]

i don't have paper so it takes a while to do calculations

[worthy falcon]

No worries!

[steady tide]

my guess is that you need to bring ln to the other side, so that you get e^C*e^(2x^2)

[worthy falcon]

is this easier to follow along with instead

i think he gone

.close

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[sweet ferry]

How to prove that

[boreal crag]

the euler macheroni constant involves a limiting process that looks more like an integral from 1 to infinity (although maybe not exactly)

and i know logs generally respond well to the substitution u=1/x

so my first thought would be to try the substitution u=1/x

[odd edge]

@sweet ferry Has your question been resolved?

[sweet ferry]

Not working well...

[odd edge]

@sweet ferry Has your question been resolved?

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[mint igloo]

what did i do wrong?

[prime pond]

You made two important mistakes while graphing the piecewise function:

Incorrect endpoint for the first piece: You have f(x) = -3 for x ≤ -2. So, you should have had a closed circle-or a filled-in dot-at x = -2. You have an open circle there, indicating that -2 is not included in this part of the function.

Incorrect treatment of the second piece at x=2: The function f(x) = 2x + 1 is defined for -2 < x < 2. That should be open circles (or unfilled dots) at both x = -2 and x = 2. You correctly have an open circle at x = 2, but you should also have one at x=-2 where this line segment begins. This part of the function should have an open circle at y=-3 when x=-2.

Here's how it will look on the graph:

For x ≤ -2, it is a horizontal line at y = -3, with a closed circle at (-2,-3);
for -2 < x < 2, it is a line with slope 2 and y-intercept 1, with open circles at both (-2,-3) and (2,5).
x > 2: A horizontal line at y = 2 with an open circle at (2, 2).
These two corrections will make your graph of the given piecewise function accurate.

[latent scaffold]

What you did looks good. Maybe they explicitly want a filled in point at (-2,3)

[mint igloo]

i figured it out, the graph is right it just wants a infinate line on the far right and i accidentally used a closed line

[latent scaffold]

Oh good

[mint igloo]

it doesnt want dots on connected lines

TT

[latent scaffold]

Also this looks like AI. As you can maybe tell, what it’s saying is off. It’s hallucinating mistakes where there isn’t.

Try and refrain from helping people out with AI answers because AI sucks at maths and just causes more confusion in a majority of cases

[mint igloo]

yeahhh i was gonna say

[latent scaffold]

That’s good sometimes those online softwares are finicky good job finding out what it wanted

[mint igloo]

i spent like a hour to figure it out TT

[latent scaffold]

Sadness

[prime pond]

You are wrong.

Thats not where the problem is try to check my comments. If you use the word maybe means that you aint sure with that answer.

[latent scaffold]

I was. The point is their graph was fine mathematically , it just needed some tweaking to fit the format the software platform uses. At this point it’s trial and error to figure out what was wrong in the first place, hence using ‘maybe’.

AI just hallucinated mistakes and provided incorrect solutions.

[prime pond]

No I rarely use AI in maths. You can confirm with any detection software.

[latent scaffold]

I don’t really care to be honest. It looks AIy and, whether you wrote it or not, it was wrong.

[mint igloo]

0 is greater than -1 right?

[latent scaffold]

Yes

[prime pond]

Yes

[mint igloo]

.close

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[ashen cradle]

I am having a problem in the 12th question
I approached it with two ways and from one of the ways I am getting wrong answer
Can someone please tell me where I am wrong (correct answer is [n-1/2])

[meager juniper]

This is probably a personal failure, but I don't understand this step. How do you go from 1 to 2? (Image uploading slowly)

[ashen cradle]

product rule differentiation

[odd edge]

@ashen cradle Has your question been resolved?

[odd edge]

@ashen cradle Has your question been resolved?

[odd edge]

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[dusty lily]

S(x) = Integral( f(t)dt) limit is 0-x and im supposed to draw S(x) while 0<= x <= 12..... i dont know where to even begin

[wooden gorge]

Could roughly estimate the curve with a parabola g(x) = (x - 4)(x - 10)

[odd edge]

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[vale vapor]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[final anchor]

1 / 2

i can show my work to show what i was thinking

but it is probably wrong

[vale vapor]

!show

[odd edge]

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

[final anchor]

so i was thinking they say use chain rule

so i must have a function with another function inside it to use the chainrule

but first i wanted to check if a function with a function inside it is always even if the two functions are even

[leaden karma]

Hint: f(-x)=-f(x) is true for all values of x

[final anchor]

Ye i do know that but i just dont know how to use the derivative for this

[vale vapor]

use g(x)=-x

[final anchor]

but that is just an example right that is not a proof?

[leaden karma]

You can differentiate both sides of this

The g(x) makes things needlessly complicated

[final anchor]

-f'(x)=-f'(x)?

wait no that cant be true right

[echo ginkgo]

almost

[vale vapor]

so close

[leaden karma]

Where did the -x go

[final anchor]

-f'(-x)

[echo ginkgo]

yea

[leaden karma]

There you go

[final anchor]

oooh then i used the chain rule for x

right in the function of f(x) sort of

[echo ginkgo]

well with the -x inside yes

[final anchor]

ye exactly

[echo ginkgo]

which was what bonk suggested

[final anchor]

ooooh

ye now i get it

i needed to use g(x) = -x in my situation so it would become f(-x)

okay thank you very much guys

[leaden karma]

A modified version of this same fact is true for even/odd functions which are not necessarily differentisble but the left and/or right derivatives exist

[final anchor]

wait

oh the left derivatives is lim x-> a- = (f(x)-f(a))/(x-a) right>

?

[leaden karma]

Correct

[final anchor]

aah okay luckily i dont have to do this stuff yet

.close

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[zenith blade]

This is a problem with its textbook solution. I've found those two x values on my own. But I'm a bit confused about the reason for eliminating the first one. I thought that I can discard it by just substituting it into the original equation and checking the result. However, the solution in the picture seems to imply that I can just take into account 6x-4>=0, which is the condition for solving the intermediate equation. But can't I ignore it because I'm solving the original equation, not the intermediate one? Or are they equivalent, so that the new condition is also relevant?

[raw palm]

no you cant ignore it

because when you are trying to solve the original equation

you will eventually reach here

if you include a value that ignores this condition

then it is not a valid solution

since it doesnt satisfy the equation which you reached

so basically what you did was like this

you started with an equation

you then said that this equation is equivalent to another equation

thus if a solution doesnt satisfy one of them then it doesnt satisfy the other

[odd edge]

@zenith blade Has your question been resolved?

[zenith blade]

But, in general, there are cases when we may lose some roots or, on the contrary, get extraneous ones. How do I know that this case is not one of them, without substituting roots into the original equation?

[jolly pecan]

hyy i need help

[raw palm]

thats exactly what x=5/12 is

thats why the conditions are necessary

[raw palm]

so can you give me an example of this

[zenith blade]

Well, that equation in the middle of the solution can be an example. In this context, I understand the condition x>=2/3.

But it's a bit harder to get it in the context of the original problem, whereas I totally see the necessity of the condition x>0.

[zenith blade]

How are they equivalent? Does this condition have something to do with it?

[raw palm]

what is the meaning of these 2 equations being equivalent to each other

[raw palm]

it is when you can start from any one of these 2 and reach the other through logical implications right ?

so here you established one way of the equivalence

that is if you start with the original equation at the beginning you can reach the equation sqrt(3x+1)=6x-4

now if these 2 equations are equivalent

then you can start with sqrt(3x+1)=6x-4 and reach the original equation

[raw palm]

is this possible ?

the answer is yes so that these 2 equations are equivalent indeed

also you can check that x_1=5/12 isnt a solution to the original equation by plugging

[zenith blade]

Well, as I implied at the beginning, that's what I did first.

[raw palm]

doesnt that show the necessity of the condition imposed in the intermediate step

i mean without this condition , x_1 will be a solution to original equation

but it is not

[zenith blade]

But the condition could be random, in general, no?

I mean it could be just a coincidence

[zenith blade]

But I'm trying to think this way

[raw palm]

is it really a coincidence ?

so for example if you reach sqrt(3x+1)=k with k<0

now if you dont just say that this doesnt have a solution

and try squaring both sides and solving

you will get into trouble

for example $\sqrt{3x+1}=-3\implies 3x+1=9\implies x=\frac 83$

[raw palm]

if $k<0$ then $\sqrt{3x+1}=k\implies 3x+1=k^2\implies x=\frac{k^2-1}3$

[raw palm]

but now try plugging

this gives $\sqrt{\frac{3(k^2-1)}3+1}=\sqrt{k^2}=|k|=-k$

[raw palm]

which is clearly neq k

so that the "solution" that you found isnt a solution

that happened because you didn't take the condition that k must be >=0 into consideration

[raw palm]

the same applies here

if 6x-4<0 you will run into the same problem

this may not be always the case but it will happen alot

it will not be the case if the quadratic that you get after squaring doesnt have any real solutions

probably

[raw palm]

i didnt give this a thought but it seems to be like this

[zenith blade]

But I've already said that I understand 6x-4>=0 if the original problem is sqrt(3x+1)=6x-4.

[raw palm]

yes but you just reached this equation from the original one

so this became the equation that you are solving

if there are no flaws in your work till you reached this equation

then solving this equation means that you solved the original

solving this equation requires the condition 6x-4>=0

so think about this logically

"if the original equation is satisfied then this equation is satisfied" is what your work says right

ok so this statement is also true : " if this equation isnt satisfied then the original equation isnt satisfied"

[zenith blade]

I'm still trying to see the equivalence(

[raw palm]

just work backwards

start with sqrt(3x+1)=6x-4, 6x-4>=0 and then reach the original equation

[spice shore]

hello Narynbek are you kazakh?

[raw palm]

here you can reuse the steps that you used to establish the implication in the first direction

[zenith blade]

yes

[spice shore]

қалайсын

[raw palm]

thats not true for any implication

but here it works

[zenith blade]

Салам, жайлап) Өзің де қазақсың ба?

[spice shore]

ия

i joined this server cause i want to improve my actually outdated math

im really bad for now

reason why iam doing is a grades

[mystic saffron]

hello nice people am new here

[zenith blade]

If you need a tutor, this server might not be for you. But talk with people in #discussion or #serious-discussion, because this and other #help channels are for asking about specific math problems.

[zenith blade]

!redir

[odd edge]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[odd edge]

@zenith blade Has your question been resolved?

[zenith blade]

I did. But what about the difference between their conditions/domains for roots?

[raw palm]

take this example: $\log (\sqrt{3x+1}-4)-\log (2x)=\log (\frac{\sqrt{3x+1}-4}{2x})$ right

[odd edge]

@zenith blade Has your question been resolved?

[zenith blade]

so?

[odd edge]

@zenith blade Has your question been resolved?

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[brittle halo]

Is the answer 8?

[brittle halo]

My reasoning is because nodes 4 and 3 are fixed.

If we consider node 1, there are 2 places for it where it is connected to node 3

If we consider node 5, there are 4 places for it where it is connected to node 4.

The other nodes are then fixed.

So 2 * 4 gives us 8

Is this reasoning correct?

Oh, wait

Maybe 16 because nodes 8 and 7 can be swapped.

I'm not sure though.

[icy kindle]

you can swap 1 and 2, 5 and 6, 7 and 8, and the edges (56) and (78)
so 2^4 = 16

[brittle halo]

Many thanks

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[steady tide]

hello i want my proof checked please

[odd edge]

@steady tide Has your question been resolved?

[wanton bison]

looks ok

[steady tide]

nice, thanks

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[surreal lichen]

can someone help me prove this:

[surreal lichen]

In induction

.help

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[surreal lichen]

.help help

[odd edge]

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[surreal lichen]

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[tame sonnet]

how do i sketch the graph of this function

[tame sonnet]

i have derived it if that helps

[nimble blaze]

get the key features
asymptotes
intercepts
stationary points
inflection points etc

and make a table of values to get more points

[tame sonnet]

how would i find the asymptote of this function

[nimble blaze]

same approach as with all other functions

[tame sonnet]

well i have only done 1 whole fraction not like this

[nimble blaze]

how'd you do it with 1 fraction

[tame sonnet]

vertical asymptote find it with denominator

and make it equal to 0

horizontal u do it by limits to - infinity and + infinity

[nimble blaze]

do the same here

[tame sonnet]

yeah but here we have 2 denominators

do i just add them?

or 3 actually

if we take 1 into account

[nimble blaze]

you can do that if you want

[tame sonnet]

so i add them all?

[nimble blaze]

if you want, it will Work

[tame sonnet]

alright

thanks i can do it from here

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[icy stirrup]

Calc 2, Volumes of revolution
I know the answer of the total volume, 76pi/3
The blue square is 16pi and the orange is 28pi/3

But by concept, why isn't the orange volume half the blue volume?

[cerulean vortex]

because the orange is more biased towards the outside where it sweeps out volume faster

[icy stirrup]

I'm struggling visualize that,
If you take a cross section at any point in the revolution it would look exactly like the graph right? So how does it sweep out more if for every cross section, the orange is half the blue?

[cerulean vortex]

that's right

[icy stirrup]

So then shouldn't the exact volume of such shape be a sum of all the cross sections?

[cerulean vortex]

what if you had two squares that were the same size, but one was close to the axis and one was far?

they have the same cross section

but the farther square sweeps out a bigger ring

[boreal crag]

One is "taller" then the other

[wooden cypress]

Which of these shapes has the highest area?

It's exactly the same concept but in 3d

[icy stirrup]

I get that part

[icy stirrup]

How is the orange triangle further away?
It can be imagined that the orange triangle is inside the blue square as well

The bottom left corner of the orange triangle is the same distance away as the bottom left corner of the blue triangle

and same for the top right corners

So shouldn't they sweep the same distance

[cerulean vortex]

but it's thicker on the outer side

or should i say, taller

hmm

[icy stirrup]

I think I got it,
using a perspective on blank area instead of solid area helps a lot

[cerulean vortex]

what about this one? orange is half of blue

but the orange ring is not half of the blue ring

this red ring has smaller volume than the other one

since they're not equal, they can't split the blue volume in half

[icy stirrup]

That helps a lot in visualization
Thank you, I think I understand it all now

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[misty gale]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[misty gale]

2

[limber glade]

ok so where you stuck

[misty gale]

simplifying it all

bcs it becomes 3^3x * 3^2 - 3^3x/...

[limber glade]

yeah then?

[misty gale]

then im stuck

[limber glade]

combine the powers

[misty gale]

3^3x+2 - 3^3x

[limber glade]

factor 3^3x out

[misty gale]

?

[limber glade]

we can write 3^3x+2 as 3² * 3^x

and then factor out 3^3x

[misty gale]

bru i had written that no offense

can u give some more explanation my brain isn't braining that much today

explanation as in next steps

<@&286206848099549185>

[limber glade]

umm so then you can write 3^3x (3²-1)

in place of 3^3x+2 - 3^3x

[misty gale]

ok

that's 3^3x * 2^3

-3^3x

won't i get 0 thne

in num.

[limber glade]

so overall you get 3^3x / (3^3 * 2^3 * y) = 1/27

as the equation

[misty gale]

oh ye nvm

[limber glade]

the 2^3 cancel

[misty gale]

yeah

so u get 3^3x/3^3*y

this is 1/27

[limber glade]

and the 3s powers after simplifying give 3^(3x-3) / y = 1/27

[misty gale]

yeah

that's 3^3x - 1/27/y=1/27

[limber glade]

so if you solve for y from here

[misty gale]

3^3x-1/27=y/27

[limber glade]

what do you get?

[misty gale]

3^3x = (y+1)/27

[limber glade]

nope

there's something wronng

[misty gale]

im mixing distributive property here

bru

mb

[limber glade]

anyway if you solve you get y = 3^3x

[misty gale]

alr

[limber glade]

and then what's y-3^3x?

[misty gale]

so 3^3x - 3^3x^3x

my keyboard is not helping me today lol

[limber glade]

bro what's this monster

[misty gale]

sht

i read y^3x

so it's 0

[limber glade]

yeah

[misty gale]

thanks a lot g

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[keen swan]

Hello, I'm just a bit confused of how to graph this

[odd edge]

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

[keen swan]

Well, finding the lower estimate, and also the upper estimate

I cant really show work, I'm not entirely sure of how to find the lower and upper estimate. I was guessing lower and upper were just the lower and higher number of the two riemann sums?

If thats correct, I'm having trouble figuring it out as there isnt exact points given on the coordinate plane

[olive needle]

The lower and upper estimates are based on the partitions you make.

[orchid torrent]

Looks like an exponential graph to me, so you can find the equation directly

[keen swan]

oh so i should try to find f(x)

[orchid torrent]

Yeah that’s what I’d do

[keen swan]

is there a way to do it without, finding f(x) as a previous question didnt want me to find f(x)

[olive needle]

f(x) * delta x

[keen swan]

I get this, it just depends on which of the two riemann sums are lower and higher

[orchid torrent]

You said yourself that the points aren’t exactly “nice” (aka you can’t just read them off)

So idk how else you would get the points to actually find the sum

[olive needle]

Yes. Because the graph is decreasing, use the right endpoint.

[keen swan]

Ah, I didn't think like that

so since decreasing use right riemann, okay

how would this result in two values though?

do i still need to take left

[olive needle]

What is the other value you are supposed to calculate?

[keen swan]

upper estimate and lower estimate

i accidently clicked the photo, and this showed

at least i have the points lol

[olive needle]

The upper estimate will be a left-endpoint Riemann Sum.

[keen swan]

L = 23.2 and R = 14. Can I imply that if the graph is decreasing the upper estimate will be the left endpoint

and vice versa, if its increasing the upper will be right end point

or should i scratch that and just get the answers

[olive needle]

Yes.

[keen swan]

and see which is which

[olive needle]

For always increasing or decreasing.

[keen swan]

?

[olive needle]

If you had a function like this