help-23

[rose plume]

cylindrical shells?

[jovial void]

integral of sqrt(x) from 0 to 1 minus the integral of x from 0 to 1

thats area tho by volume are you working with 3 dimensions?

or just a misworded question?

[dense wadi]

i am working worh volu,e

[rose plume]

im not sure what u mean u passed section of intersection

[jovial void]

one sec gotta refresh myself lol

alright so we can imagine it as a 3d shape

and if we look at it straight on from the y axis its like this hollow cone

now we can imagine we just take a cross section of it

so imagine a line that spans all y for a single x, a verticle line

this cross section is a washer so we can find the radius using this line we just imagined

the distance from the axis to a line is the radius of that function

so for y = x, the r = x

and y = sqrt(x) the r is sqrt(x)

calculate area of this washer

integrate this area from 0 to 1

[dense wadi]

so this is the shame

[jovial void]

yup

[dense wadi]

okay i acc

mentally calculated everything and

then i put in calculator

and it worked but

i have a question

like

calculator shows it like this

i did it like this

is this still good

i can put everything in abs value so its positive

[rose plume]

yes if you geometrically know the outer is... outside

[jovial void]

seems good to me

[dense wadi]

wait btu the calculator

put

[jovial void]

so long as you understand the concept and used it to get the correct answer

[dense wadi]

inner-outer

he reserved

[rose plume]

yes but he put abs

[dense wadi]

reversed

ah

so reversed is good

[jovial void]

^

[dense wadi]

as long as abs

[rose plume]

i mean

[jovial void]

with abs

i prefer without

it intuitively makes less sense

[dense wadi]

cuz sometimes i cant know graph

by function

so its good to know if i can just

abs it

[jovial void]

u dont need to tho

[rose plume]

no

u still need break the integral by when the thing of the abs is positive or negative to compute it by hand

[jovial void]

along 0-1 y=x is always less so you know thats the smaller area u subtract from the larger one

[rose plume]

it works because you already know volume outer - volume inner >=0

which implies |volume outer - volume inner| >= 0

which implies |-(volume inner - volume outer)| >= 0

[dense wadi]

ah

wait but also

he put (y-2)

[rose plume]

... ok and so on |volume inner - volume outer| >= 0

[dense wadi]

instead of (2-y)

[rose plume]

and what's the problem

its squared so its always positive

[dense wadi]

are both fine?

[rose plume]

(y-2)^2 = y^2 - 2y + 4

[dense wadi]

ah

[rose plume]

(2-y)^2 = y^2 -2y + 4

[dense wadi]

so order literally doesntmatter

idk why i spent time thinking

lmao

okay yeah

[rose plume]

for the radius it doesn't

BUT

for the volume outer - volume inner it does

... or if you fucked up the order you are just going to get the volume but with a negative sign

so yeah it kinda doesn't matter just remove the sign if that happens

[dense wadi]

ahh okay okay

tysm yall!

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[tender vapor]

Hi. I am very stuck on these questions. I need them ASAP by midnight today. I’m confused because our teacher didn’t teach us matrix/matrices and other than that I am very bad at math and have a hard time learning and memorizing concepts. If anyone could help me, I would be extremely grateful.

[tender vapor]

this is a vectors topic question, idk if this would apply

[random rivet]

omg oops

i didnt see ur msg 😭

mbb

[tender vapor]

haha it’s okay😭 I ACCIDENTALLY THOUGHT YOU WERE TRYNA HELP ME WITH THIS PIC LOL

[lavish scaffold]

Are you a bit familiar with matrices

Like some notions

[tender vapor]

not at all

my teacher hasn’t taught them

[lavish scaffold]

Thats ok lets concentrate on one of those equations

This one for example you can rewrite in matrix form

[tender vapor]

what would that look like?

[lavish scaffold]

Youd take each coefficient and put them in a matrix like this :

2 3 -1
6 -4 2
2 -10 4

You can see each line and column have corresponding coefficients right

[tender vapor]

what do you mean by corresponding?

[lavish scaffold]

Coefficient in line 1, column 1 is indeed in the 1st line and 1st column of the matrix

( you can write the matrix another way but we will just go with this for simplicity sake)

[tender vapor]

yea i see but i don’t know what importance this brings😭

[indigo wave]

Suprising you are asked to solve this with no knowledge of matrices

[lavish scaffold]

The matrix would look like this

[lavish scaffold]

It will be important for the matrix multiplication part

[tender vapor]

ohhh ok

[lavish scaffold]

Matrix multiplication goes like this:

Each line of the first matrix is "multiplied" by each column of the second

Multiplication not in the usual way of course but in a way like: first element of the line with first element of the column, second of the line with second of the column and so on

[tender vapor]

oh so like a * b?

with each line?

oh wait

like a1 * a2

[indigo wave]

If you know dot product think of it as the i,j position in the resulting matrix is the dot product of the ith row in matrix a and the jth column in matrix b

[lavish scaffold]

Take a look at this example

[tender vapor]

i think i understand

[lavish scaffold]

This also highlights a condition for matrix multiplication (number of columns of the first matrix needs to be equal to the number of lines of the second matrix)

[tender vapor]

but now i only have one box (or wtv it’s called) of matrixes, what do i do now

[lavish scaffold]

Well if you look at your equations

There are also the x y and z terms

And those can also be fit into a one column matrix (also known as a vector) to complete the left hand side of the equation like this:

Conditions are met for multiplication (3 columns and 3 lines)

[tender vapor]

okay i see

[lavish scaffold]

Now if you go on calculating this product using matrix multiplication rules itd be like:

[tender vapor]

yea i just did that and got the equations again

[lavish scaffold]

Which is exactly like those first three equations

[lavish scaffold]

Yep

Now all is left is the right hand side of the equation

Which are just those three numbers that are left on the right fit into a vector

[tender vapor]

oh yea

[lavish scaffold]

Which completes the matrix form of those equations

Now what youll be asking

Well we have to look for an intersection, aka a vector (x y z) that satisfies this equations

[tender vapor]

yea haha😭

[lavish scaffold]

Intuitevely youd think, since this is a linear algebraic equation written like Ax = b, to just multiply both sides by the inverse of A just like we do with normal algebraic equations

But with matrices an inverse is not always assured

And also multiplication is not commutative for matrices (this just means that for two matrices, the order of multiplication matters, just a fancy word)

So to find the inverse youd need to check for its determinant

[tender vapor]

what is that?

[lavish scaffold]

Well its a numeric value that gives an idea on how matrices transform a space

But simply put

[tender vapor]

im getting confused😭

[lavish scaffold]

Just a numeric value that is calculated in a super specific way lol

So dont bother yourself with that space bending stuff

Unless for later if youre interested check out 3blue1browns video on matrices but now since you dont have much time its just a little numeric value we extract from the coefficients of a matrix

[tender vapor]

yea i dont have much time rn unfortunately😭

[lavish scaffold]

Our matrix is 3x3 (3 lines by 3 columns)

So its square

So a determinant can be calculated

(Determinants are only calculated for square matrices)

For example:

You can calculate the determinant this way

https://www.desmos.com/matrix

Here is a useful website to do this for you since you have more equations to do and it takes some time for each matrix

Important is to show you understand the approach for each system of equations

[tender vapor]

okay i see

so now

[lavish scaffold]

Since we know how to calculate the determinant we can tell if there is an intersection or not in the first place before finding those x y and z

[tender vapor]

do i add the third matrix on the same line as the matrix with the equations (but ob the other side if the = ?

[lavish scaffold]

For the website ?

Just enter the matrix with the coefficients for each system of equations

And then you can use det to calculate its determinant

[tender vapor]

im lost😭

[lavish scaffold]

Ok lets backtrack

What have you gotten up to this point

[tender vapor]

this

[lavish scaffold]

You got the determinant part right

[tender vapor]

no

[lavish scaffold]

You got how to calculate it from that example?

[tender vapor]

uh

i feel like i almost got it😭

if that makes sense

[lavish scaffold]

Its okay

The determinant is useful because if its not zero we know the matrix has an inverse

And if the matrix has an inverse, the matrix equation is solvable

And if the matrix equation is solvable, that implies there is an intersection between the three planes

[tender vapor]

ok sounds good

[lavish scaffold]

And we can also find the intersection point using the inverse like this:

Its important we multiply by the inverse from the left and not from the right because theyre not the same (order matters for matrix multiplication)

So try calculating the determinant using that website

And then the inverse if the determinant is non zero

Also to clarify A is the matrix there, X and B are vectors

[tender vapor]

so would it be like 2 times 5?

[lavish scaffold]

Youre calculating the determinant manually?

[tender vapor]

oh no i was just asking if this would be the correct values to multiply

[lavish scaffold]

Oh for what

[tender vapor]

wait im starting to understand, the goal is to make likr a bottom left square of all 4 numbers being 0

[lavish scaffold]

Where?

[tender vapor]

i watched a video but maybe i got it wrong

[lavish scaffold]

No its okay tho whatd you get as the determinant, or did you understand up to this point the general approach

[tender vapor]

omg i just found my question on chegg haha😭 but looking at it, it’s not saying anything about a determinant which is weird

[lavish scaffold]

Nah it can be the better approach

Whats it say

[lavish scaffold]

Thats gaussian elimination

[tender vapor]

what😭

[lavish scaffold]

Its a way to get the solution by doing some clever operations to each line and column of the matrix until you get the identity matrix on the left

[tender vapor]

man i hate math this is so confusing😭😭😭 i have no idea how i’m gonna finish this and the other 3 questions by midnight

[tender vapor]

damn

[lavish scaffold]

Its a good way to fetch solutions but it can confuse you so leave it after midnight

Also it requires some thourough operations which can take time

Thorough*

[tender vapor]

😭😭😭 what should i do i have like an hour left for 4 questions😭😭😭

[lavish scaffold]

Its okay have you understood the general approach up to this point

[tender vapor]

a little, i got lost at the determinant

the order of which value(s) to do the operations with and everything🫠

[lavish scaffold]

Its ok all you need to understand about the determinant right now is this

There are websites to calculate the determinant for you under seconds so you dont bother with the work

Like this one https://www.desmos.com/matrix

Add a new matrix and calculate it

Its determinant*

[tender vapor]

which matrix do i write?

to calculate it

[lavish scaffold]

Lets start with this one

For the first system of equations

Rest are gonna be a breeze too

[tender vapor]

okay ill do that tn

rn

[lavish scaffold]

Youll find the determinant to be 0 normally

[tender vapor]

i write all 3 eqns?

[lavish scaffold]

All coefficients

Should look something like this for you

[tender vapor]

it shows me det (A) is -16

[lavish scaffold]

Whats the matrix you entered?

[tender vapor]

oops i forgot the negatives

it’s 0 now

[lavish scaffold]

No worries

[lavish scaffold]

Conclusion?

[tender vapor]

0 is the point?

of intersection?

[lavish scaffold]

No, determinant is zero so that means that equation we wrote earlier is not solvable

Which means no intersectiok

[tender vapor]

ohhhhh

okay

[lavish scaffold]

Intersection*

[tender vapor]

i thought 0 meant there is an intersection lol😭

[lavish scaffold]

Nah zero means no intersection

Non zero means we have to calculate the inverse of that matrix to determine the solution

But thats it for the first system of equations

[tender vapor]

okay yayyyy thank youuu🙏🙏🙏

[lavish scaffold]

No problem

[tender vapor]

would you be willing to help me with the other questions too?

[lavish scaffold]

Sure

[tender vapor]

THANK YOU SO MUCHHHH

ill resend

[lavish scaffold]

Alright

[lavish scaffold]

For a there is a bit of an exception

We only have two planes

Which means there is an intersection and its a line (not a point) or we have no intersection at all

[tender vapor]

how do you do that question?

i did it earlier and this is what i got

i feel like it’s wrong tho😭

[lavish scaffold]

Its okay if its wrong

[lavish scaffold]

Z wont cancel there youd need to sum the two equations instead

Eitherway youll get the line equation and do the same thing

[tender vapor]

i don’t wanna sound weird or anything but i’m a visual learner and it’s harder for me to understand by text😭

[lavish scaffold]

Im talking about how you calculated the difference of the two equations instead of summing them to cancel the z out here (top right)

[tender vapor]

ohhhh

how exactly do i sum the questions tho?

[lavish scaffold]

You sum each term and get x + y = 11/5

[tender vapor]

i#oh ok i just redid it

now i sum in the y value into one of the eqns?

sub*

[lavish scaffold]

Yea

[tender vapor]

ok so what im gettinf id

is

3z = 2x + (11/5) -5

[lavish scaffold]

Yea thats correct

[tender vapor]

how would i simplify it tho

cause it’ll be 2x/3 + 11/5/3 - 5/3

it’s getting too complicated

[lavish scaffold]

Itll be 3z - 2x = -14/5

[tender vapor]

okay

x = 14/5 - 3z/2 ?

[lavish scaffold]

x = 7/5 + 3z/2

3z/2 my bad

@tender vapor just rechecked and i was wrong with smth my bad:
x = 14/5 + 15z
y = -3/5 - 3z
Which will get you your parametrized plane equation

(x, y, z) = (14/5, -3/5, 0) + (3, -3, 0)*t

[safe radish]

@tender vapor Has your question been resolved?

[safe radish]

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[storm umbra]

1 and 2

.close

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[distant steppe]

solve the following equation for x:
x^5 - 10x^4 + 39x^3 - 64x^2 + 44x - 8 = 0

Im new comer help me

[covert yoke]

@distant steppe This doesn't have a closed form solution in terms of radicals.

Also this thread may lock suddenly, so please open a new help thread

See #❓how-to-get-help

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[misty glade]

how do I solve algebratic fractions

[covert yoke]

https://media.discordapp.net/attachments/359052604149465088/1228206153104752700/IMG_2284.jpg?ex=662b331d&is=6618be1d&hm=0d72fba3bd1862add17c40b1090b55ac40f2ddd7763eed4c92adc29e905ecbb9&

From her other posts elsewhere ^

I don’t get where the R(y+1) goes into 2R 2/(y+1)

[misty glade]

i mainly need steps written out as well for preperation for tommorrows quize

[covert yoke]

So if you have the same thing in both the top and bottom of a fraction then these will almost always cancel to 1.

For instance, 7/7 = 1, and 34.2/34.2 = 1. If we have even a variable x/x, this is also 1.

Except when x = 0

So when you multiply R(y+1) by (2R)/(y+1)^2 you get 2R^2 (y+1) / (y+1)^2

And you have one copy of (y+1) in the numerator and one copy in the denominator, these cancel to 1.

[vast flax]

oh wait its broken

[misty glade]

wish they had a rawing pad on disocrd

[vast flax]

wait still broken

[misty glade]

so would the steps be

[vast flax]

bruh

[misty glade]

1. cancel out duplicate facotrs

[vast flax]

$\frac{R(y+1)}{\frac{(y+1)^2}{2R}} = \frac{R(y+1)}{1} * \frac{1}{\frac{(y+1)^2}{2R}} = \frac{R(y+1)}{1} * \frac{2R}{(y+1)^2}} = \frac{2R^2}{y+1}$

[flat frigate]

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

[vast flax]

ok im done now

[covert yoke]

\begin{align*}
R (y+1) \times \frac{2R}{(y+1)^2} &= \frac{R (y+1)}{1} \times \frac{2R}{(y+1)^2} \
&= \frac{R (y+1) \times 2R}{1 \times (y+1)^2} \
&= \frac{2R^2 (y+1)}{(y+1)^2} \
&= \frac{2R^2}{(y+1)} \times \frac{y+1}{y+1}\
&= \frac{2R^2}{(y+1)} \times 1\
&= \frac{2R^2}{(y+1)}
\end{align*}

[flat frigate]

OmnipotentEntity

[misty glade]

that makes more sense

would it be the same if both top and bottom of the fraction had two different factors each

[covert yoke]

Can you give an example?

[misty glade]

Something like this

[covert yoke]

Whenever you have a fraction in the denominator you multiply the top and bottom by its inverse

So to simplify:

\begin{align*}
\frac{\frac{a}{b}}{\frac{c}{d}} &= \frac{\frac{a}{b}}{\frac{c}{d}} \times 1 \
&= \frac{\frac{a}{b}}{\frac{c}{d}} \times \frac{\frac{d}{c}}{\frac{d}{c}} \
&= \frac{\frac{a}{b} \times \frac{d}{c}}{\frac{c}{d} \times \frac{d}{c}} \
&= \frac{\frac{ad}{bc}}{\frac{cd}{cd}} \
&= \frac{\frac{ad}{bc}}{1} \
&= \frac{ad}{bc}
\end{align*}

[flat frigate]

OmnipotentEntity

[covert yoke]

Notice that at the end, we just needed to flip c/d upside and multiple it with the numerator.

But for this to be true we also needed cd ≠ 0

@misty glade does that make sense?

[misty glade]

yes it does

[timber sparrow]

Can anyone help me with this

[covert yoke]

!occupied

[safe radish]

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

[misty glade]

thank you, ill try and fit it on my index card for tommorrows quize

[covert yoke]

Best of luck

[misty glade]

thank you

[covert yoke]

You're very welcome.

Don't forget to .close if you're good to go

[misty glade]

how do closeit

[covert yoke]

".close"

[misty glade]

.close

[safe radish]

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[lime mango]

Does anyone know how we got 729 as answer
Other than 30 I can think correct answer is 64 which is from 2^6

[shell imp]

perhaps there's leaving it blank?

729 = 3^6, so theres gotta be a third choice

[lime mango]

👀

sorry, all six questions be attemted so this way how can we leave question blank

[shell imp]

you can answer true, false, or leave it blank

for each question

[lime mango]

thank you sir 😄

.close

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[tribal pecan]

I am working with two arrays, $\texttt{arr1}$ and $\texttt{arr2}$, which contain values ranging from $-7$ to $4$. I need to find a function $f: \mathbb{R} \rightarrow (0,1)$ that normalizes these values to the interval $(0,1)$. The requirement is such that if $C = \texttt{arr1} \times \texttt{arr2}$ (element-wise multiplication), I want to transform this using $f$ so that:

[ C' = f(\texttt{arr1}) \times f(\texttt{arr2}) ]

where $C'$ is the normalized form of $C$. The critical property required is that the inverse of $f$, denoted as $f^{-1}$, should allow:

[ f^{-1}(C') = C ]

This implies the following functional properties:

\begin{enumerate}
\item $f(x) \times f(y) = f(x \times y)$ for any $x, y \in \mathbb{R}$
\item $f$ and $f^{-1}$ are well-defined across the real numbers.
\end{enumerate}

Essentially, I need $f$ and $f^{-1}$ such that $f^{-1}(f(x) \times f(y)) = x \times y$.

Can you help identify such a function, or advise if this transformation is mathematically feasible given the constraints? If possible, an explicit form of this function and its inverse would be highly beneficial.

[flat frigate]

Κώστας Μήτρογλου

[tribal pecan]

Values ranging to -7 to 4 shouldn't be taken into consideration when finding the functions because these could easily change

[peak estuary]

this is not possible

you want f defined to be for all real numbers

in particular you want f(1)

lets call f(1)=a and try to find out what a can be

we notice that a*a=f(1)*f(1)=f(1*1)=f(1)=a

so a^2=a

which means that either a=0 or a=1

maybe as a more general background, this property f(a*b)=f(a)*f(b) is also called a homomorphism property and such an f is called a homomorphism (in abstract algebra). there is no such function f:R->(0,1), e.g. for the reason listed above

[tribal pecan]

oops sorry. the range should be [0,1]

f: R-> [0,1]

[peak estuary]

ok

that sadly still doesnt save you

for starters, a=0 definitely shouldnt be, cause then f(x)=f(1*x)=f(1)*f(x)=0*f(x)=0 always. so that would be bad

so lets say a=1, aka f(1)=1

lets think about what f(2)=b and f(1/2)=c could be

we notice that 1=f(1)=f(2*1/2)=f(2)*f(1/2)=b*c. so b and c should multiply to 1

but if b and c are in [0,1], then thats not possible, except for b=c=1. which would also be bad

[tribal pecan]

I would assume the issue is going from R to (0,1) (non-negatives?)
is there any more formal proof that this can't work?

[peak estuary]

I gave you a formal proof

[tribal pecan]

okay give me 2 mins before I close this to make sure

(I trust you though)

.close

[safe radish]

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[tribal pecan]

@peak estuary Thank you 🙂

[peak estuary]

yw

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[chrome cipher]

here i have to determine a region whose area is equal to the given limit without evaluating the limit

how do i interpret that as the area of the region lying under the graph from the given limit?

not sure how interval was found to be [0,2] either

i've looked at example problems and i can't seem to understand the procedure

[safe radish]

@chrome cipher Has your question been resolved?

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[chrome cipher]

.reopen

[safe radish]

✅

@chrome cipher Has your question been resolved?

[chrome cipher]

.close

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[safe radish]

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[magic hare]

Urgent HELP

[magic hare]

So it should be integrated

And then a sketch

Idk how

Pls help

[remote verge]

What are the bounds of the integral

[magic hare]

X is the axis

(X=os) translated

And the rest is as written

[remote verge]

so you are integrating from an unknown value os to 2?

[magic hare]

I need the Pd

[magic hare]

No, it just says x is the line

[remote verge]

ohhhhh

So its asking you to find the area closed between the x-axis, x = 2, y = e^x, and y = x + 1

Correct?

[magic hare]

Yes

Exactly

And to draw it in the coordinate system

[remote verge]

This is a rather strange problem

Since there isn't an area which is bound by all four lines

[magic hare]

I need the surface of that triangle in between

That's the Pd

But idk how to get to it

I need the process of integration and then addition

[remote verge]

This?

Or this?

[magic hare]

Is the black line y=e^x?

[remote verge]

Yes

[magic hare]

And green is y=x+1?

[remote verge]

Yes

[magic hare]

Yes now the surface

How

[magic hare]

Hm

[remote verge]

Im not entirely sure what you mean by surface

Do you just need the area of the black highlighted region?

[magic hare]

Surface of D and d is what was given

[magic hare]

I'm not sure either i'm thinking😭

[remote verge]

I think it might be this, since it is bound by all four of the specified lines

[magic hare]

If D is x-axis, x = 2, y = e^x, and y = x + 1

What's the surface of d

Like this?

[remote verge]

Maybe, the problem is that e^x is nowhere to be seen

I am thinking that it is the area under the curve of e^x from -infinity to 2 minus the small triangle

[magic hare]

It's probably this

[remote verge]

The equation that you would set up here is

[remote verge]

The area there is simply

[magic hare]

Aha

[remote verge]

But the problem is then you dont have x + 1

[magic hare]

Aha

[remote verge]

So I think its probably asking for this

[magic hare]

Okay then how do we solve this

[remote verge]

Which you can then solve for

[remote verge]

^

[magic hare]

Aha aha

[remote verge]

it is the area under e^x - the area under x + 1

Since that will get you the small highlighted area

Do you know how to do the integrals?

[magic hare]

I forgot

Completely

[remote verge]

The antiderivative of e^x is just e^x

[magic hare]

Aha

And x+1

[remote verge]

and to find the antiderivative of x + 1, you can just use the power rule (x^n -> x^(n+1)/(n+1))

so you would get the antiderivative of x + 1 to be x²/2 + x

Yes?

This

[magic hare]

Is this not right?

[remote verge]

This does not get you the area

You need to include the bounds

[magic hare]

Aha

[magic hare]

How

[remote verge]

The bounds are the left and right side of the area that you want to find

For the integral of e^x, your bounds would be negative infinity to 2

[magic hare]

So we get e^x - x²/2 -x+c, how did u get +x and no c?

[remote verge]

Since the tail of e^x goes to the left forever, but the x = 2 equation stops it at x = 2

?

The numbers specifying the bounds are important, you must evaluate the integrals with them

[magic hare]

How do i write it down after this?

Do i keep the bounds?

[remote verge]

You can write it down something like that

I'm not sure if this is the proper notation, but it is usually how I write it

[magic hare]

Aha

Je we just use a straight line

Like u can see here

[remote verge]

I see

[magic hare]

Now what else?

[remote verge]

Then you can plug your values into your equation, like so

After that is just simplification down to the final answer

[magic hare]

Ok let's see what it gives

[remote verge]

I believe you should end up with e² - 9/2

[magic hare]

And that's it?

[remote verge]

I believe that should be the area of this black region, yes

[magic hare]

Ok let's see what the professor says😭

Thank u!

[remote verge]

No problem

[magic hare]

.close

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[rain stump]

is it possible for me to get help in a vc cuz I am missing like a whole topic but I am a 10th grader so that'd make it easier

[fathom osprey]

Hey therr how can I help ya

I’m in school right now but if you tell me the topic I can try to sum it up

[rain stump]

bro I am missing a whole topic Ive got this project that I am doing and I moved schools

[fathom osprey]

Sorry to hear that, what topicv

?

[rain stump]

I have a like data set and they are asking me to graph this

c) Calculate the 5 values for a box and whisker plot, then draw the plot. [5]

Minimum = ________
1st Quartile = ________
Median = ________
3rd Quartile = ________
Maximum = ________

[fathom osprey]

Can i hav the data set

[rain stump]

1 sec

loading ty for helping out tho

18.660 + 25.115 + 30.415 + 32.165 + 33.640 + 34.605 + 36.835 + 57.010 + 69.905 + 94.920 + 126.055

ignore the +'s

[fathom osprey]

Minimum is the smallest # Max is largest Median is the middle number first quartile is the set of numbers before the halfway and 3rd is the set after halfway

U have 11 numbers

What number would be halfway thru the data set

[rain stump]

34 ig

34.605

[fathom osprey]

Correct

That is your median

What is the smallest number

[rain stump]

but like why are they asking for 5 values tho

[fathom osprey]

Wdym

Oh okay i see

They want the median minimum max q1 and q3

They want those 5 values

U found the median

Thats 1/5 done

Lets move on to the min and max

What is the smallest number in the set

[rain stump]

18.660 and 126.055

[fathom osprey]

Correct

So we have 3/5 values

[rain stump]

quartiles?

[fathom osprey]

So

Write two data sets for me

The first half and the second half

[rain stump]

is 34 gonna be from the first tho cuz the amount of values I have is odd

[fathom osprey]

Yeha

So everything before 34 is one set

Everything after is another

The median from the set before is your q1

The median from the set after is ur q3

[rain stump]

q1=1st set q3=2nd set?

[fathom osprey]

Correct

Everything before the whole median is q1 and everything after the whole median is q3

[rain stump]

so we are leaving 34 alone and its not gonna be in any of the sets

ok got it

[fathom osprey]

Yea

[rain stump]

.close

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[lean otter]

https://www.youtube.com/watch?v=MyQoRQiv3bc

[lean otter]

The working for pat b at 7:39

Why does he set dr/dx to 0 when differentiating implicitly?

Why is that max r

[vagrant ice]

Whenever you're looking for the maximum value or the minimum value of something, you need to differentiate and set to 0

You want the radius to be 'as big as possible'

You have a function in two variables: x and r which equals 0

You can think of r in this case as a y-value (the dependent variable: r depends on x) and x is an x-value (the independent variable)

So that's why you're finding dr/dx and not say dx/dr

[vagrant ice]

dr/dx gives you the x-value for when r is the largest or smallest

[lean otter]

Oh yes this is what I needed for clarification I forgot lol

[vagrant ice]

No worries

[lean otter]

Thanks for the explanation!

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@tepid phoenix Has your question been resolved?

[safe radish]

@tepid phoenix Has your question been resolved?

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[lean otter]

Using vectors p and q find others given.

[lean otter]

Explain why, dont just solve it.

[verbal remnant]

Note that AD and BC are equal as vectors. So are AB and DC

Notice order of letters

The rest is just parallelogram law

[lean otter]

Let me see. I will ping you once again.

[lean otter]

.close

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[median spear]

Please help, I tried using the Arithmetic formula Tn=a+(n-1)d for the sequence below, but Idk if im doing it right because I got -2 by doing T-2=3+(n-1)-2.

[median spear]

<@&286206848099549185>

[thin bridge]

!15m

[safe radish]

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.

[median spear]

Sorry

[rigid jetty]

find the gradient

and you know y intercept is 5

[median spear]

i dont even know what that is..

[rigid jetty]

the slope

[median spear]

...

no idea

[rigid jetty]

maybe watch some vids on linear functions

https://www.youtube.com/watch?v=qXX47hS2KLw

[median spear]

Alright, thanks

.close

[safe radish]

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[median spear]

.reopen

[safe radish]

✅

[median spear]

I still do not understand

[thin bridge]

you're missing parentheses / not substituting properly

you have
$$T_n = a + (n-1)d$$
you've identified $a = T_1 = 3$ \
and $d = -2$ \
to get $T_{-2}$, replace ALL $n$ in the equation with -2.

and make sure you place () where needed or use additional multiplication symbols to maintain the order of operations

[flat frigate]

ℝαμΩℕωⅤ

[median spear]

so T-2=3+((-2)-1)(-2)

[thin bridge]

use _ to denote subscript

[median spear]

what?

[thin bridge]

\verb|T-2| reads as $T-2$ \
to represent $T_{-2}$ in plain text, type \verb|T_(-2)|

[flat frigate]

ℝαμΩℕωⅤ

[thin bridge]

but the right side is fine

[median spear]

T_(-2)=3+((-2)-1)(-2)

[thin bridge]

missing ) on the right side now

T_(-2)=3+((-2)-1**)**(-2)

now simplify the right side

[median spear]

Yeah, that makes more sense than what the other guy tried to explain

[median spear]

T_(-2)=9

[thin bridge]

yes

[median spear]

Thank you

.close

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[stiff vine]

What was I supposed to do

[lavish scaffold]

One way you can do this is multiply and divide by conjugate

[stiff vine]

Wdym

Why dividing by it

[lavish scaffold]

Well you know how a^2 - b^2 = (a - b)(a + b) right

[stiff vine]

Yeah

[lavish scaffold]

Whats a - b

[stiff vine]

?

[lavish scaffold]

From this

[obsidian stream]

Basically factorize it by using it's conjugate

[stiff vine]

Sqrtx^2+1)-x

[lavish scaffold]

This

Or express in other terms

[lavish scaffold]

What would this be in terms of that remarkable identity

a - b = (a^2 - b^2)/(a + b) right?

[obsidian stream]

Yeah

But remember

x is tending to infinity

[stiff vine]

I got this

[obsidian stream]

bro

divide it by √x^2+1 + x

[stiff vine]

Why r u dividing it

[obsidian stream]

Coz you multipied it

Bruh

You have to multiply and divide with the conjugate

[stiff vine]

I’ve never seen dividing itself by conjugate

I don’t get it

[obsidian stream]

Bro see this is the question right

?

You multiplied the conjugate to it

now you have to divide to make it 1

otherwise it would be altering the question

[stiff vine]

My answer was 1

[obsidian stream]

Nah

You still don't understand

Check the answer in google if possible

[stiff vine]

I know the answer

I just don’t know how to solve this to get it

[lavish scaffold]

What you did was correct but thats for the numerator

Now theres still sqrt(x^2 + 1) + x in the denominator

Which then you can evaluate the limit normally and get 0

[stiff vine]

I don’t get how u get that

Where did that come from

[lavish scaffold]

From diving and multiplying by it

If you have a number say 3

3 = (3 * 2)/2

You can do this to perserve its value but do some operations to get what you wqnt

Want(

[stiff vine]

My problem doesn’t even have a denominator tho

[lavish scaffold]

Which is what we did with the term in our limit

Yes we added that

Because we multiplied and divided by sqrt(x^2 + 1) + x

a - b is still equal to (a-b)(a+b)/(a+b)

And in our case a is sqrt(x^2 + 1) and b is x

[stiff vine]

There’s no point in doing that tho

[lavish scaffold]

Why not

[stiff vine]

Its just (a-b)

[lavish scaffold]

Well look what it got us in the numerator

Now we dont have to deal with an indeterminate form

The x^2 cancel out

[pastel cypress]

The limit is just 0

[safe radish]

@stiff vine Has your question been resolved?

[safe radish]

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[lean otter]

Please Name 2 vectors which are parallel and have same magnitude and point towards same direction

[rustic goblet]

I don't think there are any 2 unique vectors which satisfy this tbh

[lean otter]

Hmm, can 2 oppo sides of parallelogram satisfy?

[rustic goblet]

...I wouldn't call those unique vectors tbh

that's more of like

the same vector, but translated

[lean otter]

That's what I am asking for
So will they solve the question?

[rustic goblet]

can you show the original question?

a screenshot would be nice

[lean otter]

I was reading a book
So... B is initial position of a particle and A is final position
So displacement was given
A - B vector

I got a bit confused and tried to derive

[rustic goblet]

man

don't xy me lmao

!xy

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[rustic goblet]

okay, cool.

what's the question?

[lean otter]

Here I stopped
The lower dotted line represented A - B
And upper dotted line also A - B because parallelogram
But I feel these 2 dotted lines are very different

[rustic goblet]

well it's a displacement isn't it?

all you care about is the magnitude of the vector

[lean otter]

True but direction is also important because we are dealing with vectors

[lean otter]

Kindly forgive, am just not in a very good mood

[rustic goblet]

I mean, it's the same vector, just translated

and it should point in the same direction too

(also, since I'm a linear algebra shill, I have to mention that there is a more general notion of "vector", that does not depend on direction at all)

but that's not relevant here

[lean otter]

Take a reference point, any point on plain
And look at direction they point in
Aren't they different

[lean otter]

i,j, k are that only no?

[rustic goblet]

well... no

[lean otter]

Aren't they unit vectors in a particular axis

[rustic goblet]

"vector" just means element of a vector space

[rustic goblet]

they are

but about your question... is the displacement you care about from A to B or from B to A?

[lean otter]

B to A

[rustic goblet]

right.

so there's no issue here

let me draw it

[lean otter]

Sure

[rustic goblet]

we can see that the vector from B to A is the same as the vector from O to

I'll call the point C

it's translated yes, but that's still the same vector

afaik, all vectors which point in the same direction and have the same magnitude are the same

[lean otter]

But imagine yourself standing at any point on plane and looking towards where arrows are point
Technically speaking they are different directions no

[rustic goblet]

because we always take the coordinates of a vector to start at the origin

[lean otter]

For that matter
How do u actually judge if 2-3-5-10 vectors are pointing in same direction

[rustic goblet]

I think you're talking about affine spaces lol

https://en.wikipedia.org/wiki/Affine_space

and unfortunately, I can't say I'm very well versed with them

[lean otter]

Just give the summary of it

I mean what it says

[rustic goblet]

the informal description is that an affine space is what's left of a vector space when you forget about where the origin is

in this scenario, addition of vectors isn't agreed upon because any two observers can claim they have the origin actually, this isn't relevant to your original question and might even be confusing

in your original post, one can take O to be the origin (in which case the displacement vector is OC), or one can take B to be the origin (in which case the displacement vector is BA)

[lean otter]

I didn't fully understood it but would like to continue it later
Currently in a bad mood so anyway not gonna understand anything

[rustic goblet]

I mean it's okay to be in a bad mood lmao

but I would suggest not xy'ing your problem in the future regardless of how you feel

(also, I highly recommend learning linear algebra, that subject is awesome; totally not me shilling )

"vector" can mean so much more!

[lean otter]

No that was result of my bad mood

[rustic goblet]

ah

well, just for future reference

just try not to

[lean otter]

Just shifted to new place
No friends
Life is just pissing me off

[rustic goblet]

ah that sucks

I hope things turn better for you

[lean otter]

Biggest problem is I have absolutely nobody to talk to
Not even online lmao
No family nothing

[lean otter]

Even I hope too

Less see how it goes

[rustic goblet]

I mean you can always talk about math here on mathcord lmao

[lean otter]

That's only thing keeping me sane, math!
It's now just me, math, movies kinda situation

Anyways won't bore u much

[rustic goblet]

haha

[lean otter]

.close

[rustic goblet]

don't worry about it

[safe radish]

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[rustic goblet]

you can always talk to people in the discussion channels!

[lean otter]

Naah naah it's fine
Nobody likes to listen a person just venting

[lean otter]

Let them have a good time lol

[safe radish]

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[fickle island]

Say $A \subset \mathbb{R}$, and assume $|A| > 3$. How can I show that there always exists an $a \in A$ such that $A - {a}$ is disconnected? I get it in theory but idk how to write it up. My idea so far is to break it into cases: A is countable, A is an interval, or A is a union of intervals

[flat frigate]

bot

[fickle island]

or, im trying to say you can remove "the middle point" and then you have a disconnection

[blissful hull]

No cases needed. You should be able to show it directly in the general case. Let $x, y, z \in A$ such that $x < y < z$. How can you construct nonempty disjoint two sets $U, V$ whose union is $A \setminus {a}$ using this information?

[flat frigate]

Awesam

[fickle island]

okay yeah

i got it

quite simple

.close

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[rigid imp]

conceptually, why do we divide the second derivative by x'(t) in parametric equations

[rigid imp]

why wouldnt it be y''(t)/x''(t) or (y'(t)/x'(t))'

sorry if my notation is bad

this is what im talking abt

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@rigid imp Has your question been resolved?

[rigid imp]

<@&286206848099549185>

[lean otter]

you on your own for this one bro😭 💔

[rigid imp]

its so over for my exam

[lean otter]

bro isnt this calculus

try finding the damn answer instead of foolin around on discord😭

[gilded hamlet]

its easy

[safe radish]

@rigid imp Has your question been resolved?

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[solid sequoia]

how do i use algebra to simplify this to get the limit?

[dense prism]

a^2-b^2=(a+b)(a-b)

[safe radish]

@solid sequoia Has your question been resolved?

[solid sequoia]

how do i use that in this expression

can you explain please

[rigid imp]

yk how you can turn x^2 - 4 into x+2 and x-2

radiation is saying try the same idea with both sides of the fraction

i believe

[solid sequoia]

so would a be 2 and b be x?

[safe radish]

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[stable pine]

How would you do this

[steel gulch]

I think the easiest way would be to apply MVT for integrals

[stable pine]

and how would you do that

[steel gulch]

the only problem that I think might cause some grief is with negative numbers. but that is probably going to be a result from the fact that their integral are equal.

[steel gulch]

well, have you learnt what the MVT for integrals are?

[stable pine]

No

[steel gulch]

Umm, well. If you use it, would you have to prove it? cos if thats the case then that's probably not the way to do it.

[stable pine]

no i dont have to prove it

i just need to show that the sequence diverges

[steel gulch]

yo fr? kinda crazy. they just let you use whatever thrm you want without proof?

[stable pine]

ok tbh i teach myself

im 16

i havent really touched real analysis before

[steel gulch]

another way would be to compare it to the intgral test for series which would be quite interesting.

depends if you've learnt any theorems around that?

would be more interesting that using MVT tbh

[stable pine]

so prove the series diverges?

[steel gulch]

Oh wait no, im stupid nvm.

that would only prove certain cases and not in general.

[stable pine]

ok let's say i didnt have to prove mvt

how would i use it then

[steel gulch]

haha, yeah. I got a bit sidetracked lol. So you have the integral right. Then for some c in R (in assume??) the integral $\int_{a}^{b} (\frac{f(x)}{g(x)})^{n}f(x),dx = (\frac{f(c)}{g(c)})^{n}f(c)(b-a)$

[flat frigate]

theaveragejoe6029

[stable pine]

and then what do i do with that

extreme value theorem?

[steel gulch]

since the interals of both f and g equal for all a and b, then just choose a' and b' such that $ (\frac{f(c)}{g(c)}(b'-a'))^{n}f(c)(b-a) = (\frac{f(c)}{g(c)})^{n}f(c)(b-a)$

ie that b'-a' = 1

[stable pine]

$(\frac{f(c)}{g(c)}(b'-a'))^{n}f(c)(b-a) = (\frac{f(c)}{g(c)})^{n}f(c)(b-a)$

[flat frigate]

1212312121HEYYYYYYY

[steel gulch]

ahah, thanks lol

[stable pine]

how would you know this tho

[steel gulch]

well, you are just choosing any a' and b', you can really choose anything like 2 and 1. right??

[stable pine]

yeah but like wouldn't a' or b' be 0

if you differentiate a or b since they r just numbers

[steel gulch]

sorry, i mean a' and b' as in new numbers, I can't call them a and b since they are already used. but yeah, they arent a prime and b prime lol

[stable pine]

ohhh

[steel gulch]

I was gonna use a* and b*, but I didn't think latex liked that

but yeah, then applying the MVT in reverse, you get f(c)(b-a)

which equals the integral of f which is postiive, so it diveges

[stable pine]

how would you apply in reverse

[steel gulch]

well, this is the thrm: $\int^b_af(x),dx=f(c)(b-a)$ I get its not really backwards, but just a different direction to where you normally go

[flat frigate]

theaveragejoe6029

[stable pine]

yeah ik

but how would i use the reverse on $(\frac{f(c)}{g(c)})^{n}f(c)(b-a)$

[flat frigate]

1212312121HEYYYYYYY

[steel gulch]

really just let f(c)/g(c) = h(c) which we can do since f and g are cts

shit sorry

ignor that

I mean instead that sub the integral into the f(c)(b-a)

[stable pine]

so like this? $(\frac{f(c)}{g(c)})^{n}f(c)(b-a)\int_{a}^{b} (\frac{f(x)}{g(x)})^{n}f(x),dx

then how would i go from there

Wait nvm i got it

Thank you so much dude

[safe radish]

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[tranquil vector]

This is a non homogenous PDE

[safe radish]

@tranquil vector Has your question been resolved?

[safe radish]

@tranquil vector Has your question been resolved?

[safe radish]

@tranquil vector Has your question been resolved?

[tranquil vector]

no one replied lol and the bot asks so often

[safe radish]

@tranquil vector Has your question been resolved?

[distant grotto]

cant help

i cant even do trig

[safe radish]

@tranquil vector Has your question been resolved?

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@tranquil vector Has your question been resolved?

[candid turtle]

I don’t recall stuff like this super well and I could be completely wrong

But would you not first find an equation which solves the forcing term on the RHS

And then solve the now homogenous equation using Fourier series?

[tranquil vector]

Yes. Something like that. I was told that you solve for u(x,y) = u h(x,y) + u_p(x,y) as the sum of homogenous and particular solutions.

How is this @candid turtle

[candid turtle]

Yes that looks fine, with the separation of variables stuff. I don’t remember if sinh and cosh is better as opposed to e^x and e^-x instead (I remember one or the other being much better). But I haven’t looked at this for so long I won’t be much more help than this I’m afraid 😦

[tranquil vector]

I am not sure how to handle the non homogenous part (RHS) to find a particular solution. Can you have a look at how I did that on the solution. Does it seem right? @candid turtle

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[surreal tulip]

Can someone explain the answer to this question using prime factorization? I got 5/3 but the textbook says the answer is 5/8, I checked my answer on a calculator and its correct

[frigid locust]

how are u getting 5/3?

[surreal tulip]

wait ill send my working

I turned 1500 and 2400 to 375 and 600 respectively

just halved both together until either number became a decimal number

[frigid locust]

prime factorization of 600 is
5x5x3x2x2x2

u forgot to multiply those 2s which gave u the answer 5/3

[surreal tulip]

so I'm meant to drag down all the 2's from further up the tree down to the bottom? I didn't do that for other questions and got the right answer

[frigid locust]

u are suposed to drag all the numbers

u just got lucky in the other questions

[surreal tulip]

for the other questions, if I dragged down more numbers I'd get the wrong answer, how come some questions need me to drag down and others dont? how do I differentiate between them?

[frigid locust]

well all needs to be dragged down

thats what prime factorization is

factoring a number till its in smallest prime number

like for 125
prime factorization would be 5x5x5

[surreal tulip]

I get that now. What I'm confused by is if I apply this new knowledge to the questions I got lucky with, I'd end up with the wrong answer. How come?

I'll send an example

[frigid locust]

k

[surreal tulip]

for this one, if I dragged the 2 down this would become 24/1 which is the wrong answer in the txt book

I got the correct answer which is 12/1 by only dragging the number above

[frigid locust]

by draggin down i meant these numbers

[surreal tulip]

yeah if I drag down the 2 on the right branch I'd get the wrong answer

[frigid locust]

3x3x2x5x2 would be for 180
5x3 would be for 15

5 and 3 cancels

3x2x2 would be whats left for 180
1 would be whats left for 15

3x2x2 = 12

12/1

how u got 12/1 without bringing that down is a mystery

[surreal tulip]

Wait I think I dragged it down mentally but forgot to note it down in my work. sorry

[frigid locust]

np

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[cloud pagoda]

any ideas?

[safe radish]

@cloud pagoda Has your question been resolved?

[cloud pagoda]

<@&286206848099549185>

[young nexus]

if f is a function with 3 arguments, what should f((1,1,0),(0,1,1),(0,1,0)) then mean?

[cloud pagoda]

i truly dont know

f^{-1}(af) is the set, which imagine after f is that af?

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@cloud pagoda Has your question been resolved?

[safe radish]

@cloud pagoda Has your question been resolved?

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@cloud pagoda Has your question been resolved?

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@cloud pagoda Has your question been resolved?

[cloud pagoda]

.close

[safe radish]

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[faint seal]

Find $\int\frac1{\sqrt{e^{2x}+4e^x+1}} dx$

[flat frigate]

kheerii

[faint seal]

I substituted $u=e^x, dx=\frac{du}{u}$ to get $$I=\int\frac1{u\sqrt{(u+2)^2-3}} dx$$

[flat frigate]

kheerii

[faint seal]

What do I do after this? Trig sub doesn't seem to work very well

I got the integral $I=\int\frac{\sec\theta}{\sqrt{3}\sec\theta - 2} d\theta$ which doesn't seem very favourable

[flat frigate]

kheerii

[eternal carbon]

i am confident there exists a better way to solve the original integral but i can also tell you this integral you have here is not bad to evaluate with tangent half angle substitution

[faint seal]

okay, what's the better way?

[eternal carbon]

tbh i don't know i just assumed because that's usually how nested u-subs work

[faint seal]

and yeah the integral is not hard but converting it back into x seems tedious

[calm bridge]

what if you let e^x + 2 = u at the start and then let u = sec(theta)

[eternal carbon]

that doesn't really do anything different

[calm bridge]

really?

[eternal carbon]

what i mean to say is that it just doesn't work*

[eternal carbon]

my point is that the nested u-substitutions still exist with your approach

(hence computationally the process of converting theta to x should remain the seame)

[calm bridge]

ah right

[faint seal]

is there a smarter way?

[eternal carbon]

was this a competition problem

u-substitution at the very least seems like a tacit requirement here

[faint seal]

no it's just a general problem

[eternal carbon]

i c

[faint seal]

for jee

[faint seal]

yeah

[eternal carbon]

then i guess i'd just go with the nested u-subs

[faint seal]

I don't see any other way tbh

[faint seal]

yeah

the answer is $x-\log |1+2e^x+\sqrt{e^{2x}+4e^x+1}|+C$

[flat frigate]

kheerii

[faint seal]

which seems awfully simple

[calm bridge]

i see what happened there

try taking e^2x out of the sqrt