help-39

[fading ledge]

and i want to connect it with what if terms are having same terms

xy or something

[autumn fossil]

Not sure what you mean by real life meaning, but basically, when we imagine expanding
(x+y+z)^n, its like doing (x+y+z)(x+y+z)(x+y+z)... with n factors. We can take either x, y or z from each of those factors, so each resulting term will look like x^a * y^b * z^c

There are exactly n factors, so that means that a + b + c has to be n. So counting the terms is the same thing as counting non-negative integers a,b,c with a + b + c = n

and counting a, b, c with a + b + c = n can be done with stars and bars

[fading ledge]

Okay if i changed in the equation (xy+x^2y^2+y^3)

[autumn fossil]

so like (xy+x^2y^2+y^3)^n?

[fading ledge]

Yeah

How many terms will be there?

[autumn fossil]

now this is more interesting

[autumn fossil]

im not even sure if there is some way to do this that's better than expanding it

for this specific one, we could prolly find some formula

oh, you actually did

[fading ledge]

There was a formula which tells coefficient of specific terms

Can we use it here for see different powers?

for counting total terme

[autumn fossil]

possibly, idrk, sorry. I'll also have to go soon, ill try to think about this later and let u know if i find out anything. You can keep this open in case someone else wanted to try and tackle this

Here is what i got so far: each term will look like (xy)^a * ((xy)^2)^b * (y^3)^c with a + b + c = n, the only issue is that this wont always produce unique terms. Simplifying, we get
x^(a + 2b) * y^(a + 2b + 3c). Now we when will it not be unique and maybe we would be lucky enough to get a formula for it, and then we could subtract it from stars and bars. Unfortunately, i dont really have the time to try this atm

[fading ledge]

Thanks for giving me idea

[pearl pond]

@fading ledge Has your question been resolved?

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[lone pasture]

The formula for the total number of terms in the expansion of [(x+y+z)^n] comes from the stars and bars combinatorial method, which counts the ways of distributing [n] identical items into 3 groups (the powers of [x], [y], and [z]), where each group can have from 0 up to [n] items (i.e., each variable's exponent ranges from 0 to [n]) ��.Explanation Using Stars and BarsImagine you have [n] stars to place into 3 bins, each bin representing one variable ([x], [y], [z]), where each bin's count gives the exponent for that variable in a term. To divide the stars among bins, we use 2 bars to separate the stars into 3 segments (so the configuration [||] means [x^1y^0z^{n-1}]). The number of ways to arrange these [n] stars and 2 bars is the number of terms in the expansion.Mathematically, the total number of terms is:Each term in the expansion corresponds to a non-negative integer solution to [a+b+c=n], representing the exponents in a term [x^a y^b z^c] ��.ApplicationFor [(x+y+z)^4], number of terms:There are 15 unique terms, such as [x^4, x^3y, x^2y^2, xyz^2], etc. ��.Stars and Bars VisualThink of the stars as units to assign and the bars as dividers. For three variables:Place [n] stars in a row.Insert 2 bars to split them into three groups (one for each variable).The arrangement of stars and bars tells you the exponents for each variable in every term ��.This explains how combinatorics links to the expansion formula for multinomials, using the stars and bars method.

[autumn fossil]

Source?

ChatGPT?

no, thats something else

hmm

[lone pasture]

ChatGPT + ME

[autumn fossil]

Yeah, seems like you didnt even remove the weird symbols and weird formatting

[plush bramble]

!nogpt

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[autumn fossil]

dont use AI generated content for helping others, if you need AI to generate the answer, you might not be the right person to answer the given question. AI can be wrong often

[autumn fossil]

and this part has been solved btw, the 2nd part that wasnt yet solved is
(x^2y^2 + xy + y^3)^n

[latent quail]

.solved (popking's not here)

[pearl pond]

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[verbal whale]

Of course, why bothering?

[pearl pond]

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[zenith mural]

sorta stuck

[zenith mural]

I attempted it but used a bunch of assumptions to get root 19

which I highly doubt is correct

labelled graph if that makes it easier

[vernal bear]

k what your approach?

[prisma prism]

umm like I have found a way

[zenith mural]

🥴

[vernal bear]

lol like how you got root 19

[prisma prism]

what's the answer?

[zenith mural]

yeah I was drawing it out

it's just cringing me

[prisma prism]

damn

[zenith mural]

is it that bad

[prisma prism]

I can give a hint

[zenith mural]

please do

[prisma prism]

like see the triangle IAD draw the a perpendicular to AD from I

it bisect AD at rt angle

u can find it's length

[zenith mural]

wait what

yes

yes

YES

oh

wait

[prisma prism]

like find the length of perpendicular bisector from I to AD

by pythagrous thm

[zenith mural]

but what does that make I?

I get what ym

but then what sorta triangle does that make? i can't visualize it

[prisma prism]

umm like it would be better if u join I with the pt on AD and the mid pt of BC

[vernal bear]

right triangle

[zenith mural]

what how

[zenith mural]

OH

BRO

THAT'S SO OBVIOUS

oml i need to get myself checked

thanks yall

[zenith mural]

root 8 then

[prisma prism]

yeah then add root 8 with 2

[zenith mural]

yeah ty

tytytysm

[prisma prism]

np

[zenith mural]

idk how I didn't see that

.close

[pearl pond]

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[fluid dove]

not sure how to do this

i made it 1 - (2cos^2 (t) - 1) + (2sin(t)cos(t)) / 1 + (2cos^2 (t) - 1) + (2sin(t)cos(t))

[toxic lichen]

brackets.

or TeX.

anyway:

better is $\frac{1 - (1 - 2\sin^2(\theta)) + 2 \sin(\theta) \cos(\theta)}{1 + (2\cos^2(\theta) - 1) + 2 \sin(\theta)\cos(\theta)}$

[jolly parrot]

Ann

[fluid dove]

okay

what now tho

do we take factors out?

[toxic lichen]

clean shit up first

[fluid dove]

2sin^2 (t) + 2sin(t)cos(t) / 2cos^2 (t) + 2sin(t)cos(t)

[toxic lichen]

brackets

(or, again, LaTeX)

otherwise i'll start misreading this on purpose

and now you factorize what you can.

[fluid dove]

sorry

[fluid dove]

yeah i done it now thank you

.close

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[proud kiln]

Okey so

[proud kiln]

I understand the distribution curve and that it is right skew, , however I don't really know how to estimate the average of the 10,000 policies

Sooo umm I need like the $\frac{\mu}{\sqrt{\sigma}}$ which is $\frac{150}{\sqrt{10000}}$

[jolly parrot]

Alphurion

[proud kiln]

so I get 1.5 from that............. I think

Which makes little sense... cause if thats 1.5 dollars then... maybe thats meant to be 150 dollars which makes 160 dollars outside the range

So maybe I say yes it is safe to say the average loss is under... but i not sure

maybe it means 1.5 times... something is the average

so N( 10000, 1.5)

how do I z score this? who knows...

helpppp

ok fine ummm $\frac{\overline{x} - U_{0)}}{\frac{\sigma}{\sqrt{n}}}$

whyy

[jolly parrot]

Alphurion

[proud kiln]

close enough

$\frac{160 - 150}{\frac{300}{\sqrt{10000}}} = 3.33 \overline{3}$

but idk the meaaaaaaaaaan

WAIT

160

$ = 3.33 \overline{3}$

cmooon

do ittt

$ = 3.33 \overline{3}$

hullloooo

$hellooo$

WAT

[tepid elbow]

💔

[jolly parrot]

Alphurion

[proud kiln]

whatev

[tepid elbow]

R u tryna solve smth like this

[proud kiln]

this works

[proud kiln]

og question

[tepid elbow]

Is that related to probability

[proud kiln]

It's a standard deviation question

So ummm

Er

its the confidence interval kind of, its like how sure are we that the average is below 160$

[tepid elbow]

Well till that point u r going correct

[pastel umbra]

Standard deviation is a measure of spread of the data

[proud kiln]

okay okay thats good

but what does 3.33333 represent though..

Cause I need to know how sure I am that the average will below 160$

but I'm not sure how that number helps, you know?

[pastel umbra]

I mean, I would harken to the referenced "Example 15.8" at the bottom of that screenshot

[tepid elbow]

Thought that too

[proud kiln]

hrmmm

this is like an example

so I have.... the mean... I think thats 3.3333

no.. but its not money

OH wait so do I z score the mean

er the sigma

and then itll maybe work out?

[tepid elbow]

P(z >3.33)

[proud kiln]

so I need the z score of... 160?

[tepid elbow]

Yes

[proud kiln]

so $z = \frac{x - \mu}{\sigma}$

[jolly parrot]

Alphurion

[proud kiln]

$z = \frac{160 - 150}{300}$

[jolly parrot]

Alphurion

[proud kiln]

right right

wait hmm

[tepid elbow]

Over 300?

[proud kiln]

should it be 3.333

[tepid elbow]

Yes

[proud kiln]

o

so 10 / 3.33333

[tepid elbow]

160-150 / 3

10/3 so 3.33

[proud kiln]

okay

Why did I want this number...

OH

p( 3.33 >= 3.333) ...

wel actually I got like 3.00003

because 10/3.3333

[tepid elbow]

Why 10/3.33

[proud kiln]

becauseee uhhh

why is it 3

[tepid elbow]

300/root of 10000?

[proud kiln]

oh, 3

o..

u right

so its less than, so it isnt true right

p(3 >= 3.33) is flase

[tepid elbow]

Yes

[proud kiln]

so this loss, will be greater than 160$

so bad

investment

[tepid elbow]

No bro wait

[proud kiln]

D:

[tepid elbow]

Its gonna be less than 160

[proud kiln]

Oh, because the number is less/

[tepid elbow]

Ye

[proud kiln]

like the statement is false, so it is less, therefore this is a good investment

[tepid elbow]

You can say tho

Do u have the answer key for it

[proud kiln]

This one unfortunately doesnt

[tepid elbow]

But atleast its done

[proud kiln]

Thank you for helping me out

[tepid elbow]

Np

[proud kiln]

have a good one

.close

[pearl pond]

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[pearl pond]

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[rare holly]

hello how would this loo like
The graph y=x squared is reflected in the line with the equation y=x to give me the curve C (a) sketch the graph of y= x squared and the curve C
i got something like this but idk what to do

[strong relic]

in general what do you get if you reflect a function across the line y = x

[rare holly]

the negative of it?

[strong relic]

nope

[rare holly]

then

[strong relic]

thats if you reflect it across the x axis

reflecting across the line y = x produces the inverse function

[rare holly]

wdm

how can i even draw that

is only possible to flip it oppositly

or reflect it on the sides

[strong relic]

dont think about drawing it first

what is the inverse function of y = x^2

[rare holly]

i solve for y right

when solving for inverse function

[strong relic]

solve for x

[rare holly]

x= +-root y

[strong relic]

so the inverse function is?

[rare holly]

root of the y value

[strong relic]

of the x

how would that look like on a graph

[rare holly]

?

[strong relic]

x = +/-sqrt y is the same thing as y = x^2

reflected function should still have y as the subject

[rare holly]

but u said solve for x

[strong relic]

yes thats the technique

but the inverse function will be y = +/- sqrt x

[rare holly]

yh

[strong relic]

the graph of the square root function is what your question is asking for

[regal herald]

I get what youre saying but it's slightly wrong in how youre presenting it
For a function to be invertible it has to be one-one, that's why y=x^2 in its base form has no inverse

[strong relic]

yes i guess inverse is the wrong term

but the loose idea of a square root being inverse to a square function is decent enough without needing to bring up bijections

+/- sqrt x is the reflected function @rare holly

[rare holly]

yh

ill ask my teacher tommorow

i dont understand

.close

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[vestal pelican]

can someone help me with this limit, i dont know where to start i have plugged 0 but you get 1 over infinite

[strong relic]

does the structure of the limit feel similar to anything else

[vestal pelican]

i cant think of anything

that feels similar to the structure

[strong relic]

limit defintion of a famous constant

[vestal pelican]

sorry idk

[strong relic]

do you know the limit definition of e

[vestal pelican]

damn i didnt know what that was

i just googled it

but this is such a specific case

[strong relic]

i feel like these kinds of questions are not uncommon in a school setting

[vestal pelican]

literally got this limit from last years calculus midterm

oh i thought u meant are not common

[strong relic]

yeah no ive seen this kind of limits like a hundred times

[vestal pelican]

really

[strong relic]

getting sidetracked

[vestal pelican]

because u got to remember that x is approaching 0 and then all the other bs

[strong relic]

using the limit definition can you see how it can be manipulated

[strong relic]

theres really only 2 limit definitions for e

and they can be mutually derived with a substitution

[vestal pelican]

i still dont know what to do tho

[strong relic]

basically you want the same thing in the parenthesis and the denominator of the power

[vestal pelican]

ok how can i do that

[strong relic]

so to make the 1/3x a 1/6x

you need an extra factor of 2

[vestal pelican]

btw this is the one we're talking about right

[strong relic]

yes

[vestal pelican]

ok

[strong relic]

so now you have (1 + 6x)^1/6x *2

which is just e

[vestal pelican]

yeah

[strong relic]

with the factor of 2 in the exponent

what would the limit be

[vestal pelican]

wait

but how can we just make 1/3x -> 1/6x

[strong relic]

yeah so we do 1/3x = (1/6x)(2)

thats the factor of 2 im talking about

[vestal pelican]

ohhhhhhh

but where does the 2 go

[strong relic]

it stays there

[vestal pelican]

so its 1 + 6x)^1/6x*2

[strong relic]

yes

and so

[vestal pelican]

so is this wrong?

[strong relic]

ok oops i missed the *2

[vestal pelican]

its ok

ok now what

[strong relic]

now use the limit definition

[vestal pelican]

well the definition doesnt have the *2 thing

[vagrant briar]

well think about whether the definition would still have been valid if there was any other real constant in place of the 1 in the numerator of the fractional exponent

[strong relic]

with indices rules its equivalent to ((1+6x)^1/6x)^2

[vestal pelican]

ok now im lost

[strong relic]

this would be easier if i knew how to use the latex bot thing

[vestal pelican]

i think there is a converter

online

[strong relic]

$((1+6x)^{\frac{1}{6x}})^2$

[jolly parrot]

ashy!

[strong relic]

ooh first try

[vestal pelican]

ok why ^2

now that is confusing

[strong relic]

because a^(bc) = (a^b)^c

[vestal pelican]

damn im not sure if will be able to remember all this

if this comes up in a test

[strong relic]

practice and practice 🫩

[vagrant briar]

dont try to memorize. understand! the logic behind

[vestal pelican]

it should come naturally but its complicated

[strong relic]

foundationally the indices rules show be engrained into your brain

[vestal pelican]

its easier said than one

done*

anyways

[vestal pelican]

we have this

[strong relic]

yes

using the limit definition

[vestal pelican]

still not equal to e

its ^2

[vagrant briar]

well there is a direct result for 1^(infinity) forms but this problem was so simple that the way ashy was instructing you is the best

[vestal pelican]

which the definition doesnt have

[strong relic]

ignoring the ^2

can you see that its e

[vestal pelican]

yeah

but how can i ignore it when it should be there

[strong relic]

so with the ^2 what would it be

[vestal pelican]

"simple" 😭 im trying my best

[vestal pelican]

wait HOLD UP

e^2

[strong relic]

yes 😭

[vestal pelican]

OMG

thank you so much

[strong relic]

btw you can only ignore the powers in the limit if it doesnt create an indeterminate form

and the limit exists

[vestal pelican]

meaning?

[strong relic]

if it was an ^infinity instead of ^2 you cant do that

if it looks like it wont give you any trouble (😃) its probably fine

[vagrant briar]

$lim_{x\to0}{f(x)^{g(x)}}=\exp{\lim_{x\to0}{(f(x)-1)\cdot g(x)}}$ \ where $f(x)\to1$ and $g(x)\to\infty$ as $x\to0$

[jolly parrot]

T&C

[strong relic]

yikes

quite a step up

from the limit definition of e

[vestal pelican]

lost

in the sauce

[strong relic]

basically you just do e^(6x * 1/3x)

which directly gives e^2

[vagrant briar]

haha but this is a direct result from the same definition. however it is derived by taking the natural logarithm on both sides first and then taking the limit

[vestal pelican]

i understood this better

well thank you guys so much for helping

.close

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[lunar egret]

Hi, I need help.
This notation is not mentioned in our textbook nor was it covered in class (calc 1).
I have attempted to apply the quotient rule, here, though I don't think I know how it applies here, or, how to configure that application properly.

[lunar egret]

I might be jumping the gun on needing help. But I'll keep this channel open for now, and wait for anyone to show up as I run through it again.

[unkempt yacht]

this is the leibniz notation of derivative

$\frac{\dd}{\dd x}\left(\frac{h(x)}x\right)\eval_{x = 2}$

[unkempt yacht]

this essentially means calculate the derivative of h(x)/x, then plug x = 2 in

[lunar egret]

I do not have enough tools at my disposal to do that, because I have not seen this before.
I do not know where to start.
I have found derivatives for a myriad of other problems leading up to this point, but not one where the only information given is h(x)/x.

[pearl pond]

@lunar egret Has your question been resolved?

[strong relic]

use the quotient rule

@lunar egret

[lunar egret]

I have attempted to apply the quotient rule, here, though I don't think I know how it applies here, or, how to configure that application properly.

in this situation

with a problem that looks like this

[strong relic]

show your working

what do you get after applying thr quotient rule

[lunar egret]

I'll ask someone IRL in person tomorrow. thank you.

.close

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[carmine juniper]

When doing points for parabola… and using the table method, how do I know what value for x to put?

[warped violet]

Well it depends

Do you know the vertex

[carmine juniper]

Sure.…

Give example

[warped violet]

What?

Oh, you are asking in general?

[carmine juniper]

Yes

Show më

[warped violet]

When you are given a vertex?

[pearl pond]

@carmine juniper Has your question been resolved?

[nocturne grail]

the vertex is where the parabola turns, finding then plotting the vertex first will let you choose x values around the vertex that when plotted, will give you a clear view of how the parabola should be drawn

find the vertex point of
x^2 + 2x + 1

you think what you can do after that!

the image is NOT how x^2 + 2x + 1 is drawn btw

its just an example of what a vertex is

[nocturne grail]

I'm also assuming you know how to find it...

There are other ways like finding the roots and where the parabola intersects with hte y axis

but the table method doesnt require it

actually... there are so many ways 😭 you need to choose one

or tell us what your teacher wants you to do

[light helm]

get the key values,
and use whatever values are nice to evaluate
(the more values you use the better your graph will look)

[warped violet]

Yes

But if they already know it there's no point

[pearl pond]

@carmine juniper Has your question been resolved?

[proper nova]

Is her question on how to find the vertex of a function?

[pearl pond]

@carmine juniper Has your question been resolved?

[proper nova]

@carmine juniper, is your question on how to find the vertex?

You gotta answer or else you might not be getting help

[pearl pond]

@carmine juniper Has your question been resolved?

[ivory basin]

.close It's been an hour since the bot message with no reaction so I'm guessing something went wrong

[pearl pond]

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[sharp smelt]

$\int_{1}^{y^2} \frac{2}{x^2}dx; 1≤y≤\sqrt{2}$

would that be righrt

[jolly parrot]

wai

[sharp smelt]

$1<x<2 \implies 1<√x<√2$

[jolly parrot]

wai

[inland ivy]

well

[inland ivy]

this is the CDF

[sharp smelt]

oops

so I diff it now

[inland ivy]

yes

[sharp smelt]

$2y \cdot \frac{2}{y^4} = \frac{4}{y^3}$

[jolly parrot]

wai

[inland ivy]

yes

[sharp smelt]

One more problem

so the joint pdf is firstly

$f(x,y)=\frac{(10-x)^2}{50^2}$

[inland ivy]

What is x

[sharp smelt]

A variable

Am I missing something

[jolly parrot]

wai

[inland ivy]

what happens with y

[sharp smelt]

oh, I see, this assume that both random variables are x

$\frac{(10-x)(10-y)}{50^2}$

[jolly parrot]

wai

[sharp smelt]

that's better

I plan on using this to solve it

[inland ivy]

You only have one function of X and Y though

have you ever seen convolutions

[sharp smelt]

I have

I use them here I suppose

[inland ivy]

yes

make sure to check the ranges of x and y carefully

[sharp smelt]

$\int_{0}^{t} \frac{10-(z-y)}{50} \cdot \frac{(10-y)}{50}dy; 0<t<10$

[inland ivy]

why is (10-y) squared

[sharp smelt]

typo

[jolly parrot]

wai

[inland ivy]

yes

now what's going on with the limits

what's t

[sharp smelt]

the limit of the cdf

[inland ivy]

that's z

[sharp smelt]

then what should the limits be

[inland ivy]

well $F_Z(z) = \int_{-\infty}^\infty f_X(z-y)f_Y(y) \dd{y}$

[jolly parrot]

kheer257

[inland ivy]

the thing that's limiting your integral is that f_X and f_Y become 0 in certain ranges

what is the range of Z = X + Y?

[sharp smelt]

(0,2/5)?

[sharp smelt]

.

[pearl pond]

@sharp smelt Has your question been resolved?

[pearl pond]

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[nocturne owl]

can someone help me w this please?

[plush bramble]

factor out 2 from the denominator on the left side

[nocturne owl]

can you describe me why we do that and all that please?

[plush bramble]

any algebraic expression you want to simplify as much as possible before doing anything

4 and 2 share a common factor

[nocturne owl]

okay after that?

[plush bramble]

write it out and see if you can continue simplifying

[nocturne owl]

well im very new to this "high" level of math ik this is not that high but in my country this is high level graduation in math and highschool and i wanna do that but im afraid i will be too stupid for these

so im not sure whats next actually :/

[plush bramble]

did you write out what the denominator equals

[plush bramble]

.

[ivory swallow]

Lets make this simple

Given an equation in the form a/b = c/d, you can ALSO say that a/c = b/d.

Then that gives you two linear eqns

[plush bramble]

that's actually harder than my suggestion

[ivory swallow]

Yea true

[nocturne owl]

nah i get it

[nocturne owl]

yes

[plush bramble]

write it here

and write the entire left side

[nocturne owl]

2p+q/2p+2q=10p/9q

maybe?

[plush bramble]

no

[nocturne owl]

damn

[plush bramble]

factor 2 out of 4p + 2q means 4p + 2q = 2 * (2p + q)

plug the right side into the the original fraction

[nocturne owl]

but how do i do that?

im not really sure what you say cuz my english math is very bad ngl

[plush bramble]

replace 4p + 2q with 2 * (2p + q)

[nocturne owl]

okay

[plush bramble]

and you can do that because of

[nocturne owl]

so 2p+q/2*(2p+q)=10p/9q right?

[plush bramble]

yes

simplify

[nocturne owl]

okay

q/2q=10p/9q?

[nocturne owl]

what now my friend?

[plush bramble]

no

use $\frac{c}{bc} = \frac{1}{b} \cdot \frac{c}{c}$

[jolly parrot]

riemann

[plush bramble]

c / c = ?

[nocturne owl]

1

[stable flame]

then u can simplify Q on that fraction

as its the same as multiplying by 1

[nocturne owl]

@plush bramble so the solution is 1?

[stable flame]

nope

[nocturne owl]

i write 1/2 * q/q

[stable flame]

yeah

[plush bramble]

it's not q/q

c = 2p + q

[nocturne owl]

okay and after that?

[stable flame]

then u have 1/2 = 10p/9q

[nocturne owl]

i dont understand sorry

[stable flame]

what dont u understand exactly

u simplified a fraction, u still got the rest of the equation

[nocturne owl]

so we got 1/2*2p+q right?

[plush bramble]

no

[plush bramble]

$\frac{2p + q}{2p + q} = ?$

[jolly parrot]

riemann

[plush bramble]

use this

[plush bramble]

and

[nocturne owl]

ye i write this and after that?

[stable flame]

what do u have on ur paper at the moment?

[plush bramble]

we can't tell unless you show what you have

[plush bramble]

if you didn't make a mistake you should have this

[nocturne owl]

ye well its not

[plush bramble]

[nocturne owl]

im stuck here

[plush bramble]

.

[plush bramble]

.

[nocturne owl]

1/2* 2p*q?

[stable flame]

no

what is ur thought process rn

what keeps getting u that result

[plush bramble]

no

[plush bramble]

$\frac{2p + q}{2 (2p+q)} = \frac{1}{2} \times ?$

[jolly parrot]

riemann

[nocturne owl]

but you told me to simplify dont you?

[stable flame]

yes

[nocturne owl]

and also how we got 1/2

[stable flame]

because u simplified

hes trying to explain you what simplifying is

heres a number example

[plush bramble]

you simplified incorrectly

[nocturne owl]

ye ik what simplyfying is but i dont understand why we got 1/2

[plush bramble]

answer this

[plush bramble]

using this logic

[nocturne owl]

2p/4p*1 right?

[plush bramble]

no

[plush bramble]

read this as many times as it takes

[stable flame]

how are u getting to these results?

try writting it on a paper

so we can see what ur actually doing

[nocturne owl]

okay

ill show yall how its look

my pager

[stable flame]

why did u cut stuff on one side of the equation and change stuff on the other side?

[nocturne owl]

i dont know man 😭 im confused so hard now

[stable flame]

u probably should get back to some easier problems

this is only the first step to solving ur problem and it doesnt seem like were making much progress

[nocturne owl]

like im not that kind of stupid belive me i just not always understand what you guys awnt

want

my english math is really bad

thats why i got confused and write these stupid things

this is my first time writing in this server so i dont really know how all these works

[stable flame]

no1 is calling u stupid

u have written the equation correctly, english wouldnt be an issue here

its basic algebra thats the issue

u shouldnt be afraid to take a lil step back and re-learn that

[pearl pond]

@nocturne owl Has your question been resolved?

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[quasi lynx]

my question was in fact, not resolved

[quasi lynx]

"i need to find a distance for a test mass m_2, i have the acceleration which is shown as the equation for g, which to get the v_f i multiply by time.
the problem is that to use g, i need a distance, which must be the previous distance. i need to be able to use d as a smooth function (not jumping from the d value at one second to two seconds)"

[ivory swallow]

Clarify a bit more

What are you given

What is to be solved for

[quasi lynx]

i have d_i (or d(0)), m_1, G, v_i and time is to be input, im trying to "solve" for d, however since d is in the equation for d, i cant exactly do that, if i used a regular function of d(t) then that wouldnt be smooth

this might be at a post graduate level

heeeeelp

<@&286206848099549185>

[pearl pond]

@quasi lynx Has your question been resolved?

[pearl pond]

@quasi lynx Has your question been resolved?

[ashen ivy]

assuming that d = d(t) in both places, why not multiply both sides by d^2? you would then be finding the roots of a cubic polynomial. (it appears to be solvable assuming G = G, and that velocity and mass are >0)

that being said, i am a little unsure what the setup is exactly here, and this formula strikes me as weird if im honest. are you averaging the velocities to get distance?

[quasi lynx]

this is just /\d= (v_i + v_f)/2 but with v_f being the gravitational acceleration multiplied by time

and multiplying by d^2 doest get rid of the d

[ashen ivy]

i thought it was $\Delta d = \frac{(v_i + v_f)}{2}t$, i.e. $d = \frac{v_i t + \frac{Gm_1}{d^2}t^2}{2}$? but i digress

[jolly parrot]

ηασιβ ♥

[ashen ivy]

it doesn't get rid of the d, but it gives you a cubic polynomial you can solve for 3 "possible formulas" for d

right now this is some kind of recurrence relation, but those 3 solutions will depend solely on v, t, G, and m. they would be messy, though

[quasi lynx]

i feel as though my eyes have been burnt, looking it up with some ai (not the bad kind) assistance, it seems very complicated

oh no did i write it down wrong.... is the original displacement equation multiplied by time...

shit

oh god its worse now, oh lord

ahhhh, thanks ok bye

.close

[pearl pond]

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[fluid pewter]

help plsss

[ashen ivy]

hence why i was wary about the formula 😭 but if this is it, then i would solve for 3 potential formulas and eliminate incorrect ones (e.g. a formula that assumes distance is less than zero)

[ashen ivy]

this channel hasn't opened yet - how about #help-30 or any channels in the available category?

[pearl pond]

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[ember zinc]

Hello! I am just reviewing a linear algebra exam I had a couple days ago, here is the question.

I thought both (a) and (b) would be valid diagonalizations because they just have their eigenvalues and associated eigenvectors swapped, but the answer key only says answer (a). Are both options diagonalizations?

[ember zinc]

I also plugged them in, and they both worked

[tulip ore]

yes theyre both valid

[ember zinc]

alright, I will just have to talk to the professor then I fear, thank you for the help!

[tulip ore]

np

[ember zinc]

.close

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[pearl pond]

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[mortal wolf]

For part b isn’t the range where f’(x)> 3/4? I thought it would be just above the y-intercept why is it 1/2? thanks

[hollow harness]

Help me

[rough meteor]

can u restate yr query?

[mortal wolf]

Why is the range of the inverse f’(x)>1/2? idk where they got 1/2 from and I thought it would be instead >3/4 (because that’s the y-intercept)

[rough meteor]

wait wait im rechecking

yes u r right it should b 3/4

i cant think why would it b 1/2

[mortal wolf]

Ah ok thank you

.close

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[void grail]

彡+开

I can see 5 shapes, before I saw just crumples

4 disks in a disk in in a disk in a space?

[pearl pond]

@void grail Has your question been resolved?

[pearl pond]

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[vagrant briar]

Do you know about the remainder theorem?

[eager jewel]

uh no i dont think so

i mean i might

could u tell me what it is

[vagrant briar]

hm i don't think this question will be solvable if you dont know that. not much information has been given

[eager jewel]

could u tell me what it is

i might have just forgotten it

[vagrant briar]

well simply put, the remainder theorem states when a polynomial P(x) is divided by a linear of the form x-b, the remainder is simply the value of the polynomial at x=b, i.e, the remainder when P(x) / (x-b) = P(b)

[eager jewel]

oh ok ..how do we use that here

[vagrant briar]

from this theorem, a result can be derived which sates that when P(x) is a polynomial with integer coefficients and at least one integral root for any integers a and b, P(a)-P(b)=P(a-b) which is divisible by (a-b)

[eager jewel]

oh ok

so P(8)=P(2)-P(10)=8

[vagrant briar]

yes, but this wont help us much. lets assume a=2 and b=the integral root of the polynomial

so P(b)=0

now P(2)-P(b)=13-0

so 2-b divides 13

can we say that?

[eager jewel]

uh one sec

nvm

yes

[vagrant briar]

P(2) is in your question

[eager jewel]

oh ok now i can cross check w the options and get the ans from this only

[vagrant briar]

now what is one special property of the number 13

[eager jewel]

prime no.?

[vagrant briar]

yes

[eager jewel]

so what can we say from that

[vagrant briar]

you tell me if 2-b divides 13 and 13 is a prime, what can we say about 2-b

[eager jewel]

its prime?

[vagrant briar]

if a number is prime, then only 1 and the number itself can divide it

so 2-b has to be 1 or 13

got it?

[eager jewel]

yea

[vagrant briar]

but here we're dealing with a polynomial so it may have negative roots too, so we've also consider -1 and -13

so 2-b= -13 or -1 or 1 or 13

now can you find the possible values of b?

[eager jewel]

yeah got it

thank u

[vagrant briar]

similarly you've to repeat this process with 10

[eager jewel]

oh alr alr was just abt to ask what about P(10)

[vagrant briar]

you've to repeat the same process with 10, again assume the root to be 'b'

you'll find another set of values

the root 'b' will be the one which is common to both the sets

[eager jewel]

thanks

.close

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[desert patrol]

It's asking for the positive of tanx - cotx and the answer is 3 but I want to know how it's done

[desert patrol]

Also I accumulated some of the questions I couldn't solve is it okay if I can ask all of them here?

[vernal bear]

,rotate

[light helm]

consider what
(tan(x) - cot(x))^2
expands to

[rugged iron]

Yoo guys Hlp me with an question

[ruby cargo]

!occupied

[pearl pond]

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).

[ruby cargo]

You can't close someone else's thread

You don't get to decide if someone else has a doubt or not

[desert patrol]

11 - 2tanxcotx = (tanx-cotx)²?

[rugged iron]

If,X=2(3-x) and Y=2(2-x)
Then,
X-Y=?

[ripe lantern]

That seems good to me

[rugged iron]

What?

[rugged iron]

Help me with this pls

[desert patrol]

Why don't you use another room instead?

[ripe lantern]

It was said to get your own channel

[rugged iron]

Ok

[ripe lantern]

Do you see where to go from here?

[desert patrol]

Not really

[ripe lantern]

Notice the 2tanxcotx

[desert patrol]

OH

[ripe lantern]

Can you do something to simplify this?

[desert patrol]

So 2?

[ripe lantern]

Yep

[desert patrol]

(tanx-cotx)² = 9

So it's 3?

[ripe lantern]

Yeah, the positive

Which corresponds to the solution right

[desert patrol]

Yeah

Can you also check this one out?

It's asking for tan but I don't know how to solve it

[ripe lantern]

The solution is sqrt(7)/3?

You can draw a perpendicular from A, which bisects 2beta

[desert patrol]

Yeah

[desert patrol]

So a + b = 90

Oh

[ripe lantern]

Yeah, but you know sin(beta)

[desert patrol]

I got it thank yoy

[ripe lantern]

Great 🙂

[desert patrol]

The answer is E

I need to find cotx to find tanx here?

I tried to find |AT| but it's too complicated

[ripe lantern]

Actually, finding AT seems to be a good move

[desert patrol]

But the variables make it way harder

is it sqrt(3r) - r

[ripe lantern]

If angle TAO’ is 90-alpha, what is the angle AO’T?

[ripe lantern]

Not that much

Write down your Pythagorean theorem

[ripe lantern]

You forgot to square the 3r

[desert patrol]

Wouldn't -r² be -r and 3r sqrt(3r)

[ripe lantern]

No, you would have $(3r)^2=r^2+x^2$

[jolly parrot]

FriedRiceKing

[desert patrol]

Oh...

x = 2r

I found D but it says it's E

[ripe lantern]

x=2r?

[desert patrol]

Oh wait...

8r

No😭

It should be smaller than 3r

[ripe lantern]

$(3r)^2=r^2+x^2 \ 9r^2=r^2+x^2$

[jolly parrot]

FriedRiceKing

[desert patrol]

8r² = x²

For some reason I keep ignoring the integer

x = 2sqrt(2)

[ripe lantern]

Ha I see, that happens

[ripe lantern]

Yep

[desert patrol]

Yeah thank you

[ripe lantern]

You know where to go from here?

[desert patrol]

Yeah

[ripe lantern]

👍

[desert patrol]

.close

[pearl pond]

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[desert patrol]

How do you solve these questions?

[jovial ember]

what does br mean

unit?

[desert patrol]

Yeah

[jovial ember]

there’s a formula for this

[desert patrol]

sinus theorem idk

[jovial ember]

is there any angle given?

[desert patrol]

prob cos

No

[jovial ember]

you need the area no?

[desert patrol]

It asks for cotB + cotC

So we need BD and DC

[jovial ember]

I see

[tacit saffron]

Solve linear algebra

5,9

[flint fulcrum]

Is BAC 90 degrees?

[jovial ember]

If that’s the case it’s quite easy

[flint fulcrum]

Actually isn't cotB+CotC

Just BD/AD+CD/AD= BC/AD

So it's just 8

I was overthinking it

[desert patrol]

You are right😭

Can you help me with this one as well?

It's asking the solution for cos³x/1+sinx

[vernal bear]

what do you think?

[desert patrol]

I couldnt't figure it out

[vernal bear]

you know cos^2x =1-sin^2x?

[compact ridge]

I'd do $\cos x = \frac{1}{1 - \sin x} \frac{1 + \sin x}{1 + \sin x} = \frac{1 + \sin x}{\cos^2 x}$

[jolly parrot]

south

[compact ridge]

omg I just gave you the answer

try to think and fill in the intermediate steps yourself

[pearl pond]

@desert patrol Has your question been resolved?

[pearl pond]

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[rapid wagon]

Is it possible for a singular square matrix to have LU decompositon without the Piovting matrix P?

[rapid wagon]

Let's say that the linearly dependent row is the last one, so I get a 0 for the last row of U ?

[pearl pond]

@rapid wagon Has your question been resolved?

[tropic saddle]

well just pick literally any singular triangular U you want and pick L=I

and then A=LU

[rapid wagon]

Oh..

What if A is symmetric?

Singular square symmetric matrix can have a LDL^T decomposition?

So how I know is that symmetric PD matrices for sure always have a LDL^T, so if i broaden the range and go to symmetric, it doesn't always work, because if the matrix is singular then i will have 0's in the diagonal of U so I can't do U=DL^T

Then again what if the zero is in the last row of U?

[tropic saddle]

LU decomp is literally just gaussian elimination but fancy

of course there are symmetric matrices that you can do gauss on without having to switch rows at some point

wdym you cant take U=DL^T

under what def of LU decomp are you working

[rapid wagon]

I mean if the matrix is symmetric and PD then the decomp can look like LDL^T where U=DL^T and D is a diagonal matrix with the diagonal of U, L^T is unitary upper triangle but the elements are "normalized" because i divide by the diagonal element for each row, and if the matrix is singular and I have 0 on the u_ii then i cant divide the row ii by 0 so LDL^T decomp doesn't exist for it

[rapid wagon]

I see, that answers my question ig

If i think of it like that

Buhh

If u dont have anything to add i will close ok?

.close

[pearl pond]

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[pearl pond]

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[thick musk]

Can I ask a question?

[thick musk]

Is C1 and C2 correct?

May I ask a question?

[ashen ivy]

you seem to have already asked a question :)

injectivity looks good, surjectivity is correct, but i think the proof could be clearer.
something like "let a in A" at the start, then at the end you say "f(a,b) = a for any a in A, so surjective"

[thick musk]

?

oh

haha

[ashen ivy]

i.e. just making it clear that you are picking an input for all possible outputs

[thick musk]

which proof, c1 or c2?

[ashen ivy]

both of c1 and c2, specifically about surjectivity

[thick musk]

where do I say "for any a in A"?

[ashen ivy]

right at the start should be good

something like: \
For any $a \in A,$ pick $(a,b) \in A \times B.$ Then $f(a,b) = a$, so $f$ is surjective

[jolly parrot]

ηασιβ ♥

[thick musk]

isn't mine saying the same thing? "if a is in A" implies not just one element but all elements in A?

[vagrant briar]

what i think they're trying to tell you is that proofs are better and more clearer to read and understand if you use a bit of language

[ashen ivy]

yeah, the way you wrote it, it felt like you were setting the input as arbitrary, as opposed to picking an input based on the output

[thick musk]

according to an online solution C2 i. is not injective

?

this is the online proof

[ashen ivy]

um? the proof basically says that f is injective and then the last sentence is the opposite

[thick musk]

yea I don't understand

[ashen ivy]

you see that it ends with a=c and b=d which is like the definition of injectivity, right?

maybe im losing it but that seems like a typo to me

[thick musk]

lol

can anyone else chime in?

[fluid axle]

indeed a typo

[thick musk]

ok

[thick musk]

so in this example, you are including all possible inputs of A x B, whereas in my solution it was choosing arbitrary a in A and b in B which only covered one input?

[pearl pond]

@thick musk Has your question been resolved?

[thick musk]

.close

[pearl pond]

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[thick musk]

Is D 1-3?

I think 1 and 2 are correct

3 I am having a bit trouble fully understanding

with 3 g(f(x)) => g(1/x)

^ at that step we need to evaluate the domain (0,1) to ensure every g(1/x) is defined?

then we go from g(1/x) => ln(1/x)

^ at that step we still could only evaluate x with the domain (0,1)?

so basically my question is if both F and G have different domains, we actually only evaluate the composition of both F and G within the domain belonging to the function that is being composed with? so in the case of G composed with F, we evaluate 'x' within the more restrictive (0,1) domain and if F composed with G, we evaluate 'x' within R->R?

[thick musk]

.redo

.reopen

[pearl pond]

✅ Original question: #help-39 message

[thick musk]

Is this true?

[pearl pond]

@thick musk Has your question been resolved?

[humble root]

Yes

If you are finding F(G(x)), the function is only defined when:

1. x satisfies the domain of G(x)
2. The values of x where the image of G satisfy the domain of F.

[humble root]

Also are you sure the question is right

g doesn't look well defined

[pearl pond]

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[thick musk]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[thick musk]

Why does it not look well defined?

[humble root]

Because the domain of g is all real numbers

But

That's definitely not so, if g(x) = ln(x)

[thick musk]

But it’s g(1/x) such that x is in this interval (0,1) making it defined for g(x)

1/x defined in the interval (0,1) results in a value that is in R (domain for g(x)

[pearl pond]

@thick musk Has your question been resolved?

[pearl pond]

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[dapper kraken]

AB=5, BC=6, AC=9, incircle tangent to BC and AC at D and E, AD and BE intersect at F, find CF

how do you do this without bashing like every single length

[compact ridge]

barycentric coordinates?

[dapper kraken]

havent studied that

[pure rapids]

bary bashing 💀

[pure rapids]

menelaus spam 😭

[dapper kraken]

huh i forgot menelaus existed

[wary siren]

is it to scale?

[solemn kernel]

hello fellas

[dapper kraken]

i made it so yeah

[pure rapids]

Welcome to mathcord @solemn kernel

This is a help channel, go to #discussion for casual conversation

[pure rapids]

also bro did you take amc 12

[dapper kraken]

no

[pure rapids]

pity

ok

[dapper kraken]

BD=1, CD=5, AE=4

[dense jasper]

I think you can do ||mass points|| followed by ||Stewart's if you extend CF to meet AB||

[dapper kraken]

extend CF to meet AB at X

if im not mistaken X is the other tangent (?)

ye gergonne point

[dense jasper]

ok yeah this works

[dapper kraken]

from stewarts 6^2×4+9^2×1=5(CF^2+4×1)
CF=sqrt41

put B=4 A=1 which gets C=4/5 i think?

then X=5

[dense jasper]

Ngl I didn’t actually crunch the numbers lol

[dapper kraken]

so CX=(5)/(5+4/5)×sqrt41=25×sqrt41/29

ok which is in the options

okay, tyyy

[pure rapids]

orz

[dapper kraken]

.solved

[pearl pond]

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[dapper kraken]

whar

[dense jasper]

[pure rapids]

search it up

[pearl pond]

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[gentle lichen]

I tried

• using Bernoulli's equation
• substituting y = tanθ

[gentle lichen]

The latter got me to this equation-

(x - e^θ)dθ + dx = 0

But idk how to proceed from there

[vagrant briar]

i can give you a hint

try bringing the expression in terms of dx/dy instead of dy/dx

try playing around

you'll see that type it is

@gentle lichen

[gentle lichen]

A linear differential equation

Thank you

.close

[pearl pond]

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[gentle lichen]

Trying to make dy/dx gives me - 2k²/y dx

What do I do with dx in the denominator

Is the equation even correct?

I feel like there should be a d(something) in the RHS too

<@&286206848099549185>

[drowsy adder]

try to make d(y/x)

[drowsy adder]

and then substitute it as t or something

[gentle lichen]

What about the RHS

[drowsy adder]

you want to solve the differential equation?

or what?

[gentle lichen]

Yes

[drowsy adder]

it will adjust accordingly

try

[gentle lichen]

$d\left(\frac{y}{x}\right) = - 2\frac{k^2}{yx^2}$

If I integrate both sides

There's no dx or dy on the right

[jolly parrot]

Dhruv

[drowsy adder]

hmm i think youre right

theres something off with the equation

ill try solving it as they originally intended and lyk

[gentle lichen]

👍

[drowsy adder]

yeah crossmultuply and divide by y^2

itll convert to a linear

[pearl pond]

@gentle lichen Has your question been resolved?

[pearl pond]

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[short orchid]

guys am i doing something wrong here? why is my 5/2 negative?

[serene sentinel]

what's f(x) there?

$-\frac{4}{1+2x} -\frac{3}{2-x} + \frac{5}{2-x^2}?$

[jolly parrot]

donkey

[short orchid]

this onee

[leaden wadi]

You mathed wrong somewhere because it should be positive.

[rotund ferry]

Hint, the mistake is here

Something wrong with your u-sub

[short orchid]

5(2-x)^-2
we’re supposed to bring the (-2+1) down right?

[rotund ferry]

Wdym

You're supposed to divide it by -1

[short orchid]

wait yes thats what i meant

[rotund ferry]

Anti derivative x^n = x^(n+1) /(n+1)

You forgot to divide by -1

[short orchid]

But i have multiplied it by -1

Oh wait so multiply by -1 is for the derivative of -x?

[leaden wadi]

That's from the -x, but you disregarded the negative sign from the exponent.

[rotund ferry]

Yes

[short orchid]

Ohhhhh

[rotund ferry]

So it's supposed to be positive