help-19

[next bobcat]

Ok

[cold sage]

It takes a minute to lay 10ft^2, so how long for this many

[next bobcat]

So times that by 60?

Or times it by the seconds

775,001.52

Oh wait now I get what you were tryna say

This is a yd or ft?

[cold sage]

Neither

Its ft^2

[next bobcat]

Oh

[next bobcat]

So was this affirmative?

[cold sage]

Multiply what by 60

[next bobcat]

60 minutes

Or do we multiply by the seconds an hour has

[cold sage]

No idea what you mean, if you mean multiply the 10 by 60 then sure

[next bobcat]

Wait so this is just a minute

So wouldn’t we multiply that number by 60 or the 10 by 60?

If the 10 by 60 that’s 600

[cold sage]

No that is not a minute

That is the area

You lay 10ft^2 per minute

[next bobcat]

Yeah

10ft^2 per minute so times it? No?

[cold sage]

Then yeah 600ft^2 per hour

[next bobcat]

Alright

So that’s the answer?

[cold sage]

No?

You havent found how many hours it takes to lay the floor

[next bobcat]

So we have it’s 600ft^2 per hour

What do I even do like after that.?

[cold sage]

You know the total area, you know how much is layed per hour

Think about it

[next bobcat]

Wait what even is the total area again

?

[cold sage]

Yeah ft^2

[next bobcat]

It would take 3 hours

[cold sage]

Wait no

That ones in yd^2

This is ft^2

[next bobcat]

And it’s 600ft^2 per hour

So 2 hours?

[cold sage]

No

Do the calculation

You seem to just be eyeballing it

[next bobcat]

Yeah I was

Wait so how much would 600ft^2 even be?

So this is 1ft^2 or 10ft^2?

[cold sage]

You know its neither of those

Its 12916.692ft^2

[next bobcat]

So that’s what

[cold sage]

The area??

My man, cmon

[next bobcat]

Alright

[cold sage]

This is the area, they lay 600ft^2 per hour

[next bobcat]

Ima be honest I am uh lost 😭

But

From what I know

OH

They lay 600ft per hour so

21ish hours

600*21

[cold sage]

We come full circle

Actually get the number though

[next bobcat]

So 21.5

Or 21.527

Alright

FINALLLY

But one last question…..

[cold sage]

where are these numbers from

[next bobcat]

From previous questions

That give the conversion of the usd to the currency used

[odd edge]

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[next bobcat]

.close

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[misty bane]

can someone explain why 26c+4=1

is the 1 the standard form of the equation

or is it specific to thep roblem

I understand how he did everything just that is somehting I am curious about

[chrome lynx]

i think its just a calculation step

the endgame is just c=-3/26

[misty bane]

so wait

why is the inital one set equal to 1

like

is that standard form?

@chrome lynx

[chrome lynx]

i mean i would have to see the whole problem

it seems you have a point on an affine plane

[misty bane]

this is the inital equation

or question

I am curious if i can set future equations equal to 1 with the same steps

or was it specific to this question

@chrome lynx

[chrome lynx]

i mean the =1 seems to come from the equation of the plane

itself

[misty bane]

ok so that is standard

so if the numbers changed i could still do the same thign?

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[scenic mesa]

hlep

[scenic mesa]

i need help starting the problem

nvm

.close

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[onyx solar]

anyone can explain where the minus sign comes from?

[onyx solar]

I kept putting +

since its product rule but its wrong

maybe i did somthing wrong

[fair prism]

Chain rule from derivative of a^(-t)

[onyx solar]

wym

isnt it just ln (a) * a^-t

oh

times -t = -1

[fair prism]

Yeah

[onyx solar]

and u distribute

ok got it thx

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[mystic saffron]

Can someone explain how to factor in this problem to get the derivative?

[mystic saffron]

Here is my work (problem 25). How did the factor in the numerator become x+h-1 for f(x)?

[odd edge]

@mystic saffron Has your question been resolved?

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@mystic saffron Has your question been resolved?

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@mystic saffron Has your question been resolved?

[odd edge]

@mystic saffron Has your question been resolved?

[lofty spade]

Which line?

[odd edge]

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[molten basin]

I'm stuck on this, I forgot the difference between quadratic and exponential

[mystic saffron]

have you created the scatter plot

and what does it look like

[molten basin]

No, I do not have a graphing calculator

[mystic saffron]

you can use ,w i forgor the prompt tho

,w (1911, 12) , (1919, 158)

[clever fjord]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by senaeon

[mystic saffron]

continue

[molten basin]

Ok

,w (1927, 203) , (1942, 223)

[clever fjord]

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Requested by livingsacrifice121

[mystic saffron]

i mean for more points like so

(x1,y1), (x2,y2), ....

in one command

[molten basin]

,w (1927, 201) , (1942, 223) , (1951, 198) , (1958, 161) , (1971, 9)

[clever fjord]

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Requested by livingsacrifice121

[mystic saffron]

yess some points are missing tho

first 2

[molten basin]

,w (1911, 12) , (1919, 158) , (1927, 201) , (1942, 223) , (1951, 198) , (1958, 161) , (1971, 9)

[clever fjord]

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[mystic saffron]

okay good

now what function, quadratic or exponential takes this shape

[molten basin]

uhhh wdym? I'm sorry im a slow thinker

[mystic saffron]

so all of those points give this shape right

[molten basin]

yes

[mystic saffron]

have you seen how exponential and a quadratic functions look like

[molten basin]

I think exponential are kind of scattered right?

[mystic saffron]

hmm?

can you draw for me

[molten basin]

It's a quadratic function ?

[mystic saffron]

i cant tell you that yet since im not sure you understand the distinction

[mystic saffron]

can u sketch one for me

[mystic saffron]

and this too

[molten basin]

Ok

exponentials look like this ? I drew on my pc since i dont have a piece of paper at the moment

[heavy hazel]

HEY

CAN SOMEONE PLZ HELP KE

[mystic saffron]

uh

[mystic saffron]

!help

[odd edge]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[molten basin]

I think I understand now-

.close

[odd edge]

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[heavy hazel]

HOW DO I MAKE MY OWN TESTELATION WITH 2 SHAPES

THING

[mystic saffron]

no

reopen [rip]

[mystic saffron]

exponential functions look something like not that

[odd edge]

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[mystic saffron]

.close

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[stable path]

Hi can I have some help with this quetion

[mystic saffron]

!status

[odd edge]

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

[stable path]

1

[mystic saffron]

okay

what are given so far

[stable path]

its more 2

[mystic saffron]

what have you tried

[stable path]

ive got d = 0

and then i have two equations from the points provided

[mystic saffron]

idk what technique you used but im gonna go through with you another one

[stable path]

-2 = a + b + c

and 3 = −8a + 4b − 2c

[mystic saffron]

so for any given polynomial function like this f(x) and lets day you know the turning point (c,d) what would f'(c) be

[mystic saffron]

.

[stable path]

👍

[mystic saffron]

when you differentiate this what function are you getting

[stable path]

3ax^2 + 2bx + c

[mystic saffron]

what is that called

the f'(x)

[stable path]

yeah

[mystic saffron]

if you havent seen the term tell me

[stable path]

ive seen f'(x) before

its basically dy/dx

[mystic saffron]

but why do we f'(x)

[stable path]

gradient?

[mystic saffron]

yess

so f'(x) is the gradient function

and what do you know about the gradient at the turning points

[stable path]

= 0

[mystic saffron]

very good

[mystic saffron]

you have a turning point R(-2,3)

so this would mean P'(-2) would be?

[stable path]

ill do the working now

[mystic saffron]

okay

[stable path]

12a -4b-2

not -2

• c

[mystic saffron]

the equation please

[stable path]

-c

[mystic saffron]

the full work

[stable path]

3a(-2)^2 +2b(-2) + c

[mystic saffron]

.

p'(-2) = ?

the value

[mystic saffron]

.

[stable path]

not sure

= 0

ohh

yeah

[mystic saffron]

(-2,3) is a turning point

yess

good

[mystic saffron]

so you differentiate p(x)

and do p(-2) = 0

can you work that out for me

and please show all your steps

[stable path]

12a -4b + c = 0

[mystic saffron]

what is p'(x)

[stable path]

the gradient function

[mystic saffron]

this

so p'(x) = 3ax^2 + 2bx + c

right

[stable path]

3a(-2)^2 +2b(-2) + c = 0

[mystic saffron]

the format i want

p'(x) = 3ax^2 + 2bx + c
p'(-2) = ....
0 = ....

[mystic saffron]

but yes

[mystic saffron]

go on

[stable path]

12a -4b + c = 0

[mystic saffron]

okay now we're going to pick a variable to make the subject of the formula

[mystic saffron]

pick one

[stable path]

a

a

[mystic saffron]

hmm will that be easy to work with

[mystic saffron]

id take out c

[stable path]

ok

so c = -12a + 4b

[mystic saffron]

okay good we're gonna keep that on the side

[mystic saffron]

next piece of information we have is Q(1, -2)

[stable path]

and then u input c back into the first two equations and then use elimination?

[mystic saffron]

no that will just give you zero

we're gonna solve simultaneously

[mystic saffron]

can you write that in the form

p(x)=....
?=....

[stable path]

that ends up being -2 = a + b + c

and the other point is 3 = −8a + 4b − 2c

[mystic saffron]

this format please

makes it easier for me to see

and for the person grading your working

[stable path]

sorry i have to go to dinner can i private message you about this question in 20 mins

[mystic saffron]

.close and ping when you come back i'll help if im still around

[stable path]

ok thank you

.close

[odd edge]

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[dawn mirage]

Hi, im working with complex numbers and eigen values and eigen vectors i need urgent help

[dawn mirage]

here is my work so far, and the question

getting very stuck here, as im unsure what to do now

[echo ginkgo]

I mean you pretty much have the same equation twice when you look at it a bit carefully (that's what you'd expect with eigenvalues)

y = -ix and x = iy

[dawn mirage]

i think its because i have complex eigen values

im getting confused

[echo ginkgo]

even with real eigenvalues

[echo ginkgo]

if you take that first equation and multiply both sides by i you get the second one

so one of them is redundant

[dawn mirage]

wait sorry one second im gonna watch a video on odes with complex eigenvalues and then come back if i still have more questions

i think thats my issue as i havent dealt wit hthem before

[echo ginkgo]

it's honestly the same as real eigenvalues

so now we know the eigenvectors for lambda = -1+i satisfy the equation x=iy

now just pick one of them

for example [i 1] works

[dawn mirage]

okay

[odd edge]

@dawn mirage Has your question been resolved?

[dawn mirage]

.close

[odd edge]

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[wooden python]

!status

[odd edge]

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

[restive current]

i dont know where to begin

[wooden python]

do you have the definition of an order relation in front of you?

[restive current]

no i dont

i dont think ive even learned this yet so maybe some resources or a general explanation of how to do the question would be really helpful

[wooden python]

your textbook should have it

alternatively look at intro to relations in discrete math idk

[restive current]

i dont have a textbook

[restive current]

ill look into that

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[fathom ingot]

I don’t understand how to find the vertex form

[cold sage]

do you know how to find it generally

[cold sage]

or is it just this specifically thats confusing you

[fathom ingot]

I missed over an week of school

Last I was here we were doing factored quadratic formulas

The task is “rewrite each equation in Vertex form by completing the square and list all transformations and parent functions”

[odd edge]

@fathom ingot Has your question been resolved?

[lofty spade]

,rotate

[lofty spade]

can you complete the square here?

[odd edge]

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[fallen remnant]

f(x) is function defined from Z to Z
It's given f(x+1)=C.f(x)+1 where C is a constant
Can we conclude f(x) is a linear function

[viscid flint]

what's the general form of a linear function?

[chrome lynx]

try to make an example of such a f

like lets say C=2 and f(0) = 0

what do you get for the first integers?

[odd edge]

@fallen remnant Has your question been resolved?

[fallen remnant]

f(x)=ax+b

[chrome lynx]

try to get an intuition for this kind of function

make an example first

to guide you

[odd edge]

@fallen remnant Has your question been resolved?

[odd edge]

@fallen remnant Has your question been resolved?

[odd edge]

@fallen remnant Has your question been resolved?

[odd edge]

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[hearty prawn]

how did they get -2pi/5

[hearty prawn]

wouldnt it be -0.4pi ?

lol its the same thing

[brave flint]

Hmm both are same

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[pastel dew]

what did you do ?

[thorn briar]

i was trying for similarity

but cant find any clue

@pastel dew are you still there?

[pastel dew]

yes

[thorn briar]

thanks

[pastel dew]

i always here

so...

[thorn briar]

solving it?

[pastel dew]

almost

[thorn briar]

ohh

asking school school grade physics is allowed here?

[pastel dew]

maybe no

but geometry is allowed 😉

[thorn briar]

work,power,energy?

[pastel dew]

i think there is a discord server for that stuff

[thorn briar]

ohh thanks

[pastel dew]

now you have to stay focus on this problem

[thorn briar]

ok

[pastel dew]

mhhh,,,

[thorn briar]

HMM

[pastel dew]

i have a solution that it nice for olympiad math

[thorn briar]

ohh

this my school book question

i thought it was going to be easy

[pastel dew]

do you now that two triangle are similar iff they have all three side in proportional

now if you use that onefrom that one

(its AD's length in your problem i.e. the legth of the median)

you'll find that also CB is proportional to RQ

and then your done

[thorn briar]

sorry bro explain more deliberately

[pastel dew]

do you now that two triangle are similar iff they have all three side in proportional?

[thorn briar]

yes

[pastel dew]

ok, if $\frac{AC}{PR}=\frac{AB}{PQ}=\rho$ then we have to prove that $\frac{CB}{RQ}=\rho$

[clever fjord]

everg

[pastel dew]

right ?

[thorn briar]

yes

[pastel dew]

ok so

[thorn briar]

yeah nesxt

[pastel dew]

from that one

[thorn briar]

hmm

[pastel dew]

you knwo that $a=\sqrt{{b^2+c^2}-2m_a^2}$

[clever fjord]

everg

[pastel dew]

if we know that $\rho b=b'$, $\rho c=c'$ and $\rho m_a=m_{a'}$ then

[clever fjord]

everg

[pastel dew]

$\rho a=\rho \sqrt{{b^2+c^2}-2m_a^2}=\sqrt{\rho^2{b^2+\rho^2c^2}-2\rho^2m_a^2}$

[clever fjord]

everg

[thorn briar]

here 'a ' stands for what?

[pastel dew]

CB

the side opposite to A

its a common notation

[thorn briar]

ohh

[pastel dew]

b is AC

[thorn briar]

ok

[pastel dew]

c is AB

ok so

$\sqrt{(\rho{b)^2+(\rho c)^2}-2(\rho m_a)^2}$

[clever fjord]

everg

[pastel dew]

$=\sqrt{(b')^2+(c')^2-2( m_{a'})^2}=a'$

[clever fjord]

everg

[pastel dew]

then we proved that $\rho a=a'$ so we are done

[clever fjord]

everg

[thorn briar]

ohh

[pastel dew]

two line proof

$\rho a=\rho \sqrt{{b^2+c^2}-2m_a^2}=\sqrt{\rho^2{b^2+\rho^2c^2}-2\rho^2m_a^2}=\sqrt{(\rho{b)^2+(\rho c)^2}-2(\rho m_a)^2}=\sqrt{(b')^2+(c')^2-2( m_{a'})^2}=a'$

[clever fjord]

everg

[thorn briar]

h,m

[pastel dew]

human method ?

[thorn briar]

yup

[odd edge]

@thorn briar Has your question been resolved?

[odd edge]

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[prime scroll]

using the vector product rule, e.g cross product

[zenith tartan]

!show

[odd edge]

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

[prime scroll]

can someone please tell me why the bottom column of the matrice is -2,-2,2

[zenith tartan]

it should be the cross product of b and c

[prime scroll]

ah

so you do the cross product twice?

b and c

then the result of that, calling it f then f*a

[zenith tartan]

yea first of b and c, and then what ever you get, cross product with a

[prime scroll]

.close

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[tepid lodge]

How exactly do you implement the gauss seidel method? What happens if you only have a single linear equation? I'm quite confused as to how the rearrangements for different variables work, does each variable need to be derived from a different equation?

[odd edge]

@tepid lodge Has your question been resolved?

[tepid lodge]

more specifically, i'm asking because i implemented it but it seems to converge after only 1 iteration.

This is using a single linear equation

<@&286206848099549185>

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@tepid lodge Has your question been resolved?

[tepid lodge]

Nevermind, i figured it out.

.close

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[untold turret]

help how would i do this :D

[lyric marlin]

I (assume) if you find bounds $m,M$ for $0< m\leq f(x,y)\leq M$, where $f(x,y)$ is the integrand, then $$\int\ind_DmdA\leq\int\int_Df(x,y)dA\leq\int\int_DMdA$$

[clever fjord]

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

[lyric marlin]

and those upper and lower integrals are just m and M multiplied by the area

This is the easiest possible bounds

you might also need to find easily integrable functions $m(x,y)$ and $M(x,y)$ that act the same way

[untold turret]

wait like solve the double integrals?

[clever fjord]

Edward II

[lyric marlin]

well m and M are constants so they're really easy to integrate, it'll just be 8m and 8M because 8 is the area of $D$

[clever fjord]

Edward II

[lyric marlin]

this is probably the 'bit of thought' bit though

[untold turret]

is there another way to do it other than solving the double integrals

im pretty sure i dont have to do that

or im not suppose to

[lyric marlin]

I'm... not asking you to

[untold turret]

OHH

[lyric marlin]

I'm saying that if $m\leq \frac{xe^10y^2}{\dots}+\dots\leq M$

[clever fjord]

Edward II

[lyric marlin]

then the double integrals also are bounded

[untold turret]

like you take things that are similar and obviously smaller than that integral and can prove that its in bounds

then we can just state its true

[lyric marlin]

I've not actually done it, so you might need m and M not simple constants, but if you pick them as the right functions of x and y they might still intergate really easily

[untold turret]

hmm ic

i can see that that will be lowkey kinda hard lmfao

ik how to do it if it wasnt double integrals

but

ok thanks

.close

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[flat mortar]

i am not understanding x^3

nothing is working

[gleaming coral]

synthetic division?

[flat mortar]

yes

@gleaming coral

[gleaming coral]

i was gonna suggest u substitution but i dont think that would work either

[flat mortar]

i tried 2

[gleaming coral]

it does not seem like u can factor with grouping either

soooo

?

[flat mortar]

since next one shows two

still nothing so now i don’t know

we can group by factoring

@gleaming coral

have you found annansee?

<@&286206848099549185>

[gleaming coral]

don't ping me

[gleaming coral]

and no

[flat mortar]

my bad

i am still not understanding

[odd edge]

@flat mortar Has your question been resolved?

[zenith tartan]

@flat mortar you there?

[flat mortar]

yes

[odd edge]

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[zenith tartan]

what?

[flat mortar]

accidentally closed it

[zenith tartan]

.

reopen it then

[flat mortar]

accidentsl

yes just reopened

can you help pls?

[zenith tartan]

no i didnt reopened it

[flat mortar]

i alrifready reopened

[odd edge]

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[ember wind]

What rules would be needed to solve this?

[cold sage]

the first one doesnt really have rules beyond
$$\frac{d}{dx}\left(af(x)\right)=a\frac{d}{dx}\left(f(x)\right)$$ and the second one requires the chain rule for the $(g(x))^2$

[clever fjord]

AℤØ

[cold sage]

where a is some scalar

[ember wind]

I am not sure how to use that rule

[cold sage]

which one

[ember wind]

The first one

actually both 😅

[cold sage]

ig i should also say that $$\frac{d}{dx} (f(x)+g(x))=f'(x)+g'(x)$$

[clever fjord]

AℤØ

[cold sage]

the first just means for example that the derivative of 3f(x) is just 3f'(x)

the chain rule is that if you have some y=f(g(x)) and i let u=g(x) then dy/dx=dy/du * du/dx

[ember wind]

That doesn’t make sense to me, sorry

[cold sage]

which one

the chain rule?

[ember wind]

..everything?….🥲

[cold sage]

oop

[cold sage]

alright lets go to this first one

you can factor constants out of integrals and derivatives without affecting the result
for example, if f(x)=6x+2
then 6f(x)=36x+12 right
[the derivative of 6f(x)]=36=6*[the derivative of f(x)]

[cold sage]

thats all this is saying, a is just some constant

[ember wind]

Ohh okay, that makes sense

[cold sage]

this one, is just saying the derivative of the sum of functions, is the sum of the derivatives of said functions

[ember wind]

Kinda makes sense

[cold sage]

as for the chain rule

https://youtu.be/HaHsqDjWMLU?si=weQnARXJB6ae1ijr

this may be worth a watch

[ember wind]

Okay, will be watching this now

Thank you

[cold sage]

no worries

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[odd edge]

@hidden bloom Has your question been resolved?

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@hidden bloom Has your question been resolved?

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[red pier]

I think g(x,y) = R here

but im not sure if I'm simplifying it correctly

can someone walk me through rq?

What I have so far

[odd edge]

@red pier Has your question been resolved?

[red pier]

<@&286206848099549185>

<@&286206848099549185>

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[hazy furnace]

This is the proof of the chain rule

not all of it.

But, it`s the first part which the bit I highlighted is where I dont understand as they say the derivative of u is da da da, then the next line they say note as well that the derivative of u minus the derivative u is equal to 0. Whats the point of saying that.

[odd edge]

@hazy furnace Has your question been resolved?

[swift rock]

sorry its going outside the image

but i hope this is what u meant

[hazy furnace]

yh thats what it says in the proof

[swift rock]

yeah , so why are u confused about derivative of u

[hazy furnace]

why say derivative of u - derivative of u

[swift rock]

they just simplified the expression

to show that both are equal to one another , pretty sure

[hazy furnace]

wait so derivative u is equal to derivative u - derivative of u?

[swift rock]

no

derivative of u - derivative of u = 0
derivative of u = derivaitve of u

[hazy furnace]

ok

[hazy furnace]

for this?

[swift rock]

well u need limit h -> 0 on the left expression for them to be equal

[hazy furnace]

.close

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[candid tangle]

Hi I need help with aii) I tried using the cosine rule but I think it just results in multiple answers

[swift rock]

this is good

keep solving

[candid tangle]

but ill have multiple answers tho?

[swift rock]

1 of the answer would be imaginary or a extraneous solution

only 1 would be real

[swift rock]

btw why is 5.1^2 0?

nvm

u subtracted it from 5.6^2

[candid tangle]

my i got two answers that are real

[swift rock]

what are ur 2 answers?

send ss of ur working

[candid tangle]

3.7289 and 1.4347

[candid tangle]

i typed them into a calculator

[swift rock]

solve it

u somehow made a mistake

im getting 2 answers , either -7.01 or +7.01 , and length cant be negative so that is the extraneous solution

[candid tangle]

Is something wrong over here?

[candid tangle]

i still cant find a mistake, did i mess up on the previous question?

[swift rock]

oh right i messed up

thats a quadratic equation , yeah u get 2x solutions

[candid tangle]

then which one do i use?

[swift rock]

hm

do u have a guide or key book?

only cosine law is suitable but u are getting 2x the solutions

[candid tangle]

no

[swift rock]

ig ping the helpers role

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[grizzled tide]

hilbert spaces?

should give u enough to go off of, I'm sure you can find implementations online

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[serene anvil]

how to solve this

[serene anvil]

OC=3 so area can't be 5*3?

[wheat bramble]

Do you know how to find the area of a kite?

[serene anvil]

no

so we have to use that for finding irregular trapezoid

you mean two tringles right

[wheat bramble]

Yeah you could cut it into two triangles

[serene anvil]

FROM B TO O AND A to some point on that line is height ?

like this

[serene anvil]

ohk

[zenith tartan]

also i know a trick through which you can solve these area questions like these in minutes

can i tell if youre intrested?

[serene anvil]

but can you tell why base *height does n;t work here?

[serene anvil]

please

[lean willow]

It's not a restaurant

Rectangle*

[wheat bramble]

Base x height is only going to work for regular quadrilaterals like rectangles or squares

There's a modified version of base x height for trapezoids but only regular trapezoids

[serene anvil]

and parallelogram / rhomus ?

[serene anvil]

avg of bases right

[wheat bramble]

Yes

[serene anvil]

isoceles trapezoid case

[zenith tartan]

yea i got the answer so, do you know how to solve a determinant?

[serene anvil]

yeah

matrices you mean

[zenith tartan]

like | | these one

[serene anvil]

yeah let me see

i forgot how we solve using those

[zenith tartan]

| 0 0 |
| 0 3 |
| 4 5 | and after solving this multiply it by 1/2
| 3 0 |
| 0 0 |

so what i did is

i wrote four vertices in order

then at last i wrote the first one again

[serene anvil]

okay

and rows*coulmn?

[zenith tartan]

and also this whole determinant is unnder a modulus cuz area cant be negative

[serene anvil]

right

[zenith tartan]

okay

wait

gimme a min

[serene anvil]

okk

[zenith tartan]

okay so lets ignore the half on the left, (okay i messed up saying "determinant") its just a big modulus, we have to multiply diagonally the black ones first and thesse are going to be added, once we are done then we are going to mulitply diagnonally the green and we are gonna subtract them, whatever comes the final answer goes inside the modulus and then multiplied by 1/2 $$ \frac{1}{2}\times |0+0+0+0-0-12-15-0|=\frac{1}{2}\times |-27|=13\frac{1}{2}$$

[clever fjord]

Yajatjamal

[zenith tartan]

you can use this to find area of any shape with whatever sides you have

[serene anvil]

bro but this will take time to solve

[zenith tartan]

it legit took 15 seconds for me

okay so how were going to do it anyways?

[serene anvil]

ok so whever coordinates are given we can use it

[zenith tartan]

i mean you can do the same for any triangle you have

[zenith tartan]

how?

[serene anvil]

okay got it

[serene anvil]

say 3 coord of triang are given then

formula is same

[zenith tartan]

yes

write all three coords then repeat the first one and do the same

[serene anvil]

okay

thanks

@zenith tartan can help with this

[odd edge]

@serene anvil Has your question been resolved?

[serene anvil]

@zenith tartan bro is this different way

[zenith tartan]

yea ik

[serene anvil]

is it different

?

[zenith tartan]

i dont know about the parallelogram one but i know the triangle one, this is another way to calc area of triangle

both can be used

[serene anvil]

okay

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[viral harness]

I am studying Brownian motion, and what does it mean that the increment is normally distributed?

[viral harness]

W(t) is defined over [0,T], and depends continuously on t in [0,T]

for 0 < s< t <= T, the increment W(t) - W(s) is normally distributed with mean zero and variance t-s

WHAT

<@&286206848099549185>

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@viral harness Has your question been resolved?

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@viral harness Has your question been resolved?

[viral harness]

For any additional information

[late dust]

It means that to go from W(s) to W(t), you take a value from that normal distribution

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[rotund mural]

I have to use Fermat's little theorem to solve mod(19^32, 17)

I thought since 19^(32 mod 16) = 19^0 the answer would be 0, but the correct answer is 1. What did I do wrong?

[rotund mural]

.close

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[naive hazel]

Assuming you mean the x values so y is 0 (idk the english term intercept):
That can depend on various factors: For example, x^3 and 3*x^3 have the same intercepts but follow a different path.

A function can't be defined by its intercepts only

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[fluid kindle]

So

If I follow this,
a / 1 - r -> 4/(1-0.1) = 4.444444 but it's saying it's wrong. I'm assuming now that it might be divergant.

But I'm not sure why it would be.

[wooden python]

4/(1 - 0.1) is not equal to 4.4440 as you wrote

keep it as a fraction.

but simplify it

[fluid kindle]

?

I doubled checked

I just rounded

[wooden python]

well don't round!

you can do basic arithmetic, right?

[fluid kindle]

indeed

[wooden python]

yeah so do it

your homework system doesn't want any repeating decimals.

4/(1 - 0.1) = 4/0.9 = 40/9, it's simple as that

[fluid kindle]

hmmm

oh yeah

you could multiply by 10

sure

to get rid of the decimal

ty

.close

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[warm pivot]

how can i factor this?

[signal crown]

https://tenor.com/view/heart-locket-quadratic-formula-formula-quadratic-gif-17389447399699544686

[warm pivot]

oh yah, forgot about that thing

using it i get the numbers inside of the square root 0

so would it not be

-9/2?

aha it is thanks

.close

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[young sedge]

i dont understand b and c

the entire function is greater than zero

[late dust]

It's not

[young sedge]

wdym

[tall veldt]

try x=1/2 for example

[late dust]

Try P(0.5)

[young sedge]

everything is either on or above the x axis

[tall veldt]

dont be fooled by a zoomed out picture

[young sedge]

are you shitting me

how was i supposed to know this?

[tall veldt]

you have the equation of the function

[young sedge]

well what do i do with it

[late dust]

Finding zeroes is helpful

[tall veldt]

in its factored form the squared term is always positive, you need to find x values that make only one of the other two factors negative

[young sedge]

well my zeros were -1 and 1

i didnt factor in the (x+1)^2?

[late dust]

You have another zero at zero...

[young sedge]

0 counts as a zero?

[late dust]

P(0) = 0, so yes x=0 is a zero of the function

[young sedge]

cause for b i did

[late dust]

Are they wrong answers? I'm not sure what you're asking

[young sedge]

just pointing out my own error

[late dust]

But that's c

[young sedge]

right cause i understand why i got b wrong

so i moved on to c

how do i know the graph goes below 0 just by looking at the zeros?

[late dust]

If the function is continuous (which is the case here), you can just test a value between your zeroes

[young sedge]

wait lets go back to b im confused.

(-inf,0) dips below the x axis

[late dust]

Here this might be useful

It does not, it touches it at -1

[late dust]

I don't know why you have the y axis so stretched out here

[young sedge]

how did you get this graph?

cause when i put the given equation in it gave me this

[late dust]

It's not what it did for me, you might have saved a setting somewhere

Try clicking the wrench

[late dust]

Actually, click the home icon

It should reset

[young sedge]

wait what

did the home icon "fix" it?

what was the issue

cause i just got the same graph

[late dust]

This is the issue

[young sedge]

oh

[tall veldt]

if youre on touchscreen you can scale the axes by pinching them, thats maybe what happened

[young sedge]

i am not

[tall veldt]

otherwise no idea why your y axis was scaled so much

[young sedge]

so i could see the whole thing

i dont know what to set my window to in these problems

[tall veldt]

well if you want to see behaviour happening at very small y values then a small y axis range is a good idea

[late dust]

You don't really need to see outside of [-2,2] here

[young sedge]

thanks guys

wouldve gotten it right if i didnt scale it out so much

[late dust]

That's not the fault of the graph though, you should never trust a representation to give you precise or even correct answers

[young sedge]

thats true

i have another question

i dont understand any of this

all i know is that the area has to be 2x + 2l

why is the length 130 - x?

[late dust]

You have a rectangle whose perimeter is 260, what does that say about its sides?

[young sedge]

i dont understand what you're asking

its sides are 2x and 2l

[late dust]

No, x is the width and L is the length

[young sedge]

there's two widths and two lengths no?

[late dust]

The question is why L = 130 - x given that the perimeter is 260

[late dust]

Yes but each side isn't 2x or 2L

[late dust]

That's how I read this

[young sedge]

cause x = the total width and l = the total length?

[late dust]

Do you agree this is what you have

[young sedge]

yes

[late dust]

So what's the perimeter

[young sedge]

260

[late dust]

Ok, but can you express it another way?

[young sedge]

thats not 2x + 2l = 260?

[late dust]

It is

Can you solve for L?

[young sedge]

yea

l = 130 -x

[late dust]

Is it still unclear?

[young sedge]

no

b and c make sense

but i dont understand how they found d

[late dust]

The area must be 2200

[young sedge]

oh right

[young sedge]

2200 = x(130-x)
2200/x = 130 - x

2200/x + x = 130

the x cancels

2200 = 130?

[late dust]

$\frac{2200}{x} + x$

[clever fjord]

Nel

[late dust]

The x does not cancel

Don't try moving x to the denominator, that just makes it more complicated

[young sedge]

hm then how do i isolate it

[late dust]

Do you know how to find the solutions to a quadratic equation?

[young sedge]

yea the quadratic formula

[late dust]

So just develop the right hand side and move everything to one side to equal zero

[frigid sable]

you have 2L+2x = 260,
then L+x=130

and L=130-x

it's algebra

does this clear things up

[young sedge]

yup

appreciate it

.close

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[civic folio]

T

[civic folio]

If the negative is inside the function we reflect it over the y axis correct or incorrect?

if its f(-x)

[civic folio]

thx

[odd edge]

@civic folio Has your question been resolved?

[hallow cove]

basically find the natural number N so that

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[dusky patio]

can someone just tell me how to solve this

[next crest]

Use the concept of similarity

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@dusky patio Has your question been resolved?

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@dusky patio Has your question been resolved?

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[warm pivot]

could someone explain where in the world (+-3, 0) and (0, +-2) come from

i genuinely cannot tell

[pastel dew]

i think there is a formula based on a and b where

$x^2/a^2+y^2/b^2=1$

[clever fjord]

everg

[warm pivot]

is that not what x^2/9 + y^2/4 = 1 is

[warm pivot]

ohh uh

to find c its a^2 - b^2 = c^2 right?

yah

[pastel dew]

yes ...like pythagora

but it is dark pythagora

[warm pivot]

is that not

sqrt(9^2 - 4^2) which is 8.0622...

[pastel dew]

?

[warm pivot]

which is quite far from 3 and 2

to find c

[pastel dew]

no

you have to do sqrt 3^2-2^2

because here a=3 b=2

[warm pivot]

ohh

thats what mixed me up

[pastel dew]

i know ..it s night for everyone

[warm pivot]

so a is x and b is y?

[pastel dew]

i don t know these details ...

[warm pivot]

fair enough

thanks for ur help

[pastel dew]

😉

[warm pivot]

.close

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[vivid coral]

Is there a formula I am missing?

[vivid coral]

The answers are all crazy rational roots

[fringe bough]

pythag identities

in this case tan^2(t) + 1 = sec^2(t)

you can also try drawing a right triangle with an angle t, opposite leg 1 and adjacent leg 9

then use the pythagorean theorem and calculate the other trig ratios

[vivid coral]

Gotcha ty!

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[sage bear]

A ratio to correspond the radian measure to 57 degrees

[sage bear]

Is it 19pi/60 rads

[cold sage]

57/180=19/60

seems correct

strange wording though imo

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[mystic wigeon]

Can someone explain how to factor number 12

[sly badger]

,roate

,rotate

[small tiger]

I would use box method

[mystic wigeon]

box method?

[small tiger]

I’ll show you

One sec

[fervent peak]

What level of math are you at? Have you heard of complex numers or the quadratic equation?

Because if not, this equation is unsolvable

[mystic wigeon]

algebra 2 highschool

[fervent peak]

Have you ever worked with complex numbers before?

[mystic wigeon]

Nope

[small tiger]

Are you working ahead

You should learn complex numbers in a bit

[mystic wigeon]

I barely know how to factor its a miracle I passed algebra 1

[fervent peak]

Well without using complex numbers you have factorized this as far as you can

[small tiger]

factoring is necessary for literally any math going forward

😭😭

you gotta get used to it

[mystic wigeon]

i just barely get it

[small tiger]

Are you planning on taking precal

[mystic wigeon]

Yes in college I think

Taking college alg senior year

[fervent peak]

You basically need to find the roots of a polynomial. So find solutions for x for which p(c)=0

[mystic wigeon]

Idk what p or c are

[fervent peak]

I meant x sorry

And p(x) is what I used to describe a polynomial, sorry I should have clarified

[fervent peak]

For quadratic formulas, ones with the generic formula p(x)=ax^2+bx+c there are a few ways to find roots. My favourite one is the sum-product method. Basically if you have a polynomial where a=1, we have that c is the product of two numbers, of which b is the sum. This however can be trivial. So for harder questions you may need to use a different method

[mystic wigeon]

So like

X^2+5x+6

(X+3)(x+2)

Because c is product of 3 and 2 and b is the sum of 3 and 2?

[fervent peak]

Yes!