help-39

[lunar atlas]

im so cooked i asked my teacher

abt 13

even though its holiday

[quartz yoke]

a holiday?

[lunar atlas]

ya

[quartz yoke]

wait where do you live?

i probably could have worded that better

[lunar atlas]

around asia

i suspect youre not from asia

[quartz yoke]

ohh thats why

im from an east asian ethnic though

[lunar atlas]

yea here we have a lot of religious holiday

[quartz yoke]

o rly

[lunar atlas]

mostly islamic and buddhist

well idk the way but i still think its 7

[quartz yoke]

ok i think ill just write that down

my teachers gonna go over it tmrw

i think ill sleep soon

cya

[lunar atlas]

cya too

[quartz yoke]

.close

[pearl pond]

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[pearl pond]

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[fossil jacinth]

I did l'hopital

but then got stuck

still undefined and doing l'hopital again does not seem to do anything, also makes the expression more complicated

[frozen elk]

!show

[pearl pond]

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

[fossil jacinth]

[plush bramble]

If you can do L'Hopital does that mean you can do series expansion too

[fossil jacinth]

yes

[pearl pond]

@fossil jacinth Has your question been resolved?

[fossil jacinth]

<@&286206848099549185>

[pearl pond]

@fossil jacinth Has your question been resolved?

[fossil jacinth]

apparently l'hopital should work here somehow

[plush bramble]

did you try $\cos(1/x) \approx 1$, $\sin(x) \approx x$ and $\arctan(x) \approx x$

[jolly parrot]

riemann

[fossil jacinth]

I am not sure how to do that

[plush bramble]

yea i'm not surprised l'hopital works. just the derivatives with product/quotient/chain rule just sounds awful

,w taylor series cos(1/x)

[fossil jacinth]

they poorly taught us to do it with taylor

[plush bramble]

wait i have this

,tex .maclaurin

[jolly parrot]

riemann

[plush bramble]

hmmm no arctan though

,w taylor series arctan(x)

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rie.mann

[fossil jacinth]

[plush bramble]

you just need to keep up to order 1 from them

[plush bramble]

and arctan(x) ~ x here

[fossil jacinth]

how do I implement it

[plush bramble]

plug the three into your fraction

[fossil jacinth]

like I choose a certain level of polynomiyal accuracy + O(x) and plug it in it?

[plush bramble]

yea it takes practice to know how many terms you need to keep

and it doesn't always work.

[fossil jacinth]

yea this makes the method really confusing, I guess that's why they didn't dive too much into it this semester for solving limits

so basically when you plug the right expression I should get what I need when x is 0?

and the practice you talk about is how many terms I need? until it works out basically

[plush bramble]

yea start low, then go up

same fuzziness with l'hopital too. sometimes you need to differentiate the quotient multiple times before you get something not indeterminate

[fossil jacinth]

I wonder if there is a trick somewhere because both lhopital and tylor here are a bit annoying to do

taylor for arctan i'm required to prove too

[pearl pond]

@fossil jacinth Has your question been resolved?

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[rapid marten]

Idk how to do it

[hot canyon]

[rapid marten]

Wat does that mean

[hot canyon]

Median?

Given values are: Minimum, Q_1, Q_2, Q_3, Maximum

For outlier, you want < Q_1 - 1.5IQR or > Q_3 + 1.5IQR

[pearl pond]

@rapid marten Has your question been resolved?

[rapid marten]

I got 19.5 and 89.5

The 19.5 was wrong idk what it is

@hot canyon

[pearl pond]

@rapid marten Has your question been resolved?

[pulsar shore]

try it again cuz i got 21.5

[rapid marten]

How

[pulsar shore]

Q1 - 1.5IQR

IQR is 64-47 right

and Q1 is 47

[vapid parrot]

yup

[rapid marten]

Oh I thought Q1 was 45

[pulsar shore]

well IQR is Q3 - Q1 and u got that bit so maybe u just mixed them up

[rapid marten]

How to do this

[pastel umbra]

Are you familiar with stem-leaf diagrams? @rapid marten

[rapid marten]

No

[pastel umbra]

ah

Each digit on the right column represents one entry

The underlined entry
2 | 1113__3__9
is the entry "23" (where the 2 is from the left)

[rapid marten]

So like is the answer 370000

[pastel umbra]

yee

[rapid marten]

TY

[pastel umbra]

Sometimes the data's written like this since the alternative is literally writing out
60,000 60,000 70,000 70,000, ...

(you can see that that might get tedious at best and torturous at worst 🤣)

[rapid marten]

Plz help

[pastel umbra]

(ugh I hate quartiles 😅)

Try 28 and 56?

My thinking being that, given 9 values, the median of those values should be the 5th number (and apply that logic to both the first and last half of the data)

[pearl pond]

@rapid marten Has your question been resolved?

[pearl pond]

@rapid marten Has your question been resolved?

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[lethal mango]

im not looking for math help can anyone help me with literature and like script stuff. i need to cut down my script

[lethal mango]

uh

.close

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[vale tapir]

Hello can someone help. I know that I have to use the small change formula but im unsure of the values

[iron stream]

NGL the spheres too shiny

[vale tapir]

Fr

[iron stream]

Error approximation formula using derivatives?

[vale tapir]

whats that formula?

[iron stream]

(4/3 pi 5³)±3(0.1)

Like the powers become the coefficients, I don't remember the exact name !

[vale tapir]

Let me check answers

[vale tapir]

This gives the V in the V cm^3 ± b cm^3 but not b

[finite geode]

yeah this is correct

the +-b will be your delta V

plug in values of r and delta r

[vale tapir]

r = 5 and delta r would be 0.1?

[finite geode]

yes

[vale tapir]

ah ok

thanks

[finite geode]

using r=5 you can get V

[vale tapir]

Yeah

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[mystic wraith]

is there an ambigous case for cosine law? i dont understand when u use 360 - answer or smth

[mystic wraith]

and what was the formula for how to find the 2 different angles
sin(x)= θ
cos(x)= θ
tan(x)= θ

[pearl pond]

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[mystic wraith]

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[lapis sluice]

I’m confused on two step equations

[karmic fern]

list out what you know and what you need to find out

then try writing an equation relating everything

[pearl pond]

@lapis sluice Has your question been resolved?

[lapis sluice]

Sorry I was cooking something

I know the answer is top right because I tried it

But I thought it was top left

Because my teacher said to not switch the numbers orders

[karmic fern]

why do you believe that 265f + 8 = 305 is true?

[lapis sluice]

Because you have 265 first so you pht it first then you have 8 guests and you don’t know the cost of each favor so you use a variable then the answer is 305

<@&286206848099549185>

[karmic fern]

none of what you just said explains why it's true

you're just repeating something that someone told you or what you believe someone told you

[lapis sluice]

That’s what my teacher said

[karmic fern]

okay but I asked you why you believe that it's true

you should be able to come up with some reasoning besides "so-and-so told me it was this way"

[pearl pond]

@lapis sluice Has your question been resolved?

[pearl pond]

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[nocturne night]

Oh shi

[rain river]

Lol

[nocturne night]

Sorry lmao

[rain river]

Np

[rain river]

I can't understand what exactly is happening when we divide the permutation by K!

Nvm

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[vale tapir]

How would I sketch f(x) from this?

[hot canyon]

That looks like a good enough sketch ngl

[calm wing]

context?

[vale tapir]

Black is f'(x) we need to sketch f(x)

[calm wing]

ok

[hot canyon]

That looks like y = -1/|x|

So taking f'(x) = -1/|x| and solving should give you appropriate f(x)

[vale tapir]

how would I do this? This is calc free

[hot canyon]

Uff

[fierce totem]

f'(x) = -1/x^2, f(x) = ?

[calm wing]

pretty certain this is -1/x^2

[calm wing]

you have a function that is decreasing everywhere

because the derivative is negative everywhere

[vale tapir]

something like this?

[calm wing]

close, but notice that it is increasing for negative x

[vale tapir]

like this?

[calm wing]

yep

indeed

that's the sketch

[vale tapir]

Ah ok thanks

.close

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[pearl pond]

Please don't occupy multiple help channels.

[summer imp]

<@&268886789983436800>

[pearl pond]

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Please don't occupy multiple help channels.

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[olive copper]

how do I parametrise the trace of the curve of a hyperbolic parabola of the equation : z = x - 2y^2 and the plane x = 2

[rough forge]

A parameterization looks like p(x,y,z) = (x,y,z) so you insert x = 2, y = y and z = x - 2y^2

[olive copper]

wouldnt it be in terms of a parameter t

?'

[rough forge]

you can replace y with t yes

[olive copper]

oh I see

thanks

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[low skiff]

idk where to start

[pearl pond]

@low skiff Has your question been resolved?

[bitter herald]

would u know how to do if it was like

3x+y = 4
4x + 5y = 3

or something random like that

[low skiff]

yes

that's easy

[bitter herald]

yeah

so its kinda the same really

its just u r hiding the numbers behind a,b,c, and d

[low skiff]

hm ok

[bitter herald]

so

[bitter herald]

what do u do to solve this

[low skiff]

i'd multiple the top by 5 then eliminate 5y

[bitter herald]

also can i friend u rq, the spammer flag is annoying

[low skiff]

sorry but friending me doesnt change anything, only goes away if ur on mobile

[bitter herald]

not even that? damn ok

anyways

[bitter herald]

ok so how would u do it for the equations they gave u

you just have to replace 2 things in what u said

[low skiff]

multiple top by b then eliminate by ?

[bitter herald]

yea

so try it

multiply top by b, what do u get

[low skiff]

abx + by = bc

[bitter herald]

yea

so, substract second equation from what u just got

to eliminate by

[low skiff]

abx - x = bc - d

then i take out x?

[bitter herald]

and lastly, solve for x

yes

[low skiff]

x = (bc - d) / ab - 1

[bitter herald]

(ab-1) but yipee

[low skiff]

oops yeah

[bitter herald]

so now

plug that in one of the equations

and solve for y

and u r done

[low skiff]

ok then i just chuck it in

yay

[bitter herald]

woooooo

[low skiff]

ok i did smth wrong

my signs r flipped

[bitter herald]

nah

its exactly the same solution

[low skiff]

oh.

[jolly parrot]

Aero

[low skiff]

im plugging in (bc - d) / (ab - 1) into x + by = d and it's just not working 😭

[bitter herald]

good thing is that this server is all about making it work

what went wrong?

[low skiff]

im just getting a rlly big equation and the answer is just y = (c-ad) / (1-ab)

[bitter herald]

ok lets walk thru it then

[low skiff]

i scribbled out my answer before i finished bc i knew it was gonna be wrong in the end lmao

[bitter herald]

chug in the x in the equation

what form is ur equation in afterwards

[bitter herald]

lmao 😭

[low skiff]

(bc - d) / (ab - 1) + by = d

[bitter herald]

yeah

[low skiff]

a fraction

[bitter herald]

your goal is to isolate the y right

[low skiff]

yes

[bitter herald]

so

try that

subtract off that chungus from the left

[low skiff]

first i get by = d - (bc -d) / (ab - 1)

[bitter herald]

yea

[low skiff]

and then i tried making the rhs into one fraction

[bitter herald]

yea

[low skiff]

and then it's (abd - d - ba - d) / (ab - 1)

and this is looking wrong already lmao

[bitter herald]

i mean i think u made that 'bc' a 'ba' there

[low skiff]

ou

[bitter herald]

also ur signs are a bit fucky

it needs to be +d not -d

[low skiff]

whoops

let me try again

okay i see

[bitter herald]

whatcha get

[low skiff]

wait idk 😭

[bitter herald]

😭

[low skiff]

when i move (bc - d) / (ab-1) do i need to flip any signs

[bitter herald]

how do u go from okay i see to idk 😭

[bitter herald]

u just make it -(bc-d)/(ab-1)

u already did that tho

so u gucci

but like, expand -(bc-d)

what is thart

[low skiff]

so im getting by = (abd - d - bc +~~ d~~) / (ab -1)

[bitter herald]

yippeeee

[low skiff]

so by = (abd - bc) / (ab - 1)

[bitter herald]

yuh

[low skiff]

y = (abd - bc) /( ab^2 - b )
?

waot

[bitter herald]

yeah but u didnt need to distribute it in the denomiunator

bcuz like

[low skiff]

oh

[bitter herald]

abd - bc

factor this guy

[low skiff]

b ( ad - c)

ohhhhh

[bitter herald]

mhm

which is ur original answer

so g'job!

[low skiff]

so then [ b(ad - c) ] / [ b (ab -1 )

then y = (ad - c) / (ab - 1)

[bitter herald]

exaaaactly

[low skiff]

hooray

[bitter herald]

the answer they provided is also the same like i said earlier

just switch em up

[low skiff]

I think im gonna struggle with the next question tho 😥

[bitter herald]

uh oh

wot is it

[bitter herald]

yea more fraction nonsense

same procedure tho

so like

here's ur objective

eliminate the y's first and foremost

[low skiff]

um, how

[bitter herald]

so like

u wanna make them the same right

[low skiff]

yes

[bitter herald]

try to make both equations aby for example

what do u multiply by with to get aby

[low skiff]

a

[bitter herald]

yea

[low skiff]

and then multiple the bottom by b

?

[bitter herald]

yea

tell me whatcha get

[low skiff]

a^2 * x - aby = a^3
b^2 * x - aby = b^3

[bitter herald]

awesome sauce

so subtract first from second

[low skiff]

so a^2 * x - b^2 * x = a^3 - b ^3

[bitter herald]

or other way around doesnt matter

[bitter herald]

forgot an x there but yea

= not -

ok ill stop being mean LOL

[low skiff]

oops

now what

[bitter herald]

well u r like

pretty much done arent u

solve for x

[low skiff]

um

how

[bitter herald]

factor x from a^2 x - b^2 x

[low skiff]

oh whoops yeah

x = (a^3 - b^3) / (a^2 - b^2)
then x = a - b ?

[bitter herald]

no

[low skiff]

oh

[bitter herald]

thats not how that works

[low skiff]

lol

[bitter herald]

ur answer is correct

[low skiff]

oh yeah

[bitter herald]

but like

u cant just divide like that

[low skiff]

bc it's - not *

[bitter herald]

no

its not htat either

like

[low skiff]

eh

[bitter herald]

you were thinking of [
\4{(a-b)^3}{(a-b)^2}
]
in which case yeah its equal to $a-b$

[jolly parrot]

Aero

[low skiff]

oh ok

[bitter herald]

but a^3 - b^3 isnt the same as (a-b)^3 so keep that in mind

but yeah that part is done

do the same

plug in x, solve for y

[low skiff]

so now x = (a^3 - b^3) / (a^2 - b^2)

[bitter herald]

yeah

[low skiff]

uhh so ik the bottom is (a+b)(a-b)

i forgot how to do the top

[bitter herald]

you need to use difference of squares and difference of cubes formulae

[low skiff]

i forgot the difference of cubes

[bitter herald]

its a^3 - b^3 = (a+b)(a^2+b^2 -ab)

[low skiff]

ok so i got x

[bitter herald]

yea

[bitter herald]

altho im pretty sure this should be (a^2 -ab + b^2)/(a-b) in our case

so either the solution is wrong or we're wrong

[low skiff]

um

idk

[bitter herald]

let me check

oh yeah

[bitter herald]

this should be its a^3 - b^3 = (a+b)(a^2+b^2 +ab)

not -ab

my bad on that

[low skiff]

okk

so rn I have (a^2 b + ab^2+ b^3) / (a+b) - ay = b^2

oops

wait

ok

[bitter herald]

sounds legit

yeah

drag the chungus over

like last time

[bitter herald]

it should be a^2 b

u were correct

[low skiff]

hollup

[bitter herald]

yeah

[low skiff]

yeah mb

so - ay = b^2 - (a^2 b + ab^2+ b^3) / (a+b)

[bitter herald]

ye

[low skiff]

then I make it - ay = (ab^2 + b^3 - a^2 b - ab^2 - b^3) / (a+b)

[bitter herald]

yes

shit cancels as u can see

[low skiff]

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

[bitter herald]

yes

[low skiff]

so -y = (a^2 b ) / a(a+b)
then -y = ab / a+b
then y = - ab/a+b

😥

[bitter herald]

You likw

Erased a minus sign

[low skiff]

oh no

[bitter herald]

It's -y = -a^2b/(a(a+b))

[bitter herald]

Check with this

[low skiff]

ohhhh

ok

i got it

thanks

[bitter herald]

Yippee

[low skiff]

that's enough for today ty for ur help

[bitter herald]

Ofc good luck

[low skiff]

.close

[pearl pond]

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[pearl pond]

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[nocturne night]

Plane passes through line of intersection of planes x-z=1 and y+2z=3 and is perpendicular to the plane x+y-2z=1

[nocturne night]

Find plane

dunno where to start

[rough forge]

perpendicular to the plane x+y-2z=1
that should tell you something about the normal vector of your plane

[pearl pond]

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[lofty raft]

how would i go about this? i tried to rewrite the statement in the terms of a congruence statement where (n-2)^6 is congruent to 33 mod n

[lofty raft]

this is chat gpt...

[brave sluice]

sure

[lofty raft]

okay, you say n =7 is a solution but (7-2)^6 congruent 33 mod 7 is clearly not true

[valid peak]

don't pass GPT solutions as your own.

[lofty raft]

neither is n = 19 a valid solution. (19-2)^6 congruent 33 mod 19 is not true

so lol

<@&286206848099549185>

[dusty jungle]

this is quite simple

First try to write the question statement in mod notation

$(n-2)^{6}\equiv33(mod n)$

[jolly parrot]

CherryMan

[lofty raft]

yes i did that

[dusty jungle]

then use the binomial notation to expand (n-2)^6

the cool thing using the binomial theorem here

is that any term which has n as a coefficient is congruent to 0 mod n

so try to continue from here

[lofty raft]

so i expanded it and got

$n^6-12n^5 +60n^4-160n^3+240n^2-192n+64$

[jolly parrot]

jess

[lofty raft]

so the only term that would natter is the 64?

[dusty jungle]

exactly

try to continue from there

$64\equiv33(mod n)$

[jolly parrot]

CherryMan

[lofty raft]

yes

[dusty jungle]

that imples

$64-33\equiv0(mod n)$

[jolly parrot]

CherryMan

[rough forge]

Little add on: I think you wouldn't need expansion you could just write (n-2)^6 = (-2)^6

[dusty jungle]

yeah i actually wanted them to skip the middle terms

but its fine if you wrote the whole thing. only it gets a little messy for higher powers

[dusty jungle]

pls try to solve it from here

[lofty raft]

so just 1 and 31?

[dusty jungle]

yeah!

those are your answers

[lofty raft]

is it sufficient to say that since 31 is prime, that proves 1 and 31 are the only possible positive solutions?

[dusty jungle]

yes by the definition of prime numbers, as they only have two positive divissors

[lofty raft]

okie! thanks so much

.close

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[keen frost]

<@&268886789983436800>

[pearl pond]

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[mighty basalt]

Im supposed to have sqrt(5) i guess?

but not sure how

Guessing the answer is D
It's D so yeah, how are you supposed to find the sqrt(5)

[rough forge]

you mean the lower bound -sqrt(5)?

[mighty basalt]

yes, I just got -5

[rough forge]

x²-5 = 0

x² = 5

now square root both sides

[mighty basalt]

oh in the video it made it look like it was supposed to be

[rough forge]

well they factored

you can do the same

difference of squares

[mighty basalt]

no like you combine both x's

y = 4x & y = x^2 -5

$0 = x^2 - 4x - 5$

[jolly parrot]

Ousel

[rough forge]

that's for finding the common intersection

which seems to be at x = -1 but you asked about -sqrt(5)

[mighty basalt]

Thats what im trying to understand yes.

[rough forge]

,w 4x=x^2-5

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[rough forge]

ok I see

x = 5 is the second intersection

which is irrelevant

[rough forge]

it's pretty right

[mighty basalt]

So for that area to find the intersection there I need to find that = 0 alone

[rough forge]

you solve the equation 4x=x^2-5 which is the same as solving x^2-4x-5=0

[mighty basalt]

which gives the left intersection that I need

but dont need right

[rough forge]

well if you integrate in terms of x you need to split it into two integrals

[mighty basalt]

which is why I need 0 = x^2-5

[rough forge]

so you need to find their intersection between -sqrt(5) and 0

which is at x = -1

the x = 5 is a solution yes, but here not relevant

[mighty basalt]

Ok, I think I got what I was missing now. Thanks. Can actually make out what im doing for the most part now. 😅

[pearl pond]

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[hard cipher]

Trying to find the area of the enclosed triangle. Have successfully translated this geometry problem into an algebra one:

Find

$\frac{(l_1w_1)+(l_1w_2)+(l_2w_1)}{2}$

Subject to
$l_1w_1=12$

$l_2w_1+l_2w_2=10$

$l_1w_2+l_2w_2=12$

But I was unsuccessful in solving for the four variables (there are only 3 equations given, not to mention they are nonlinear nor homogenous).

[jolly parrot]

pianist

[hard cipher]

<@&286206848099549185>

[dapper moth]

you could use vectors...

to find the area of the triangle

[hard cipher]

It’s equivalent

Won’t make a difference no?

[dapper moth]

are these numbers the areas of the triangles

[hard cipher]

Yes

[dapper moth]

Cool

[hard cipher]

Ok

[dapper moth]

Is there more information ?

[hard cipher]

No, thats why i struggled

I dont think personally that 6 5 5 is any special choice

There must be some rule for arbitrary areas for the three triangle that can be derived and proven

[dapper moth]

The key, i think, is that we are not trying to find w1 w2 l1 or l2 we are trying to find w1l1 which we know and w2l1 and w1l2

[hard cipher]

Yep

We have one of them in terms of the other as well

Since $l_1w_2-l_2w_1=2$

[jolly parrot]

pianist

[dapper moth]

Yes, I agree

[hard cipher]

Meanwhile I found a particular solution (l_1,l_2,w_1,w_2)=(4,2,3,2) which yields 13 as the area of the enclosed triangle

I might want to try different values for the three smaller triangles

[dapper moth]

I also found that l2/l1 = w1/w2 - 1

I don't know if that helps though

I assume we don't know the area of the rectangle

[hard cipher]

We dont

[dapper moth]

I found that the area of the rectangle is 12 + l2w1, I don't know if that is helpful tho...

[hard cipher]

Then we have that the area of the triangle is l_2w_1 - 5

Still need to find l2w1

[dapper moth]

yes....

[hard cipher]

I’ll try to do some angle chasing

[dapper moth]

also I run into a bit af paradox: if A, the area of the triangle, is A = l2w1-5 then because l2(w1+w2) = 5 <=> l2w1 + l2w2 = 5 <=> A = -l2w2 < 0 ??!!

[hard cipher]

Oh god

I’m working towards something @dapper moth I’ll let you know if i make any progresss

I have got a function f such that w_2=f(l_1)
And then another function g such that l_2=f(l_1)

<@&268886789983436800>

Get out of my channel

👍

[pearl pond]

@hard cipher Has your question been resolved?

[pearl pond]

@hard cipher Has your question been resolved?

[pearl pond]

@hard cipher Has your question been resolved?

[hard cipher]

@dapper moth I wrote a minipaper regarding the problem and have a read if youre interested like me

[pearl pond]

@hard cipher Has your question been resolved?

[pearl pond]

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[wintry wave]

I had said that for there to be a unique solution, Ax = z must be consistent and contain zero nontrivial solutions.

[west sapphire]

what can you say about the dimensions of the kernel and image of A?

[wintry wave]

It does not span all of R3 because of y?

[west sapphire]

right

i'm not sure what background you know, do you know about injectivity and surjectivity?

[wintry wave]

I'm taking an intro linear algebra class

[west sapphire]

sure, but do you know those terms?

[wintry wave]

No

[west sapphire]

or maybe their equivalent terms, 1-to-1 and onto?

[wintry wave]

Haven't covered those yet

[west sapphire]

hmm

[wintry wave]

The chapter goes over parametric vector form and homogenous/nonhomogenous solution sets

.close

[pearl pond]

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[subtle chasm]

I need help

[open rivet]

!da2a

[pearl pond]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[subtle chasm]

Can someone explain why the answer is B for #3

[open rivet]

substitute in x=-3

[subtle chasm]

Im confused

[open rivet]

$\left.\begin{aligned}y&=3x+1\x&=-3\end{aligned}\right}\implies y=3(-3)+1$

[jolly parrot]

Bonk

[open rivet]

does this make sense?

[lavish nimbus]

yea, perfect

[subtle chasm]

No it does not make sense to me

[light helm]

which part

[subtle chasm]

Aren’t x’s the missing numbers? Wouldn’t I need to figure out those missing numbers then do the math with the other numbers?

[open rivet]

you are given the x

[light helm]

you're being asked in the question what happens

when x = -3

[open rivet]

so you can just plug that in

[subtle chasm]

I am not understand these examples

[lavish nimbus]

x is -3, so put it in y

[plush moss]

You are asked to evaluate the value of y when x = -3

y = 3x+1

In English this means that y is equal to the value of x multiplied by 3 plus 1

Since the value of x is -3

To find y we simply take -3 multiplied by 3 to get -9 and add 1 to get?

[subtle chasm]

-8

[plush moss]

Yea

So y=-8,when x=-3

You got your answer

[subtle chasm]

Ok but what was the x for next to the 3 on the y equation

[light helm]

that's part of the equation telling you how x and y are related

[plush moss]

3x means 3 multiplied by x

[subtle chasm]

Sorry if you explained this and I’m asking this question, but if the variable is “x” how would I know what to multiply by?

[plush moss]

You mutliply by the value of x which they have stated is -3

Maybe I can try to explain this ig

[subtle chasm]

Ohh ok

[plush moss]

Do you get it like fully now

[subtle chasm]

By multiplying the x variable which is -3 x 3 which you said was 9

Wait

[plush moss]

-9

[subtle chasm]

Nvm I understood it

I was gonna ask

If you added 1 to -9 wouldn’t that be -10. . .

[open rivet]

does it make sense like this?

[subtle chasm]

But by adding 1 it subtracts the negativr

[plush moss]

I am 9 dollars in debt

I work my ass off to get one dollar

I pay my one dollar of my debt off

How many dollars in debt am I in now @subtle chasm

[plush moss]

Oh nvm you got it

[subtle chasm]

-8

I have one more

It’s gonna be the same as last time but in the y values

I’m not very good at functions I’m practicing and learning

So let me try to break this down first

And I’ll type it here

Then when I make an error I would like to be enlightened on where I went wrong please

So it’s saying what is the value of y in this function when x = -2

Meaning x is -2

And at the bottom we see
y = (x-4)/(2 -x)

[open rivet]

please dont write a fraction like this

write it as y=(x-4)/(2-x)

[subtle chasm]

Ok. Yea I don’t know where to begin on this one

I’m having the same issue where I don’t know what to do with the x values

[open rivet]

once again, plug in your x value

every where you see x, replace it with -2

[subtle chasm]

Ok (-2 - 4) = 2 / (2 - -2) = 0

The answer has to be 2 which is C

I was wrong, it was D.

[open rivet]

how are you getting 0?

you are making mistakes with addition and subtraction

$-2-4\neq 2$

[jolly parrot]

Bonk

[subtle chasm]

Where did I go wrong?

[open rivet]

i have honestly no clue what you did

[subtle chasm]

I’ll tell you

You say replace all the x’s with -2

[open rivet]

you did $\frac{-2-4}{2-(-2)}=\frac{2}{4}?=0?$

[jolly parrot]

Bonk

[subtle chasm]

Oh thats not what I meant to do

[subtle chasm]

I ment to replace all the x’s you see in the equation at the middle

And do the math from there

[open rivet]

yes, this one

[subtle chasm]

I thought (-2 - 4) = 2

By subtracting -2 from four or even adding four to -2 that would still = 2

[open rivet]

thats wrong

(-2+4)=2

you should really practice some addition and subtraction with negative numbers if you are struggling with this

what grade are you in?

[subtle chasm]

10th

[open rivet]

you really should not be struggling with this

honestly

[subtle chasm]

Sorry

[open rivet]

its okay

its something you can practice

[subtle chasm]

I overlooked the subtraction sign in the middle

[open rivet]

maybe check out khanacademy or smth similar

so $\frac{-2-4}{2-(-2)}=?$

[jolly parrot]

Bonk

[subtle chasm]

-2 - 4 =-6

2 - (-2) =-4

[open rivet]

wrong

[subtle chasm]

Yikes

I must be over thinking this really hard then

Because if they are both minus signs

[open rivet]

you have a positive number, you subtract a negative number, and then you end of with a negative number?? that doesnt make sense

[subtle chasm]

Oh so then it’s 0

[open rivet]

no...

$2-(-2)=2+2$

[jolly parrot]

Bonk

[subtle chasm]

I’m sorry what

[open rivet]

honestly, this is like grade 6 or 7 maths

if you dont know this, go practice it

[subtle chasm]

Can you explain how that = 2 + 2

[open rivet]

otherwise you will struggle much much more in the future

[open rivet]

subtracting a negative number is the same as adding a positive number

[autumn fossil]

https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-negative-numbers-add-and-subtract/cc-7th-add-negatives/a/negative-numbers-addition-and-subtraction-faq

Might be best if you just check khan academy, read some articles / watch some videos / try some exercises

[subtle chasm]

Where did the + sign come from though?

[open rivet]

its from the two negative

[subtle chasm]

What you typed completely and has me in utter confusion

[open rivet]

it shouldnt...

[subtle chasm]

And you said this is basic subtraction and addition?

[open rivet]

you know $a-b=a+(-b)$?

[jolly parrot]

Bonk

[subtle chasm]

No I’m a beginner at functions.

[open rivet]

for negative numbers, yes. grade 7

[open rivet]

wdym function?

this isnt a function

[autumn fossil]

I think that it's not the idea that confuses them, it's the letters in this case

[open rivet]

how?

i think in that case we have abigger problem than just addition and subtraction...

[autumn fossil]

bc they connected it to functions which is where letters appear

checking khanacademy is probably the best thing to do in this case

there is plenty of material on these topics

[subtle chasm]

Addition and subtraction is not my issue here I can assure you that

[open rivet]

lots and lots of practice

[open rivet]

it definitely is

otherwise you wouldve had the answer already

[autumn fossil]

addition and subtraction with negative numbers is though

[autumn fossil]

this link covers exactly that

[subtle chasm]

I appreciate the feedback

Thank you

[open rivet]

goodluck

[subtle chasm]

Thanks

[pearl pond]

@subtle chasm Has your question been resolved?

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[subtle chasm]

I’ll be back to redeem myself

[pearl pond]

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[split steeple]

ive figured out that the height from a is 12 but nothing else

[errant fable]

well did you draw the height

can you draw it and give the point a name, then share the triangle again?

[split steeple]

ill try to draw but its on the computer

[errant fable]

can you calculate |DE|?

[split steeple]

no

how is that possible

[errant fable]

you know |AE|=12, |AD|=13 and <DEA=90°

[split steeple]

idk

wait

i might be able to

[errant fable]

a²+b²=c² if you remember

[split steeple]

yh that was what i was thinking of

13^2 = 169 and 12^2 = 144

so a^2 is 25 and DE =5

[errant fable]

nice job

now you should calculate |EC| to use pythagorean theorem again

[split steeple]

alright

wait but isnt EC = 1 since AD is median

[errant fable]

yep

[split steeple]

alright

1^2 + 12^2 = 1+144 =145

so AC is root of 145

[errant fable]

👍

[split steeple]

and how do i figure that out?

or is the answer just root of 145

[errant fable]

just leave it as is

[split steeple]

oh ok

[errant fable]

or use a calculator to figure out some first digits

it'll be pretty close to 12

[split steeple]

calculator not allowed 😭

tysm

[errant fable]

the exact answer is sqrt(145)

you can close the channel by typing .close

[split steeple]

.close

[pearl pond]

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[tulip cradle]

How did they get from the first differential to the second differential (f’’(C1))

[plush bramble]

looks like product rule

might help to simplify f(C1) before differentiating at all

[pearl pond]

@tulip cradle Has your question been resolved?

[tulip cradle]

Oh ok thanks

.close

[pearl pond]

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[cunning pulsar]

Hi can someone double check my work? Thanks!

[regal herald]

remember your x values are negative

[pearl pond]

@cunning pulsar Has your question been resolved?

[pearl pond]

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[alpine prairie]

geometry, can someone PLEASE explain what formulas im meant to be using or steps im meant to take for "similarity and altitudes in right triangles" because my teacher did not :(

[cobalt hinge]

Do you know trigonometry?

[alpine prairie]

uhhh

i dont think so?

[cobalt hinge]

Ok, what about the Pythagorean Theorem?

[alpine prairie]

absolutely

[cobalt hinge]

Okay.

Apply the Theorem on XY.

[alpine prairie]

but id need two of the sides to do that wouldnt i?

[cobalt hinge]

Or, hmm, since this is similar triangles, I'll take a different approach.

Which ones would be similar here?

[alpine prairie]

wxz & xzy

wait letter order matters doesnt it

one sec

[cobalt hinge]

Yes.

[alpine prairie]

wxz & xyz

[cobalt hinge]

Better.

We know WX=22 and WY=26.

How would the ratios go here?

[alpine prairie]

lemme pull out desmos rq

[cobalt hinge]

Oh, by the way, there are more similar triangles.

[alpine prairie]

there are 😭 ???

[cobalt hinge]

This is a nice problem.

[cobalt hinge]

Yes.

[cobalt hinge]

I remember doing something like this at a competition.

Note the outermost triangle, WXY. What would it be similar to, and how?

[alpine prairie]

why would the outermost triangle be similar to another triangle when the other two are right triangles and the outermost one isnt ?

genuine question

the sides wouldnt be proportional

i mean at least in comparison to the other two triangles inside of the outermost one

and theres only those three as far as im aware

[cobalt hinge]

It is.

The outermost triangle is a right triangle.

[eternal tulip]

hopefully this is more clear :)

[cobalt hinge]

Look at angle WXY.

[alpine prairie]

OHHH WAIT YOU RIGHT

mb

[cobalt hinge]

Yeah.

So, for the sake of this question, which triangles would be similar?

[alpine prairie]

all three?

[cobalt hinge]

Yeah.

More specifically?

[alpine prairie]

WXY & WZX?

that may or may not be the right order

[cobalt hinge]

Yeah.

[cobalt hinge]

So now, make a ratio that would relate the side lengths given, and the ones you want.

[alpine prairie]

wx/wy=wz/wx ???

[cobalt hinge]

Nice.

Now solve for WZ and you're good to go.

[alpine prairie]

betttt

thank you :)

[cobalt hinge]

👍

[alpine prairie]

.close

[pearl pond]

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[pearl pond]

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[noble scaffold]

could someone give me the first step to solve this?

[rough forge]

have u tried chain rule

[noble scaffold]

what

[noble scaffold]

idek what im solving for

[pulsar lark]

Wdym?

Do you understand z_v?

[noble scaffold]

no

thats what i mean by first step

[plush bramble]

have you learned partial derivatives

maybe try watching the video first https://www.youtube.com/watch?v=HaHsqDjWMLU

and
https://www.youtube.com/watch?v=XipB_uEexF0

[noble scaffold]

yea

ill have to watch the chain rule for partial tho

[pearl pond]

@noble scaffold Has your question been resolved?

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[outer hare]

Im trying to understand how to prove this:

[outer hare]

Why does x existing on both sides mean (AUB)' is a subset to A'intersectB'

[versed mica]

do you know the definition of what it means for A to be a subset of B

[outer hare]

kinda, i just understand why x being part of both sides means its a subset

[versed mica]

state it

[outer hare]

By my understanding, a set A is a subset of B if all elements of A are also part of B

[versed mica]

yep

[versed mica]

so why is this confusing

[outer hare]

I dont understand how x being a part of both implies that A is a subset

[versed mica]

you assume x is in (AuB)’ then show that it is in A’ n B’

for some arbitrary element in (AuB)’

[prisma elk]

because all of its element are in the other side too

[warm current]

It's kind of axiomatic, really.

Proving it really depends on what you can already assume is true

[versed mica]

yea you sort of wiggle around with english definitions and common sense

[warm current]

btw you don't need to do a separate case for $A\cup B=\emptyset$.

[jolly parrot]

SWR

[outer hare]

just to state im new to set theory and profs, but im still failing to understand two things:

1. Why is this proving that (AUB)' is a subset of A'intersectB' and not the other way around
2. How does this proof eliminate the idea of an element being a part of (AUB)' and not A'intersectB'

I understand how the rule works logically and through ven diagrams im just new to understanding the proof

like it makes sense, but if i were to prove it im unsure as to how this proof addresses the 2 points above

[warm current]

This is always an unfortunate proof that gets given to new students, because it's not really "provable". Most books "prove" it by just kind of fudging the words around with "and"s and "or"s

[outer hare]

Oh aw ok

[warm current]

so, don't feel bad that you are not understanding this one.

[outer hare]

I still kinda wanna get it because it makes sense to me logically

im just not grasping how it takes care of certain cases

like, ignoring this, if i needed to prove that one set was a subset of another how would I do it?

Because just finding one x that seams to exist in both doesnt seem that convicing to me

if that makes sense

[warm current]

Like I mentioned earlier, this law is typically an axiom

It's just given as true because it's so basic

well, to be more specific, it is an axiom in boolean algebra

but what you are trying to prove is just the set-theory version of the same thing

[outer hare]

hmmm ok

[warm current]

which cases?

[outer hare]

But in geniral for proofs, how would I go about prooving one set is a subset of another

[outer hare]

these two :

1. Why is this proving that (AUB)' is a subset of A'intersectB' and not the other way around
2. How does this proof eliminate the idea of an element being a part of (AUB)' and not A'intersectB'

[warm current]

To prove $A\subseteq B$, you go "if $x\in A$, then blah..blah.blahh.. Therefore, $x\in B$. But, the choice of $x\in A$ was arbitrary, implying $x\in B$ for every $x\in A$. Thus $A\subseteq B$."

[jolly parrot]

SWR

[outer hare]

ohhhhh

so x represents every element in A

not like a specific object

[warm current]

as long as you attach no added qualifiers to x, yes

[outer hare]

ok sorry that clears a lot up

[warm current]

Like "choose some number" vs "choose the number 2" or "choose a prime number"

[outer hare]

Ok so what this is saying is

x is every element in (AUB)'

[warm current]

This is the book proof, or yours?

[outer hare]

book

i reproved it myself but just kinda got lost on why its a subset

but i think im getting it

so

if we assume x is every eleement in (AUB)'

then we know x does not exist in AUB

meaning x is nor in A or B because union just combines all the elements in both

If x is not in A then it is in A', and the same logic for x existing in B'

So x exists in A' and B'

and because intersect is just what objects exist in both

x exists in A' intersesepts B'

meaking all objects in (AUB)' exist in A' intersesepts B'

making it a subset

[warm current]

kinda? We are saying "$x$ can be $\textit{any}$ element in $(A\cup B)'$". This is different from $\textbf{every}$, which makes it sounds like $x$ is somehow multiple things at once.\
\
You are saying two things\

1. If $x$ is $\textit{any}$ element in $A$, then $x$ has some property $P$\
2. The choice of $x$ in step 1 was arbitrary. Therefore, $\textbf{every}$ $x\in A$ has the property $P$

[jolly parrot]

SWR

[outer hare]

So its a specific point but it represents any specific point in (AUB)'

So if it cannot be x, it cannot be inside (AUB)'

which answers the question i had about "How does this proof eliminate the idea of an element being a part of (AUB)' and not A'intersectB'"

[warm current]

Not quite. We are not considering anything not in $(A\cup B)'$. We make no assertion one way or another in that regard

[jolly parrot]

SWR

[outer hare]

hmm

So like lets say set A = {1,2,3,4}

and I was to say

[warm current]

This is what is known as a material implication

[outer hare]

for x exists in A

x therefor represents either 1, 2, 3, 4. Any property that applies to x applies to 1, 2, 3, 4

if its not 1, 2, 3, 4 we cannot disribe it as x

[warm current]

if it's not in 1,2,3,4, then we make no claims one way or the other

[warm current]

back to what I said here

[outer hare]

ok im a little lost again, so if we were to taking about lets say 5 which exists in U

[warm current]

You are saying that something is true on the condition that something else is true. If that "something else" is not true, then you make no claim whatsoever

[outer hare]

ah ok

So anything implied to x will work on any elemet in A

[warm current]

you have it kind of backwards

[outer hare]

but we cant say anything about the elements outside of A

oh

[warm current]

err.. I guess how you are describing it is okay

How I think of it "I am showing that x has some property P if x is some element of A"

[warm current]

exactly right

Not so much "can't", but "don't"

we don't care

[outer hare]

ah ok, os the reason this proof works is because any element inside of A will take property P of x

[warm current]

this is an important distinction, because it gets you into what are known as vacuous truths

[outer hare]

meaning if we prove that any abstract x existing in A is also in B

[warm current]

exactly

[outer hare]

then A is a subset of B

Ok this makes more sense now

[warm current]

cool

[outer hare]

thanks for your patience lmao

[warm current]

no worries

I like talking about this stuff

btw, that vacuous truth idea, it's pretty wild

"if false, then <any statement>" is always true. This is known as a vacuous truth

[outer hare]

so just a fancy way of saying common sense?

[warm current]

That's because our "if" condition is false, so we make no claim whatsoever. So, if we make no claim, we cannot be wrong. Therefore, we must be right.

[warm current]

It's less common than you think

[outer hare]

tbh im really new to proofs

[warm current]

"If I have two heads, then mayonnaise tastes good on spaghetti" is a true statement

[outer hare]

its kinda interesting but its weird to prove things i just kinda understand

[outer hare]

hmm

[warm current]

More mathematically,

• "if 2 is odd, then the twin prime conjecture is false"
• "if 2 is odd, then the twin prime conjecture is true"
  Both of these are true.

[outer hare]

so if something is always false, then something thats usually false is true