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[pearl pond]

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@wary heron Has your question been resolved?

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[sharp smelt]

I'm trying to find the 2nd partial derivatives of $e^{xe^y}$
\
$\pdv{e^{xe^y}}}{x} = e^ye^{xe^y}$
\
$\pdv{e^ye^{ye^x}}{x} = e^{2y} e^{xe^y}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[inland ivy]

too many }

[sharp smelt]

$\pdv{ e^{xe^y}}{x} = e^ye^{xe^y}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

$\pdv{e^ye^{ye^x}}{x} = e^{2y} e^{xe^y}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[inland ivy]

the function changed

[sharp smelt]

huh

where

[sharp smelt]

this is the double derivative

[glacial sequoia]

what's the problem?

you seem to know how to do it

[sharp smelt]

Just wanted it verified

[cursive wraith]

the exponent here is xe^y

[cursive wraith]

and here it's ye^x

which one is it?

[sharp smelt]

oops

$e^{xe^y}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[cursive wraith]

ok then all good

[sharp smelt]

thanks

.close

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[quick star]

coin w is tossed once and coin X is tossed twice. Assume both coins are fair the what is the probability that coin W returns heads and coin x returns tails twice

[quick star]

1/2 * 1/2 * 1/2 = 1/8

but this is the probability of one arrangement, no?

why don't we multiplying it by 3C2 = 3?

dumb question but like that pivots me to asking why we can't treat this as a binomial problem?

[brave sluice]

they're 3 independent events

[quick star]

hmm could you elaborate?

so okay let me get this straight

u can't merge 3 independent events into 1

and well a binomial only works on one single event at a time?

[rich pecan]

the probability values involved could be the exact same, but that doesn't necessarily mean that you can generalize that entire system into one

because i think the way the question was phrased, coin w flip and coin X flip are two independent systems

[urban sun]

why tf?

wait mb

[rich pecan]

maybe it can be written as P(w) & P(X)

[urban sun]

by what logic are u saying *3c2?

[rich pecan]

so by product rule (if AND, multiply probabilities, if OR, add probabilities)

so (1/2) * (1/2*1/2) = (1/2) x (1/4)

yes the value is the exact same if you treated it as just flipping a coin three times, but if you look at it aside from the number only aspect

for example if you made coin X so that it's slightly more likely to return heads, then you can't treat the two systems as one anymore

[pearl pond]

@quick star Has your question been resolved?

[quick star]

okay but why?

i know how to get the right answer

it would just be 1/2 (1/4)

[rich pecan]

yes

[quick star]

but tbh i never really could justify why u shouldn't multiply by the number of arrangements

someone asked me this

idk the answer though lol

[rich pecan]

ohh

like, in general?

i mean mathematically that is true because of the general counting principle

but in real life of course there aren't perfectly fair coins or a 100% sure predictable outcome

[solid pier]

when the coins are distinct from each other like this, this is the only arrangement that gives you a good outcome (any other arrangement of one head and two tails would mean coin W returned tails, which wasn’t what we wanted. so multiplying by 3 leaves us with two cases that we counted that have coin W on tails)

the probability of just getting one head out of three is actually 3/8 like you’d suspect.

[quick star]

i'm confused 😭

maybe i'm being dumb

[solid pier]

like what’s the purpose of multiplying by the combinations of H/T in the binomial model?

it’s because we have not counted the other arrangements of coins, and those arrangements are ones that give us an outcome that we’re asking about

but in this case, there are no other arrangements of H/T that have one head in it that give us the outcome “head on coin W and tails on the other two”

[quick star]

hmm right but like how do we know that right off the bat?

(W_H)(X_T)(X_T)

[solid pier]

we could just reason about it binomial style, like we ask if there are any other arrangements that work, and in trying to do so, we realize that we must first have exactly one head, and also that head must go on coin W, so we’re done.

[quick star]

mmmmm okay so

exactly one head and it's in the first position

H _ _

[solid pier]

yeah

and then two tails

so it must be HTT

[quick star]

which is 1/2 * 1/2

🤔

but we treat the tail as distinct objects

uh wait

[solid pier]

we don’t, if we require that one head and one tail be flipped among the X coins then we would multiply by 2

just like you’d expect

[quick star]

oh yeah it's

exactly one head: (1/2) (1/2)^2

but since H is fixed and TT has only one arrangement

then it's just 1 * 1/8 = 1/8

[solid pier]

one could also think of it like: exactly one head gives us 3/8 from the binomial model and then correcting for counting arrangements that don’t work gives us (3/8)/3

if you don’t like that then dwai

[quick star]

wait where did you get the 3/8 again

😭

[solid pier]

it comes from the following line of reasoning:

let’s say i have three coins, and i flip them and i get HTT. that had a 1/8 chance of occurring. but i also could have gotten THT and TTH and would have been just as happy to say i got exactly one head. since these are all disjoint outcomes all with probability 1/8, then the probability of flipping heads is 3/8

[quick star]

oh lol

the 3C1 in the binomial thing yeah

whoops i completely forgot about that

[solid pier]

so in the new scenario you don’t have the last bit of reasoning because i would not be happy with the outcome THT

for example

[brave sluice]

i might have been wrong about this...

[quick star]

yeah so the answer to my main main question was actually just:

it can be treated as a binomial where you roll for exactly 1 head or exactly 2 tails

but then since the positions are fixed

you don't deal with the combinations/permutations bla bla

wow okay

[solid pier]

the positions are not fixed among the last two coins

but that’s fine because 2C0 is 1

[quick star]

yeah but TT and TT are the same so i guess it all checks out

thanks!!

wait @solid pier

my question doesn't suggest that coin W is tossed first

and then coin X

or the other way around

so wouldn't 😭 it actually be 3/8

i could flip X first then W then X again

or X twice then W

or W first then X twice

[solid pier]

so, this is one nuance where you have to have a model for what you’re doing. one way is to flip them in order, the other is to just pretend you have three coins and they’re labeled 1-3 and then not care about the labels in the last two once you flip them. but HTT refers to coin W getting heads, coin X1 getting tails, and coin X2 getting tails.

[quick star]

but isn't it the question's responsibility to establish the model?

😭

so would you say the question is ambiguous?

cuz like from our point of view, we're suggesting that the probability of W returning head and X returning tail twice depends on the order these coins are tossed

which is maybe not an unfounded assumption?

btw really sorry for the delay

[solid pier]

no because they’re all the same

[solid pier]

we aren’t doing that

we’re assigning outcomes supposing that we toss them in some specific order. if we choose a different order we would shift the outcomes and events to reflect it in a way

like if we flipped the W coin last then the new “good” outcome is TTH

but like that’s only a change in perspective and doesn’t affect the numbers

and even if you had a model where you had three coins and flipped them all at the same time

[quick star]

hmmm right

i think this whole thing is messing with my brain a bit 😭

[solid pier]

and even better it doesn’t affect the reasoning (up to some permutation of the symbols on the page)

[quick star]

it feels like philosophy at this point 😭

[solid pier]

some of this is philosophy, some people really despised probability back in the day

[quick star]

omgg

loll yeah i think i can see why

[solid pier]

but actually it has solid foundations now and when you start talking about how to model something like coin tosses you now have unobjectionable ways to talk about it

like if you don’t want to do any philosophy

but then you’d talk about sigma-algebras and probability spaces in the abstract and then you’d translate it to that language

but it’s easier in this case just to use our words

[quick star]

wait so if i'm flipping three different coins then philosophically getting tails on the first coin (the rest are tails) is the same as getting coins on the second coin (the rest are tails)

ughhh

tbh this doesn't make sense to me because like let's give some values to flipping heads and tails:

heads = 2 points and tails = 1 point

in this case, we'd consider 2 + 1 + 2 different from 1 + 2 + 2, no? Like this is a different case, no?

[solid pier]

they are different, but requiring one of them to occur is indistinguishable from requiring the other to occur in some sense

it’s very feels-based for when one is different from the other

sort of like saying “speed is relative”

[quick star]

😭

[solid pier]

like no you’re still gonna get a speeding ticket if you go 70mph in a residential zone

[quick star]

true true but hmm okay

[quick star]

with this you'd eliminate the philosophical bit?

[solid pier]

uhhh i’m not super sure. when you try to apply math to real-world scenarios there’s always going to be some mental gymnastics that you have to do

but doing probability nowadays it’s common to use this language, and when you have something in that language of probability spaces then it’s easier to do math, because you don’t have a lot of ambiguous terms

or like, not even that

but like you don’t have to use your physical intuition of “i’m flipping coins, but wait how am i doing that? and does the answer to this question matter?”

[quick star]

hmmmmmm

so i'm just supposed to intuitively know when the

permutation doesn't matter and when it does, i guess?

that's what i'm gathering from this

i'm sorry if i'm bothering you with this philsoophical argument though 😭

i genuinely didn't know there was that aspect to it

[solid pier]

it doesn’t bother me, i’m a bit distracted with other things atm though so i’m gonna be slow to respond

[solid pier]

in a sense yeah

you’ll build this skill over time

one might turn to a combinatorics book

instead of a probability one

because that’s actually the problem, not the actual probability that comes after the combinatorics

but i don’t really know any good references unfortunately

i just use my gears that i polished over a long time :P

[pearl pond]

@quick star Has your question been resolved?

[quick star]

ohhh no worries

[quick star]

okay makes sense 😭 this is a bit mind bending

but i get the gist

thank you for the help

[quick star]

yeah i think the combinatorics part is fine

it's just the philosophy part

but okay i think i understand it now

[solid pier]

the philosophy is part of the combinatorics of flipping coins

i would argue

i enjoy answering your questions

they are very insightful.

[quick star]

aww thank you 😭

[quick star]

right okay

thanks again

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[pearl pond]

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[torn lodge]

im confused for all parts of this question. I had a look at the mark scheme but i dont understand where the answers value come from?

[bitter herald]

so what does f'(x) < 0 mean to you as a statement?

[torn lodge]

that the solutions need to be less than 0

[bitter herald]

what are the "solutions" though

[torn lodge]

the roots

[mystic bison]

do u know what a derivitive is

[torn lodge]

nope, havent been taught that yet

[mystic bison]

erm

[bitter herald]

it's not the roots of the function f; it is the derivative of it

that's what f'(x) means

[mystic bison]

well thats a derivitive like aero said

[torn lodge]

oh- so basically i cant do this question?

[bitter herald]

no

well

it's not like f'(x) < 0 is particularly crazy of a statement

in other words, it's saying where is the function decreasing?

[torn lodge]

here?

[bitter herald]

yeah

thats it

[mystic bison]

derivitive is just the rate of change

[torn lodge]

icl my initial interpretation of this question was that i would need to see where the x values/roots are below the x axis which is the type of questions ive been used to seeing but in this case, only one part of the function is below the x axis so i was confused

[bitter herald]

you're confusing f(x) < 0 with f'(x) < 0 which is why

[torn lodge]

ahh i seee I'll keep that in mind for when im taught it

[torn lodge]

exactly, ive never seen f'(x) used in any of the questions ive been given but yea thanks anyways!

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[bitter herald]

part b and c dont need it btw

just as an fyi

[pearl pond]

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[final hinge]

can anyone help me calculate $\dv{t} x^T(t) A x(t)$ where $A$ is a matrix and $x(t)$ a vector, which in this case is $x(t) = c/t + d$

[jolly parrot]

π=√g

[plush bramble]

product rule

[final hinge]

is it $-\frac{2}{t^2} c^T A(c/t +d)$ ?

[jolly parrot]

π=√g

[plush bramble]

omg

[final hinge]

lol

sorry

[lethal pond]

why does it look so.... 😩

[final hinge]

????

[lethal pond]

uh mb

[final hinge]

brother is thirsty for mathematics

[lethal pond]

are you not

[final hinge]

anyways, if anyone could confirm this it would be very nice!'

[plush bramble]

show your work but first test it in #latex-testing if you insist on latex'ing. otherwise just write it on paper and take a picture or type it out

[final hinge]

:/ my 10 years of latex all for nothing

show my work uh ok

[plush bramble]

latex is a different skill from math

[final hinge]

i just need someone to confirm pls lmao, trying to implement interior point optimisation and the shit aint working and its driving me mad😭

[plush bramble]

you saw this right?

[final hinge]

Aw man

lol

anyways, i will assume it is correct

thanks

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[desert solar]

Consider an orthonormal reference system in space ( S ). Two planes are given, respectively:

[
\pi_h : hx - hy - (2h + 1)z + 2 = 0
]
[
\pi' : 2x - y + 3z + 1 = 0
]

where ( h \in \mathbb{R} ) is a parameter.

(a) Determine for which values of ( h ) the two planes are perpendicular.

(b) Given the value of ( h ) from the previous point, let ( r ) be the line of intersection of the two planes. Determine a skew line ( s ) with respect to the line ( r ).

(c) Determine the line ( t ) that is perpendicular to the line ( r ) and also intersects the line ( s ).

(d) Determine the point of intersection between the line ( t ) and the line ( r ).

[jolly parrot]

Task Bot

[desert solar]

a) aa'+bb'+cc'=0 -> h=-1

I'm stuck at point b

I don't know where to start from that point

[pearl pond]

@desert solar Has your question been resolved?

[desert solar]

<@&286206848099549185>

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[sleek plover]

this is where i got to i am absolutely stumped pls help 😭

i was thinking mb expanding the rhs of the 4th line

by e^-x/e^-x

f this ima go sleep bruh

if any of u solve ts dm me

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[muted moth]

I have a problem with this derivative exercise:
Determine the increments of the following functions for the given intervals.

[midnight haven]

whats the function

[muted moth]

(sorry, I was trying to do in the formula writer)

I have this rn

(im taking the photo sorry)

[midnight haven]

increment is just difference

[pearl pond]

@muted moth Has your question been resolved?

[muted moth]

<@&286206848099549185>

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[arctic hazel]

whatd i do wrong here ?

[barren tide]

It can't be seen

[arctic hazel]

mb lemme send another

[barren tide]

You calculated S_5

that is, the last partial sum they asked you for

u must do the same for s_1, s_2, s_3 and s_4, separately

[pearl pond]

@arctic hazel Has your question been resolved?

[arctic hazel]

oh

brahhh

so it’s basically going to be 5 different equations ?

ok makes sense

ty

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[fair venture]

I'm trying to use conservation of energy to solve this problem. would that work? and also how would I find the moment of inertia?

[drowsy adder]

How would you conserve energy?

[fair venture]

uhhh

idk

maybe angular momentum?

instead?

[drowsy adder]

the angular accelereation is askedd initially, no w will be developed

[fair venture]

would torque work?

[drowsy adder]

yep!

[fair venture]

ok

thx

[drowsy adder]

And since these are point objects, their moment of intertia will just be md^2

[fair venture]

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[fair venture]

.reopen

[pearl pond]

✅

[jolly parrot]

Gen

Gen

[fair venture]

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[steep pier]

how is it 44?

[steep pier]

D is 68 right because corresponding angles

[valid kettle]

yes

[sharp smelt]

Notice triangle BDE is isoceles

isosceles

[steep pier]

yes

so two same angles

[sharp smelt]

yes

[steep pier]

ohhhh got it

180 - 68 - 68

= 44

[valid kettle]

yes

[steep pier]

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[steep pier]

how is x = 70?

[steep pier]

on my teachers answer key it says 70

i need help understanding how she got that

[gaunt surge]

FE and CD are parallel

[barren tide]

angles ECD and EDC are equals

[rich pecan]

since triangle cde is an isoceles, the two angles missing in triangle cde have to be the same

( 180 - 40 ) / 2

[steep pier]

so 40, 40, and 100?

[rich pecan]

40 70 70

[steep pier]

ohhhhh

i get how you got 70 and 70

[rich pecan]

since cd and fe are parallel as Precalc hater said, therefore angles ecd, edc, and therefore fec are equal like sapoperro said

[steep pier]

ohhh got it tysm!!

[pearl pond]

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[steel canopy]

What's the problem?

@steep pier

[barren tide]

A+C=90

[steep pier]

i need help on what the next statement should be

wait i got it tho tysm

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[sharp smelt]

I'm trying to use the chain rule to differentiate this ( This chapter is on multivar calc)

[sharp smelt]

I'm confused, can't I just sub in what x and y are

If not it would be something like

$2x \frac{dx}{dt} + 2y \frac{dy}{dt} + y \frac{dx}{dt} + x \frac{dy}{dt}$

[jolly parrot]

ƒ(Why am. I here)= I don't know

[sharp smelt]

Which feels overly messy for what is essentially a single variabel function

[plush bramble]

It's about the same amount of work

[sharp smelt]

thanks

got it

.close

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[sharp smelt]

I can do it from here

[pearl pond]

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[sharp smelt]

I'm rtying to find $\pdv{z}{s}$

[jolly parrot]

ƒ(Why am. I here)= I don't know

[sharp smelt]

so what I got was

$\pdv{z}{s}=e^rcos(\theta) \frac{dr}{ds}+ e^r \cos(\theta) \frac{d \theta}{ds}= e^rcos(\theta) t+ e^r \cos(\theta) \frac{2s}{\sqrt{s^2+t^2}}$

which feels wrong

[rotund yoke]

Hey

[viral oracle]

You have [s} in there

[jolly parrot]

ƒ(Why am. I here)= I don't know

[surreal relic]

Cos gotta turn into sin

$\pdv{z}{s}=e^rcos(\theta) \frac{dr}{ds}+ e^r \cos(\theta) \frac{d \theta}{ds}= e^rcos(\theta) t - e^r \sin(\theta) \frac{s}{\sqrt{s^2+t^2}}$

Shit -sin

[jolly parrot]

Lex

[sharp smelt]

oops

[surreal relic]

Also s

Not 2s

[rotund yoke]

Middle school math is difficult. What is the answer to question 3?

[surreal relic]

!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).

[sharp smelt]

Got it

thanks

.close

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[midnight haven]

isnt this set empty?

[midnight haven]

from my understanding, the union of integers with a interval thats only 1 wide, would only at most net 2 integers

or is it saying that the resulting integer from this union needs to = 3 itself

this might just be me failing to understand this notation

[sharp smelt]

I think it's saying the cardinality of the set is 3

[midnight haven]

so what i said first?

meaning it would be an empty set?

i said union but i meant intersection

mb

[sharp smelt]

Well yes, but can you prove it

[midnight haven]

i can attempt, I just wanted to confirm im understanding this notation even

proving its an empty set

anyway, thankyou

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[vital herald]

hi can anyone help or check my answer?

[pearl pond]

@vital herald Has your question been resolved?

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[quick star]

just wanted to know a convention but are all prisms assumed "right" unless stated otherwise?

[quick star]

if i say triangular prism then it's not assumed to have rectangular faces?

Like here i googled "surface area of triangular prism" and it shows assumes right thus giving square faces

i think my question is how we know that the edges connecting bases are perpendicular to the bases

<@&286206848099549185>

[warm current]

I wouldn't assume a right triangle from "triangular prism"

[quick star]

hmm okay but then how can you infer that the above faces are rectangular? 😭

[warm current]

That comes from the definition of prism itself

[quick star]

In this image

the definition of prism allows for faces to be parallelograms

[quick star]

the right makes it known that the parallelogram is a rectangle

[warm current]

All have rectangular sides

[quick star]

isn't this the defintiion?

[quick star]

and this too

[warm current]

I believe this is called an oblique prism

[quick star]

but in this book, if you look at the part of "right prism"

and then now see this part

[warm current]

(btw that's a super uncommon name, I don't think you'll be expected to memorize it)

[quick star]

"infer from the problem that the prism is indeed a right prism"

and the book is in accordance to wikipedia

wiki says the same thing

[warm current]

correct. Most of the time, when you read "prism", a right prism is assumed

[quick star]

oh

why didn't they just write assume

😭 i'm confused

since they said "infer"

[warm current]

They wrote "infer" here, which is basically "assume"

[quick star]

infer means logically conclude, no?

sorry i'm not trying to troll 😭

i'm genuinely confused

it sounds like i'm trolling

[warm current]

close

infer = deduce without explicit or complete information

[quick star]

so here's another question (i don't need help with solving it)

we assume all the faces are rectangular?

the book said they will leave out the "right" part bc they want you to infer it

[warm current]

I would, yes.

[warm current]

correct

So you infer that it would be a right rectangular prism

[quick star]

which means there's no justification for this other than just assuming?

[warm current]

justification for what?

[quick star]

why all the faces must be rectangles

[warm current]

correct

without the inference, you would not be able to know for sure

[quick star]

yeah okay fair enough

but this book makes it seem like

😭 they've "solved" something

bc the answer starts with "all of the faces are rectangles"

[warm current]

I believe it says this as a start for the remainder of the solution

like "we are going to use the fact that <blah blah blah>"

[quick star]

hmm okay i guess

when will i know when it's not right?

[warm current]

From what your book is telling me, if you read only "prism", then assume right prism. Therefore, if it is not a right prism, you will be told explicitly.

[quick star]

hmmm 😭 okay i can live with this i guess

[warm current]

I understand and respect your preference to have certain definitions be very explicit, and to be as wrote and comprehensive as possible. However, relying on inference to reduce reading is a handy tool, and something mathematicians like to do.

[quick star]

yeah fair enough but like 😭 idk to me it feels a bit like guessing

because we didn't even need the pictorial

we didn't even need the question

to make this "inference"

[warm current]

It is some mindfulness required on your end.

Gotta not let yourself forget it, which can be easy to do at times

[quick star]

okay fair enough, thanks

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[lone fox]

can someone tell me what the real name of the "golden cycle" of a trig function is called, everywhere i search i cant find it and im struggling to find a video to walk me through it

[lone fox]

is it a fundamental period?

and how would i be able to make it for a graph like y=2csc(4x-pi)-5

[pearl pond]

@lone fox Has your question been resolved?

[pearl pond]

@lone fox Has your question been resolved?

[steel canopy]

U mean "period" of a function?

[lone fox]

i dont think its just the regular period of it

[strong relic]

ive never heard of that term, i dont think its in standard usage

use of the word cycle does make me think its related to the period

[lone fox]

it says to graph the full golden cycle of the trig function

this is what i've done so far

but im just not sure what to do from here

or is that it

idk if i graphed this even right since hte period is 2pi but idk what to do

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[hollow drum]

I need to understand Radius of Convergence for my Exam; support requested Calculus 2

Section is called "Detect Power Series and determine the radius and interval of convergence for power
series. Combine, integrate, and differentiate power series"

[pearl pond]

@hollow drum Has your question been resolved?

[pearl pond]

@hollow drum Has your question been resolved?

[hallow remnant]

okay so do you understand the radius of convergence?

it is the point at which the taylor series approximation stops equaling the actual function

so specifically, here we are given the taylor series approximation

locally, this is modeling some smooth function

however, this summation isn't equivalent to that function after x is big enough

this will happen when the terms start to diverge

and go to infinity

so when will $\sum_{n=1}^{\infty}(-1)^n\frac{x^n}{n^25^n}$ converge?

[jolly parrot]

Arnavutköy

[hallow remnant]

if $|x|$ is less than or equal to $|5|$, then all of the terms are $\le\frac{1}{n^2}$ right?

[jolly parrot]

Arnavutköy

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[quasi moon]

How to do this?

[rich pecan]

u can start by finding what only one square area is equal to

[random hornet]

yeah

[autumn fossil]

what's that?

[quasi moon]

Idk

[autumn fossil]

is there no context?

nothing else written there?

[random hornet]

well

find a number that =5 and is used 4 times

like 1.25

[rich pecan]

my guess is that it's looking for the area of the 3x3 grid based on only the 2x2 grid's area

[random hornet]

that equals 5 x 4

[autumn fossil]

the "pattern" could be anything

[autumn fossil]

how do we know it's area?

[random hornet]

true

[quasi moon]

Yes

[random hornet]

but

[autumn fossil]

it could be circumfence

or number of squares + 1

[random hornet]

am guessing its that

else its not possible

[rich pecan]

its impossible to know without context :')

[autumn fossil]

exactly

[random hornet]

exactly

[rich pecan]

i was just guessing haha

but u are right

[autumn fossil]

if there is nothing else written there, this question is completely meaningless

[random hornet]

am just guessing that if its solvable one square is equaled to 1.25

[autumn fossil]

or maybe one edge = 5/12, who knows

[random hornet]

sooo the answer is 1.25 x 9

[random hornet]

yeah

[autumn fossil]

or maybe the 5 is just number of squares + 1

[rich pecan]

is there any other context for this perhaps ?

[random hornet]

noo i dont think soo

noo.

she said no

[rich pecan]

ohh it says on the question / instruction ?

[random hornet]

ohhh

then what

wait

if theres no instructions then its 9 x 1.25

and 9 x 1.25=11.25

sooo thats the answer am guessing

does any one agree?

i got to go byeee

[pearl pond]

@quasi moon Has your question been resolved?

[urban hound]

I don't know if this has been stated before but couldn't it also be the points where 2 lines intersect?

Discounting the corners

[daring copper]

i'm a bit late

but my guess is the amount of total squares in the figure

like there are 4 1x1 squares in the first figure and 1 2x2 square

so in the second figure it would be 9 1x1 squares + 4 2x2 squares + 1 3x3 square = 14 total squares

[pearl pond]

@quasi moon Has your question been resolved?

[pearl pond]

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[clever briar]

In a triangle ABC, with obtuse angle Â, let D, E, F be the feet of the heights of A, B and C, respectively. DE is parallel to CF and DF is parallel to the bisector of BÂC. Determine the angles of triangle ABC.

Gauge: 108, 18 and 54 degrees.

I saw a solution using cyclic quadrilaterals but this subject was not yet taught in the book I got, so is there an alternative solution that does not use this?

[tame oxide]

hey

is there 3 or just one triangle?

[midnight haven]

what does iscolese triangle mean?

[plush bramble]

memes go in #chill

[clever briar]

yes

[tame oxide]

yes to which one?

[tame oxide]

aaaa thx for the pic

[pearl pond]

@clever briar Has your question been resolved?

[pearl pond]

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[broken quarry]

Hi

[broken quarry]

Hello! I have this integral and I need to rewrite it to make it look like the other image. I really don't know where to start, except that I already know that the derivative of sin(x) is cos(x).... Thanks.

[jolly terrace]

Easy

[broken quarry]

Not for me xdd

[merry carbon]

(also assumedly you/"they" meant f'(g(x)) * g'(x) for that second picture?)

[broken quarry]

no, it goes without " f' ":

@merry carbon

[merry carbon]

Well, that wouldn't be very helpful (as you should know that f'(g(x)) * g'(x) is the derivative of f(g(x)), which is what you want to recognise to do the integral)

Unless you wanna introduce another letter for the antiderivative of f, but then, like, why

[broken quarry]

I don't know my friend, that's bc i'm asking : ) They only give me these hints

And I think that I must not introduce another extra letter

[leaden wadi]

Try substituting cos(3x), see what happens, and then try sin(3x) and see what happens.

[broken quarry]

I don't think I understand, should I try a change of variable with those two functions?

[leaden wadi]

Yes.

[pearl pond]

@broken quarry Has your question been resolved?

[quasi moon]

[broken quarry]

Finally f(x)= 1/x g(x)=sin(3x) g'(x)=3cos(3x). I had doubts with f(x)=1/x. But I see that works.

Thanks for your help @merry carbon @leaden wadi

.close

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[quasi moon]

Paweł rode a bike along a route 700 meters long in 2 minutes.

Complete the sentence.

The average speed Paweł achieved on this route is equal to:

[quasi moon]

700 / 2

,calc 700/2

[jolly parrot]

Result:

350

[quasi moon]

350 m/m

,calc 350/60

[jolly parrot]

Result:

5.8333333333333

[quasi moon]

,Calc 5,83 m/h

[jolly parrot]

The following error occured while calculating:
Error: Unexpected operator , (char 2)

[supple knoll]

This is per second

I guess…

[quasi moon]

Nooo…

It’s per minute

I’m diving 2 minutes

@supple knoll

[supple knoll]

I mean 5.8

[dreamy wind]

5.8 should be per min

Oh wait no ur right

It is per sec

[supple knoll]

[quasi moon]

350 m

[supple knoll]

Ok

[quasi moon]

Divided by 60

It’s by

Seconds

5,8 m/s

,Calc 5.8/1000

[jolly parrot]

Result:

0.0058

[quasi moon]

0.0058 km/s

,Calc 0.00586060

[jolly parrot]

Result:

20.88

[quasi moon]

20,88 km/h?

<@&286206848099549185>

Closest answer is 21 km/h…

,Calc 0.005.83333333333336060

[jolly parrot]

The following error occured while calculating:
Error: Unexpected part ".8333333333333" (char 6)

[quasi moon]

,Calc 0.00583333333333336060

[jolly parrot]

Result:

21

[quasi moon]

ok

Gotcha

Babes

[modest crater]

if you're done please type .close

[quasi moon]

.close

[pearl pond]

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[quasi moon]

.reopen

[pearl pond]

✅

[quasi moon]

@supple knoll @dreamy wind how am I supposed to calculate that without a calculator?

It’s from an exam…

[supple knoll]

Idk…

[modest crater]

just do normal arithmetic

[quasi moon]

What?

What’s normal arthmetkc

[quasi moon]

Unprofessional…

[modest crater]

like long division

[supple knoll]

It’s ((700/2)/1000)*60 I guess it is not that hard

…

[acoustic pivot]

😂

[modest crater]

700/2000*60
7/20*60
420/20
21```

[dapper kraken]

((700/2)/1000)×60
700/2000×60
7/20×60
420/20
21

[supple knoll]

Four zero top, four bottom, Cross out 2 from bottom and from six, leave us 3*7…

[pearl pond]

@quasi moon Has your question been resolved?

[pearl pond]

@quasi moon Has your question been resolved?

[pearl pond]

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[river lodge]

in the equation V = kQ/r
where V is voltage
If we get a negative answer as V, what does that mean?
and what is the differnece compared to when we get a positive answer

[sharp vigil]

voltage can be either positive or negative

generally positive charges will move from areas of positive voltage toward areas of negative voltage, and negative charges will do the opposite

and generally positive charge creates positive voltage, whereas negative charge creates negative voltage

[pearl pond]

@river lodge Has your question been resolved?

[river lodge]

but when u get an answer for V using the equation V = kQ/r, if V is negative, what does that indicate?

[sharp vigil]

that would indicate Q is negative

[river lodge]

my teacher said it has to do with if the charges are opposite, in that case V would be negative as the field does work, but if they are two similar charges then V would be positive

Is that true?

[sharp vigil]

V only depends on the source charge

[river lodge]

so if the source is negative, then u would get negative V but if source is positive you would get +V?

[sharp vigil]

yes

the potential energy U on a charge q is given by U = qV, and U is positive when they have the same charge and negative when they have opposite charge

which might be where you're getting it from

[river lodge]

but what does that negative actually indicate in terms potential difference

like what does a negative value as the potential difference indicate

[sharp vigil]

generally we consider that the voltage is 0 infinitely far away, so a positive voltage will be higher than that and a negative voltage will be lower than that

and generally charges will go to wherever their potential energy will be lowest

so the overall effect is that opposite charges attract and like charges repel, as you already know

[river lodge]

right, i understand that, but im trying to understand what the signifcance of a - vs + is for V, like for example an object's displacement can either be negative or positive and that indicates whether the object is moving in the positive direction or negative direction. In the case for potential difference, is there any signficance like this?

[sharp vigil]

the sign of the electric potential/voltage reflects the sign of the charge that created it

for the displacement case, the sign indicates a direction, because displacement is a vector quantity. but voltage is a scalar quantity, so it doesn't have a direction in the same sense

[river lodge]

oh i see

that makes sense

thanks for ur help

i really appreciate it

.close

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[inland moon]

How can we find the critical points and ashmptote

[inland moon]

Is it asking for vertex?

[plush moss]

Critical points are where the derivative is equal to 0 or undefined

[inland moon]

What is a derivative

[plush moss]

Never mind

Firstly just graph the function

Using the transformations

[inland moon]

So y is at -3

I got some like this

now how do we find the critical points @plush moss

[plush moss]

Well critical points in graphically is where the function has a horizontal or vertical tangent

[inland moon]

What’s a tangent

[plush moss]

Actually nevermind I think I'm not going to be able to teach this to you

[gleaming moss]

have you not been taught derivatives?

[inland moon]

no my teacher never mentioed it

[plush moss]

Anyway I don't think this function has any critical points

Unless their school defines critical points differently

[inland moon]

they make us plug a point

to find it

if that helps

[gleaming moss]

based on the function, I'd guess they are looking for the y and x intercepts

so you want to find the points where the function intersects the y axis and the x axis

[plush moss]

Bruh

[gleaming moss]

?

[plush moss]

Ive never heard of the term critical points used that way

That's kinda weird

[gleaming moss]

neither, but they said they have to plug in values and the extreme points are both infinity

so I can't imagine it's anything else

[plush moss]

I guess you're right

[crude aspen]

i think critical points might be like the "important" or easy points

like in a quadratic its 0,0 1,1 2,4

then u can just put the transofmraotions after to graph it

[gleaming moss]

definitely not

[inland moon]

srry guys ion need help no more

[crude aspen]

fr

[inland moon]

.close

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[stoic imp]

Three consecutive prime numbers, each less than $100$, have a sum that is a multiple of $5$. What is the greatest possible sum?

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[viscid sierra]

there's a conjecture that says except for 2 and 3 all primes are of the type 6k±1

[stoic imp]

with k in Z

?

[viscid sierra]

ofc

Z+

[stoic imp]

but

have a sum that is multiple of 5

[viscid sierra]

well find all possible sums of these three primes and then find a k such that it is divisible ny 5

[stoic imp]

a,b,c is our three consecutive primes

primes except 2,3, can be found from 6k ± 1

with k a positive integer

(a+b+c) mod 5 = 0

[stoic imp]

but is a lot of possibilities still? correct?

[viscid sierra]

wait

I missed the consecutive part

that makes our problem easier

[stoic imp]

how, exactly?

[viscid sierra]

consider a<b<c
let b=6k-1, this fixes a must be 6(k-1)+1 and c must be 6k+1

similar for b=6k+1

note that you must check individually that for a certain k a,b,c are primes since all primes are of the form 6k±1 but the vice versa is not true

[stoic imp]

why a is 6(k-1)+1

[viscid sierra]

"consecutive"

guve it a think

[stoic imp]

let b = 6k-1
then a = 6(k-1)-1

why did you summed 1

[viscid sierra]

no

a=6(k-1)+1

[stoic imp]

because we want the greatest sum possible or?

[viscid sierra]

no

consider b=6k+1=37

what would be the greatest number less than 37 of the form 6k±1

[stoic imp]

6(5)+1 = 31

[viscid sierra]

or wait, I think this won't work so good

since for a=6(4)-1=23 b would be 6(4)+1=25

I don't know if there is a analytical solution to this

anyways the first thing that struck my mind was just summing group of primes from the right end

[stoic imp]

yeah

but there are multiple possibilities lol

[viscid sierra]

that will be some pain

but i don't see another way out

[stoic imp]

ok

6(17) ± 1

[viscid sierra]

erm?

Just open a prime table chart from 1 to 100

[stoic imp]

,calc 6*17+1

[jolly parrot]

Result:

103

[viscid sierra]

and start summing them from the rught end

I checked it , won't take so long

also you only need to sum their units place

[viscid sierra]

if the sum of units place ends in 0 or 5 the sum is divisible by 5

[stoic imp]

is that divisibility rule of 5

3 + 9 + 7

[viscid sierra]

may be tagged as a rule but it's just common sense

[stoic imp]

9 + 3 + 9

3 + 9 + 3

[viscid sierra]

wait

not like that

you left a better sum

83+89+97
79+89+97
79+83+97
79+83+89
and so on

[stoic imp]

but you said sum of the units place ends in 0 or 5

unit is the right most digit right?

[stoic imp]

this

[stoic imp]

is not consecutive primes

79 + 89 + 97 is not a sum of three consecutive prime numbers

[stoic imp]

or maybe I am tripping hard

what does consecutive even mean

?????????

,w define consecutive

[jolly parrot]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

ok good im not crazy I think

3 + 9 + 3 = 15

(73 + 79 + 83) mod 5 = 0?

,calc 73 + 79 + 83

[jolly parrot]

Result:

235

[stoic imp]

ok thank you b-ears

you helped me a lot, I couldnt have done it without you

.solved

[pearl pond]

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[tawny yarrow]

Can someone help me with this

[tawny yarrow]

I don’t remember how to do step functions

[plush moss]

Well let's just say someone asked you to calculate the cost of 2 days

How would you do it

[tawny yarrow]

I would think that it's just 38(2)+15 but it's a fixed rate for 1 day. So like if I rent it for 2 days and a half it's not gonna be 38(2.5)

[plush moss]

Well that's a good concern I guess

[tawny yarrow]

so a step function y=⌊x⌋

that rounds down right?

[plush moss]

Technically if x is only ever an integer 38x+15 works too

[tawny yarrow]

answer key is C(x)=-38⌊-x⌋+15

is that because - goes down, so ⌊-2⌋=-3

it rounds to the nearest day

and then the -38 cancels out that negative

[plush moss]

Yea i guess so

[tawny yarrow]

ok

thank you

.close

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[brave sluice]

did i answer these right?

i'm nervous, i don't want to assume too much, but i also don't want to over-explain

also cosets are just confusing

[gritty tulip]

I see

[pearl pond]

@brave sluice Has your question been resolved?

[pearl pond]

@brave sluice Has your question been resolved?

[pearl pond]

@brave sluice Has your question been resolved?

[brave sluice]

.close

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[sharp smelt]

So What I would do is :
\
$2x+y^2 = 2z \pdv{z}{x} \implies \pdv{z}{x} =\frac{1}{2z} (2x+y^2)$
\
$x^2+2y = 2z \pdv{z}{y} \implies \pdv{z}{y} = \frac{1}{2z}(x^2+2y)$

[pearl pond]

Please don't occupy multiple help channels.

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[sharp smelt]

Now here, this was what I was thinking

I prove that $z = \pdv{z}{x} x+\pdv{z}{y} y$

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[sharp smelt]

which feels very sus

I suppose I could substitute $x^2=z^2-y^2$?

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[sharp smelt]

or do I consider an arbitrary tangent plane and work with that

[autumn trellis]

Yes

[sharp smelt]

$(z-z_0)=. \pdv{z}{x} \cdot (x-x_0) + \pdv{z}{y} \cdot (y-y_0)$

$(z-z_0)=. \pdv{z}{x} \cdot (x-x_0) + \pdv{z}{y} \cdot (y-y_0)$

[jolly parrot]

ƒ( wai ina teacup)= I don't know

ƒ( wai ina teacup)= I don't know

[autumn trellis]

Use the symmetry to reduce calculations

Wlog you can assume x=y i.e you only need to check tangent planes of pts on y=x

[sharp smelt]

how can I assume what WLOG

[autumn trellis]

Because of the symmetry

[sharp smelt]

so $\pdv{z}{x} = \frac{1}{2z} (2x+x^2)$

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[sharp smelt]

and $\pdv{y}{z} = \frac{1}{2z} (x^2+2x)$

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[brave sluice]

this is wrong

some terms disappear in the partial differentials

[sharp smelt]

oops

yean

I forgot

[brave sluice]

the x^2 and y^2

[sharp smelt]

$\pdv{z}{x} = \frac{x}{z}$; $\pdv{z}{y} =\frac{y}{z}$

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[sharp smelt]

well, then it's pretty straightfoward

$\frac{x_0^2+y_0^2}{z_0} =\frac{-z_0^2}{z_0} = -z_0$

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[sharp smelt]

as desired

does this work

[brave sluice]

what did you do?

[sharp smelt]

I set x=y=z=0

and proved the rhs yielded -z_0

[brave sluice]

uh

i feel like a sign got flipped somewhere

but then flipped back

[pearl pond]

@sharp smelt Has your question been resolved?

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[cloud wing]

I'm looking for a website or file that includes the integral questions from easiest to somewhat hardest

[pearl pond]

@cloud wing Has your question been resolved?

[sharp quest]

https://openstax.org/books/calculus-volume-2/pages/1-introduction

https://tutorial.math.lamar.edu/classes/calcII/calcII.aspx

[pearl pond]

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[uncut crest]

Ummm how do i prove
31 | ((3²⁰×2²⁰)⁷-5×6¹³⁸)

[uncut crest]

Idk how to do it

[autumn fossil]

I'd try calculating (3²⁰×2²⁰)⁷-5×6¹³⁸ mod 31

if it's 0, then it's divisible

[uncut crest]

How do i do that 😭

[autumn fossil]

do you know modulo?

oh wait there might be a simpler way

just apply exponent laws

im dumb lol

[uncut crest]

Aaaaa

Ok

[autumn fossil]

use this stuff to simplify it

[uncut crest]

Thank youuuuu

[bitter herald]

i believe you'd still need to use fermat's little theorem, no?

[autumn fossil]

no

it will simplify really nicely

[uncut crest]

Omg it does

6¹⁴⁰-5×6¹³⁸

And then 6¹³⁸(6²-5) right?

And that's 6¹³⁸(31)

[autumn fossil]

yep

[uncut crest]

Ty againnnnn

.close

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[hollow cobalt]

What's your question?

[pearl pond]

@lethal shore Has your question been resolved?

[pearl pond]

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[hollow cliff]

how do I do this?

[midnight haven]

what

[hollow cliff]

Here's the question

So I had to make a graph and graph it on desmos

It's about a car model

From the years 2015-2024

And how the prices changed over time

I used a trigonometric function

And got this

y=13793.8241sin(0.585195(x-1.28605))+51317.0035

[midnight haven]

ok..

[hollow cliff]

Now I need to find the price of the car in 2030

How do I do that?

That's the graph

[midnight haven]

imma write it

[hollow cliff]

Ok

[midnight haven]

i got around $46,863 point somethin g

oh, .36

you gotta substitute 2030 in

[hollow cliff]

Chatgpts answer is different

Wanna see it?

[midnight haven]

yeah

[hollow cliff]

To estimate the price of the car in 2030 using your equation:

[
y = 13793.8241 \sin(0.585195(x - 1.28605)) + 51317.0035
]

we need to calculate (x), the years since 2015:
[
x = 2030 - 2015 = 15
]

Now substitute (x = 15) into the equation:

[
y = 13793.8241 \sin(0.585195(15 - 1.28605)) + 51317.0035
]

Step-by-Step Calculation:

1. Calculate (x - d):
   [
   15 - 1.28605 = 13.71395
   ]

2. Multiply by (k = 0.585195):
   [
   0.585195 \times 13.71395 \approx 8.025
   ]

3. Find (\sin(8.025)):
   [
   \sin(8.025) \approx 0.989
   ]

4. Multiply by (a = 13793.8241):
   [
   13793.8241 \times 0.989 \approx 13641.96
   ]

5. Add (c = 51317.0035):
   [
   13641.96 + 51317.0035 = 64958.96
   ]

Final Answer:

The estimated price of the car in 2030 is approximately $64,959.

This estimate assumes the trigonometric model is a reasonable approximation for the data. If the model doesn't fit well, this value might not be accurate.

[jolly parrot]

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

[midnight haven]

ask gpt to reevaluate

[hollow cliff]

Alr

[midnight haven]

ohhh

[hollow cliff]

Upon reevaluation, the estimated price of the car in 2030 using the given equation is approximately $64,909.

[midnight haven]

i know why

[hollow cliff]

Why

[midnight haven]

Gpt is right

is x in 2015 = 0

if

[hollow cliff]

I got no idea

Yk anyone else who knows?