help-33

[smoky marsh]

imma write it out and post it ok

a bit messy but yea

[marsh citrus]

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[smoky marsh]

t

[glacial hedge]

Didn't you know you can just read the normal vector off the equation ?

[smoky marsh]

i'll be honest here bub, no.

[glacial hedge]

1x+1y-1z = 0 so it's (1, 1, -1)

[smoky marsh]

oh, that.

[glacial hedge]

so it's the basis of the orthogonal vector space because it's 1D

[smoky marsh]

ok so what does that mean in regards to (1,-1,-1)?

[glacial hedge]

where does that vector come from ?

[smoky marsh]

perp Basis of W?

[glacial hedge]

you found a basis and it doesn't contain (1, -1, -1) so where does that vector come from ?
also what's perp ?

[smoky marsh]

perpendicular

[glacial hedge]

orthogonal. They're vectors

[glacial hedge]

are you implying you did gram schmidt for no reason at all ?

[smoky marsh]

im just following what i did for question 1 and i got the right answer for that

alright, i'll look further into. clearly im lacking a bit of understanding so imma watch a few videos. thanks for your help and i really do appreciate it .

.close

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[twilit notch]

Hi. I don't understand this solution after "and consequently"... I understood the part before, and noticed that the numbers in [] are solutions of the equation A^2 - A(a^2 + 2) + 1 where A = |z^2|, but didn't manage to get further.

[marsh citrus]

@twilit notch Has your question been resolved?

[twilit notch]

<@&286206848099549185>

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[daring hornet]

hey 🙂 I've got an operator A from sequence spaces ell1 to ell2 defined like this. two questions:

• if we start with an element in ell1 (blue a) and apply A to it, by definition of A the result (orange a) is in ell2. but does it stay in ell1 as well ?

• is Aa = b <= a = b true ?

[marsh citrus]

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[inland bay]

Hey!! So I have a pretty simple question

[inland bay]

I’m just wondering if you infinitely get 0*infinity (an indeterminate form I know), does the limit exist? It was the only question on my test I was confused on because the derivative was 1/x^2 * -26(x^(3/2)

It was asking for the limit as x approaches O from the right, so for the infinity section, the square roots just keep going up for every derivative, and for 0 it keeps dividing by zero obviously

I’m scared now because after wording it out I feel like the limit of that AS IT APPROACHES 0 FROM THE RIGHT is 0, but the limit just from x=0 is DNE

DAMNN please tell me I’m wrong

[gilded smelt]

pardon, but what limit exactly were you trying to evaluate?

[\lim_{x\to{0^+}}\frac{1}{x^2}-26(x^{\frac{3}{2}})]

or

[\lim_{x\to{0^+}}\frac{1}{x^2-26(x^{\frac{3}{2}})}]

?

ah wait that's the derivative

well there are ways to solve limits that do not involve l'hopital's rule

I can't tell you more than that without seeing the question, really

[inland bay]

Yeah sorry it was the limit as x>0+ of like 113sqrt(x) + ln(1+x)

Or something close to that

[gilded smelt]

huh?

that's just 0+0 though

[inland bay]

Damn idek if I can remember like what the question was I’m so sorry

Thanks for your help anyways 😭❤️

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[gilded smelt]

no problem lol

[marsh citrus]

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[warm delta]

I'm at a loss for how to even attempt this optimization question

[warm delta]

this is grade 12 calc so no multi variable stuff

[leaden monolith]

What’s a sphere

[warm delta]

all i really know is that volume = pi*r^2*h

[warm delta]

like a circle but in 3d? 😬

[leaden monolith]

Ok what’s a circle

[warm delta]

like its a shape defined by all the points in 3d space that have like the radius

so like the x y and z are a distance of r from the center

[leaden monolith]

Ok very close

All the points on a sphere are equidistant from the centre

Now if we want to fit a cylinder inside

[warm delta]

yea

[leaden monolith]

That means the cylinder must not have any measurement greater than the diameter of the sphere

Otherwise, it wouldn’t fit!

[warm delta]

yes that makes sense

[leaden monolith]

Ok

What is the largest dimension of a cylinder?

[warm delta]

it depends i guess

or do you mean like

if i drew a cylinder from the side

i could draw the diagonal or something

[leaden monolith]

exactly

That diagonal is always the largest measurement of any cylinder

So

We know the sphere has radius r

10 cm

[warm delta]

so the diagonal has to always be 2r

i think

[leaden monolith]

Yes

Now

Use Pythagoras to relate the radius of the cylinder, the height of the cylinder, and the diameter of the sphere

[warm delta]

ah

so r^2 + h^2 = d^2

where r^2 is radius of cylinder not the sphere

[leaden monolith]

Not quite

r² is only half the side of your triangle

Try do the 2D version of your problem

[warm delta]

oops

yea

i just realized

[leaden monolith]

Maximise area of a rectangle inside a circle

[warm delta]

alright tysm

big help!

ill close this room now

[leaden monolith]

[warm delta]

.close

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[marsh citrus]

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[marsh citrus]

@tulip folio Has your question been resolved?

[marsh citrus]

@tulip folio Has your question been resolved?

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[still temple]

No idea

[marsh citrus]

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[turbid frost]

Hello does anyone know how to do part b

[green sky]

Use part a?

[marsh citrus]

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[shell hill]

hi,
how to solve :
-8+6i=z^2

[teal hearth]

!status

[marsh citrus]

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

[shell hill]

the whole thing man it says that it has two solutions

don't know how to solve it

[teal hearth]

Ok

[shell hill]

i need just some help

[teal hearth]

Here is the first step

[shell hill]

please don;t give me the answer

directly

[teal hearth]

You have to let z=a+ib

[shell hill]

okay

[teal hearth]

Because a complex number is defined by its imaginary part and its real part

So if you solve for a and b you got z

[shell hill]

okay

[teal hearth]

Tell me when you need help

[shell hill]

just don't close this channel i'm trying to solve it

[teal hearth]

Yep

[shell hill]

i got a^2-b^2 = -8

correct?

[teal hearth]

Yep

[shell hill]

b = 3

found it!!

[teal hearth]

Maybe i didn't do the math

[shell hill]

z = 1 + 3i

[teal hearth]

But you got 2 equations from the imaginary parts and the real parts

[shell hill]

two solutions

so z = -(1+3i)

[teal hearth]

Yes it works

I should be it

So the whole point here is to think of z = a+ib

[shell hill]

yeah and the rest it easy

tnx man you i really appreciate it

[teal hearth]

Np that's great !

[shell hill]

have a good day

bye

[marsh citrus]

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[marsh citrus]

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[craggy tusk]

what would be theta =<

[broken forum]

@craggy tuskthis is the hard part

its easy to count the petals and the radius

to figure out the theta you have to trace out the graph with theta

see how long it would take to complete the entire graph

[craggy tusk]

how would you approach it

[broken forum]

i just told you

read what I said again

[craggy tusk]

=v how do I actually trace it by subbing number in

I have no clue how to trace

[broken forum]

calculate the r value at each theta (0, pi/4, pi/2, ...)

get a feeling for how fast the shape is being drawn

[marsh citrus]

@craggy tusk Has your question been resolved?

[marsh citrus]

@craggy tusk Has your question been resolved?

[broken forum]

@craggy tuskI explained it to you

[marsh citrus]

@craggy tusk Has your question been resolved?

[light bridge]

where school you at why the subject is hard to solve

[open relic]

,w integral from -pi/4 to pi/4 of (41/2(4+4cos(4x))^2)

[elfin berry]

Results provided by WolframAlpha

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

Requested by ELeonardo#2239

[open relic]

what limits did you try first?

your answers looks close, maybe you forgot 1/2

$A=\int_{\theta_1}^{\theta_2}\frac{1}{2}r^2, d\theta$

[elfin berry]

ELeonardo

[marsh citrus]

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[smoky marsh]

for number 1

would my answer still be right? this is the textbook answer below

can you have multiple answers for q?

[marsh citrus]

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[tender mantle]

How to calculate this limit

[modern osprey]

e raised -infinity is 0. As one can rewrite it as 1/e^infinity which is 1/infinity which equals 0.

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[wise jackal]

<@&268886789983436800>

[floral stump]

.close

[marsh citrus]

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[bright parrot]

ty

[marsh citrus]

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[azure plaza]

hi

[azure plaza]

how?

[celest wyvern]

so

that main figure is a square

so

BC is also a

use the pythagorean theorem on ABC

[azure plaza]

yep and AC = 2a^2 i wrote that

still dont understand how i can prove what is DF

if that is what u mean than here it is i dont understand how to countine

(WROTE CE AND DC)

help someone

[thick junco]

that's not right

AC is a*sqrt(2)

AC^2 = 2a^2

[azure plaza]

why

[thick junco]

because of pythagoras

it's not c = a^2 + b^2, it's c^2 = a^2 + b^2

[azure plaza]

true my bad

than AC = 2a right?

[thick junco]

no

it's sqrt(2) a

you have to root the 2 and the a

[azure plaza]

wait what sorry i am not good at geometry in english can you explain it more simple (i use other langue in geometry)

what sqrt mean?

[thick junco]

"Square root"

$\sqrt$

[elfin berry]

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

[thick junco]

this thing

[azure plaza]

oh yeah so

AC^2=2a^2 / sqrt(2)
AC = 2a?

?

help

DF=a\left(\sqrt{2}-1\right)

help

[thick junco]

@azure plaza sorry for that, my internet broke and i had to restart my computer

my point was that if $(AC)^{2} = 2a^2 \implies AC = \sqrt{2a^2} = \sqrt{2} \cdot \sqrt{a^2} = a\sqrt{2}.$

[elfin berry]

OutOfNosh

[thick junco]

אתה רואה?

[azure plaza]

i guess yes

than where the -1a

[thick junco]

@azure plaza well, can you work out the distance AE?

the wifi keeps dropping in and out

it'll probably happen again

[marsh citrus]

@azure plaza Has your question been resolved?

[spark berry]

Use this as reference @thick junco

[thick junco]

why are you pinging me

i didn't ask the question

[spark berry]

Ah mb

@azure plaza

Use the fact that <fec is 90 and cd= ce. That's a kite. So DF = DE
AC = sqrt(2a^2)or a(sqrt(2))
EC + AE = AC so AC - EC = AE
asqrt(2) - a = a(sqrt(2)-1)
Via similar property of similar trianlge
AE = FE = FD
So AE = a(sqrt(2)-1)

[marsh citrus]

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[sinful burrow]

how do i do this

[sinful burrow]

im not good at reading probloms

[marsh citrus]

@sinful burrow Has your question been resolved?

[sinful burrow]

?

[uneven bane]

let y=mx+c be the equation describing the graph
points (34,65) and (62, 42.6) are on the line

@sinful burrow

[sinful burrow]

ok

then what?

?

[fair kindle]

then plug in the two points and get m and c

[sinful burrow]

i have never done this kind of problom i need a step by step

[fair kindle]

(34,65) is a pair in the form (x,y) . you place 34 in place of x and 65 in place of y.

you then get 65 = m*34 + c

this is an equation in two variables

you do the same for the second point

42.6 = 64*x + c

then these two equations need to be solved algebraically together

do you know how to solve equations like this ?

[sinful burrow]

no

[fair kindle]

you can simply subtract the top equation from the bottom equation

C cancels

and you get 1 equation in m.

65-42.6 = (64-34)*m

does that make sense or not ?

[sinful burrow]

ig so the answer would be 56/75?

[fair kindle]

which answer ?

[sinful burrow]

for miles

[fair kindle]

how did you get the answer, we haven't even finished this step xD

[sinful burrow]

oh i did 65-42.6 = (64-34)*m and sloved

my b

[sinful burrow]

then what do we do from here then?

[fair kindle]

you found m

you plug it in one of the equations to get c

the equation you got is miles = time * m + c

you plug in the time and get the miles and finish the question

[sinful burrow]

i am dumb and need you to show me how to do sorry

.close

[marsh citrus]

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[mild kayak]

Hey, I asked about this problem 2 days ago but never got the help i needed with it, so i am reposting the question below

[mild kayak]

Can someone look over my work and tell me if this is correct, or if my logic at least makes sense? Im having a hard time with Power Series and and I had to spend a lot of time watching supplemental videos to even get this far, so I want to make sure im doing it correctly at least.

Apologies that my work is a bit messy, i tried to keep it somewhat organized but it didnt go so well. If i need to explain what my thought process was for any part of it let me know

It was suggested that I do it a different way, but I dont understand how to do it, and I am fairly certain I messed it up, Im including that work with this message

Please help I hate power series so much 🙏

[marsh citrus]

@mild kayak Has your question been resolved?

[mild kayak]

<@&286206848099549185>

[compact rivet]

This looks right

[compact rivet]

Same lol

[mild kayak]

Which version looks right? because tbh I barely understand what i wrote in the second one

[compact rivet]

The second one looks right

[mild kayak]

alright nice nice thank you

[compact rivet]

Np

[mild kayak]

.close

[marsh citrus]

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[torpid swan]

what am I doing wrong?

[still temple]

Seems correct... What makes you think you're wrong?

[torpid swan]

so the code I get on the bottom is 130 which if I check the form I am given to check the code shows that it is wrong

[still temple]

Oh wait I haven't checked the graph yet my bad

Didn't notice that was part of the answer too

[torpid swan]

ya its a code work sheet

[still temple]

,calc 675/7.5

[elfin berry]

Result:

90

[still temple]

,calc 675/12

[elfin berry]

Result:

56.25

[torpid swan]

where did I go wrong?

[still temple]

,calc 5012+107.5

[elfin berry]

Result:

675

[still temple]

,calc 12+7.5+675+150

[elfin berry]

Result:

844.5

[still temple]

Wait no that must mean the digits

,calc 1+2+7+5+6+7+5+1+5

[elfin berry]

Result:

39

[torpid swan]

I used to think it was that but I think its 5 and 8 on the options

[still temple]

... How'd you get 130?

Oh right obviously

[torpid swan]

no worries i did the exact same thing for like 30 min

[still temple]

So your mistake is that you didn't include 7

Notice equality 5 has the greater than or equal to symbol

[torpid swan]

mhmm

[still temple]

Meaning the line is included as well

[torpid swan]

so its 221?

[still temple]

And to avoid ambiguity, I checked manually if 7 was included using its coordinates on the graph

,calc 17*13

[elfin berry]

Result:

221

[still temple]

Yep

[torpid swan]

I did earlier and its right on the line

okay lemme check on the form

Hrmm it says its wrong

[still temple]

scheiss

,calc 1012+607.5

[elfin berry]

Result:

570

[still temple]

,calc 3012+207.5

[elfin berry]

Result:

510

[still temple]

As expected...

[torpid swan]

wait so the money doesnt corollate correctly?

[still temple]

Oh no I was just checking that 9 and 1 weren't included

[torpid swan]

okay

[still temple]

I can't figure it out

[torpid swan]

okay so im not the only one

[still temple]

Everything up to that point makes absolute sense

So I don't know how 221 could not be the right answer

[torpid swan]

I dont really want to message my teacher because I just did on about another page

[still temple]

Do the answers to the other questions matter or is it just the code?

[torpid swan]

So each page has a code and i have to basically show the work and how i got the code

[still temple]

Okay but does it automatically check every answer you gave or does it only check the code and then your teacher comes along to check how you found it?

[torpid swan]

so I have a google form where i can put the codes I have in and check if they are correct and then once I get them all I will send the whole file to her to review

[still temple]

Hmmm

.close

[marsh citrus]

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[still temple]

.reopen

[marsh citrus]

✅

[still temple]

Tried helping this guy, couldn't figure out what's wrong with his answer

Long story short, we worked together to get 221, and according to him the Google form rejects it as incorrect

[zinc onyx]

i tried it myself and 221 seems correct

[torpid swan]

Fine I'ma message my teacher about it

Thx for the help

.close

[marsh citrus]

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[marsh citrus]

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[nimble junco]

What happens if you don't substitute n for n-1

You get $s_{n+1} = s_n + 2n - 1$

so can't you just put in s0 still?

[elfin berry]

Aorliei

[marsh citrus]

@nimble junco Has your question been resolved?

[marsh citrus]

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[hollow pelican]

not sure where to start

[sweet pawn]

consider only the magnitudes of the terms

then just solve the inequality

1/n^6 < 0.00005

[hollow pelican]

oh ok

so I just solve for n

log_6(1/0.00005) terms rounded to the nearest whole?

[tight furnace]

that's not the inverse function of n^6

you're looking for 6th root

[marsh citrus]

@hollow pelican Has your question been resolved?

[hollow pelican]

oh, mb

how would I calculate this without a calculator

[tight furnace]

n^6 > 1/.00005

n^6 > 20000

n=10 works

[hollow pelican]

just dont be an idiot

got it

[tight furnace]

so does 6^6 for that matter

[hollow pelican]

ok imma go do some practice problems

ty

[tight furnace]

np

[hollow pelican]

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[thorny onyx]

so i divide the line into 3 rectangles to give us a width of 1 for each area

and then something about the endpoints

i will have x = 0, 1, and 2

for my right endpoints

from -1 to 0, my endpoint is 1

from to 0 to 1, my endpoint is 5

from 1 to 2, my endpoint is 17

what do i do next?

[marsh citrus]

@thorny onyx Has your question been resolved?

[thorny onyx]

<@&286206848099549185>

i figured out how to do left and right endpoints

now how do i do my midpoints

[marsh citrus]

@thorny onyx Has your question been resolved?

[thorny onyx]

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[pallid frigate]

What have you tried?

[serene heart]

i meant how do i determine the bounds

or set up the double integral

[pallid frigate]

Have you tried plotting the points and visualizing the region?

[serene heart]

yeah, isnt it like a triangle

[pallid frigate]

indeed

have you tried using that to set up the bounds?

[serene heart]

im not sure how to use the triangle to set the bounds tho

[pallid frigate]

please stand by; let me acquire the image real quick

[still temple]

you want to determine the bounds?

Graph your 3 points first

[pallid frigate]

this is the region we're working with

for each x, what do you think y is ranging between?

@serene heart

[serene heart]

0 to 3

[still temple]

Yes

[pallid frigate]

that's for x=0. it won't work for, say, x=0.5

because y is changing with x

[pallid frigate]

try answering this question again

[still temple]

isn’t a dydx integral easier here

[pallid frigate]

the purpose of the exercise is to carry out a double integral

[still temple]

i mean first dy, then dx in the double integral

[pallid frigate]

no. in this particular case, the complexity will be equivalent.

@serene heart, are you following?

[serene heart]

im still thinking

[pallid frigate]

okay let me help you out a bit

[pallid frigate]

let's begin here: can you tell me the equation of each of the lines you see in this image?

namely:
what is the equation of the black line?
what is the equation of the green line?
what is the equation of the red line?

i would appreciate it if you responded more frequently, @serene heart

[pallid frigate]

please accept my apologies. indeed, a dydx integral would be easier. and that is in fact what we're doing. i'm sorry for not properly reading your question earlier before responding.

[still temple]

np

@serene heart

[pallid frigate]

i assume you've figured out the problem, kikaru. good job! have a nice day.

[marsh citrus]

@serene heart Has your question been resolved?

[marsh citrus]

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[vapid quiver]

help

[odd turret]

on

[vapid quiver]

do horizontal dilations affect the range?

i’m pretty sure it doesn’t affect the horizontal asymptote right?

for exponential functions

[silk sable]

No.

[vapid quiver]

alright thank you sm

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[gentle owl]

D? This would be a revolution about the x axis so it would be integral from 0 to 2, pi(ln(3-x)^2) dx?

[marsh citrus]

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[marsh citrus]

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[still temple]

@stark tartan

[marsh citrus]

Please don't occupy multiple help channels.

[stark tartan]

ty how do i get help?

[still temple]

.close

[marsh citrus]

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[still temple]

past ur thing in here

paste it

[stark tartan]

ok thx

[still temple]

and then it will open under ur name

[tall umbra]

he already have a channel

#help-22

[still temple]

i closed it

[stark tartan]

oh

[still temple]

it souldn't be under my name

.close

.reopen

[marsh citrus]

✅

[still temple]

.close

[marsh citrus]

Channel closed

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[tall umbra]

it doesn't close instantly

and he already have another channel

[still temple]

omg

what

[tall umbra]

#❓how-to-get-help

[still temple]

o

no

whatever

all g

[tall umbra]

it's closed just wait

[still temple]

this is my question

part bi

.close

[marsh citrus]

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[jolly sigil]

I'll be honest here, I don't think doing another LH rule is gonna help me

[jolly sigil]

Hahaha

Is there something else I should do

omg

figured it out

after step 8

it's 16

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[analog edge]

if a question says that we have a function in the first octant what is my theta?

[analog edge]

pi/2 or pi/4?

i thought it would be pi/4 since its the first octant

1/8 of a circle

[marsh citrus]

@analog edge Has your question been resolved?

[stoic coral]

and it is wrong?

theta in [0,pi/4]?

[analog edge]

okay I checked again, the question asked for a sphere, my bad, its supposed to be 4pi/8 which is pi/2

.close

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[dense pollen]

Given a function f(x,y), would it make sense to calculate the partial derivative of x with respect to y?

[stark trail]

What do you mean by , "would it make sense to" ?

[dense pollen]

Can I do it?

I know partial f with respect to x exists

Partial f with respect to y exists

[fair fulcrum]

dy/dx is something you control

f(x,y) doesn’t tell you anything about that

[dense pollen]

Hmm, alright thanks

That makes sense

[fair fulcrum]

it’s like saying which angle to slice that function of x and y that plots on z axis

partial derivative means slicing parallel to x or y axes

[dense pollen]

Ahhh alright so it can be anything then basically

[fair fulcrum]

ye

[dense pollen]

Okay thanks

[marsh citrus]

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[marsh citrus]

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[woven harness]

what does that swiggly line above B mean?

[tall night]

complement of B

[woven harness]

what does that mean

ok i see

.close

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[marsh citrus]

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[mystic viper]

Determine the radius and center of a circle using the equation (x-6)^2 +y^2 = 1

[feral agate]

Do you know the general equation of a circle?

[mystic viper]

yes

[feral agate]

Well can you compare this equation with that one?

Maybe try writing one below the other

[mystic viper]

So what b is 0?

[feral agate]

Yes if by b you denote the y coordinate of the center

[mystic viper]

(x-a)^2+(y-b)^2=r^2

[feral agate]

Yes b = 0

[mystic viper]

ok

[feral agate]

Can you solve this now?

[mystic viper]

yes. Check if the point (0, -5) lies on the circle (x-3)^2 + (y+1)^2 = 25

How do i know if it's on the circle

[stoic saddle]

hint: this has nothing to do with circles

and instead it has to do with knowing how to translate between a curve and an equation that defines said curve

[mystic viper]

What do you mean

[stoic saddle]

i mean exactly what i said

to say that a certain curve is defined by a certain equation is to say that "point (x,y) lies on <curve>" and "point (x,y) satisfies <equation>" are one and the same

this is the cartesian plane's whole THING really

[mystic viper]

can you elaborate

[stoic saddle]

i have two options here: either reply with a snarky "there's nothing to elaborate on" (because there really isnt. this is the basics) or i can show you some examples

are you familiar with straight lines and going between graphs and equations for those?

[mystic viper]

I am not

[stoic saddle]

💀

... are you in school at the moment?

from what i've seen it looks like you are missing a lot of knowledge

[mystic viper]

Can you show me some examples

[stoic saddle]

of what

[mystic viper]

What you were talking about earlier

[stoic saddle]

ok alright so like

lets backtrack a bit further

just to ensure you arent stupefied by whats about to come

are you familiar with the cartesian plane ITSELF?

[mystic viper]

no

[stoic saddle]

what.

oh my god.

what the shit.

what grade are you in???

do you know how to plot a point on the cartesian plane given its coordinates, and how to read the coordinates of a point plotted on the plane?
no?

[mystic viper]

I do not

[stoic saddle]

yikes!!!

okay, seriously. what grade are you in?

[glossy moth]

Bruh grandpa how are you survive the ww2

[stoic saddle]

why are you coming here with problems which are like one or two grade levels above what you're able to do??

[mystic viper]

Can you just help me

[stoic saddle]

meter

do you realize that trying to "help" you

would be like trying to "help" someone read a novel when they don't even know the alphabet

[mystic viper]

Show me the examples

[stoic saddle]

trying to "help" you
would be like trying to "help" someone read a novel when they don't even know the alphabet

theres no point in me showing examples

you are missing prerequisite knowledge

like ok i could try to idk show the line y=2x+3 and tell you that the point (-1, 1) lies on it while (1, 9) doesnt and try to explain why

but you would not get any of it

and it would be a waste of both of our times

[feral agate]

Oh damn what happened here

[mystic viper]

The former lies on it because for every x it goes up 2y's

[stoic saddle]

bad.

[mystic viper]

-1, 1 0, 3 1, 5

[mystic viper]

Check if the point (0, -5) lies on the circle (x-3)^2 + (y+1)^2 = 25
How do i know if it's on the circle

[feral agate]

Well you have the equation of a circle and you have been given x and y values

You need to check if those satisfy the equation

[mystic viper]

so if it equals 25 or lower?

[feral agate]

If it exactly equals 25

[mystic viper]

but it's on the circle right?

[stoic saddle]

it says (x-3)^2 + (y+1)^2 = 25, not (x-3)^2 + (y+1)^2 ≤ 25.

so none of that "25 or lower" business.

[mystic viper]

Why does it have to be exactly 25

[stoic saddle]

because the equation says (x-3)^2 + (y+1)^2 equals 25.

[feral agate]

If its greater than 25 it would be outside the circle

If its less than 25 it would be inside the circle

[stoic saddle]

in mathematical usage "circle" refers to only the curve

the bit inside it is called a disk

[mystic viper]

inside is the same as on right?

[stoic saddle]

no

[jagged dagger]

It isn't unfortunately, in math

[feral agate]

It refers to just the circumference

Not the area

[mystic viper]

Alright i got it

.close

[marsh citrus]

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[marsh citrus]

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[lilac pine]

The lines L1 and L2 have only one shared point with coordinates (4,-1) and they tangent the ellipse x^2+2xy+2y^2=5. Find the tangentpoints with the ellipse and the equations for the two lines.

[lilac pine]

I'm lost, I don't know what to do. Can someone help me?

I did partial derivative with respect to x and y to find the gradient, but what's the next step?

[marsh citrus]

@lilac pine Has your question been resolved?

[marsh citrus]

@lilac pine Has your question been resolved?

[lilac pine]

I found a similar problem online but I need help explaining some steps:

can someone explain this part?^

[broken forum]

what of that do you not get

p^0 is the point

weird notation but w/e

[lilac pine]

I dont get how he get the solutions for x and y

[broken forum]

plug in the values for the point and then its just a basic system of equations

[lilac pine]

I must be slow today.. I just don't see how he gets x=0 and y=3 to begin with. He has the point (12,3), pluggin in the values into the first equation gives me 144+36-36=0? What am I missing

oh wait nvm

i got it, embarrasing..

.close

[marsh citrus]

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[hard grail]

how to do this

[broken forum]

Do what

That’s a statement

[real glacier]

that is not a statement

[broken forum]

Uh

What is it then

It’s not a question

[real glacier]

a number if it converges

[broken forum]

We don’t know that

We don’t know what the question is

You’re assuming they’re asking that

[real glacier]

the question is to solve to summation

[broken forum]

that’s your assumption

[real glacier]

what else could do mean in this context

[broken forum]

could be right, could be wrong

don’t assume

you inquire

[real glacier]

?

[broken forum]

you ask them

[real glacier]

i think you have to approximate this one, it seems transcendental

if its asking about convergence, you need to do a p series convergence test

because you can show this is bounded by two p series

[hard grail]

its asking converge or diverge

[marsh citrus]

@hard grail Has your question been resolved?

[rare nexus]

Do you have an intuition for what the answer is ?

[hard grail]

no

[main idol]

That's all kefir was asking you. Don't make it so hard next time

Use one of the comparison test or ratio test or root test or a combination

[hard grail]

what?

yea i dont know how thats why Im asking

[main idol]

.

[main idol]

This is a starting point

[hard grail]

ok so what test do I use and how do I do it

[main idol]

I listed a few for you to try

Have you not learned them yet?

[hard grail]

i have but i forgot

im asking if someone can just do the problem for me

[main idol]

Not how the server works

It's against the rules to give away answers

If you have an attempt, share it

If not, read
https://tutorial.math.lamar.edu/classes/calcII/seriescomptest.aspx

And
https://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx

[marsh citrus]

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[earnest swallow]

question might be a bit uncommon but i have a spreadsheet for some damage calculation in a game and the game's wiki states this:

In Overwatch 2, Armor reduces damage by 30% (as compared to flatly reducing it by 5 points in Overwatch 1). The reduction occurs after all other modifiers (damage boosters, headshots, etc)
If an instance of damage breaks armor, the amount of damage is dynamically reduced at the rate of 43% of remaining armor health. Simply, every 7 armor health reduces 3 damage.

so basically heroes have normal hp and armor hp and if the damage dealt is greater than the armor hp its said to be "dynamically reduced at the rate of x%" and i dont quite understand how those "dynamically reduced at rate of 43%" relate to the "every 7 armor reduce 3 damage"

whats the correlation here?

[earnest swallow]

and tbh i dont even understand what mathematical concept a "dynamic reduction at rate of x%" refers to

[sleek lake]

well 0.43 is almost 3/7

so that's what they meant, the reduction equals 43% of armor instead of what they started with (30% of damage)

i don't know

[earnest swallow]

oh i see, i still dont see what dynamic reduction at a rate means, since that rather sounds like the reduction decreases or something and the decrease speed is .43 per unit or something

and ye what you said is correct, i was able to doublecheck in the game with some damage benchmarks

maybe a weird wording then i guess

.close

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[amber meadow]

hi i wanna ask how to solve for max and min of x 😭

[distant peak]

the hardest way maybe, differentiation

that minus makes me look dumb

[amber meadow]

oh my god what’s differentiation

[distant peak]

ah you dont know

[amber meadow]

😭💔

[distant peak]

then forget it

for now

[amber meadow]

💔

[distant peak]

-1 <= -cos(x+20) <= 1

that means
-1 <= cos(x+20) <= 1

[amber meadow]

uhuh

[distant peak]

[amber meadow]

HELP

[distant peak]

thats what im doing

[amber meadow]

i’m so grateful actually

i’ve been sobbing over this for two hours 😭

[distant peak]

now here come the weirdest part

[amber meadow]

uhuh

[distant peak]

we only want x at the end

how to get rid of that cos ?

I dont want to say things that will sound weird to you so huh ill let you think

[amber meadow]

do i change it to sin

[distant peak]

You will make this even harder than its already is

[amber meadow]

UM

[distant peak]

we just want to get rid of cos to only have x+20

[amber meadow]

OMG

do i divide by -cos

[distant peak]

you just insult mathematics rn

[amber meadow]

😭😭😭😭😭😭😭

IM GOJNG TO FAIL TMR 💪💪

[distant peak]

but you have the idea

actually, its not a division you will do

but something called reciprocal function

[amber meadow]

what

what is tgat 💔

[distant peak]

yes you heard me

welcome to calculus

[amber meadow]

SOBBING

[distant peak]

So in simple words, thats something that will get rid of the cos

but it almost works like division

[amber meadow]

is it the graph with the fractions

[distant peak]

no

[amber meadow]

☹️

[distant peak]

reciprocal function is inverse of a function under the law of composition of function

[amber meadow]

uhuh

[distant peak]

yeah im saying weird shit

[amber meadow]

no i sort of get it

[distant peak]

I dont know how to dumb that for your level

[amber meadow]

gasp

ur right tho

but i still cant

figure it out

bc my teacher taught me to just add the numbers as i go

[distant peak]

so, there is a function you maybe have heard of

[amber meadow]

uhuh

[distant peak]

$\cos^{-1}$

[elfin berry]

Ｈｅｒｅｌｓ

[amber meadow]

OH MY GOF

🧍🏻‍♀️

[distant peak]

do you know that ?

[amber meadow]

yes💔

[distant peak]

nice, we will use this

[amber meadow]

uhuh

[distant peak]

$-1 \leq \cos(x+20) \leq 1$ right ?

[elfin berry]

Ｈｅｒｅｌｓ

[amber meadow]

uhuh

[distant peak]

yes or no, do you agree ?

[amber meadow]

yes i do

[distant peak]

do you understand why ?

[amber meadow]

yes

[distant peak]

ok

[amber meadow]

the graph thing 💪💪

[distant peak]

we will use our special function now

like this :

$\cos^{-1} (-1) \leq \cos^{-1} (\cos(x+20)) \leq \cos^{-1} (1)$

[elfin berry]

Ｈｅｒｅｌｓ

[amber meadow]

wait question

does the cos^-1 thing get rid of the -cos

[distant peak]

it get rid of cos, not -cos

[amber meadow]

oh so what’s left is -(x+20) (?

[distant peak]

not -(x+20) but x+20

i have already taken care of that -

[amber meadow]

OH

I GET IT

pls continue 🤲

[distant peak]

$\cos^{-1} (-1) \leq (x+20) \leq \cos^{-1} (1)$

[elfin berry]

Ｈｅｒｅｌｓ

[amber meadow]

uhuh

[distant peak]

you said it yourself, cos^-1 delete cos

[amber meadow]

wait

is this correct

[distant peak]

you forgot to write cos^{-1} (1)

[amber meadow]

oh oops

ok wrote it💪

[distant peak]

the rest is kinda wrong fr

[amber meadow]

🧍🏻‍♀️

💔

[distant peak]

maybe I forgot to tell you that cos^-1 is a decreasing function, so huh the <= should change to >=

[amber meadow]

um wait

so like this(?

[distant peak]

yes

$0 \leq x+20 \leq 180$

[elfin berry]

Ｈｅｒｅｌｓ

[amber meadow]

uhuh

wait um

is this still trig

bc the previous questions were like

[distant peak]

?

[amber meadow]

like idk

should the cos and sin still bc there

[distant peak]

Ahhhhh wait a sec

[amber meadow]

i’m so so sorry

[distant peak]

Maximum value of the function ? damn

i thought you were searching for the maximum and minimum values of x

[amber meadow]

i’m so sorry omg i should’ve mentioned it

[distant peak]

wellllll

[amber meadow]

but i’m so grateful i’m actually crying rn

[distant peak]

npnp

normally, the only thing to write is :
-1 <= -cos(x+20) <= 1

[amber meadow]

mhm

[distant peak]

and its done

[amber meadow]

what

[distant peak]

yes

[amber meadow]

why don’t i need to do anything to the -cos(x+20°)

[distant peak]

the question is asking for the max and min values of this function

[amber meadow]

mhm

[distant peak]

no matter what input cos have, it will always be between -1 and 1

[amber meadow]

my jaw is on the groudn

so the previous questions i sent you were different right

[amber meadow]

like this one

[distant peak]

wdym by different ?

[amber meadow]

bc the max and min are larger and smaller than 1 and -1 respectively

[distant peak]

sin(x) and cos(x) alone, without a coefficient or an additional term are always between -1 and 1

[amber meadow]

oh my god

so if i get another question like the -cos(x+20) one it will always be 1 and -1(?

[distant peak]

you should think, its not a formula

I could definitely ask you what are the max and min values of -50cos(x+20)+9

[amber meadow]

i think i get it

so i only need to take into the acc the -50 and the 9 right

[distant peak]

first thing to say,
-1 <= cos(x+20) <= 1
thats the least to know

[amber meadow]

uhuh

[distant peak]

the rest, you should know how to deal with inequalities and you are good

so you will have :
-50 <= 50cos(x+20) <= 50

-50 <= -50cos(x+20) <= 50

[amber meadow]

oh my god you are an actual saviour

[distant peak]

-41 <= -50cos(x+20) + 9 <= 59

[amber meadow]

i’m actually crying sobbing rn u are so 🙏🙏🙏😭😭😭

thank you so so so much

u have no idea how much this helped thank you sm😭

[distant peak]

no problemo 👌

[marsh citrus]

@amber meadow Has your question been resolved?

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[full comet]

The solution is x = 9, y=8
0 = (10 + (x) + (-y)) % 11
3 = (4 + x + y) % 9

How do I find this solution using arithmetic?

[narrow barn]

what's the work that you have for it?

[marsh citrus]

@full comet Has your question been resolved?

[full comet]

The work is that I tried every possible x,y combination

And 9,8 is the only solution

I would like to solve it using modulo arithmetic

If possible

But idk how

[narrow barn]

so you need to solve,
$$0=\frac{10+x-y}{11}$$ $$3=\frac{4+x+y}{9}$$

[elfin berry]

Mudkip

[narrow barn]

yes?

oh wait

[full comet]

No I don’t think so

[narrow barn]

iti's mod 11 and 9

[full comet]

Yea

[narrow barn]

I don't remember modular arithmetic sorry

[full comet]

The solution is x = 9, y=8
0 = (10 + (x) + (-y)) % 11
3 = (4 + x + y) % 9

How do I find this solution using arithmetic?
<@&286206848099549185>

[marsh citrus]

@full comet Has your question been resolved?

[full comet]

<@&286206848099549185>

[spark siren]

you know that there is not only one solution but there is an infinite number of solutions?

or are there some additionel criteria you didnt mention?

[full comet]

oh i forgot to mention that 0<=x<=9 and 0<=y<=9

sorry about that

[spark siren]

ok, then the solution is a little bit simpler.

lets start.

[full comet]

Someone said this. "CRT just reduces the system to x = -10(y+1) (mod 99)"

but im not sure what that means

[spark siren]

te first equation 10+x-y has remainder 0 by division by 11 means that x=y+1

[full comet]

You dont do (-x)?

i am very new to modulo arithmetic/modulo congruence

[spark siren]

if 10+a has remainder 0 by division by 11 then a has to have remainder 1 by division by 11.

a has to be 1, 12, 23, ....

[full comet]

oh i see now

[spark siren]

so x-y has to have remainder 1. as x and y < 10 so follows x=y+1

[full comet]

i see ok that makes sense

and i can do the same for the second equation

[spark siren]

you can use x = y +1 in the second equation.

[full comet]

ohh ok

3 = (4 + (y + 1) + y) % 9
then what?

[spark siren]

5+2y has to have remainder 3 by division by 9. -> so 2y has to have reaminder -2 which is equivalent to remainder 7, as y < 10, the only possible value for that is 16. so y = 8. therefore x = 9.

as 2y is even, 7 is no possible value for 2y.

[full comet]

ok this makes sense

thank you for the help

.close

[marsh citrus]

Channel closed

Closed by @full comet

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hidden citrus]

Hello, I need an explanation how to solve this.. It is about quadratic functions. (La° or Lo should be "L zero" and is for the start value of the size)

[hidden citrus]

Please sb help me

[marsh citrus]

@hidden citrus Has your question been resolved?

[hidden citrus]

No

[marsh citrus]

@hidden citrus Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[still temple]

I'm so confused

I get everything they have

except (7+0.5x)

not 7 + x

So for area of triangle

I get root(9-x^2)(1/2)(x)

(x)(root(9-x^2))

[frosty hatch]

Area of a trapezoid = $\frac{a + b}{2} h$ where a and b are the parallel sides' lengths and the h is the height

$h = \sqrt{9 - x^2}$

$A = \frac{14 + x + x}{2} \sqrt{9 - x^2} = (7 + x) \sqrt{9 - x^2}$

[elfin berry]

Oğuzhan

[still temple]

(14 - 2x)[root(9-x^2)]

[still temple]

ah gotcha

I'll use that method next time

but could you tell me what was wrong with my method

[frosty hatch]

If you want you can create individual triangles and prove the formula yourself

[still temple]

yeah

It's what I ended up doing

problem is I don't see where I went wrong

[frosty hatch]

it's 14 - x

and there is a 1/2

Because it's a triangle

[still temple]

oh yeah i got it

thank you sir

enjoy your day

c.lose