help-23

[sturdy swallow]

how

[lean otter]

Literally just picture a vehicle getting to the turn but not actually taking it and just continuing forward

[sturdy swallow]

thats 110?

[lean otter]

Yep

[sturdy swallow]

wait what b4 did i say was 110 lol

[lean otter]

The turn at that 290° angle

[sturdy swallow]

co interior angle with 70?

[lean otter]

whatever that means 👍

oh

I think I remember this term

[sturdy swallow]

is there extra resource for smth like this?

[quiet juniper]

I think this way is the easiest if you notice you can do that, but if you're very used to using like, corresponding angles, vertical angles, angles in a n-gon add up to 180(n - 2), that sorta thing then there's a way to do it with those theorems too

[sturdy swallow]

how do i do it lool?

we have to give reasons

thats the annoying part ay

[quiet juniper]

Uhhh hold on I'll draw a diagram

[sturdy swallow]

lol ty

[lean otter]

I doubt that

I mean it's by far one of the easiest methods I've had to explain

Just continue the line, draw the angle, calculate the angle, and add/remove it to the total angle

[sturdy swallow]

but how will that constitue to "alternate angles in a parellel line"?

[lean otter]

hmmm what

We've calculated a total angle of 125° up to the 3rd turn, and that 4th turn is 110°. What's the total angle?

what the fuck

[quiet juniper]

Oh mb

Thought he'd gotten the solution

I'll wait for that then

[lean otter]

He did, it was given

[quiet juniper]

Oh

[lean otter]

I was what the fucking your diagram

[quiet juniper]

lol

[sturdy swallow]

lmfao

[sturdy swallow]

wait bro can u put that up again?

[quiet juniper]

Ah wtf

[sturdy swallow]

lmfaoo

[quiet juniper]

So starring from the top you do vertical angles

Although

I could've skipped that step and just did alternating angles

To find the 45 degree angle in the triangle at the bottom

[sturdy swallow]

ohhhhhh shit

wtffffffffffff

[quiet juniper]

120,70, 110 come from supplementary angles and 290 + 70 = 360

[sturdy swallow]

yea yea yea sheesh ok

[quiet juniper]

60 comes from quadrilateral angles adding up to 360

at that point you could do exterior angles of a triangle to find a but I found the supplementary angle first

[sturdy swallow]

bro i got everything except the answer lol

howd u get 15

[quiet juniper]

45 + 15 = 60

Or 45 + 120 + 15 = 180

[sturdy swallow]

ohhhhhhhhhhhhhhhhhhhh

[lean otter]

@quiet juniper You tripping cuh

[quiet juniper]

[sturdy swallow]

thats right i reckon?

[quiet juniper]

Yeah

[lean otter]

@sturdy swallow C'mon now don't quit at the very end

[sturdy swallow]

tysm boys preciate youse❤️

what u delete lmfao

[lean otter]

The bot deleted my message because I mentionned eye cue if you see what I mean

[sturdy swallow]

ag 🫡

[lean otter]

"I know you got the eye cue to do addition and subtraction"

[sturdy swallow]

i got to write the reasoning

so i have alternate angles

bro i spent so long i got to get ready lmfao

thx tho preciate it

love lads ❤️

.close

[safe radish]

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[visual moss]

hi can someone explain this to me please?

[quasi bison]

how much do you know about square roots in general

[visual moss]

multiplication and division

we havent done that much surds tbh

[quasi bison]

so then you do know that sqrt(a) * sqrt(b) = sqrt(ab), and the same for division?

[visual moss]

so sqrt 140 / sqrt 2 is the answer?

figured it out thanks

.close

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[copper sparrow]

hi

[copper sparrow]

so i'm doing LU factorization and i understand how to derive L and U and everything but (a) is confusing me

because if what the question is saying is true, shouldnt X be an upper triangular matrix ?

if R_j + aR_i -> R_j

[safe radish]

@copper sparrow Has your question been resolved?

[copper sparrow]

<@&286206848099549185>

[safe radish]

@copper sparrow Has your question been resolved?

[peak estuary]

a single row operation doesnt turn a matrix into an upper triangular matrix

[safe radish]

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Channel closed

Closed due to the original message being deleted

[crystal mist]

$\mathbb{F}_{\text{un}}$

[flat frigate]

sakka ismail

$\mathbb{F}_{\text{un}}$

[lean otter]

No

[lean otter]

Asalamualaikum

[crystal mist]

Wa alaykom asalam

[lean otter]

Wanna give a brother a hand

[crystal mist]

Yes

Can you help me with this first exercise

[lean otter]

No

[gray flume]

@crystal mist is this a practice test or an actual one?

How time sensitive is this?

[crystal mist]

It is an actual test

[edgy cloud]

It's Drey's channel, is what I think the point was.

[crystal mist]

From 2006

[gray flume]

Okay so banned

Or wait

No you confused me

Anyway sorry drey I was trying to bait this guy 😛

@crystal mist don't ask in other people's channels

[crystal mist]

I don't understand how this server works

[crystal mist]

Whre i ask?

[gray flume]

Read #❓how-to-get-help

[edgy cloud]

(It was Drey's channel before he deleted his initial post, I should say).

[lean otter]

np

i'll forgive u if u help

[gray flume]

I figure as much, I deleted drey's modping so that if this guy was asking for help on a test he'd feel free to say it so I'd knock him

[lean otter]

:)

[gray flume]

Lmfaooooo

Alright (also salaam to you too Drey)

[lean otter]

:0

aleykum selam

[edgy cloud]

Drey, you'll want to open another channel; this one will auto-close in a few minutes.

[gray flume]

I can help somewhat quickly, the idea is to use the chain rule

[lean otter]

i have my question pinned

[gray flume]

You have z(x(u,v,w), y(u,v,w))

[safe radish]

Available help channel!

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Remember:
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[lean otter]

noo

[gray flume]

Lmfao that's what got pinned

[lean otter]

XD

[gray flume]

Anyway do you agree with this?

[lean otter]

yes

[gray flume]

So we wanna compute derivative of z with respect to v

We could expand it out if we were nerds

But we can do something better can't we

[lean otter]

shortvut

i forgot it

[gray flume]

Chain rule

[lean otter]

oh ive been doing thast

[gray flume]

So what does the chain rule say in this setting?

[lean otter]

on the right side

[gray flume]

Well let's do it slowly

[lean otter]

u are a sloth afterall

[gray flume]

truuuuuuuuu

[gray flume]

Wow okay 😦

[flat frigate]

Sloth King Daminark

[gray flume]

(Right? I'm not just asking rhetorically here my multi is barely existent)

[lean otter]

mhm

[gray flume]

So dz/dx is what?

[lean otter]

4x^3_y^2

+*

dx/dv=v^4

dz/dy=2xy

dy/dv=1

[gray flume]

Careful

dz/dx = 3x^2 + y^4

[lean otter]

hmm

[lean otter]

oh shoot wrong one

i meant this s

[gray flume]

Oh

[lean otter]

ss

mb

[gray flume]

So dz/dx = 4x^3 + y^2

dz/dy = 2xy as you say

dx/dv = 4uv^3

dy/dv = e^w

You see why?

[lean otter]

yes

[gray flume]

Okay so

When u=3, v=2, and w=0

What are x and y?

[lean otter]

3times 16 so uhh 48

then y is 5

[gray flume]

Alright so now let's plug stuff in

dz/dx = 3x^2 + y^4 = 3(48)^2 + 5^4

And so on

I'll have you take the reigns for a sec

[lean otter]

7537

480

96

1

[gray flume]

So plug the stuff in, does it work?

[lean otter]

that's such a large number

hmm

[gray flume]

(I'm not quite checking your arithmetic here)

[lean otter]

oof

[gray flume]

Though if you have limited tries

I will

[lean otter]

it is limited

😰

i used up most

[gray flume]

Okay

[lean otter]

i get 724032

looks so wrong

[gray flume]

Ah I was looking at old stuff.
So dz/dx = 4x^3 + y^2 = 4(48^3) + 5^2 = 442393
dz/dy = 2xy = (2)(48)(5) = 480
dx/dv = 4(3)(8) = 96
dy/dv = e^w = 1

So we want 442393(480) + 96 = 212,348,736

[lean otter]

WHATTT

[gray flume]

Yeah I was looking at an old message when I calculated dz/dx lol

[lean otter]

OMG

i HAVE one last try

[gray flume]

Shit

Oh oh

[lean otter]

[gray flume]

Wait hmm, dx/dv = 4uv^3 right, so that's 4(3)(2^3) = 96

That was right

This is weird

Let me see if someone who's better at multivariable calculus can see what happened

[lean otter]

alr

[white umbra]

lemme see

[lean otter]

my iranian savior

bry my teachers iranian too

yall smart as hell

[white umbra]

Huh. I got the same thing as Dami.

212 348 736 should be right

[lean otter]

WHATT

get somebody else

[white umbra]

It might be a bug with the software

Unless you somehow entered the number wrong, but it looks right to me

[gray flume]

Maybe I'm secretly Iranian too

Nah but jokes aside this is strange

[clear blade]

I got 42470208

dz/dv = (4x^3 + y^2)(4uv^3) + (2xy)(e^w)

?

,calc (4(48)^3 + (5)^2)(4(3)(2)^3) + (2(48)(5))(e^0)

[flat frigate]

Result:

4.2470208e+7

[clear blade]

@lean otter

[lean otter]

TUSHARRR

[lean otter]

Would y bet ur life on this answer

I used up all my chances

Unless I refresh the page I can get more but do u wanna redo it

[clear blade]

listen to me at your own risk

[lean otter]

Lool

Okay I’ll get more chances

That looks very wrong comepared to the old answers the software gave me

[clear blade]

nah, I'm 99.9% sure

[clear blade]

rip, there goes all of our work

[lean otter]

Sorry

I didn’t wanna risk losing my chances

But now Atleast I have more

[white umbra]

oh I'm dumb, I multiplied the wrong numbers together in the calculator

@gray flume you multiplied together dz/dx * dz/dy instead of dz/dx * dx/dv

lol

[white umbra]

So yeah, it's the same exact formula, just plug in your values of u, v, w

what are x and y equal to now?

[gray flume]

Ahhhhh

Wait is it the same formula? It's xy^3 instead of xy^2

[clear blade]

also w^5 -> w^3

don't think that changes anything

[white umbra]

oh yeah it's slightly different

[gray flume]

Well we're also taking d/dw now

[white umbra]

but same technique of dz/dx dx/dw + dz/dy dy/dw

@lean otter have you calculated what x and y are equal to?

[clear blade]

true

[lean otter]

mhm

[white umbra]

What are they equal to?

[lean otter]

-162

for x

y=-5

iget 724032

idk

[white umbra]

lemme check that answer

hmmm I'm getting a different answer, can you write down your steps for 724032?

[white umbra]

your values for x and y are correct btw

[lean otter]

What did u get

[white umbra]

Well I wanna lead you to get to the answer yourself, I'm just trying to figure out which step you may have made a calculation error

[lean otter]

My camera roll is full :(

[white umbra]

😭

okay well let's start with, did you use the formula dz/dx dx/dw + dz/dy dy/dw?

what value did you get for dx/dw and dy/dw?

[lean otter]

,calc (4(-162)^3 + (-5)^3)(-3)^4) + (3(-162)(-5)^2)(e^0)

[flat frigate]

The following error occured while calculating:
Error: Unexpected operator ) (char 27)

[lean otter]

,calc (4(-162)^3) + (-5)^3)(-3)^4) + (3(-162)(-5)^2)(e^0)

[flat frigate]

The following error occured while calculating:
Error: Unexpected operator ) (char 21)

[white umbra]

Hm where did you get the (-3)^4 from?

That might be tripping you up

[lean otter]

The v

[white umbra]

But you're trying to find dx/dw, right?

[lean otter]

Dz/dw

[white umbra]

Right, I mean, is (-3)^4 meant to be dx/dw?

[lean otter]

Yup

No

[white umbra]

I think you mighta calculated dx/dw wrong then

Oh

[lean otter]

Dx/do

Do

Dv*

[white umbra]

why are you taking dx/dv?

[lean otter]

But probably needs to be dw

[white umbra]

yeah

[lean otter]

Rip

[white umbra]

it should be (dz/dx)(dx/dw) + (dz/dy)(dy/dw)

[lean otter]

So 3w^2

[white umbra]

yea, and since w=0, what is that?

[lean otter]

0

[white umbra]

👍

so the first term cancels out

and then what's dy/dw?

[lean otter]

Great

Ve^w

[white umbra]

Exactly!

and what number is that?

[lean otter]

-3

[white umbra]

perfect

so now

you already said that dz/dy was 3(-162)(-5)^2

,calc 3(-162)(-5)^2 * (-3)

[flat frigate]

Result:

36450

[white umbra]

I think that should be your answer ^

[lean otter]

:DDDDDDDD

Now this whole shabang (old puc)

Pic*

But I will try it first

Wajt

What does the w come from

[white umbra]

w is the name of the function

[lean otter]

Oh yeah

[white umbra]

So w depends on x,y,z, but each of x,y,z depends on r and t

if that makes sense

[lean otter]

So for the dw/dx part do you just ignore the radical

And then make it 2x

[white umbra]

No, you have to account for the radical

Try using the chain rule

[lean otter]

What is the chain rule for this

Uhh I can’t find dw/dx

Is it (2x+y^2+z^2)^1/2

Nvm it’s x/root x^2+y^2+z^2

elp

[white umbra]

yea

okay lemme see what you put in

what did you get for dx/dr?

[lean otter]

7te^r

[white umbra]

I think that's dy/dr?

I think you're missing a negative sign for dx/dr

and you're missing dz/dr

[lean otter]

Ohh

[lean otter]

E^rt

[white umbra]

not quite

remember to use the chain rule

[lean otter]

Idk

Te^rt

[white umbra]

yup

[lean otter]

i just get the same

\frac{-5e^tx+7te^ry}{\sqrt{x^2+y^2+z^2}}

\frac{-5re^tx+te^{rt}y}{\sqrt{x^2+y^2+z^2}}

for dw/dr and dw/dt

$$\frac{-5re^tx+te^{rt}y}{\sqrt{x^2+y^2+z^2}}$$

[flat frigate]

Drey 🔍

[lean otter]

$$\frac{-5e^tx+7te^ry}{\sqrt{x^2+y^2+z^2}}$$

[flat frigate]

Drey 🔍

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

come back

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Hellp

Hello

But also help

[lean otter]

.close

[safe radish]

Channel closed

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[royal arch]

im wondering if i can get some help on this

i tried differentiating both sides but i just got sec^2(theta) = (1/x)(dh/dt)

[safe radish]

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[royal arch]

im wondering if i can get some help on this
i tried differentiating both sides but i just got sec^2(theta) = (1/x)(dh/dt)

[safe radish]

@royal arch Has your question been resolved?

[royal arch]

!help

[safe radish]

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

[royal arch]

<@&286206848099549185>

im so confused where is my error

thats not even one of the solutions

<@&286206848099549185>

my work is above

idk wtf i did wrong

Nvm i got it

.close

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[dull sequoia]

can anyone think of a way to do this

[dull sequoia]

without just degenerately splitting the left hand side to look like the right hand side

like currently i have $e^{i\frac{\alpha-\beta}{2}}-e^{i\frac{\beta-\alpha}{2}}=2i\sin\left(\frac{\alpha-\beta}{2}\right)$

[flat frigate]

Frosst

[dull sequoia]

but that's so degenerate

[light shoal]

try factoring something out of e^(i alpha) - e^(i beta) to get a factor of that form

[dull sequoia]

i mean that's what this is doing

you just force things out of the power until you get what you want

[light shoal]

yes

what are your concerns with that

[dull sequoia]

is there not a better method

[light shoal]

better in what way

this is only like two lines

[dull sequoia]

maybe im not factoring the right thing

or not seeing the biggest thing to factor

lol it's so dumb

[light shoal]

take what you currently have, and multiply both sides by e^(i(a+b)/2)

the right hand side will then match the problem statement

simplify the left hand side until it does, too

[dull sequoia]

,rotate

[dull sequoia]

this is so degenerate imo

[light shoal]

hint: what's (a+b)/2 plus (a-b)/2 ?

[dull sequoia]

yeah but this is what i mean

you're splitting a into (a+b+a-b)/2

[light shoal]

you don't want to be factoring the exponent, factor the whole exponential

e^x e^y = e^(x+y)

[dull sequoia]

oh wrong way

i want to end with the (a-b)/2 factor

i mean i just end up starting at $e^{i\alpha}-e^{i\beta}$

[flat frigate]

Frosst

[dull sequoia]

then writing down $k\left(e^{i(\alpha-\beta)/2}-e^{i(\beta-\alpha)/2}\right)$

[flat frigate]

Frosst

[dull sequoia]

then trying to figure how what k is

but maybe there just isn't a better way

it feels really cheap to just go "oh i know it looks like that so im just gonna force it"

[light shoal]

it may feel like a cheap trick the first couple of times you see it, but it's common enough that it becomes a technique

you see e^(ia) - e^(ib) and say, what would it take so that instead of a and b as exponents, i had two numbers that are symmetric around zero, so that becomes a sine

well the difference between a and b is a-b, so i would need (a-b)/2 and -(a-b)/2 as exponents

then work out what you need to factor out to get those

[dull sequoia]

mhmm i see

ok thanks guys

.close

[safe radish]

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[split bear]

i got y'' as (-2y+2x)/4y^2

but it marked it as wrong

[royal ferry]

whats your first derivative?

[split bear]

wait i figured it out

.close

[safe radish]

Channel closed

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[alpine bough]

How does number 3 make sense?

I can't seem to wrap my head around how the addition of friction would result in the mass reaching a greater height because of an increase in the angle. I would imagine two things that lead to a decrease in height would result in a greater decrease in height.

[quiet juniper]

Friction force is proportional to normal force so...

[safe radish]

@alpine bough Has your question been resolved?

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[safe radish]

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Send your question here to claim the channel.

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[gaunt forge]

Hey everyone. why the mid-interval of this:
98 < V ≤ 104
would be 101 and not 100.5?
Is it because the symbol before V is < and not ≤???

[fallow cairn]

Assuming V is an integer, yes, the equality sign is causing that

[gaunt forge]

ok thanks!!!

[fallow cairn]

Or actually

[gaunt forge]

??

?

[fallow cairn]

It's if V is not strictly an integer that this is true

[gaunt forge]

ok

got it

?

[fallow cairn]

If we count them out
\begin{center}
99, 100, 101, 102, 103, 104
\end{center}
If V is an integer, the mid value is 101.5

But if V may be a non-integer, we instead find the mid-interval with $\frac{104+98}2 = 101$

[flat frigate]

@fallow cairn

[gaunt forge]

Alright

Thanks

so much

!

.close

[safe radish]

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[empty yew]

Did I do this correct?

[thin bridge]

no

can you show your work

[empty yew]

https://www.youtube.com/watch?v=IBKk7eHove8&t=337s&ab_channel=MrBdubsMathandPhysics

i kinda follow this guy

[thin bridge]

do you have a clearer image

don't know what you're doing at the very start

[thin bridge]

where's sin^2(t) = 8.9 coming from

[quasi bison]

^

[empty yew]

Oh don't mind that I forgot to erase that part

[quasi bison]

ok but then $\sin(\theta) = 19.21 (1 - \sin^2(\theta))$... where is \textbf{that} coming from?

[flat frigate]

Ann

[thin bridge]

thing that's supposed to be 79.21

[empty yew]

yeah

[thin bridge]

but still

[empty yew]

sry about my writing

[thin bridge]

ok but then $\sin(\theta) = 79.21 (1 - \sin^2(\theta))$... where is \textbf{that} coming from?

[flat frigate]

ℝam()n()v

[quasi bison]

how about you just redo this all from scratch, and make an effort to (a) write more neatly and (b) actually write things that make sense

[empty yew]

K

[safe radish]

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[safe radish]

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[tropic flax]

its implicit differentiation but i have no clue how to find the equation of the tangent line to the elipse that actually hits (4,0)

equating the slope (m) that you get when you use points-slope formula and the slope (m) or rather, y' when you implicitly differentiate gives me another equation with variables x and y which i dont know how to solve for

[delicate bobcat]

Can you get y'?

Or dy/dx

As in, have you successfully solved for that already

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[frigid spruce]

A(1,3) and C(7,5) are vetices of square ABCD. Find the equaltions of AB and AD.

[frigid spruce]

isnt there any other way rather than making 4 equations and solving for vertices of B and D

[peak estuary]

which 4 equations do you want to make

[glass carbon]

you can also use distance formula, because if you have the AC you also have side of the square

(to find coords of B or D, then equation through two points)

[frigid spruce]

i got the point of intersection of digonals

will that help

[glass carbon]

is it (4,4)?

[frigid spruce]

yes

its square

so we have many relations

[glass carbon]

Idk if it helps

[frigid spruce]

we can find the equation of line of digonal BD

howw?

[peak estuary]

nvm

[frigid spruce]

Denascite

[safe radish]

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[safe radish]

@frigid spruce Has your question been resolved?

[frigid spruce]

.close

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[safe radish]

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[charred atlas]

!status

[safe radish]

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

[charred atlas]

substitude

if $g(x)=x^3-2x^2-3x+1$

[flat frigate]

TimK

[charred atlas]

then $g(-2) = (-2)^3-2(-2)^2-3(-2)+1$

[flat frigate]

TimK

[charred atlas]

formulae?

[thin bridge]

functions

[cosmic grove]

how are you doing an exercise of something you don't know what it is

[safe radish]

@visual oar Has your question been resolved?

[safe radish]

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[lucid echo]

how would i answer this stats question: "Write a data set that has 10 entires, a mean of 10, and a population standard deviation that is approximately 3. (There are many correct answers)"

[lucid echo]

i get how to find population std. dev for a data set but i dont know how to do it backwards for this question

[quasi bison]

try satisfying the mean and count first

find stdev then adjust accordingly

[lucid echo]

so it's just trial and error?

is there no faster way

also how would i be able to guess numbers reasonably... just based off of feeling?

[peak estuary]

well you have to develop a feeling somehow

trial and error works for that

[lucid echo]

alr im trying it rn

ok like

the problem im facing rn is that when i match the mean to 10, the std. dev is like 2.7

wait is 2.7 std. dev close enough to 3 to answer the question

!closed

!answered

!close

.close.

.close

[safe radish]

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[lean otter]

I have a question concerning these two investment scenarios. In question 1 they find the time to get to $15,000 by finding N and dividing it by the compounding period of 4.

But in question 2 they don't divide the number from N by the compounding period and take it as is(5.04 years)

why did they divide by compounding period in 1 but they didn't in 2?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[kind fractal]

Hello, the questions are different. In q1, they want the time converted from quarters to years. In q2, they want it from years to quarters

[lean otter]

i got that part

but why didn't they divide the N in question 2 by 4

they got 5.04 for N

but isn't N = compounding period * years

so you have to divide by the compounding period in order to get the number of years

but what they did is they just took the number as is to get 5.04 years

[kind fractal]

N = (compounding periods per year) x years

[lean otter]

yes

[kind fractal]

= C/Y x years

In q2, C/Y = 1

As it's compounded at the end of each uear

[lean otter]

no?

in q2 c/y = 4

p/y is 1

[kind fractal]

Yeah you're right, mb

[lean otter]

so why didn't they divide by 4?

[kind fractal]

It looks like the units of N depend on P/Y

In q1, P/Y = 4, so N comes from the calc in quarters

In q2, P/Y=1, so N is given in years

Another possible option is that they've omitted a part of the solution

[lean otter]

how do you mean

[kind fractal]

I mean, 1.b and 2.b are supposed to be the exact result that comes out of the calc, right?

[lean otter]

oh wait I think it is the former

N is the total number of payments $$=payments per year×number of years. P/Y is number of payments per year. C/Y is the number of times interest is calculated per year. See below to work through an example involving a reduce balance loan with differing payment and compound periods with the TI Nspire calculator. TI ...

but how does this make sense?

what if you don't put in any payments and let it increase with just the compound interest

then wont p/y be 0

but in this case(question 1) they set p/y = c/y

[kind fractal]

Maybe when PMT = 0, you can set P/Y to any number you want I guess

It depends on what units you want for N

[lean otter]

but in question 1 when they made no payment they set it to 4

which was the compounding period

[kind fractal]

Yeah but they could've chosen any number I think

It's probably usual to set P/Y = C/Y when PMT=0

[lean otter]

hmm lemme just test that real quick

[kind fractal]

Focus on q1, PMT=0
They choose P/Y=4, so N is given in quarters
If you try P/Y=2, you get N in semesters

Yeah please try that

In this last case, you should get N = 20.25 semesters (?

[lean otter]

if I set c/y =4 but p/y=2 I get N=20.25

[kind fractal]

Ok great

[lean otter]

what does this mean though

[kind fractal]

Then P/Y defines the units of N, and P/Y must always be a positive integer

[lean otter]

is this correct

[kind fractal]

When PMT is not 0, then P/Y is defined by the problem's statement

[lean otter]

but if PMT is 0?

[kind fractal]

If it's 0, then you can choose it depending on what you want

It's probably usual to set it equal to C/Y

[lean otter]

but when I chose 2 it gave 20 quarters right?

as opposed to the 41?

so it gave a different ans

[kind fractal]

No, it was 20.25 semesters

P/Y sets the units of N

[kind fractal]

It's confusing I know, that's why it's easier if you just set P/Y=C/Y

when PMT=0

[lean otter]

wait so if I set PMT=1 it should give N in years?

[kind fractal]

Yess

That's why in q2 N is in years

[lean otter]

ohhh

this application a bit confusing

[kind fractal]

🤯

Yes it's always been this way

[lean otter]

🤣

[kind fractal]

I haven't used that in a lot of time

It's so counter intuitive that it's easy to forget haha

[lean otter]

mhm yeh loo

ight thanks

.close

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[chrome parcel]

i dont understand the following word problem: "The sum of two numbers is 6, and the sum of their squares is 28. Find these numbers exactly."

[chrome parcel]

i mean, i got

$x + y = 6$

[flat frigate]

GMDennis

[chrome parcel]

and $x^2 + y^2 = 28$

[flat frigate]

GMDennis

[delicate bobcat]

Ok great

You have a system of equations now

[chrome parcel]

yesssir

[delicate bobcat]

Both must be true

[chrome parcel]

right

our teacher told me something about making y = 6x

and then substituting for the second equation

[delicate bobcat]

Basically

[chrome parcel]

sorry, $y = 6 - x$

[flat frigate]

GMDennis

[delicate bobcat]

Solve for one of the variables

Either x or y

[chrome parcel]

yeah ik but it's not that easy

what i got was

[delicate bobcat]

In the first equation its easy tho right

y = x - 6

Thats a oneline solving for y

[chrome parcel]

$x^2 + (6 - x)^2 = 28$

[delicate bobcat]

Yeppp

[flat frigate]

GMDennis

[chrome parcel]

is that right?

[delicate bobcat]

Yess

Thats exactly right

[chrome parcel]

right

[delicate bobcat]

Now you can expand the second term

[chrome parcel]

so now do i just FOLD everything out and add the x^2 and subtract the 28 to get a = 0?

[delicate bobcat]

Yes although you may have stuff cancel out

It may not end up as a quadratic

[chrome parcel]

ok stop

let me right it out

$x^2 + 36 - 6x - 6x + x^2 - 28 = 0$

[flat frigate]

GMDennis

[delicate bobcat]

Ok good

[chrome parcel]

we add like terms, etc

to get

$2x^2 -12x + 8 = 0$

[flat frigate]

GMDennis

[delicate bobcat]

Looks fine so far

Now you have a quadratic equation you can solve in 1 variable

[chrome parcel]

wait a sec

i have to use quadratic equation?

why not just create the binomials

like (x - )(x - )

[delicate bobcat]

The equation is of the class quadratic equations

There is a quadratic formula

[chrome parcel]

fuck meeeee

ok fine

so

how do i do the + _ symbol?

[delicate bobcat]

In here?

Idk lol

[chrome parcel]

ye

fuck

lemme google it

[delicate bobcat]

+/- i would write

[chrome parcel]

i think i got it

±

LETS GO

ok

$-b ± √b^2 - 4(a)(c)/2(a)$

[flat frigate]

GMDennis

[chrome parcel]

ok well that didn't work out

anyway

you get it

right

[delicate bobcat]

Yea

[chrome parcel]

ok

lemme plug the shit in

working on it

what would be a radical for the square root of 80

2√ 20?

bcuz 40 * 2, then 20 * 2 * 2

so 2√ 20

YES

I GOT IT

AND GOT IT RIGHT COMPARED TO THE BOOK ANSWER

LETS FUCKING GOOO

the answer is 3 ± √ 5

.close

[safe radish]

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[chrome parcel]

.reopen

[safe radish]

✅

[chrome parcel]

.close

[safe radish]

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[chrome parcel]

.reopen

.close

[safe radish]

✅

[chrome parcel]

.reopen

[safe radish]

Channel closed

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✅

[chrome parcel]

.close

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[sudden sparrow]

can someone help me with this

[safe radish]

@sudden sparrow Has your question been resolved?

[celest ember]

Familiar with the formula
area =1/2 absin(c)?

[sudden sparrow]

yes

[twilit flame]

Rearrange it

[safe radish]

@sudden sparrow Has your question been resolved?

[safe radish]

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[misty arch]

i am having trouble with part 2

[misty arch]

This is what I have so far

[safe radish]

@misty arch Has your question been resolved?

[misty arch]

<@&286206848099549185>

[vagrant jolt]

Alrighty

Do you understand what it means by rate of change?

[misty arch]

it means i should take the derivative right?

[vagrant jolt]

Wait a sec I just saw this

Did you get the answer wrong?

[misty arch]

yea, the answer is -0.003

[misty arch]

any idea on how i get -0.003?

[vagrant jolt]

Ah try implicit differentiation

Go from tan(theta)=261/x

@misty arch

[misty arch]

ok, let me try that

[misty arch]

Something like this?

ah the cos should be cos(theta)

[vagrant jolt]

Almost there

[vagrant jolt]

Yes and you forgot to differentiate the 261/x

[misty arch]

to get the derivative of 261/x i use the quotient rule right?

[vagrant jolt]

You could

Or you could bring the x up

So it becomes x^-1

And then differentiate normally

[misty arch]

yea i got to the part where i moved sec^2(theta) over, what do i do from there?

@vagrant jolt

[vagrant jolt]

We know theta from the formula from part a of the question

And we are told x is 128

Be careful here because you can’t just divide by 1/cos(x) in a calculator

You either turn it into multiply or you find what cos(x) and then do 1/value

[misty arch]

This should work to right?

From here do I just put it in a calculator?

[vagrant jolt]

Does your calculator have a sec^2 function?

If so then yeah

I would rather convert it into cos and do it separately to make it less compact

[vagrant jolt]

By this I meant this

[misty arch]

hmmm, im getting -0.03

[vagrant jolt]

You should get -3.088574641x10^-3

Which is roughly -0.003

[misty arch]

here let me input in in again

[vagrant jolt]

Send a pic of what you’re typing in calc if you’re still getting smth else

[misty arch]

ah, i left the one on top when multiplying it

[misty arch]

yup got -0.003, thanks!

[vagrant jolt]

All good 🫡

Don’t forget to type .close

[misty arch]

.close

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[civic bramble]

Hello, I'm trying to find the limit of the pigeon hole principle and I really need your insight on this. (

Scenario is: let there exist 367 people. A year can be 365 days or 366 days. Then there are at least 2 people with the same birthday. Let set A contains people and B contain days. So surjectivity between B and A are established.

My hypothesis, let there be 3 people with the same birthday. Then there exist, although slim, for a day that has no people to be born in it. Thus, surjectivity between B and A break.

My problem being trying to generalize this. Let n +x, m where n+x and m belong to N. All results of n + x are elements in set A. All objects m are in set B. At what point does the surjectivity does not hold?

This can be achieve if I take |A| - |B| such that this result in a negative x. But I feel like i did something wrong. I tried to ask my professor and he said my understanding of the principle is wrong. Any help is appreciated

[safe radish]

@civic bramble Has your question been resolved?

[civic bramble]

!status

[safe radish]

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

[civic bramble]

3 or 4

[safe radish]

@civic bramble Has your question been resolved?

[civic bramble]

<@&286206848099549185>

please

[marsh walrus]

I'm not sure I understand the last part of the question

This can be achieve if I take |A| - |B| such that this result in a negative x. But I feel like i did something wrong. I tried to ask my professor and he said my understanding of the principle is wrong. Any help is appreciated```

this

can you explain more?

[safe radish]

@civic bramble Has your question been resolved?

[civic bramble]

So, I'm finding that if n + x - m = x. Somehow if |A - x| then the surjectivity between set B and |A-x| no longer hold if that make sense even though |B| to |A| was previously surjectived.

Extend this into limit as N is not bound above. So |B| never has any surjectivity to |A| for all size

If that make sense.

[civic bramble]

[safe radish]

@civic bramble Has your question been resolved?

[civic bramble]

Pleased?

[marsh walrus]

@civic bramble what is x?

what do you mean if |A-x| then ...

it doesnt even seem like youve written a valid implication here

you may need to explain more broadly the bigger picture thing youre trying to accomplish

and stop posting sad emotes

[safe radish]

@civic bramble Has your question been resolved?

[safe radish]

@civic bramble Has your question been resolved?

[civic bramble]

.close

[safe radish]

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[lean otter]

Hey, how can I evaluate cos(2) in degrees using a triangle ?

[lean otter]

<@&286206848099549185>

[native ledge]

!15min

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lean otter]

Ok but 15 mins have passed

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[surreal fable]

how is the square root of 4 times 2 equal to 2 times the square root of 2?

[wise junco]

sqrt(4 * 2) = sqrt(4) * sqrt(2) = 2 * sqrt(2)

[surreal fable]

Isnt that 4? -> 2 * sqrt(2) -> sqrt(4)

[wise junco]

no? how

n * sqrt(m) = sqrt(n * n * m)

[surreal fable]

interesting.

[wise junco]

i mean put m = 1, and its fairly obvious property

n = sqrt(n^2)

[surreal fable]

I just don't know why, I'd expect 2 * sqrt(2) to be sqrt(4).

[wise junco]

2 * sqrt(1) isn't = sqrt(2) :D

[surreal fable]

hmm

what is that then?

[wise junco]

well 2 * sqrt(1) is 2

which is sqrt(4)

[surreal fable]

huh

got it

so for example here:
4 is a perfect square so I can evaluate it, and then just keep the square root of 2.

[safe radish]

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[dim moon]

Can somebody help with this and show the working out?

[little mesa]

!status

[safe radish]

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

[dim moon]

2. I know the answer to question a) but have no idea how to calculate or what is B)

[finite dew]

ok so how many grams is one mol of methane

1.008x4 + 12.01

[dim moon]

that would be 16.042

[finite dew]

so how many atoms in one mol

[dim moon]

4 atoms of hydrogen and 1 of carbon

[finite dew]

wait hold on

yea i think this is the method i could be wrong

[finite dew]

no how many molecules are there in one mole of methane

avogadros number right

have you learnt this?

6.022x10^23 atoms in one mole

ahhh so sorry i have to go, maybe someone else can explain it

gllll

[little mesa]

[dim moon]

thanks for your help anyway

[little mesa]

what'd you get for a

[dim moon]

2.658x10^-23

[little mesa]

so if one molecule is 2.658*10^-23, how many molecules in a gram...

think of a simpler example, if you have a molecule of 0.1g, then there's 10 molecules in a gram, because 1/0.1=10

[dim moon]

oh alr

[little mesa]

does that make sense or nah

[dim moon]

yeah that does

[safe radish]

@dim moon Has your question been resolved?

[safe radish]

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[lean otter]

I need help understanding this

[dim moon]

you have mathswatch haha enjoy it when you can before ur teacher switches to sparx

this channels being used

[dim moon]

Ill try and help ya out

[lean otter]

thank you

[dim moon]

W= 7,3
T=6,2
U=4,3
V=6'4

Think of it as a mirror

@lean otter is it correct?

[lean otter]

lemme check

its wrong

lemme send a pic

[dim moon]

Oh wait

[dim moon]

I think i know where i went wrong

[lean otter]

U forgot about the y=2 I believe

[dim moon]

my bad bro im sorry

[lean otter]

Nah u good

i got to my percentage cuz it switched up my questions

so i got a 100 now

so its all good

that wont even be on the test

[dim moon]

so would you say this is all done?

[lean otter]

Ig so

Thank you for your attempt

[dim moon]

.close

@lean otter say .close

[lean otter]

.close

[safe radish]

Channel closed

Closed by @dreamy bridge

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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.

[sour mica]

could someone explain why the solution set is empty in this case

[jolly shard]

If you convert these equations into slope intercept form, what do you get?

[sour mica]

uhh

2x^2 + -14y^2 = 0^2?

no

i dont remember slope intercept form

[jolly shard]

No, simpler than that

Y=mx+b

[sour mica]

-14y = -2x + 0

[jolly shard]

And divide so you get y alone for both equations

[sour mica]

y = 1/7x + 0/14 or error?

[jolly shard]

0 into 14 groups is zero in each, so 0

[sour mica]

ahh okay

so just

y = 1/7x

[jolly shard]

And

For the other

[sour mica]

63y = 9x + 16
y = 9/63 + 16/63

[jolly shard]

Which simplifies

[sour mica]

1/9x + 16/63?

[jolly shard]

Try the division again

1/7x

[sour mica]

ahh yes i mistook the result for how much u divide it by

[jolly shard]

Yeah, so if you look at the graph of these equations, when will they cross?

[sour mica]

i dont know how to look at the graph of them but i assume given theyre the same thing but different Bs they never do

[jolly shard]

Exactly

[sour mica]

wait doesnt this mean u can using ur graphing calculator to find the answer to a system of linear equations without having to do it by hand?

[jolly shard]

So if you know the solution to the system of equations is when they cross, what is the solution to a system that never crosses?

[sour mica]

nothing

[jolly shard]

You can also just look at the slope, if they’re the same they either never cross or always ‘cross’

And if the slope of two lines is different they will cross exactly once

[sour mica]

im more asking cause im somehow notirously bad at doing these by hand

just

the worst

i dont understand how i cant figure it out

usually the ones with 3 systems

[jolly shard]

3 equations and/or 3 variables?

[sour mica]

yah

[jolly shard]

Well trying to visualize that gets trickier than 2 variables, but it’s the same idea

[sour mica]

yah... i just

suck at them

[jolly shard]

Do you have one of those?

[sour mica]

can try and find one

[jolly shard]

Okay so how do you think you’d try to solve this?

There’s a few different ways

[sour mica]

first add equation 1 and 3 together to get rid of x

[jolly shard]

Good idea

[sour mica]

then multiply equation 3 by 2 and add 2 and 3 together

[jolly shard]

Uhh you don’t need to scale it, just add row 3 to row 2

[sour mica]

how would u get rid of x then?

[jolly shard]

Oh ohh yeah you’re right

I thought you were doing something else

[sour mica]

nono lol

[jolly shard]

So add 2*R3 to R2 and that’s your new R2; then add R1 to R3 to become your new R3

[sour mica]

get
10y - 2z = 56 and
23y - 4z = 127

[jolly shard]

Then you can do more row operations to solve one of them

Try adding -2*R2 to R3 and solving for y

[sour mica]

get (4, 5, -3)

which is correct in this case

cant believe i picked one where i get it right

[jolly shard]

Good job!