help-39

[tulip ore]

(1) different inputs must lead to different outputs
(2) the same output can only come from the same input
(1) and (2) are identical ways of saying the same behavior (they are contrapositives)

a function that goes up then down (has a hill or valley) would not be one-to-one, since you can get the same output on either sides of the hill

[swift olive]

ohh so it has to pass the horizontal line test

[tulip ore]

brainrote

[tulip ore]

you can do that for functions that can be graphed but its going to be better later on to learn (1) and (2) as the primary way of knowing one-to-one

[swift olive]

so just make a table with the given function right

[tulip ore]

brainrote again

[tulip ore]

making a table is very tedious, youre not going to find the same output for two different inputs by testing a lot of them manually

you need to think about what youre doing here

[swift olive]

the textbook says to just plug in f(a) and f(b) and find if they're equal to each other

[tulip ore]

yea

[swift olive]

is that it?

[tulip ore]

yea

thats not a method btw

thats just the definition but laid out

if they tell you a and b, you can test if f(x) isn't a one-to-one function by seeing if f(a) = f(b)

you need to have a good picture of whats happening instead of just memorizing methods

[pearl pond]

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[bitter locust]

So I have the problem (√3+√2)²

[bitter locust]

I assumed that the square would cancel out and I would have (3+2) = 5

[last summit]

no

(a+b)^2 = a^2 + 2ab + b^2

(a+b)^2 does not equal a^2+b^2

remember to FOIL, or box method, or whatever

[bitter locust]

I haven't done either yet! So what roughly do I need to do to solve this (I need the answer to be a surd in its simplest form)

[last summit]

you have

(a+b)(a+b)

multiply them

that's all

people call it different names to help you remember how to multiply them

sometimes it's called the foil method, or the box method, but it's just multiplication

$$(\sqrt{3}+\sqrt{2})^2=(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})$$

[jolly parrot]

Austin

[last summit]

then multiply

[bitter locust]

Ok. so instead of distributing the exponent across all terms, I instead multiply the brackets by itself

[last summit]

It is not true that (a+b)^2 = a^2 + b^2

so you cannot "distribute the exponent across all terms"

[bitter locust]

Why aren't they equal? Like obviously now that you have pointed that out that it isn't, it isn't true, but why not?

[last summit]

because multiplication

doesn't work like that

[last summit]

is obviously true right

when you multiply the right hand side, it wont equal sqrt(3)^2+sqrt(2)^2=5

so the left hand side also doesn't equal 5

(1+1)^2 = (2)^2 = 4
(1+1)^2 does not equal 1^2 + 1^2 = 2

[bitter locust]

Yea that makes sense.

So I tried again and I got 5+2√6

is that correct?

[last summit]

looks good

[bitter locust]

Thank you so much! is there a way to verify this (like on a calculator or website) when I need to have my answer in surds, not an irrational number?

nvm

Thanks for your help @last summit you deserve an award! That put me in the right direction

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[pearl pond]

@prisma terrace Has your question been resolved?

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@prisma terrace Has your question been resolved?

[velvet thorn]

What do u have so far @prisma terrace

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[robust stump]

how do i start?

[minor bramble]

find what the cost is in terms of x and y

[robust stump]

i still dont know how to really start idk

[minor bramble]

try finding the lengths of RQ, QR, and RS in terms of x and y first

[robust stump]

im sorry i have to go but ill come back

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[modest pagoda]

how do i find x and y please help its due in less than 2 hours

[modest pagoda]

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[upbeat cypress]

Someone help pls for trig math

[zinc idol]

Is there a particular question?

[upbeat cypress]

I need help with practically every question, I have to submit it due tmr for a grade so I need to make sure the assignment is correct

Well it doesn't have to be correct

Just finished

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[gritty warren]

A water tank with a volume of 9 is to be made by welding two hemispheres to the center of a right circular cylinder. The material in the hemispheres is 4/3 times as expensive per unit area as the material in the cylinder. Dimension the tank so that the material cost will be minimal.

[gritty warren]

I used wolfram on the function i got at the end, and there were no solutions for r, and by the looks of it i seem to have done something wrong

How can i solve this?

[unkempt yacht]

did you try sketching out the shape

[gritty warren]

Yea

Looks like a pill

[gritty warren]

But i dont know how the shape will help

Only the formulas for the shape

[unkempt yacht]

its always best to sketch out the shape to make sure you are going the right direction

i assume the question meant the radius in order for minimal material consumption?

[gritty warren]

Yea

For V i took the volume of a sphere + volyme of a cylinder

With the same radius

And i used that to get the length L in terms of radius r

[unkempt yacht]

did you take account for the fact that you did not also account for the area of the two bases of the cylinder and two bases of the hemispheres?

[gritty warren]

Yea i removed the bottom areas of the cylinder

I really dont see why my solution doesnt work

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[fathom skiff]

Just want to check would x0 be a cluster point in this situation

[fathom skiff]

since if you take the open interval to not include the gap in the set there would be infinitely many points of A

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[desert solar]

Draw the graphs.The limit function is f=0
on [0,1). The ϵ fence will be given by y=±ϵ. Do the graphs all fit inside for large n?

[desert solar]

$x^n$ is the sequence of function

can someone show me the graph

[jolly parrot]

pocoyo

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<@&286206848099549185>

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[smoky saffron]

does this method actually work ?

bc i dont understand it

how did they get the discriminant?

[pearl pond]

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<@&286206848099549185>

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[vivid pilot]

how would i do this? not sure where to start

[pearl pond]

@vivid pilot Has your question been resolved?

[vivid pilot]

Anyone???

<@&286206848099549185>

[somber mason]

I don't know how to do it but since nobody is helping have you tried visualizing it on a smaller scale?

i.e. smaller gridsize?

[vivid pilot]

I might try

But I don’t know the general method

[somber mason]

could it potentially be 30+30+50 for the two shorter sides of the rectangle plus one of the longer side, plus 30x50 for 50 30 street long roads?

it's not that effective lol

[vivid pilot]

How did you get that?

[somber mason]

but can it be less?

I removed all the unneccessary roads I guess but idk if it's right

[vivid pilot]

I’m not sure tbh

[somber mason]

this way each road is only connected to two other roads, except for 50 of them which would be connected to 3 roads

[vivid pilot]

I’m really confused

[somber mason]

imagine 50 roads down, instead of just 6

if the box was filled up lol

then there's also this, I didn't feel like completing it

Idk if it would lead to less roads though

or maybe there is some mathematical way to find out

I can't find anything on google 😦

But I have to go to sleep I'm interested to know the answer I hope someone comes along

Good luck

[somber mason]

Hey man, I think I got it... start in one corner of your grid and have a road all the way to the other end, that's 50 roads, one road down; 50 roads left; one road down; 50 roads right and so on until you have hit each intersection

there's no road on a grid like this that is connected to any more than 2 roads

[somber mason]

it would be 1530 roads, but a design like this is the same, so

ok holy shit I have to sleep

[vivid pilot]

Ok I’ll look over this

Thank you for your help!!

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[iron oracle]

hi

i would like to dbl check if the bearing is 060

[pearl pond]

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[vestal venture]

how'd you got?

the calculations are pretty nasty

[pearl pond]

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[floral terrace]

hemlo, are both jus the same? or not

[floral terrace]

<@&286206848099549185>

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[warm horizon]

i have showed 2a, how do i show 2b?

[west sapphire]

well part (a) shows that conjugation by tau sends a k-cycle to a k-cycle

express sigma as a composition of disjoint cycles

and apply (a) to each of those cycles

[warm horizon]

what about this

let sigma be the product of disjoin cycle

o = ab

T(ab)T^-1 = TaT^-1 T b T^-1 = (TaT^-1)(TbT^-1) then apply rule a

cuz the T^-1 times T is the identity right so I can keep the equaloity the same?

does this prove it for abc.... too tho...

[west sapphire]

yep

just need to show that the new cycles are also disjoint

but that's fairly obvious

[warm horizon]

Oh just do like

Tx1 .... xn T^-1 and i say I can put T^-1 T in there

Like

not just one

but like

in between each x1, ... xn

lke T x1 T^-1 T x2 T^-1 T ... T xn T^-1

does that prove it?

[cursive wraith]

Yes and by question 2a, (TxiT^-1) are still disjoint cycles

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[lilac coral]

Please someone help explain to me how i could find the co-ordinate of A

[lilac coral]

i know that it is 4, but i got to it in quite a long way. Is there a quick simple way that i missed out?

[wooden ice]

there's not

u just have to put y=0 to find x value

[lilac coral]

i rearranged it to get (x−4)(x^2 +4x+16)=0

[lilac coral]

thought i was just being stupid and there was an easier way to solve

.close

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[cyan coral]

Ok so ive used matrices with the vector product but i dont understand what their role in math is nor how they otherwise are used

[cyan coral]

Are they a way to simplify/visualize?

[pearl pond]

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[marble sigil]

there's a sort of wild way to do a cross product where you multiply along diagonals and subtract the left diagonals from the right diagonals

in math the cross product is basically saying "multiply these two numbers with directions and also incorporate how much the directions are different"

it shows up in things like the arrea of a parallelogram because a larger angle means a wider shape and bigger area, or questions like "if I hit a door that's in this direction with a force in this direction, how much does the door actually turn?" because you want to hit a door at a right angle to be effective

(typo in the image: the 7k at the bottom should be -3k because 2-5=-3)

oh and the biggest use really is that this calculation gives you a vector that's perpendicular to both original vectors, and it's the easiest way to find something like that

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[drowsy leaf]

I know how to solve this but how would I solve what the question is asking?

I could put all of them into my calculator to see which graph matches

But I want to know how

[hollow condor]

can you find the roots of the given equation?

[crystal lion]

You just expand $(x - 5)^2$

[jolly parrot]

casework

[crystal lion]

And put 16 to the other side

[hollow condor]

you don’t even need to expand

[drowsy leaf]

oh so you turn (x-5) squared into

[hollow condor]

take the square root of both sides

[drowsy leaf]

x squared -10x+25

[crystal lion]

Not the optimal sol

[drowsy leaf]

its x=9 and x=1

[crystal lion]

Not just subtract 16 from both sides

[hollow condor]

wdym

[drowsy leaf]

thats what x-5 squared is

[crystal lion]

$$x^2 - 10x + 25 = 16$$

[jolly parrot]

casework

[crystal lion]

Just put everything to one side

It doesnt ask you for sols

But rather which 2 eqs are the same

[proper tartan]

subtract 16 from both sides

[hollow condor]

oh

[drowsy leaf]

so its A

[hollow condor]

nvm

yes

you’re right

[drowsy leaf]

what if the answer choices was x^2-10x+9 or x^2-10x+25=16

Which one would be right

[crystal lion]

Both

[drowsy leaf]

Is this right

I put them all in my y= in my calculator

all 4 had 2 zeros

[crystal lion]

Well them its right if you say so

[drowsy leaf]

is it??

[pearl pond]

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[smoky saffron]

I dont understand this question

Why did he solve for time in the original equation?

why cant he plug 0 for t in the derivative equation

[versed mica]

ivy

this is hard to read

@smoky saffron

[smoky saffron]

read harder

[versed mica]

ok

so

what’s ur question

[smoky saffron]

this

:p

[versed mica]

because

he needs to use it for the velocity

he needs to first solve how long it takes to get to the dock

then use that same time to calculate the velocity of the boat

[smoky saffron]

why would he plug in 0 to find out how long it takes

[versed mica]

because

s

is the distance

TO THE DOCK

so if it’s at the dock

then the distance to it is zero

[smoky saffron]

oh ok

[versed mica]

studious

[smoky saffron]

ik

[versed mica]

math let’s

mathlete

[smoky saffron]

I Need u seriously im so screwed for this test on thursday

[versed mica]

nah ur chillin

related rates r easy

[smoky saffron]

yea for you

[versed mica]

fair

💀

it’s ok

[smoky saffron]

I fumbled the quiz i did

[jade umbra]

dw math is something that'll eventually just click

[smoky saffron]

Its been 17 years its not clicking

[versed mica]

interesting name

[jade umbra]

it's been almost the same and it's now just clicking

[smoky saffron]

I lost 16% of my quiz grade because first i thought 2x3 = 8 and because i didnt add like terms

[jade umbra]

damnn

[versed mica]

2+2=5

[smoky saffron]

fr

[versed mica]

1984 reference lol

[smoky saffron]

ur not even that old

[jade umbra]

hey can you two help me with this equation

[smoky saffron]

I did all the calculus right

[versed mica]

🤦🏼‍♂️

1984 is a book

[smoky saffron]

Yea im smarter than knief u can coutn on me

[versed mica]

george orwell

[jade umbra]

#help-25 message

[smoky saffron]

Omg two of my friends are reeading that

[jade umbra]

this problem has me stumped

[versed mica]

skill issue

[versed mica]

it’s a classic

[smoky saffron]

I thin k the answer is dy/dx = 5

[versed mica]

wrong

[smoky saffron]

okay buddy

[versed mica]

ok pal

do you have a question

[smoky saffron]

atleast u dont have to memorize 3 calculus proofs for the test on thursday

[jade umbra]

did you carry the 1?

[versed mica]

why would i need to memorize them😹😹😹

[smoky saffron]

Not yet can you come back to me in 10 minutes when im solving questions

[smoky saffron]

🧠

[versed mica]

everything gets memorized no need to try

[smoky saffron]

Oh so your one of those mother

fucker

[versed mica]

debatable

[jade umbra]

bros got that photographic memories

[versed mica]

doesn’t exist

it’s a myth

[smoky saffron]

my friend says she has it

[versed mica]

she’s lying

[smoky saffron]

i LOATH people with photographic memory

[jade umbra]

ongg

[versed mica]

well great because they don’t exist😹😹

[smoky saffron]

I thinik ur delusional

[versed mica]

there are eidetic memories

but not quite photographic like mike ross

[versed mica]

💀

not the same

common misconception

[pearl pond]

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[rough stream]

The wording is pretty bad. I think we're left to assume:

• They're six sided dice
• Both are rolled exactly once.

[raw axle]

ok

[pearl pond]

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[smoky saffron]

nvm

i get it

[pearl pond]

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[sonic geyser]

The trick is to notice that the common denominator is designed to be the denominator of the original expression. So
[\frac{f(x)}{x} + \frac{g(x)}{x^2} + \frac{h(x)}{x + 2}]
has a common denominator of $x^3 + 2x^2$. so you need to put them all in that form and then you can figure out what $f(x), g(x), h(x)$ have to be to get the numerator to be $x^2 + 37$

[jolly parrot]

Awesam

[prisma terrace]

I don’t know how to do that 😅

[sonic geyser]

do you know how to make a common denominator?

[prisma terrace]

No

If you can guide me I’ll follow along

[sonic geyser]

$\frac{f(x)}{x} \cdot \frac{x^2 + 2x}{x^2 + 2x} = \frac{x^2f(x) + 2xf(x)}{x^3 + 2x^2}$

[jolly parrot]

Awesam

[sonic geyser]

it's like that

[prisma terrace]

Ok ok

So what’s next

[sonic geyser]

$\frac{g(x)}{x^2} \cdot \frac{x + 2}{x + 2} = \frac{xg(x) + 2g(x)}{x^3 + 2x^2}$

[jolly parrot]

Awesam

[sonic geyser]

I'll leave the h term up to you

[prisma terrace]

I don’t understand. I’m a bio major

This is my last question 🥲

[sonic geyser]

oh

what can you multiply $x + 2$ by to get $x^3 + 2x^2$?

[jolly parrot]

Awesam

[prisma terrace]

X^2 ?

[sonic geyser]

yup

[prisma terrace]

Ok now what

Fuck it I ran out of time

[pearl pond]

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[dark jackal]

Hey guys, could someone tell me the answer to this question please?

[midnight haven]

,rotate

[midnight haven]

did u try it

[pearl pond]

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[pale iris]

Hello mathematics discord community,
I am a student trying to solve a physics problem on a TI-Nspire CAS calculator, and am encountering an issue when using the solve function that I'm hoping I can get some input regarding either my setup or perhaps an obvious pointer to where my mishap in the input may have been.

I have the equation arranged where mass is multiplied by acceleration on the left and the weight of the woman is calculated on the right as well as the tension force subtracted from the product of the weight of the woman.

When I checked the answers, I could see that I had the right arrangement for my equation, but I was getting an answer that does not appear to match the answer key.

I have attached both the answer key and my calculations any input would be greatly appreciated.

[marble sigil]

Tension is already a force so you don't need to multiply it by m here

[feral leaf]

tension force subtracted from the product of the weight of the woman.
Why would you do this? The equation is T - mg = ma
So if you wanted ma on the left, then you would have ma = T - mg so in the calculator, you have m * 4 = 630 - m * 10

[pale iris]

I must have said the wrong thing, it looks like you took the weight of the women and subtracted it from the tension force, which would make sense because the tension force is holding the women up meaning it must be greater or equal to the force of the weight of the women, that's why it must be positioned there.

In my calculator I definitely did express it as the weight of the women minus the tension force, which doesn't make sense.

[feral leaf]

It's the tension minus the weight of the woman

Not weight of woman minus tension

m * 4 = 630 - m * 10
That is the appropriate equation you are suppose to type in

[feral leaf]

You're rearranging the equation wrong which results in a different answer

[pearl pond]

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[tall prawn]

For part a I wrote ${dP/dt} = -r*P+k$

[jolly parrot]

alani

[tall prawn]

I'm lost with part b

[pearl pond]

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[pearl pond]

@tall prawn Has your question been resolved?

[pearl pond]

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[stuck widget]

could someone guide me on how to do this?

[humble lintel]

I would start by setting lambda2 = 0 and study the line a + lamdba1 v1 to help you find a.

[pearl pond]

@stuck widget Has your question been resolved?

[stuck widget]

Sorry for late reply, could you elaborate especially line a

.close

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[urban cypress]

Hi, basically I'd like to find a function (ig it would be exponential) where I know 2 points and a vertical asymptote. Is that possible, and if it is how would I find it? Thanks in advance

[light helm]

wouldn't really be exponential

[pseudo oxide]

wdym "i know 2 points"?

[urban cypress]

Okay, I was guessing, I haven't done anything remotely like this in a long time

[pseudo oxide]

an exp passing through two given points with a vertical asymptote?

[light helm]

the domain of standard exponential functions is the set of reals and don't have vertical asymptotes

were you given a specific question?

[urban cypress]

This has just come from my head, I'm trying to graph a line that goes through 2 points I have and has a vertical asymptote

[pseudo oxide]

what are the two points?

[urban cypress]

(70, 16) and (80, 8) and y would approach infinity as x approached 0

[pseudo oxide]

f(70) = 16
f(80) = 8

hm

one sec

[void yew]

what

[urban cypress]

Thanks

[pseudo oxide]

ok got it

[urban cypress]

I feel like it should be possible with the information I have, but I haven't the faintest idea of how to do it

[pseudo oxide]

i think i got it

1/0.625(x - 60)??

,w plot 1/{0.625(x-60)}

oh god not what i meant

brb

[urban cypress]

I presume you mean 1/(0.625(x-60)) since that would have a vertical asymptote

[pseudo oxide]

back

[pseudo oxide]

yes

[jolly parrot]

Results provided by WolframAlpha

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

Requested by justarenegade

[urban cypress]

But the vertical asymptote would be at 60 instead of 0

[pseudo oxide]

,w 160/{(80-60)}

[pseudo oxide]

i know

[urban cypress]

okay

[pseudo oxide]

well, that was unintended

hang on

[urban cypress]

I appreciate the help btw

[pseudo oxide]

i aint helping very much, am i

thanks anyway

[urban cypress]

No you are, anything is helpful

[pseudo oxide]

omfg

WA stop

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by justarenegade

[pseudo oxide]

ykw

[urban cypress]

See in my head the horizontal asymptote shouldn't be at 0, which is messing with me, that would be lower but idk where

[pseudo oxide]

frick WA

[pseudo oxide]

horizontal?

[urban cypress]

Like uhhh the limit as x approaches 0 y would approach infinity, but the reverse of that wouldn't be true, like where x approaches infinity in this imaginary line, y would be some negative value but idk what

[pseudo oxide]

negative?

shouldn't it be positive

[urban cypress]

1 sec, Imma draw something on paint

[pseudo oxide]

okay

[urban cypress]

I think the line that fits my points and the limit would look sorta like this, so it would go below 0 on the y axis

At least visually that makes sense to me

[pseudo oxide]

okay

um

that...

wha-?

[pseudo oxide]

well, ramonov is a genius i guess

[urban cypress]

Damn

[pseudo oxide]

idk where that came from but ok

[urban cypress]

https://tenor.com/view/the-goat-gif-4115569673926897153

[pseudo oxide]

LMAO

[light helm]

goats don't like gifs

[urban cypress]

sorry

[pseudo oxide]

...?

why is bro apologizing

[light helm]

for what you're describing, with vertical asymptotes, rational functions can give you what you want

[urban cypress]

Sure, yeah. Sorry Idk terminology for things anymore, I haven't done maths in like 5 year aside from some basic stuff

I just didn't know how to find said rational function with the information I had

[light helm]

the simplest one that has a vertical asymptote at x=0
is the form y = k/x isn't sufficient to get you what you want, so proceed to
y = (ax + b)/x^2
from there solve a system of equations to determine the values of a,b and you'll be able to reach what i obtained

[urban cypress]

Okay I'll give it a go, thanks

Okay I got it, thank you

That was actually very simple, but I'm far too rusty at this sort of thing. Appreciate the help

.close

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[barren nymph]

hello

[barren nymph]

could u solve ex 3??

just need to find a and b

and u have the point x10 y5 and x20 y0

[pseudo oxide]

translate

[pseudo oxide]

!noans

[pearl pond]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[barren nymph]

i alr translate it is the ex c there is an equation you need to fina a and b

[barren nymph]

you getting it?

[pseudo oxide]

wat??

[barren nymph]

someone could solve it??

o mean i dont know what to do

[pseudo oxide]

bro

we are here to help not solve for you

and it would help if you translated the ENTIRE question

i've had too many times where a small omission in translations has led to a big misunderstanding

[barren nymph]

it is what i told you

[pseudo oxide]

ich spreche nicht deutsch

[barren nymph]

i alr tried to solve it didnt got it

ahaha

i gave you all the information you need

[pseudo oxide]

i only speak very little

barely

well, i don't think i can solve this

[barren nymph]

ahah ik its difficult

thatts why i asked for help

[pearl pond]

@barren nymph Has your question been resolved?

[iron basin]

hmm can you translate the qn

[pearl pond]

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[languid jungle]

I need help with this word problem pleaseee

[pearl pond]

@languid jungle Has your question been resolved?

[pearl pond]

@languid jungle Has your question been resolved?

[pearl pond]

@languid jungle Has your question been resolved?

[trim current]

Hello do you still want help?

[languid jungle]

please

I have a test on this in an hour

😭

I still don’t get it

The % are messing me up i can do ones without % but im confused with this one

I just don’t think I’m doing it right

[trim current]

what do you not understand

[languid jungle]

Like I get how to put it into an equation but I don’t get how to actually solve this

[trim current]

ok show me your equation

[languid jungle]

Okay I had

0.16y+ 0.05x= 10,000
And
X+ 865= y

I feel like these are very wrong and that’s why im not getting it

[trim current]

yeah

well we can think about it like this so the combination of both is equal to 10865

x + y = 10865

and so we know that stock A is 16% more than itself than in the past

[languid jungle]

Yeah

[trim current]

so we can say x2 = x+16%

and y2 = y - 5%

wait

i might be going wrong

[languid jungle]

It’s linear equations so you are trying to solve and find the missing x and or y

But it should pretty much equal whatever the number is after the = sign

[trim current]

ok

then couldnt we write it out as a linear equation?

[languid jungle]

Yeah that’s what i tried doing above but I think it’s wrong

[trim current]

so it would be y = (-0.05x + 10000)/0.16

that would be the linear equation

[languid jungle]

So what would I put in x though that’s where I’m stuck like how do I solve to get what y equals

[trim current]

ohhhh

ok dont worry

you just need to place y as the amount increased

then solve from there

[languid jungle]

So I just 865 where y is and solve for x?

[trim current]

yes i think...

im also learning this

[languid jungle]

Fr?😭

[trim current]

yea

[languid jungle]

Omg

[trim current]

im in highschool

[languid jungle]

LMFAO

IM IN COLLEGE

[trim current]

oh mb

[languid jungle]

I’m re learning stuff I learned in high school but I don’t remember literally any of this 🥲

[trim current]

i swear this is just a simple interest equation

not a linear one

[languid jungle]

It is

I think

It’s just one of the questions if gives me for my linear equation chapter

It could very well just be a basic interest question

[trim current]

i swear it is

we just need to figure out the basics

[languid jungle]

How would u solve it like a basic internet question then?

I can get the answer one sec

Okay so it says stock A should be 6500 and B is 3500

But idk how they got that

[trim current]

hmm

[languid jungle]

yes :(

[trim current]

sorry

let me write this out

im also very confused

[languid jungle]

It’s okay take ur time

[trim current]

i got 10865 = A(1.16) + B(0.95)

I GOT IT

WOOOHO

LETS GOOO

Ok so first things first we find the individual values of stock A and stock B

Using the Simple interest formula

So stock A would be P(1+0.16(1) and stock B would be P(1-0.05(1)

so it would turn out A = P(1.16) and B = P (0.95)

then we get 10865 = A(1.16) + B(0.95)

we know that B = 10000 - A

so we place that as its value

so we get 10865 = A(1.16) and ((10000 - A)0.95)

[languid jungle]

😭😭

[trim current]

10865 = A(1.16) +9500 - A(0.95)

subtracting we get 1365 = A(1.16) - A(0.95)

so we get 0.21A = 1365

so we get 6500

as A

then we just substitute

[trim current]

sorry it took so long

[languid jungle]

Nahh ur good ur helping me a lot

[languid jungle]

Wait sub what

[trim current]

A as 6500

into B = 10000 - 6500

and thats how we get 3500

uhm so you understand?

[languid jungle]

Yes

[trim current]

great

good luck in college

[pearl pond]

@languid jungle Has your question been resolved?

[languid jungle]

Thanks!

[pearl pond]

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[midnight haven]

I did something wrong but I don’t know what

[pearl pond]

@midnight haven Has your question been resolved?

[vestal sable]

What makes you think you've gone wrong?

[pearl pond]

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[urban adder]

How would you finish simplifying this trigonometric expression?

[urban adder]

I was able to advance this far, but I don't know how to continue.

[old marsh]

nahhh

hold on i think i have smth simpler

[urban adder]

okk

the result is 1

[old marsh]

yo

i just made the most beautiful solution known to mankind

@urban adder

thank me bro

the fact u get to witness this

are u ready?

ping me when u ready

[urban adder]

yes

[old marsh]

this some greatness right here

ok

lets start

i immediately saw that a+b and was disgusted

so i knew we had to get rid of it

let x = a+b

y = a-b

can u write 2a in terms of x and y

[urban adder]

by adding both equations?

[old marsh]

exactly, and what does that get

2a = ?

[urban adder]

2a = x+y

[old marsh]

good

now can u do the same thing, except for a-b?

[urban adder]

hmm no

[old marsh]

sorry i mean 2b

[urban adder]

?

[old marsh]

sorry

[urban adder]

a

subtracting both equations

[old marsh]

yes

2b = ?

[urban adder]

x-y

[old marsh]

good job

now lets start

$\cos^2 (x) + \cos^2 (y) - \cos(x+y) \cdot \cos (x-y)$

[jolly parrot]

Stephen

[old marsh]

this is what we have now correct?

[urban adder]

yes

[old marsh]

ok, can you expand cos(x+y)?

[urban adder]

cos(x).cos(y) - sin(x).sin(y)

[old marsh]

perfect

can u expand cos(x-y)?

[urban adder]

cos(x).cos(y) + sin(x).sin(y)

[old marsh]

good

so now we have

$\cos^2 (x) + \cos^2 (y) - [\cos (x) \cos (y) - \sin (x) \sin (y)][\cos (x) \cos (y) + \sin (x) \sin (y)]$

[jolly parrot]

Stephen

[old marsh]

yes?

or no?

[urban adder]

yes

[old marsh]

ok good

now look at that big multiplication

do u notice anything special?

[urban adder]

difference of squares

[old marsh]

YES

so when u apply difference of squares here, what do u get

[urban adder]

[cos(x).cos(y)]^2 - [sin(x).sin(y)]^2

[old marsh]

yes

so now we have

$\cos^2 (x) + \cos^2 (y) - [[\cos (x) \cos (y)]^2 - [\sin (x) \sin (y)]^2]$

[jolly parrot]

Stephen

[old marsh]

ok now we are near the final steps

distribute the minus

what do we get

[urban adder]

cos^2(x) + cos^2(y) - cos^2(x).cos^2(y) + sen^2(x).sen^2(y)

[old marsh]

a

hablas espanol

[urban adder]

si

cómo sabés?

[old marsh]

aprende en la escuela

o

vi q escribiste "sen" en lugar de "sin"

[urban adder]

xd

[old marsh]

re fachero

ok

so now lets look at this

$\cos^2 (x) + \cos^2 (y) - \cos^2 (x) \cos^2 (y) + \sin^2 (x) \sin^2 (y)$

[jolly parrot]

Stephen

[old marsh]

look at those middle 2 terms

cos^2(y) and cos^2(x)cos^2(y)

can u factor anything out from them?

[urban adder]

podes sacar factor comun cos^2(y) y queda multiplicando a (1 - cos^2(x))

[old marsh]

bien

[urban adder]

y 1 - cos^2(x) es sen^2(x)

[old marsh]

bien

y ahora tenemos

$\cos^2 (x) + \cos^2 (y) \sin^2 (x) + \sin^2 (x) \sin^2 (y)$

[jolly parrot]

Stephen

[old marsh]

sip?

[urban adder]

si

[old marsh]

ok

look at the last two terms

can u factor anything out>

[urban adder]

si sacas factor común sin^2(x) queda multiplicando a (cos^2(y) + sin^2(y))

[old marsh]

sip

[urban adder]

cos^2(y) + sin^2(y) = 1

[old marsh]

sip

siii

entonces?

[urban adder]

cos^2 (x) + sin^2(x).1

= 1

[old marsh]

yay

[urban adder]

hermoso

[old marsh]

claro q si

me diverti mucho

adiossss

[urban adder]

yo tambien, buena explicacion

gracias

[old marsh]

gracias

[urban adder]

.close

[pearl pond]

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[pearl pond]

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[rain flame]

This is telling me to sketch each angle in standard position and what quadrant it is in

[rain flame]

And it’s saying pi/6

[jade umbra]

can you show your paper

[jade umbra]

convert the radians to degrees and then sketch

[rain flame]

Like this??

[jade umbra]

if i'm not mistaken that would be correct

[rain flame]

How would that work for 3/4 one?

180•3pi?

[jade umbra]

3pi/4 • 180/pi

[rain flame]

Does that just become 540pi? / 4pi?

[jade umbra]

the pi cancels out

and simplify

[rain flame]

Yeah

So 540/4?

[jade umbra]

135°

[pearl pond]

@rain flame Has your question been resolved?

[rain flame]

When this question says “3 radians” it means 3pi, right?

[pearl pond]

@rain flame Has your question been resolved?

[pearl pond]

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[willow ginkgo]

i need to prove DE = 1/2AB

i just dont understand how I can just fill in 2/1 for BC/DC

well I do know why ig, since given is that D is the exact middle

wait, so since I know the ratio, I can just fill in the ratio and use that to prove it

.close

[pearl pond]

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[pearl pond]

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[sonic kelp]

Why would this be wrong?

[regal herald]

cos is a function, you cant just apply the power rule

it requires the chain rule

[sonic kelp]

Why though?

[autumn trellis]

You apply the chain rule even when you do the power rule

[white juniper]

beacause x^n are different functions than cos^n(x)

thats it

[jolly parrot]

smidgin

[sonic kelp]

Thank you

how do I mark that my question has been solved?

[white juniper]

".close"

[sonic kelp]

.close

[pearl pond]

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[vivid pilot]

this is what i did so far, but its wrong

[vivid pilot]

$\frac{\pi}{16}\left(f\left(\frac{\pi}{32}\right)+f\left(\frac{3\pi}{32}\right)+f\left(\frac{5\pi}{32}\right)+f\left(\frac{7\pi}{32}\right)+f\left(\frac{9\pi}{32}\right)+f\left(\frac{11\pi}{32}\right)+f\left(\frac{13\pi}{32}\right)+f\left(\frac{15\pi}{32}\right)\right)$

[jolly parrot]

talk_less

[vivid pilot]

where $f(x)=\pi\cos^6{x}$

[jolly parrot]

talk_less

[vivid pilot]

anyone??

<@&286206848099549185>

[pearl pond]

@vivid pilot Has your question been resolved?

[pearl pond]

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[wanton sequoia]

A standard die is rolled twice. Then a card is chosen at random from a standard deck. How many outcomes are possible?

[elder plaza]

what have you tried

[wanton sequoia]

well i know that if i roll the dice twice, the probability of getting any number will become 1/36

but i get stuck after that

i know the answer is 208

i just dont know how. i know i am missing something

the card is chosen once.... so the probability of any card being picked is 1/52 ....

[elder plaza]

yes

so

if we roll two dice the possible amt of outcomes are 6*6

36

[wanton sequoia]

yes ma'am

but i need to find the possible outcomes with the dice and the card.

[elder plaza]

are u sure it’s 208?

give me a second

[wanton sequoia]

umm well it's a question from my Son's study guide. the teacher gave out the answet sheet and it says 208

when i did it in my way, i got 312

thank you, please take your time. i am pretty ok with middle chool math but its been a while ... so i got stuck

[elder plaza]

hmm

i got 1872.. here's what I did

6 outcomes for rolling one die, we roll it twice
6 * 6

52 outcomes for a standard deck of cards, we add that to our multiplication series

6 * 6* 52 = 1872

[wanton sequoia]

thats what i did the first time ... but the key says 208 😦

[elder plaza]

i'm not sure why it would be 208 unless i'm reading the problem wrong

[wanton sequoia]

thats the first answer i came up with ... because they only asked for outcomes

[elder plaza]

mm

[wanton sequoia]

not a specific combination

I think I can challnge the teacher on that ... cuz thats what i got too the first time

[elder plaza]

yes

[wanton sequoia]

its not like they said a 2 and a 5 of spade

[elder plaza]

take one more opinion here maybe and if its the same thing

you can likely challenge it

[wanton sequoia]

if i may ask, do you teach or study math?

i am pretty ok with math and not getting the answer is killing me haha

i am this close to buying a math app

[elder plaza]

im taking calculus 3 and linear algebra so mostly just second year college math courses

im a math major

[wanton sequoia]

which brings me to ask, if you are someone in the math field ... is there an app you would recommend

[elder plaza]

gauthmath

photomath works until calculus as well i believe

[wanton sequoia]

i think i have become pretty confdent with your answer cuz that was my first answer too

I got Middleschoolers, so i think Gauth math is fine

thats the one i downloaded

but maybe its a paid one

ill pay if it saves me time

[elder plaza]

AI is pretty reliable too if you use two engines at once and cross check them

[wanton sequoia]

AI ... an app?

[elder plaza]

like i use chat gpt and google gemini and if they get the same answer im pretty confident

im sure chat gpt has an app

artifical intelligence

[wanton sequoia]

i know its artificial intelligence

[elder plaza]

oh yeah

chat gpt is an app im pretty sure

[wanton sequoia]

just asking that is that in an app or how do i access it

a site?

[elder plaza]

and google gemini might be an app but i know for sure you can access both on a website

either works

and they are free resources

[wanton sequoia]

ram, honey, thank you so much ....

i really appreciate you taking the time out

[elder plaza]

ofc GL

im just taking a study break from bio so no worries

[wanton sequoia]

aha! i am studying for my USMLEs

[elder plaza]

oh brother

[wanton sequoia]

bio is fun. n with the resources these days, you will love learning it

we did it from books, and imaginations

[elder plaza]

epithelial lining i just stopped on

yeah

epithelial lining of the intestines and stuff

[wanton sequoia]

oh that you need to remember forever

so don't cruise by it

[elder plaza]

well i think ill forget it after this semester 🤣 im a computer science and math double major

[wanton sequoia]

well then CRUISE BY IT haha

would it be ok if i add you? for math sometimes?

[elder plaza]

yes for sure

if i dont reply quickly feel free to ask in this server some of these people are a lot smarter than me

oops

did not mean to send that

[wanton sequoia]

sounds great 🙂 thank u added

well if u ever need to talk about epithelial lining ....

haha

let me know.

thank u 🙂

[elder plaza]

of course ill reach out

[wanton sequoia]

take care. bye

[elder plaza]

you too

type .close

[pearl pond]

@wanton sequoia Has your question been resolved?

[pearl pond]

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[pearl pond]

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[plush otter]

The gradients C1, C2, etc are vectors here. I am confused on how to read the vertical bars followed by the superscript. My current understanding was that this is the squared magnitude of the gradient but I'd like confirmation as I am struggling to notice an observable difference in the simulation I am building

[unborn musk]

what are those symbols, what am i getting myself into

def college lvl right?

[plush otter]

Hah, -C is a scalar. w1 is the inverse mass for particle 1, and what happens to it I have no idea. Multiplied by the squared magnitude of its respective gradient was my guess

[distant goblet]

i would assume its the magnitude squared as well

[unborn musk]

lamda?

the first one

[plush otter]

Ok, I was curious if the L2 norm was ever depicted with single vertical bars and a superscript

[plush otter]

correct

[unborn musk]

wavelength of a particle?

[distant goblet]

i think sometimes the euclidean norm is vertical bars but that has a subscript 2 not superscript. i think if it was to represent a certain type of norm or magnitude it would be subscript. i think the fact that it is superscript implies that it must be magnitude squared

[plush otter]

Thank you for the confirmation. This makes the most sense to me as well! I appreciate the help

!close

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

is thid right

[normal depot]

Yea

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

ty

.close

[pearl pond]

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[grizzled dust]

so for part b) of finding the slope of the tangent at theta = 3pi/4
i kind of thought of it as a parametric equation as suggested to me by a helper here yesterday where x(t) = r cos t = (1-cos t)cos t, y(t) = r sin t = (1-cos t)sin t
etc. then got dy/dx by using (dy/dt) / (dx/dt). After that I just used regular cartesian equation of a line etc. and this seems to have worked.

I was wondering, is there a way to solve this differently \ in a more simple way without thinking of it in cartesian terms or is this kind of the standard method for this type of problem?

[pearl pond]

@grizzled dust Has your question been resolved?

[sharp vigil]

https://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx

that seems to be a fairly common approach

the site i linked also has a general formula but it's derived from the same approach

[grizzled dust]

.close

[pearl pond]

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[tidal otter]

How should I go about proving this?

[merry python]

Do you know a way of 'factoring' this?

[tidal otter]

Geometric series?

[merry python]

Yes, basically

[tidal otter]

Alright so $\frac{x^{n }- 1}{x-1}$ is a prime number

[runic zephyr]

x^n-1

[jolly parrot]

Normed

[merry python]

Yes

Have you ever factored x^n - 1 btw?

that's the way I recognized this

[tidal otter]

hmm makes sense

[merry python]

You can factor anything of the form $a^k - b^k$

[jolly parrot]

Jelle

[merry python]

so, what if n is not prime?

(proof by contradiction)

[tidal otter]

Ahh I see we are trying to contradict that 1 + x + x^2 + ... + x^{n-1} is not prime?

[merry python]

Yes, that would mean that: if n is not prime then this can be factored

How could you factor the numerator if n isn't prime (so n has some factor > 1)

[tidal otter]

(x-1)(1+x+x^2+...+x^{n-1})

[merry python]

Yes, but in another way, if n = ab: $$x^n - 1 = (x^a)^b - 1^b$$

[jolly parrot]

Jelle

[tidal otter]

Ohh right

(x^{a}-1)(1+x^{a}+...+x^{a}{b-1})

[merry python]

Yes

now it's correct

[tidal otter]

And both these factors are greater than 1

[merry python]

And (x - 1), the denominator, divides the first factor

[tidal otter]

Yeah