help-13

[austere lantern]

hi can i please have help

[austere lantern]

with vectors

not sure how to do vectors, and youtrube isnt helping 😦

[dull oxide]

Do the tops and bottoms individually

So for 1.6, the top element would just be -6 plus 5

[austere lantern]

so

-1

over 4

for 1.6

and 4 over 7 for the 1.5

[dull oxide]

correct

[austere lantern]

thankyou

how many questiins can i ask?

i have a moduel to do

can i keep this chat open for any hellp i need

[dull oxide]

you may want to double check on this one

[austere lantern]

okay let me check

4 over 17 my bad

[cedar kiln]

@austere lantern Has your question been resolved?

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[fair copper]

can someone explain where the -46 came from

[radiant scaffold]

did they sub x back into f(x) to find y

[fair copper]

idk what they did to find y

but when u sub x back in it gives u a positive

[radiant scaffold]

what did they give f(x) as

[hearty arch]

they subbed x into f(x)

x=3

because that's the point of inflection from your second derivative, so to find where that is on your original function you just need to plug in the x-value

[fair copper]

why do we plug it into the original function

and not the second derivative function

[radiant scaffold]

because the second derivative function doesn’t give you the y value on the graph

if that happened, then we’d be saying f”(x) = y and that’s just wrong

[fair copper]

ohh ok

[fair copper]

also is there another way of doing this instead of the table or is this the most efficient way already

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[coral jewel]

A box contain 21 balls including 9 blue balls numbered from 1 to 9, 7 red balls numbered from 1 to 7 and 5 yellow balls numbered from 1 to 5. Randomly choose 3 balls from that box, calculate the probability for those 3 balls to have all 3 colors, and the number on each ball is unique.

[coral jewel]

there are 1330 ways to choose 3 balls, 315 ways to choose balls with 3 different colors

but im not sure how to calculate number of ways to choose balsl with 3 diff colors and 3 diff numbers

[cedar kiln]

@coral jewel Has your question been resolved?

[coral jewel]

<@&286206848099549185>

[icy quartz]

hope this helps

[coral jewel]

!nogpt

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[icy quartz]

no gpt

[coral jewel]

this is literally chatgpt

[icy quartz]

idk my friend sent me it

[coral jewel]

your friend sent you chatgpt

[icy quartz]

ok mb

sorry

lemme solve it one sec

[coral jewel]

i appreciate your enthusiasm, but i would prefer you helping if you know what you are doing

[icy quartz]

i think it will be fav outcomes/total pos outcomes

i.e. 315/1330

[coral jewel]

315 is only the outcomes of picking 3 balls that are different in colours

[icy quartz]

what answer do you need

[coral jewel]

what im stuck is finding the outcomes of picking 3 balls that are different in colours and numbers

[icy quartz]

specify

ok

AND

alright

[slate lintel]

let's assume you managed to pick three balls that are different colors, now arrange them yellow-red-blue. what is the chance that they are unique numbers?

first consider yellow, it's unique

then red, there's a 6/7 chance it's unique

then blue, there's a 7/9 chance it's unique

so it should be whatever your previous result is * 6/7 * 7/9?

[coral jewel]

9/38 * 2/3 = 3/19

but like, how did you come up with this so fast?

[slate lintel]

uhhhh

[coral jewel]

oh rethinking this, it made sense

idk why i was thinking i should create each case of diff colors, diff numbers, and then use venn

but anyway, thanks!

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[slate lintel]

if you had tried to do it in the other order it would have been harder

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[lavish bobcat]

Does anyone know how to do this?

a) Simplify the function f so that it does not contain angular functions.
b) Determine the domain of the function f and sketch its graph.
c) Sketch the graph of the function ln(8f /7).

[cedar kiln]

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[lavish bobcat]

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[merry nacelle]

Can someone see where i went wrong here? My answer is different from the answers i see online but i dont get how its wrong. The question is find the equation of the plane passing through P( 3, 5, -1) and contains L : x = 4 - t , y = 2t - 1 , z = -3t

[merry nacelle]

I got vector v and point Q from the given line equation

Found vector PQ and crossed it with v to find the normal vector

[cedar kiln]

@merry nacelle Has your question been resolved?

[cedar kiln]

@merry nacelle Has your question been resolved?

[merry nacelle]

<@&286206848099549185> calc 3 plane question

[cedar kiln]

@merry nacelle Has your question been resolved?

[cedar kiln]

@merry nacelle Has your question been resolved?

[glad peak]

What is the equation you are expecting? is it -5x - y + z = -21?

[merry nacelle]

I figured it out ty

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[dense hatch]

HOW IS THIS A GEOMETRIC SERIES

[fiery whale]

common ratio is -1/e

hence geometric

[dense hatch]

but the -1 is to n+1 and the e is to n

[royal loom]

all terms have a factor of pi

[hollow trail]

,, (-1)^{n+1} = (-1)(-1)^n

[wraith dagger]

cloud

[royal loom]

yes

[dense hatch]

NOOOOOOOOOOOOOO I FORGOT THAT EXISTED

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[merry shadow]

i'm trying to justify that E(t) and L(t) are continuous so that I can use EVT for part (c). would it be sufficient enough to say E(t) is continuous because it is a sinusoidal function?

[hidden mural]

yes sin is continous

power series are continous

in fact smooth

[merry shadow]

gotcha gotcha so for L(t) would i say it's a power series function hence continuous?

[hidden mural]

yes

cuz technically for any real number x you would define 2^x = e^(x log2)

and u define e^x by power series etc

i mean yeah it requires effort to set this all up

[drifting fable]

you can say it is continuous because it is differentiable over all t

[merry shadow]

oooo okay okay

[merry shadow]

oh fr?

[drifting fable]

yes a function is differentiable over k only if it is continouus over k

[merry shadow]

i wasnt sure if we had to prove continuity before differentiability. is that implied by saying the function is modeling a rate?

[drifting fable]

i mean, i have never heard of a rate being undefined at some segments

[merry shadow]

okay wait that makes sense

thank you both so much <33

[drifting fable]

ty

[merry shadow]

sobbing how do i close this

.close

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[errant wing]

I need 6 more equations to fulfill the requirement of my desmos school project, does anyone have suggestions on what to add?

[dim birch]

Would you like to be an art major or a math major?

“Yes”

[candid marsh]

whoa, that rose is amazing

Give him a top hat
and a bow

[slate lintel]

give him a cutie mark

[cedar kiln]

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[errant wing]

thanks! 🐧

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[drifting fable]

show work

[solar tendon]

I got number 1 but number 2 is the one im having issues with.

Imma show solution wait a bit

[drifting fable]

youre having trouble finding big R?

[solar tendon]

I think so because in the answer sheet it says 3.9 but i got 2.9

[drifting fable]

ok

[solar tendon]

This is just a practice question btw

20=5(1.1+r)
20=5.5+5r
14.5/5 =r
r=2.9

Idk whats wrong

Or the ans sheet is wrong idk

[drifting fable]

your first equation is wrong

your first equation says the voltage across the 1.1 ohm and the r ohm is 20 v

when in reality the only the voltage across big R is 20v

[solar tendon]

What do i do?

[drifting fable]

assuming r is the internal resistance

the question tells you that the voltage across r is 20v

using V=IR, you can then find r

20=Ir

thats wrong

the question tells you that 25.5-20=Ir

from the diagram one can see that

25.5=I(r+R)

you just solve the equations

[cedar kiln]

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[fleet remnant]

is a^1/2 = sqrt of a?

[earnest geyser]

Yes

[fleet remnant]

thanks

.close

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[neon fiber]

How do I solve these type of questions?

[sand cradle]

,w arccos(0.25)

[wraith dagger]

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Requested by kepe_5235

[sand cradle]

How did you get your first result of 1.311? It should be 1.318

And the second should also be different:

[neon fiber]

I seriously dont know, thats why im here

[sand cradle]

Ah, it looked very similar to the correct result, so I thought you just miscopied from your calculator

[neon fiber]

Probably, these thing used to give a demonstrate video, for some reason these question dont

what about second one

[sand cradle]

Which happens to be 1.318 in this case

Though, there is another value too

For that, you should look at the unit circle

You covered the unit circle, right?

[neon fiber]

yea

[sand cradle]

Now, the question is, at which angle is cos(alpha) the same

[neon fiber]

the same?

[sand cradle]

Yeah

Well, for the green triangle, right?

And the angle of that is 2pi - alpha

Basically you go alpha in reverse direction (not rotating to the left, but to the right)

[neon fiber]

So for my question it will be something like cos^-1(2pi-0.25)?

[sand cradle]

Your calculator gave you alpha

Now the second angle is 2pi - alpha

So 2pi - 1.318

,w 2pi - 1.318

[wraith dagger]

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[neon fiber]

Ahhh I see now

[neon fiber]

What about the second one

[sand cradle]

$2 \cos(\theta) = 1$

[wraith dagger]

Kepe

[sand cradle]

I.e. how do you get the left side to be only cos(theta)

[neon fiber]

cos^-1(1/2) maybe...?

[sand cradle]

Exactly, that's what you'll have for theta later, but I was talking about the middle step right now

You divide both sides by 2

[neon fiber]

And we apply similar method for first question?

[sand cradle]

Yeah, we do the same as for the first

[neon fiber]

let me try: cos^-1(1/2)=pi/3 and get second answer from cos^-1(pi/3)?

hmmm, calculator got error for second answer

[sand cradle]

cos^-1(1/2)=pi/3

Yep, pi/3 is the first correct angle

Well, to find the second correct angle, you look at the unit circle again, the same thing as before: instead of rotating in counter-clockwise direction, we get the same length and so the same cos for the triangle if we rotate clockwise

So the second angle is 2pi - pi/3

[neon fiber]

oh right forgot the 2pi part

[sand cradle]

"as exact values"

So yeah, you can say pi/3 and the simplified version of 2pi - pi/3

[neon fiber]

5pi/3 got it

[sand cradle]

Yep

[neon fiber]

Can you help me with another question?

[sand cradle]

Yeah, just ask

[neon fiber]

wait wrong one

This one confuse me so much Dx

[sand cradle]

What is cot(x)?

/how can you express it differently?

[neon fiber]

cot is similar to tan, but up side down, so express it will be something like 1/tan(x)?

[sand cradle]

Yeah

Now how can you express tan(x) differently?

Hint: it'll involve both sin and cos

[neon fiber]

uhhh is it sin(x)/cos(x)?

[sand cradle]

Yeah

[sand cradle]

Ok, so what is $\cot(x) = \frac{1}{\tan(x)} = \frac{1}{\frac{\sin(x)}{\cos(x)}}$? Can you do a simplification?

(A simplification to the rightmost term)

I mean in general, what would you do with $\frac{1}{\frac a b}$

[neon fiber]

flip it?

so like cos(x)/sin(x)?

[sand cradle]

Yep

$\cot(x) = \frac{1}{\tan(x)} = \frac{1}{\frac{\sin(x)}{\cos(x)}} = \frac{\cos(x)}{\sin(x)}$

Ok, now replace x with -x and plug the rightmost term into the original expression instead of cot(x), see what we get

[neon fiber]

Ok, in calculator I got -csc(x)

[sand cradle]

Yeah, we would've gotten that this way too

Now, you might notice an identity in the numerator

[neon fiber]

ummm Im having trouble identify the identity

[sand cradle]

Have you seen $\sin(x)^2 + \cos(x)^2 = 1$?

[neon fiber]

Yeah

[sand cradle]

Yeah, we can use that to make the numerator 1

So we get $= \frac{1}{\sin(-x)}$

We can simplify that a little bit more, though not sure if your class wants that

This would already be $\csc(-x)$

[wraith dagger]

Kepe

[neon fiber]

Oh thats how it turn to csc(-x)

[sand cradle]

But we can say that $\sin(-x) = -\sin(x)$

[wraith dagger]

Kepe

[sand cradle]

And so $\csc(-x) = \frac{1}{\sin(-x)} = \frac{1}{-\sin(x)} = - \frac{1}{\sin(x)} = -\csc(x)$

[wraith dagger]

Kepe

[sand cradle]

The rightmost term is what your calculator gave you

[neon fiber]

Yeah I can see it, the question want me to answer in term of sine and cosine

Do I convert csc to sine or cosine?

[sand cradle]

Depending on your class' convention of pulling out the - from the sin(-x), it will be $\frac{1}{\sin(-x)}$ or $- \frac{1}{\sin(x)}$

[wraith dagger]

Kepe

[sand cradle]

They're equivalent but the thing you're typing it into might not detect that

[neon fiber]

I got it, thank you

Also, can you check on why my answer is wrong here? I know the second answer is off due to it greater than 2pi, but I dont know why first one is wrong too

[sand cradle]

$0 \leq \theta < 2\pi$

[wraith dagger]

Kepe

[sand cradle]

You gave a negative value

[neon fiber]

Oh shoot, forgot about that too

But using the method you taught it become something like this: sin^-1(-sqrt2/2) which give me the -pi/4

what did I go wrong?

[sand cradle]

Yeah, your calculator generally spits out values in the range $[- \frac \pi 2, \frac \pi 2]$. An angle of $- \frac \pi 2$ is equivalent to an angle of $2 \pi - \frac \pi 2 = \frac{3 \pi}{2}$ (look at the unit circle for this)

If the exercise wants angles in the range [0, 2pi) from you, you need to convert

[neon fiber]

Ahhh so for any negative value in this type of question, just go for unit circle

[sand cradle]

Yeah, you need to convert to positive angles

I.e. adding 2pi to it

[sand cradle]

Also, 9pi/4 is larger than 2pi

There is something wrong with that too

(as in, it might be correct, but the exercise only wants angles in the range of 0 to 2pi)

You should look at where your angle of 3pi/2 is on the unit circle

And what its corresponding sin is

Sry, not 3pi/2

I meant 2pi - pi/4

= 7pi/4

[neon fiber]

Yeah I know about that part, I got 7pi/4 for one answer

the other one is 5pi/4 I think?

[sand cradle]

Drawing the angle of 7pi/4 first (red triangle)

Then you see the angle it encloses with the x-axis is pi/4

So the other angle is pi + pi/4 = 5pi/4

[sand cradle]

So yes

[neon fiber]

I see

[neon fiber]

last bit of question on this one, if the number given is negative, what should I do?

[sand cradle]

So e.g. find all $0 \leq \theta < 2\pi$ with $\sin(x) = -0.5$

[wraith dagger]

Kepe

[sand cradle]

,w arcsin(-0.5)

[wraith dagger]

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Kepe

[neon fiber]

Got it

Thank you so much man

[sand cradle]

Now you need to find the other angle too

Generally there will be two angles (with a few exceptions like when one theta is pi/2)

[neon fiber]

Wait...

Do i need to add the little 0 thing in front for the calculator to work?

[sand cradle]

-46.88 is in degrees

Not in radians

2pi is in radians

[neon fiber]

oh ok, so I add 360 instead?

[sand cradle]

You need to get them to the same unit

Yeah

Or convert -46.88 to radians

[sand cradle]

This should be preferred since in the end you'll need to give the answer in radians

[sand cradle]

Switch your calculator to radian mode

[neon fiber]

Oh Im doing answers in decimals

[sand cradle]

That doesn't matter, if you switch to radian mode, it will give you -0.818 instead of -46.88

[neon fiber]

yeah I do get -0.818

Thank you for your patience, have a great day!

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[urban mason]

I need help in french

[unique valley]

i can help in polish if you want

[urban mason]

thx

[ionic elm]

I think this a channel related to maths rather than languagess???

[urban mason]

yes I meant math which in french

[ionic elm]

ohhh

kkk

[cedar kiln]

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[brazen topaz]

hello

[brazen topaz]

these questions are apperantly no calculator

but im unsure

i tryed to find limit from right and left

im just confused overall

nvm

.close

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[native spindle]

Good morning, I need help doing series and parallel circuits.

[whole yarrow]

i can try and help if youd like though im not great at circuits

[native spindle]

This isn't a test, just general worksheets buti can't seem to fully grasp it

Always appreciate any help. Worst case we can figure out what we're doing wrong together lol

[whole yarrow]

that looks a bit more complicated than what ive seen

i dont think i can help sorry

[native spindle]

That's okay, thank you anyways. I'm going to post the one with all the resistances listed to make the sample a littler easier, then can move on to this one.

Resistances are given. If I've done my calculations correctly the total resistance should be 39.673ohms

[whole yarrow]

do you know the correct answer?

or what it should be i mean

i think it should be 17.5

[native spindle]

I'd like to start there. To find this I took the parallel legs that are in series

(r7-5r, r8-15r, r9-20r: (40ohm)

r5-10r, r6-30r:(40ohm)

40^-1=40^-1 = 0.005
0.005^-1 = ((20r))

r4-10r. r3-30r

10^-1+30^-1 = 0.133
0.133^-1 ((7.5r))

r10-15r, r11-10r (25ohm)
r2-10r

25^-1+10^-1 = 0.14
0.14^-1 = ((7.143r))

r1-5r

Leading me to 39.643ohm

[native spindle]

Can you explain how you got to 17.5?

[whole yarrow]

i started out the same way you did but in the r4 and r3 one i added the equivalent resistor from the previous one (20ohms) to get (10 + 20)^-1 + 30^-1

then continued in a similar pattern down the circuit

[native spindle]

So, I would carry the 20 ohms into the next branch before continuing? Not add all after?

[whole yarrow]

yeah you have to add before continuing

the idea is that the parallel circuit can be replaced with one equivalent resistor placed with in series with r4

if that makes sense

[native spindle]

I get it now. It makes sense that way running from a parallel to a series you'd have to add in the resistance

Hmm, gonna work on this real fast and see what I come up with and go from there

[whole yarrow]

alright

[native spindle]

Oh yeah. 17.5, I think this calls for a celebratory stress break

That makes a lot of sense though as the current would flow regardless

[whole yarrow]

yeah

fr

[cedar kiln]

@native spindle Has your question been resolved?

[native spindle]

I'm going to come back to this after break but will be an hour, I appreciate the help so far! Is it alright if I bug you later with a message? @whole yarrow

[whole yarrow]

sure its no problem!

[native spindle]

Sweet, bother you later then 🙏

[cedar kiln]

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[untold sable]

could someone explain how i would find the phase portraits?

[untold sable]

this is the solution but idk how they got it

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[trim hull]

Guys it's bad

Idk if im actually stupid but I cannot do this for the life of me

[trim hull]

I need help😭

[royal loom]

, rotate

[dim birch]

saving that for later

[royal loom]

Which question are you stuck on, and what step

what have you tried

[trim hull]

Any where I can find a place where I can factor stuff online? A free factored firm calculatior?

[dim birch]

quadratic equation

[royal loom]

you should be able to factor just from seeing the graph in these cases, no need for a calculator

or the quadratic equation

[trim hull]

I just don't know how to factor and I've been trying to do it for 3 days and I've resorted to searching onlins

[royal loom]

and factoring is a skill you will need to learn

[trim hull]

I know, it's bad😬

[royal loom]

I have to write an algebra midterm but I can help you later, if no one else does in the time being

[dim birch]

https://www.youtube.com/watch?v=9MUDAr1PnlE

[royal loom]

Khan Academy I know has videos on factoring

or that

[trim hull]

I need help on #3

[royal loom]

What are the roots of number 3

[dim birch]

watch the video first and make sure you understand what you're doing, solving it without understanding won't get you anywhere

[trim hull]

What is meant by roots?

[royal loom]

Where is the function 0

[trim hull]

Thanks

[royal loom]

the x-intercepts

are the roots

[trim hull]

So I need a number that multiplies to 6 but adds to 15?

How

Factoring doesn't seem possible

[dim birch]

You should double check the function you came up with

$3.75x^{2}+15x+6$ doesn't look right

[wraith dagger]

Cyphercrit

[trim hull]

I did, after I did it I double checked with google

I tried factoring it

Got
Y=3.75x(x+4)+6

[dim birch]

The green one is the function you wrote down, the blue one is the one its supposed to be, right?

[cedar kiln]

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[tired inlet]

Could anyone help me out with this question?

[trim hull]

Yes

Thank you

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[ripe valve]

need help to understand why this is correct, i only got it when i plugged theta=0 (y=0), r will be 10 (x=10)

[ripe valve]

when i plug in theta = 5 (y=5), i get 5(1+cos5) which is some random number and not x=0

[slate lintel]

it doesn't seem like you understand how polar coordinates work

[ripe valve]

pretty much

[slate lintel]

theta is an angle from the positive x axis, and r is radius or distance away from the origin

[ripe valve]

i just know (x,y) is (r, theta)

[slate lintel]

it's not

[ripe valve]

dont i just plug in values for theta to output r?

[slate lintel]

i mean yes you do plug in values for theta to output r

but in terms of graphing it, it works entirely differently than the (x,y) graphs you're used to

so let's plot a few values

so if theta = 0, then as you said, r = 10

[ripe valve]

yeah so i translated that to (x,y)

(10,0)

which corresponds to graph 2 and which was the right answer

[slate lintel]

that... doesn't make any sense hold on

[subtle harbor]

https://tutorial.math.lamar.edu/classes/calcii/polarcoordinates.aspx here is a useful resource as well

[ripe valve]

but idk why other numbers i plug in doesnt output the same number

[slate lintel]

you need to re-learn polar coordinates from the start

you don't understand them at all

[subtle harbor]

(x,y)->(sqrt(x^2+y^2)=r,arctan(y/x)=theta)

[slate lintel]

we can do this here or you can watch videos or look at online resources

[solid juniper]

hayley table

[ripe valve]

oh so r =/= x and theta =/= y

and i just got lucky with (0,10)

[subtle harbor]

no

[slate lintel]

that's correct, r is not x and theta is not y

[subtle harbor]

do you know metric system v.s. imperial system

its like that

they use different numbers to measure the same thing

[ripe valve]

so i just plug theta and r into this to get the x and y

[subtle harbor]

yes

[ripe valve]

ok thanks i got this now

[subtle harbor]

think of it like this

[subtle harbor]

just like how 1 inch=2.54 cm, (2,pi/4) (polar)=(2cos(pi/4),2sin(pi/4)) (cartesian)

[slate lintel]

you can convert to x and y but i'd recommend staying in polar

and thinking about angles and radii

[ripe valve]

this is just an HW excercise in conversions

so i gotta do a lot of converting

[slate lintel]

looks like it's about graphing to me?

or does it go on to ask you to convert?

[slate lintel]

okay yeah you'll have to convert then

[slate lintel]

[solid juniper]

[cedar kiln]

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[viscid copper]

need some help understanding how this works.. copied from a review sheet but still do not understand (question 18)

[cedar kiln]

@viscid copper Has your question been resolved?

[viscid copper]

<@&286206848099549185> 🙏

[radiant scaffold]

do you know tan and cot graph?

[viscid copper]

unit circle?

[radiant scaffold]

the actual graph

[viscid copper]

i didnt learn that, no

[radiant scaffold]

ah

[viscid copper]

oh nevermind

ive seen that

[radiant scaffold]

yea

the red is tangent graph

see where it has asymptotes

[viscid copper]

yea

[radiant scaffold]

x cannot be at those points

[viscid copper]

what about sec and csc

[radiant scaffold]

1/cosx

the denominator cannot equal 0

so when cos =0, 1/cosx doesn’t have real values

[viscid copper]

that makes sense

[radiant scaffold]

same for cosec

[viscid copper]

alr ty

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[coral jewel]

Given the fourth degree polynomial $f(x)$ with the following graph. The area of the region confined by $f(x)$ and $f'(x)$ is $\frac{428}5$. Evaluate $\int_{-2}^1 f(x)\dd x$

[coral jewel]

ignore the blue handwriting, its me sketching f'(x)

not sure what to do here

[kindred venture]

notice you have 2 repeated roots

or 'double' roots

so we can actually find the function f(x)

[coral jewel]

uh no i cant

i would still need at the very least the maximum of f(x)

[kindred venture]

true

thats why they give you the 2nd piece of info

that the area ocnfined between f anf f' is 428/5

[coral jewel]

my experience with these type of problems is that they usually dont give enough information to find f(x), or finding f(x) is just unnecessarily long

[kindred venture]

so we can write f(x) = A(x-1)^2(x+2)^2

[kindred venture]

perhaps it's long. im not sure of another method

but there's only 1 unknown we need to calculate

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[crimson sedge]

i am lost for (i) i need a hint

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

someone knows??

[granite knoll]

You could ask in #discrete-math or #probability-statistics . Seems like a combinatorics problem

[crimson sedge]

ye it is

very very difficult

[granite knoll]

Yeah, ask in those channels

[cedar kiln]

@crimson sedge Has your question been resolved?

[orchid panther]

you can go about it this way: let's start by finding the number of surjections from an r element set {1,2,3,...,r} to a two element set, so essentially we are trying to find how many ways we can cut the set {1,2,3,4,...,r} into two different subsets I and J such that these two are disjoint and not empty, so wed pick how many subsets I are of length of 1 and J of length r - 1 (it would be r = C(r, 1)), and Card(I) = 2 & Card(J) = r - 1 ( it would be r(r-1)/2 = C(r,2)) ... and so on

so wed be summing over C(r, n) with n going from 1 to r-1, we are not counting n = 0 and n = r in the sum because we want to keep I and J non empty

the sum would yield 2^r - 2

now one last step

the two boxes are identical

so since the problem is symmetric ( you can find Card(I) = k & Card(J) = r - k, the same way you can find Card(I) = r - k & Card(J) = k ), we can just the divide the sum by two and get what they asked for

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[uneven crest]

We induct on the number of Klein bottles considered in the connected sum. When there is but one bottle

I follow Munkre's drawing for the CW-complex representation of the connected sum for the two tori, even if I use the ordering of the edges in the Klein bottle (instead of the torus) and alternate the orientation of the edge upon which we are performing the attachment. How do I conclude the induction for a general n?

should i use Van Kampen's theorem to prove my induction instead?

<@&286206848099549185>

[crimson sedge]

wait 15m

please

[uneven crest]

oh sry

[brittle terrace]

probably would take longer to find a helper for more difficult problems

[rough terrace]

ye my brain is too small to handle that

unfortunately

[slate lintel]

i'm confused. why do you have pre-university math?

[rough terrace]

lol

wait what

???

bro is trippin

[uneven crest]

sry i have my ra with me hes helping me out

[slate lintel]

see #get-advanced-access to get access to channels for undergraduate+ math like this

[cedar kiln]

@uneven crest Has your question been resolved?

[sacred grail]

the g-fold connected sum of K with itself is the non-orientable surface of genus 2g

[uneven crest]

ohhh the orientable double cover must have twice the Euler characteristic and this pins down the homeomorphism type of the cover.

got it

i think thats enough

for me so far

.close

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[unique crane]

how do i find the derivative of $e^y$ with respect to x?

[wraith dagger]

guy

[unique crane]

would that just be $e^y*\frac{dy}{dx}$?

[wraith dagger]

guy

[rough terrace]

It would be 0

[long swan]

No

y is a function of x

[rough terrace]

If you consider its partial derivative

Is it a partial tho

[unique crane]

i dont know what a partial is haha

[rough terrace]

So basically we are treating e^y as a constant

Since we are differentiating with respect to x.

[unique crane]

how do you know y is a constant though

im assuming its not because in every other problem ive done we havent treated it as one

[rough terrace]

Since we are differentiating with respect to x any other variables would be treated as a "constant"

d/dx (e^y(z))=0

For example something like this

[unique crane]

this is the equation, and i was trying to use product rule for the first part of it and couldnt figure out e^y

is that a partial?

[rough terrace]

Right. No its not a partial then

Use implicit differentiation

[unique crane]

ok i see

so we treat y as a constant then?

[rough terrace]

I kept thinking of taking its partial lol

I mean taking partials on both sides would result in e^y(e^x)-sin(x)=0

[unique crane]

oh okay lol

alright thanks

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[lean blaze]

can i get help with some understanding on a question here

[rough terrace]

Sure

[lean blaze]

i managed to get through a bit but i still need help on these last few

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[covert gorge]

Can I get help on this!?? my teacher said its wrong And i dont understand and he wont help. This is what he said...

"You almost had it on the second part. The gazebo diamond shape has 4 coordinates - one for each corner, make sure to convert each of these. Also, you have forgotten to add a new axis with the origin at the x point. You need to do all your rotations from that point so therefore, you need to convert each coordinate of the gazebo to new coordinates based on the new axis with the origin at the x point. You can add the new x and y axis by using the line tool in canva."

[cedar kiln]

@covert gorge Has your question been resolved?

[cedar kiln]

@covert gorge Has your question been resolved?

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@covert gorge Has your question been resolved?

[covert gorge]

😢

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[hardy obsidian]

Convert the following to a polar equation:
x^2 - y^2 = x

The answer key given by my teacher states its r = cosx/cos2x. However no matter what I do I keep getting r = cos2x/cosx.
Thanks.

[kindred venture]

can we see your work?

[hardy obsidian]

yeah getting the ss from my ipad one sec :)

[kindred venture]

how did you cancel those r^2 ?

[hardy obsidian]

r^2-r^2 is what i thought it was unless its not

[kindred venture]

4x - 2x != 4 - 2

that's the logic that's going on here

just keep looking at that 2nd to 3rd line and you will see the problem

[hardy obsidian]

i just thought that r^2-r^2 was 0 cause thats what google said 😭

[kindred venture]

that's not the LHS tho

its r^2cos^2 - r^sin^2

two different things

[hardy obsidian]

ah

i see

wait so would it be r^2-r^2(cos2x)?

[kindred venture]

$r^2\cos^2(\theta)-r^2\sin^2(\theta) = r^2(\cos^2(\theta)-\sin^2(\theta))=r^2\cos(2\theta)$

[wraith dagger]

Obotron

[kindred venture]

it's just factoring out the r^2

[hardy obsidian]

oh my god thank you so much

been stuck on this for a minute and google was not helping

[kindred venture]

no worries

[hardy obsidian]

appreciate you have a good day bro 🙏

[kindred venture]

you too!

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[bleak harbor]

is this correct

[cedar kiln]

@bleak harbor Has your question been resolved?

[cedar kiln]

@bleak harbor Has your question been resolved?

[autumn heath]

What is t? Is it some parameter?

[bleak harbor]

yeah

I just wanted to check the first expression matched the final one

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[valid fiber]

i dont understand these solutions why ahve they done aoc acute and obtuse?

[deft thunder]

i might be because angle at the center is double the angle at the circumference

[valid fiber]

ohh wait i seee ty

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[humble sonnet]

Hello, i need help with this question and i have no idea where to start. any help is very much appreciated. Thank you.

[dim birch]

a ≅ b^2, a ≅ sqrt(c), hence sqrt(c) ≅ b^2

[humble sonnet]

i have tried something like that but that was not correct

[dim birch]

can you send your work?

[humble sonnet]

okay give me a second

[dim birch]

$\frac{\sqrt{162}}{9} = \frac{\sqrt{c}}{25}$

i believe

[wraith dagger]

Cyphercrit

[lyric widget]

a = m1 b^2 = m2 sqrt(c)
<=> (m2/m1)sqrt(c) = b^2 (1)

From the information

9m1 = m2 sqrt(162)
<=> m2/m1 = 9/sqrt(162)

Plug that into (1) and you can solve for c

[humble sonnet]

for that i got 1250 which was correct

thank you very much for the help. very much aprriciated

[dim birch]

no problem

just remember that this is basically just ratios

[humble sonnet]

okay will keep in mind!

once again thank you so much, have a nice day

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[honest torrent]

help with extended euclidean algorithm

[honest torrent]

whats the purpose of the algorithm?

[deft thunder]

The Extended Euclidean Algorithm is an extension of the Euclidean Algorithm that not only finds the greatest common divisor (GCD) of two integers but also finds the coefficients of Bézout's identity.

[honest torrent]

can you help me with finding the coefficients?

i dont know a way to find the numbers after a, b

[gritty viper]

how did you get 12

[gritty viper]

.

[cedar kiln]

@honest torrent Has your question been resolved?

[honest torrent]

300 mod 18

then 18 mod 12 for 6

and 6 gcd because last non zero remainder

[gritty viper]

300 mod 18 hmm

It's better to think of it a different way

Find the largest multiple of 18 smaller than 300 and subtract it

300 - 18*16

,calc 300-18*16

[wraith dagger]

Result:

12

[honest torrent]

i mean the row of numbers is for finding gcd

[gritty viper]

So then, if 300 is x and 18 is y, how can we write 12 in terms of x and y

[honest torrent]

thats what i need help with

i found a pattern to bring the prevoius y to become that term x, and then solve it like an equation

[gritty viper]

How did I find 12

How would you calculate 300 mod 18 without a calculator

[honest torrent]

i would do 18 times something that wont get me over 300

and then i will get the remainder

[gritty viper]

Right

18*16 as I said

300 - 18*16 = 12

[honest torrent]

yup

ok i get that

[gritty viper]

And 300 is x, and 18 is y

[honest torrent]

oooooo

so

x is 1

y is 16?

[gritty viper]

12 is x - 16y

so the x row you put a 1, and the y row you put a -16

those aren't the values for x and y though, those are the multipliers for x and y to get 18

The 18 column is just saying 12 = x-16y

[honest torrent]

i thought i would have to express 12 like 12 = 300x + 18y

[gritty viper]

ok ig that's also true but that makes it harder to see

[honest torrent]

oh it works

x = 1, y = -16

but would i just switch the signs?

[gritty viper]

no

300(1)+18(-16)

Remember that's how we got 12 in the first place

[honest torrent]

but we did 300 - 18(16)

[gritty viper]

yeah exactly

300 is the same as 300(1)

[honest torrent]

how would i do 6?

[gritty viper]

how would you find 6 (without using the word mod)

[honest torrent]

18 - 12(1) = 6

[gritty viper]

Right

and 12 is 300-16*18

So

6 = 18 - (300 - 16*18)

[honest torrent]

how would i express that in terms of x and y

[gritty viper]

well going with your system

distribute it out

6 = 18(17) + 300(-1)

[cedar kiln]

@honest torrent Has your question been resolved?

[honest torrent]

where did 17 come from?

[gritty viper]

Distribute the subtraction

[honest torrent]

yeah once i solve it i just get 6 = 18 - 12

how did you get 18(17) and 300(-1)

[honest torrent]

?

[gritty viper]

Don't get rid of the 300 and 18

Only distribute the subtraction

[honest torrent]

6 = 18-300+16*18

?

[gritty viper]

Yes

Now combine like terms

[honest torrent]

dont you ahve to multpliy first

[gritty viper]

Remember, we want to treat the 18 and 300 like variables, lest we lose track of what combination of 18 and 300 we have

Right now we have 6 = 18-300+16*18

[honest torrent]

oh ok

[gritty viper]

If we simplify the right side too far, we will just get 6=6 which is useless

[honest torrent]

yeah

idk how to get 17

[honest torrent]

.

[gritty viper]

You have 18

And you have 16*18

And you're adding them together

and you're supposed to treat 18 like a variable

[cedar kiln]

@honest torrent Has your question been resolved?

[cedar kiln]

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[north stirrup]

I REALLY need help with some cram calculus study guide before my final

[north stirrup]

My professor gave me an old exam he used a few years ago and i realize i know nothing on it

nvm i realized how, im an idiot

.close

[cedar kiln]

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[cedar kiln]

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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.
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[obtuse coral]

How do I calculate this?

[cedar kiln]

@obtuse coral Has your question been resolved?

[obtuse coral]

Apparently’s the answer’s 23.1, but I have no idea how

What’s the expected value?

[granite mantle]

Bro is this AP bio 2020 mcq practice 1

[obtuse coral]

Yeah lol

[granite mantle]

I do not know how that shit worked

I didn’t get the question right answer

Either*

[obtuse coral]

It literally doesn’t even explain it

[granite mantle]

All g

[obtuse coral]

Makes no sense lol

I have no idea how it’s 23.1

[granite mantle]

Yeah I also have no clue

[obtuse coral]

I feel like the expected value’s smth other than 100

But idk what

[cedar kiln]

@obtuse coral Has your question been resolved?

[cedar kiln]

@obtuse coral Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[wheat rain]

another silly question
what is (2/5) to the -2 power
and why is it not 1?

[granite knoll]

why do you think it should be 1?

[prisma gull]

did you mean $\left(\frac{2}{5}\right)^{-2}$?

[wheat rain]

yeah

[wraith dagger]

.doc

[prisma gull]

^

[wheat rain]

idk cuz my friend says it should be one because
negative powers is just dividing and 2/5 divided by 2/5 is 1

[granite knoll]

your friend is wrong

[dusk goblet]

i think what your friend means is that subtracting one from the exponent divides by the base

but if you start at one which is (2/5)^0

to get to -2 you subtract one twice

or divide by (2/5) twice

[wheat rain]

sorry what

[dusk goblet]

which part was confusing

[wheat rain]

umm all of it

[dusk goblet]

ok so i’d start by thinking about the patterns with exponents to understand the exponent rules

4^3 is 64

4^2 is 16

4^1 is 4

4^0 is 1

notice

[wheat rain]

yeah

[dusk goblet]

each time the power decreases by 1

[wheat rain]

yeah

[dusk goblet]

it divides by base

which is 4

same goes for increasing the exponent by 1 you are just multiplying by the base

so if you’re at 4^0

and you want to find out 4^-1 you’d take 4^0 which is 1 and divide by 4

then to get to 4^-2 you divide by 4 again to get 1/4 divided by 4 which is 1/16

and you can obviously tell this is the reciprocals of the powers of 4

[wheat rain]

yeah

[dusk goblet]

so what your friend was saying was that when you subtract from the exponent all you’re doing is dividing what you had originally by the base however many times

in this case

(2/5)^-2 is like taking (2/5)^0 then dividing by the base which is 2/5 twice because to get to -2 you subtract 1 two times

but if you know your exponent rules it can be much simpler and less deep

if the exponent is negative you can rewrite the base as the reciprocal of it

so (2/5)^-x is just (5/2)^x

[wheat rain]

oh

wait why does that work

[dusk goblet]

because dividing by a number is the same thing as multiplying by its reciprocal

similarly multiplying by a number is the same as dividing by its reciprocal

[wheat rain]

that makes sense

[dusk goblet]

i like to think of it with real cases

like with 4

if you had

4^- anything

it’s really 1/4^anything

because you keep dividing by 4 when you decrease the exponent by 1

but this can also be seen as multiplying by 1/4

so (4)^-2 is just 1/16 but that’s just (1/4)^2

hope that makes sense

[wheat rain]

that actually does

why cant you be my math teacher

she sucks

[dusk goblet]

yea i know most math teachers i’ve encountered haven’t been great lol it always seems like they don’t understand why what they teach works they just know enough to tell you what to do

you should always try to get an intuitive approach

[tropic jay]

Realest stuff, they teach cuz they have no other option

[dusk goblet]

especially high school

or anything lower than that as well

[tropic jay]

yep

fundamentals are the most important

they ruin it

[wheat rain]

wait i had another question but it evaporated and i have to go find it

[dusk goblet]

go ahead

[wheat rain]

Please simplify the following:
3f/3f^7

is it not 3f^-6

[dusk goblet]

do you mean

(3f)/(3f^7)

if you mean this then the 3s would cancel

and you’d get f^-6

which is 1/f^6

[wheat rain]

oh

[cedar kiln]

@wheat rain Has your question been resolved?

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• 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>.
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[crimson sedge]

I really need help with this, I don't really understand how to do this question

[crimson sedge]

I've been trying to figure it out but it's just confusing me

<@&286206848099549185>

[crimson sedge]

hello :D

[tough dock]

It is asking you to find two different functions

One that outputs the longitude and one that outputs the latitude

[crimson sedge]

I'm confused on how to go about doing that

[tough dock]

You can use a graphing calculator to do this

You need to input the values that you have in your table into your graphing calculator

You do not need to do both latitude and longitude at once; you can do each separately

[crimson sedge]

I put in all the stuff from the tables on desmos and got a graph, but besides that I don't really have a function

[tough dock]

That is where equation fitting comes in

What kind of graph do you think would best fit the points?

[crimson sedge]

well this is the graph that I have rn for both of the points

[tough dock]

It is a bit hard to work with this kind of scale

You can fix this by changing the scale values in the top right

This will give you a more clear shape for your data points

[tough dock]

And also the bounds of your graph

[crimson sedge]

wdym

[tough dock]

These

We do this so our data takes up more of our graph, and we can see the "shape" of the data points better

[tough dock]

Yes

You can change the values of those inequalities

To "zoom in" on the data points

Your time values go from 0 to 4

So I would recommend choosing a range that is just slightly bigger than 0 <= x <= 4

[crimson sedge]

Idk what that is xd

all i can do is show what the graph shows

[tough dock]

This

Change these numbers

[crimson sedge]

I know but like idk how to change it properly

i'm not so much confused about the graph i'm confused on how to make the equation

[tough dock]

it is easier to see the kind of equation you need to make if you can change the scale

[crimson sedge]

but even if i see the graph with a certain scale i still won't know what type of equation i need

[tough dock]

You will

[crimson sedge]

changing how i see the graph doesn't magically make me know something that i don't have knowledge of

ok then tell me what to change the numbers to

[tough dock]

The shape will become more apparent

Try setting the x-axis bounds to -1<=x<=5

[crimson sedge]

but i'm telling you that i don't know what shape needs what equation

[tough dock]

And for your longitudes, your values range from -83.3 to -79.4, so an appropriate range would be something like -84 <= y <= -79

This will hide your latitude values, but we can deal with those later

[tough dock]

Okay

So now that we can see the shape of the data points more clearly

[crimson sedge]

yes

i still don't know what equation to use

[tough dock]

We can go back to the document and see that they give you the shapes of exponential, quadratic, and linear

Linear is just a line, which doesn't fit your shape too well

An exponential looks like this, which isnt too great either

[crimson sedge]

so it's quadratic?

[tough dock]

A quadratic I think would best fit your data here

yes

[crimson sedge]

ok

[tough dock]

To fit data points in Desmos, you can use ~

The general form for a quadratic is ax²+bx+c, so we can use this to fit our points

[crimson sedge]

didn't do anything in desmos

[tough dock]

To tell Desmos to fit our points, you can do something like this

[crimson sedge]

nvm

[tough dock]

Since the x coordinates are x1 and the y coordinates are y1

[tough dock]

That's strange

Oh

I see

Sorry

You have your points labeled slightly differently than what I expected

[crimson sedge]

this is the x1

[tough dock]

Your x-values are just named t