help-26

[ivory wave]

well the word problems im doing is all about quadratic functions

it's algebra 2

[vale furnace]

Im not familiar with American system

But if you want to maximise the area then

[ivory wave]

it's canadian question

i mean canada math is just lame

also what's derivative? I think this belongs in calc right?

[acoustic pecan]

if they get a parabola by the substitution they could find the vertex of A(x) or A(y)

[vale furnace]

Basically the rate of change of a function

Oh yeah

[ivory wave]

yup never learned that yet

[ivory wave]

sorry im so lost rn

[vale furnace]

Find the function of area in terms of x or y

A = xy sub in x=600-2y say

[ivory wave]

[vale furnace]

[ivory wave]

so umm

A(x) = (600-2y)(y) and A(y)=(600-2x)

right?

[vale furnace]

You only need one

[ivory wave]

ok

[vale furnace]

Okay let’s say you get A(y) = (600-2y)y

So A(y) = 600y -2y^2

Yeah?

[ivory wave]

yeah

so now start with vertex form?

[vale furnace]

It’s a negative parabola so if you find the vertex that’s the maximum

[ivory wave]

okay

got it

thanks

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[cedar sinew]

competition math

anyone want to help?

i can start with the fact that with a quadruplet of 4 points on the octagon, there is one pair of diagonals with an internal intersection

so there are 70 internal intersections, but are not distinct

i want to remove all the double/mutiple counts

obviously the center is one

there are 6 possibilities to remove

so i am thinking it is 49 just from the fact that we alr have at most 64 as the answer

but im not sure

ok no yes 49 is correct but i need to know how

wait i just realized that doesnt work 😅

so i just guessed it correct

wait no im confused

there are 4 directly opposite diagonals

4 choose 2 yields 6 possible diagonal intersections that we overcount

so we must subtract 5

to get 65

now we have to show that there exists at least one double count to arrive at 49

this would mean that we have three concurrent lines

other than the longest diagonal

the diagonal cannot skip over only one point

beacuse then you cannot have a triplet of concurrent lines

[strong citrus]

well you did guess it

[cedar sinew]

yes

but i need to know

why

[strong citrus]

but that's kinda how I would approach it

[cedar sinew]

what happens if there was another option?

[strong citrus]

we have 4 diagonals intersecting at the center right

[cedar sinew]

like 52?

yes

that gives 6 intersection points that we count as the same

so we subtract 5

you understand how i got 70 for the total right?

[strong citrus]

70 is the maximum, we subtract 5

we get 65

[cedar sinew]

yes

but that is wrong

there are more

double counts

[strong citrus]

there will be more double counts

and

there is only one option that's less than our number, which is 65

being 40

[cedar sinew]

49*

[strong citrus]

49 mb

[cedar sinew]

but how do you know there are more double counts

[strong citrus]

try to picture it

or

well

[cedar sinew]

i drew a picture

[strong citrus]

drawing is the most clear option

[cedar sinew]

obviously one of the three diagonals cannot skip over only one point, right?

and we cannot have two longest diagonals right?

the longest diagonal being the one that goes directly oppostie

when i say that a diagonal skips over a point i just mean that it connects two points with a single point between them

in the picture above the diagonals skip two points

if we have two D_3 (diagonal that skips three points) then we intersect in the middle whihc we alr counted

and if we have 0 D_3 then we must have 3 D_2 but as pictured above that doesnt work

so then there is 1 D_3 and 2 D_2

wait

OHHHH IM SO STUPID

[strong citrus]

lol i was trying to draw it up for you

[cedar sinew]

i can just reflect one D_2 diagonal over a D_3 diagonal to yield another D_2 diagonal that together forms a triple of concurrent lnies

either consec D_2's or not

so we do have more double counts

but how do i get to 49

well lets see there are 4 potential lines of symmetry

and then there are two possibilites for each of them

so i get 8 triples

each triplet has three intersections

i subtract two as to now overcount

so i do

65 - 8 * 2

which is 49

yayyy

woohoo

i did it

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[cedar sinew]

thanks for your help @strong citrus

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[hardy steppe]

need someone to check my work on related rates

[hardy steppe]

didnt simplify eyt

yet

[topaz sinew]

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[hardy steppe]

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[dim yew]

Hello can somebody help me solve this, or walk me through it

[pearl peak]

have you encountered the quadratic formula

[dim yew]

yes but do you know nay other ways i can solve it

[pearl peak]

think that way is the best

as there are irrational coefficients

[torpid matrix]

you wont be able to factor it in any other way, as ||it has no solutions||

[dim yew]

I had a problem on my quiz with irrational coefficients and use the quadratic formula on it when we had 3 minutes per problem and ended up getting √5000 something. (I failed the quiz)

[dim yew]

💀

[torpid matrix]

you can do the discriminant test to determine it has no roots though

[hollow drum]

No real solutions*

[torpid matrix]

if you do the full quadratic formula you just end up with imaginary solutions

[torpid matrix]

true, but I assume someone asking this question isn't at a level of math such that they are expected to find imaginary solutions

[dim yew]

oh ok thanks

[torpid matrix]

regardless, note that $\sqrt{b^2-4ac} < 0$ and therefore there are no real* solutions

[dim yew]

ok :>

thanks

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[sonic bone]

help🥲

[sonic bone]

just need guide through question

[radiant tapir]

so the first thing first, what is the period of sin(x)?

[sonic bone]

is it 8pi

[radiant tapir]

well I meant in general, but yes for your fucntion it's 8 pi lol

[sonic bone]

o

[radiant tapir]

what is the period of a normal sine function?

[sonic bone]

is it 2pi

[radiant tapir]

bingo

so we need to see how we translate for 2pi to 8pi

we can do that with the following: $2 \pi = x8 \pi$. Solve for x. That'll give you your scalar for your sin function

no wait

[sonic bone]

4?

[thorny flame]

MellowDramaLlama

[radiant tapir]

sorry I had it backward lmao

[sonic bone]

so(x)*8pi=2pi?

is that what im figuring out

[radiant tapir]

yeah so solve for x just like you did before

[sonic bone]

1/4?

[radiant tapir]

bingo!

hey would you look at that? 😮

since you didn't have to consider vertical stretch, vertical translation, or horizontal translation, you're good to go 🙂

[sonic bone]

so for this one

2pi = x2pi/3?

[radiant tapir]

yep!

so what would x be?

[sonic bone]

is it 3

[radiant tapir]

bingo

see? nothing to it

[sonic bone]

ohh okay this isnt that bad

[radiant tapir]

where it gets trickier is when you have vert/hor translations, vert/hor stretches all at the same time haha

[sonic bone]

lemme do this rq and see if i need help again

[radiant tapir]

sounds good

[sonic bone]

i understand the period of this is 2pi

right

this wavelength is built a lil different tho

it doesnt go through the middle perfectly like the other

[radiant tapir]

yeah so this one is a horizontal translation

ummmm

I guess you can take your best guess? I hate when they do that to you though

it looks like y = 0.75, would you agree?

[sonic bone]

watchu mean

[radiant tapir]

it's not crossing at y = 1, it looks a little bit less

but not as small as y = 1/2

so it's somehwere between that

best guess I would say is that it's crosses the y axis at y = 0.75

(or 3/4)

[sonic bone]

are u talking about like here when it first goes through

[radiant tapir]

I'm talk about where it crosses the y axis, aka the horizontal translation

[sonic bone]

?

[radiant tapir]

yeah exactly

I'm arguing that since it's not a nice number and we can't see the value we have to guess

which sucks

I hate when they make you do that

buuut I'm saying it's not 1, it's not 1/2, it's somewehre in between

[sonic bone]

yeah i gotchu now

[radiant tapir]

so I was saying my best guess would be 3/4

[sonic bone]

so im trying to transform this to the like orginal way sine is?

[radiant tapir]

omg I"m in an idiot I"m so sorry

I was looking at the graph like a vertical translation 😦

horizontal translation would be somthing like y = sin(x - d)

so we have to find d

[sonic bone]

how i do that

[radiant tapir]

where does sin(x) normally cross the x axis?

[sonic bone]

0?

[radiant tapir]

sorry okay after that

how about from 0 < x < 2pi ?

[sonic bone]

pi/2?

[radiant tapir]

nope

pi

so we can see in your graph that it crosses the x axis at 3pi/4, which means we need to translate the graph over pi/4 units to the left

so how can we do that?

[sonic bone]

-pi/2?

[radiant tapir]

no too much

we only need to shift by pi/4

ok let's take something more simple

something like y = 2x + 4

how would we shift it 1 unit to the right?

[sonic bone]

-1?

+1

[radiant tapir]

-1 where though in the equation?

[radiant tapir]

good start 🙂

it'd be y = (2x - 1) + 4

[sonic bone]

oh ok

[radiant tapir]

every function, no matter what, will abide by these properties

so if you want to shift something to the left, you add a value into where x is (aka within the parentheses). If you want to shift something to the right, you subtract a value similarly

so we're shifting to the left by pi/4 units, so what should we do?

we have y = sin(x - h) here

Here's another chart if it helps

[sonic bone]

so im moving my graph to the right?

so its gone be y = sin(x-d) like u said?

[radiant tapir]

in this case to the left

it's crossing the x axis at x = 3pi/4, which is less than pi

[sonic bone]

so we going this way?

[radiant tapir]

yes correct

[sonic bone]

so its gone be sin(x+d)

[radiant tapir]

yep!

so what's d?

[sonic bone]

pi/2?

[radiant tapir]

nope

normally it crosses the x axis over at pi, we're not crossing it over by 3pi/4,
so what's pi - 3pi/4?

[sonic bone]

pi/4

?

[radiant tapir]

bingo

[sonic bone]

oh i said that before i edited it

[radiant tapir]

so what's the equation? 🙂

[sonic bone]

sin(x+pi/4)

how do i find the cosine one now🥲

[radiant tapir]

there's a general rule that `sin(x) = cos(x - pi/2)

so instead you'll just do cos(x - pi/2 + pi/4) and simplify

[sonic bone]

3pi/4?

[radiant tapir]

postiive or negative?

[sonic bone]

postive?

[radiant tapir]

,w pi/2 - pi/4

[thorny flame]

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[radiant tapir]

it's positive pi/4

double check your work

did you just guess or did you work that out?

[sonic bone]

i thought we were adding pi/2+pi/4

[radiant tapir]

it's -pi/2 + pi/4

[sonic bone]

ohh

i see what i did

[sonic bone]

postive?

[radiant tapir]

there ya go, that's how to find the cos equiv 🙂

[sonic bone]

i got negative pi/4

[radiant tapir]

youre right my b

it should be cos(x - pi/4)

[topaz sinew]

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[spiral yarrow]

taking ap calculus and this is supposed to be relatively easy but i’m not sure on what to do after i take the derative of the original problem😭

[radiant tapir]

do you recall what happens when we have a horizontal tangent? What does the slope equal?

[spiral yarrow]

slope = derivative

that’s all i know

and OHH

DO WE EQUAL IT TO 0?

we dooo right

but idk what after lol😭🙈

[radiant tapir]

you're on the right track

so let's take problem 12

so we set f'(x) = 0

so what do we have there?

[spiral yarrow]

uhh

u want me to

replace all the x’s w 0?

[radiant tapir]

no

that'd be f'(0)

[spiral yarrow]

oh yeah

u can’t take the derivative of 0

cuz that’s just like. nothing right

[radiant tapir]

well you can, it's just 0

[spiral yarrow]

yeah

[radiant tapir]

no what I mean is this

0 = 2x^3 + 3x^2 + x

[spiral yarrow]

okayy hmmmm

[radiant tapir]

like literally set f'(x) = 0 = 2x^3 + 3x^2 + x

now we have to solve for x

so we have this:
0 = 2x^3 + 3x^2 + x

what's something we can do to simplify this?

[spiral yarrow]

factor?

[radiant tapir]

very good

factor what?

[spiral yarrow]

an x

or can we do more

[radiant tapir]

yeah that's what I was implying 🙂

so what do we have?

[spiral yarrow]

if i factor out an X?

we have

$$x(2x^2 + 3x+ 1)$$

[thorny flame]

eve:P

[spiral yarrow]

waittt

$$x^2(2x+3+1)$$

[thorny flame]

eve:P

[spiral yarrow]

nvm that wouldn’t work cuz of the one

[radiant tapir]

This is correct

Finally, 2x^2 + 3x + 1 looks pretty factorable to me

so far we have 0 = x(2x^2 + 3x + 1)

[spiral yarrow]

it does? lemme think

yes

[radiant tapir]

coolio, waht did you get?

[spiral yarrow]

(2x+1)(x+1)

[radiant tapir]

yep looks good to me!

so we have x(2x + 1)(x + 1) = 0. What are our solutions?

[spiral yarrow]

if@is

x = -1 and

x = -1/2

[radiant tapir]

there's one more 🙂

[spiral yarrow]

howww there is??

andd uhh

[spiral yarrow]

that’s all to me

Oh

[radiant tapir]

what about this little fella?

[spiral yarrow]

UHHHHH x= 0?

[radiant tapir]

bingo

[spiral yarrow]

dangggt

[radiant tapir]

oh hey look at that

[spiral yarrow]

that was it and i was stressing lol

[radiant tapir]

fire

[spiral yarrow]

ooooo i didn’t even know that could happen

[radiant tapir]

exactly what we expected

[spiral yarrow]

the way i alr got one horizontal tangent randomly 😭

just not in the right way cuz i did twice the derivative .. but hey i still got x= -1/2

[radiant tapir]

well what you found there was an inflection point

[spiral yarrow]

i just didn’t get the other too

Ohhh

[radiant tapir]

that's the 2nd derivative

[spiral yarrow]

well

[radiant tapir]

wait

hol don

[spiral yarrow]

thank you !!

oop

okay

[radiant tapir]

oh that's the 3rd derivative lol

[spiral yarrow]

omg it was

what was i even doing i couldn’t even fully do the first derivative

😭😭🙈

anywaysss thank u sm for ur help :)

[radiant tapir]

yeah no problem. Also what would help is notation:
f'(x) = 1st derivative
f''(x) = 2nd derivative
f'''(x) = 3rd derivative

write the same number of ticks for which derivative you're taking

[spiral yarrow]

how far does it go Omg😭😭 i’m scared for when that comes up later

[radiant tapir]

anything past the 4th derivative you just write the number in parentheses

[spiral yarrow]

ahhh

[radiant tapir]

for example like if you had the 5th derivative it'd be f^(5)

Her'es a simple example

[spiral yarrow]

so f’(5)

okk

[radiant tapir]

$f(x) = x^6\\f'(x) = 6x^5\\f''(x)= 30x^4\\f'''(x) = 120x^3\\f''''(x) = 360x^2\\ f^{(5)} = 720x \\ f^{(6)} = 720$

[thorny flame]

MellowDramaLlama

[radiant tapir]

after about 3 to 4 ticks you can switch notation

[spiral yarrow]

OHHH

that is insane but also simple at the same time??

[radiant tapir]

you can technically use the (a) notation for any of them, but by convention the ticks are easier

[radiant tapir]

well when you get into oscillating functions like sin and cos it gets crazy

for example

so we know the following facts:

$f(x) = sin(x)\f'(x) = cos(x)\f''(x) = -sin(x)\f'''(x) = -cos(x)\f^{(4)} = sin(x)$

[thorny flame]

MellowDramaLlama

[radiant tapir]

make sense, right?

[spiral yarrow]

uhh

yes

i did that in class

[radiant tapir]

or did you not know this yet ? 😛

ok bet

[spiral yarrow]

LOLL NOO

i got that

[radiant tapir]

ok cool

well judging by this pattern

[spiral yarrow]

well until the 2 line yes

no i mean 3

[radiant tapir]

what would the 5th derivative be? At the 4th derivative we come back to the original function

[spiral yarrow]

so it would be

[radiant tapir]

ah ok

[spiral yarrow]

$f^(5) = cos (x)$

[thorny flame]

eve:P

[radiant tapir]

Bingo!

[spiral yarrow]

OHHHHH

SO ITS A FR PATTERN

[radiant tapir]

it is

want to know something even crazier? 😮

[spiral yarrow]

that doesn’t seem too bad but i don’t wanna jinx it

yes

[radiant tapir]

off the top of your head, how would you know something like the 385th derivative of sin(x)?

[spiral yarrow]

ummmm i wouldn’t know😭 maybe knowing it’s divisible by 5 it wojld make it cos (x) ?? idk

[radiant tapir]

oooh you are so close

[spiral yarrow]

AGRGAGDH

add them all up?

[radiant tapir]

it correlates to the remainder when you divide the derivative number and divide it by 4

so what is the remainder of 385/4?

[spiral yarrow]

96.25

[radiant tapir]

no sorry just the remainder

[spiral yarrow]

ohh

so um

.25?

I don’t even know what remainder means tbh☠️☠️

[radiant tapir]

not quite, remember a reaminder has to be a either a postive integer or a 0

[spiral yarrow]

ohh

0

uhhhh

[radiant tapir]

like this

here the remainder is 5

[spiral yarrow]

OHHH

[radiant tapir]

it does correlate to .25 however 🙂

[spiral yarrow]

it’s 1/4? uhh I’m sorry I don’t know what the remainder is😔

[radiant tapir]

it's the circle thing when you do long division

[spiral yarrow]

ohhhh

96

OK umm

i’m just guessing rn cuz like like

[radiant tapir]

lol ok

hold on

[spiral yarrow]

i’m pretending to get it kinda but long division left my head ages ago

😭😓😅☠️whoops

[radiant tapir]

it comes back around to you so I would suggest refreshing on it

ok Imma try to manually do this real quick

[spiral yarrow]

omg what

[spiral yarrow]

okay no rush you can take your time

[spiral yarrow]

is geometry gonna come back too cuz

😗😗…

[radiant tapir]

lol oh yes

integrals are basically a geomtetric proof, but it's not too bad

if you don't go past calculus then you probably won't use it too much

but in advanced math it rears its head

ok there we go

[spiral yarrow]

😔😔😔i don’t know a spec of geometry

ooo okk

so the remainder would be one?

[radiant tapir]

so you see how our remainder is 1?

[spiral yarrow]

yes

[radiant tapir]

yep exactly

also notice that 1/4 = 0.25

like you said earlier

[spiral yarrow]

yes

Is that wrong?

[radiant tapir]

so that's how that's related

it's not wrong

but 0.25 is not teh remainder

1 is

[spiral yarrow]

Oh, cause we’re dividing it by four and the remainder was one

ohhhh

[radiant tapir]

bingo

[spiral yarrow]

But how does that relate to trigonometry and cosine?

[radiant tapir]

the reason we divide it by 4 is because 4's remainders can only be 0, 1, 2, 3

it just happens that the derivative correlates to the remainder

get a remainder of 1? It's the first derivative

remainder of 2? 2nd derivative

remainder of 3? 3rd derivative

[spiral yarrow]

ohhhhhhhh

[radiant tapir]

remainder of 0? original fucntion

(which is also the 4th derivative

[spiral yarrow]

so it’s sin x?

[radiant tapir]

close

it'd be cos(x)

[spiral yarrow]

ARGH

how

[radiant tapir]

our original funtion was sin(x)

[spiral yarrow]

You said it’s the 1st derivative

ohh

IHHH

OH

Lightbulbs are flickering in my mind right now

[radiant tapir]

excellent, that's why I wanted to take the extra time 🙂

ok so one more example

[spiral yarrow]

okiiii

[radiant tapir]

let's say the 571st derivative

we get a remainder of 3

[spiral yarrow]

yes

[radiant tapir]

so what's our derivative?

[spiral yarrow]

-cos x

I had to school all the way up to see your picture for that, but I still got the answer right .., :) ?

[radiant tapir]

yep exactly!

see?

super cool right

BUT

this only works if it's the original parent function sin(x) or cos(x) or if it has a scalar like 2sin(x)

but if you were to have something like sin(2x), then you can't use this trick

[spiral yarrow]

yes but that means I have to find out the remainder through long division every time, right

[radiant tapir]

yeah but only for higher order derivatives

it's waaaaaaaaaaay faster

lol

[spiral yarrow]

So nothing inside the X

[radiant tapir]

yeah exactly

even something like sin(x) + 1 won't work

becuase you lose that + 1 in the derivative

[spiral yarrow]

this is what I like to hear😊😊😊😊😛😛

ohhh

[radiant tapir]

so it's kind of a special case thing

[spiral yarrow]

But the pattern is easy than it being something different every time and you having to actually calculate it

[radiant tapir]

yep exactly

really what I wanted to show was that there are sometimes patterns that make your life so much easier

it also shows up in some parts of higher math but it's complex and I don't feel like getting into it right now haha

[spiral yarrow]

loll don’t worry u don’t need too but i do appreciate you teaching me this little trick cuz if my teacher ever mentions it imma feel ahead and lowkey like a smarty pants LOL and it’s fun so thank u !

[radiant tapir]

yeah no problem! :)O

best of luck to you

[spiral yarrow]

thank u! you too😊

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[rose sandal]

I need help regarding bounds in a u and v substitute for a double integral im in calc 3 and dont get this part help !!

[rose sandal]

i will include screenshot of given solution work and circle the part im confused about

does anyone get how he gets the bounds for v???

plzzzz

!status

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[rose sandal]

5

!status 2

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5. I have a question about someone else's work/solution.
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7. None of the above

[rose sandal]

!2

!help

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[soft escarp]

Calculator help - idk how to exit this

I just want to exit so i can do general calculations

[reef fjord]

2nd+quit?

@soft escarp

I mean 2nd+mode (which will quit)

so press 2nd, then press 'mode'

[soft escarp]

Umm that didn’t really work for some reason

But then i removed the battery

And then put it back

And it reset like a charm

Lol

Well anyhow the problem has been solved!

Thank you for your help : )

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[tiny yew]

Could someone check my work on this proof?

[tiny yew]

This is Corollary 2.61

[topaz sinew]

@tiny yew Has your question been resolved?

[tiny yew]

<@&286206848099549185>

[reef fjord]

if no one answers this, may have better luck in #numerical-analysis

[tiny yew]

Yeah I might try that

Or just wait until tomorrow since it's late tonight

[reef fjord]

i would try to answer but i am very eepy

[tiny yew]

Lol I'll take anything

But it's fine if you don't want to

[odd pagoda]

yo @noble laurel someone from your class?

[tiny yew]

Yes

[odd pagoda]

answer looks good

[tiny yew]

Ok great

Thank you

[odd pagoda]

4th order should be enough btw

at least thats what austin got

[tiny yew]

Yeah I saw that but I kind of did it out of order and went overkill with the order 5 and didn't bother to go back and change it

[odd pagoda]

yeah a bit out of order

usually better to first check how much you have to do before doing it

[tiny yew]

It wasn't really any extra work because all the summation was on a calculator

I almost did like order 10 before revising down lol

[odd pagoda]

yeah of course. just a general note

[tiny yew]

👍

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[deft fulcrum]

Hey 😃Is this red market step allowed to do ?

[deft fulcrum]

p(x) = -2x^2 - 2x + 4

[topaz sinew]

@deft fulcrum Has your question been resolved?

[deft fulcrum]

<@&286206848099549185> maybe

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[olive edge]

im stuck on the taylors remainder theorem part

ive found (f^3(c))/3! (x^3) but i dont know what to do after

[strange whale]

how do you get that ?

what's taylor remainder theorem for you ? do you have a statement of it at hand ?

[olive edge]

thats what ive used

[strange whale]

yeah ok

first we know that x should be between 0 and 1/2

[strange whale]

does that allow you to simplify anything in that expression ?

[topaz sinew]

@olive edge Has your question been resolved?

[olive edge]

i can divide by 3! but thats it i think

which gets me 3c^2(1-c^2)^-5/2 (x^3)

[strange whale]

I mostly think about the x and c

you want the answer to be a plain old number

so you have to manage to get rid of those

by using some inequalities you know (or find) on x and f^(3)(c)

[olive edge]

im not sure

x<1/2 maybe but idk about the c

[strange whale]

yeah right so we know $|R_n(x)| = \left\lvert \frac{f^{(3)}(c)}{3!} x^3 \right\rvert \leq \frac{1}{48} |f^{(3)}(c)|$

[thorny flame]

aPlatypus

[strange whale]

that 1/48 is just (1/2)^3 * 1/3!

now the problem is that we have to bound f^(3)(c) for c between 0 and 1/2

it's time to get a few more derivatives of f

[olive edge]

how did you get to this??

[strange whale]

x is between 0 and 1/2

so x^3 is between 0 and (1/2)^3

and the 3! was already there

[olive edge]

oh okay

[strange whale]

you'll prolly have to get 4 derivatives

[olive edge]

what do i do after that?

[strange whale]

well try to find what the minima and maxima of f^(3) are

[olive edge]

min (0,0)

im not sure about the max though

[strange whale]

what have you done then ?

[olive edge]

found f^(4) and found x when f^(4) equals 0

would the max just be when x=1/2 for f^(3)

[strange whale]

if f^(4) is always positive then yea

[olive edge]

i have no idea what im doing

[strange whale]

well what have you got then ?

like post your working if you're unsure

@olive edge

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[thorn terrace]

i wanted to find the intervals for which f(x) = 4sin^2x + 8sinx + 3 is concave up or down in the intervals 0 <= x <= 2pi, and i found x = pi/6, 5pi/6, and 3pi/2. when i tested it, i found that 5pi/6 < x < 3pi/2 is concave up, and 3pi/2 < x <= 2pi is concave up.

[thorn terrace]

does this mean 3pi/2 is not an inflection point?

0 <= x < pi/6 is concave up
pi/6 < x < 5pi/6 is concave down

so pi/6 and 5pi/6 are the inflection points

but is 3pi/2 an inflection point? since the concavity doesnt change

[ruby tree]

According to Wikipedia this would be an undulation point instead

[topaz sinew]

@thorn terrace Has your question been resolved?

[ruby tree]

It's because the lowest-order non-zero derivative above the second has even order

In other words, at x=3pi/2, the second derivative of f is 0, the third derivative is also 0, and the fourth derivative is not

The fourth derivative is of lowest even order after the second

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[jovial cape]

Question: Take a shuffled deck of 52 cards and throw the first top 10 cards in the bin. what is the probability that the new top card is red?

Is it just 0.5?

[jovial cape]

As all the scenarios of takiing more red than black should balance out with the scenarios of taking more black than red

[torpid matrix]

yes

"what's the probability the 11th card in the deck is red"

[jovial cape]

that's a good way of thinkiing about it I didn't consider lol

Thanks!

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[deft fulcrum]

Can I repoen this one :D

[restive inlet]

claim a new available channel and repost the question

[deft fulcrum]

Arl, thanks!

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[topaz sinew]

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[quaint fulcrum]

can somebody help with this?

what i did:

4+4+9+9 = 26
20/4 = 5
26x5=130

[balmy roost]

that's not how you do it

area is length x width

[quaint fulcrum]

im so stupid im thinking perimeter

thank you for that hahah

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[brisk geyser]

not sure where to go from this, can someone help?

[acoustic pecan]

how come you found the magnitude of the normal vectors?

[brisk geyser]

tbh i dont even know

[acoustic pecan]

write the normal vectors

[brisk geyser]

(2, -1, 4) (-1, 5, -1)?

[acoustic pecan]

yeah

do you know what to look for when checking if theyre parallel?

[brisk geyser]

if the angle between them is zero or pi

[acoustic pecan]

i mean, sure but theres a more direct and simple approach

just check if they are multiples of one another

[brisk geyser]

whats the more simple approach

sorry?

[acoustic pecan]

whats the point of confusion?

[brisk geyser]

do you mean if u x v = v x u

is that what you mean if they are multiples of one another

[acoustic pecan]

no thats just multiplying vectors

[brisk geyser]

oh no

[acoustic pecan]

what i mean is, if you have some vector v
if another vector w = av where a is a scalar, then w and v are parallel

check if they are multiples (scaled versions) of eachother

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[errant flicker]

hello i need help about differantial equations

[eternal spoke]

Do you have a specific question

[errant flicker]

i separeted x s and y s than i integrated x's polynoms but i cannot did -y. y' d/dy

[eternal spoke]

Can you show your work

[errant flicker]

thats the answer

but as i said i didn't do -y.y' part

couldn't***

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[slow cape]

help me

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[tight rivet]

Help

[ancient shoal]

use the formula arc length = R theta

where theta is in radians

[tight rivet]

What about bottom one

[ancient shoal]

use formula for area

of sector

[tight rivet]

Idk it

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[green dock]

How do I find the roots

[magic cedar]

Could you think of another way to rearrange the last but one step? where you have -108cos^3+256sin^3=0

[green dock]

no

[magic cedar]

if you move the cos^3 on the other side, then?

[bitter hemlock]

how did cos^3 suddenly become sin^3 from second to last line to last line?

oh nm

[green dock]

idk if its right

i just thought u could use complementary angles

[bitter hemlock]

so you have
-108 cos^3 theta + 256 sin^3 theta = 0
am i reading that right?

on 2nd to last line?

[green dock]

yes

[bitter hemlock]

instead of last step:
256 sin^3 = 108 cos^3
sin^3 / cos^3 = 108 / 256
tan^3 = 108 / 256
then you can take cube root of both sides and take inverse tangent of both sides to solve for theta

[green dock]

oo

okay let me try

omg ty

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[serene dagger]

I'm completely at loss with how to do these questions (I've missed the lesson and the internet isn't really helping),

[serene dagger]

that's my attempt so far, if you could even call it that.

can someone please explain what it means?

[topaz sinew]

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[serene dagger]

<@&286206848099549185> sorry to tag you

[neon iron]

What are mean reactions at c and d ?

[serene dagger]

i assume seeing as this model only has "downward" motion (6kg pulling downwards), i only need to look at A and B that's going down respectively?

[neon iron]

I will try

[neon iron]

I think this is very easy, but I can't help you enough because my English is kind of weak. Use YouTube, that's a good option too

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@serene dagger Has your question been resolved?

[serene dagger]

<@&286206848099549185> sorry to tag

[foggy ginkgo]

ken la o kepeken ni, ni li lukin pona
https://skyciv.com/docs/tutorials/beam-tutorials/determine-the-reactions-at-the-supports/

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[marble igloo]

how did they go from first line to second line

[marble igloo]

btw this is for vectors

<@&286206848099549185>

for more context

[dense ridge]

Because vector a cross itself is zero(outer product of two vectors which are parallel to each other is zero)

[marble igloo]

so this is equal to 0?

[dense ridge]

Yes

Vector a is parallel to itself

[marble igloo]

so how did 1/2 a × 1/13 b become 1/2 × 1/13 |a × b|

[dense ridge]

The coefficient can be extracted out the product
Because we add the absolute sign,just care its magnitude

[dense ridge]

Sorry, more corretly,zero vector

[marble igloo]

oh yeah you're right

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[regal kiln]

I just wanted to confirm 10pi was the answer

I can ofc explain why

Since its only asking for the differential, we only need to find f'(x)*h correct?

The formula for area of a circle is A=pi*r^2, and the derivative is A = 2pir

Since we need to find f'(x), x being 10, I plugged that in to our derivative

That gives us 20pi. H is 0.5, bc 10.50-10 is .5. We multiply 20pi by h, or 0.5

That gives us 10pi

Given all that, I just wanted a confirmation that was correct

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[strange mortar]

i need to write this as a fraction. i tried turning it into a series of 35/1000+35/100000......... then into a geometric series but i got it completely wrong

[strange mortar]

ill recreate my work in ms paint rq

i converted that to a geometric series after but it didnt work

im probably screwing smth up from the very beginning lol

[hybrid fractal]

so that last line isn't right

[strange mortar]

oof

[hybrid fractal]

10^(n+2) starts at 1,000 then goes to 10,000

[strange mortar]

oh

im dumb

2n+1 then?

that makes 3, 5, 7, blah blah blah right?

[hybrid fractal]

yeah

i think you want to write it where n starts at 0, so you can use the formula for the summation of a geometric series

actually that might not matter

[strange mortar]

didnt matter

i got it

thanks dude. i really gotta stop speed running this stuff lmao

i even thought of 2n in my mind but it didnt cross my mind just to like, be smart and not dumb (manually writing the series)

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[hybrid fractal]

ok nice

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[modest frigate]

this is more of an inquiry... but for the first one how did they come up with those two points in the blue there... i know it uses the equation but i cant figure out how, and for the second one i kinda get it but why was x just replaced with 0

[thorn wing]

for the first one they came up with those two points using the y intercept and the gradient

the first point as you can see is the y intercept, 4

[modest frigate]

i see that

[thorn wing]

the second point is just the gradient, -2/3 because they moved 2 down and 3 across from the y intercept

[modest frigate]

i thought that meant it was two thirds of 4

which shouldn't be possible...

[thorn wing]

ah it doesnt mean that

-2/3 is a gradient

[modest frigate]

i see now

what about the second question then

[thorn wing]

ok so

whenever there is y intercept

x always = 0

thats why they substituted it

so it can be like

(0,5)

(0,4)

so when you substitute x for 0 you can find y intercept

and likewise when you substitute y for 0 you can find x intercept

[modest frigate]

i see they basically just did algebra to solve the equation for x or y

right

[thorn wing]

yes

anything else?

[modest frigate]

dont think so

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[twin juniper]

i dont understand this equation, how can i proof it with induction ?

[keen raptor]

which part are you having trouble on

do you know what the inductoin hypothesis should be

[twin juniper]

i always had something like this

now there isnt a "="

does kk! mean k * k! ?

[keen raptor]

probably yeah

[twin juniper]

and what do i have to do with the <

[keen raptor]

do you know how a proof by induction works

[twin juniper]

yes

with n+1

[keen raptor]

you want to prove this statement for every n, right

[twin juniper]

yes

[keen raptor]

and you want to do this using induction on n

[twin juniper]

yes

[keen raptor]

do you know what the inductive hypothesis should be

[twin juniper]

i dont know the english word for it, do you mean n+1 ?

[keen raptor]

no

describe to the best of your ability how a proof by induction works

and i'll explain it in the terms you use

[twin juniper]

i already did this

[keen raptor]

oh

you want to assume the inequality is true for n right

and then use that to prove that it's true for n+1

[twin juniper]

yes

[keen raptor]

so, what's the assumption, specifically

and what do you need to prove

[twin juniper]

that k * k! is smaller than (n+1)!

[keen raptor]

i think you're missing the sum part there

and also you didn't tell me which one that is

[twin juniper]

i would start like this

okay nvm the last part is wrong

i dont know how to do the inductive step

[keen raptor]

(n+1)! is not equal to (n+1)*(n+1)!

as you seem to have indicated in the second line

[twin juniper]

i put n+1 in for all k's and n's

[keen raptor]

i dont think the logic of your work was thought through very far

can you write the statement you assume is true (for the induction step) and the statement you need to prove from it

if not, you don't know how induction works

[lean meteor]

For this proff

Proof

You rewrite the

K.k!

As (k+1)! -k!

[twin juniper]

why ?

[keen raptor]

it's not needed

[lean meteor]

Bwcause

When you put

K+1

Terms cancel

[keen raptor]

can you not

[lean meteor]

Ok

Ill let you carry on

[twin juniper]

i tried to use n = n +1 in the second step like in my previous induction.

i dont know why that is false

[keen raptor]

well

the statement n = n+1 is universally false for all values of n

so that's probably why

[twin juniper]

does it have to be k * k! < (n + 1 + 1)! ?

[keen raptor]

when you get rid of the sum entirely that statement makes no sense

[twin juniper]

can i pretend like the < is a = ?

[keen raptor]

sort of

if you put an actual sum there

but uh

i think the problem is you're trying to form-fit this to a different problem

you need to take the time to actually understand what a proof by induction is

[keen raptor]

rather than just trying to play fill-in-the-blank with a problem copied from someone else doing an induction proof

[keen raptor]

stop using n and n+1 interchangably

(n+1)! is not equal to (n+2)!

the sum on the left is not less than (n+1)!

you know all of these statements are just universally false

always

all the ones circled in red

pretty much the entire "outline" just consists of mathematically false statements

[twin juniper]

IF the < would be and =, this would be how i do it

all my inductions worked like that before

[keen raptor]

do you understand why they work

[twin juniper]

what do you mean ?

[keen raptor]

do you know why proof by induction is valid

why did someone first decide this was a good way to do mathematical proofs

how do we know it works

[twin juniper]

we are trying to prove if this equation is true for 1 number (n) it will be true for all other number ( n+ 1)

[keen raptor]

that sentence makes zero sense

[twin juniper]

thats how i learned it

why does it make no sense

okay i dont mean it will be true, i mean we try to prove that its true for all other numbers

[keen raptor]

n+1 is not "all other numbers"

if n is "1 number" then n+1 is also "1 number"

https://math.libretexts.org/Courses/Monroe_Community_College/MTH_220_Discrete_Math/3%3A_Proof_Techniques/3.6%3A_Mathematical_Induction_-_An_Introduction

i would recommend reading something like this

and comparing it to the proofs you've seen that use induction to see if you can understand what's happening

[twin juniper]

isnt that what i did there ?

[keen raptor]

no, that's not a proof by induction

it might be one small step of a proof by induction

[twin juniper]

yes thats the beginning step

[keen raptor]

but it doesn't really have to do with the overall meaning of a proof by induction

[twin juniper]

i just wanted to know if my start is correct

[keen raptor]

ok yes your proof will have something similar in it

[twin juniper]

i just need to simplify this equation till its the same or am i wrong again ?

[keen raptor]

no, I'm not sure what you mean by the same

[twin juniper]

right part needs to be the same as the left part

[keen raptor]

no

it's already clearly a true statement

you don't need to prove that it's true

[twin juniper]

my teacher told me that if i can show that this statement is true, my induction is done

[keen raptor]

https://www.youtube.com/watch?v=GdM_iA1Zek4&pp=ygUScHJvb2YgYnkgaW5kdWN0aW9u maybe something like this?

[keen raptor]

not this statement

that statement is always true

[twin juniper]

https://www.youtube.com/watch?v=dMn5w4_ztSw

he i doing the induction like i said

by showing that the statement is true

what else should induction be

[keen raptor]

I assure you induction is not about "proving" this obviously true statement is true

[keen raptor]

otherwise there would only ever be one "proof by induction" and it would be the proof that this statment is always true, regardless of what your a_i values or k is

because it is

the proof is of the proposition P(n+1)

which in the case of the video is the statement that the sum of the first n+1 integers is equal to (n+1)(n+2)/2

[keen raptor]

this is just a tool sometimes used to accomplish that proof

[topaz sinew]

@twin juniper Has your question been resolved?

[topaz sinew]

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[topaz sinew]

Please don't occupy multiple help channels.

[lunar gust]

.close

[topaz sinew]

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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mint perch]

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