help-13

[mortal ether]

i don't understand how to answer the last two problems

how many items do i use to calculate the average with?

and i know how to find the average and marginal cost. but it looks like the last problem looks like it's asking for an exact cost?

[cedar kiln]

@mortal ether Has your question been resolved?

[mortal ether]

<@&286206848099549185>

[mortal ether]

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[crude lagoon]

hello

i understand it wants inequality notation

so x>0

does that mean interval notation (current input)>0

i dont really understand what it's asking for really

[vestal bear]

thats the correct symbol for union?

[crude lagoon]

yes

but im pretty sure it wants a > there

since it says inequaltiy notation

im just not sure if it expects the polynomial on the other side

like the unfactored polynomial

anyone

[obsidian coral]

Wouldn't inequality notation be something like -10 < x < -3 OR x > 1?

[crude lagoon]

that makes sense

[obsidian coral]

Where you use AND or OR

[crude lagoon]

but im not sure how to type or in

nvm it has a thing

thanks

wait

and or or

it can be that and it can be that

it can be that or it can be that

can't it be both?

so it's and?

one more question same topic

i have x^3-x

which factors to

x(x^2-1)

wait

x(x-1)(x+1)

i can identify the positive and negative values for x+1 and x-2

but idk what to do with the x that goes before those 2 factors

does it matter at all

the whole problem is x^3-x<0

[cedar kiln]

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[foggy mason]

Hey everyone, I got this question from an old previous midterm and I am not sure how to get the answer, any ideas? Sadly it did not come with a key either

[foggy mason]

I was considering since they were both planes, the normal vector of x + y + z = 2 should be the same as the tangent plane of the paraboloid

[slate lintel]

you'd want the two unit normal vectors to be equal

[foggy mason]

Yeppers

Having a hard time working with that algebraicly

I know the equation of a tangent plane but Im not sure really what goes where to say the least

[slate lintel]

I'd work with the vectors themselves

but feel free to show what you have

[foggy mason]

Do I need to do something with level curves

So like when the gradient vector = the normal vector

plug and chug for an x,y,z because idk how else to really do that besides i hat = i hat etc

[slate lintel]

that seems reasonable

[foggy mason]

I dont understand how that woluld give me a point

Well I can say it out loud

[slate lintel]

what's your normal vector / gradient to that surface?

[foggy mason]

I found an x y z such that it has the same direction as a normal vector

hmm

[slate lintel]

seems right to me, what point?

[foggy mason]

if i plug the found x,y,z into f again, then set it equal to what i got

then thats the level curve right?

[slate lintel]

well it should be 0

[foggy mason]

so like if i found for example x, y,z to be (1,1,1) if i did f(1,1,1) got like idk 2 then set f(x,y,z) = 2 then thats the level curve at p

[foggy mason]

sorry im kinda talking about other stuff now

unless?

thats kinda a tangent but im pretty certain thats generally how you find a level curve for a function that contains a point

anyways

yeah is the whole purpose just to set grad f = normal vector of the tangent plane

[slate lintel]

like half of these points aren't even on the surface

[foggy mason]

sure

sorry ignore my random rant about **level surfaces

im kinda just speaking out loud to re inforce the key ideas ya know

[foggy mason]

but this is true yes?

[slate lintel]

normal vectors are perpendicular to all level sets which is why the gradient is perpendicular to the surface

[foggy mason]

okay sweet

yeah

ok im just trying to wrap my head around this tho

[slate lintel]

am I tripping or is only one of those points actually on the surface

[foggy mason]

lmao would not be suprised i wish i had that prof

he would be chill af and reward people who actualyl found that out

probs intentional

lemme see

ok yeah lol

hahahha wow thats funni

anyways but how does this work

like finding x,y,z of <2x,-2,-2z>

[slate lintel]

I'm confused though because the gradients don't match at (1,1,1)

[foggy mason]

that it equals <1,1,1>?

[slate lintel]

should be <2x, -2, 2z> I think

[foggy mason]

yes sorry

[slate lintel]

and yeah it needs to be a scalar multiple of <1,1,1>

there is a point that does that and is on the surface

[foggy mason]

y is a free variable right?

[slate lintel]

yeah

at least in terms of the gradient

[foggy mason]

well it cang be e d or b

or well

c

but 4,-2,4 is not a scalar multiple of <1,1,1>

but ok like what the heck

[slate lintel]

oh I know what happened

[foggy mason]

my main concern is

[slate lintel]

once you find the correct point I'll tell you (it's none of the ones listed)

[foggy mason]

wut

ok he would not do that

it has to be one of the answers on the thing

[slate lintel]

I can see what went wrong

[foggy mason]

otherwise the question would be null

what went wrong

[slate lintel]

I'll tell you once you find the right answer

you know the gradient is a scalar multiple of 1,1,1

and you should be able to figure out what that scalar is

[foggy mason]

oh wait there is a missing negative

its supposed to be (-1,y,-1) or something

[slate lintel]

and what's y?

[foggy mason]

-2

[slate lintel]

is that point on the surface?

[foggy mason]

neigh

[slate lintel]

okay well pick y such that it is

[foggy mason]

but isen't y forced to be -2

in our gradient problem

[slate lintel]

no why would it be

the gradient component is -2 for any y

as you said it's free

[foggy mason]

how does that work

sorry wut da heck

i need a picture

:3

if u can/want

i did say that but idk now my mind changed/deicded not to accept that anymore

[slate lintel]

we are looking for a point (-1, y, -1) that is on the surface x^2 + z^2 - 2y = 0

[foggy mason]

well y = 1 works

but i mean

[slate lintel]

the gradient at all points is given by <2x, -2, 2z>

so y doesn't affect the gradient at all

[foggy mason]

hm.

[slate lintel]

it's like how a line has a constant slope

yeah 1 works, that's the only solution

yeah
so what happened was that surface eqn is written out of order
so the question author listed x and z coordinates first by accident, and then y

[foggy mason]

ohhh boy hahaha now i remember the person who gave thsi to me

was malding about this quesiton

[slate lintel]

as well they should have been lmao

[foggy mason]

welp

i think it got nulled so theres that

[slate lintel]

yep
your intuition was right (look for coinciding normal vectors) the answers were just bad

[foggy mason]

hmmm so any point along y

has value - 2

well partial derivative of -2

slope of -2 along y

[slate lintel]

eh
the gradient has a component of -2 in the y direction

the gradient isn't really like the derivative you're used to since it's perpendicular

[foggy mason]

mmmmm the gradient still is like a sorta mystery for me

like idk its real inner workings

[slate lintel]

it's just the normal vector

[foggy mason]

besides it got derived from a directional derivative

when u = 1 and cos = 0

[slate lintel]

you can compute it in 2d and that may be educational

if you take like $y - 2x^2 = -3$ and compute the gradient

[wraith dagger]

Hayley

[slate lintel]

then plot it at certain points on that curve

or make a more elaborate curve

[foggy mason]

i tried doing something similar sorry to cut you off

[slate lintel]

eg find the gradient of a circle, x^2 + y^2 = 9

[foggy mason]

but like i found the partial derivative of a circle in x and y

oh yeah haha

and i was like hmmm wow if i add these two

partial derivate vectors

i get a line perp to the circle

[slate lintel]

yep! same thing in 3d

or any dimensions

[foggy mason]

wowie yeah that made sense

but okay

sorry to go backkkk

we have <2x, -2, 2z> = <1,1,1>

need xyz to be a linear combination of the normal vector

and also for it to exist on the plane

the plug and chug method i suppose is the only real suitable way for this

[slate lintel]

uh you need <2x, -2, 2y> to be a linear multiple of <1,1,1>

and yes the point needs to exist on the surface

[foggy mason]

linear multiple and linear combo is the same thing ye?

[slate lintel]

yeah (I'd call it scalar multiple if it's just one)

[foggy mason]

okay

but heres the kicker right

[slate lintel]

but like
technically it's a linear combination of the one lol

[foggy mason]

this is where the confusion is

wdym "the one"

or dare we get there

goodness sorry i keep bringing us off course hahahaa

i get it needs to be dependent on the normal vector

i think thats plenty for my understanding in regards to that part

anyways like the gradient is only perpindicular to a level curve

but the gradient we have is not from a level curve

wait is it

i think it is wait okay nvm

uhhh what if it was not a level curve

is a plane a level surface?

oh boy sorry for spam

i think it is because it = c

[slate lintel]

"the one" being the one normal vector we care about

a plane is a level surface yeah

it's a very level surface xd

[foggy mason]

hahahahahah

sorry i was just overthinking it

but what if it was not a level surface

what if they just gave us some function how would this problem change

[slate lintel]

hmm then it'd be harder to phrase

since here we just had the one

you could say like

here's two level sets

at what points does a ray from the origin pass perpendicularly through both

[foggy mason]

oh boy

uhh never heard the term ray really before in math

but im curious :3

[slate lintel]

ray is just a line with one end capped

[foggy mason]

oh boy

[slate lintel]

so like uh
$x^2 + y^2 + z^2 + 2xy + 4yz + 6xz = 20$ and like $x^2 + y^2 + z = 6$

[wraith dagger]

Hayley

[slate lintel]

first one is an ellipsoid second is a paraboloid

at what x and y are their tangent planes parallel?
(it won't be at a spot they actually overlap unless I'm very lucky)

,w graph x^2 + y^2 + z^2 + 2xy + 4yz + 6xz = 20

[wraith dagger]

Results provided by WolframAlpha

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[slate lintel]

oh that's not an ellipsoid

well that's cool anyway

[foggy mason]

funky

would u say that this is a type of question a calc 3 prof would ask

cus i mean

🙂

im interested but midterm is tmrw and i mean

[slate lintel]

yeah this is calc 3 stuff and really hard too

I'd expect much closer to your original problem

[foggy mason]

well i get 50 minutes and 5 long answer questions

and the hardest part is to firstly understand what the hell my prof wants aka deciphering the question

sorry that was kinda rude but

[slate lintel]

this problem would take me a solid 10 minutes without a computer

at least

[foggy mason]

ah

[slate lintel]

yours I did in my head in about 2 which is about right, I should be about 3-4x faster than you

[foggy mason]

that was taken from another profs midterm

where u got like

3 long answer and

idk like 7 mc

so

i think

if i had 5-6 minutes ur probs right i could get er done

for that type of question but

i doubt i would even see it cus i get long answer ones yaay?

anyways sorry continue with your problem :3

[slate lintel]

oh if you think I'm solving that shit you are mistaken lmaooo

I just made up those numbers there are guaranteed demons hiding in there

[foggy mason]

hahahaha fair enough

ah makes sense

w

this is a good one

i dont really know how to apply the sandwhich theorm here

thats such a random calc 3 problem like tbh why is that there

that should be calc 1 or 2

there should be a 2 path test or something instead of squeeze theorm what the heck

[slate lintel]

sorta except the multiple variables

[foggy mason]

well i mean its kinda the same thing

the hard part is the squeeze

only thing i can think of

is ima call taht thing f

-1(f) <= f <= 1(f)

where cos decided to dissapear

hehehe

does this factor nice lets see

okay well if my process is correct then u can break apart the denom and then u would get some -# that i think would not be 0 <= f <= #

but then that does not help me i dont think cus like thats not a squeeze is it

[slate lintel]

I'll be honest I have no idea how to do this one lmao

[foggy mason]

do u think symbolab will

my beloved symbolab

its become less and less useful but

its always there

[slate lintel]

give it a good 5-10 minutes of trying stuff

cosine is very squeezable

[foggy mason]

indeed

but you get some -# <= f <= #

and idk if thats proper

like it would 100% work if it was # <= f <= #

[slate lintel]

you need the outer limits to be the same

[foggy mason]

yeah but then what the heck

cus cos is between -1 and 1

so you would have -3x^4/... <= f <= 3x^4

[slate lintel]

it's not between -1 and 1 near the origin

I mean it is

but it's much tighter than that if you want it to be

[foggy mason]

ayo wdym

[slate lintel]

like you could say 0 < cos <= 1

[foggy mason]

wut how

[slate lintel]

because we're talking about points near (0,0) right?

[foggy mason]

yeah but hmm cos is never 0 near 0

[slate lintel]

yeah it's even higher

it's definitely not -1

[foggy mason]

i can see 1 but 0 and -1 nvm

maybe its just 1

cos is just 1

then eveyrthing is happy 🙂

to be fair this question will probs not even be near the midterm it would be a 2 path probs if there is a limit

because like we never went really too deep into squeeze

there was a squeeze on the hw but i just used polar coordinates like a chad cus aint nobody doing squeeze if polar is an option

aka this question im kinda fine without

... but might still be good to know?

[slate lintel]

like I said idk how I'd do this so probably? but I hated series

[foggy mason]

no way this is a series question

well you have been super duper helpful so far

i really appreciate it

so we can count that one as a pass

damn we are almost near the bottom of the help channels

i almost made it

ur cool af and helpful tho

ill let you be free and help others

thank you very much

.close

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[viscid pendant]

Am I allowed to take the negative 1 out of the (-3)^n

[tropic oxide]

not like you did you aren't

[viscid pendant]

or do I make it (-1)^n * (-3)^n

[tropic oxide]

(-1)^n * 3^n

[viscid pendant]

oh alright

thanks

so that series converges absolutely right?

[cedar kiln]

@viscid pendant Has your question been resolved?

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[viscid pendant]

how would I find the convergence or divergence of this series?

[royal loom]

what is (1)^n

for all n

[viscid pendant]

1

[royal loom]

ok

[viscid pendant]

1/sqrt n^2+3

I'm thinking of doing integral test but like what integration method would I use

ik that comparison test doesn't work

[crimson sedge]

Comparison definitely would work

[viscid pendant]

how come

1/n is bigger than 1/sqrtn^2+3

would I use limit comparison test?

[crimson sedge]

Yes

[viscid pendant]

wouldn't the limit be 0?

when using comparison test

limit comparison test

that yields inconclusive no?

[leaden otter]

Are you sure that's (1)^n and not (-1)^n?

[viscid pendant]

yeah its (1)^n

I can sohw you the full problem

show*

[leaden otter]

Also, try again

[viscid pendant]

just want to make sure

I'm taking the limit of sqrt(n)/sqrt(n^2+3)

right

[crimson sedge]

No

Wrong

[viscid pendant]

oh

its 1/n

right

[crimson sedge]

Yes

[viscid pendant]

n/sqrt(n^2+3)

[crimson sedge]

How did it become sqrt(n)

[viscid pendant]

?

[viscid pendant]

I made a mistake

[crimson sedge]

How is n = sqrt(n)

Okay

[viscid pendant]

that was the mistake lol

I got it mixed up

but yeah its supposed to be 1/n

the limit is 1 correct?

[leaden otter]

Indeed

[viscid pendant]

thank you guys!!

[cedar kiln]

@viscid pendant Has your question been resolved?

[cedar kiln]

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[sand cradle]

Out of $n$ national teams in soccer, the best one should be determined. For that, a qualification round consisting of a first and a second round is organized. Each nation plays against each other exactly once in the first and the second round, but once as home country and once as guest country. How many rounds will there be in total?

[sand cradle]

So if we have $3$ teams for example, $1, 2$ and $3$, then in the first round, we could have $12$, $13$ and $23$. In the second round, the home countries would become the guest countries, so we'd have $21$, $31$ and $32$, right?

[tropic oxide]

yes by the looks of it

[sand cradle]

For n teams, the first team will have n - 1 possibilities, the second n - 2, ...
So there will be 1 + 2 + ... + n - 1 possibilities for the first round, right?

So there will be (n-1)n/2 rounds for the first round and just as many in the second, so we'll have n(n-1) rounds, right?

[tropic oxide]

would seem so yes

[sand cradle]

Thank you!

.close

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[thin hedge]

how is b wrong

[thin hedge]

i dont get it

-r=3sin(negative theta)

multiply both sides by -1

sin(-0)=sin0

so you get r = -sin(0)

0 is j theta btw

so how the hell is that symmetric?????

yeah

wouldn't i have to get 3sin0

for it to be symmetric

mathematically

im expected to show the work

ik u can graphically do it but

[cunning forum]

huh

it is symmetric though?

[thin hedge]

can you tell me how

it would be

im j confused lowkey

i think i did my algebra wrong

[cunning forum]

whered this come from

[cunning forum]

this

[thin hedge]

from b

like

i didnt get the original expression

from doing that math

do yk what im saying

[cunning forum]

i dont get what youre saying

huh

[cunning forum]

whered this come from?

show your work

[thin hedge]

ok

bet

[cunning forum]

~rotate

huh no paradox

,rotate

[thin hedge]

line = b

which is the pi/2

OHH WAIT

SIN(-0)=SIN0

-SIN)

-SIN0

[cunning forum]

i think i see the thing

lol

its -sintheta

[thin hedge]

-3 and -sin0

[cunning forum]

not sintheta

[thin hedge]

its then

3sin0

oml

[cunning forum]

@thin hedge .close

[thin hedge]

.close

ty

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[glossy jewel]

how to solve this plz help

[vagrant elbow]

can you tell me what indeterminate form the limit is of

(or should be)

[tropic oxide]

so fucking ugly oh my god

[vagrant elbow]

nah it's not that bad

JEE has a reputation for this kind of shit

[glossy jewel]

0/0?

[vagrant elbow]

yes

Can you set the numerator to 0 and get a relation between p and q

You'll get a good looking one

[glossy jewel]

wait

[vagrant elbow]

After subbing in x = 2p

[glossy jewel]

the mod will open with +ve na?

i have got

[tropic oxide]

sure does

[glossy jewel]

x(x-4p)+(q+4)^4

[tropic oxide]

annoying symbolpushing with many places to make mistakes in

[glossy jewel]

ma'am itz just jee mains

you have to see alot in jee adv

😂

[glossy jewel]

@vagrant elbow

[vagrant elbow]

You're meant to sub x = 2p

Also power 4?

[glossy jewel]

in that piece wise fxn or wot?

[vagrant elbow]

no

In the numerator of the limit

[glossy jewel]

after getting relationship?

[vagrant elbow]

uh we're getting a relationship right now

[glossy jewel]

sorry bro No gay

😂

[glossy jewel]

man i am talking about this

[vagrant elbow]

yeah why do you have a 4th power

and sub in x = 2p

We want the numerator to be equal to 0

Because only then can the limit stand a chance of existing

[glossy jewel]

i got 4p^2=(q+4)^2

or 2p=q+4

[vagrant elbow]

well technically you need to have absolute values but yes

$\lim_{x \to 2p^+} \floor{\frac{2\sin^2 (x^2 - 4px + 4p^2)}{(x - 2p)^2}}$

[wraith dagger]

neonperseus

[vagrant elbow]

Have you understood what was done here

[glossy jewel]

wait a sec

did you sub q with 2p-4?

[vagrant elbow]

I subbed (q + 2)^2 with 4p^2

so yes

[glossy jewel]

ok

ok

got it

[vagrant elbow]

Do you still need help or do you think you can do it from here

[glossy jewel]

na i can

thanks man

[vagrant elbow]

alright

[glossy jewel]

the approch i folow was

to use the

identity of 1-cosX/x^2

but yeah thats wrong b'coz that should tends to zero

not some random bullshit

Thanks Man Once again @vagrant elbow

.close

[cedar kiln]

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[vagrant elbow]

You're Welcome, Glad I Could help

[cedar kiln]

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[cedar kiln]

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[fallen plaza]

Hi. Could someone help me with this question? I haven’t done anything with P(x) yet

[crude gale]

im assuming the "break-even quantities" means when P(x)=0?

So the question is just asking you to solve 0=-x^2+12x+28

[cedar kiln]

@fallen plaza Has your question been resolved?

[fallen plaza]

So should I substitute P(x) as 0?

[crude gale]

yes

If you think of P(x) as profit, when P(x) is negative, you are at a loss. When P(x) is positive, you have well. made a profit. When P(x)=0 is when you have just broken even, no loss nor gain.

[cedar kiln]

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[crude lagoon]

i dont understand how -2 doesn't include -2

if that makes sense

since it's <=

[iron cave]

hi

[crude lagoon]

img above is my answer

the one below is the key

[iron cave]

so basically, it's because the denominator cannot be 0

so x+2 is not 0, so x is not -2

thats why

[crude lagoon]

ohhhhhhhhhhh

what does the box want??

[autumn sorrel]

I would assume the box wants x isolated

So something like -1 < x < 7/2

[crude lagoon]

forgot that existed somehow

[cedar kiln]

@crude lagoon Has your question been resolved?

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[inner needle]

im trying to solve this for my math homework but cant get the correct answer i am a bit stuck on it need some help

[inland fossil]

multiply

Now

[untold torrent]

okay, so what's the formula for the area of a triangle

[untold torrent]

yes

Now

[inland fossil]

[untold torrent]

lol

[inner needle]

1/2 base hight

[untold torrent]

yes

and in the case of a right triangle?

[inner needle]

1/2 base hight aswell

[untold torrent]

well yes

but in a right triangle,

you can think of the 2 sides adjacent to the 90 degree angle as base and height

[inner needle]

yeah

[untold torrent]

so area here is just ab/2

[inner needle]

i see

[untold torrent]

now try to apply that to your problem

[inner needle]

so (2x+5)*4(x-1) and all of that divided by 2

[untold torrent]

yes

$\frac{(2x + 5) \cdot 4 \cdot (x - 1)}{2}$

[wraith dagger]

redstoneplayz09

[untold torrent]

simplify it

[inner needle]

oj

ok

[inner needle]

should i simplify this one

[untold torrent]

thats what I wrote

i just wrote it nicely

its the same thing

[inner needle]

ok

8x^2+12x-20 divide by 2

which is 2(x-1)(2x+5)

hello @untold torrent

[untold torrent]

why are u factoring it back

u expanded it

and then factored again?

just expand

[inner needle]

no because i had to divide by 2

[untold torrent]

divide this by 2

what do you get

[inner needle]

8x^2+6x-10

[untold torrent]

no

$\frac{8x^2 + 12x - 20}{2}$

[wraith dagger]

redstoneplayz09

[inner needle]

1/2 (2x+5)(4x-4) = 30

(2x+5)(4x-4) = 60

8x^2-8x+20x-20 = 60

[untold torrent]

dont factor..

[inner needle]

[8x^2+1x-80 = 0] / 4

[untold torrent]

$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$

[wraith dagger]

redstoneplayz09

[untold torrent]

do you understand why this is always true

whenever you have addition in the numerator, you can "split" the fraction

[inner needle]

yeah

[untold torrent]

do the same here

[inner needle]

2x^2+3x-20 = 0

[inner needle]

is this it

[untold torrent]

its nice that you got the right answer and all

but factoring there was completely unnecessary

[inner needle]

1/2 (2x+5)(4x-4) = 30

(2x+5)(4x-4) = 60

8x^2-8x+20x-20 = 60

[8x^2+1x-80 = 0] / 4

2x^2+3x-20 = 0

[inner needle]

its the way i was tought

by my teacher

[untold torrent]

im sure he didnt teach u to do unnecessary thing

[untold torrent]

all you need is to divide by 2

yet you bring everything back into parenthesis

[inner needle]

he showed us both ways but he told us to use this because it is easier to understand and is apparently more reliable

[untold torrent]

huh

okay..

you do you

[inner needle]

yeah

[untold torrent]

do u have anymore questions?

sorry just hard to believe u were taught it this "way"

[inner needle]

part b

[untold torrent]

ok NOW you need to factor

[inner needle]

my favorite thing to do

[untold torrent]

yes I saw that

[inner needle]

(2x-5)(x+4)

2x-5=0

x+4=0

x = 5/20 or x = -4 i think

would that be correct @untold torrent

[untold torrent]

x = 5/2

not 5/20

other than that, yes

[inner needle]

thats what i meant idk why i put 20

final part

[inner needle]

thanks last part. then im done

[untold torrent]

one the solutions here needs to be eliminated

[inner needle]

its asking for both

[untold torrent]

for (c)

you need an x value that makes sense

[inner needle]

oh

[untold torrent]

x = -4 doesn't make sense because then your sides would be of negative length

[inner needle]

do i substitute it in to check

[inner needle]

yeah that makes sense so its x=5/2

[untold torrent]

yeah

and now plug it in to find the sides

[inner needle]

ok

[inner needle]

would the bottom side be 10

[inner needle]

and the standing side be 6

[untold torrent]

yes

[inner needle]

so now i work out the hypotenuse with those values

BC = √6^2+10^2 = 2√34 = 11.66?

is that it @untold torrent

[untold torrent]

yeah

[inner needle]

ok bro thanks for your help and time i reall appreciate it a lot thanks a lot i hope you have a wonderful day my g

[untold torrent]

np

[inner needle]

u can end this help now

[inland fossil]

only you can

.close

[cedar kiln]

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[inland fossil]

wtf

ok i guess i can

[inner needle]

how

[inland fossil]

@mossy mango gave me the power

[inner needle]

innit

[cedar kiln]

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[dull swan]

i dont understand what im doing wrong

[dull swan]

heres my work

[bold vine]

you typed 26 not 36

@dull swan

[dull swan]

RIP thank you lol

[bold vine]

np np

[dull swan]

.close

[cedar kiln]

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[bold vine]

always check your webwork inputs 🙂

[cedar kiln]

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[bold vine]

hang on i just need wolfram for a sec

[bold vine]

,w solve x(x+7)^2 (x+4)^5 (x-2)^6 (x+9) < 0

[wraith dagger]

Results provided by WolframAlpha

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

Requested by shenny17#0

[bold vine]

.close

[cedar kiln]

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[buoyant latch]

There’s a bot channel for that lol

[bold vine]

Lmao I didn't know XD

Where is it?

[cedar kiln]

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[marsh pond]

can anybody help me with this pls?

[marsh pond]

I have 4 options, A,B,C,D
it may include 1 or more than 1

not sure where this 12 comes from..

b - a
5 - 1 = 4

(b - a )/n
4 / (1/3) = 12

on the right track?

[violet night]

do you need help with part a) or b)?

[marsh pond]

part a to start with

i guess it would be
(2 - 0) / n

(upper limit - lower limit) / intervals

2/n

but the options are only

so it would be C?

not sure where n^6 is coming from

I think it's kinda like saying 2^6 = 64?

so maybe the answer is actually D

i don't think B counts because it's just i, not i^5

[violet night]

have you tried working it out yourself, could be quite long considering its x^5 , I don't know if there is a general thing to look for sorry about that

[marsh pond]

no, I'm not sure how to solve this

[dire geode]

n^6 = n^5 * n

Then put things under the 5th power

[marsh pond]

for these?

[cedar kiln]

@marsh pond Has your question been resolved?

[cedar kiln]

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[lucid tartan]

Can someone walk me through the process of getting started on this?

[lucid tartan]

That’s where I got up to, but I keep mentally stalling out here because of the fractional exponents I’m like overthinking this

[rose rose]

The limit should be as t->1 not h

[lucid tartan]

Oh it’s AS t = 1 right

[rose rose]

but it does indeed look exceedingly annoying

Since it's asking for a limit for rate of change I'm guessing l'hopital is out of question, I don't really know how to do this one ngl, you could maybe expand it as binomial series but not even sure that's helpful at all

[lucid tartan]

Gotcha gotcha gotcha no problem! Thanks for the attempt

Is there anybody else who would want to take a shot at this <@&286206848099549185>

[rose rose]

I have one idea, let's take $p = (t+7)^{\frac{1}{3}}$, as $t \to 1 \implies p \to 2$, now we have $$\lim_{p \to 2} \frac{3p^2 - 12}{p^3 - 8}$$

[wraith dagger]

learath2

[rose rose]

That actually works, wow

@lucid tartan I think you can figure out what to do from there

[lucid tartan]

How did you get the 8 In denominator, shouldn’t it be 1?

[rose rose]

Note that p^3 = t + 7 as per my definition, but we wanted t - 1 at the bottom, so need to subtract 8

That is to get the -1

[lucid tartan]

But wouldn’t that mean that t = p^3?

[rose rose]

I mean look at how I defined p, how could t ever be p^3?

[lucid tartan]

It’s supposed to be t - 1 in the denom

[rose rose]

Maybe just solve my definition of p for t and put that in instead of t? You'll see you get the -8

[lucid tartan]

Okay I think I got it

That’s pretty wild how you did that

[rose rose]

I couldn't sleep if I didn't figure it out

[lucid tartan]

Yeah wow

[rose rose]

The inspiration was that I really wanted to get rid of the 1/3 in the exponent

[lucid tartan]

So fundamentally so I know for the future

Yeah stuff like that

I can just regroup it into other variables to substitute it into other chunks so I can solve everything around it?

[rose rose]

You can always make substitutions in limits

[lucid tartan]

Also why did you choose p > 2 specifically

[rose rose]

Well if t approaches 1, I took the limit of my definition of p as t approaches 1

[rose rose]

Oh one exception is if the function is non continuous iirc, so be careful with that one

[lucid tartan]

I’m still a little lost

On how you got 2

[rose rose]

Well let's think about it intuitively, I want to write my limit in terms of some new variable p that depends on t

I need to know what value p approaches as t approaches the original limit point

[lucid tartan]

Because that will let us know how t behaves?

[rose rose]

I want to know how p behaves, I want to get rid of the t

but p depends on t

so I need to see what happens to t and find what that does to p

Another way to think of it is $t \to 1$ becomes $p^3 - 7 \to 1$ through a bit of notation abuse you can see how that would lead to $p \to 2$

[wraith dagger]

learath2

[lucid tartan]

It’s hard because I get why you did it

I just don’t know how I would be able to recreate that

[rose rose]

If you are looking for a formal explanation 4 am is a little too late for me to go through a proof of it

[lucid tartan]

Like it all makes sense in hindsight but I don’t know the process of why you made those choices

[rose rose]

The only choice I made was $p = (t + 7)^{\frac{1}{3}}$

[wraith dagger]

learath2

[lucid tartan]

Gotcha! Yeah it’s 9pm here by me but maybe not even a proof just a thought process of what you wanted to do

[rose rose]

The rest were not choices but just directly followed from that choice

[lucid tartan]

Yeah yeah

And the intuition in the denominator was really just rearranging to accompany for the original value?

[rose rose]

I plugged in $p^3 - 7$ everywhere I saw $t$

[wraith dagger]

learath2

[lucid tartan]

Okay right right

[rose rose]

Personally I did that one instinctively, but rigorously it was just replacing the t's with p^3 - 7

[lucid tartan]

Okay that makes sense

[rose rose]

Like I knew cubing p got me t + 7, but I needed t - 1, so I subtracted 8

[lucid tartan]

Ohhhhhh

Yeah yeah that’s sick how you thought that way

That makes sense

Wait wait

I thought p > t+7 ^ 1/3

So the square follows to p

Where does the cube come from

[rose rose]

Not p > (t+7)^(1/3). = not >

Just cube both sides

[lucid tartan]

This

[rose rose]

You mean for the top how I got the square?

[lucid tartan]

For the bottom how you got the cube

[rose rose]

I just cubed both sides for my definition of p

[lucid tartan]

The square is there because of the substitution

[rose rose]

The cube is also there because of the substitution

The numerator has $(t + 7)^{\frac{2}{3}} = ((t+7)^{\frac{1}{3}})^2$

[lucid tartan]

So here’s what I’m getting:

-you sub p for t+7^1/3
-p on top gets squared and multiplied by 3

[wraith dagger]

learath2

[lucid tartan]

Okay yeah that’s that part

I understand that

Now for the bottom

It starts out as t - 1 right?

[rose rose]

Yes

[lucid tartan]

So now how does our p sub change what that becomes

[rose rose]

Ok there are two ways to do that

[lucid tartan]

Okay

[rose rose]

Normal way is just solving p for t and figuring out what t is in terms of p

Then you can just plug in what you get instead of t

[lucid tartan]

Now wouldn’t that go to 0 or

Oh in terms of p

So cubing it and adding 7

Which is what you did

And got the answer

Wow

[rose rose]

Well I didn't add 7

[lucid tartan]

Okay correct me on that bit

[rose rose]

$p = (t + 7)^{\frac{1}{3}} \iff p^3 = t + 7 \iff p^3 - 7 = t$

[wraith dagger]

learath2

[rose rose]

I thus wrote t in terms of p, right?

[lucid tartan]

….. yess

OH

AND THATS WHY ITS MINUS 8

[rose rose]

Yes

[lucid tartan]

THAT ONE CARRIES

wow

[rose rose]

I did not really solve for t in my head because as you can see it's very simple, just it cubed looks enough like t - 1

but that is how you substitute normally, you pick a substitution, you solve for the variable you are trying to subsitute away, then you just plug it in

(it'll also be the same when you get to integrals, same exact idea)

I have to go get some sleep so hopefully I've been of some help, good night

[lucid tartan]

You’re a legend

Thank you so much Learath

[rose rose]

Did you manage the new limit btw?

[lucid tartan]

I didn’t yet, about to try it now

[rose rose]

|| It's just factoring out a (p - 2) from the numerator and the denominator || <- open if you can't figure it out

Good night

[lucid tartan]

You’re the best, goodnight!

.close

[cedar kiln]

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[cedar kiln]

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[magic hedge]

or anything close to that

im not sure if its even possible

<@&286206848099549185>

[cedar kiln]

@magic hedge Has your question been resolved?

[cedar kiln]

@magic hedge Has your question been resolved?

[cedar kiln]

@magic hedge Has your question been resolved?

[magic hedge]

.close

[cedar kiln]

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[jolly sable]

If A ∩ B = B holds for all sets B of the universal Set S, then
Options :

1. A′ = B
2. A = S
3. A = B
4. A = ø

[jolly sable]

I didn't understand the question

[south tundra]

We are working with some universal set S and are given that A ∩ B = B is true for all B that you could pick

Any ideas about what A is so far?

(You can arrive at the right answer using elimination method as well though)

[jolly sable]

A is totally exist in B

[latent pond]

can there be multiple correct answers

[south tundra]

No, there's just one

(Note that A has to be fixed, it can't be dependent on B)

[south tundra]

You mean A is a subset of B?

[jolly sable]

So here A and B are subsets of S

[south tundra]

Yes

[jolly sable]

Yes

[south tundra]

That can't be true because of this

Let me show you

[jolly sable]

It is a little bit weird and not a clear question

[south tundra]

If we pick B as the empty set, then A will be the empty set according to your hypothesis, right?

[jolly sable]

Yes

[south tundra]

Right, so A = ∅ and ∅ ∩ B = B has to be true for all B

[jolly sable]

B=phi

[south tundra]

But that is false, because ∅ ∩ B = ∅

So we have that B = ∅ for all B?

Clearly a contradiction

[jolly sable]

A∩∅=∅

[jolly sable]

Yes

[south tundra]

So we have to resort to the other case: B is a subset of A for all B

[jolly sable]

Then a, b intersection will be b

[south tundra]

But, since B could be any subset of S, it would mean that A has to contain all elements of S so that this is true

Meaning A = S

[jolly sable]

But i didn't understand this question

[south tundra]

I think your confusion comes from the fact that A has to be fixed meanwhile B could be any subset of S

[jolly sable]

So how does it get solved?

If A is S then all the A intersection B will be B

Ohh i just checked all options

We have to assume for maximum subsets it can produce and check the property holds or not

So only A=S will be safe

This is what you want to explain?

[south tundra]

Pretty much, yeah

[latent pond]

if A = B, then A changes when B changes

[jolly sable]

A=B can satisfied once

[south tundra]

And, like I said, you could just eliminate some options as follows:
A can't be the empty set for obvious reasons (I also mentioned this above)
A can't be B because A has to be independent of B and the goes for the option where A' = B

[south tundra]

Yes, but we want A to be fixed, that's the issue

[jolly sable]

Yes true

[uneven forum]

Let A = (2m, 3m+2) B=[-1,3]. Find m so that A ∩ B =∅ help me

[latent pond]

i guess its a matter of "assume whats in the question and check if the options are true" instead of "assume if the options are true and check if the question is true"

[south tundra]

Open your own channel

[latent pond]

so the phrasing of the question precludes 1 and 3 from being possible answers

[south tundra]

Yup yup

[jolly sable]

.close

[cedar kiln]

Channel closed

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[jolly sable]

Thank you

. reopen

.reopen

[cedar kiln]

✅

[jolly sable]

U wanna say something?

[uneven forum]

yeah i open it

pls help me ;-;

[gentle flower]

[cedar kiln]

@jolly sable Has your question been resolved?

[jolly sable]

.close

[cedar kiln]

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[fringe magnet]

I need to find the limit of ((n+2)/n)^3n as n goes to inf. so i started by making it (1+2/n)^(2/n)^3n/(2/n) to get e^2^3n/(2/n) but after that idk what to do

[restive hedge]

i can't tell what you're rewrite is meant to be

could you do it on paper or latex?

[dreamy sleet]

initially you could have rewritten it as (1 + 2/n)^(3n) and then used the special limit relating to e

[fringe magnet]

ye but to use the special limit dont i need 2 do it like this?

[dreamy sleet]

the fourth line is wrong

[fringe magnet]

oh it has 2 be 1/n

[dreamy sleet]

the function is not equal to (the limit as the function goes to infinity)

[fringe magnet]

ah yes i dident write lim

[turbid mesa]

,rotate

[dreamy sleet]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[fringe magnet]

so imagine i put lim what is my next step or what did i do wrong?

[restive hedge]

If you've taken a limit there shouldn't be an n's left

Basically the form you got it in wasn't quite right to get the limit and use the identity

You can start from here (like even order group said)

(1 + 2/n)^(3n)

I suggest writing he inner bit as

(1+1/(something))^(3n) first

then whatever (something) is will determine how you should rewrite the exponenent

[fringe magnet]

ah ok i got it e^6

thx

.close

[cedar kiln]

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[cedar kiln]

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[sand cradle]

Out of $n$ national soccer teams, the best one should be determined. \[5pt] To do that, a round where each team plays against each other in a first round and the same teams play but with switched sides (guest and home) in the second round , in the same order, is organized. \[5pt] Why is it impossible that for $n > 2$ guest and home plays are always played interchangedly?

[sand cradle]

We can't save that though. How do we prove or show that we will end up with something like this for all n > 2?

<@&286206848099549185>

[cedar kiln]

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[jolly sable]

I guess this question is wrong....it is asking what will be angle C?

[jolly sable]

My answer is 63

[vague rapids]

did you draw a diagram

[jolly sable]

Yes

,rotate

[vague rapids]

angle E is 63 degrees

[vague rapids]

Texit is broken

[tropic oxide]

angles B and E correspond to each other, not C and E as you drew.

when stating two triangles are similar, we always write their vertices in the same order, so saying ABC ~ DEF means A <-> D, and B <-> E, and C <-> F.

so in this case the mistake is yours

[jolly sable]

Ahhh my drawing mistake

Rest the question is easy

[tropic oxide]

drawing/reading mistake i'd say

[jolly sable]

Yes true

[tropic oxide]

it is subtle but important

[cedar kiln]

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[crimson sedge]

I need help with this and what trig laws to loook out for in these questiond

[south tundra]

What's 1 - sin^2(x) equal to? (recall the Pythagorean identity)

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[crimson sedge]

well sin^2 +cos^2 = 1 but Idk how to make this equation into the identity

[south tundra]

Try moving sin^2 to the other side

[crimson sedge]

wait a minute

you can turn the 1 into a sin^2 + cos^2?

right?

[south tundra]

Yup

[crimson sedge]

and then

cross out the cosines?

[south tundra]

Yes

[crimson sedge]

now it's sin^2 - sin^2x/cos^2x, now what's next

[south tundra]

You should have gotten cos^2/cos^2 which cancels out to just 1

Because 1 - sin^2 = sin^2 + cos^2 - sin^2 = cos^2

[crimson sedge]

cross out just means take out of the picture to me so I was confused

[south tundra]

Did you just simplify $\frac{\sin^2{x} + \cos^2{x}}{\cos^2{x}}$ as $\sin^2{x}$?

[wraith dagger]

alonelybean

[crimson sedge]

yeah

[jaunty mural]

how

[south tundra]

No, that is not how cancelling out works

[jaunty mural]

,,\frac{1+2}{2}=1

[south tundra]

Try instead rewriting that as $\frac{\sin^2{x}}{\cos^2{x}} + 1$

[wraith dagger]

alonelybean

[jaunty mural]

i feel like thats a step too far

[jaunty mural]

First, do you see why this is false

[crimson sedge]

Kind of?

[jaunty mural]

Do you see why this is false