help-17

[flint idol]

so we can ignore that cuz we’re looking for global max

can you now try to do the other 3 critical points on your own?

[solemn falcon]

is 4 and 6 are at ends of a curve does that make them both mins?

[flint idol]

4 is a min but not 6

[solemn falcon]

and 0 is in the middle of both curves does that make it a max to?

too?

[flint idol]

yep

[solemn falcon]

so 0 and 6?

[flint idol]

yeah

[solemn falcon]

but isnt there only one global maximum?

[flint idol]

yeah so that’s why we need to compare them

we need to find what F(0) and F(6) is, and see which is greater

[solemn falcon]

putting them into the integral equation?

[flint idol]

yep

we don’t really need to antidifferentiate it

we can just use the graph to calculate the area

[solemn falcon]

how would I do that? does the 0 annd 6 go into the x and the t? or jsut one of them?

[flint idol]

it goes into x

so $F(0)=\int_{0}^{0}f(t)dt$

[twin meteor]

y0shi

[solemn falcon]

thats what I was imagining but I dont know what to do with that

[flint idol]

well if we have the lower bound and the upper bound as the same number

the integral always evaluates to 0

regardless of what f(t) is

basically $\int_{a}^{a}f(x)dx=0$

[twin meteor]

y0shi

[solemn falcon]

that makes sense but what about tthe 6

[flint idol]

so we need to look at the graph above

there is a circle

and a triangle if we go from 0 to 6

we need to get the area of each and then add them together

remember that anything under the x axis would be negative

[solemn falcon]

making 0 the Maximum

[flint idol]

yep you got it

remember when it asks for the maximum, we always use the y value

not the x value

[solemn falcon]

phew lol thank you so much ill write it all down and practice it a bit because I never whould have gotten there

[flint idol]

but in this case they’re the same so it doesn’t matter

[flint idol]

yw!

[solemn falcon]

okay okay

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[vast shale]

I cannot seem to solve this problem at all

[vast shale]

i don't even know where to begin

[wicked shard]

we can do triangulation here

we are given 2 angles and a side

so we can use sine law

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[stone pecan]

i'm not sure how to find out the bounds for the integral. any help?

[agile wolf]

that's one of those polar coordinate problems right?

do you know the formula for dV?

[stone pecan]

is it polar? i'm not really sure when to use cartesian and when to use polar

thats the entirety of the question, i thought we had to freedom to choose the order of integration

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@stone pecan Has your question been resolved?

[stone pecan]

<@&286206848099549185>

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[frank iris]

How do I write this function interms of y

[frank iris]

Is it even possible to isolate x?

[sweet flower]

I don’t think so, check x=8 and x=-8, unless there’s a condition on x, y can take multiple values for the same x

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[vast shale]

I know what it's asking

I'm just not getting the right answer

I could go about this the long quotient law way

but instead I used derivative shortcuts

to get

$m=x^2-5+\tfrac{4}{x^2}$

[twin meteor]

Remlis

[vast shale]

setting m=0 would make a horizontal tangent

$0=x^2-5+\tfrac{4}{x^2}$

[twin meteor]

Remlis

[vast shale]

there seems to be a way to solve for x

but I just don't know of it

$5=x^2+\tfrac{4}{x^2}$

[twin meteor]

Remlis

[vast shale]

$5=x^2(1+4x^{-1})$

[twin meteor]

Remlis

[vast shale]

mulitply the equation by x^2

and then let x^2=t

then it's just a simple quadratic

$5x^2=x^4(1+4x^{-1})$

[twin meteor]

Remlis

[heavy yoke]

a useful strategy is often to turn an equation like this into a polynomial

[vast shale]

$5t=t^2(1+4\tfrac{1}{t^2})$

[twin meteor]

Remlis

[vast shale]

it seems pretty much the same

$5t=t^2(1+\tfrac{4}{t^2})$

[twin meteor]

Remlis

[vast shale]

i meant multiply this by x^2

$5t=t^2+4$

[twin meteor]

Remlis

[vast shale]

and then sub x^2=t

wait this is working

$0=t^2-5t+4$

[twin meteor]

Remlis

[vast shale]

good job

$0=(t-4)(t-1)$

[twin meteor]

Remlis

[vast shale]

$$t=4$$
$$t=1$$

[twin meteor]

Remlis

[vast shale]

$$x^2=4$$
$$x^2=1$$

[twin meteor]

Remlis

[vast shale]

$$x=2$$
$$x=1$$

[twin meteor]

Remlis

[vast shale]

@vast shale am I missing something

$x^2=y \implies x=\pm y$

[twin meteor]

yajat

[vast shale]

ohh right

$$x=\pm 2$$
$$x=\pm 1$$

[twin meteor]

Remlis

[vast shale]

there we go

thanks

[vast shale]

also just check if you're getting m=0 by putting all 4 solutions in this equation

[vast shale]

i am

ok

you got your answer then

thx

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[hushed atlas]

Don’t get

[sly sierra]

hint:

try multiplying them together

[desert hornet]

Try multiplying those two matrices

[hushed atlas]

U get I^2 - A^2?

[desert hornet]

Yes, and what’s A^2

[hushed atlas]

0

[desert hornet]

So?

[hushed atlas]

Ok so ur left with I^2 uhh

[desert hornet]

And what’s I^2?

[hushed atlas]

Mmm there’s some theory im missing

[desert hornet]

well, I times any matrix is that matrix

So I^2=I*I=I

[hushed atlas]

Wtf

lol

Seriously, so I squared is just i

Bruh

[desert hornet]

Yes

[hushed atlas]

What so does that mean I^3 - I?

[desert hornet]

In fact, any power of I is just I

[desert hornet]

Yes

[hushed atlas]

=**

Oh

[desert hornet]

It’s kind of like 1

How 1 acts with numbers

[hushed atlas]

I thought that only applied to other things

[desert hornet]

1n=n1=n

[hushed atlas]

Didn’t realise it applied to itself

lol

How would u know to multiply them in the first place

Didn’t cross my mind

And like why do u, why do u care if it equals I

Like i get it i just don’t get why

[desert hornet]

It means that they are inverses of each other

[desert hornet]

Well, (a+b)(a-b)=a^2-b^2

And the property we’re given tells us about A^2

So it’s pretty intuitive

[hushed atlas]

Ohhhhhh

Shit yeah

Hm. So if you multiply two things and get I they’re always teh inverse of each other?

[desert hornet]

Yes

Always

[hushed atlas]

Ok thanks

A lot

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[crisp zenith]

What is the formal defintion the two sides of an open sentence?
3x + 3 = 5x - 3

[crisp zenith]

I was thinking of just saying the left side of the equation but I was wondering if there was a better definition to describe it

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[austere plover]

Did I do this correctly? My friend got a different answer so I want to be sure.

[austere plover]

This is the question

<@&286206848099549185>

[vocal sleet]

@austere plover Has your question been resolved?

[austere plover]

<@&286206848099549185> :nyoron:

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@austere plover Has your question been resolved?

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@austere plover Has your question been resolved?

[austere plover]

<@&286206848099549185>

[novel iris]

You can substitute the values of u,r & t for what they’re equivalent too and then differentiate like that

[austere plover]

I did that #help-17 message. Just want to make sure the answer is correct

[novel iris]

du/dx looks right. I just did it and got the same answer

@austere plover

Wait, no scratch that

[austere plover]

🫠

[novel iris]

@austere plover

[austere plover]

What happened to the y^3 at line 4

Which is at the end of line 3

[novel iris]

Nice catch

Yep
Looks like I siked myself out the first time

[austere plover]

Lol

[novel iris]

Your du/dx looks good

Im just dumb lmao

[austere plover]

So is my friend

Good to have the assurance

[novel iris]

[austere plover]

Thanks

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[uncut mist]

Can someone just confirm that what I did to get the answer was right?? The part above the red arrow is just the right side of the product rule

[fading willow]

u seem right

[uncut mist]

Okay great that’s all I needed ty

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[merry umbra]

Ok so i need to create and equation that gives the volume of water in the troph based on the current water level x. I think i figured it out but it just seems wrong so was after some help.

[merry umbra]

Im sorry my writing is a hot mess 🙏 also sorry its in pen my pacer broke 😦

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@merry umbra Has your question been resolved?

[fading willow]

seems correct

[merry umbra]

ok well thats good new it just felt wrong to be doing sin(cos^-1())

[fading willow]

but in the final answer u need to put parentheses around the area of segment B

[merry umbra]

okok that easy enough ty 👍

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[fading willow]

nah, just because u are dealing with area of sections of a circle, thats why its kinda off

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[fading willow]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[zealous hill]

1 ig

[scenic agate]

well the paths width is 75 cm all the way around

actually wait

[zealous hill]

2w + 2L = 2000cm

[scenic agate]

wait what

[zealous hill]

?

[scenic agate]

where did you get 2000 cm

[zealous hill]

perimeter of the pool is 20m

[willow pike]

one min

will draw diag

[scenic agate]

oh ok

[zealous hill]

per of pool is 20m and thats w+w+l+l

[scenic agate]

yeah get that

so solve ig

[zealous hill]

i was thinking of doing something along the lines of (150+w) x (150+l) = 4000

[scenic agate]

that sounds about right

[merry umbra]

re arrange your equation in terms of either length of width your choice so you are only dealing with 1 unknown

[zealous hill]

you mean instead of 2w+2l = 2000

2w = 2000-2l?

[merry umbra]

ye

then divide by 2

so w = 1000-l

then wherever you have to use w just substitute 1000-L

[zealous hill]

I can use the same equation?

2(1000-l)+2l=2000

[merry umbra]

eg 2w + 2l = 2000
would be 2000 = 2(1000-L) + 2L

[zealous hill]

that came out to 0

nvm

actually ye it did

[merry umbra]

yes because you need to use the information in the rest of the problem to create a new equation

[zealous hill]

I'll try (150+w) x (150+l) = 4000?

[willow pike]

here you go @zealous hill

[zealous hill]

oh wow

[willow pike]

now you know lw and l + w

now substitute for w = 10 - l

l(10-l) = 22.75

10l - l^2 = 22.75

l^2 - 10l + 22.75 = 0

or you can multiply by 4 to make it easier

4l^2 - 40l + 91 = 0

you should get pretty normal answers

what are you getting

[zealous hill]

what does the 2.25 mean?

[willow pike]

i just expanded (1.5+l)(1.5+w)

1.5 * 1.5 = 2.25

[zealous hill]

ah

[willow pike]

all of it is in metres

anyways what did you get from the quadratic

[zealous hill]

I re-worked it following your steps but in cms and it just messed up idk why

the150 x 150

[willow pike]

did you convert 40 m^2 to 40 * 10^4 m^2

i did meters because cms is easy to mess up with

[vocal sleet]

@zealous hill Has your question been resolved?

[zealous hill]

am here now

crazy to think all of this for just 6marks lol

[willow pike]

true

what did you get

[zealous hill]

l=3.5
w=6.5

however i didnt do the quadratic lol

I just trial and erorred my way into it

[zealous hill]

cause this part messed me up

[willow pike]

oh

how did that happen

[zealous hill]

since the quadratic wouldve started with a - no?

as -l^2

[willow pike]

just take it to the other side

0 = l^2 - 10l + 22.75

[zealous hill]

ye right lol

so simple bruh why am i overcomplicating thingssss

but yeah thanks for the help

especially with the diagram n such

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[vast shale]

help

[vast shale]

im back again @narrow eagle

x^2+9=0

hwo do

[narrow eagle]

u can find x by using the quadratic formula

[sharp briar]

complex numbers

[narrow eagle]

although in that case you'll be getting an imaginary number

[vast shale]

no work

tf is a imaginary number

[mental falcon]

if you don't know how to solve x^2 = c i'm guessing you aren't doing complex numbers

[vast shale]

im in grade 8 😭

[narrow eagle]

yep

[mental falcon]

x^2 = -9 so no solutions (a square is always 0 or positive)

[vast shale]

ok

what are complex numbers

x^2-36 is also complex numbers?

@mental falcon

[mental falcon]

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:complex/x2ec2f6f830c9fb89:complex-num/v/complex-number-intro

you probably won't learn that for 1-2 years though

[gritty hemlock]

should've been a - sign

I can sense the underpaid teacher vibes

[vast shale]

nah id win

[gritty hemlock]

bro is not gojo

[gritty hemlock]

ew byjus

[gritty hemlock]

check if this is really the question

[narrow eagle]

its a photo from google

xd

[gritty hemlock]

that company is in server debt

and used to downplay students in front of their parents

so that the parents would buy them the subscription

[vast shale]

it is

@gritty hemlock

help me

x^2-36=0

how do

[gritty hemlock]

mostly it was like "your kids base is weak" which might be the case but then they proceed to teach the level which they are at instead of making the base strong

[mortal holly]

there is no way to teach this. this counts as common sense dude

[mental falcon]

probably just stick to helping people in the help channels, not random rants @gritty hemlock

[vast shale]

x= +-6?

[gritty hemlock]

do you know difference of squares identity

[vast shale]

no

[gritty hemlock]

bruh

[mortal holly]

[gritty hemlock]

but yes that is correct

[vast shale]

Im grade 8

chiil

[gritty hemlock]

dawg it's in 7th grade

[vast shale]

🤫

[gritty hemlock]

$$a^2-b^2=(a+b) \cdot (a-b)$$

[twin meteor]

The Prophet Of The Damned

[vast shale]

ooh ik that

[obsidian vigil]

Do you know how to find the zeroes of a factorised polynomial

[gritty hemlock]

I never got the chance before

[obsidian vigil]

@vast shale

[vast shale]

no

[vast shale]

no

[obsidian vigil]

Oh

[vast shale]

maybe

idk what that big word means

[mental falcon]

$x^2-36=0\
x^2=36\
\sqrt{x^2}=\sqrt{36}\
|x|=6\
x=\pm6$

[twin meteor]

∫ooshⁱˣ

[gritty hemlock]

ok so not by name is fine
36=6^2 right

[obsidian vigil]

Factorised is expressed as a product od linear factors

So for example

[vast shale]

ye

[vast shale]

what does the lines next to x mean

[obsidian vigil]

(x-6)(x+6 ) is the same as x²-36

[vast shale]

ye

[obsidian vigil]

But the left is the factorised version

[mental falcon]

absolute value

[gritty hemlock]

so what will will x^2-6^2

[vast shale]

((x-6)(x+6))

[twin meteor]

html hater

[mental falcon]

if you take an even root of an even power, since it's always positive but the expression inside the even power could have been negative, you need to add an absolute value

[vast shale]

so

[gritty hemlock]

great now since that is equal to zero that means one if them is zero
so the first case becomes
x-6=0
and the second becomes
x+6=0

[vast shale]

x^2=-81 not possible without complex numbers

[vast shale]

k i get it

[gritty hemlock]

squares are always positive

[mental falcon]

or 0

[gritty hemlock]

it's on the edge

[vast shale]

k

[gritty hemlock]

(dont dare you talk about complex numbers)

[mental falcon]

what? that has nothing to do with complex numbers

[gritty hemlock]

so surely there exists no number whose square is -81

[vast shale]

ye

[gritty hemlock]

x^2=-81 has imaginary roots wdym

[mental falcon]

i'm talking about the squares being positive or 0 comment

[vast shale]

https://tenor.com/view/a-train-happy-couple-the-boys-a-train-the-boys-meme-gif-26392949

[gritty hemlock]

#help-17 message

[gritty hemlock]

bro is zoro

get back to the question

[vast shale]

mb

[gritty hemlock]

dont get lost

[vast shale]

kk

[gritty hemlock]

what was it again

[vast shale]

idk but i got it

thx

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[frail maple]

I need help with this question

[frail maple]

This is my work. I don’t k is if it’s correct or not

[fading willow]

he needs a unit vector

here n is just a normal vector

[frail maple]

How can I get it

[fading willow]

its the vector over its amplitude

[frail maple]

Do I use AB or AC?

[fading willow]

to get a unit vector in the same direction of the original vector, u divide that vector by its amplitude

which vector do u want to get its unit vector here?

[frail maple]

The normal vector?

[fading willow]

yes

[frail maple]

How about the angle do you think it’s correct?

[fading willow]

yep

[frail maple]

Alright thanks

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[formal bronze]

how did they go from here

[formal bronze]

to here

to this

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[gleaming owl]

@open sundial

[gleaming owl]

<@&286206848099549185>

sorry

I need to find the derivative of x^5ln(x) and what I've done is I've done the product rule f prime of x times g of x+ f of x times g prime of x and I got ((5x^4)(ln(x))) and I don't know if I should leave it like that or simplify by using the chain rule and then something else

but then i can't simplify cause I don't know how do do the chain rule with a function like that

the chain rule is the inside function evaluated at the outside function times the outside functions derivative

[lime gorge]

i dont see the other part of the product rule

5x^4 * lnx is just part of it

[gleaming owl]

so the product rule is the derivative of f times g plus the f times the derivative of g right?

isn't that what I did?

[sage dew]

that applies when there exists a product of two functions, this is just one function raised to a function power

you could try implicit differentiation

[lime gorge]

is the function: $x^5 \cdot \ln(x)$

[twin meteor]

Stephen

[lime gorge]

or is it not

[gleaming owl]

so how do I find the derivative?

yes

[sage dew]

oh I thought it was all in the exponent

then yes it would be a product

[lime gorge]

so what did u get for f' * g

@gleaming owl

derivative of f times g

[gleaming owl]

that is the problem

so for f' times g 5x^4*ln(x)

[lime gorge]

ok good

now how about f times g'

[gleaming owl]

x^5*1/x

[lime gorge]

which is?

simplify that

[gleaming owl]

one moment

[lime gorge]

what's up

[gleaming owl]

i'm stuck

[twin meteor]

Stephen

[gleaming owl]

yeah

would It just be that?

[lime gorge]

$\frac {a^b}{a^c} = a^{b-c}$

[twin meteor]

Stephen

[gleaming owl]

well actually it's x^5 times 1/x

any ideas?

wait

it is just that!

[lime gorge]

im not too sure what you're saying

[gleaming owl]

the simplication was correct I did the math

[lime gorge]

what did you simplify it to

x^5 / x

[gleaming owl]

x^4

[lime gorge]

correct

so our answer is

5x^4*ln(x) + x^4

[gleaming owl]

and I just leave it like that?

thanks! btw for the help

[lime gorge]

yea, or u could factor out an x^4 and get x^4 (5lnx + 1)

[lime gorge]

yea no problem

[gleaming owl]

alright gimme a while I'm gonna do the problem

i have to find where it's increasing and decrasing this was just a small part of it but I know how so thanks!

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[gleaming owl]

so....How do I find where it's increasing and decreasing?

[gleaming owl]

the previous problem? sorry again

[snow trout]

where what is increasing/decreasing?

[gleaming owl]

x^5ln(x)

[snow trout]

consider the values of f'(x) as f(x) is increasing, decreasing, or neither

e.g. what can you say about the slope as f(x) is increasing?

[gleaming owl]

so I got the derivative and it is x^4(5log(x)+1)

I need to find the critical numbers before I get the second derivative and I don't know how to do that

[snow trout]

critical points are where f'(x)=0 or dne

but that's different from what you were asking initially

[gleaming owl]

oh wait yeah

one second

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[hoary raven]

I just want to check my answer

[snow trout]

👍

you can always check your work by computing the inverse of the inverse lol

[hoary raven]

my bigger issues was if t-9 was right I googled a couple of similar answer and a lot of people were saying v-3 so I want to make sure using the t is correct

but if I reverse my x and y values then it has to be t and not v

[snow trout]

why change to x and y 🤔

[hoary raven]

its just something I do..might be better to just stick with V and T

[snow trout]

try to get comfortable with other vars if you can, you may come across more problems using different letters in the future

as far as independent vars go t is quite common actually

[hoary raven]

yeah I'm starting to notice that

I'll switch it to V and t before I submit it. Thanks for the help

[hoary raven]

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[snow trout]

letters are universal anyway (coping hard)

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[prisma kite]

can someone help me solve this?

[wicked shard]

lets do part a

first

its easier

[prisma kite]

k

so for a i think u just plug in -1

and a

and then divide by a+1

right

so since f(-1) is 0

[wicked shard]

yep

[prisma kite]

it would be sqrt(a^2-1)/(a+1)

[wicked shard]

change in y over change in x

[prisma kite]

b was the part that was mainly confusing me

[wicked shard]

alright

we need to do some algebra

write down the expanded limit thing of this function

while you do that ill do the same thing but in latex

[prisma kite]

this is what i got

lim h->0

[wicked shard]

$\lim_{h \to 0} \frac{\sqrt{(x+h)^2-1} - \sqrt{3}}{h}$

[twin meteor]

hired

[wicked shard]

oh you expanded it nice

so the challenge is somehow factoring out an h from that

[prisma kite]

yea

this is where i got stuck

cuz i didnt know what to do from here

[weary sentinel]

you can evaluate that limit using a conjugate

[wicked shard]

shit forgot about tha

t

[prisma kite]

but then how would you get rid of the square root in the denominator?

[wicked shard]

been a while since i did limits

[weary sentinel]

why get rid of it? the conjugate is strictly positive >0, what do you care that it's there

h is the problem

[prisma kite]

but h is in the square root

as well

[wicked shard]

yeah but we dont care about that

[weary sentinel]

yeah but right now we're fixing the numerator

[prisma kite]

ok

[weary sentinel]

we can't fix it all at once haha

[prisma kite]

lemme try that

[weary sentinel]

one thing at a time

also you need to replace x with 2

it's (2+h)^2 not (x+h)^2

[prisma kite]

ye

this is what i got

after doing that

what now

Also when someone sees this ping me so I get a noti

Huh?

How do you isolate h

So how do you do that

O

K

Thx

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[errant thicket]

Can someone pls help me get figure M onto figure N (I already know that you will need translation + rotation but i dont know the direction, units, and the rotation it goes.)

[errant thicket]

Translation goes: UP, DOWN, LEFT, or RIGHT ++++ AMOUNT of units. 90 degrees clockwise, 90 degrees counterclockwise, 180 degrees clockwise, or 180 degrees counterclockwise?

-fill in blanks

[marble hemlock]

You can think of the problem in reverse: how to go from figure N to figure M, and then reverse the transformations that you get.
This helps because it's easier to start with the rotation: how should we rotate figure N to get a figure that is in the same orientation as figure M?

[errant thicket]

oh, well im using deltamath.com and in the examples it has it like that and i have to have it this specific way as it is in the examples. Going from translation to rotation. If thats what your saying?

[marble hemlock]

No I'm saying that you can start with thinking about how you should rotate the figure and then translate it, because this makes it easier than thinking how to translate the figure then rotate it.

In the end you will write the transformations in the correct order of course.

[errant thicket]

ohh

okay i get it now tysm

[marble hemlock]

No problem

[errant thicket]

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[keen tangle]

Can someone help me approach this math problem?
So, I have $u_{n+1} = \frac{1}{n+1} + \sqrt{u_n}$ a sequence and I know that $u_0 = a$ where $a \in \mathbb{R}^+$

I need to find
$\lim_{n \to \infty} (\frac{u_n - 1 - \frac{2}{n}}{\frac{1}{n}})$
How would you guys try to solve this problem?

We know that $\lim{n \to \infty} (u_n) = 1$
And that $t_n$ which is a special case of $u_n$ where $t_0 = 4$, has for limit when $n \to \infty$ the limit: $\lim{n \to \infty} (t_n) = 1$ and $\lim_{n \to \infty} (\frac{t_n - 1 - \frac{2}{n}}{\frac{1}{n}}) =0$

[twin meteor]

Smile!

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@keen tangle Has your question been resolved?

[keen tangle]

<@&286206848099549185> Please

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@keen tangle Has your question been resolved?

[ancient horizon]

why yall doing math

lets talk about roblox

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[proper igloo]

is the region the tip of the cone??

i can't think it would be anything else but its not above the cylinder so im lost

i think its super simple and im just losin braincells but

[snow trout]

👍

bounded below by the cone tip, above by the cylinder surface

[proper igloo]

it would look something like that right?

wait

lemme fix

like that?

[snow trout]

i think they mean this?

[proper igloo]

ahhh

man i hate my textbooks wording lol

[proper igloo]

yea u rite

cool thankss

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[silk garden]

I have the transformations and everything but I don't understand how to apply them to these questions

[vocal sleet]

@silk garden Has your question been resolved?

[bronze osprey]

The first column of the standard matrix is where the basis vector (1, 0) ends up after the transformation

The second column is where the basis vector (0, 1) ends up

Does that help?

[silk garden]

Not exactly

Like what am I supposed to do when I have both transformations

And also it says dilation, does that mean both a horizontal and vertical dilation?

[bronze osprey]

Yes

A dilation is where all the directions get scaled by the same amount

[silk garden]

Okay

[bronze osprey]

Just see where each basis vector ends up

[silk garden]

So then if I have both transformations as matrices

[bronze osprey]

So for question a), where does (1, 0) end up after the two transformations?

[bronze osprey]

You could also do it like that

Then just multiply them

First matrix goes on the right, second matrix goes on the left

The order of matrix composition is from right to left

[silk garden]

So it's the (reflection x dilation)

As opposed to (dilation x reflection)

[bronze osprey]

Yes, so if A is the dilation matrix and B is the reflection matrix

It would be BA

[silk garden]

Ooooh

And that's just it?

[bronze osprey]

Yes

[silk garden]

Just curious, going about it like this would also yield the same answer as just multiplying the separate transformations right?

[bronze osprey]

Yeah, they would both give the same answer

[silk garden]

Okay, cause I think my professor may want me to do it the first way you mentioned

And I just wanna know if I'm getting the right idea with that

[bronze osprey]

The reason why is that if you do this, you're actually multiplying by the identity matrix {{1, 0}, {0, 1}} in disguise

[bronze osprey]

Yeah both methods are absolutely fine

I just prefer my way obviously

[silk garden]

So I'm going 3[1,0] and 3[0,1] I get [3,0;0,3] and then I go -1[3,0] and 1[0,3] and get [-3,0;0,3]

?

@bronze osprey

[bronze osprey]

Yeah so (1, 0) ends up at (3, 0), and then is reflected to (-3, 0)

(0, 1) ends up at (0, 3) and it is unchanged if you reflect across the y-axis (cause that's just x -> -x)

So yeah, the matrix with columns (-3, 0) and (0, 3)

[silk garden]

So I don't technically multiply by anything doing reflections, it's just indicating a reflection

[bronze osprey]

Yeah, well there is a reflection matrix also

It's $\begin{pmatrix} {-1 & 0} \ {0 & 1}} \end{pmatrix}$

[silk garden]

Yeah I know there's a reflection matrix

[twin meteor]

south
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[bronze osprey]

If you reflect across the y-axis

And yeah multiplying by this matrix will give you the same as doing each basis vector individually

[silk garden]

Cool cool

And only reason I think I should do it the first way is because my second question in the homework wants us to check if each part is commutative

By multiplying the matrices

[bronze osprey]

Right

[silk garden]

But yeah this makes a lot more sense now

Thank you!

[bronze osprey]

No worries!

[silk garden]

.close

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[silk garden]

.reopen

[vocal sleet]

✅

[silk garden]

@bronze osprey I'm sorry I have another question

In part c, I'm not sure how to apply the transformations to individual vectors

[bronze osprey]

Right, so if you reflect (0, 1) across the line y = x, it becomes (1, 0)

You just swap x and y around

And then we have a reflection which is similar to what we did in part a

[silk garden]

Yeah so my new vectors were [0,1] and [1,0]

[bronze osprey]

Remember that the first column represents where (1, 0) ends up

Ah ok

[silk garden]

Yeah I did the y=x reflection

[bronze osprey]

And then after you do the reflection...

[silk garden]

But it says the reflection is represented as [1,0;0,-1]

[bronze osprey]

Yeah that's the matrix

[silk garden]

So is mine just [0,-1;1,0]?

@bronze osprey

[bronze osprey]

Yep

$\begin{pmatrix} {0 & 1} \ {-1 & 0}} \end{pmatrix}$

[twin meteor]

south
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[silk garden]

Ah I see I see

@bronze osprey On part d, I have my 45 degree rotation

$\begin{pmatrix} {√2/2 & -√2/2} \ {√2/2 & √2/2}} \end{pmatrix}$

[twin meteor]

✿ 𝓢𝓬𝓱𝓲𝓪𝓯𝓯𝓲𝓷𝓸 ✿

$\begin{pmatrix} {√2/2 & -√2/2} \\ {√2/2 & √2/2}} \end{pmatrix}$
```Compilation error:```! LaTeX Error: Unicode character √ (U+221A)
               not set up for use with LaTeX.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.49 $\begin{pmatrix} {√
                          2/2 & -√2/2} \\ {√2/2 & √2/2}} \end{pmatrix}$
You may provide a definition with
\DeclareUnicodeCharacter```

[silk garden]

$\begin{pmatrix} {sqrt2/2 & -sqrt2/2} \ {sqrt2/2 & sqrt2/2}} \end{pmatrix}$

[twin meteor]

✿ 𝓢𝓬𝓱𝓲𝓪𝓯𝓯𝓲𝓷𝓸 ✿
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[silk garden]

Okay idk how to fix that

But I have this

I just don't know how to apply the y=-x transformation

[twin meteor]

✿ 𝓢𝓬𝓱𝓲𝓪𝓯𝓯𝓲𝓷𝓸 ✿
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vocal sleet]

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[drifting patrol]

Please help! I dont get it and my teacher haven't taught us anything. If you can explain please do! Thank you!

[elfin coyote]

do you know what "similar" means in the context of geometry?

[drifting patrol]

Yes

Similar triangles are triangle with sides proportional and angles congruent

[elfin coyote]

When a line is draw from the right "corner" of a right triangle to the hypotenuse, forming a 90 degree angle, it creates two smaller triangles that are similar to the larger one

For question 5, Triangle XWZ is similar to triangle XZY

[drifting patrol]

Oh so like the original triangle is similar to both the triangles formed?

[elfin coyote]

yes

does that help?

[drifting patrol]

Very much

But how do you get the mean

Like in 19-25

[elfin coyote]

the mean?

[drifting patrol]

The middle line

[elfin coyote]

oh i didnt look at those gimme a sec

[drifting patrol]

Oki, thank you

[elfin coyote]

ah ok

so if two triangles are similar, the ratios of the lengths of two corresponding sides are the same for both triangles

[drifting patrol]

Ohh i get it now, thank you so much

[elfin coyote]

the ratio of the length of the short leg to the length of the long leg is the same for all similar triangles

so for 19, 4/x= = x/16

and solve from there

alr glad i could help

[drifting patrol]

Oki, thank you!

.close

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[heavy terrace]

quick question

[heavy terrace]

for leading coefficient test, how do we determine which way it falls and rises?

i dont have a specific problem im trying to solve, but rather just general info

[heavy yoke]

for a polynomial, you have two main types of behavior

either the leading coefficient is odd, in which case it goes to -infinity on the left and infinity on the right (e.g., y = x)

or the leading coefficient is even, in which case it goes to +infinity in both directions (e.g., y = x^2)

if you have a negative sign in front of the leading coefficient that just flips the signs

[heavy terrace]

Im gonna nap rq but can u give example of both scenarios?

[heavy yoke]

for y = x^3 we follow the first scenario, it goes to -infinity on the left and +infinity on the right

for y = -x^3 it does the opposite, going to +infinity on the left and -infinity on the right

for y = x^8, it follows the second scenario, going to +infinity on both sides

for y = -x^8, it does the opposite, going to -infinity on both sides

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@heavy terrace Has your question been resolved?

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[potent stirrup]

does a ring homomrophism send unity to unity?

[river minnow]

Depends on which definition/type of rings we are talking about

For homomorphisms of unital rings, the unity is preserved

For homomorphisms of (possible nonunital) rings, the unity need not be preserved

[potent stirrup]

I want to determine all the rings homomorphism between Z_12 to Z_30

I found the possible group homorphims,

i want to proceed from there

[river minnow]

Does your definition of ring require unity?

[potent stirrup]

it's abelian under addition and x is a binary operation which is associative and satisfy distributivity

non commutative and doesn't have unity

[river minnow]

I see

Take the group homomorphisms that you have discovered and see which ones preserve multiplication

That should be it

[potent stirrup]

explicity check the preservation ?

for group homomorphism, i used the fact if phi(1) = a, then phi(x)=ax

then rule out the possible values of a using lagrage's theorem and homomorphism order division property

The possibilities are a=0,15,10,25,5,20

[river minnow]

So to preserve multiplication you need axy = a^2xy for all x, y in Z12

For x = y = 1 you get a^2 = a, so a should be idempotent in Z30

[potent stirrup]

a is fixed for me

[river minnow]

Right

[potent stirrup]

is it, $\varphi(a \cdot y) = \varphi( 1a \cdot y)$

[twin meteor]

Dubs

[river minnow]

What I wrote there is $f(xy) = f(x)f(y)$ simplified given $f(x) = ax$

[twin meteor]

A Lonely Bean

[river minnow]

Anyway the only values of a that satisfy a^2 = a are 0, 10, 15, 25

[potent stirrup]

is it x or cdot?

[river minnow]

x

[potent stirrup]

ohh x

mb

[potent stirrup]

i didn't get this sorry

[river minnow]

The homomorphisms determined by these values of a should be these

[potent stirrup]

axy=a^2xy

[river minnow]

If $f(x) = ax$, then how would you simplify $f(xy)$ and $f(x)f(y)$?

[twin meteor]

A Lonely Bean

[potent stirrup]

f(xy)=f(x).f(y)=axay

[river minnow]

And f(xy) = axy, right?

[potent stirrup]

since Z_30 commutes, it's f(xy)=a^2xy

axy=a^2xy

[river minnow]

Right

a = a^2 is yielded when we let x = y = 1 in that equation

Find which values 0, 5, 10, 15, 20, 25 satisfy a^2 = a

[potent stirrup]

but just verifying for x=y=1 will imply for any x,y?

[river minnow]

Well yes because any other x or y will be a sum of 1's anyway

[potent stirrup]

it works since its integers right?

in integers multiplication can be seen as repeated addition

but i don't think this is the case in arbitary rings, arbitary rings addition and mulitplication are fundementally different i suppose

[river minnow]

Yes, axy = a(1 + ... + 1) = a + ... + a = a^2 + ... + a^2 = a^2(1 + ... + 1) = a^2xy

[river minnow]

Of course

[potent stirrup]

= a + ... + a = a^2 + ... + a^2

can you explain this?

how did it become a^2+ a^2 + ... +

as a=a^2 ?

i see

makes sense

helpful

[river minnow]

Yes, given that

[potent stirrup]

how is this all related to the number of homomorphism is the gcd of two groups?

a hint maybe helpful than giving out answer :)

[river minnow]

Is it though? We found 4 homomorphisms whereas gcd(12, 30) is 6

Ah are you talking about the group homomorphisms?

[potent stirrup]

group

homomorphism i mean

[river minnow]

In general, when the domain of a homomorphism is cyclic, you can determine it by just the image of the generator

[potent stirrup]

between Zm and Zn, the choices of a are really the common divisors of m, n

[river minnow]

So given a homomorphism phi: Zn -> Zm, you need to just find all values of phi(1) such that the order of phi(1) in Zm divides n

Do you know how order of an element in Zm in found?

[potent stirrup]

the smallest the additive powers of a can go is the gcd (a,m)

so it reaches identity in m/(a,m)

[river minnow]

Right, so you need m/(a, m) to divide n

Hmm

[potent stirrup]

i think this is a homomorphism property

[river minnow]

The order of the image divides the original order, yes

[potent stirrup]

ie if a is mapped to phi(b), then order phi(b) divides |a|

in the previous example

the possible orders are 1,2,3, or 6

Z12 to Z30, by lagrange a must divide 30 and by image of homomorphism property a must 12

ie a is a common divisor 12 and 30

[river minnow]

Right

[potent stirrup]

now to collect a,

[potent stirrup]

i need to manually look for elements in Z12 whose gcd are these

0

15 has order 2

i don't think any other number in Z12 can have order 2

how do i find the number of elements in Z12 having a specific order?

just take phi(order) ?

i don't recall

[potent stirrup]

similarly i found others, but i don't know how this all connected to number of homomorophism being the gcd

@river minnow number of elements of order d in a cyclic group is phi(d) right?

where phi is totient function

the possible orders are 1,2,3, or 6

so number of homorophism are phi(1)+phi(2)+.._phi(6)

idk lol

im trying

[potent stirrup]

this worked tho

it's 1+1+2+2=6

not sure why its gcd

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[woeful rover]

this is the graph of which function ?

[woeful rover]

can answer in terms of x and y

[potent stirrup]

,w plot -1/x

[twin meteor]

Results provided by WolframAlpha

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Requested by dubleyou

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[lapis robin]

hello im studying for a test and how do i get the outside numbers for example on 30 degrees its radical 3/2 and 1/2 how do i get that?

[bronze osprey]

There are only two triangles you have to remember

This is one of them

The base has length a/2

So $(a/2)^2 + h^2 = a^2 \implies a^2/4 + h^2 = a^2 \implies h = \frac{\sqrt{3}}{2} a$

[lapis robin]

im still lost

[twin meteor]

south

[bronze osprey]

Lost on which part?

[lapis robin]

can you like use a calculator to find it?

[bronze osprey]

Well yeah but that'd be missing the point a bit

If the point is to know these without a calculator

[lapis robin]

ohhh

yeah my test is no calc

[bronze osprey]

So if we take cos 60 for example, cos 60 = adjacent/hypotenuse = $\frac{a/2}{a} =\frac{1}{2}$

[twin meteor]

south

[bronze osprey]

and similarly sin 60 = $\frac{\sqrt{3}a/2}{a} = \frac{3}{2}$

[twin meteor]

south

[lapis robin]

a is adj right

[bronze osprey]

a is the side length of the equilateral triangle

[lapis robin]

oh

[bronze osprey]

So what we did was divide it into two smaller triangles

[lapis robin]

where did you get the 3a/2 from

[bronze osprey]

This

Pythagoras

[lapis robin]

wait so every time like the denominator is 2 for both

bc for all of them is /2 on each for everything

[bronze osprey]

No, so the base is a/2, the hypotenuse is a

And the height is h (for now)

[lapis robin]

i get that but how do i get these now without a calc

[bronze osprey]

Ah so with the equilateral triangle cut in half you can get pi/3 and pi/6

For pi/4 you need a different triangle

[lapis robin]

45 45 90

[bronze osprey]

Yep

Actually you can just let the bases be 1 cause it's only the ratios that matter

So if you have the base being 5 and the hypotenuse being 5 sqrt(2)

cos 45 is 1/sqrt(2), which is the same as 5/(5 sqrt2)

So it might be easier to think of the 30-60-90 triangle as being 1, sqrt(3), 2

The hypotenuse is still twice the base

[lapis robin]

hm ok

wait so all those numbers are always the same right

i can just memorize the first quadrant then just copy

[foggy arch]

have you learnt circular trigonometry

[lapis robin]

nah

lol

i dont remember a thing from geometry

[foggy arch]

man if you havent learnt circle trig

[lapis robin]

nah i get it tho

[foggy arch]

just learn 2 tirangles

remember these 2

thats all you need

[lapis robin]

i know those

30 60 90 then 45 45 90

[foggy arch]

then thats all u need for the test

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[jovial vale]

what is the connection between field extension and the degree of the minimal polynom? for example [L:K]=17 , what do i know about the degree of the minimal polynom?

[hard atlas]

its 17

because if alpha generates L, then 1,alpha,alpha^2,...,alpha^16 are linearly independent (otherwise it wouldnt generate L and instead a subfield (which could actually only be K itself because 17 is prime)) and then alpha^17 is linearly dependent on the previous ones, so you can write down a polynomial of degree 17 which has alpha as a root

[jovial vale]

does it make a difference if the degree is non prime?

[hard atlas]

then there could be fields between L and K

if there is a field K<=F<=L, then [L:K]=[L:F]*[F:K]

[jovial vale]

but the degree of the polynomial is still only 16 as an example, if [L:K]=16?

[hard atlas]

well the polynomial is the wrong way to say it

unless you are talking about a specific element already

there could be several elements which generate L

and each of them will have a minimal poly of degree 16

but these minimal polynomials can be different

gtg

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[wanton shuttle]

Why only there are two legs the are equal and not all of them?

[hardy vector]

Do all the sides look equal?

[wanton shuttle]

nope

[hardy vector]

Then there is your answer

[wanton shuttle]

i mean this

@hardy vector

[hardy vector]

I think they mean BD = AC.

I don't understand what your question is?

[wanton shuttle]

nope

[hardy vector]

You can use congruency testing

[wanton shuttle]

the question in why they are only two legs that are equal and not 4

its a class question

[hardy vector]

So the first step would be to determine what sides you are referring?

[wanton shuttle]

i need to send to the teacher why

oh sorry yeah

diagonals

[wanton shuttle]

the AB and DC

[hardy vector]

What you mean by not 4?

What are the 4?

[wanton shuttle]

diagonals

all diagonals are not equal

so the class assignment is to tell why there only two diagonals that are equal and not all off them

@hardy vector

[hardy vector]

To provide an intuitive explanation.:

Do you understand the sine rule

[wanton shuttle]

what is the sine rule?

[hardy vector]

Forget it

[wanton shuttle]

we didn't learn in class the sine rule

[hardy vector]

Basically since the side AD is smaller than BC we end up with diagonal not being equal.

[wanton shuttle]

right

[hardy vector]

Let E be the intersection point of the diagonals