help-13

[cedar kiln]

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[kindred storm]

No problem.

[cedar kiln]

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[slender jay]

Hi

[slender jay]

How would I approach this q1

v

This is the q

[slender jay]

I got to here

[deep heath]

Let u = sqrt(m), u^2 = m. So you end up with a polynomial for left hand side.

[slender jay]

wait

so -10u^2-8u-3<0

[deep heath]

Yep.

This looks nicer to deal with.

[slender jay]

quadratic formula ?

lmao

is it nicer though

since we still have to find a value less than 0

if you set that to 0

you get complex roots

bear in mind this is a non calc question

[deep heath]

You need to first figure out when
-10u^2 -8u - 3 = 0,
So the discriminant is
64-4(-10)(-3) = 64-120 < 0.
So no real solution.
We have all values to -10u^2 -8u -3 is either positive or negative, so choose a given values to test it out.

[slender jay]

but we want to find a value

of m right?

[deep heath]

Pick something nice like we have u = sqrt(m), choose m = 1 so that u = 1. We have -10u^2 -8u -3 = -21, so all non negative real number is you solution. The reason for non negative is u = sqrt(m).

[slender jay]

so how would we go about doing that

its a value

like a number

thats the value of m

but idk how they got it

1/4

[deep heath]

Wait, I was doing the first part of the problem. The first screenshot.

[slender jay]

oh the -10m-8sqrtm-3<0

thats the simplified version

of the question

i simplified it

[deep heath]

Yeah.

[deep heath]

The thing is since we have no real roots, either the graph of -10u^2 -8u -3 is above the x axis or below it.

[slender jay]

yeah

[cedar kiln]

@slender jay Has your question been resolved?

[cedar kiln]

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[slender jay]

.reopen

[south juniper]

@crimson sedge What are you asked to do?

[slender jay]

you want to find the volume of both shapes

and add t hem together

so the bottom one is a cone right?

whats the top one

a right cylinder i think

yes

so find their volumes

and add em up I believe

you're given all the info

so now its a matter of plugging the values in properly

yes

whats the radius ?

do you know?

yes

[harsh relic]

4 is the radius of the cone in fact

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[ivory marten]

What are the steps to solve this? I don't just want the answer obviously. This is for an online class (calculus 3) that doesn't have any lectures so I usually learn this stuff through youtube. This time around, I cannot find a video for this type of problem.

[zinc wraith]

Do you know how to approximate the values?

[ivory marten]

I don't think so. No

[zinc wraith]

What ab finding tangent plane at a point?

[ivory marten]

...No

Is it two separate things

in one question

[zinc wraith]

https://tutorial.math.lamar.edu/classes/calciii/TangentPlanes.aspx

Consider this resource

Paul’s online notes are really useful

[ivory marten]

thank yoi

[cedar kiln]

@ivory marten Has your question been resolved?

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[lucid maple]

how do i divide a non isosceles triangle into an equilateral triangle, an isosceles triangle, a sharp triangle, an obtuse triangle and a right triangle?

[cedar kiln]

@lucid maple Has your question been resolved?

[lucid maple]

<@&286206848099549185>

[bold hinge]

what is a "sharp" triangle

i've never come across that term before

and googling doesn't turn up anything either

[cosmic steppe]

Knife

[bold hinge]

?

this is weird terminology you're using

[cosmic steppe]

A sharp triangle is a way to describe a kitchen knife

[bold hinge]

ok sure

but what is it

[cosmic steppe]

I'm assuming acute

[bold hinge]

what is that supposed to mean lmao

[cosmic steppe]

Acute triangle idk

You never know if there's a triangle that's just all obtuse smh my head

It could be a 5D triangle that has all right angles

[cosmic steppe]

What even is "sharp"

@bold hinge uh

So scalene with no equal sides?

This is the closest I can find but like

[lucid maple]

sharp means all angles all below 90 deg

[cedar kiln]

@lucid maple Has your question been resolved?

[lucid maple]

nobody?

[cedar kiln]

@lucid maple Has your question been resolved?

[rustic siren]

you want to divide a triangle into 5??

why tho

@lucid maple

[cedar kiln]

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[fallen vine]

I don't get d)

[fallen vine]

I have the domain of f

It's D: {x E r | -1 < x <1}

^ I don't really get how I did it

but I just know that it's in between these two numbers

[flint plinth]

1 - x^2 equals (1-x)(1+x), not (x+1)(x-1)

1 - x^2 > 0 if and only if (1-x)(1+x) > 0

[fallen vine]

Right

[flint plinth]

that last inequality says that the product of two numbers is positive, which can only happen if both numbers are positive or both are negative

so there are two possibilities:

(1) 1-x > 0 and 1+x > 0

[fallen vine]

because they have to both be bigger than 0

in order for that to be true

I see

[flint plinth]

or (2) 1-x < 0 and 1+x < 0

[fallen vine]

oh right

yea

they could both be negative too

[flint plinth]

notice that (2) is impossible because it's equivalent to x > 1 and x < -1

so (1) must hold, and (1) is equivalent to -1 < x < 1

[fallen vine]

oh yea

I see

Ok that makes sense

what about d)

I was thinking about it

but how can I prove that the second derivative is < 0

do I put in the end points of the domain?

[flint plinth]

well first compute the first derivative

then compute the derivative of that, and hopefully it'll be clear why

[fallen vine]

oh wait

Ok so

Lightbulb moment

here's the second derivative

but anything I put, will be positive in those brackets

[flint plinth]

yes that looks helpful

[fallen vine]

which means

that the -2

will apply to the whole equation

and therefore make the whole equation negative

[flint plinth]

yep (x^2 + 1) is always positive, and so is (x^2 - 1)^2

[fallen vine]

yea

[flint plinth]

exactly

[fallen vine]

aw yea

ty :)

[cedar kiln]

@fallen vine Has your question been resolved?

[cedar kiln]

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[orchid sleet]

say that symmetric linear operator $S$ on $\mathbb{R}^2$ has a matrix representation $A=\begin{bmatrix} a & b \ c & d \end{bmatrix}$ in some orthonormal basis ${u_1,u_2}$ such that $ab+cd=0$ (which is the same as $S(u_1)\cdot S(u_2)=0$). what other properties does $S$ have?

[wraith dagger]

Beous

[cedar kiln]

@orchid sleet Has your question been resolved?

[orchid sleet]

ok i was able to show that the eigenvalues of S must have the property that h_1^2=h_2^2

.close

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[obsidian sinew]

does anyone know why adding one more 4 cent stamp makes it k + 1

[cedar kiln]

@obsidian sinew Has your question been resolved?

[cedar kiln]

@obsidian sinew Has your question been resolved?

[amber ridge]

I'm not sure tbh

so I looked up strong induction and apparently you have to use the fact that it works for all values below the value you're on

so what I have is: assume for all 30 <= n <= k, that there are values a>0, b>0 that satisfy 4a+11b=n

but I'm not sure how to use that fact to show that there are values a>0, b>0 that satisfies 4a+11b=k+1

any hints would be appreciated

[amber ridge]

oh I believe this should be a>=0, b>=0

but yeah I'm still not sure how to do the k+1 case

[cedar kiln]

@obsidian sinew Has your question been resolved?

[obsidian sinew]

yeah k + 1 for strong induction is still confusing to me

[amber ridge]

I think I get it now

basically you want a solution for 4a + 11b = k+1

you first have to find solutions for n = 30, 31, 32, and 33

from then on, you can look at the solution for k-3 (which is greater than 30), and then increase the value of a by 1

the resulting equation will give you the solution for k+1

so basically you're looking at the solution that was 4 steps previous and using it to prove your current solution (of k+1)

[cedar kiln]

@obsidian sinew Has your question been resolved?

[obsidian sinew]

so by adding 4 , k -3 becomes k + 1?

[cedar kiln]

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[normal crescent]

trig proofs
I swapped out the bottom sin(2A) for 2sinAcosA
not too sure what to do to the top 2cos(2A) to make it equal the right hand side, do I have to do something like 2(cos^2A-sin^2A)?

[cedar kiln]

@normal crescent Has your question been resolved?

[normal crescent]

nvm i got it

.close

[cedar kiln]

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[crimson sedge]

Hey i am coding a discord bot to calculate profit shares inside my team.

My team share percent is:
Person 1: 10%
Me: 40%
Person 2: 50%

However, on some sales, i have to recieve 60% of the shares.

How can I calculate the new rate of Person 1 and Person 2 after applying my new rate

[crimson sedge]

(I am terrible at maths, and I won't show any work, it's a math not coding server 🥴👍

[reef venture]

if you want to get 60% of the profit how will you give 50% of it to person 2

[crimson sedge]

This is my question how can o equally sink the other two shares to reach 100%

Shrink*

[reef venture]

ah i see

so person 1 gets 10% of the total, person 2 gets 50%

what is the ratio of person 1's profits to person 2's profits?

im asking this to you, i already know it

[crimson sedge]

Im pretty bad at maths but i would say the ratio would be
.1 and .6 by default

If this is what ratio means x)

[reef venture]

do you know what ratios are?

its a fraction basically

a/b

[crimson sedge]

No i don't else I would have it in my code haha

[reef venture]

so just divide .1/.6

or 1/6

[crimson sedge]

So round.16

[reef venture]

so person 1 gets 1/6th of the profit that person 1 and person 2 together make

which would be 40% of the total profit if you take 60% from it.

[crimson sedge]

Yes this is right

[reef venture]

in other words if the total profit is x.
person 1 gets 1/6 × .4 × x

person 2 gets 5/6 × .4 × x

[crimson sedge]

Correct but this is assuming both always get the same rate

But sometimes I may get 80%

Or 20%

[reef venture]

just change this

1-your rate

if you get 80% then this becomes 1-0.8 = 0.2

person 1 gets 1/6 × .2 × x

[crimson sedge]

Multiplicator

Ok got it

Thanks for your help i will try to apply this formula later !

.close

[cedar kiln]

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[shadow lynx]

Hi

i dont get this thing

[torn hare]

Have u tried solving 7?

With an area formula?

[shadow lynx]

what the formula again?

i forgot

[torn hare]

Pi r squared

[shadow lynx]

452.16?

wrong?

[sick ruin]

it's right

you don't have to write decimals tho

just write 144pi

[shadow lynx]

ohh

oke

[torn hare]

Wait

We got the diameter

Not radius in the question

[shadow lynx]

7. 144π

?

[torn hare]

36 pi

Since it’s the radius squared. The question is the talking about diameter which is 2r

[shadow lynx]

ohh

i dont get it

[torn hare]

What part?

[shadow lynx]

7

[torn hare]

The area of a circle is pi * r^2. Right?

[shadow lynx]

yes

[torn hare]

And in the question we get diameter

Not radius

Correct?

[shadow lynx]

ohh

yes

[torn hare]

So to get the radius we just divide diameter by 2

12/2

=6

[shadow lynx]

6

6^2 is 36

[torn hare]

Yes

[shadow lynx]

ohhhhhhhhhhhhhhh

[torn hare]

Yes

[shadow lynx]

36 cm^2?

[reef venture]

forgot the pi

[torn hare]

Just 36pi

But yes, with units it’s 36 pi cm^2

Make sense?

[shadow lynx]

yep

Sector COL

[torn hare]

This also has a formula.

[shadow lynx]

(60/360°) x pi*36^2

?

[torn hare]

How did u get theta-360?

Not a minus

[shadow lynx]

OH

mb

[torn hare]

Very close

It’s just the radius squared

Not the radius squares squared

So just 36

Not 36^2

[shadow lynx]

ohhh

(60/360°) x pi*36

[torn hare]

Yes

[shadow lynx]

1/6 x 113.04

18.84

[torn hare]

Looks right

[shadow lynx]

18.84π cm^2

[torn hare]

Wait, idk if u need an extra pi

Because to get the 113.04 didn’t u already multiply the radius by pi? To get that?

[shadow lynx]

i see

Sector LOQ

[torn hare]

You can use a sine formula

[shadow lynx]

sine?

[torn hare]

This one takes a little diagraming for easier visualization

1/2absineC

[shadow lynx]

i think i havent study sin yet

[torn hare]

There may be another way, I just thought of this first

[shadow lynx]

oh ok

[reef venture]

there's a formula for the area of an equilateral triangle

even if you do not remember that you can derive it

[torn hare]

It’s an isosceles triangle right?

[shadow lynx]

this?

[reef venture]

60deg

[torn hare]

How do u know it’s equilateral?

[reef venture]

yes

[torn hare]

It’s LOQ

[reef venture]

oh sorry didn't see that

[torn hare]

All good

[torn hare]

I think this is easiest

[reef venture]

they haven't learn sin yet

you cant expect them to use trig

yes it is the easiest though

[torn hare]

True, agree. I think it’s simple to teach?

Or explain

[reef venture]

yeah but what if they need to show the working

[torn hare]

@shadow lynx what math class is this assignment for?

[shadow lynx]

Circles and arcs

[torn hare]

But what’s the overall class?

The class in general

Or if u are okay sharing, your grade?

[shadow lynx]

grades?

[torn hare]

Grade level?

Year?

[shadow lynx]

grade 6

[torn hare]

Wait, and your learning sectors?

[shadow lynx]

ye my school is advanced

[torn hare]

U in the US? Or somewhere else?

[shadow lynx]

im from malaysia

anyway lets skip the LOQ

[torn hare]

I think I could show u

The rule it’s very simple in this context

[shadow lynx]

oh ok

[torn hare]

However all your choice

[shadow lynx]

show me an example

for LOQ

[torn hare]

First off

We can establish that the triangle LOQ is an isosceles triangle

Do u understand how?

[shadow lynx]

no

[torn hare]

Since two of the sides are just the radius and the radius is the same always, it can be established it’s isosceles

Just by the rule of an isosceles triangle

[shadow lynx]

bh/2?

[torn hare]

We can’t use that because we don’t know b or h

But does it make sense why it’s isosceles?

[shadow lynx]

there sides are equal?

[torn hare]

Two sides are the same and since the third side is not a radius it can’t be equilateral

[shadow lynx]

a + b + c = 180

?

[torn hare]

Well we can use that next

To find the three angles of LOQ

So we have established it’s an isosceles

Now, we can find the angles

What is angle LOQ?

Using the information we already have in the question

[shadow lynx]

60?

OH

12

[torn hare]

12?

[shadow lynx]

nvm

CQ is 12

36

[torn hare]

All good

I’m asking for angle degree

The angle of O

Which we can find using the idea that 60 + angle O = 180

Does that make sense?

We can assume this because CQ is the diameter

[shadow lynx]

60 + 36 = 180
96 = 180
1.875

[torn hare]

???

[shadow lynx]

wrong?

[torn hare]

That is not an accurate statement

And without a variable idk what your solving for

[shadow lynx]

hmm

[torn hare]

Going back to this. Can u solve for angle O?

Think of it as X if that helps

[shadow lynx]

x = 120

[torn hare]

Yes

That is the angle

[shadow lynx]

ohh

[torn hare]

Does that all make sense?

[shadow lynx]

yes

[torn hare]

Alright

[shadow lynx]

120 cm^2

[torn hare]

Now we need to use the formula

I mentioned awhile ago

Here is an example

It has the exact same information (different numbers obv) as we have

I would make a diagram of your triangle and the info u know

And then try plugging in the numbers into the eq 1/2abSinC

[reef venture]

trig in 6th grade 💀

[torn hare]

I know right

I was doing like basic algebra

[reef venture]

we were introduced to algebra in 6th

[torn hare]

Exactly. Same

[shadow lynx]

where did 6 came from

[torn hare]

It was when they did 3 * 4 * .5

[shadow lynx]

ohhh

[torn hare]

12 times .5

[shadow lynx]

3.45?

[torn hare]

Can u show me your eq?

[shadow lynx]

i dont have a camera rn

[torn hare]

Type?

[shadow lynx]

ok wait

6 sin (120°)
6 (0.5806)

3.4836

[torn hare]

Why did u use 6?

[shadow lynx]

ohhhhhhhhhhhhh

i forgot

60 and 120?

ahhhhhh

wiat

wait

i dont get it

[torn hare]

All good

So u want to focus on the 1/2 ab portion of the formula

A and b represent two sides of the triangle

[shadow lynx]

yes

[torn hare]

In your triangle, u know two sides

What are the lengths of your two known sides?

[shadow lynx]

60 and 120?

[torn hare]

???

Length

[shadow lynx]

LMFAO

sorry

[torn hare]

All good

Look at the picture

[shadow lynx]

this

?

[torn hare]

Your problem

Mb

I’ll give u a hint. They are the same

[shadow lynx]

60?

[torn hare]

That is an angle

No length can be that large because think about your diameter

It’s only 12

[shadow lynx]

6?

[torn hare]

Yes!

[shadow lynx]

i think i need to study more LOL

[torn hare]

So you will have 1/2 * 6 * 6

Or in full

[shadow lynx]

12

[torn hare]

1/2 * 6 * 6 * sin 120

[shadow lynx]

12 sin (120)

[torn hare]

Hmmm

Try that multiplication again

[shadow lynx]

3

xd

[torn hare]

6 x 6?

[shadow lynx]

36

[torn hare]

Times 1/2

[shadow lynx]

18

my braincells are gone

[torn hare]

18 sin 120

[shadow lynx]

10.4510

[torn hare]

U in degrees on your calc?

[shadow lynx]

10.3923

[shadow lynx]

RAD

[torn hare]

You need to be in degrees

Because your angle is in degrees

[shadow lynx]

this

[shadow lynx]

this is in deg

[torn hare]

Is that a final answer?

[shadow lynx]

yes

[torn hare]

I didn’t get that answer either in degrees or radians

But either way, your calculator needs to be in degrees not RAD because your angle (120) is in degrees.

[shadow lynx]

OH

I TIMES IT AT 12

NOT 18

[torn hare]

All good.

No worries

[shadow lynx]

15.5884

[torn hare]

Yes

[shadow lynx]

15.5884 cm

[torn hare]

I got that as well

Cm^2

[shadow lynx]

okkeee

[torn hare]

Next one?

[shadow lynx]

wait

[torn hare]

And do u generally understand? Ik it’s new material and such. But generally

[shadow lynx]

i do

[torn hare]

So youre good?

For that one?

[shadow lynx]

ye

i can do 10 and 11

for now

Thank you ellie

[torn hare]

No problem

[shadow lynx]

should i close it or a bot will appear

[torn hare]

Close it if you’re done

[shadow lynx]

and he will asked some question if my question is done

[torn hare]

Just use the command .close

[shadow lynx]

ohh

Ok thanks have a good day or night

.close

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[tulip orchid]

for this.. i have a question
For ratios...
is it always smaller/bigger?

[tulip orchid]

ps. i might not reply cuz i needa sleep

[glad kestrel]

as long as you’re consistent, it doesn’t matter

[cedar kiln]

@tulip orchid Has your question been resolved?

[tulip orchid]

i mean like

if i have to find the larger bowl.. is it always gonna be
smaller/ bigger x the opposite of what we are looking for

ok for example

we know what the height for the small one is but we dont know what the height for the larger bowl is

like um

4/7.8 = 9.2 / x

to find the larger bowl's height

will it always be
x= 4/7.8 x 9.2

its hard to explain

ill deal with this tomorrow

[cedar kiln]

@tulip orchid Has your question been resolved?

[cedar kiln]

@tulip orchid Has your question been resolved?

[glad kestrel]

as i said, as long as you're consistent, it doesn't matter

4 / 7.8 = 9.2 / x will yield the same result as 4 / 9.2 = 7.8 / x

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[hidden marsh]

Hello

[hidden marsh]

I was wondering if anyone can help me with the range of T

What would be the range of T?

Wouldn’t it be everything in that matrix?

(a+b)c, a+b+c+1, ab+1

[dire geode]

No

The range is the subset of R^3 that can be mapped to by T for some vector (a,b,c). It's not asking for the definition of T, it's asking for what values T can take

E.g. y=x^2 for x real. The range isn't x^2. The range is [0, infinity)

[hidden marsh]

Hmmm

The question actually asks if the range of T is a subspace of R^3

@dire geode

Wouldn’t the range go to infinity on this one too?

a = (a+b)*c

The range would be (-inf, +inf)

I guess?

It’s in R^3

Or am I wrong 😅

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[west dome]

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[wanton glacier]

don't get the last line

[wanton glacier]

$|7|>5$

[wraith dagger]

Hyperlix

[wanton glacier]

that implies

$7<-5$ or $7>5$

[wraith dagger]

Hyperlix

[wanton glacier]

ohhhh

or

okay

get it

[broken mist]

It's an or statement yeah

SO only one of those has to be true

You got it

[wanton glacier]

yeah

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[fierce laurel]

I do not understand C.

They did 300 + (80/0.7)

I know where they got every number from, but i do not understand the logic from the math they did

[cedar kiln]

@fierce laurel Has your question been resolved?

[grim moth]

1 + x + x^2 + x^3 + x^4 ... + x^infinity = 4

[fierce laurel]

What?

[cedar kiln]

@fierce laurel Has your question been resolved?

[dire geode]

What step of what math are you stuck on

[crimson sedge]

Well

I don't understand why they put the numbers in that fashion

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[crimson sedge]

.reopen

[crimson sedge]

Like why did they do 300 + (80/0.7)?

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[crimson sedge]

Uh

[crimson sedge]

[crimson sedge]

My question is up here

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[sour cedar]

Hey, I have a quick question

[sour cedar]

It's about normal distributions in stats

[deep heath]

Ask away.

[sour cedar]

I think I'm mistaken, but can I take the mean and standard deviation of any dataset and put it on a bell curve, or should the data set be normally distributed by default?

[wraith dagger]

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

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[formal nest]

how do i find the zeros ?

[dire geode]

,w factor (2x^3 - 6x^2 + 4x + 1)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by riemann#2529

[dire geode]

disappoint

,w zeros (2x^3 - 6x^2 + 4x + 1)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by riemann#2529

[dire geode]

well that doesn't look fun

[formal nest]

nope

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[cloud sphinx]

How to do question 7 and 9

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<@&286206848099549185>

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[untold salmon]

Hi! What does it mean for a set to coincide with a plane exactly?

[untold salmon]

Please someone help I have a meeting soon and I want to be prepared any input would be appreciated

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[crimson sedge]

Hey i got some issues with a matrix , im suposed to calculate A^H and afterwards (A^H )*A , but the issue is its an 2x3 matrix . How can i calculate A^H ?

[sweet stag]

I assume it means hermitian transpose

Which is just the standard transpose but you take complex conjugate of entries aij

[crimson sedge]

Like for example we got ( from wikipedia ) {1 -3i 5-i , 2+i 4+2i 6i } so i build A^T wich would be { 1 2i , -3i 4+2i , 5-i 6i} but how can i change the addition to subtraction from the i

Which is just the standard transpose but you take complex conjugate of entries aij
@sweet stag
So can i just change like 2+i to 2-i?

[sweet stag]

That's just the complex conjugate

this is what you mean right

[crimson sedge]

Jep

[sweet stag]

this is the conjugate transpose

its valid to do that

[crimson sedge]

Cuz thats what ive gotten atm

@sweet stag ok i guess i can do that then , thanks

[sweet stag]

yeah the complex conjugate of each the entries reverses the sign on the imaginary component

https://en.wikipedia.org/wiki/Complex_conjugate

[crimson sedge]

Ok i think i got it now thank you very much

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[tulip light]

what would last be?

28 was incorrect as well

anyone?

@upper abyss

what would last part be?

[upper abyss]

I don't know what |Δ| is supposed to mean, but they haven't asked you what the area under the curve is

So 7×4?

[tulip light]

28?

i tried

It was 0 lol

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[tame current]

This first bit is just the prompt.

Suppose that data are assigned to some particular objects $\Omega_1$ and $\Omega_2$ as follows. For every coordinate system $\textbf{x}=\widehat{x}^i(u^j)\textbf{e}_j$ on $\mathbb{R}^2$, the object $\Omega_1$ is assigned two values (indexed by $i\in\lbrace 1,2\rbrace$), and $\Omega_2$ is assigned four values (index by $i,j\in\lbrace 1,2\rbrace$). For example, with respect to the $u^j$-coordinates system given by $\textbf{x}=x^i\textbf{e}_i=((u^1)^2-u^2)\textbf{e}_1+u^1\textbf{e}_2$, the object $\Omega_1$ is assigned $-2$ as its first value, and it is assigned $1$ as its second value. Also, the $(1,1)$, $(1,2)$, $(2,1)$, and $(2,2)$-values assigned to $\Omega_2$ are $2$, $0$, $1$, and $-1$, respectively.
\
\A group of mathematicians and physicists discover that, for any other coordinate sytem $\textbf{x}=\widehat{x}^i(v^k)\textbf{e}_k$, the $i$the value assigned to $\Omega_1$ is $-2\frac{\partial v^i}{\partial u^1}+\frac{\partial v^i}{\partial u^2}$ and the $(i,j)$-value assigned to $\Omega_2$ is $2\frac{\partial v^i}{\partial u^1}\frac{\partial u^1}{\partial v^j}+0\frac{\partial v^i}{\partial u^1}\frac{\partial u^2}{\partial v^j}+\frac{\partial v^i}{\partial u^2}\frac{\partial u^1}{\partial v^j}-\frac{\partial v^i}{\partial u^2}\frac{\partial u^2}{\partial v^j}$, where the partial derivatives are evaluated at the point $\textbf{x}_0\sim (-1,1)$.
\
\Realizing that this is precisely how components of vectors and linear transformations behave (indeed, the above formulas are transformation laws for vectors and 2nd-order tensors!), mathematicians and physicists can study these objects by assuming that $\Omega_1$ is a vector in $\mathbb{R}^2$, and $\Omega_2$ is a linear transformation on $\mathbb{R}^2$ (with $\textbf{x}_0$ regraded as the origin of $\mathbb{R}^2$). In particular, they have all of the mathematical theory about vectors and linear transformations to study the properties of these objects!

[wraith dagger]

glomswamp

[tame current]

What I'm trying to show is what Omega_1 and Omega_2 are with respect to the standard basis

My gut instinct is to say they are the coefficients on the transformation law given respectively, but I'm not sure how I would go about proving that.

I guess maybe using just the transformation laws themselves, but maybe then the coefficients represent the components with respect to the u^j system instead of the standard basis

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<@&286206848099549185>

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[woeful hill]

does anyone know where the -8 is coming from?

[dire geode]

completing the square

they added zero to y=2x^2 - 8x + 2. specifically, 0 = 2 * 4 - 8

[woeful hill]

could you elaborate a little more

[mossy flower]

do you want to draw a graph ?

like on the graph sheet ?

[dire geode]

https://www.storyofmathematics.com/completing-the-square

[mossy flower]

instead of that

we can use quadratic formula ryt ?

[dire geode]

quadratic formula works i guess. just a lot more work

where as adding and subtracting a constant to get your equation in vertex form is way simpler

[mossy flower]

but for me it looks easier than completing the square

[mossy flower]

hmm

[dire geode]

nah i generally don't recommend doing things the hard way when there are more obvious simpler ways.

simpler means fewer room for mistakes

[mossy flower]

yeah i agree

thank you for enlightening me

[woeful hill]

I still don't understand

ugh

[dire geode]

read

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[crimson sedge]

I'm having a hard time finding these limits

I get that its spherical coordinates so the last limit is 0 - 2pi

[dire geode]

set them equal?

it's definitely not spherical coordinates

[crimson sedge]

That May be obvious to you lol, I'm teaching myself here.

It's my hobby man

[dire geode]

Read textbooks?

[crimson sedge]

I am dude why are you so condesendeijg

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[dire geode]

just giving helpful suggestions.

[crimson sedge]

Next time skip over me

[dire geode]

sure

[tropic oxide]

.reopen

[cedar kiln]

✅

[tropic oxide]

@crimson sedge spherical? not cylindrical?

given that z=r is described as a cone i would think the coordinate system is cylindrical

[crimson sedge]

Yes that makes sense

[tropic oxide]

you can integrate this in polar coordinates by noting that the height of your solid is sqrt(1-r^2) - r

and you integrate wrt r from 0 to whichever value makes sqrt(1-r^2) = r

$\int_0^{2\pi} \int_0^{1/\sqrt{2}} (\sqrt{1-r^2}-r) r \dd{r} \dd{\theta}$

[wraith dagger]

Ann

[crimson sedge]

Okay, lemne take that in

Where did the -r

In the sqrt come from?

Nm

Z = r

Ann, thank you so much. I love math, i wish i had a teachers and textbooks are hard for me to understand.
Æthank you for your help.

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[lone dome]

can anyone help me understand the workflow here? the only line i dont understand is how we went from the line with the yellow + green to the next line
ive tried multiplying stuff for over an hr and im so lost at this point
i feel like its something super obvious that im missing so if anyone could drop a hint that'd be really appreciated!

[dire geode]

they didn't?

i think you're just misinterpreting

unless there's some dependence between a, b, and p you omitted

[cedar kiln]

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[lone dome]

hmm yeah i think im misinterpreting
makes sense that my random multiplying/factoring didnt work
thanks!

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[hearty basalt]

I need help on Part B

[wispy talon]

Did you get part A?

[hearty basalt]

I think Part a is 2x + 5

[wispy talon]

Yes

So what's the formula to find the perimeter of a rectangle?

[wispy talon]

In terms of length (l) and width(w)

[hearty basalt]

Perimeter = 2(width + length)

[wispy talon]

Yup

So you have the perimeter, the width and the length

Put in the values and you'll get your answer

[hearty basalt]

But the width has to be double the length + 5 and I dont know the length

[wispy talon]

Length is 'x'

Width is '2x+5'

We know this from part A of the question

[hearty basalt]

so P = 2(x + 2x+5) ?

[wispy talon]

Yes

[hearty basalt]

and then I use SADMEP to find x?

[wispy talon]

Yeah

[hearty basalt]

x = 89?

Thanks so much dude I was stuck on that for awhile

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[last sundial]

What do I have to do here? How do I multiply the thing in [] with the 2 dimensional vector?

[last sundial]

I have to solve it for x ∈ R |z|=5

but how do I start?

<@&286206848099549185>

[cedar kiln]

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[last sundial]

can someone help me? I need no complete solution just a way to start

pls

i dont know this is all i got

ok but can i multiply the solution with the 2 dimensional vector?

ok thx so much

❤️

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[crisp obsidian]

Looking for a math tutor in undergraduate topology, analysis, and abstract algebra. I was told to post here

[upper abyss]

I'm sorry someone mislead you, haha. These are for individual questions.

[crisp obsidian]

^ No worries. I'll close it up

[upper abyss]

#math-discussion is likely better, but in general this isn't a common server service

[crisp obsidian]

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[gusty arrow]

I don't understand how anything can be proven when n could be anything.

[upper abyss]

Well, try proving things for n = 3 then.

You'll realize before long that these proofs still work even without setting n

[gusty arrow]

Ok, I’ll try. Thanks

.close

[cedar kiln]

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[late drum]

How to a rearrange y=Mx+B for a fc vs v^2 graph

[upper garnet]

fc?

[late drum]

For cuticular motion

[green hornet]

fc=mv^2/r right?

whats the problem then

[late drum]

I have to write an equation for the line on my graph

[green hornet]

i assume you are talking about centrifugal force

[late drum]

Where Y= inward force and X = speed

Yeah centripetal force

[green hornet]

yeah so you do know this right?

[late drum]

Yeah

[upper garnet]

so u want to lineairize a power relation function?

[green hornet]

fc vs v^2 is a linear graph tho?

[late drum]

No. We did a lab and had to write a graph for the results. Have to write an equation for the line.

[green hornet]

the y intercept would be 0 yes

what would the slope be?

[late drum]

Ik it will be Fc = _ _ _ + V^2

I just don’t know about the rest

In between Fc and +V^2

[crimson sedge]

I need help with this question

[green hornet]

mv^2/r right

you just told me

so the slope would be m/r

[crimson sedge]

it is: calculate 8.31²

[green hornet]

bruh stop trolling

[late drum]

So the equation would be Fc = mv^2/r + V^2

?

[green hornet]

wait why add the separate term?

[late drum]

Cause it needs to be a rearranged version of y=mx+b

[green hornet]

b can be 0

[late drum]

I just need the variables

[green hornet]

centripetal force is mv^2/r

compare it to mx+b when x=v^2

[late drum]

What about the m

[green hornet]

ok m in centripetal force is mass
in the line eqn it is slope

im sorry i used the same variable to denote both

but you could compare and see

[late drum]

Okay

I don’t think x = v^2

The teacher said B = v^2

[green hornet]

whats the plot

you are to graph

what are your axes?

[late drum]

Wdym

Like what units?

[green hornet]

like in a straight line

we do y vs x

here its fc vs v^2 right

[late drum]

Yeah

[green hornet]

so wouldnt v^2 be your x?

[late drum]

Oh yeah

Holy

I have to be the dumbest guy

[cedar kiln]

@late drum Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[dire geode]

do you know pythagorean theorem

can you draw a triangle where the yellow line is the hypotenuse?

[devout radish]

draw a line straight down from the x axis to the middle of the right angle

so now you can do pythagorean theorem

you know all sides but the hypotenuse which is the length you are looking for

yeah thats right

idk what you are doing here, so I cant help much

[cedar kiln]

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[cedar kiln]

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[glad kestrel]

what have you tried

[cedar kiln]

@serene tendon Has your question been resolved?

[serene tendon]

@glad kestrel

im not sure if this is correct and what to do from here onwards

<@&286206848099549185>

[dire geode]

you have the values for z_0 and z_1 and the formulas for what to do with the two residues.

[serene tendon]

@dire geode for this question do i just do the limit test a_n+1/ a_n?

[dire geode]

it's called the ratio test

that would be a good first one to try

root test might also work

[cedar kiln]

@serene tendon Has your question been resolved?

[serene tendon]

is there a way to keep simplifying from here @dire geode

<@&286206848099549185>

<@&286206848099549185>

[dire geode]

Yes. Look up exponent rules for algebra

[cedar kiln]

@serene tendon Has your question been resolved?

[serene tendon]

i finally managed to do it and i got ROC = 1 because the ratio approaches z

i hope im correct

[hot garden]

Fish

[cedar kiln]

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[cedar kiln]

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[rotund mason]

Hello, I wanted to ask, how can we prove the following inequality?

[cedar kiln]

@rotund mason Has your question been resolved?

[rotund mason]

<@&286206848099549185>

[cedar kiln]

@rotund mason Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

I'm clueless on how to get limits for this in spherical coordinates

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

I'm Sad Haha

[crimson sedge]

.close

[cedar kiln]

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[crimson sedge]

in this question

Let triangle ABC be an isosceles triangle with BC=AC. Points M and N are on BC and AC, respectively, so that AM and BN are angle bisectors (that is, angle BAM= angle CAM and angle ABN= angle CBN). Show that AM=BN.

[crimson sedge]

angle ABN= angle CBN, and angle BAM= angle CAM always be true no matter what?

[tropic oxide]

$\angle ABN = \angle CBN$ and $\angle BAM = \angle CAM$ are literally the definitions of the statements that $AM$ (resp. $BN$) are angle bisectors for $ABC$

[wraith dagger]

Ann

[tropic oxide]

so yes they are true

[crimson sedge]

but the values of position of M and N dont change whether the equations are true or not

[tropic oxide]

??

values of position? what are you talking about?

[crimson sedge]

like where M and N are on BC and AC

[tropic oxide]

well, no, the positions of M and N along the sides of the triangle are determined by their being the endpoints of the angle bisectors...

i'm not sure what point you're trying to make

[crimson sedge]

me neither

im confused

[crimson sedge]

but, we are just comparing the angle of B, which is always itself

[tropic oxide]

what?

"the angle of B"?

did you even draw a diagram for this or are you trying to reason this using nothing but symbolpushing?

do you see how trying to talk about ``the angle of $B$'', or $\angle B$, makes zero sense in this diagram?

[wraith dagger]

Ann

[crimson sedge]

Yes

[tropic oxide]

so can you please restate the point you were trying to make?

[crimson sedge]

Nvm it’s cleared up

But how should I approach this wuestion

[tropic oxide]

perhaps recall what you know about isosceles triangles

i've already kind of dropped a hint in the diagram by placing some marks that imply it (which i would not have had the right to place if the fact i was hinting at was to be considered as yet unknown to you)

perhaps also recall something about congruence of triangles

[crimson sedge]

well in this picture AM and BN are not angle bisectors, right?

[tropic oxide]

they are, the picture is just not drawn to scale

but i have marked as equal the angles that are supposed to be equal

[crimson sedge]

how can i be certain tho

[tropic oxide]

certain of what?

[crimson sedge]

nvm

you said to use smth about the congruence of triangles

[tropic oxide]

i pointed at some concepts from geometry that you should recall

it is up to you to think about how these apply, if they apply at all

[crimson sedge]

is it the fact that triangle NAB and triangle MAB are congruent?

[tropic oxide]

are they congruent?

are you able to prove that they are congruent?

[crimson sedge]

i think i recall seeing smth like this before

[tropic oxide]

are you able to also justify my placement of 4 identical angle marks instead of two pairs as i would be required in any other circumstance?

[crimson sedge]

you mean these?

[tropic oxide]

yes i mean these

formally, we don't yet know that these angles can all be marked the same - unless you are allowed to treat the theorem "The angles at the base of an isosceles triangle are congruent" as known, in which case i have been speaking about a non-issue.

[crimson sedge]

we know for certain that these two angles at the bottom are equal (as a whole i mean)

[tropic oxide]

you really should not have written that "(as a whole i mean)" thing

and that's not my point of (possibly misguided) concern

the thing to be concerned about is why the ones on the left are equal to the ones on the right

[crimson sedge]

but isnt that already proved

idk

im sorta confused

we want to prove the angles are equal

cant we just try to counter the opposite, and try to show why it wont work if AM /= BN

but then we still need to show that it will work

forget that then

[crimson sedge]

im stuck on how to solve this

[crimson sedge]

if you labeled the point of intersection X, we know that triangle AXB is isosceles

since AM and BN are equal

so MAB and NAB are equal

bc of the "The angles at the base of an isosceles triangle are congruent" theorem

proving that triangles NAX and MBX is a differentr story

any ideas?

[tropic oxide]

i think i made a mistake by drawing your attention to the angle mark thing.

[crimson sedge]

what am i supposed to do then

[tropic oxide]

you were supposed to note that angles CAB and ABM are known to be equal due to our triangle being isosceles

and thus their halves are also equal

and then you were supposed to use that to establish the congruence of triangles ABN and BAM, taking note of their shared side AB

but no, you overthought it all