help-26

[carmine pelican]

but the ans √2 -1

[sonic hemlock]

lemme try

[carmine pelican]

wait can u do something ggeometrically

r-i is basically the median or smthg

of r, r-2i and 2i

there a triangle forming

[sonic hemlock]

huh, i'm getting something else too xD

[carmine pelican]

Can we do something by coordinate way

[sonic hemlock]

lemme check my calculations

[carmine pelican]

oh okay

[sonic hemlock]

lemme see

[carmine pelican]

[sonic hemlock]

yup i made a calc error like usual

im getting sqrt(2) - 1

[carmine pelican]

nicenice

[sonic hemlock]

uh i just plugged in the assumption of u into the condition of a GM

[carmine pelican]

i mist hav calc error too then

[sonic hemlock]

then simplified and squared both sides to get
r^2 + 2r - 1 = 0

[carmine pelican]

[sonic hemlock]

maybe

[carmine pelican]

imma try once mor

[sonic hemlock]

alright o7

[carmine pelican]

this helped thanks

got it

[sonic hemlock]

my pleasure o7

[carmine pelican]

i was suqaring both sides once should've done twice

Tyty

ye

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[mossy garnet]

,tex Find all the eigenvalues and eigenvectors of ( A = \begin{pmatrix} 1 & 3 \ 1 & -1 \end{pmatrix} ). If ( f: \mathbb{R}^2 \to \mathbb{R}^2 ) is the linear transformation defined by ( f(\mathbf{x}) = A\mathbf{x} ), decide if there exists a basis ( B ) of ( \mathbb{R}^2 ) such that ( M_B(f) ) is diagonal.

[thorny flame]

renato

[mossy garnet]

,w det {{1, 3}, {1,-1}}

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[mossy garnet]

,w det {{1-lambda, 3}, {1,-1-lambda}}

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[mossy garnet]

,w solve lambda^2 -4

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[mossy garnet]

,, \lambda^2 - 4 = (\lambda -2)(\lambda + 2)

[thorny flame]

renato

[mossy garnet]

,, \lambda_1 = 2 \
\lambda_2 = -2

[thorny flame]

renato

[mossy garnet]

,w rref {{-1,3,0},{1,-3,0}}

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[mossy garnet]

,, \begin{cases} x1 -3x2 = 0\ x1=3x2 \ x2 = x2 \ \end{cases}
\bm{Span_{\lambda_1 = 2}} = x2 \begin{pmatrix} 3 \ 1\end{pmatrix}

[thorny flame]

renato

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[raven plank]

Find the rest of f:g

[raven plank]

the options

I was thinking maybe I could use the rest theorem and divide by x-2 twice

but I don't know how would that work

since I dont have clear coeficients in f

nvm

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[muted needle]

Can anyone solve this?

[muted needle]

Ignore the Orange

[whole oracle]

Just count the square
And simplify it

[outer ledge]

Just expand it.

[muted needle]

Answer?

[outer ledge]

10

[muted needle]

So, not 2√3?

[outer ledge]

Sorry bout the shadow thing....

[outer ledge]

I got this....

[muted needle]

Ok, thx dude

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[visual panther]

I’m doing a math project to collect data but I don’t have anyone to collect data from and I don’t know what to do

[visual panther]

It’s a project to find bias based on questions we ask and I have a good question and created the Google form

I just need submissions

Teacher told us to make a Google form: https://docs.google.com/forms/d/e/1FAIpQLSfX-WUmOoW_9QQQE0dE1Ex4HHnN97rIrJVlB4zQWuNt3sQTqw/viewform?vc=0&c=0&w=1&flr=0

[sweet shard]

can you not spam

thanks

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[visual panther]

Sorry

[fallow igloo]

Place it in #1021175428326633542

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[quasi bone]

I'm just not able to figure out the process of going about this

Normally I do F'(X) = 3x^2

then I do F'(a) = 3a^2 -> 3 * 6 = 18

but it's labeled as wrong

[odd pagoda]

what are you even asked to do

[radiant sparrow]

send the full question

[quasi bone]

Oh apologies

I have to find F`(A)

Here's the whole question

Do I need to use the chain rule to figure this out?

[sweet shard]

yes try it

$F'(x) = f'(x^3) \cdot 3x^2$

[thorny flame]

riemann

[quasi bone]

Alright, let me give this a try real quick

I got 36

I substituted x with a

f'(a^3) * 3a^2 to then get 2 * 3 * 6

Thank you for the help

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[quasi bone]

.reopen

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[quasi bone]

I forgot to ask about G'(A)

Do I need to find the derivative of both x^3 and f^3?

I know the derivative of x^3 = 3x^2, what about F^3?

I normally end up getting 432 because of the chain rule

[sweet shard]

Find G'(x) first, then plug in a

[quasi bone]

G'(x)?

I'm wondering because I need the derivative of F^3

Since G(x) = (f(x))^3

or unless it's worded as

(f(x)^3)^3

[sweet shard]

$(f(x))^3 = h(f(x)), \ h(x) = x^3$

[thorny flame]

riemann

[quasi bone]

Wait, what's H here?

[quasi bone]

Oh

Shoot, thank you

So it would be

(f(x))^3 = x^3(f(x))?

[sweet shard]

no

[quasi bone]

I don't understand how to get the derivative of f^3

[sweet shard]

h(f(x)) means f(x) plugged into h(x)

h(f) = f^3

,tex .diff rules

[thorny flame]

riemann

[quasi bone]

Ooooh, this is interesting

So this would be G'(a) = (x^3)^3 times 3x^2?

[sweet shard]

no

[quasi bone]

or would it be h(f(x^3)?

[sweet shard]

Find G'(x) using the second bottom row

G(x) = h(f(x))

[quasi bone]

Okay, so G(x) = x^3(f(x)) right?

[sweet shard]

no

[sweet shard]

what's h'(x) ?

[quasi bone]

3x^2?

[sweet shard]

yes

what's h'(f) ?

[quasi bone]

Oh dear

3x^2(f)?

[sweet shard]

no

h'(f) is the derivative of h evaluated at f

[quasi bone]

Evaluated at f?

Is it 3f^2?

[sweet shard]

yes. then you multiply by the derivative using chain rule

[quasi bone]

Okay!

[sweet shard]

this one

[quasi bone]

Wait, did we find the g(x) already or?

or was it 3f^2

[sweet shard]

g(x) in this image acts as f(x) in your problem

[sweet shard]

read

[quasi bone]

Okay

So does that mean that f(x) is equal to 3f^2?

[sweet shard]

no

f(x) = f(x)

[sweet shard]

h'(f) = 3 f^2

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@quasi bone Has your question been resolved?

[quasi bone]

I'm not sure I understand

Hm, I'll go do some research, thanks for the help

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[wooden osprey]

a shape has an object base with radius 1

[wooden osprey]

cross-dissecting the object with a plane perpendicular to the x-axis at any given $-1\le x\le1$ obtains an equilateral triangular cross section

[thorny flame]

FungusDesu MSC2020 34A05

[wooden osprey]

find the volume of that object

[kindred linden]

The base is a circle of radius 1 ?

[wooden osprey]

not sure what to do here, i tried finding the graph of the parabola-like to integrate but to no avail, so i guess 2

[wooden osprey]

yes

[steady quarry]

The triangle is equilateral so we may derive the height of the object and thus find an equation for the parabola

[wooden osprey]

hm

the maximum is 1

so the maximum height sqrt(3)/4

[steady quarry]

And we know height is proportional to distance from the x axis

[wooden osprey]

distance from the altitude's base to the x axis?

[steady quarry]

Yes

As it’s a triangle, the height decreases linearly

[wooden osprey]

wait, even after finding the equation of the parabola, integrating it would only gets the area of the biggest cross section perpen to y

right?

[steady quarry]

Yes

You need an equation for the cross sectional area at any plane y=c and then integrate over all the cross sections

This is essentially a double integral

[wooden osprey]

uh this is calc 2

[steady quarry]

Then integrate once to get a general form of the area, and then integrate again over all the areas

[wooden osprey]

hm right

[steady quarry]

(Double integrals are just sticking integrals inside integrals)

[wooden osprey]

first i need to find the cross section triangle side given x

uhh

[steady quarry]

Oh wait I’m being stupid, you can just integrate over cross sections for planes of the form x=c

Which is significantly easier lol

Just take the area of a triangle, bh/2, and find expressions for base and height given x

No double integrals needed

[wooden osprey]

alright got it, let me do that rq

[steady quarry]

(In fact given that the triangles are equilateral you don’t even have to worry about the height and parabolas and whatnot, and you can just use a single trig function)

[wooden osprey]

$f(x)=2\sqrt{1-x^2}$

[thorny flame]

FungusDesu MSC2020 34A05

[wooden osprey]

this correcr yes?

this is the base

[steady quarry]

Yes

[wooden osprey]

so the height should be

[steady quarry]

You should probably use only that to find the height since I’m not even sure if it’s a parabola anymore

[wooden osprey]

$f(x)=(1-x^2)\sqrt{3}$

[thorny flame]

FungusDesu MSC2020 34A05

[wooden osprey]

and now i integrate this from -1 to 1?

[steady quarry]

That’s the full area term

So yes

[wooden osprey]

so integrating this gets me $\frac{4\sqrt{3}}3$

[thorny flame]

FungusDesu MSC2020 34A05

[wooden osprey]

which seems to be the correct answer

nice

thanks guys

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[summer yoke]

I would like help for this question please

I would the answer

[granite hull]

great reaction image

[keen venture]

You're close. You've forgotten the definition of "independent" though

If A and B are independent:
P(A n B) = P(A)P(B)

[summer yoke]

so what can we conclude about the two events

p(a)p(B) =0?

@keen venture please help if you could, im desperate

[worthy storm]

what does the fact that they are mutually exclusive tell you about P(A∩B)?

[summer yoke]

P(AnB)=0

please get back to me when you guys have time. ik you guys are busy :)

[worthy storm]

yep correct

so combining that with P(A n B) = P(A)P(B), you indeed have P(A)P(B) = 0

which means what?

[summer yoke]

theyre both 0, meaning that theyre non-existent?

[worthy storm]

if the product of two things is zero, does that mean they both have to be zero?

[summer yoke]

atleast one of has to be 0

[worthy storm]

yes correct

at least one of A or B has zero probability

that doesn't necessarily mean non-existent/impossible

[summer yoke]

so thats the answer??

that at least one is 0?

[worthy storm]

yep

[summer yoke]

thank you for helping me!

[worthy storm]

sure, cheers

[summer yoke]

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[marble quartz]

hi

[marble quartz]

is 2nd answer correct?

[teal gust]

yes

[marble quartz]

thanks

[gleaming reef]

You should get honorable

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[teal gust]

wat

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[trim totem]

<@&286206848099549185>

the thing is like my calculations are weird

[neon iron]

yo

its proportionate

[trim totem]

yo

wsg

[neon iron]

right

[trim totem]

yeye i think

[neon iron]

how far do u know abt what u have to do

[trim totem]

like

what do i know or what i have to do

[neon iron]

what do you have to do

[neon iron]

what calculations did u get

[trim totem]

lemme get an example

one sec

well ill do the work and then ill be like 2 off

every time

im so lost

i been doing this for two hours

[neon iron]

hmm

[trim totem]

u think 58 makes sense

[neon iron]

so those triangles are proportional to each other

let me check

not rlly?

ok so if u have 2 triangles

the angles are all the same

but 1 is 3x bigger than the other

js multiple the other by 3 right?

[trim totem]

i mean ion know cz is that 3x

[neon iron]

not in this one

im js givng a general example

of the concept

so u'd make a an equation

[trim totem]

aight

yea

imma js answer and see what it tells me

imma go down tho

[trim totem]

soo

45.5 or am i wrong

[neon iron]

yes i think that is right

[trim totem]

it is

ty

[neon iron]

same concept here

use that right angle to decide which side correlates with which

[trim totem]

this is wha i was talking ab

73 makes sense right

i got it right but like idk sometimes i js get it wrong for no reason

[neon iron]

yh thats the proportion

u multiply it to the other number

51.1

[trim totem]

45.5 make sense for this

or nuh

[neon iron]

yes?

yes it does

[trim totem]

alright

ty ty

hwo do i close i forgor

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@marsh mural Has your question been resolved?

[steady quarry]

How many eigenvectors does the matrix have?

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[marsh mural]

i figured it out

ty though

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[versed widget]

how do i even begin

do i have to consider the case where x and y are on the same side of 0 and the one where they are on different ones?

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@versed widget Has your question been resolved?

[opal vault]

yes

[versed widget]

this just means that i have to show that cosh(x) is at least 1 and goes to infinity

right

[opal vault]

yes

you have to show that cosh reaches 1 though

and then if you have that cosh is continuous and goes to infinity it works

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[uneven vector]

The angle of depression from the top of the mountain to the horizon is 2 degrees. The angle of elevation from viewpoint number 1 and viewpoint number 2 to the top of the mountain is 10 degrees and 20 degrees respectively, and the distance between the two viewpoints is 12 km. What is the height of the mountain and the true radius of the Earth? There is another mountain roughly with the same height 5000 km away on the ground. Give the arc distance between these two mountains and give the straight-line distance between these two mountains.

[compact zodiac]

you can use trig for the 1st part, as of the latter parts, i can't say

[uneven vector]

Yea ive found the height of the mountain

but I cant find the radius

[uneven vector]

mountain height is 4.1km from what ive found

<@&286206848099549185> anyone?

I cant seem to understand how with the height of the mountain I can find the radius, ive tried drawing but still I do see a right triangle but not a value to any sides

[real barn]

Find hypo I'm pretty sure that's the radius

[uneven vector]

im not sure if my drawing is right

can i send a photo?

[real barn]

Sure

Don't expect too much tho I'm just a highschooler

[uneven vector]

drawing is not that uh straight but

i think its something like that?

[real barn]

Ok so the base isn't at the centre it's over there huh

[uneven vector]

yeah because from what I know the line should be tangent

and that would be 90degrees

not sure how the height of the mountain ties into this though

[real barn]

Hmm for the triangle it's trigonometry

I've never seen this

Mah bad

Sorry man

[uneven vector]

np

thanks for trying haha

been stuck with this for a few days now

[real barn]

Well hope u get the solution

[uneven vector]

anyone?

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[opal violet]

<@&286206848099549185>

[sweet shard]

!15m

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[neon iron]

what have u tried?

do u know what's the defn of stationary value?

[opal violet]

ive managed to get the answer but i dont get how it proves that f(x) has a stationary value

hang on i think i understand

[opal violet]

bc i got (x-1)e^2x -1 = 0 when differentiating and simplifying f(x) which is equal to the statement that we know is equal to 0

therefore it has a stationary value, is that correct?

[neon iron]

yeah

[opal violet]

alr thanks

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[raven crag]

Lim(x -> 0) x/(x+sinx)=1/2

[raven crag]

By method ephsilon delta

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[raven crag]

.reopen

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[raven crag]

4 and 5

[noble laurel]

!show

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[neon iron]

just need clarification on the question's logic

[elder tendon]

think u gotta use chainrule for a and product rule for b

[neon iron]

I just want to understand why you'd do so

[elder tendon]

like in a think of it as g(x)=f(k(x)), where k(x)=x^2. then g'(x)=f'(k(x))*k'(x)

in a you wanna find a derivate of a function which has another function inside of it. this wouldnt be easy if you didnt use chainrule

[neon iron]

$g(x)=f(k(x))$ where $k(x)=x^2 \rightarrow g'(x)=f'(k(x))*k'(x)$

[elder tendon]

in b you have the product of two functions where the first, lets call it b(x)=2x and the other is just f(x)

[thorny flame]

wyldinwilliam

[neon iron]

are you using liebniz notation?

[elder tendon]

no its lagrange i think

[neon iron]

let $x^2=u , y=f(u)$ needed: $\frac{dy}{dx}$

yeah I'm seeing how you couldn't use liebniz here

or maybe you could idk

[thorny flame]

wyldinwilliam

[neon iron]

$d'u= 2x, d'y=d'f(u)$

[thorny flame]

wyldinwilliam

[neon iron]

$d'f(u)\cdot 2x$

[thorny flame]

wyldinwilliam

[neon iron]

$d'f(x^2)\cdot 2x$

[thorny flame]

wyldinwilliam

[neon iron]

$f(x)\cdot 2x$

[thorny flame]

wyldinwilliam

[neon iron]

$2xf(x)$

[thorny flame]

wyldinwilliam

[neon iron]

^^ okay so this is right but I have no idea why

[neon iron]

is from here #help-26 message to here #help-26 message correct?

[neon iron]

ok looking back at this

and I'm kinda confused

$g(x)=f(k(x))$ where $k(x)=x^2 \rightarrow g'(x)=f'(k(x))*k'(x)$

[thorny flame]

wyldinwilliam

[elder tendon]

this is what i would do

youre not given an expression for f(x)?

[neon iron]

nope

ans is $f(x)^2 \cdot 2x$

[thorny flame]

wyldinwilliam

[neon iron]

wait this might just be really easy

no wait you can't do that tho

or can you idk

[elder tendon]

when you have to differentiate a function that has another function inside of it you can say "theres and outside and an inside function". The derivate of the whole thing is the "outside function" differentiated with the "inside function" inside of it times the "inside function" differentiated

[neon iron]

okay I get this now

[elder tendon]

in a) your outside function is "f of something" with the inside function being x^2

[neon iron]

yeah that makes sense to me now

[elder tendon]

😎

[neon iron]

this is what they have

[elder tendon]

what the answer?

[neon iron]

yea

$$g(x)=f(x^2)$$
$$g'(x)=f(x^2)\cdot 2x$$

[thorny flame]

wyldinwilliam

[neon iron]

but how come, is it because f(x^2) has no power?

but then that wouldn't make sense either

[elder tendon]

im not sure how their answer makes sense

[neon iron]

because f(x^2) would disappear since it gets reduced to 0th power

me neither lol

oh wait nvm @elder tendon

it's on the sheet of

errors

lol

was suspecting it

[elder tendon]

bruh moment

[neon iron]

ok now to understand this

1st and 2nd look okay

is the third one simply just a derivative principle

seems so

actually it is nvm

yeah okay I get this then

thanks lol

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[strange wagon]

how do i do this

[strange wagon]

missed the lesson

i know it has something to do w related rates

but i believe there's something i am missing

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@strange wagon Has your question been resolved?

[strange wagon]

<@&286206848099549185>

<@&286206848099549185> ?

[crimson crag]

so

look at the right triangle fromed by the light pole , tightrope and the ground

lets say that x represent the distance from A to B (midway point

then we have the proportion based on similar triangles : ,, [ \frac{x}{50} = \frac{70}{30} ]

,, [ \frac{x}{50} = \frac{70}{30} ]

[thorny flame]

Lowie

[crimson crag]

,, [ 30x = 50 \cdot 70 ] [ x = \frac{50 \cdot 70}{30} = 116.67 ]

[thorny flame]

Lowie

[crimson crag]

(used the cross multiply)

[strange wagon]

ok

oh then i can find the speed from that

ok thanks

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[fallow chasm]

How do I find the normal subgroups of a Dihedral group (specifically D_12)?

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@fallow chasm Has your question been resolved?

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@fallow chasm Has your question been resolved?

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@fallow chasm Has your question been resolved?

[neon iron]

what have you tried so far doing?

[fallow chasm]

I don't even know where to start other than the identity and the whole group are normal subgroups of it

[topaz sinew]

@fallow chasm Has your question been resolved?

[tough nest]

there are a lot of equivalent statements for normal subgroup

the one i came across the most in my course on groups was gng^-1 in N

for all n in N and g in G

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[keen vessel]

in linear algebra, Ax=b where A is a matrix and x and b are vectors. From what I can interpret from this textbook, it says the set of all possible x's is the union of the null space and the row space. Am I interpreting that right? If so, I am confused why that is so

[keen vessel]

I can understand why the null space would be part of the possible x's when b=0, but Im confused about why the row space is

[topaz sinew]

@keen vessel Has your question been resolved?

[radiant marlin]

well x can be anything right, so it's saying the basis of the null space and the basis of the row space unioned together span the full n-dimensional space

like if x was a 3d vector one scenario would be the row space being a 2 dimensional plane and the nullspace being a line orthogonal to that plane, and it's saying to get any 3d point you can add something from the plane and line together

[keen vessel]

thanks that makes sense

[radiant marlin]

the "why" is that the nullspace is all the vectors that are orthogonal to all the rows, and since everything's a vector space with n basis vectors in n dimensions, the row space can be made with some of those basis vectors and the nullspace is all the rest of them

[keen vessel]

would the span of the row vector always cover some R^m where m<n?

like instead of the row space being a plane, it will just cover a small part of that plane?

[radiant marlin]

yea there's cases where it's any dimension between 0 and n

and the nullspace is just the rest of the dimensions

[keen vessel]

if its a 2x3 matrix with 2 pivots, would the row space always be a plane?

[radiant marlin]

yes, the pivot number should be the row space dimension

[keen vessel]

ok i think it all makes sense now

thanks for the help

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[wispy island]

can anybody please explain how can he partially apply limits when it is in favour like is it the rigth way to do it?

[neon iron]

I have a question is there is any formula for (a+b)^r, where r is ncluded in Q

[wispy island]

how is this related to the above question

?

[neon iron]

it's not relted to it at all, I just didin't' knew where I should ask

[wispy island]

please occupy an avaliable room this is alreasdy occupied

[neon iron]

ok I'm sorry

[wispy island]

<@&286206848099549185>

.close

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[unkempt vapor]

how do i even approach this question ;-;

only thing i got is the square side, 10

[outer carbon]

Ok thats some complicated geometry

I didnt evne have this type of sum on olympiads

[outer carbon]

how did u find this?

if this is correct I can probably help

[keen rivet]

yeah this isnt easy

[outer carbon]

my mind cooking something

I found a way

[keen rivet]

i can also see something but im trynna check a more efficient way

[outer carbon]

Im waiting for his response

then I'll explain

and exactly what area is the question talking abt

[unkempt vapor]

uh practice question grade 11

the topic is circular measure

[outer carbon]

what area is it talking abt

im a bit confused

[unkempt vapor]

i have no idea 😭

sorry

[outer carbon]

ok its what im thinking

[unkempt vapor]

i was thinking this

[outer carbon]

I think its the outer circle and the annulus

4-3.14 : 2.14

0.86 : 2.14

[outer carbon]

u can find the

diagonal

of the square

it will give u the radius

of the outer circle

[unkempt vapor]

7.071

[outer carbon]

now area

of the bigger circle

[unkempt vapor]

oh

50pi

157.08

[outer carbon]

does that ratio work now

we gonna have to try all types of ratio ig

[unkempt vapor]

how do i find the annulus

what is it 😭

[outer carbon]

I think

its area of bigger circle : area of annulus

[unkempt vapor]

is it big circle - small circle

[outer carbon]

yeh

thats annulus

[unkempt vapor]

50pi - 25pi

wouldn't that js be 1:2

[outer carbon]

we need 2.48 : 1

whered the 2.48 go

what areass

damn

[unkempt vapor]

wtf does "square divides the area of annulus" means

[outer carbon]

yeah im confused abt that too

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@unkempt vapor Has your question been resolved?

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@unkempt vapor Has your question been resolved?

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[steel drift]

I need help trying to convert this. They say I have to convert the fractions but never show how to

[thorny flame]

! What the hell am I doing here?

[abstract wadi]

Can you add it and find what becomes of it?

@steel drift

[steel drift]

Sure

[steel drift]

Hold on? Where did the 2/1 come from?

[abstract wadi]

Well, what's $2 \div 1 ?$

[thorny flame]

! What the hell am I doing here?

[steel drift]

2

[abstract wadi]

It means you can write 2 = 2/1

Doesn't it?

Just helps you add those fractions.

[steel drift]

Ok ok

[steel drift]

But 2 divided by 1 is 2 tho. And there is in the bottom

I still don’t get it(sorry if I’m being slow)

[abstract wadi]

Well look at the middle.

We had 2 + 1/4

[steel drift]

Ok

[abstract wadi]

We didn't touch the 1/4

But we wrote 2 = 2/1

And 2 divided by 1 is 2.

Which is something you said is true as well.

If it is the same thing, then what's wrong with writing it that way?

[steel drift]

Ok ok. So basically bring down the number that at the top of the side?

[abstract wadi]

I'm not sure what that means.

[steel drift]

You brought down the one to the bottom of 2

[abstract wadi]

I didn't bring anything, anywhere.

2 IS 2/1

This 1 doesn't come from anywhere.

It's always there. Always has been, always will be.

[steel drift]

Alll

[abstract wadi]

When someone says 2, they mean 2/1. It's a pain to write it that way, so just 2.

But it always means 2/1 as well.

[steel drift]

Ahh I see

I see

Ok

Let me find out the answer

[steel drift]

Is this right?

[abstract wadi]

...

It is but I want you to write your answer ad a fraction for a moment.

[steel drift]

Ok

[steel drift]

This?

[abstract wadi]

I KNEW IT

[steel drift]

?

[abstract wadi]

You don't know how to add fractions.

But worry not.

[steel drift]

👍🏿

[abstract wadi]

https://youtu.be/5fbzZEK77DQ?si=yTfKFMMcOg4qS0uB

[steel drift]

I’m watching it on my pc

[abstract wadi]

You can't separately add numerators and denominators.
What you've written now (3/5) is actually 0.6, but I already told you 2.25 is correct. So I hope you understand you're wrong somewhere at the very least.

[steel drift]

I do

[abstract wadi]

I said you CAN'T do that.

But you went out of your way to do it anyway and sent an image as well?

[steel drift]

What did I do?. I did what the video said?

[abstract wadi]

The video said add the numerator and denominator separately?

[steel drift]

Are you talking about the one at the top?

I forgot to erase that one

[abstract wadi]

And what might (6+2)/5 be?

Which one's that?

[steel drift]

I followed the step on the video

[abstract wadi]

But for what question

[steel drift]

This

[abstract wadi]

yeah that's wrong, I'm not even sure how you got numbers like 6, or 5

I'm pretty sure the video asked you to make sure the denominators are same before you simply add the numerators.

[steel drift]

Oh crap I see

I added

I was supposed to multiply

💀💀

My fault

[abstract wadi]

Well do it again. Hopefully you'll get it right now?

[steel drift]

Yeah

[abstract wadi]

OH 6 = 2+4, 5 = 1+4?

[steel drift]

Yeah

[abstract wadi]

Yeah too bad, it was meant to be multiplication.

[abstract wadi]

(8+1)/4

Is correct

But that's not 8/4

[steel drift]

Oh crap, I was still thinking it was times

I had to actually add by then

[abstract wadi]

Correct.

Now that's 2 1/4

[thorny flame]

! What the hell am I doing here?

[steel drift]

Thanks

Also

Can I keep this channel up

I’m studying for the asvab and I have a lot of other questions I need help on to

Like I need a good understanding of the equations I need to do

[abstract wadi]

Well, you can close the channel for the time being and by the time you have another problem, open another. This also moves your new channel to the top. Also it's easier for others to help you with what problem you're doing exactly if you're doing multiple.

[steel drift]

Alright

appreciate it

This actually helped a lot

Thanks

How do I close the channel?

[abstract wadi]

.close

[steel drift]

.close

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[shy pivot]

Hi here is a problem i struggle with

[shy pivot]

Let f be a differentiable function from R into C

We know f + af' -> 0 where a is a complex with real part > 0

Prove f -> 0

This is a tough exercise for 2nd year uni just to give a level of difficulty

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[burnt ocean]

Can I

[snow nimbus]

No

[burnt ocean]

So if I have 14sqt2 / 21
Can I make that 2sqrt2 / 3

[snow nimbus]

Yes

[burnt ocean]

I can simply still?

[snow nimbus]

No

[burnt ocean]

What

[snow nimbus]

You can't simplify further than what you already did

[burnt ocean]

I gotcha

Thank you

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[vernal vale]

I'm confused here

I've only uploaded the entire page because if context is missing; I'm really only confused about one tiny piece, specifically: the second line from the bottom.

[thorny flame]

jan Niku

[vernal vale]

but I'm not really following this step

the other term changing in this way is obvious but i am having a hard time seeing it here

[thorny flame]

jan Niku

[pseudo jetty]

this is probably coming from the fact that [ \dv t f(\vec r(t)) = (\grad f) \cdot \dv [\vec r(t)] t ]

[vernal vale]

but where is the function of r(t) here?

oh is it the derivative itself

[pseudo jetty]

here you have [ f = -\f p\rho - gz, \quad r = r(\sigma, t)] and the derivative is with respect to $\sigma$ instead

[vernal vale]

i guess it doesnt seem like anything is a function of r(t)

[pseudo jetty]

[ \dv \sigma f(\vec r(\sigma)) = (\grad f) \cdot \dv [\vec r(\sigma)] \sigma ]

[pseudo jetty]

i think it's implicitly assumed that [ f(\vec r) = -\f {p(\vec r)} {\rho(\vec r)} - z(\vec r) g(\vec r) ]

[vernal vale]

oh geezus

thats horrible

[pseudo jetty]

those might not all be dependent on r i haven't read the context

[vernal vale]

idk anything about physics he says its ideal incompressible fluid moving under gravity

so maybe its just z

[pseudo jetty]

but i'm assuming the integrand depends on r

[vernal vale]

either way i guess it has to be something

or that term wouldnt be created

[pseudo jetty]

yeah

[vernal vale]

okay lemme work forward from here

thank you snow

.close

[pseudo jetty]

np

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[cursive sparrow]

hi, im trying to find f'(2) of the equation

[cursive sparrow]

The answer key says the answer is supposed to be 2/3 but im getting -2/3

[shut spindle]

show your work

[cursive sparrow]

the 1/2 was 2 instead of 1/2 when I moved 2(1+x^3)^1/2 up

.close

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[marsh mural]

help pls

[keen venture]

Similar procedure to the other one. Let's start with a change of basis matrix from B to E

The weird part is that B is a 2 dimensional space so your matricies won't be square

[marsh mural]

oh hello again

[marsh mural]

yeah thats whats confusing me becasue i cant use the inverses

i got a like thsi
but idk b and c

[keen venture]

Good!

Now, the change of basis matrix between B and C will be square, since it's just two different 2 dimensional spaces

[marsh mural]

ohhh

so i should probably find [T]c first

but thats part c

aa

idk what im doing wr4ong

this is the answer

so i found [T]c and [T]b correctly

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[hasty surge]

Can someone help with both ?

[trim jasper]

Do you know how to find the maximum and minimum value of a function

[hasty surge]

Nope

[trim jasper]

then how are you solving the question

[hasty surge]

I originallt thought itd be abs min 0 and abs max 3/10 but it said incorrect for that one too

i was j guessing 😞

[trim jasper]

you just plugged in

yeah

[hasty surge]

So is the abdolute minumum 0 and max 3/10?

idk why it said incorrect

[trim jasper]

it is

[hasty surge]

oh 😭 shi

[trim jasper]

well it cant be negative by any chance beause the function itself is "positive" and the minimum value is 0

[hasty surge]

Oh so minimum is 0, and mad id 1/2?

[trim jasper]

yes

obviously..

how would 1/2 be the minimum and 0 be the maximum

[hasty surge]

Just making sure 😭

[trim jasper]

on a serious note, you should revise your notes or watch some lecture with theory

you cant just guess it

[hasty surge]

yeah ik, they did this lecture ok friday i haven’t watched it yet due to another exam i have

Do you by chance know how to solve the other one ?

.close

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[brazen breach]

help understanding b

[worthy storm]

the answers look ok

what is your doubt?

[brazen breach]

those are the answers

[worthy storm]

oh i see

[brazen breach]

but dont really get how

i get a

[worthy storm]

ok suppose 1 is the smallest element

that means you have to choose two elements from {2,3,4,5,6}

to complete the set

does that make sense?

[brazen breach]

why 2?

[worthy storm]

you already picked 1

[brazen breach]

oh

ok ok

[worthy storm]

you want a subset with 3 elements

so you need 2 more

and they have to be larger than 1

[brazen breach]

the rest choose 2 basically

[worthy storm]

yep

and it works the same way with the others

for example

if 2 is the smallest element in the subset

then 1 can't be in the subset

so the remaining two elements have to come from {3,4,5,6}

[brazen breach]

oh ok ok

wow that was simple

[worthy storm]

yea

just have to think about it the right way

[brazen breach]

how does all them added up = a

[worthy storm]

ah because those are all the possibilities for forming a 3 element set

it has to have a smallest element

and that smallest element has to be 1, 2, 3 or 4

[brazen breach]

oh wow lmfao

[worthy storm]

it can't be 5 or 6 because you need 3 elements

[neon iron]

Wassup

What the fuck is this

1+1 is 2

[brazen breach]

w explaination

wait can you help explain the logic here

[worthy storm]

hmm well you can calculate how many subsets of S do not contain 1

and subtract that from the total number of subsets of S

can you see how to count how many subsets of S do not contain 1?

[brazen breach]

isnt it 2^n-1

since we are removing 1

[worthy storm]

do you mean (2^n)-1 or 2^(n-1)

[brazen breach]

2nd

[worthy storm]

yes

that is correct

basically how many subsets are there of {2,3,...,n}

so subtract that from the total number of subsets

2^n - 2^(n-1)

[brazen breach]

2^n - 2^n-1

oooo i get that logic

[worthy storm]

as for how many contain 1 and have size k

you have 1
now choose k-1 other elements from {2,3,...,n}

[brazen breach]

what does have size k mean

[worthy storm]

contains k elements total

including 1

size of a set = the number of elements it contains

[brazen breach]

wait why does it say

[worthy storm]

because the number of subsets of {1,2,...,n} that contain 1 is the same as the number of subsets of {2,3,...n-1}

and the latter is 2^(n-1)

reasoning here is that you're forced to include 1 but then you can include or exclude any of the other elements

[brazen breach]

so the amount of subsets that exclude 1 is the same thing as subsets that include 1

[worthy storm]

yes that's true

[brazen breach]

why is it k-1 here is it because not including 1 in the subset, which is the same as how many contain 1?

[worthy storm]

basically it's "take any k-1 elements from {2,3,...,n} and then add 1 to the set"

your freedom of choice is the k-1 elements from {2,3,...,n}
the 1 is forced so it doesn't contribute to the total number of ways

[brazen breach]

mmmm

ty!!

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[whole nimbus]

finding the tangent by drawing

[whole nimbus]

help

[eternal spoke]

A drawing would help

[whole nimbus]

oh my focus is on the graph showing the angles along the horizontal and tangents along the vertical

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[split rover]

Example of latice without maximal element

[split rover]

I tought of Z or N

But is there a simpler example

?

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[glass rain]

how is this wrong?

.close

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[unborn edge]

I have a sphare K and a plane E that tangents with the sphare in the point f. I have the value for the middle of the sphare (-13, 20, 0), I have the radius of the sphare which is 18, and the vector N with the value (4,-8,1). I want to find f

I currently made the vector go through both the middle of the sphare and the plane at a 90° angle creating "h" but im completley stuck in what to do, this is what i've done so far. Where to go next?

[scenic flare]

du weißt ja, dass die Kugel den Radius 18 hat, also alle Punkte, inklusive F, eine Distanz von 18 zum Kugelmittelpunkt haben müssen

bzw Distanz(M,F) = 18

wenn du also die Länge von dem Normalenvektor der Ebene kennen würdest, nennen wir sie L

dann gilt: M + nE * 18/L = F

weil nE ist ja die Normalenvektor der Ebene, nE/L ist derselbe Vektor auf die Länge 1 normalisiert

und nE/L * 18 wären dann 18 "Schritte" in die Richtung nE

naja und da M und F auf einer Gerade dieser Richtung von nE liegen, ist M + nE * 18/L = F

@unborn edge

[unborn edge]

Ahhhh, vielen Dank, das macht Sinn

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[scenic flare]

np 🦇

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[visual panther]

How do I solve this?

I have work but it gives me a totally different answer

The cos was from a different problem

[fallow igloo]

there are multiple ways to solve this question

which one do you know?
cosine law
abc/4R
sine area law
sqrt3 a^2/4 = area
sqrt3 a / 2 = height

[visual panther]

Uhhh

I have no idea what you just said

I don’t know what I did wrong

[fallow igloo]

all of them?

[visual panther]

Cosine law?

[fallow igloo]

is this your progress?

[visual panther]

Adjacent over Hypotnuse

[visual panther]

Yeah

I got to a dead end tho

I’m not getting any matching answers

[fallow igloo]

they seem to be irrelevent

what is 1/2(26)(3)(90) ?

[visual panther]

That is my formula to find the area

1/2(a)(n)(s)

A=apothem

N=number of sides

S=side length

[fallow igloo]

but where did you get n?

If I weren't mistaken, the formula is rs

s stands of the half magnitude of perimeter

[visual panther]

From the problem

[visual panther]

There is 3 sides here

I a triangle has 3 sides

[visual panther]

What does R and S stand for

[fallow igloo]

r = apothem
s = half the magnitude of perimeter

@visual panther

[visual panther]

<@&286206848099549185>

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[restive steeple]

Hello, how can I go about solving this?
I have tried splitting the fraction but due to the 1/x^2 being positive i am unable to use the property for integrals.