help-13

[teal magnet]

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[cedar kiln]

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[next spruce]

the topic is factoring perfect square trinomials and I'm learning it from a ppt. What happened to 20ab?

[dusk finch]

x^2 + 2xy + y^2 = (x + y)^2

in this case, 20ab is the 2xy

4a^2 = (2a)^2, 25b^2 = (5b)^2 and 20ab = 2*(2a)*(2b), so overall it's equal to
(2a)^2 + 2*(2a)*(5b) + (5b)^2

= (2a + 5b)^2

[next spruce]

ohhh

thanks

.close

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[spring coyote]

P(A_) with line above A is probability that A wont happen = 1 - P(A)

[spring coyote]

what if A and B both not happen

Is it then 1 - P(A) - P(B) ?

[dusk finch]

whats the probability that both A and B happen???

assuming independence

[spring coyote]

yes independent

w8

w8

i actualy calculate they are dependent

[dusk finch]

can you show the original question

[spring coyote]

Probability that family has Personal computer is event A, 0.53 . It has laptop (event B) = 0.42. It has both is 0.25

P(B|A) = 0.47 =/= P(B) = 0.42 which proves they are dependent

[dusk finch]

yeah

[spring coyote]

so question was

whats proability we pick a household that has no PC and no Laptop

so I assume its 1 - P(A) - P(B)

0.05

[dusk finch]

so if probability of having computer was 0.5 and laptop 0.5 as well, would the probability of having no PC and no Laptop be 1-0.5-0.5 = 0?

Try something different, since this doesnt work well

[spring coyote]

PC is 0.53 and laptop 0.42

so comes to 0.05

[dusk finch]

No, see my point above? If the probabilites were 0.5 and 0.5, then it would be impossible to not have laptop and not have PC

Try calculating P(not A) and P(not B)

then P((not A) and (not B))

and keep in mind they are not independent

so you cant just multiply them

[spring coyote]

hmmmmmmmmmm

i rly dont understand

where 0.5 cme from

[dusk finch]

It was just a random example showing that your calculation method doesnt make sense

because if 50% of families have computer and 50% have laptop, then it's not true that 100% - 50% - 50% = 0% have nor computer, nor laptop

[dusk finch]

Just try doing this instead

[spring coyote]

🤔

I got 0.2726

w8 no

cant multiply

[dusk finch]

try using venn diagrams

Just fill out the missing areas, alternatively you can use inclusion exclusion

[cedar kiln]

@spring coyote Has your question been resolved?

[cedar kiln]

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[paper gull]

This is what I've done so far to solve this question...

I need help with part C (without giving the answer), and i've also posted how I got the ans. to part A and B

By the way
_--------

[stiff totem]

when you measure the distance between a point and a plane, you want to measure perpendicular to the plane

[paper gull]

right, and taking that into consideration

what i've thought of is

point P is given, so i can make a vector equation of line passing through it

with direction vector as coefficients of the cartesian form of the plane equation

and then i find the intersection by substituting the line into the plane equation.

i then take the distance between the 2 points

but there's a little problem

Since the coordinates of point P is given as x_0, y_0 and z_0

it will make me waste a full page or something

i want to do it in a shorter way if possible

[stiff totem]

you could normalise the direction vector so that the parameter is exactly the distance travelled

then start from P and see how large you need to make the parameter until you hit the plane

but your orginal way works

[paper gull]

so i take the unit vector ?

[stiff totem]

yeah

[paper gull]

divide by its magnitude 3?

[stiff totem]

yes

I feel like it'll be a similar amount of work in the end though

[paper gull]

so you mean i try for different values of the parameter?

[stiff totem]

P + t(2/3, -2/3, 1/3)

and see what value of t intersects the plane

[paper gull]

hmmmmmmmmmm

okkkkk

ill give it a try just cause i've never done this, even if it is slower

[paper gull]

oh, i shouldve asked earlier, how will i know when i reach the plane?

since P is given as unknown constants

[stiff totem]

it reaches the plane when the three components of the vector satisfy the plane equation

I didn't mean to literally vary t, because P is in unknowns, as you say

but you can plug in the points into the plane equation

so 2(x0 + t2/3) - 2(y0 - t2/3) + 1(z0 + t/3) = 0

and solve for t in terms of x0, y0, and z0

[paper gull]

ok. And the absolute value symbols, i need to argue with words saying distance is an absolute measure?

to justify it being equal to the given expression

[stiff totem]

yeah

remember that t is positive in the direction of (2,-2,1)

so if the plane is on the other side of P than that direction

that'll give a negative displacement

[paper gull]

isn't it the other way around?

but i get the point anyway

[stiff totem]

it's P + t(2,-2,1) right?

it'll be positive in the direction of (2,-2,1)

since that's the direction the vector goes as you increase t

[paper gull]

oh, i thought u meant for the point (2, -2, 1)

t is -1

[stiff totem]

ah right

[paper gull]

ok, this approach is much better. Really i should stop and visualize myself but i dont have the time rn 😭

thanks @stiff totem !

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[steady wind]

how to find the intersection of sin2x=cot2x?

[steady wind]

i equate two functions and have sin2x=cos2x/sin2x

not sure where to go from here

[south tundra]

If you multiply both sides by sin(2x) and use a certain trig identity, the equation turns into a quadratic in cos(2x)

[steady wind]

sin^2(2x)=cos2x?

[south tundra]

Yes

[steady wind]

sin^2(2x)=1-2sin^2(x)

[south tundra]

No, you are confusing it with cos(2x) = 1 - 2sin^2(x)

Replace sin^2(2x) with 1 - cos^2(2x)

And move everything to one side

What do you get?

[steady wind]

oh ok

cos^2(2x)+cos2x-1=0

[south tundra]

+cos2x *

Yeah

Before solving this, would be nice to check if the quadratic has any solutions at all

Try computing the discriminant

Ah it's positive

[steady wind]

5?

[south tundra]

Discriminant is 5 yeah, so there may be two solutions to this

Try solving the quadratic equation now

[steady wind]

so cos2x=(-1+sqrt5)/2 or (-1-sqrt5)/2

then i go 2x=cos^-1((-1+sqrt5)/2) or cos^-1(-1-sqrt5)/2?

but it asks for a point of intersection

after i get my two x values i can compute in using one of the functions? either one should work since they are intersecting

thus i get my x and y coords

.close

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[cedar kiln]

Channel closed

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[unreal zinc]

I don’t know how to do this one

[tropic oxide]

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[native lagoon]

the answer is A, you first sub in x=-1 and x=2 to get 2, we know that they intersect at these points so these coordinates also represent points on line l so you just make a linear equation

and then you found out that m=-2, answer A

[cedar kiln]

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

[junior dome]

oh you aint a helper

[unreal zinc]

Mb sorry

[tropic oxide]

it applies to everyone not just peolpe with the helper role

[junior dome]

the first three words

as a helper

[tropic oxide]

the role just subscribes you to a pingstorm

"helper" as in "anybody who helps" not "person bearing the Helper" role

you may notice i lack it

this is on purpose

[unreal zinc]

Thanks btw

[junior dome]

ooh sorry mb

misinterpreted

[tropic oxide]

just sayin'

[native lagoon]

i am getting crapped on rn for helping someone smh

that aint a bannable word right

[tropic oxide]

you are not being crapped on.

nor shit on, nor wailed on, etc.

you are simply being reminded of how things work around here.

[unreal zinc]

It’s just warning

[native lagoon]

you just stabbed me in the back

[unreal zinc]

Man I just asked a question it’s your choice that you told me the answer

[cedar kiln]

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[native lagoon]

get your head outta ur bum, i was being nice, no need to be an ass

[unreal zinc]

I wasn’t being mean, I did what I supposed to do

Alr let’s end this argument

[native lagoon]

bro, don't take it too deep it's a maths server lmfao

finish your maths hw

[unreal zinc]

Close this channel btw

[native lagoon]

.close

[cedar kiln]

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[native lagoon]

pce

[unreal zinc]

Alr bro

Cya

[native lagoon]

it opened it again smh

.close

[cedar kiln]

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[wheat moss]

a line is inclined at theta degrees to the x-z plane
and at phi degrees to the x-y plane

if you are positioned somewhere along +z, looking in the negative z direction

how long does the line appear, if its actual length was 'a'

i thought the apparent length might be sqrt[ (sin(theta) * a)^2 + (sin(phi) * cos(theta) * a)^2 ] but that doesn't seem to be right

[wheat moss]

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[dusk finch]

will the ping work when you edit it?

[junior dome]

они покажут путь <@&268886789983436800>

[wanton sail]

seems like they already left

.close

[cedar kiln]

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[wanton sail]

banned them anyways

[cedar kiln]

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[carmine hare]

idk how to do this

[lunar lynx]

Do you know what is mean defined as?

[carmine hare]

yes

[lunar lynx]

What's that?

[carmine hare]

average of the data

summation of fx by summation of f

[lunar lynx]

Average is just sum of all observations divided by number of observation. Right?

[lunar lynx]

Right.

So, aren't you already given summation of observation?

[carmine hare]

idk what is w

[lunar lynx]

w is wind speed.

Your question clearly says that.

[carmine hare]

31

yes

[lunar lynx]

No. That's number of observation.

[carmine hare]

n=summation of f

[lunar lynx]

n always denotes number of observation/number of data points/sum of frequencies.

[carmine hare]

yes

[lunar lynx]

If f is frequency, only then.

[carmine hare]

yes f is that

[lunar lynx]

you are also given summation of f(x).

[carmine hare]

where

[lunar lynx]

What do you think $\sum {z}$ is ?

[carmine hare]

summation of random something

[wraith dagger]

Enemagneto

[lunar lynx]

Not random something. 🤦‍♂️

What is z in your question ?

It's summation of z values.

[carmine hare]

wind speed

[lunar lynx]

Come on. Read the question properly first.

Is z wind speed ?

[carmine hare]

it doesnt say anything about z

its only in the formula

given

[lunar lynx]

Really?
What do you make of Z = (W-3)/2 ?

[lunar lynx]

Yes, so it's a metric defined for wind values.

[carmine hare]

but idk what w is

or mean is the w

[lunar lynx]

For example, we can say that Temperature of any day is z, where z = (w-3)/2

Similarly, z is any metric.

Its exact name doesn't matter.

[carmine hare]

okay

so fx is z?

[lunar lynx]

Yes. So, for every wind speed w, z is defined via the formula given.

[carmine hare]

how will I understand it if the case was different

[lunar lynx]

Different how?

[carmine hare]

like similar

question

[lunar lynx]

What's your issue with this question ?

[carmine hare]

Im newer to this type of questions anyway so fx is z

then was is fx^2

and it said coded data

[carmine hare]

to calculate sigma

[lunar lynx]

Coded in the sense that you can assume z to be an encoding function. Basically, whenever you give a w value to your function, it returns a z value which is just like a coded version of w-value.

As, for every w-value, there is a unique z value.

That's what they meant. z value is being obtained from w-values using the z definition.

Let's say w=3

For this, we have z = (w-3)/2 = (3-3)/2 = 0

So, z=0 is just a coded value of w=3.

You won't get z as 0 for any other w value.

[carmine hare]

we do trial and error?

[lunar lynx]

Also, given a z, you can always go back and find w value.

[lunar lynx]

No. Why would we do that?

I'm just trying to explain via example.

[carmine hare]

ok so i calculated mean

[lunar lynx]

I was explaining what encoding means.

[carmine hare]

oh

[lunar lynx]

Now, since n is 31, you have thirty one values of w.

[carmine hare]

how 13

[lunar lynx]

Z value was calculated and you were directly given sum of z values.

I said 31.

[carmine hare]

13

[lunar lynx]

Oh. Sorry

Fixed

[carmine hare]

ok

[lunar lynx]

makes sense?

[carmine hare]

so put 31 in w>?

[lunar lynx]

No.

31 is number of values you have for wind.

[carmine hare]

so what is the value of w

[lunar lynx]

Something like this:
Day Wind speed
1 5
2 12
3 9
.
.
.
30 17
31 4

Values are only exemplary.

THEY AREN'T ACTUAL VALUES IN YOUR QUESTION.

[carmine hare]

idk how the wind speed will be from day 1 to day 31

[lunar lynx]

On day 1, wind speed was 5 Km/h.

On day 2, 12km/h

Like that

[carmine hare]

cant even average because idk the values

[lunar lynx]

Also, i have given an example.

They don't necessarily have to be day wise even.

They might be 31 readings taken 5 minutes apart.

Or anything for that matter.

I just gave you an example.

[carmine hare]

yes i see so how will I calculate z

or w

[lunar lynx]

You don't have to calculate either.

You have to calculate their mean and SD.

[carmine hare]

i did calculate mean

for sd i need fx^2

[lunar lynx]

Firstly, to ensure that you understood all that i said - Calculate z value when w = 9.

[carmine hare]

sigma=root over varaince

[carmine hare]

yes

[lunar lynx]

Calculate z value when w = 9.

[carmine hare]

3

[lunar lynx]

Good

Now, show me how you calculated mean.

[carmine hare]

106/31

[lunar lynx]

Okay. Yes. Good

So, we are done with first one.

[carmine hare]

what is sigma

idk fx^2

[lunar lynx]

It signifies summation.

[carmine hare]

I need z

not that

sigma means sd

[lunar lynx]

Oh. That sigma.

[carmine hare]

yes that

[lunar lynx]

They gave that in question.

$S_{zz}$

[wraith dagger]

Enemagneto

[carmine hare]

idk what is this

[lunar lynx]

It also represents standard deviation, I think.

I have only seen S or SD usually though.

[carmine hare]

what is Szz

[lunar lynx]

Let me confirm. a moment.

Ah. It's standard deviation only.

So, you have standard deviation for z.

That's all you need to solve it for w now.

[carmine hare]

uh what

i neeed to find sd

[lunar lynx]

S stands for SD only. 🤦‍♂️

[carmine hare]

szz

Szz

Sz^2

[lunar lynx]

You shouldn't be so recalcitrant about your terminology. Statistics has bunch of different terminology for same stuff.

[lunar lynx]

Lol. No

[carmine hare]

what is Szz and Sz

[lunar lynx]

Okay. Alright. New development - S is standard deviation. But Sz and Szz might be different. I'm unable to find anything substantial on internet.

[carmine hare]

I hate Statistics

[lunar lynx]

https://www.reddit.com/r/HomeworkHelp/comments/jqhni5/a_level_maths_statistics_what_does_s_subscript_zz/

They aren't sure either.

@carmine hare Tell me what you think it is. We can move ahead with that one.

[carmine hare]

I found something

There

So

[lunar lynx]

So Szz is sum of squares of (Zi -Zmean) ?

[carmine hare]

n is 31

[lunar lynx]

Yeah. For standard deviation, just divide Szz by 31 and take sqrt.

,rotate

[lunar lynx]

Let me know when you are done.

[carmine hare]

done

Sigma is 1.61

[lunar lynx]

Cool.

Now, on to second part.

This is mainly the issue.

[lunar lynx]

So, to find mean of w, we need $\sum w$.

[wraith dagger]

Enemagneto

[lunar lynx]

We have $\sum z$.

[wraith dagger]

Enemagneto

[lunar lynx]

$$\sum z$$ $$=\sum \frac{w-3}{2}$$
$$= \sum \frac{w}{2} - \sum\frac{3}{2}$$

[wraith dagger]

Enemagneto

[lunar lynx]

@carmine hare That should be enough to work it out. Try

[carmine hare]

I dont understand

[lunar lynx]

Which part?

[carmine hare]

this

interms of w

oh ik summation of z

wait

can you cacel sigma?

[lunar lynx]

What do you mean by "cancel sigma"?

[lunar lynx]

I just put value of z from the function.

[carmine hare]

idk but Sigma w/2 - sigma 3/2 = 106

[lunar lynx]

Yes. That's correct.

Now, just simplify to find $\sum w$.

[carmine hare]

now there is sigma

[wraith dagger]

Enemagneto

[carmine hare]

how can I make w the subject

oh wait

[lunar lynx]

I have to go. Be back in 5 minutes roughly.

[carmine hare]

how to simplify this

[lunar lynx]

Well, firstly, $\sum \frac{w}{2} = \frac{1}{2}\sum w$.

[wraith dagger]

Enemagneto

[lunar lynx]

Do you get that?

@carmine hare

[carmine hare]

idk how todo this

first of all how to remove the sigma

[lunar lynx]

You don't need to remove the sigma for w one. You need the value of Sigma(w).

To find mean.

For the $\sum (3/2)$, use definition of sigma.

[wraith dagger]

Enemagneto

[carmine hare]

summation of something

idk the value of sigma w

[cedar kiln]

@carmine hare Has your question been resolved?

[cedar kiln]

@carmine hare Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[limber snow]

How would I prove its injective

[upper abyss]

Hey calc

As was said, this is basically impossible to do it this way

You want to show that N×N is countable, and so every equivalence relation over N is countable

[limber snow]

Is there no other way because Ann was helpin me earlier with that and it seemed so confusing

wait

I just prove that N x N is countable?

and thats good enoiugh?

[mental trail]

An equivalence relation is a subset of N×N, and so if you prove that N×N is countable, the proof is over

[upper abyss]

There's an easy diagonal method to show that the Cartesian product of any two countable sets is countable

I'm trying to look it up atm

[mental trail]

Yes, but here's a more rigorous proof

[upper abyss]

But you've likely covered it in class

[limber snow]

ye we did the diagnol thing to prove that |(0, 1)| = |R|

[upper abyss]

Different diagonal proof lol

[mental trail]

If you do the 'diagonal argument', you still have to construct the function which is complicated to show. However, consider $f:\begin{cases}\bN × \bN \longrightarrow \bN \ (m,n) \longmapsto 2^m(2n+1) - 1 \end{cases}$

[upper abyss]

There's a few ig

[wraith dagger]

rafilou2003

[limber snow]

Could I like list all of the pairs in a way where the nth pair contains all the pairs whos elements summed up return n

wait no fuck that wont work because 0 isnt a natural

[limber snow]

how did u come up

wit

2^m(2n + 1) - 1

oh wait could I

[mental trail]

Oh is 0 not a natural integer for you ? You poor thing

[crimson delta]

(you also have to show that an equivalence relation on N cant be a finite set)

[limber snow]

do this but where the sum of them returns n + 1

[limber snow]

ye not in this course

but this is what im thinking

lets say

[mental trail]

Okay so :

Consider $f:\begin{cases}\bN × \bN \longrightarrow \bN \ (m,n) \longmapsto 2^{m-1}(2n-1)\end{cases}$

[upper abyss]

This is my argument, but applied to fractions. It's easy to change this to N×N though

[wraith dagger]

rafilou2003

[limber snow]

1st pair = (1, 1) => 1 + 1 = 2
2nd pair = (1, 2), (2, 1) => 1 + 2 = 3
3rd pair = (1, 3), (3, 1), (2, 2) => 1 + 3 = 4 and 2 + 2 = 4

..

uptil n

where the nth will contain pairs that sum upto n + 1

wait is that even right

[limber snow]

I dont get how u came up with that doe

2^(m - 1)(2n -1)

of what

[mental trail]

Yeah this is factoring by 2 as much as possible

For every natural integer, you take the biggest power of 2 that divides it

So you can write this power of 2 as 2^(m-1)

And what you're left with has to be an odd number, which can be written as 2n - 1

[mental trail]

Also, if you want another bijection that relies on this technique you used by regrouping all couples that have the same sum :

$f:\begin{cases}\bN × \bN \longrightarrow \bN \ (m,n) \longmapsto \frac{(n+m-2)(n+m-1)}{2} + n \end{cases}$

[wraith dagger]

rafilou2003

[mental trail]

So f(1,1) = 1

f(1,2) = 2

f(2,1) = 3

[limber snow]

ah

[mental trail]

Oops I made a small mistake in the function

[limber snow]

wait so I could use

either of the functions u stated

[mental trail]

I'll rewrite this second one in a few moment

[limber snow]

ill just use the firts one

first

[mental trail]

Yes the first one I know works

[limber snow]

and now I prove that its bijective right

[mental trail]

Yes

[upper abyss]

It's definitely good to get a few ways to prove this. I disagree that mine is not rigorous, and I personally believe that the function is trivial to create with my method.

Just can't be written in a f(x) = ... manner, which is not important to the proof

[mental trail]

It can be written, here it is

[upper abyss]

With the diagonal argument, I mean

[mental trail]

Yeah this is the diagonal argument It's not quite the same as this but you just reverse everything so that every diagonal is read from left to right

[upper abyss]

Or keep it the same as that haha

[limber snow]

but how would I use the diagnol in my proof like do I just draw that out and say that every pair could be counted starting at (1, 2) then (1, 2) then (2, 1) then (1, 3) then (2, 2) and then (3, 1) ....

[mental trail]

This is where some profs will not like the drawing of the diagonal argument as proof

[limber snow]

ye so idt the diagol arguement would b good in my case

diagnol

[upper abyss]

Imo, your class has already proven that the Cartesian product of any two countable sets is countable, and probably used the diagonal argument to prove it

[limber snow]

oh wait lol I j remebered we could use anything thats proved in class for free

[mental trail]

I know that a lot of profs don't like the diagonal argument without an explicit bijection because I made an entire course on set theory and countability and my "supervisor" specifically told me not to allow proofs by drawing

[limber snow]

so if this question was to appear on my tests i could just say that since we know cartesian product of two sets is countable then N x N is countable n done lol

[upper abyss]

Mind you, I am not in your class so I can't be SURE this happened

[limber snow]

ye ik but i remember we proved it

also rafilou how would I prove the surjectivity

Im stuck on that

[mental trail]

The surjectivity is not that important, but if you want it : just remember that for a natural integer N, the set ${m\in \bN, 2^{m-1}|N}$ is bounded

Latex bot has failed us :(

I'll copy and paste it again, hopefully latex bot will wake u0

[limber snow]

also one more thing

after proving that

N x N is countable

how would I relate it to the question

[stiff totem]

it's because there's a space before the last $

[limber snow]

that |e_1| = |e_2|

[wraith dagger]

rafilou2003

[mental trail]

Thanks

Ok so after saying that N×N is countable, the next step is to show that an equivalence relation over N is infinite

After that, since an equivalence relation is a subset of N×N, it will be countable

[limber snow]

wth how do I prove that

Ive never seen anything like that in my book

[mental trail]

Use one of the properties of equivalence relations :)

If you wanna guess, you have 1 chance out of 3 to get it right

[limber snow]

imma try solving it brb

is it the transitive property

[mental trail]

No :(

Still 2 guesses

[limber snow]

reflexice?

reflexive*

[mental trail]

Yes!

[limber snow]

how so

[mental trail]

Can you tell me what reflexivity is over $E_1 \subseteq \bN^2$?

[wraith dagger]

rafilou2003

[limber snow]

(a, a) in E_1 => (a, a) in N^2 ?

[mental trail]

No, this is just inclusion

Reflexivity is : a in N => (a,a) in E_1

[limber snow]

and N is infinite so its infinite?

[mental trail]

Yes

[limber snow]

ah so that was it?

and the same goes for E_2

so both are countable

[mental trail]

Yes

If you're interested in similar exercises over countability, I can hand you over my exercise sheet

[limber snow]

yes pls

[mental trail]

Alr I'll hand it to you in DMs

[limber snow]

bless u.

.close

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[limber snow]

.reopen

[cedar kiln]

✅

[limber snow]

@mental trail

how would I do this

[mental trail]

(Already Answered in DMs)

[cedar kiln]

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[calm totem]

So this is a solution my prof posted on an example about the summation of series. I can follow what he does fine until the middle section where he plugs in the values of n but its divided by 2.

[calm totem]

We're supposed to use this formula but its divided by 6 instead of 2. And why is the 2x10 turned into 200?

[autumn fox]

They probably mixed up some things when using the closed form of the summation

[calm totem]

I've been trying to understand where I went wrong but the more I think about it, the more I want to say he made an error

[autumn fox]

The summation of k from k = 1 to n is n(n + 1)/2, so they might have mixed up those two in terms of the denominator

[calm totem]

That makes sense. And how would u explain the 2x10 turning into 200?

[autumn fox]

My best bet is that your professor was tired when doing the calculations

[calm totem]

Alright that makes me feel good. I've been going over this for 1.5hrs.

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[fervent crag]

is this a proper evolution? or did i mess a step up.

2(x^2 +4x -12) = 0
2(x^2 +4x + (4/2)^2 - 12) - 8 = 0

[fervent crag]

i am trying to make it into vector form

[crimson sedge]

evolution?

[crimson sedge]

vector form?

[fervent crag]

yes

like is the first step equal to the second

[wanton sail]

.close

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[wanton sail]

(user got muted)

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[oblique raven]

sorry im copyrighting lolmao

[oblique raven]

please remove this

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[hoary chasm]

What

[hoary chasm]

.close

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[mild carbon]

i need help with an inference question

[mild carbon]

i don't get the 4th inference is valid

could someone explain to me

?

[cedar kiln]

@mild carbon Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

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[exotic tartan]

hey i need help for something idk if this is the right channel for that but the question goes like this, find a problem in your day-to-day life that you can solve For example,
suppose you are coordinating a youth group event to make sub sandwiches to sell as a fundraiser. You might have tuna subs, regular subs, and vegetarian subs. Some of the
ingredients will be used in several or all three types, while other ingredients will be unique to one kind of sub. use set theory to think through how many portions of each ingredient you
will need.

[jade charm]

Uh I'm not sure if set theory is the thing you want here

[tropic oxide]

instructions that conflict with the problem 🤡

"use a hammer to dig out a hole in the ground"

[exotic tartan]

wtf

[cedar kiln]

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@exotic tartan Has your question been resolved?

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[rotund vigil]

Is the answer right ?

[vague rapids]

seems right

[cedar kiln]

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[iron dagger]

The angle of depression from the top of a 20m building to a worker standing on the ground below is 40 degreees. Find the distance of the worker from the base of the building to 2 decimal places

[iron dagger]

I'm so confused, how would i work out the distance from base to worker?

[exotic furnace]

Draw a figure first

[iron dagger]

I did but the question doesn't provide any info on how tall the worker is

[exotic furnace]

We assume the worker is a point

Because the worker is clearly not significantly tall compared to the building

[iron dagger]

ok but then it says find the distance of the worker from the BASE of the building? how does that make sense?

nvm i figured it out thx for ur help anyway!

!close

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[dim tiger]

how is the number of deleted rows and columns deleted from the original matrix to obtain the minor M k

shouldnt the minor M be a kxk matrix

because it is obtained by choosing k rows and k columns then the elements consisting these rows and columns will be the elements of the minor M

[celest ledge]

Yeah so it’s a mistake

[dim tiger]

oh ok a mistake in the book

i have another question

[celest ledge]

Go ahead

[dim tiger]

how is rotating 180 deg about principle diagonal change the angle with the row from $\alpha$ to $90-\alpha$

[wraith dagger]

calculus is fun

[celest ledge]

No idea, I don’t think transposition is a rotation at all, in my mind it’s a reflection

[dim tiger]

the x-axis is the row the grey line is the principle axis of equaton y=-x the red line is th segment joining 2 elements of the matrix anf the brown segment is the result of rotation

is this rotation correct ?

[celest ledge]

I don’t know, sorry I can’t help you, I just don’t think transposition is a rotation at all. To me it’s like reflection by the line y=-x

[dim tiger]

oh ok but i mean if i am willing to just rotate the red segment 180 deg about y=-x in general will the result of rotation be the brown segment ?

[celest ledge]

I understand rotation by an angle. I have no idea what rotation about a line means

[dim tiger]

oh ok

tysm for your time

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[keen nova]

i want to ask some prob v basic questions about factorization in hope that it can make me do it more intuitively

[keen nova]

$x^4 + 3x^3 + 2x^2 = x^2(x^2 + 3x + 2)$

[wraith dagger]

marty

[keen nova]

$x^4 + 3x^3 + 2^2 = x^2(x^2 + 3x) + 2^2$ ?

[wraith dagger]

marty

[magic solar]

no

when you expand

well you didnt expand it

[vagrant elbow]

*yes but that doesn't get you anywhere

[magic solar]

i think you're missing a x^2 right

in the 2^2 on lHS

[keen nova]

no, the idea is that if 2x^2 instead was 2^2

[vagrant elbow]

There's no x^2 term

[keen nova]

and thus x^2 is no longer a common factor of 2x^2

is it then no longer possible to factor at all

[magic solar]

yea

you cant apply the same idea from first latex to second one

[keen nova]

so to factor, the common factor needs to apply to all terms of the polynomial

[magic solar]

yes, or it could be something like

2x^2-x + x^3-

well

nvm bad example

[vagrant elbow]

We'd be unsure of that

[keen nova]

unsure if it is possible to factor?

[vagrant elbow]

It may or may not be

,w factorize x^4 + 3x^3 + 4

[wraith dagger]

Results provided by WolframAlpha

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Requested by neonperseus

[keen nova]

yes so you effectively try to factor by 1 which is obv the same as it was

because 1 is the only common factor

of all the terms

[magic solar]

well just because all terms dont have a common factor does not mean it isnt factorable

[vagrant elbow]

Yeah sorry I was hit with a lagspike

[keen nova]

and to investigate if it is factorable you do what?

[vagrant elbow]

you hope and pray

[magic solar]

yeah but

there obviouslt are some ways

[vagrant elbow]

and throw in a few small integers

[magic solar]

mostly just visually obvious, or yeah sub in some numbers which equate that to 0

usually integers or else too tedious

[vagrant elbow]

If you have something like $x^6 - 3x^3 +2$ you can sub $x^3 = y$ and continue factoring

[wraith dagger]

NEON

[keen nova]

$y^2 - 3y + 2$

how does this help ?

[vagrant elbow]

y^2 - 3y + 2

[keen nova]

right

[wraith dagger]

marty

[keen nova]

still

[vagrant elbow]

You know how to factorize a quadratic surely

[keen nova]

2 has no common factors

oh

shit

yeah

okay

yes i know but this is the type of stuff i dont think about doing

how did you think of that, i want to understand better

[vagrant elbow]

Even in your previous thingie

$y^3 -3y + 2$

[wraith dagger]

NEON

[vagrant elbow]

This is still factorable

[vagrant elbow]

It's just noticing the powers you'll get used to it

[vagrant elbow]

Note how the sum of the coefficients is 0

It implies y - 1 is a factor

which means you can long divide

[keen nova]

i would not think of that

[keen nova]

and why

[vagrant elbow]

it's the first thing you should do when working with powers higher than 2

[vagrant elbow]

Well substitute y = 1 and see what happens

For a polynomial $P(x)$, the remainder upon dividing by $x - r$ is $P(r)$, by the remainder theorem, if this remainder is 0, i.e. $P(r) = 0$, it implies that $r$ is a root of $P(x)$.

[wraith dagger]

NEON

[vagrant elbow]

Which means it can be expressed as $(x - r)Q(x)$ for some polynomial $Q(x)$ which can be found using long division.

[wraith dagger]

NEON

[crimson sedge]

add and subtract y^2 works too

[vagrant elbow]

2y^2 I think

[keen nova]

there are many things here i need to intuitively understand, even when you say long division i need to make sure i know 100% what you mean by that

[vagrant elbow]

Polynomial division?

Synthetic division?

https://youtu.be/_FSXJmESFmQ?si=2Ye4MFd_5FanHN0L

[keen nova]

polynomial long idivision

yes i just opened this video lmao

[vagrant elbow]

lol

[vagrant elbow]

If it hasn't been taught to you, then you can safely ignore everything after this point

I explained it since you said you wanted methods of factorization

[keen nova]

the extent of my knowledge is apparently "look for common factors, if none, don't factorize". And factorizing quadratics ik how to do mostly, but i never think of it

[keen nova]

but for this example

i was trying to do it while we spoke and i couldn't get it right first try. I tried with

$(y-1)(y^2+2)$

[wraith dagger]

marty

[keen nova]

but thats wrong

we want to get y^3 and 3y and 2. How can i think of this to solve quickly

[crimson sedge]

try adding and subtracting y^2 and it becomes quite straightforward

[keen nova]

what do you mean by that exactly

[keen nova]

adding y^2 to this?

[crimson sedge]

adding and subtracting the same term has no effect on the equation, but it may help you in factorization

[keen nova]

i still cant see it

[wraith dagger]

marty

[crimson sedge]

rearrange it and split -3y into -y and -2y

$$y^3-y^2+y^2-y-2y+2$$

[wraith dagger]

dimpledoink

[cedar kiln]

@keen nova Has your question been resolved?

[keen nova]

theres some knowledge im lacking because still i cant see

i dont see a common factor of 1 term, so obviously it has to be of at least 2 terms. But idk how to look for that

should i do polynomial division of y + 2?

[crimson sedge]

you can either do y-1 or y+2

[keen nova]

so

the strategy is to find the common factor aside from the constant/non-factorable term, then add the non-factorable term

so y is the common factor, plus 2, the non-factorable term

then do long division

[crimson sedge]

ok that doesnt always work

[keen nova]

ok well

at least i managed to do it now

so long division is cool

i never did it before

do you have any tips in finding what polynomial to divide with

[crimson sedge]

once you have something that looks like this you can try and start "pairing" the terms

like y^3-y^2 is one pair, y^2-y is another, and -2y+2 is another

then you take the common factors out which here would be y^2 in the first pair, y in the second and -2 in the third (not +2)
and you end up with something like this $$y^2(y-1)+y(y-1)-2(y-1)$$

$$y^2(y-1)+y(y-1)-2(y-1)$$

[wraith dagger]

dimpledoink

[crimson sedge]

now the pairing is not always going to work out, and if it doesnt, try rearranging and splitting terms in different ways

you'd ask why -2 and not +2, and that's because we need the same terms in the parentheses so we can divide by the same term
if you took +2 as the common factor in the last pair you would've ended up with something like

$$y^2(y-1)+y(y-1)+2(1-y)$$

[wraith dagger]

dimpledoink

[crimson sedge]

after this just divide by y-1, and you get a normal quadratic equation which is easily factorized

[keen nova]

for these type of examples its v easy, ik 12 is a sum and 32 is a product, so find compatible numbers

so when i hear "how to factorize a quadratic" this is what i imagine

can i relate this somehow with what you're now saying

[crimson sedge]

have you been taught the quadratic formula yet

[keen nova]

yep

it just seems like i have knowledge gaps

[crimson sedge]

ok that's the easiest way to factorize a quadratic

use the quadratic formula and find the roots

[keen nova]

okay

yeah makes sense

[crimson sedge]

whatever the roots are, you flip the sign
for example, in the above example, the roots will be 8 and 4
and then you flip the sign and get -8 and -4
and then add x to both and get x-8 and x-4

essentially reverse engineer the factors using the roots, normally you'd find the factors of the quadratic, equate each of them to zero, and then find the roots

[keen nova]

yeah, got it

so thats for a quadratic equation

[crimson sedge]

yeah

not for any other degrees

[keen nova]

thank u

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[plucky barn]

How do I solve this?

[tropic oxide]

what does $n = 2x - 13$ have to do with the line above?

[wraith dagger]

Ann

[tropic oxide]

as far as i can read, the top line says: $$(h^{-1} \circ h)(b) = ; ?$$ please confirm whether i have read it correctly

[wraith dagger]

Ann

[plucky barn]

Not b, 6

But otherwise, yes

[mental trail]

Well, what do you know about functions composed with their reciprocal

[plucky barn]

Same values?

But x and y are flipped

So they cancel out?

[stiff totem]

yes, a function and its inverse cancel out like that (kind of by definition)

I think you have a sign error on h^-1 though

[cedar kiln]

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[mental trail]

!status

[cedar kiln]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[tidal wasp]

1

[mental trail]

have you tried computing directional derivatives ?

[tidal wasp]

yes

but the function at 0,0 is just 0

[mental trail]

yes but the directional derivative might not be 0

have you tried computing the derivative with $y = \lambda x$ ?

[tidal wasp]

teh gradient vector would be (0,0)?

[wraith dagger]

rafilou2003

[tidal wasp]

what?

[mental trail]

the gradient is not 0 I think

[tidal wasp]

what do you get when you differentiate 0?

[mental trail]

aight, bet with a counterexample

[tidal wasp]

the partial derivative would just be 0

[mental trail]

Let $g(x) = x$ the function over the real numbers $\bR$

[wraith dagger]

rafilou2003

[mental trail]

$g(0) = 0$ right ?

[wraith dagger]

rafilou2003

[tidal wasp]

yeah

[mental trail]

what about $g'(0)$ ?

[wraith dagger]

rafilou2003

[tidal wasp]

may or may not be 0

[mental trail]

may not

$g'(x) = 1$

[wraith dagger]

rafilou2003

[tidal wasp]

how?

[mental trail]

what's the derivative of x

[tidal wasp]

1

[mental trail]

so g' = 1

so evaluate it in 0

[tidal wasp]

but how do you know that the function is g(x) = x

[mental trail]

because I created it that way to show you a counterexample

[tidal wasp]

oh right

well g'(0) = 1

[mental trail]

So f(x,y) = 0 doesn't imply the gradient is 0

[tidal wasp]

i see

so what can i do here?

[mental trail]

The way the directional derivative works, is that if you want for example a derivative along the vector (1,2), then you let (x,y) = (1t, 2t) and you compute f'(t)

so in our example :

if we have $(x,y) = (1t, 2t)$, then f(x,y) = ...

[wraith dagger]

rafilou2003

[tidal wasp]

$\frac{2t^3}{t^2+16t^4}$

[wraith dagger]

Bedsheat

[tidal wasp]

?

[mental trail]

you got it!

[mental trail]

And then to find the directional derivative from the vector (1,2), you just differentiate that

[tidal wasp]

would the derivitive of that be $\frac{2-32t^2}{1+256t^2}$

[wraith dagger]

Bedsheat

[mental trail]

you're differentiating a quotient

[tidal wasp]

yeah i used the quotient rule

then cancled out t^4

[mental trail]

at the denominator I think is (1+16t^2)^2

[tidal wasp]

but the denominator is t^2+16t^4

did you cancel it out?

at the start?

[mental trail]

cancel by t^2 before differentiating

So $\frac{2t}{1+16t^2}$

[wraith dagger]

rafilou2003

[tidal wasp]

i see i did that before differentiating

[mental trail]

differentiates into $\frac{2 + 32t^2 - 64t^2}{(1+16t^2)^2} = \frac{2-32t^2}{(1+16t^2)^2}$

[tidal wasp]

should it not be -32t and not -32t^2

[mental trail]

Oh lemme correct that

[wraith dagger]

rafilou2003

[mental trail]

because -2t * 32t

[tidal wasp]

right

[mental trail]

This is well defined in 0 and values 2, so $D_{(1,2)}f(0,0) = 2$

[wraith dagger]

rafilou2003

[mental trail]

oh I just realised a small hiccup

$f(1t,2t) = \frac{4t^3}{t^2+16t^4}$

[wraith dagger]

rafilou2003

[mental trail]

and not 2t^3

but it's alright, the derivative is just multiplied by 2

and so $D_{(1,2)}f(0,0) = 4$

[wraith dagger]

rafilou2003

[tidal wasp]

how

the function is xy^2

t*2t^2

ohhhhhh

right

bruh

i see

[mental trail]

Alright so we're starting to get some intuition

$D_{(1,2)}f(0,0) = \frac{2^2}{1}$

[wraith dagger]

rafilou2003

[mental trail]

we're still left to prove that in the general case

so now we just have to do the same thing but with $(x,y) = (u_1t,u_2t)$ with $(u_1,u_2)$ a unit vector as the question suggests

[wraith dagger]

rafilou2003

[mental trail]

Now that we know exactly what to do (and we've done it before) it should go faster

So what's $f(u_1t,u_2t)$ ?

[wraith dagger]

rafilou2003

[tidal wasp]

lemme do it

[mental trail]

I'll be away for the next 10 mins, if you're stuck sometime in between you can ping helpers

[tidal wasp]

k thx!

$f(u_1t,u_2t)=\frac{u_1u_2^2t^3}{u_1^2t^2+u_2^4t^4}$

[wraith dagger]

Bedsheat

[tidal wasp]

$f'(u_1t,u_2t)=\frac{-u_1^3u_2^2-2u_1u_2^6t^2}{(u_1^2+u_2^$t^2)^2}$

[wraith dagger]

Bedsheat

[tidal wasp]

idk i messed up somewhere

i think its $\frac{u_1^3u_2^2-u_1u_2^6t^2}{(u_1^2 + u_2^4t^2)^2}$

[mental trail]

the denominator should be $(u_1^2 + u_2^4t^2)^2$

[wraith dagger]

rafilou2003

[tidal wasp]

okay i think this time im correct

[tidal wasp]

it is tho isnt it?

ho wait

[mental trail]

(a+b)^2 ?

[tidal wasp]

its whole square

[mental trail]

yes

this is not the first time I see someone writing (a+b)^2 = a^2 + b^2 so I don't judge ;)

[tidal wasp]

there

now it should be fine

[wraith dagger]

Bedsheat

[tidal wasp]

not it wasnt

NOW it is

[mental trail]

yes that should be right

now, is this defined in 0?

[tidal wasp]

sure is

[mental trail]

almost

[tidal wasp]

bruhhhh

[mental trail]

there's a value for which it's not defined

(depends on u)

[tidal wasp]

0,0

[mental trail]

oh it's just not defined in 0,0 when u_1 = 0

because of the denominator cancelling

[tidal wasp]

yeah i was about to correct myself

when u_1 = 0

[mental trail]

yep

and so when it's not everything's fine!

[tidal wasp]

its not defined at the point 0

[mental trail]

so let's assume $u_1 \neq 0$

[wraith dagger]

rafilou2003

[mental trail]

in that case, what is $D_{(u_1,u_2)}f(0,0)$?

[wraith dagger]

rafilou2003

[tidal wasp]

$\frac{u_1^3u_2^2}{u_1^4} = \frac{u_2^2}{u_1}$

[mental trail]

yes, and simplified ?

[wraith dagger]

Bedsheat

[mental trail]

👍

[tidal wasp]

phew

[mental trail]

Alright so which ones do we now know is true/false ?

[tidal wasp]

1 is false

2 is true

we dont know 3 or 4 yet

[mental trail]

I'll just do this for myself

[tidal wasp]

alr

[mental trail]

So looking at question 3

or statement 3 sorry

[tidal wasp]

it seems false

there is no maximum

[mental trail]

"the maximum of directional derivatives occurs in the direction of the vector (5,5)"

yes there is no maximum !

can you see why?

[tidal wasp]

yeah i can

the higher u_2 and closer to 0 u_1 is the more the value of hte directional derivitive

[mental trail]

yes that's kinda the idea

though you do have to be careful, since $(u_1,u_2)$ is a unit vector you have that $u_1^2 + u_2^2 = 1$

[wraith dagger]

rafilou2003

[tidal wasp]

ohhhhh

[mental trail]

So while we can't make u_2 "bigger and bigger" as it's bounded by 1, we can make u_1 closer and closer to 0

[tidal wasp]

i forgot about that

i see now

[mental trail]

alright, now left with the statement about tangent lines

[tidal wasp]

yeah

[cedar kiln]

@tidal wasp Has your question been resolved?

[cedar kiln]

@tidal wasp Has your question been resolved?

Channel closed

Closed by @tidal wasp

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[noble perch]

how do i identify the pivot columns

[noble perch]

im confused because row 2 eliminates row 3 so im not sure if columns 2 and 3 count as pivot colums along with column 1

[royal loom]

continue row reducing

right now you have a 2 along the diagonal

a zero row from row reduction tells you that one of your columns of your matrix is linearly dependent on the other two. i.e you will not have full rank, or full pivot columns (you won't have all 3)

[noble perch]

oh okay

i didn't know that

[noble perch]

im having trouble understanding why (1,2,0) is linearly dependent on (1,0,0) and (0,1,0)

[royal loom]

<1,2,0>= 1 * <1,0,0> + 2 * <0,1,0>

it can be represented as a linear combination of the other two

[noble perch]

ohhhh

[royal loom]

therefore it is linearly dependent on them

but

that is not the take away you should be getting from your row reduction

[noble perch]

yeah

[royal loom]

a pivot is the the first nonzero element in a row

what is the pivot of row 1?

[noble perch]

1

first row first column

[royal loom]

which is in the first column

so column 1 is a pivot column

what is the pivot of row 2?

[noble perch]

i wanna say 2

[royal loom]

yes

2 is the first nonzero element in row 2

so it is the pivot

and 2 is in which column?

[noble perch]

second column

[royal loom]

so the second column contains a pivot

so the second column is a pivot column

[noble perch]

oh

this makes a lot of sense now

so im also asked to find the column and row space

[royal loom]

the column space, is the space spanned by the columns

do you know what I mean by "spanned"?

[noble perch]

yeah it means a set in which the vectors are linear combos of each other

[royal loom]

sure

so the column space is all possible linear combinations of the columns of our matrix