help-33

[dreamy lintel]

(with other tutors too)

incorrect angle?

but I used the directions given

[near pecan]

OK

That sounds frustrating. So, I have two questions:

1. What has your overall experience with this math class been like?
2. Would you like to explain (briefly) why I brought up the focus on vectors having directions, but not lines?

[dreamy lintel]

overall my experience in the class is fine

I understood nothing from the first 4 modules on the test until like the night before looking at the steps for solving the problems

examples I mean

and then applied those steps to the test

and did very well despite not understanding much theory

I'm kind of behind actually

I didnt learn this module in lecture because I was absent

[near pecan]

did very well despite not understanding much theory
Congrats!

I'm kind of behind actually
That can be challenging. Math is very cumulative.

I didnt learn this module in lecture because I was absent
I'm sorry to hear

[dreamy lintel]

basically my goal right now is to just learn the steps in solving these problems on this homework assignment

I also have a quiz on it tomorrow

[near pecan]

Sounds good

At some point you might want to understand the theory and catch up. If there's something from the past you've not understood, it will likely catch up with you.

So, let's go through this a bit faster OK?

[dreamy lintel]

yeah I'd definitely like to learn the theory but at the moment Im a bit crunched on time

sounds good to me

[near pecan]

But please dive in deeper later

[dreamy lintel]

will do

[near pecan]

The one on the right is the correct one

Vector V is the same vector in both. It has a single direction and length

The projection is supposed to make a right angle with the line.

The line goes on forever

[dreamy lintel]

gotcha thats really simple

[near pecan]

Cool

[dreamy lintel]

weird that i missed that

[near pecan]

All g.

[dreamy lintel]

ofc it forms a right angle

its a shadow essentially

[near pecan]

Only the direction of the vector matters

[dreamy lintel]

right

[near pecan]

Next we're looking at c - projkc

Lemme draw this out for you

[dreamy lintel]

sounds good

[near pecan]

It'll be different than your work, but we'll go over the concept quickly

[dreamy lintel]

sure

[near pecan]

OK.
Picture 1: we have two vectors C and D

2: We make D negative (literally reverse its direction)

3. We can move -D. -D=-D regardless of where it is.

4. C-D = C + (-D)

5. We can move the vector C-D around too

[dreamy lintel]

gotcha

[near pecan]

That should help w question 2, right?

[dreamy lintel]

yes it adds context

[near pecan]

Actually, can you post the questions again?

[dreamy lintel]

yes lol

so its c - the projkc

but isnt it already in both directions because its a line?

[near pecan]

ProjKC is not a line. K is. ProjKC is a vector that's the shadow of C on the (infinite) line K

[dreamy lintel]

rightright

so would we just make a shadow in the inverse direction?

[near pecan]

Hold on. More drawings incoming

[dreamy lintel]

gotcha

[near pecan]

Yes. You could, but look at picture 5. Shortcut for vector subtraction

C-D = from tip of D to tip of C

[dreamy lintel]

right

so we just create a parallel k at the tip of c going in the inverse direction?

[dreamy lintel]

so just inverse the green line at the tip of c?

[near pecan]

If you mean what I think you mean, then yes

[dreamy lintel]

so how do we use that to solve the problem?

is it just an explaination or is there math to solve?

[near pecan]

Well you should be able to draw a vector for c-projkc

then

[dreamy lintel]

right

[near pecan]

New drawing incoming

[dreamy lintel]

gotcha

[near pecan]

OK. If we project C onto D, we get proj_dC

I just moved it down a little so it's not on top of D

Good?

[dreamy lintel]

uhh

where is d from

[near pecan]

Just showing two vectors

We're developing the dot product

[dreamy lintel]

gotcha

[near pecan]

Note that D is on a line (this is called a span, but you don't have to know that)

We can measure the length of this projection

the dot product of c and d is the length of the projection times the length of d

[dreamy lintel]

what is the dot product

[near pecan]

in your question it says (c-proj) * [1, 2]

[dreamy lintel]

gotcha thats dot prod form?

[near pecan]

that's (c-proj) dot [1, 2]

yes

it's a way of multiplying vectors that results in a scalar

[dreamy lintel]

gotcha

[near pecan]

(5,4)*(-1,8) = (5)(-1) + (4)(8)

= 27

[dreamy lintel]

so thats the length?

[near pecan]

Not quite. Technically, it's the product of the length of the projection with the length of the other vector

[dreamy lintel]

I see

[near pecan]

Hold on

[dreamy lintel]

sure]

[near pecan]

If you project a vector onto itself, it's the perfect noon shadow. You get the vector back

So, proj_cC = C

then the length of that projection times the length of c is
the length of C * the length of C

But that's the dot product of C with C!

In general dot product is a measure of how aligned vectors are and their lengths

Good video here: https://www.youtube.com/watch?v=LyGKycYT2v0&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=9

OK

[dreamy lintel]

I see

so what are we trying to get with this problem?

[near pecan]

It looks like they're having you derive an identity

You got c-projC?

[dreamy lintel]

no im a bit confused now

unfamiliar with these steps

[near pecan]

You already have projC

Now you just need to do the subtraction as shown above

[dreamy lintel]

so how do I do the subtraction

[near pecan]

C-D = from tip of D to tip of C

Connect the arrows in that order

[dreamy lintel]

right but what are the measurements

[near pecan]

It doesn't matter

[dreamy lintel]

when I look on chegg it has weird calculations and stuff

[near pecan]

They're asking you to draw it

[dreamy lintel]

it says "What should it be"

bit of a vague question lol

like

theoretically? mathematically?

[near pecan]

Hold on. Lemme review this

[dreamy lintel]

sure

[near pecan]

Yeah. That's definitely wrong

Oh wait

hehe

[dreamy lintel]

the chegg?

[near pecan]

No. Sorry it's all right. They're good. I misread the question

[dreamy lintel]

oh ok

[near pecan]

OK. That makes sense

[dreamy lintel]

gotcha

[near pecan]

C = (1,3), right?

[dreamy lintel]

yes

[near pecan]

OK. Yeah. The projection of c onto K is 7/5K, not 7/5C. They are wrong

[dreamy lintel]

so chegg is also wrong gotcha

where do we get 7/5k from

[near pecan]

There'a a formula for the vector projection. It would be (c dot k)/(k dot k) all times k

But it's not necessary

The question asks you about c - projkc, right?

That's the orthogonal projection

Yeah. Don't worry about measurements. I think I did read it correctly.

Check this out. This should be the last drawing then we'll review

[dreamy lintel]

ok

[dreamy lintel]

no shadow?

[near pecan]

Note that the projection of purple vector onto the blue (or any vector along blue)

Yes. Has no shadow

That should be all you need

Do you have a text book?

[dreamy lintel]

yes

the hw is from the textbook

im ngl

im very confused

[near pecan]

Oh nice

OK. Let's review

[dreamy lintel]

I understood like the concept of the inversion of projkc since its negative

but thats all

[near pecan]

That's it!

Can we do voice calls?

[dreamy lintel]

sure

[near pecan]

I'm not sure if this works in this room

[dreamy lintel]

yeah its a text room only i think

[near pecan]

OK. You get that a projection will be positive if the two vectors point the same way?

And negative if they point in opposite ways

[dreamy lintel]

yes

[near pecan]

What's between positive and negative? 0

So if they are perpendicular, the projection is 0

[dreamy lintel]

and no shadow

[near pecan]

Exactly

[dreamy lintel]

so the question just asks I explain that concept?

wb the *[1,2} attatched?

[near pecan]

The book is trying to get you to derive the orthogonal projection

[dreamy lintel]

how do i derive it?

[near pecan]

Can you show me your drawing of c - projc?

Dude, we are so close!

[near pecan]

OK. A few notes. Your first drawing was (almost) perfect:

That dark arrow under C is the projection of C onto K

(almost)

But the angle is a little wide. projc should be longer so that the shadow makes a right triangle

[dreamy lintel]

it just needles to be straight yeah

[near pecan]

Not that C land K point in the same direction

Sorry. C and projc

[dreamy lintel]

right

[near pecan]

that black arrow is your projection

it's projc

[dreamy lintel]

yep

[near pecan]

so c - projc goes from the tip of projc to the tip of c

it's the right angle of that triangle

does that make sense?

(we're just doing the subtraction)

[dreamy lintel]

a little confused there

just not vizualising the words

[near pecan]

So to correct this drawing.

1. Make it a bit bigger to give yourself more space.
2. That black arrow is (almost perfect), just make it have a right angle

(not like the one on the left)

Show me that much, even a rough sketch and you'll be well on your way

This is the (almost) correct drawing:

[near pecan]

Almost!

It looks like that arrow is going from C. We want it to go to C

Unless I'm seeing it incorrectly

But still, that's great

[dreamy lintel]

I thought it comes off the k line in the direction 1,3?

so I did 1,3 on 1,0 where k starts

[near pecan]

No. I'm not talking about C

[dreamy lintel]

oh

[near pecan]

That's perfect

Please add labels too!

It might help to move the label of K away since the line should be looooong

[dreamy lintel]

ok just did

[near pecan]

OK. Nice!

[dreamy lintel]

now what

[near pecan]

Hold on

Can you see this all right?

[dreamy lintel]

yes

[near pecan]

Turquoise is?

[dreamy lintel]

c?

[near pecan]

yes!

dotted blue line is?

[dreamy lintel]

K

[near pecan]

perfect

green is?

[dreamy lintel]

projkc

[near pecan]

Nice.

And what's the purple vector, from the tip of projkc to the tip of c?

[dreamy lintel]

projcc?

[near pecan]

Nope

[dreamy lintel]

-projkc?

[near pecan]

Remember:

C-D is the vector from the tip of D to the tip of C

[dreamy lintel]

oh ok so its c-d

[near pecan]

Well in our case, we don't have d

[dreamy lintel]

so what do we call it?

[near pecan]

(in the image with the purple vector)

so what do we call it?
You've already told me

[dreamy lintel]

still c-d even though we dont have a d?

[near pecan]

from the from (the tip of projkc) to (the tip of c)

what do we have in place of d

[dreamy lintel]

c-projkc

[near pecan]

Exactly!

[dreamy lintel]

im silly you even have it labeled as d as a substitute

my brain isnt braining

[near pecan]

it's ok. take a break after this and congratulate yourself, you're literally trying to learn vector addition, subtraction, scalar multiplication, dot products, parameterized lines and projections in one night.

Last part is: what's the shadow of c-projkc on the line containing (1,2)?

[dreamy lintel]

fair

no shadow?

[near pecan]

Of course!

That's why you don't need to calculate or measure

[dreamy lintel]

ahhh ok so the answer is to just say that it is a vector that casts no shadow?

[near pecan]

If you want. Your teacher might have a different metaphor or explanation. They might prefer saying that the vectors are perpendicular so the dot product is 0. Or that cos(pi/2) = 0. But the gist is the same

Once you establish that, your last question is asking you to solve an equation set to zero.

Let the first parts sink in, take a break, and you should be able to get the last one.

[dreamy lintel]

ok

I understand

ive got it

[near pecan]

Nice. Go chill for a bit, then review and try the last problem.

[dreamy lintel]

the funny thing is

everything we are doing rn

is the first problem

of 2

[near pecan]

Well, we've reviews a lot of concepts:

• What is a vector in space?
• Subtraction: C-D is the vector from the tip of D to the tip of C
• Vector projection: The shadow of one vector onto the line containing another (line connecting should be perpendicular)
• Dot product: the length of the vector projection of C onto D times the length of D. Also a number that relates how much two vectors point in the same direction (positive means similar, negative means opposite, zero means perpendicular)
• Orthogonal projection (orthogonal also means perpendicular)

[dreamy lintel]

yeah I do feel like im studying at the same time

which is good for the quiz tomorrow\

Im ready for the next one when you are

[near pecan]

I gotta drop soon. But please give your mind a rest. A short break will do you good

[dreamy lintel]

oh gotcha how long you have left?

[near pecan]

Just a few minutes

[dreamy lintel]

oh ok

Well I appreciate the help, definitely one of the more dedicated sessions I've had on here

wayyy better than the net tutors as well

[near pecan]

No problem!

[dreamy lintel]

spent an hour graphing 1,2 and 1,3 with them lmao

this was helpful

preciate it

[near pecan]

Thanks

Take care! Also, if you missed a whole module, you might want to ask your teacher for more time to catch up. And they might say "no" in which case, you might want go above their head

Ciao!

[dreamy lintel]

You too thank you

[marsh citrus]

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[marsh citrus]

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[humble lichen]

Looking for some help with this question, my steps to get here were:

[naive stump]

it is a bit hard to read your handwriting from a small image but here is what i would do

rewrite it as
x dy/dx - y = 8xlnx

which turns into

dy/dx - y/x = 8lnx

turning into

d/dx (ylnx) = 8lnx

then integrate both sides iwth respect to x

ylnx = 8(xlnx -x), i did integration of lnx by inspection because it is a common integral to do but u can do it by parts just to make sure

so therefore ylnx = 8xlnx -8x + c

[amber birch]

your integrating factor should be $e^{\int -1/x \ dx}$

[elfin berry]

higher's secret brother

[naive stump]

oh i also made that mistake

[amber birch]

oh ok you got that

[naive stump]

yikes

(disregard what i wrote above i also forgor minus sign)

[humble lichen]

yea I could try to take a better image but after integration my rho(x) is e^-lnx which I said simplified to 1/x

I did xy` = y + 8xlnx
y' = y/x + 8xlnx
y' - y/x = 8xlnx
rho(x) = e^integral(-1/x) = e^-lnx = 1/x
y/x = 8*intregral(lnx) + c
y/x = 8xlnx - x + c
y = 8x^2 lnx - x^2 + cx
Initial condition is y(1) = 8 so
8 = 8ln(1) - 1 + c
8 = -1 +c
c = 9
y = 8x^2 lnx - x^2 + 9x

[marsh citrus]

@humble lichen Has your question been resolved?

[humble lichen]

Hmm, I realized that I should have distributed the 8, corrected it, and it's still wrong 🫠

.close

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[stiff yarrow]

4x + 10 = -26

[stiff yarrow]

i dont how to algebra at all i started a month late

[grand radish]

subtract 10 in both LHS and RHS

[stiff yarrow]

what does that mean

[grand radish]

isolating variables and numbers

so you subtract 10 from both sides

4x +10 -10 = -26 -10

[stiff yarrow]

oh

[grand radish]

4x = -36

[stiff yarrow]

-4?

[grand radish]

no

you have to divide both sides by 4 now

[stiff yarrow]

oh

x = 9

[grand radish]

-9

because its -36/4

[stiff yarrow]

i keep forgetting negatives 😭

[grand radish]

you can do like -1* (36/4)

which is -9

[stiff yarrow]

oh

ty

[grand radish]

np

[stiff yarrow]

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[frail violet]

im so confused how am i supposed to get exact points or zeros from this

[sand fable]

you kind of can't, so what i would do is assume the roots are integers and use visual inspection to find them in accordance with the fact that you know the locations of two points with x-coordinates 0 and 2

[marsh citrus]

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[hazy dragon]

difference between

1. $Mat_{m \cross n} (\mathbb{F})$
   and
2. $\mathbb{F} ^ {m \cross n}$
   is nonexistent?

[devout mauve]

the first one is very false

[elfin berry]

Ayanokoji (ALWAYS PING ME)

[devout mauve]

ok now its the same

[hazy dragon]

so just notation?

[devout mauve]

yes

[hazy dragon]

ty

.close

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[still temple]

<@&286206848099549185>

[slate yarrow]

!15mins

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[cunning oak]

is (x-a) multiplied with (x-b)(x-c)?

[still temple]

yes

i even have

an identity

but i dont know how to

use it to this sum

[slate yarrow]

hint a^3+b^3+c^3-3abc=0,iff a+b+c=0

[still temple]

but a +b + c is not equal to zero

a (x-a)

a = (x-a)

[slate yarrow]

||a+b+c=3x ||
||(a-x)+(b-x)+(c-x)=0||

[still temple]

not it is not zero

how it is zero

b = (x-b)

and c =(x-c)

[slate yarrow]

do you know how subraction works?

[still temple]

yeah

everyone knows it

[slate yarrow]

yeah so i can write 3x=x+x+x right

[still temple]

ok

[slate yarrow]

a+b+c-x-x-x=0

(a-x)+(b-x)+(c-x)=0

[still temple]

ok

[slate yarrow]

now use this

[still temple]

the answer is zero

[slate yarrow]

yes

[marsh citrus]

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[marsh citrus]

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[near elbow]

Hi guys, how to solve 4cos^2 x = 1? I’m new to this topic

[vernal forge]

isolate cos^2(x) first

[near elbow]

What is isolate? Is it making cos^2(x) = 1/4?

[vernal forge]

yep

[near elbow]

Oohh okok

What should i do next?

[vernal forge]

now notice that cos(x) is either +1/2 or -1/2

[near elbow]

Ohhh we can move the squares?

[near elbow]

Yess

[spark otter]

after that all you need to do is solve cos(x) = 1/2 and cos(x) = -1/2

(important values of trig circle can help)

[near elbow]

Ohhhh

[near elbow]

Okayyy thank you so much

[near elbow]

Thank youuu so much

[marsh citrus]

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[snow plank]

In part(c) it says to find tanACB which should be -8/15, but the answer is 15/8?

[snow plank]

The answer

[agile iron]

The answer in the textbook is correct, Check the steps carefully again.

[marsh citrus]

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[marsh citrus]

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[floral tide]

Why am I getting two different answers?

[floral tide]

I put the answer from my physical calculator first and got it right, but when I used the calculator on my phone I got 8/3 instead of 1/24

[naive stump]

your phone is being dumb

it did multiply instead of divide

[floral tide]

I’ll stick with the physical one

Thanks

.close

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[novel juniper]

if $W_1$ and $W_2$ are vector subspaces of $V$ show that $W_1+W_2$ is a vector subspace of $V$

[elfin berry]

Veni, vidi, perii is not f(wai)

[novel juniper]

I'm having trouble proving closure under addition

like $v_1 \in W_1$ and $v_2\in W_2$ from that how do I conclude that $v_1+v_2 \in W_1 + W_2$

[elfin berry]

Veni, vidi, perii is not f(wai)

[devout mauve]

by definition of what W_1+W_2 means

but just considering v_i in W_i is not enough to show closure

[novel juniper]

yeah, $v_1 \in W_1 + W_2$ too

[elfin berry]

Veni, vidi, perii is not f(wai)

[naive stump]

try to find au + v in W_1 + W_2

ie:
Let u be element of (W_1 + W_2), then scalar multiplication au is still element of W_1 + W_2
Then there exists u1 in W_1 and u2 in W_2 s.t. u1+u2 = au
then do the same for v where v1 element of W_1 and v2 element of W_2

since subspaces are closed under addition, then we know au1 + v1 element of W_1, au2 + v2 element of W_2, then au1 + v1 + au2 + v2 = a(u1+u2) + (v1+v2) = au + v is element of W_1+W_2

im not sure if what i said is rigorous because my textbook im studiyng off didnt come with written solution for exercises

i did a similar exercise from a few weeks ago

[novel juniper]

Thanks!

[devout mauve]

well you skipped a couple steps

which is not good

[naive stump]

yeah true

for a proof question

thats bad

[devout mauve]

you dont have to bring any closure under scalar multiplication in here

[novel juniper]

I was thinking of something along the lines , by definition the sum of two subspaces is the set of the linear cobination of every vector in W_1 with every vector in W_2

So as of now in our course,we're only dealing with $R^n$

[elfin berry]

Veni, vidi, perii is not f(wai)

[novel juniper]

so $W_1 = {(x_1,x_2 \dots, x_n) x_i \in \R }$

and $W_2= (y_1,y_2 , \dots, y_n)$

[elfin berry]

Veni, vidi, perii is not f(wai)

[devout mauve]

no, W_1 is not a vector

the set on the right is R^n

[elfin berry]

Veni, vidi, perii is not f(wai)

[devout mauve]

what you want to say is that if x is in W_1, then x=(x_1,...,x_n) for some x_i in R

[novel juniper]

so $W_1+W_2 = { (x_1+y_2 , x_2+y_2 \dots , x_n+y_n \ | x_i, y_i \in \ R}$

[devout mauve]

which is really just W_1 subset R^n rephrased

[elfin berry]

Veni, vidi, perii is not f(wai)

[novel juniper]

so $v_1+v_2 \in W_1+W_2$

[elfin berry]

Veni, vidi, perii is not f(wai)

[novel juniper]

Atleast for $R^n$ vector spaces

[elfin berry]

Veni, vidi, perii is not f(wai)

[devout mauve]

you have shown that R^n + R^n = R^n

[novel juniper]

how come, depending on the subspaces, some of $y_i$ for instance, can always be zero

[elfin berry]

Veni, vidi, perii is not f(wai)

[devout mauve]

your notation doesnt reflect that at all

[naive stump]

actually reminds me i skipped 0 vector

[novel juniper]

How do I make my notation reflect that?

[devout mauve]

you basically just dont

restricting yourself to R^n doesnt help here

it makes you try things which lead you down wrong paths

[naive stump]

also denascite, if u dont mind writing a more rigorous proof if u have spare time that would be appreciated as well, im self learning off textbooks because sitll in highschool so i dont really have a grasp of whats considered sufficient

there aren't solutions to most of the exercise questions either with the textbook

[devout mauve]

open your own channel

[naive stump]

i think i wll just follow the conversation in this channel

sry for botheirng above, that was not appropirate in hindsight

[devout mauve]

I mean I can help you, I would just prefer it if you opened your own channel

you will have different questions than wai

better to separate it

[novel juniper]

If this question comes in the exam, I will HAVE to restrict myself ot R^n as we haven't done arbitrary vector spaces in class

[devout mauve]

the point is the following: thinking that all vectors have to look like x=(x1,..,xn) and trying to work with that representation will not help

at least not in the way you tried

it is very much enough to just leave the vector called x and work with only that

x can still be in R^n

[novel juniper]

okay, sure

$x \in W_1, y \in W_2$, by definition $X+W$ contains $x+y$ thus $x+y \in W_1+W_2$

[devout mauve]

ok, but to show that W_1+W_2 is closed under addition you need to take x,y in W_1+W_2 and show that x+y is in W_1+W_2

[novel juniper]

I just have

[elfin berry]

Veni, vidi, perii is not f(wai)

[devout mauve]

no you didnt

[novel juniper]

But this shows $x+y \in W_1+W_2$, right

[elfin berry]

Veni, vidi, perii is not f(wai)

[devout mauve]

but x isnt a general element of W_1+W_2

[novel juniper]

x+0 does , so it does

[devout mauve]

x is a very special element of W_1+W_2 is what I meant

it is an element of W_1+W_2, sure

but not every element of W_1+W_2 looks like that

[novel juniper]

So what am I missing

[devout mauve]

take an arbitrary element from W_1+W_2

not one that is already in W_1

similarly for y

and then show x+y is in W_1+W_2

[novel juniper]

Okay, thanks

.close

[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[knotty trellis]

what does the "school use only" mean?

Is this an exam?

[gilded ember]

mock

i got the answers but i dont understand what theyre doing

[knotty trellis]

u should either stick to your previous channel or close it btw

[gilded ember]

im doing 2 at once

mb

[marsh citrus]

@gilded ember Has your question been resolved?

[marsh citrus]

@gilded ember Has your question been resolved?

[marsh citrus]

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[near elbow]

Guys im stuck

[red nimbus]

very neat work btw

you came so far

dont let that quadratic make you give up

[near elbow]

Alrighty

Imma try quadratic formula 😭😭😭

[marsh citrus]

@near elbow Has your question been resolved?

[near elbow]

Guys i think im doing it wrong

[glass silo]

[near elbow]

Oh 😁😁

Thank youuuu

[glass silo]

(seems you got the numerical value right actually )

Oh, actually here too seems like for the value of x, you used the wrong one, but the numerical line above it is fine

[near elbow]

Yesss i got the correct answer alreadyy

[near elbow]

Thank you soso much!!

[near elbow]

Thank youuu

[glass silo]

Well done

[tawdry rampart]

hey dino

[near elbow]

Hii

I have another question

Is sin^2(x)/cos^2(x) = tan^2(x)?

Since tan x = sin x/ cos x 😀

Okayyy thankyouuuu

Gonna close this channel

Thanku sm everyone!!

.close

[marsh citrus]

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[red nimbus]

The dino didn't go extinct after

[marsh citrus]

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[chrome coral]

I'm so confused

I tried expanding idk if I'm doing it right tho

[cold pecan]

you can take 3 common in numerator

3(9u + 8)

[chrome coral]

Ohh alright tysm!

[cold pecan]

[marsh citrus]

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[marsh citrus]

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[flat bear]

How does one solve this? No matter how if i do the determinant the ordinary way or do gauss elimination before getting the determinant do i get an expression i can solve? Any tips/solutions? 🙂

[copper raven]

just compute these two dets, whatever way you want to

then you'll get two polynomials on both sides

[flat bear]

like this i get using the ordinary method

like how does one solve that? you cant move stuff between the left and righthand as that might break the inequality right?

[copper raven]

you completely can

as long you just do additions or substractions it doesn't break the inequality in any way

[flat bear]

really?

I can get to this top step then but how does one get to the factorization below?

How do i know what the factors will look like?

or is there a better way to solve it

than factoring it

[copper raven]

yea I tried finding a proof of that online, but I found nothing interesting :/

[copper raven]

there's the rational root theorem that can help you factor

[flat bear]

Thx for looking 🙂

[flat bear]

ok will look that up

[marsh citrus]

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[marsh citrus]

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[unreal fern]

Anyone knows how to do this? Can you show me the ways pls?

[unreal fern]

<@&286206848099549185>

[whole panther]

For the first one A will be 17

[crisp folio]

the second is D

if you want i can explain

@unreal fern

3 C

[marsh citrus]

@unreal fern Has your question been resolved?

[unreal fern]

<@&286206848099549185>

[crisp folio]

what?

[marsh citrus]

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[marsh citrus]

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[queen swan]

Some help on this would be appreciated

[marsh citrus]

@queen swan Has your question been resolved?

[lofty gyro]

did you try to draw it out?

[queen swan]

Yes I just don’t understand bearings in general

[lofty gyro]

oh i see

for bearings, it's always start from 0° as the North, and the angle will become larger and larger going clockwise, until i reaches back the North

[lofty gyro]

can you show it here?

[marsh citrus]

@queen swan Has your question been resolved?

[marsh citrus]

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[humble river]

periodic table is needeD?

[unique basin]

Yes

Find atomic number of Ti

[humble river]

so just to be sure

the proton number is 73

right

[unique basin]

They said

[humble river]

wait atomic number

[unique basin]

Titanium

[humble river]

Oh

oops

Ok so

mass - protons = neutron

.close thhnxs got it 26 right

[marsh citrus]

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[marsh citrus]

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[fossil badger]

Can someone check what I am doing wrong?

[fossil badger]

For part a

Nevermind my bad

Thanks

.close

[marsh citrus]

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[still temple]

question 1:
How to find the coordinates of the circumcentre? Coordinates of vertices are given.
Question 2: How to find the circumradius? Length of sides are given.
Question 3: How to find the coordinates of the orthocenter? Again, the coordinates of the vertices are given.

[grand radish]

you know that the altitudes of triangle are perpendicular to the base

[still temple]

yah

[grand radish]

and the condition for perpendicularity is m1*m2 = -1

both are slopes

say AB is perp. to CD

find the slopes

we have m1 (slope of AB) and you have to find m2(slope of CD)

once you get slope find the equation of line (CD)

Repeat this process for AC perp. to BF

you will get two equations of line

Solve those two equations of line in terms of x and y to get the coordinate of orthocentre

Circumcenter is easy to find. Circumcenter is equidistant to all the three vertices of a triangle

you can use OA = OB = OC

distance formula

you can find just two for the coordinates

[marsh citrus]

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[marsh citrus]

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[velvet yarrow]

a

[spark otter]

treat the 7 pieces of card as separate things first, how many ways can you pick 4 cards out of those?

[velvet yarrow]

7C4

[spark otter]

yes

[velvet yarrow]

btw order doesnt matter in part a

[spark otter]

yup

[velvet yarrow]

and there is one way to get different letters

M I N U

[spark otter]

there is only one "letter" choice

but how many "card" choices does that make?

[velvet yarrow]

card choice

?

[spark otter]

like

we already said each piece of card is a separate unique piece

(at least we treat it that way)

[velvet yarrow]

wdym each pice of card?

[velvet yarrow]

ok

[spark otter]

we can call them cards instead of pieces of card

you have 7 different cards

so if you tell me you got the letters M I N U

does that tell me which cards you picked exactly?

[velvet yarrow]

oh right

so each M isn the same

i thought what whole point of the q was that we remove any choicse like MMII

also i think 7C4 is wrong

[spark otter]

why?

[velvet yarrow]

cause that doesnt account for repetitions?

[spark otter]

which repetition?

we already said each card is different

[velvet yarrow]

so 2 Ms are not the same?

[spark otter]

picking cards 1 2 3 4

is not the same as picking cards 2 3 4 5

even though you get the letters MINI for both

we treat each card as a unique card

[velvet yarrow]

so thats 3 choices for an M, 2 choice for I and 1 for N and 1 for U

3 * 2 * 1 * 1

= 6

[spark otter]

yes

so probability?

[velvet yarrow]

so we get 6/35

answer meant to be 1/11 tho

its written at the bottom

the first number on the bottom right

[spark otter]

wait

[velvet yarrow]

yeah so like 7C4 double counts, cause choosing an M, not matter what M should just be 1

i think?

[spark otter]

but that doesn't really make sense

[still temple]

thank you very much @grand radish

[spark otter]

because each letter configuration has a different probability

[velvet yarrow]

uh

cases?

do we have to do that?

[spark otter]

no but like

[velvet yarrow]

its shoudnt be that bad

[spark otter]

I understand how the exercise correction reasoned

like

[velvet yarrow]

wdym exercise correction

[spark otter]

I understand how one can think 1/11 is the answer

[velvet yarrow]

how

[spark otter]

out of the 11 possible "4-letter configurations"

only one has each letter different

M I N U

so you would think the answer is 1/11

[velvet yarrow]

i dont reall yunderstand wahts wrong with 7C4

how did they get 11 total

[spark otter]

me neither

I'm trying to explain how one MIGHT come to the conclusion

[velvet yarrow]

so 7 C 4 is just saying choose any 4 letters from the 7 right

[spark otter]

I'm not saying said conclusion is correct

[spark otter]

choose any 4 cards from the 7

[velvet yarrow]

so say we have M1 M2 M3 I1 I2 N U

just fo rthe sake fo the argument

[spark otter]

yep

[velvet yarrow]

right so 7 C 4 accounts for M1 M2 I1 I 2 and M2 M3 I1 I2

rghjt?

but we conisder the Ms to be the same

so thats a double count?

7 C4 thinks they are all different

[spark otter]

not in our possibility space

[velvet yarrow]

but in really they are not

[spark otter]

like the EFFECT will be the same

[velvet yarrow]

wdym

[spark otter]

but if you number the cards 1 to 7

[spark otter]

think about this:

two dice are rolled

fair dice

we wanna know the probability the sum is 7

[velvet yarrow]

right

[spark otter]

even though

there are 11 possible sums

2 to 12

it doesn't mean

the probability you roll sum 7

is 1/11

[velvet yarrow]

so how does this tie into the previous example?

the question

i mean

[spark otter]

well

[velvet yarrow]

they seem like different scenarios to me

not really analogous

[spark otter]

even though M I N U is one of 11 possible letter configurations

[velvet yarrow]

uhuh

[spark otter]

there are actually 6 instances where you get M I N U

M1 I1 N U

M2 I1 N U

M3 I1 N U

etc...

[velvet yarrow]

yeah but all Ms are the same

and all Is are the same

[spark otter]

nooooo

[velvet yarrow]

so all those 6 instances are teh same

[spark otter]

they bring the same letter

but remember

each card is DIFFERENT

[velvet yarrow]

u just need M I N U

[spark otter]

it's like you number them

[velvet yarrow]

u dont tho

cards dosent matter

just think u are picking 4 letters from 7

[spark otter]

think smaller example

[velvet yarrow]

all that matters are the letters

[spark otter]

think smaller

3 cards

M M I

[velvet yarrow]

seperate doesnt mean didfferent

[spark otter]

what's the probability of picking an M?

[velvet yarrow]

2/3

[spark otter]

"No it's not 2/3"

"the Ms are the same"

"it's 1/2"

[velvet yarrow]

is its 2/3

what

how did u get 1/2

[spark otter]

"well I can either pick M

or I"

"so I pick an M 1/2 of the time"

[velvet yarrow]

yeah u have 3 things altogether

[spark otter]

"but the Ms are the same"

"so whether you pick M1 or M2 it actually doesn't matter"

[velvet yarrow]

ok ok no

how many ways can u arragen them

if all Ms are the same

thats all its asking

[spark otter]

no that's not what it's asking

it's asking the PROBABILITY

[velvet yarrow]

just cause they are the same doesnt mean they are one thing

[velvet yarrow]

u just put that number over the total

[spark otter]

exactly

[spark otter]

so

[autumn agate]

What class is this?

[velvet yarrow]

M I N U is one thing

[spark otter]

no

M1 I1 N U is one thing

[velvet yarrow]

i swear it isnt like that

how is that even possible

[spark otter]

it is EXACTLY like that

[velvet yarrow]

the cards arent numbers

[autumn agate]

Is this statistics?

[spark otter]

do you wanna go back to the previous example?

[velvet yarrow]

perms and coms

[spark otter]

please explain to me

when there are 3 cards, M M I

[velvet yarrow]

and we pick 2 distinct cards

[spark otter]

why the probability of picking an M is 2/3 and not 1/2

[spark otter]

why though?

aren't they the same card?

[velvet yarrow]

thats a smaller version of the quetsion

u get I M

theres only 1 way

assuming order doesnt matetr

If u have a clone

[spark otter]

so the probability of picking M would be 1/2?

[velvet yarrow]

are u two teh same ppl or different

[velvet yarrow]

no in ur case its 2/3

how is it 1/2

[spark otter]

why though?

[spark otter]

And I'm asking how is THE PROBABILITY 1/11 in your original problem

[velvet yarrow]

probability its how many ways can u get what yu want/ total choices

[velvet yarrow]

im assuming there are 11 different ways to pick 4 cards out of 7

in total

M I N U is one

I I M N or U

[spark otter]

BUT

[velvet yarrow]

I I M M

M M I N or U

[spark otter]

is every letter combination with the same probability?

[velvet yarrow]

M M M I or U or N

[spark otter]

let me give you an example here

[velvet yarrow]

no these are sepcific cases

of ways tottal ways to arrange

[spark otter]

yes

so you agree

that this doesn't mean each way to arrange has the same probability

[velvet yarrow]

no bro ur misundertanding the probability part

its just ways to get different/total ways

ur thinking about th eprobabiilty too much

[spark otter]

no.

[velvet yarrow]

yes

how many ways can u make M I N U

1

[spark otter]

(number of ways that work)/(total number of ways) only works when each way has the same probability of happening

[velvet yarrow]

thats why in the cases

we are setting some letters in stone

[spark otter]

please listen

[velvet yarrow]

o go on

[spark otter]

for a quick moment

what 1/11 IS, is the number of letter choices that give different letters (only MINU), divided by the total amount of letter choices

that I agree

what I DON'T AGREE on

is that 1/11 is the PROBABILITY

of getting different letters

[velvet yarrow]

just in words, how do u suggest we find the probability

whats like ur oevrall plan

[spark otter]

if you want to get the probability

you need to find a way to describe the problem

so that each outcome has the same chance of happening

this way, the formula

"(number of ways that work)/(total number of ways)"

WILL give you the probability

and guess what

[spark otter]

we get this

[velvet yarrow]

thats why for the cases we set some letter in stone

and only look at the probabilty of 1 letter

which all haev equal chance

[spark otter]

ok but do you agree that in real life

the probability will not be "setting letters in stone"

[velvet yarrow]

no thats how we think about it to solve the problem

most of maths hs thinking "not nreal life"

[spark otter]

what I'm telling you is that your exercise is wrong

at least I'm 99% sure about it

now take what you will

[velvet yarrow]

im telling u taht there there is only on way to get M I N U

should we get third part

i just wanetd to understand why 7C4 was wrong

[spark otter]

I'm telling you 7C4 isn't wrong...

I'm telling you the exercise is wrong

1/11 is wrong

[velvet yarrow]

i can giev u the sol

if u want

[spark otter]

I'm telling you the solution you have is most likely wrong

if they got to 1/11

[spark otter]

yeah it's wrong

[velvet yarrow]

do u agree that thats all the possible cases we can get

we can get all differnet

2 Is

2 Ms

3 Ms

[spark otter]

okokok listen

[velvet yarrow]

or 2Ms and 2Is

[spark otter]

remember the 2 dice rolls problem

wanting to find probability of the sum being 7

even though there are 11 cases for the sum

[velvet yarrow]

yeah tehre are differnet distinct sums

[spark otter]

doesn't mean the probability is 1/11

[velvet yarrow]

for the two die problem u dont look at sum u want/a ll possibble sums

u dont look at sums

[spark otter]

look. I don't know why you're fixated on saying your solution is correct. I'm telling you that what they computed is NOT the probability

[velvet yarrow]

u look at outcomes

[velvet yarrow]

no its is correct

but i want to know why 7C4 is wrong

[spark otter]

it's not. It's not a probability

[velvet yarrow]

how is not a probability

[spark otter]

ok

[velvet yarrow]

for the die problem u do it over 36

not 2

[spark otter]

let's go past this and just go general

[velvet yarrow]

12

ok

[spark otter]

probability means

if I repeat the experiment a near infinite amount of times

what's the proportion, on average, of the times it did work the way I wanted

right

[velvet yarrow]

yh

[spark otter]

so If I'm asking what's the probability the sum is 7 on two dice rolls

I'm asking

If I repeat my two dice rolls infinitely

how many times on average will the sum be 7

[velvet yarrow]

yes if u roll ur diec

ho wmany possible sums can uget

or how many possible sums

[spark otter]

no. that's not the same thing

[velvet yarrow]

go on

[spark otter]

ok, imagine something else

There are two events that could happen:

• either nothing happens

• or a plane crashes into my house

there are two cases here

[velvet yarrow]

yh

[spark otter]

Does that mean that if I repeat the experiment an infinite amount of times

that half of those times a plane is gonna crash into my house?

[velvet yarrow]

depends on the probability of plane crashign

[spark otter]

exactly

[velvet yarrow]

like for a die

its 1/6

for a number

[spark otter]

okok so follow me for a sec

just because there is 1 case where a plane crashes into my house

and there are 2 cases in total

doesn't mean the probability a plane crashes into my house is 1/2

[velvet yarrow]

yes

[spark otter]

ok

so, do you understand

that just because there are k cases that work in my favour

and there are n cases in total

doesn't mean the probability that it works in my favour is k/n

[velvet yarrow]

no thats excperimental probs

theoeritcal prob is different

yes

[spark otter]

it depends on the probability of each case right?

[velvet yarrow]

yeah

[spark otter]

ok

so

just because there is one case where all letters are different (M I N U)

and there are 11 cases in total

doesn't mean the probability that all letters are different is 1/11

it depends on the probability of each case

[velvet yarrow]

is that no the theoretical probaility

[spark otter]

so just one last thing to recap

[velvet yarrow]

we didnt do the experiemtn

[spark otter]

it's not even the theoretical probability

either it's the probability or it isn't

what your solution computed

I will agree that 1/11 is (cases that work)/(total number of cases)

but this is NOT a probability

[velvet yarrow]

is that not he definiton of aprobaility

on average

[spark otter]

no

[velvet yarrow]

if u keep repeating

u get 1/11

[spark otter]

I'm telling you

if you do the experiment as stated

meaning pick 4 letters at random out of your 7 letter bag

[velvet yarrow]

yes

[spark otter]

and you repeat the process an infinite number of times

I guarantee you

the number of times you get M I NU

is not gonna be 1/11

[velvet yarrow]

this the def right

[spark otter]

this is the "definition" that applies everytime

[fleet fossil]

I agree with rafilou btw.

You have 3*2*1*1 combinations of cards to draw that give MINU, in 4! Different orders. That divided by the total number of arrangements you could draw (7*6*5*4) is 6/35

[velvet yarrow]

so why is the solutio wrong

[spark otter]

are you asking why the reasoning is wrong or are you asking "how could my teacher ever ever be wrong"?

[velvet yarrow]

the reasoning

cause like if this is true