help-19

[true tartan]

il have a look and see what i can find

[odd edge]

@mystic saffron Has your question been resolved?

[uneven osprey]

I think there is a centre of symmetry in this

[odd edge]

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[subtle sleet]

I don‘t think we can get the correct answer

[subtle sleet]

cuz tooooo hard!!!!!!!!!

[woven kelp]

?

[uneven osprey]

bro's question is that hard we can't even see it

[woven kelp]

Our eyes can't process it

[subtle sleet]

hahah

[inland rover]

it ia imaginary obviously

[odd edge]

@subtle sleet Has your question been resolved?

[odd edge]

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[mystic saffron]

Did you find something?

[true tartan]

Not yet

[mystic saffron]

There isn’t, my teacher said so

Something to do with double tetrahedral structure. I have no idea what he meant. I mean he never considered something of that sort for any of the other problems and all the other problems also had sp3 hybridized C

How do I even picture what the double tetrahedral structure would look like

[uneven osprey]

btw, does bond length affect centre of symmetry?

[mystic saffron]

Yes

I think

Wdym

[uneven osprey]

I mean the bond length of C-H and C-Br will be different

[mystic saffron]

Like the way he defined “centre of symmetry” was similar groups attached on opposite sides at equal distances

[mystic saffron]

Why?

[uneven osprey]

There's no comparison between H and Br😶

the one thing I am sure of is that cis isomer of that compound will definitely have a plane of symmetry

But that centre of symmetry is doubtful

[mystic saffron]

Why is the mid of C-C bond not the centre of symmetry is what I’m confused about

[sonic widget]

Doesn't the molecule have a degree of freedom at the bond between the second and third C?

[mystic saffron]

Wdym

[sonic widget]

It can rotate around that bond.

[uneven osprey]

yeah but what's that to do with centre of symm?

[uneven osprey]

oh

I think I am getting what you mean

[mystic saffron]

What do they mean

[sonic widget]

If you divide the molecule in two identical parts at the middle bond, these parts can freely rotate around that axis. The picture you have is just one of many possible arrangements in space.

[mystic saffron]

Okay

So how does that help us with this problem?

[sonic widget]

If it is rotated around the bond, there is no symmetry.

[mystic saffron]

Why not?

[sonic widget]

Because the Bromine atoms do not have to be symmetrical to the centre of the molecule.

If you want to have a better visualisation, go to google and enter: "2_3-Dibromobutane pubchem". The first hit should take you to the National Library of Medicine. There is a nice visualisation of possible 3D confomeres under point 1.2.

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[remote token]

'

[remote token]

I know that I am supposed to find the asymptotes

Which would be y=0, x=2, x=-2

then what?

I need the set of all x values where y<0 right?

[elfin zodiac]

Do a table with the sign in the key points

You have 3 key points

You already found 2

[woven kelp]

^

[remote token]

huh

what are the key points

[elfin zodiac]

Factor the denominator

What do you have

[remote token]

(x+2)(x-2)

[elfin zodiac]

Ok so we have

(x+5)/(x+2)(x-2)

[remote token]

yup

[elfin zodiac]

We exclude obviously 2 and -2

[remote token]

yup

[elfin zodiac]

Now you have to check numbers that might change the sign of this fraction

We start with for example 1

When x=1 what sign u have

[remote token]

2

[elfin zodiac]

We are checking signs

[remote token]

so just if the outcome is positive or negative

[elfin zodiac]

Yes, you have 3 factors here

[remote token]

positive then

[elfin zodiac]

x+5 in the numerator

U sure?

x+2 and x-2 in the denominator

When x=1 you have

[remote token]

oh wow

negative then

[elfin zodiac]

Possitive in the numerator

[remote token]

positive and negative

[elfin zodiac]

But negative times positive in the denominator

[remote token]

positive/negative

[elfin zodiac]

Exactly

So you have to do now with -1 for example

What do u have

[remote token]

(-1+5)/(-1+2)(-1-3)=-4/-4=neg/neg

[elfin zodiac]

Check better

-1+5 is not -4

[remote token]

4

oops

so pos/neg

sorry i have been doing a lot of math lately

[elfin zodiac]

You already found the inner one

Now lets check with the denominator

x+5, we already know that

[remote token]

yup

[elfin zodiac]

For x>2 it will be positive

Because we get pos/pos

So we check for x=-5

[elfin zodiac]

What do you have with x=-5

[remote token]

0/21

[elfin zodiac]

Is that negative?

[remote token]

no

[elfin zodiac]

Perfect and finally

A number < -5

For example -6

Check this one

[remote token]

it will always be neg/pos

[elfin zodiac]

So we know that the intervals are

Tell me

[remote token]

when x is less than -5 we get negative so that is in the solution

and then greater than 2

[elfin zodiac]

Greater than 2 is positive or negative?

[remote token]

if x is greater than 2 then then both numerator and denominator are positive

[elfin zodiac]

So we dont include any number >2

And between 2 and -2?

[remote token]

yes

then numerator is positive and denominator is negative

so negative

[elfin zodiac]

Do we include the 2 and -2 themselves?

[remote token]

NO

[elfin zodiac]

And finally

[remote token]

sorry

no

[elfin zodiac]

Between -2 and -5?

[remote token]

it is positive

so we dont include that

so from below -5 and between -2 and 2

[elfin zodiac]

So we found all intervals

[remote token]

?

[elfin zodiac]

Yes, do u know how to write those intervals?

[remote token]

I think i can do that

[elfin zodiac]

Show me

[remote token]

wait

i know we use U for union between 2 pairs of curly braces

{}U{}

then the intervals go in there

[elfin zodiac]

Dont worry too much about that

You can write like this

[remote token]

{x<-5}U{-2<x<2}

[elfin zodiac]

That is not an interval

The interval is written like this

(-2,2) for one of them

[remote token]

yup

[elfin zodiac]

And (-inf, -5)

And now the solutions that is what is being asked

[remote token]

normal braces show that they arent included right

[elfin zodiac]

It is written like this

-2<x<2

And

x<-5

[remote token]

that was harder than i thought

let me try the other one

so we have a horizontal asymptote at y=1

and a vertical asymptote at x=5
and we need to find the intervals where the result is negative

so far so good?

[elfin zodiac]

y can never be equal to 1

[remote token]

and x cannot be 5

[elfin zodiac]

Yes

[remote token]

great

so lets see

we have the intervals where x is over 5 and x is under 5

seperated by the vertical asymptote

if x<5

maybe 4

then we get a negative answer

so that is in the answer

if x<-1

then then negative/negative=positive

for us to get a negative result, we must make either the numerator or denominator negative, but not both

to make the numerator negative, then -1>x

that also gives us a negative denominator, however, so that wont suffice

the denominator will be negative and numerator positive in the interval -1<x<5

so the answer is -1<x<5

correct?

[elfin zodiac]

Yes, but u forgot to check

x=-1 and x>5

In general terms when you have this

((x+a)(x+b))/((x+c)(x+d))

You check both sides of each one

For the numerator

Bigger and smaller than a, and b

[remote token]

ohh so we include x=-1

[elfin zodiac]

For the denominator than c and d

No

We dont incluse -1

But we check to prove

[remote token]

but at -1 we have 0/negative

[elfin zodiac]

Yes

It is obvious but I would check anyways to write in the paper

And teacher sees you realized it is not negative

One thing is you know is 0/neg and the other is that you show it is 0/neg

The teacher can’t read your mind

[remote token]

but how is it not included then

[elfin zodiac]

Because 0 is not negative

[remote token]

I would be more rigorous if this was for school yes

ohhh wow i am so stupid

if x>5 then we get a positive result as pos/pos=pos

correct?

[elfin zodiac]

Yes

So the solution u already said before

[remote token]

but if this was for school i would show all intervals

and the asymptotes

?

[elfin zodiac]

Verticals or horizontals?

[remote token]

vertical

[elfin zodiac]

Tell me which one is vertical

[remote token]

x=5

we set the denominator equal to zero and solve for x

x-5=0, x=5 so theres a vertical asymptote at x=5

[elfin zodiac]

When x->5 from the right

y goes to what

From the right means (a number >5 but very close to 5, something lile 5.00000001)

[remote token]

pos inf

[elfin zodiac]

And from the left?

[remote token]

yes ive learned about limits

neg inf

[elfin zodiac]

So we have 2 there

And now the horizontal

[remote token]

wdym we have 2

does it count as 2 if they the right and left limits dont equal each other?

[remote token]

y=1 because the leading terms have the same exponentiation and the same coefficient

[elfin zodiac]

When does y->1?

[elfin zodiac]

No, bad wording, dont think about this

[remote token]

at pos inf?

it sounds right but i cant tell you why exactly

[elfin zodiac]

And at negative inf what happens

[remote token]

wdym

[remote token]

no clue

[elfin zodiac]

Calculate the limit

When x->inf (x+1)/(x-5)

And same for x->-inf

[remote token]

it is 1

and x->-inf (x+1)/(x-5)=1

[elfin zodiac]

So we found that y->1 when x->(+/-) inf

[remote token]

yes

[elfin zodiac]

There is the horizontal one

[remote token]

great

thanks man

you're a great teacher btw

[elfin zodiac]

Thanks

[remote token]

.close

[odd edge]

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[remote token]

.reopen

[odd edge]

✅

[elfin zodiac]

Nothing i wanted to say

[remote token]

oh ok

[elfin zodiac]

!done

[odd edge]

If you are done with this channel, please mark your problem as solved by typing .close

[elfin zodiac]

But u knew it

Lol

[remote token]

lol

.close

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[junior dove]

How do you integrate the first one

[slender turtle]

Switch order of integration ig

Graph it on a paper and change y to x to x to y

The domain i mean

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@junior dove Has your question been resolved?

[odd edge]

@junior dove Has your question been resolved?

[odd edge]

@junior dove Has your question been resolved?

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[tranquil hatch]

ABC = [(b – a) ∙ AB + (a – b) ∙ AC + (a + b) ∙ BC] / 2

[tranquil hatch]

My doubt is, if AC = AB, can I say that ABC = [(a + b) ∙ BC] / 2

[viscid flint]

does a mean the same thing as A?

[tranquil hatch]

no

a is radium of a circle, A is a point

[viscid flint]

yes

(b-a) = -(a-b)

so it cancels

[tranquil hatch]

The question is asking : " if AB = AC, show that h = a + b "

So I can say that the are of the triangle is h * BC / 2

And if AB = AC, ABC = a + b * BC / 2

[tranquil hatch]

well, that makes sense but why does it works? It was + ( a - b )

[viscid flint]

(b-a)AB + (a-b)AC
(b-a)x + (a-b)x
(b-a + a-b)x
0x
0

[tranquil hatch]

great, now i understood

thank you very much, have a great day

.close

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[tiny sigil]

A = 10,653 * (1 + 0.043/365)^(365*4)
A = 10,653 * 1.012

10780.836?

I keep getting different answers

<@&286206848099549185>

Can you please @ me when you see this

[open blade]

@tiny sigil

The formula your using seems really right

Are you sure you typed it in right

Because when I typed it I got something else

Maybe you may have missed some paranthesis

[tiny sigil]

Hi sorry

[open blade]

Hi

[tiny sigil]

Let me try once more

[open blade]

This was my result

It seems you may have messed up on the exponentiation part

[tiny sigil]

I got 15,729.88 haaha

[open blade]

Hmmmm

Can you send a picture of what you typed

[tiny sigil]

10,653(1+0.43/365)^(365×4)

I left out a zero oops

[open blade]

Ohh

[tiny sigil]

I got same as you now

Ty sorry

[open blade]

Its fine

[tiny sigil]

Do I just round it up or take the decimals away cause final answers shouldn't have decimals

[open blade]

I think in the instructions it says to round to two decimals places

[tiny sigil]

12652?

[open blade]

12652.20

Two decimals places means two digits after the dot

[tiny sigil]

Yay that's correct thanks

[open blade]

Your welcome

[tiny sigil]

.close

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[tiny sigil]

Oops

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[teal maple]

why does the x^(1/2) end up in the denominator of the second term when multiplying?

[teal maple]

im gonna be honest that doesnt help me much

is it just replacing the x?????

[wanton bison]

,,\frac{x^{\frac{1}{2}}}{x} = x^{\frac{1}{2}-1}

fuck latex

x^1/2 / x^2/2

use power rule

[wanton fossil]

yea

[teal maple]

OH i see

[wanton bison]

= x^(1/2 - 2/2) = x^(-1/2) = 1/sqrt(x)

[teal maple]

so its the same as 1/2-1 for the exponent

[wanton fossil]

yea

[teal maple]

which makes -1/2, putting it under the denom

[clever fjord]

𝔸dωn𝓲²s

[teal maple]

ah okay okay thanks

[wanton bison]

yea anyeay

[teal maple]

i love taking diff eq and upper level math and being too stupid to do 8th grade algebra

lol

[wanton bison]

hahaha

[teal maple]

thanks guys

.close

[odd edge]

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[teal maple]

.reopen

[odd edge]

✅

[teal maple]

hey while I have people looking, do you guys know why/how the left side of the equation just becomes d[v/x^1/2]?

[cold sage]

reversal of the product rule
try differentiating vsqrt(x) and youll see

[teal maple]

whats with the brakets and d? idk whats happening here

[cold sage]

i would write it as d/dx [ vsqrt(x) ]=4x^3/2

[wanton bison]

divide by dx

[wanton fossil]

it means if we differentiate vx^1/2 we will get the same expression as written above... it just reverse of product rule just as said by @cold sage

[teal maple]

ah okay i gotcha

thanks again guys

.close

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[sweet glacier]

i need help on #70

i got 3 by using intuition

but idk how to solve using numbers

i tried 6!/(6x2!x4!)

but i got 2.5

[odd edge]

@sweet glacier Has your question been resolved?

[latent scaffold]

I don't think this actually works though. If you place the 2 white chips you get 15 combinations, and then you would divide by 6 and this would give you back 2.5.

I think the issue is that some arrangements in a circle do not give 6 distinct "circular symmetries".

I feel like this gets into the realm of necklaces, and I doubt there's a very clear way of doing this other than working it out trying to place the 2 white chips.

[orchid torrent]

true I'm high

wait am I double high

[latent scaffold]

I mean I did that at first as well hahah it's just I got 2.5 and though it was odd like OP

[orchid torrent]

yeah fuck it

fix one position as white

label the blank positions anticlockwise as 1,2,3,4,5

put the other white in 1,2,3 -> you win

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[royal pilot]

i need some to pls check my work i think the answer is 1.2 hours

[orchid torrent]

,w \frac{1}{0.21}ln(22/17)

[clever fjord]

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Requested by civil_service_pigeon

[orchid torrent]

looks good to me

[royal pilot]

awsome thanks i alo have one more if mine checking as well]

i say its 7.3

7.3 seconds

[orchid torrent]

,w \frac{1}{0.18} ln(85/23)

[clever fjord]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

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[orchid torrent]

[royal pilot]

thank so much @orchid torrent

.close

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[grave mason]

can someone help em solve this

[grave mason]

<@&286206848099549185>

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[silent solar]

Not sure what to fill in

.close

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[silent solar]

my fault

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[weary tusk]

any recommendations on how to understand putting in the limits when we change variables in double integrals

[unkempt bluff]

are you talking aboout u-substition or changing the coordinate system youre using?

[weary tusk]

this stuff

[unkempt bluff]

oh i dont know about this stuff tbh😭

i only know about changing limits with u sub and going to another coordinate system

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[echo fable]

how to do

part iv

because P(Y=1/2) in continuous distr. is 0 no?

[weary pelican]

you have joint density, so you can use the condition pdf you found

not necessarily

[echo fable]

damn that gives me right answer thank you

[violet topaz]

Then i leave it to you ahah o7

[weary pelican]

even if f_X,Y is defined up to "almost everywhere", the probability that f_X|Y is not well defined at y = 1/2 is null

[echo fable]

thanks guys for responding quickly

should i close it ?

[weary pelican]

you can if it's solved

[echo fable]

.close

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[wide drift]

I thought i knew how to do this but I keep getting it wrong

(y + 1)^2 + (x + 2)^2 = 4

y = mx + c

mx - y + c

[wide drift]

So a = m, b = -1, and c = c

y = mx + c

7 = m(-2) + c

c = 7 + 2m

[wide drift]

(ax + by - c)/sqrt(a^2 + b^2) = 2

((m)(-2) + (-1)(-1) - (7 + 2m)/sqrt((m)^2 + (-1)^2) = 2

(-2m + 1 - 7 - 2m)/sqrt((m)^2 + (-1)^2) = 2

(-4m - 6)/sqrt((m)^2 + (-1)^2) = 2

-2m - 3 = sqrt((m)^2 + (-1)^2)

4m^2 + 12m + 9 = m^2 + 1

3m^2 + 12m + 8

I can already tell this will not be correct...

[ebon harness]

Your work is a little confusing, so instead i will just tell you what you need to do instead of verifying; tangent line you are seeking shall be perpendicular to a line that passes through center of circle and point P.

[wide drift]

Point P is not on the circle though...

.close

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[ebon harness]

I see

[ebon harness]

Almost there, go on (omitting the fact you didnt use absolute value, as you need only one tangent line for the solution)

You just need to solve the quadratic to find value of m

And thats it

That gives you one of the two tangent lines

[odd edge]

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[loud grotto]

what are some steps to solving a trig equation?

[ebon harness]

Depends on type of the question

[split gyro]

We ball

[loud grotto]

yeah i thought so

welll

with an example like

[loud grotto]

sin(x) = 1/2 would this be a good example for this question

[ebon harness]

Are you searching the x value?

@loud grotto

[loud grotto]

yeah i guess so

[ebon harness]

Well first off, sinx is a periodic function so there are infinitely many many values that satisfy this

To find the most basic one, you need to know 30 60 and 90 right triangle

And definition of sine

[loud grotto]

okay

[ebon harness]

Just draw the triangle and use definition of sine, and you will get the first solution

Which is 30 degrees

Sin function has peroidic of 2pi

[loud grotto]

okay

[ebon harness]

So if you add 2pi to 30 degrees, you will get a new solution

Add 2pi = 360 degrees to new solution and you will get another

[loud grotto]

okay yeah

is this a good example ofhow most trig equations go?

[ebon harness]

30 degrees is the solution on the first quadrant

There is one more solution on second quadrant

And that is 150 degrees

[ebon harness]

There is a lot more to cover

A lot lot more

[ebon harness]

And this is because sin function satisfies
$$ \sin x = \sin ( 180^{\circ} -x) $$

Also im not using the degree symbol, pardon me for that

[loud grotto]

nah its fine

[ebon harness]

Its necessary

[loud grotto]

oh

[ebon harness]

Typo one sec

[clever fjord]

Cyrenux

[ebon harness]

Finally

Ok so 3 and 3 degrees is different when it comes to talking about trigometric functions

When i say 3, i mean 3 radians

$\pi$ radians $= 180^{\circ}$ degrees

[clever fjord]

Cyrenux

[ebon harness]

So $\sin 4 < 0$ while $\sin 4 ^{\circ} > 0 $

[clever fjord]

Cyrenux

[loud grotto]

oh dang

i thought this one would be simple

[ebon harness]

Nah im adding extra stuff

You need to learn basic trig stuff first to understand

And need to know quadrants where sin is positive and negative

As well as cosine

Sin function is positive on cartesian coordinate system where y >0

So first and second quadrants

Its negative on third and fourth quadrants as y < 0 there

Cos function is positive on coordinate system where x > 0
So first and fourth quadrants
Its negative where x < 0 so second and third quadrants

To find signs of tanx and cotx functions on quadrants, use their definitions

Consider first quadrant for tanx: tanx = sinx/cosx
And sinx, cosx > 0 on first quadrant so
tanx > 0 for first quadrant

On second quadrant, sinx>0, cosx < 0 so
tanx < 0 and cotx < 0

There is so much to cover about them, i cant mention them all here

But most iconic one is cos²x + sin²x = 1 from pythagoreas

[loud grotto]

oh my gosh

thats a lot one sec

thank you for such detailed explanations

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[amber junco]

breh

this is stupidly easy but so hard at the same time

[fossil hawk]

Understandable

[fossil hawk]

1,2,2 is one tuple that satisfies it

[odd edge]

@fossil hawk Has your question been resolved?

[summer mountain]

0,0,0 is also one of em ig

well i got a general way

1/a+1/b+1/c=4/x {Where x is product of repeated factors}

idk now wat to do

[fossil hawk]

How did you reach to this conclusion?

[summer mountain]

abc/x = [abc(1/a+1/b+1/c)]/4

thus 1/x=[(1/a)+(1/b)+(1/c)]/4

so we can say

[fossil hawk]

Hmm nice that's some progress towards the sol ig

[summer mountain]

nvm

well a,b and c all 3 cant be distinct primes

or all 3 cant be product of different primes

as then ab+bc+ca will be odd

[fossil hawk]

Ok

[summer mountain]

all cant be same also (except 0,0,0 case)

idk if (0,0,0) counts

so amongst a,b,c , two of them must be product of same prime , with one of the factors being 2

in all cases

[fossil hawk]

Hmmm

[summer mountain]

this still doesnt guarantee this equality

we missing smthing

i got one more

all a,b,c simultaneously cant be different powers of 2, atleast one of them has a distinct odd prime factor(1 wont be counted as power of 2 a.k.a. 2^0)

bruh we still missing smthing

[odd edge]

@fossil hawk Has your question been resolved?

[summer mountain]

sry man i gotta dip

it is like 1:30 am here

if i get anything else imma dm ya

k then cya

[fossil hawk]

Ok

[fossil hawk]

Thank you for taking some time out for me gn

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[solemn plover]

Hi, I'm just stuck on how to get started on this question. I tried rearranging the equations but that didn't work :/

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[solemn plover]

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[odd edge]

@dusky goblet Has your question been resolved?

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[mystic saffron]

could someone help me with 6,7,9 and 10

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[arctic jackal]

i was wondering what a good way to remember all the properties/rules for exponential function and logarithms including natural log is

[arctic jackal]

i understand it and can solve problems if i have the rules in front of me in writing but its hard for me to just memorize it. if someone can send me some videos or something to read that would really help

[potent sphinx]

I find the best way for me is to write something over and over until it sticks

[minor crypt]

with enough practice u should remember it eventually

just keep solving more problems

[arctic jackal]

well alright

thanks

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[leaden karma]

Exponents kinda change addition to multiplication and logarithms change multiplication to addition

That’s a good way to think about it

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[hybrid summit]

2 part problem :)

[brittle notch]

Find total area

Find area of white portion

Subtract both

@hybrid summit

Ping when u reply

[hybrid summit]

for the area yes but what about the perimeter one?

[brittle notch]

They gave sides right

And u can find circumference of semi circle

[brittle notch]

They gave radius indirectly

[hybrid summit]

through what

im very tripped up about the wording

oh wait no im stupid

@brittle notch how are you splitting this up?

i can find the area of the rectangle white portion

but idk how to get the semicircles at the end

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[brittle notch]

Look

The diameter is 60

Distance between two lines

U got it from there ig

Sry for the late reply was busy

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[whole haven]

How many permutations of the letters ABCDEFGH contain the strings CAB and BED?

[torpid edge]

Fix the letters as a single block

And then permute those

[whole haven]

But B is in both

[torpid edge]

So what is the only possible way to have both those blocks

[whole haven]

ohh CABED

Makes sense

[torpid edge]

Yep

[whole haven]

thanks

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[cerulean fractal]

Help

[dark kraken]

!status

[odd edge]

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

[cerulean fractal]

6

[torpid edge]

If you’re done type .close

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[minor crypt]

How would u go on about solving the matrix equation
A^2 = I
where the identity matrix is of order 3

[desert marlin]

Can you post the original question?

[minor crypt]

Im doing lin alg and they havent introduced eigenvectors yet in the book but the solutions online somehow use eigenvectors

[weary pelican]

Yeah you have to diagonalize A

[minor crypt]

is there another method?

i was thinking

AA=I
since I is a 3x3 matrix

it holds that A is a 3x3 matrix

[weary pelican]

yes that we all agree on

[minor crypt]

we can take arbitirary letters for A

so we get 9 variables

and from AA=I we get 9 equations

[weary pelican]

oof 9 variables and 9 equations, and they're not even linear

[minor crypt]

☠️

the book is on crack

im on the first chapter

how would u do it using eigenvectors

[weary pelican]

well

[minor crypt]

ik a lil about them

[weary pelican]

A is diagonalizable

so you can write it as PDP^-1

where D is a diagonal matrix

and P invertible matrix

but A^2 = (PDP^-1)(PDP^-1)

so it's PD^2 P^-1

but that is also I

so D^2 = I

since it's veeeery easy to square a diagonal matrix

you get that D must have +/- 1 on the diagonal

so A = PDP^-1 where P is any invertible matrix and D has +/-1 on the diagonal

[heady plover]

Why

[weary pelican]

D^2 = I

compute D^2, it's the diagonal matrix of all coefficients of D squared

each coefficient squared on the diagonal is equal to 1

and well it's easy to solve d^2 = 1...

[heady plover]

Why can't it be 3*(1/3) or something like that ooh nvm i forgot It's same

[minor crypt]

ohhh that makes sense

[weary pelican]

?

[minor crypt]

i forgot how to diagonalize matrix but i know the gist of it

squaring a diagonal matrix

u square the main diagonal right

[weary pelican]

yeah

[minor crypt]

okiee

tysm

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[elder vault]

This is not the graph of W=135T

Bruh

[zinc glacier]

why not

[elder vault]

The function does not pass through the point (12,2000)

135 * 12 doesn’t not equal 2000

[zinc glacier]

it doesnt, yes

[elder vault]

So it is not W=135T

[zinc glacier]

does the line graphed go through (12,2000)

[elder vault]

The graph isn’t match

[elder vault]

It does not

[zinc glacier]

so whats the issue

[elder vault]

W=135T does not

[elder vault]

The graph drawn by this silly guy does not fit the reality

“Isabella” should have a brain check or something

[elder vault]

Damn

The one who needs a brain check is me instead

I was so aggressive

Sorry Isabella

I have no problem now

Thank you 🧄

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[terse coral]

Thats the full instructions, Im assuming, and I want to ask if they ask for a positive and a negative can I subtract both times?

Im also really confused by the work shown

It saids only go around the circle once?

So, do they only want me to subtract it my 360 once?

[slim kettle]

(17pi)/3 = (15pi)/3 + (2pi)/3

(15pi)/3 = 5pi

5pi = pi = 180 degrees

(2pi)/3 = 120 degrees

So, basically (17pi)/3 = pi + (2pi)/3

[terse coral]

Why does 5pi = pi = 180?

[slim kettle]

Okay I will explain

5pi = 4pi + pi

Okay?

[terse coral]

okay

[slim kettle]

And 4pi is just 2 times 2pi

Okay?

[terse coral]

yes

[slim kettle]

And 2pi is 360 degrees or starting point which 0 degree

Okay?

[terse coral]

yes

[slim kettle]

So, if you are at 0 degree initially, and make one full revolution(2pi), then you will be back again at 0 degree

Okay?

[terse coral]

yes

[slim kettle]

So, then going 2pi a second time will also result in coming back to starting position

Okay?

[terse coral]

yes

ohhh I got it okay

[slim kettle]

Hence we can consider 4pi = 0 degree

[terse coral]

so, we'll just ignore this then?

[slim kettle]

Yeah

[terse coral]

Alright, thank you, do you mind helping me with something else too?

[slim kettle]

Oh wait, you got the answer?

I mean what you mean by ignore

[terse coral]

Im just going with what they write on there, since dif schools and teachers take dif answers

but

I can't wrap my head around this

(Im studying these packets so they're not mine)

Im just looking at the work and stuff

[slim kettle]

Oh, so you want me to give/explain the answer of the previous question?

[slim kettle]

This one

[terse coral]

Yes

I just want explainations yeah

But for this I don't even know where to start

[slim kettle]

Yeah, I just explained you how 5pi = pi, nothing else lol

[slim kettle]

We will discuss it also after this one

[terse coral]

okk

You can type everything up, Ill read it

[slim kettle]

Yeah, so 5pi = pi = 180 degrees

And 2pi/3 = 120 degrees okay?

So, 17pi/3 = pi + 2pi/3 = 180degrees + 120 degrees = 270 degrees

So for positive coterminal angle, we just gonna add 360 to 270 which is....

[terse coral]

I can just convert it using a formula though

[slim kettle]

2pi + 5pi/3 = 11pi/3

[terse coral]

oh

wait

[slim kettle]

And for negative coterminal, you just gonna minus 360 from 270

So, 5pi/3 - 2pi = -pi/3

[terse coral]

Ohh, okay, I got it, let me write this down and recap this rq

[slim kettle]

Yeah, ask any doubts you have with this question

[terse coral]

sorry, I keep on getting confused because of the weird text thing

[slim kettle]

Yeah, writing in book and sending pics is way too better

I have to type pi lol

[terse coral]

Wouldn't that make it 300 degrees

[slim kettle]

Never mind, my bad

Silly mistake

You right

300degrees

[terse coral]

kk, Im gonna continue writing it

its okay, Im just glad someone is here to help me

[slim kettle]

hah nw

You know to convert 300 into raidans?

lol, I wrote the degree one wrong

But radian one correct

300 = 5pi/3

[terse coral]

Im a bit confused, where'd the 5pi/3 come from, since I know about the 5pi and the 2pi/3 but

[slim kettle]

Oh ok

[terse coral]

Yeah, you multiply it by radian/180 if Im not wrong

[slim kettle]

pi/180

[terse coral]

lately ppl have just been quiet and I can't find it lol

[terse coral]

oh, I thought they were the same thing, mb

[slim kettle]

Yeah, you don't need answers, just little bit of concept clarification and then you good to go

[slim kettle]

5pi/3 is just the addition of pi and 2pi/3

The angle given in our question was 17pi/3 which we converted into 5pi/3

[terse coral]

oh okk

[slim kettle]

Lemme send you a pic of why writings, it will be just lot lot more clear and easier for us

[terse coral]

yeah, that would, ty lol

[slim kettle]

Got through this in the order, and things should be clear

Feel free to ask if any doubt still

Cuz even small doubt can cause big confusions in angles things

[terse coral]

nope nope, nvm Im cleared now

[slim kettle]

cool man

[terse coral]

thank you for taking your time out to write this on calendar pages lmao, but uh, bad news you still have to decode that other thing

[slim kettle]

I am idle all day lol

[terse coral]

this mess

[slim kettle]

Hah lez get it

[terse coral]

Im pulling a whole kay chung bc Im rushing until my exam lmao

this is the last unit though

[slim kettle]

Nice

[terse coral]

alr, I'll read through it after you explain

[slim kettle]

Ok

You know how much in degrees in pi/3 ?

It is 60 degrees

[terse coral]

ohhhhhhh

I was so lost on the 45-45-90 rule and everything

[slim kettle]

Cuz pi = 180 and pi/3 = 60

[terse coral]

ohhhh

[slim kettle]

lol

[terse coral]

goddamn, you're like aligning all of my lightbulbs

okay, continue onn

[slim kettle]

So, you know value of sin(60)?

[terse coral]

Hold up, am I supposed to remember these?

[slim kettle]

Yeah

There's no way to solve it

You have to remember sin and cos values for some important angles

[terse coral]

I see

I'll note that, thank you

[slim kettle]

So sin(60) is (underroot3)/2

You just have to remember it, no other way

You will have to remember the sin, cos and tan values of these angles: 0, 30, 45, 60, 90

These 5 angles, and you can solve anything

[terse coral]

Thank you, wrote everything down

[slim kettle]

It's great that you are writing, most people would just read things like stories

[terse coral]

I gotta review, so unless I wanna fail then yeah

[slim kettle]

haha

[terse coral]

everything you say is basically photo copied and explained in my notes rn

[slim kettle]

yeah, maths is same for all

That's why I love it

Uhm, gimme a minute, cuz I have to use some brain for second part

For cos(5pi/12)

[terse coral]

no problem, you can take all the time in the world

[slim kettle]

lmao

[terse coral]

wait, wait, how'd we get from the pi/3 to sin(60)?

[slim kettle]

Oh

In question we were given to find sin(pi/3)

And pi/3 is in radians which equal to 60 degrees

Sin(pi/3) = sin(60degrees)

[terse coral]

oh, so I have to use calculator for that part

for the radian part

[slim kettle]

You mean calculator for converting radian into degrees?

[terse coral]

yeah

[slim kettle]

it's ez, 180/3 = 60

Wait are you taking pi/3 = 3.14/3 ????

[terse coral]

no no no no

just the converting radian part to get to 60 lol

[slim kettle]

Oh lol

I was panicked for a moment

[terse coral]

ohh, yeah since pie is 180

[slim kettle]

Yeah

Yeah

[terse coral]

you make me feel so smart rn, thank you okay

[slim kettle]

lol, never mention

Ok, I got the second part, but it's gonna be tough for you

[terse coral]

okay

[slim kettle]

Cuz, you don't remember values of important angles

And lemme ask you about a formula

[terse coral]

alr

[slim kettle]

If there are two angles A and B, then do you know what will be cos(A+B)?

Say yes or no lol

[terse coral]

nope

[slim kettle]

Hmm, there's an equation/formula for this

[terse coral]

oh, it wasn't mentioned in my packet so Im not sure

maybe its in the further pages

probably

[slim kettle]

So if you get this formula for in your head, then we start with solution of second part

Okay, so I am starting with explanation of solution

If we convert 5pi/12 into degrees, then it will be 75 degrees

And 75 = 30 + 45

And hence, cos(75) = cos(30+45)

[terse coral]

oh

[slim kettle]

Now you have to know that sin and cos values for 30 and 45 degrees

Then just put em up in he formula above

And you will get the answer

[terse coral]

I see okay

I have a calculator with me

Ill try to plug it in

[slim kettle]

Calculator for what?

[terse coral]

(I mean as in plug into the formula)

the values

for the 30 and 45

[slim kettle]

Nah, don't use calculator

Hell naw

Cuz it will try to give you answers in decimals form

[terse coral]

oh

no no seperate

just like cos30 or cos45

thats it

[slim kettle]

It will try to find the values of underroot3 and all

[terse coral]

oh

[slim kettle]

We need cos(30) to be underroot3/2

Not the exact value

[terse coral]

oh?

[slim kettle]

That's the whole point for remembering these values XD

[terse coral]

crap :)

how do I do this then

[slim kettle]

You gotta mug up the values

[terse coral]

do I just, remember this and move on until I get all of it memorized?

[slim kettle]

Ye

[terse coral]

how do I mug up the values without remembering or knowing they exist T-T

alr

[slim kettle]

It is what it is

[terse coral]

Ill move it on and write this formula down

[slim kettle]

Sure

There are formulas for cos(a-b), sin(a+b) and sin(a-b) also

[terse coral]

yeah spare me the details T-T

actually no

you can just tell me them

I'll write them down

they'll probably be useful

[slim kettle]

Sure

Ofc they are

[terse coral]

wait I think I found the whole sheet for it

wait nvm its not

[slim kettle]

It would be better if you write in a tabular form

[terse coral]

I don't have a clue what that is but Im guessing like in a sheet like from sin to cos to tan

[slim kettle]

Hmm, it also has sec, cosec and cot

[terse coral]

yup, that too

are you also idle for another question T-T

[slim kettle]

Sire

But see this pic first

Mug up this and you be flying through trigonometry for now

Why are some values in grey 💀

[terse coral]

thats a crazy choice

prob forgot the white background

[slim kettle]

We'll, if will be better if instead (root2)/2, you remember it as 1/root2

[slim kettle]

Yeah

[terse coral]

works too

Okay, I have the photo saved in my drive now

[slim kettle]

Hmm, okay if you clear with all till now, then lez get the next one

[terse coral]

kk

This thing

I don't even know what its about, they mentioned everything but this in the packets

It popped out of the blue

Usually it'll have partial instructions, but this doesn't

[slim kettle]

Hmm

This is basically just circumference and area of a part/slice from circle

[terse coral]

oh

[slim kettle]

For eg, in the first one,

We have to find area and length of 1/4th circle

Circle means 360degrees

They ask about pi/2 = 90degrees = 1/4 of circle

So u find the area of circle which is of radius 5m, and then divide it by 4

[terse coral]

oh, wait let me try this out

[slim kettle]

Sire

Sure

And also you do same thing for length of secotr

U find circumference of complete circle

Then divide it by 4

[terse coral]

okay for area thing I did pi(r)squared

which 25pi

then divide by 4

[slim kettle]

That's it

You got it

Now do similarly for length like I said

[terse coral]

wait but dividing it by 4

how would I show that

like write it out

[slim kettle]

25/4

And pi besides it

[terse coral]

oh ok

[terse coral]

so, then itd be 2pi(r) for circumfrence

so I'd do 2pi(5)

which is 10pi

[slim kettle]

Yes