help-42

[odd belfry]

What formula is this?

[plucky walrus]

cosine law.

[odd belfry]

I mean the name

[plucky walrus]

cosine law

[odd belfry]

In triangle ADN, x^2 / 4 + y^2 - xy cos(x)= 36

Why is it 36?

[plucky walrus]

Because AN = 6

AN^2 = 36

[odd belfry]

So In triangle ADN, x^2 / 4 + y^2 - xy cos(x)= AN^2 ??

And why do we x^2 devide by 4?

Isn't cosine rule like this:

c = root(a^2 + b^2 - 2ab * cos(angle ab))

Ohh, I now get it. Sorry for stupid question😁😁 @plucky walrus

But I still don't understand why we don't multiply xy * cos(x) by 2.

Oh, I again understood😂😂😂

[plucky walrus]

Understood?

@odd belfry

[odd belfry]

Yes, solving. Just want to finish solving and then close the chat. Sorry for keeping the chat busy.

[plucky walrus]

No problem

[odd belfry]

@plucky walrus I solve till this point by subtracting, canceling etc.

Now I am here. Can you give a hint what to do next?

First page

Second page

[odd belfry]

I am here D, I sent the first and second pages to show how I got here

[calm coral]

@odd belfry Has your question been resolved?

[odd belfry]

<@&286206848099549185>

[calm coral]

@odd belfry Has your question been resolved?

[calm coral]

@odd belfry Has your question been resolved?

[warped surge]

Multiply by 4

[calm coral]

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[limpid geyser]

i just want to double check something for part a, can i just write my solution out as f is defined as:

...
0 -> 1
1 -> 2
2 -> 3
3 -> 4
4 -> 6
5 -> 7
6 -> 8
7 -> 9
8 -> 11
...

then do i just show that this is bijective?

[bold bear]

You'll be there a very long time

[limpid geyser]

lol

[bold bear]

You have to describe it with some kind of formula, you can't list out infinitely many values by hand

[limpid geyser]

not really sure how to write that out as a formula though theres not an easy way that Z maps to S

like it gets messed up w negatve values and im kinda stuck there

[bold bear]

You can do the same thing for negatives that you did for positives

Divisibility doesn't care about positive/negative so the pattern is exactly the same

You just have to work out how to write it formally

You may find it easier to define it recursively (that may or may not be allowed but it's definitely how I'd choose to do it)

[elder pawn]

in this case it seems easier to write out the first few mappings in a way where we can easily see how it continues

as it stands u do need to rewrite the pattern to involve negative integers

[bold bear]

It does say explicitly though, I'd say writing a ... on either side wouldn't really count as explicit

[elder pawn]

meh its one of those cases where writing a formula seems really painful and takes a bit away from the spirit of the exercise

[north dome]

and you rather go to bed and stop thinking about it and hopefully it will go away

[bold bear]

It's actually really not as bad as it looks lol

I don't disagree though

[mortal orbit]

||f(n) = n + 1 + floor(n/4)|| now I think it's good

[limpid geyser]

if we let $k=4a+b, \quad a \in \mathbb{Z}, \quad b \in {0,1,2,3}$ \

then say $f(4a+b)=5a + (b+1)$ does that work?

[potent lotus]

nilo

[bold bear]

Yep I think that works

[limpid geyser]

then i just show its bijective and that completes the solution right

[bold bear]

Yeah exactly

[limpid geyser]

what does the floor() notation mean? sorry not familiar w that

[bold bear]

It's just the integer immediately less than or equal to x

so floor(1.2) = 1

[mortal orbit]

floor(x) is the integer that verifies floor(x) <= x < floor(x)+1

[limpid geyser]

ah ok

[mortal orbit]

in you're case floor(n/4) = "a"

[limpid geyser]

got it

thanks

[calm coral]

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[eternal shard]

help

[eternal shard]

i cant tell if thats right because i keep getting 0 the computational way, i wil show in a bit

[eternal shard]

I applied this where a is in C

[pseudo wedge]

I don't understand the last step

[pseudo wedge]

but you have zbar, not z

[eternal shard]

i just realized looking at the differential

https://tenor.com/view/dead-my-honest-reaction-damn-skull-skeleton-gif-15885937969029652188

[pseudo wedge]

1/zbar = z/|z|^2=z which is analytic

[eternal shard]

i thought i did some 💀

[pseudo wedge]

so the integral should be 0

[eternal shard]

aight thanks man

.solved

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[kind swallow]

I need help to comprehend this question. what is it asking you find?

[fickle flume]

They want you to find x

for each of the different diagrams a-b

[kind swallow]

What is the side of the square?

[fickle flume]

It says 1

[kind swallow]

what are some strategies i can use?

do they allhave same area?

[fickle flume]

So you are given some information about each diagram

Well the squares, individually, have an area of 1 unit squared

and then you know that the overlapping area

has the same area as the non-overlapping areas

That's what the beginning part of the question tells you

and with that alone you cannot find x

but then they give you more information

for each of the diagrams

This was the extra info for a

And the diagram is not to scale

Do you have an answer key for these problems? I'm just curious

[kind swallow]

no

[fickle flume]

Okay

Other than comprehending the question I haven't really seen a problem like this before

Since the diagram isn't drawn to scale

[fickle flume]

we want to look closely at the given information here

[kind swallow]

perhaps they're equal seections?

[fickle flume]

so other than comprehending the question i have to admit i dont exactly know how to solve this, but it looks fun

and there is a language barrier for me

im not entirely sure what "symmetric about" mean

i assume that means the diagram looks the same on both sides of the line

and we know the squares share a vertex

[kind swallow]

yes symmetric means line of symmetr

i dont think they share a vertex tho

[fickle flume]

unfortunately the ydo

that's given to us

so it is safe to say they share at least one vertex

I'm trying to imagine how

actually even though the diagram isnt drawn to scale

i suppose it would look very much like this diagram

because in the image it is symmetricaly

about that line

sorry i wrote imagine instead of image

i mean we do have a right angled triangle

and

and

we know its area

[tall mason]

it's almost to scale

[kind swallow]

thats not particularly relevant

[fickle flume]

maybe if we could describe the area using a system of equations it could be, but again i dont know how to solve this

if someone else has better ideas/know how to solve this and want to help Green

please do so

he is the original problem asker

[tall mason]

the area of all three colours is the same

[fickle flume]

^

[rocky flower]

I think for this one, based on the given diagram, each colour has the same area given by the problem statement. Because by the additional fact that it’s symmetric about the line AB, I think it’s safe to say that the line AB splits the purple area into two congruent right triangles. Then find the area of the purple region and blue region in terms of x and equate them.

[tall mason]

and you know that the squares are both area 1

[fickle flume]

that's how we can know the area of the triangle

[tall mason]

so we can try to figure out the area of the purple

The way i solved this is by finding the total coloured area in 2 seperate ways and equating them

[fickle flume]

okay so you did use a system of equations

kind of?

[tall mason]

1 equation

[fickle flume]

ok nvm

[kind swallow]

how do you do it

if its symmetrical?

[tall mason]

can you find the area of the purple as a function of x?

[rocky flower]

It’s symmetrical about the line AB. Says in the problem statement

[kind swallow]

maybe ((1-x)(1) / 2)*2

Which ends up 1-x?

[tall mason]

yup

and so the total area is?

[rocky flower]

That’s correct. Now find the blue/pink area in terms of x

[kind swallow]

do you just do 1-(1-x)*2
= 1-(2-2x) = 1+2x?

[rocky flower]

That’s not the blue area. Use the fact that the blue area + purple area = square

[tall mason]

easier is to remember the big property of the coloured areas

[kind swallow]

So they overlap?

[rocky flower]

The squares overlap to form the coloured regions of equal area

[kind swallow]

Then how do we find x just from the area of say blue? is it like trapezium?

[rocky flower]

I think you did it wrong. Blue area should just be ||1-(1-x) = x||

[rocky flower]

Use the fact that blue area + purple area = square. We already know area of square is 1, and purple area is 1-x

[kind swallow]

so you mean the blue area is just the obviously blue area, not including the overlap?

[rocky flower]

Yes

[kind swallow]

then how do we find x from that weird looking shape?

[rocky flower]

By equating the two areas. Blue area = purple area

[rocky flower]

You found the purple area here btw

[fickle flume]

This is why he is equating them

[kind swallow]

ok

now lets do b

I dont know the method here

[rocky flower]

look at the top half of the square with blue and purple. You can see that the purple area is a trapezium. We just need to find the top side of this trapezium because height and bottom side are already known

[kind swallow]

i think its more a pentagon. are you referring to just one side of the symmetry

[rocky flower]

Yes just one side of symmetry

[kind swallow]

how do we find the top side

[rocky flower]

Ok, try and angle chase. Use the fact that the line of symmetry creates an angle of 45 degrees in the bottom left corner of the trapezium

[kind swallow]

so can we just draw a virtual line down there and make an isoceles triangle so we just subtract that

[rocky flower]

Yes, we can do that too

[rocky flower]

Remember that the height is 1/2 here for the isosceles triangle and the trapezium

[kind swallow]

wdym 1/2 here

[rocky flower]

You can see that the length of the blue line is 1/2, right?

[kind swallow]

yes

So we can just use half of side of square?

[rocky flower]

Yeah and then double it back

[kind swallow]

Im still not sure for C or D. I think for c you will need diagonal but i dunno how it works

[rocky flower]

I think for c you can use the fact that the purple region is a square with its diagonal being x

[kind swallow]

how do you know?

And how do we do it knowing that?

[rocky flower]

From right angles I think

[rocky flower]

Then use Pythagoras theorem on one of the the triangles (the top or bottom one)

[kind swallow]

What do you do after that?

[rocky flower]

Also like it part b the line of symmetry cuts the purple region into two triangles with angles of 45 degrees either side

[kind swallow]

Since it is a square I drafted this plan that uses a but then im stuck

[rocky flower]

You can find what x is by equating blue area with purple area again

[rocky flower]

Use Pythagoras theorem on the top triangle of the purple square. Then you find what ‘a’ is in terms of x

The triangle is created by the line of symmetry btw

[kind swallow]

So we just find the area of the blue which is 2 trapezims gotten from
(a+1)/2 * (1-a)

will this work?

[rocky flower]

Yeah this should work, or you can even use the fact that blue area is 1-purple area. So 1-purple area = blue area = purple area and so purple area is 1/2

[rocky flower]

Actually I think this applies to all the problems lol

[kind swallow]

Now D's like really confusing

[rocky flower]

I think you should draw a straight line down from the point in the square between blue region and purple region

[rocky flower]

Or you can draw from the same point I mentioned before a segment onto the top left corner of the blue/purple square. You can three triangles, of which the bottom and left triangle have equal area. And the dimensions of the top triangle can be found from a right angled triangle

[kind swallow]

where do i draw the line?

[rocky flower]

Down the bottom left corner of the pink square

[kind swallow]

what is the best method to approach this (the easiest)

[rocky flower]

Sorry I meant bottom left corner

[rocky flower]

I think this method might be bit easier

You draw a line segment from the same point again to the top left corner of the blue square to create 3 triangles. The left and bottom triangle have same area, so we just need to find the area of one of them and the top triangle

[rocky flower]

Or maybe just this method is easier ngl

[kind swallow]

can you illustrate it. i dont understand

[rocky flower]

Im sorry. I cannot illustrate atm, but you see the point in the blue/purple square. Draw a vertical line down it

It creates a rectangle and two triangles in the blue region

[kind swallow]

Which point?

[rocky flower]

The bottom left corner of the pink/purple square

[kind swallow]

ok what do i do with that

[rocky flower]

First find the area of the rectangle, then the bottom triangle and then the top triangle. At least that’s how I did it

All in terms of x

[kind swallow]

I drew the line but i cant find the retangle

[rocky flower]

Draw the line to the top as well

Continue it

[kind swallow]

yep

So from this, how do we find the dimensions of the triangles

[rocky flower]

Did you find the dimension of the rectangle?

[rocky flower]

For the two triangles, first find the dimension of the bottom triangle and then use the fact that the top triangle is similar to the bottom triangle

[kind swallow]

no i cant find it

[rocky flower]

For the rectangle, we just need to find the top side of it. You can see that there is an isosceles right angled triangle with hypotenuse of x. From there you can find the top side

[kind swallow]

how do you know the hypotenuse is x. i dont understand

[rocky flower]

Next to the blue line, do you see that at the top there’s a right triangle with blue+purple that has a hypotenuse of x?

[kind swallow]

I'll label the parts so you can refer to them

[rocky flower]

Okay thank you so much. Look at triangle ACD first

[kind swallow]

ok

then?

[rocky flower]

You can find what AC is and then the other part of that side of the square (which is a dimension of the rectangle)

[kind swallow]

how do you find AC?

[rocky flower]

Triangle ACD is a right angled isosceles triangle with hypotenuse x, so you can find AC in terms of x

[kind swallow]

how do we know this? do we just assume CAD=90

[rocky flower]

Yes, we can assume this because we drew a line perpendicular to the base that passes through D

[kind swallow]

so we use pythagoras after that

but how do we find the area of the blue rectangles and thus find the area of the purple

[rocky flower]

Yes

[rocky flower]

You mean blue rectangle and triangles. And also we know that blue area = purple area and blue area + purple area = square = 1. So blue area is just 1/2

[kind swallow]

yes but how does the vertical line AF apply in this scenario if we already know the value of blue and purple.

in other words how do we find x

[rocky flower]

We find the individual areas of the blue region in terms of x and we add them together

The individual areas are the rectangle and two triangles created by vertical line AF

[kind swallow]

how do we do that

[rocky flower]

First find the dimension of the blue rectangle

[kind swallow]

would that be AC * 1 = AC
We know AC is in a pythagoras relationship with x right

[rocky flower]

Yes, that’s right

[kind swallow]

so we know the area of the rectnagle is x/root(2)

what do you do afterwarrds

[rocky flower]

Oh wait you mean 1-AC. Look at the rectangle left of the blue line

[rocky flower]

I didn’t mean that rectangle, I meant the other one. The left one

[rocky flower]

Find the area of the top and bottom blue triangles

[kind swallow]

I dunno what you mean. do you mean GAEF

[rocky flower]

Yes I mean that

[kind swallow]

What do I do?

[rocky flower]

Find AC, then find AG. Then you know what the area of that blue rectangle is

[kind swallow]

the whole blue rectangle?

or just GAEF

but you still dont know x

[rocky flower]

Whole blue rectangle = GAEF

[rocky flower]

You can find AC since ACD is a right angled isosceles triangle in terms of x. Then you find what GA is and this area GAEF

[kind swallow]

And then how does that help finding x

[rocky flower]

Okay, then find the area of the two blue triangles. First see that they are congruent, and then find the dimensions of the bottom triangle

[kind swallow]

how do you do that

[rocky flower]

You get an equation where you find what x is

[kind swallow]

if they're congruent how does thatt help

[rocky flower]

I’ll show you later. First we will look at the bottom triangle. What is the height of it?

What is DF?

[rocky flower]

We are looking at triangle DFG btw

[kind swallow]

?

[rocky flower]

So you first found the area of rectangle GAEF. Now look at triangle DFG

[rocky flower]

Sorry I meant triangle DFG

[kind swallow]

how do you find the area

can you show me how you would find out the area

by writing out the solution

[rocky flower]

Find DF first. Notice that DFE is an isosceles triangle

[rocky flower]

Ok, area GAEF = (1-x/sqrt(2)). Area DFG = (1-x/sqrt(2))(x/sqrt(2))/2 = area ABD. Adding all this together, we get the the total area of the blue region is 1-x^2/2. However we know that this area is also 1/2, so we get that x=1

[kind swallow]

lemme first see about these solutions.

[rocky flower]

Okay

[calm coral]

@kind swallow Has your question been resolved?

[calm coral]

@kind swallow Has your question been resolved?

[calm coral]

@kind swallow Has your question been resolved?

[kind swallow]

.close

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[brittle sail]

Limit ( z ) tends to 0 ( \left(\frac{\sin z}{z}\right) ).

[potent lotus]

Andy

[brittle sail]

Z is complex number

Sin(x+iy)/(x+iy)

[calm coral]

@brittle sail Has your question been resolved?

[tepid laurel]

do you need to do an epsilon delta proof?

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[brittle sail]

.reopen

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✅

@brittle sail Has your question been resolved?

[tacit moat]

you mean to 1?

oh nvm

are you allowed to use taylor here?

[brittle sail]

Yes

I got it 1

Thanks

.close

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[tacit moat]

nice

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[rare bluff]

0 < sin(5πx) ≤ 1 became 3/2 ≤ 1/10 + (2/5)n ≤ 2; can anyone explain how this was done?

[finite oasis]

Can you provide context

[rare bluff]

i have to find a, where a is [3/2;2]

i can send you some of the work done, but it's not on english

the writer then does this

and then whaterver happened to sinus

i know that n came from the period of sinus function but i'm confused about how everything was done

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.reopen

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✅

[rare bluff]

<@&286206848099549185>

[mental tree]

Hey

Okay so in this question

we take the square root of a number only positive and not negative

So given inside the log the term is a sin function

The range of it is between 0 and 1

But log value value less than 1 would be negative

So the sin term term has to be 1

@rare bluff

[rare bluff]

Thanks, i'm just thinking

so then we say that sin5pix = 1? and solve for that

my main question was just how 0 < sin(5πx) ≤ 1 became 3/2 ≤ 1/10 + (2/5)n ≤ 2

[mental tree]

So here, we are finding a general solution of sin 5pix=1

Here, they’ve taken General solutions of sinx as 2npi + x

Alright it has to be n*pi +(-1)^n x

The general solutions they considered would be if n is even

[rare bluff]

ok i think i got it

thanks :)

.close

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[sweet pike]

The answer is given (d)

[sweet pike]

What I got didn't match any options, but was closest to (a)

I am getting abccb

which will make the pattern acbbacacbbacacbbacac

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[plucky walrus]

Maximize and minimize:
sin(x) (2cos(x) - sin(x))

[velvet osprey]

let me guess: no calculus allowed again?

what's the range for x

[plucky walrus]

-inf to inf

[velvet osprey]

k

[plucky walrus]

I need to go now.

Thanks.

[velvet osprey]

...

[plucky walrus]

.close

[calm coral]

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[velvet osprey]

why did you open in the first place...

[finite oasis]

Well that's sin(2x) - sin^2(x) so that gotta help somehow

Oh

Welp

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[velvet osprey]

show parts a and b and the entire question

[elfin gull]

f(x)=x^3 - x^2 - 6x

[velvet osprey]

ok

but show parts a and b anyway since the answer to either one might be useful

[elfin gull]

ty @velvet osprey

.close

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[elfin gull]

thank you @velvet osprey

ah no thanks bot?

[velvet osprey]

was that sarcastic or genuine?

(i can't tell, hence why i'm asking)

[elfin gull]

I am geniune

I figured it out lmao thank you sweetie

[velvet osprey]

... not sure i like being called "sweetie" by a stranger, but you're welcome.

thank you for the clarification.

[calm coral]

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[rain rover]

@whole hinge hello i awake from logic slumber

[rain rover]

how is the suggestion true

if we do the contrapositive then \sigma must be unsatisfiable, does that get used here

[vagrant oak]

What exactly do you not get?

[rain rover]

${{\Sigma}_1};\alpha$ is never satisfiable for any finite $ {\Sigma}_1$

[potent lotus]

reg

[rain rover]

assuming you havent yet proved compactness theorem

[vagrant oak]

they are essentially going for a proof by contradiction here.

For sake of contradiction, they assume that neither Sigma; a, nor Sigma; -a have the property that every finite subset of them is satisfiable. That means that there is some unsatisfiable finite subset of Sigma; a and there is some finite unsatisfiable subset of Sigma; -a

the first subset will have to include a (otherwise it would be subset of Sigma and hence satisfiable) and the second one will have to include -a for similar reason

so they just named the subsets: Sigma1; a and Sigma2; -a

i can latex it if it's hard to read

[vagrant oak]

Oh and importantly, it's not any, it's some

[rain rover]

yes that is what they claim

[rain rover]

but couldnt it be that when we check E;alpha and E;-a for different truth values of their sentence symbols
for whatever E1 you choose
one configuration satisfies E1;a but E;a fails because the rest of E is false

like whenever E1;a is satisfied, E-{E1};a is false
for all E1

[vagrant oak]

they chose E1 such that E1; a wouldnt be satisfiable

[rain rover]

ye im asking how can you be certain such an E1 exists in E

[potent lotus]

MathIsAlwaysRight

[vagrant oak]

oh and we're negating it because we're going for a proof by contradiction

so we will assume that the result (i.e. the above statement) is false, and eventually derive a contradiction

[vagrant oak]

Once you negate this, you should be able to see why under this assumption E1 and E2 are guaranteed to exist

,texsp There exist finite $\Sigma_1 \subseteq \Sigma$; $\alpha$, $\Sigma_2 \subseteq \Sigma$; $\lnot \alpha$ such that both $\Sigma_1$ and $\Sigma_2$ are unsatisfiable.

^answer

[potent lotus]

MathIsAlwaysRight

[vagrant oak]

this should be the correct negation, now if you add a to E1 and -a to E2, they're obviously still gonna be unsatisfiable

[rain rover]

waait isnt this different from saying
either E;a or E;not a is satisfiable
if this statement is true then your statement is also true
if this statement is false then how can you be sure your statement is true (without compactness theorem)

[vagrant oak]

they never mentioned anything about E;a being satisfiable

it says every finite subset of E is satisfiable

and then you are supposed to show that the same is true of at least one of E;a and E;-a

[vagrant oak]

so you are supposed to show exactly this

[rain rover]

we have to prove
either E;a or E;~a is satisfiable
(which implies that every finite subset of E with either a or ~a satisfies)
the negation of that would be
neither E;a nor E;~a is satisfied
isnt what you wrote the same as the sentence in brackets, different from what we have to prove

wait

[rain rover]

oh yea

oof

[vagrant oak]

either E;a or E;~a is satisfiable
This is actually a stronger statement

[rain rover]

sorry i misread the question

[vagrant oak]

happens lol

[whole hinge]

xD

do you still need help

[rain rover]

yess and we take their union?

[rain rover]

noo ty i misread the question

[vagrant oak]

Yeah, the union should help with the contradiction part

Try and see if you can find a contradiction

[rain rover]

the union tautologically implies both alpha and not alpha

and we know the union can be satisfied

[vagrant oak]

Yeah, that works

[vagrant oak]

Yep, since its a finite subset of E

[rain rover]

yee

got it thanks @vagrant oak

[vagrant oak]

np

[calm coral]

@rain rover Has your question been resolved?

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[calm coral]

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[simple musk]

,w (1,2,-1)x(1,0,-3)

[potent lotus]

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[simple musk]

L1 : X = k(-6,2,-2) + (x1, x2, x3)

L : X = (k+3,1,-3k)

[glad parrot]

why are you doing these calculations

[simple musk]

every point in L1 is at a constant distance to Pi of sqrt(6), i.e. L1 is paralel to Pi

@glad parrot

Pi: x + 2y - z = 3

,, \frac{|k+3 + 2 +3k|}{\sqrt{1 + 4 + 1}} = \sqrt{6}

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

,w sqrt(6)*sqrt(6)

[potent lotus]

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[glad parrot]

and so the directional vector of L is lineraly dependant of the normal vector of the plane PI

?

[simple musk]

let me explain it

[glad parrot]

bro please you don't need wa for this

[simple musk]

lmfao

[simple musk]

sir, if every point in L1 is at a constant distance to Pi

then L1 is parallel to Pi, agree?

[glad parrot]

yep

[simple musk]

ok, now

if the line L1 is parallel to plane Pi

then the normal vector of Pi

where will it point towards?

and the direction vector of the line where will it point towards?

dont you see they are clearly orthogonal?

[glad parrot]

its not for me lmao, maths is a matter of writing also, not only calculations

i just want to be sure your paper isn't just full of numbers

but also text explaining why you doing this calculations

[simple musk]

ok srry, I just have done a lot of similar problems like this and I directly link some ideas together, if you guys havent practiced as many of this problems it may seem like I am jumping without actual reasoning

[glad parrot]

i see, but keep the redaction as part of your work cuz its the most important either when you are in a math channel or when exam

[simple musk]

ok, for a sec I thought nobody was watching tho

|4k + 5| = 6

i) 4k + 5 = 6 ==> k = 1/4
ii) -4k - 5 = 6 ==> -4k = 11 ==> k = -11/4

L : X = k(1,0,-3) + (3,1,0)

L : (k+3,1,-3k)

Pi : x + 2y - z = 3

,, \sqrt{6} = \frac{|k+3 + 2 + 3k - 3}{\sqrt{1 + 4 + 1}}

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[simple musk]

6 = |4k + 2|

i) 6 = 4k + 2 ==> 4 = 4k ==> k = 1
ii) 6 = -4k - 2 ==> 8 = -4k ==> k = -2

L : X = k(1,0,-3) + (3,1,0)

L : (k+3,1,-3k)

k=1:
(k+3,1,-3k) = (4,1,-3)
k=-2:
(k+3,1,-3k) = (-2+3,1,6) = (1,1,6)

L1 : X = k(-6,2,-2) + (1,1,6)
L1' : X = t(-6,2,-2) + (4,1,-3)

.close

[calm coral]

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[ornate cape]

Assume that the probability of having a boy or a girl is equal.
You know that the Janssen family has three children and that one of them is a boy.
Calculate the probability that the other two children are girls.
Draw a tree diagram

[ornate cape]

Assume that the probability of having a boy or a girl is equal.
You know that the Janssen family has three children and that one of them is a boy.
Calculate the probability that the other two children are girls.
Draw a tree diagram

[sweet stag]

!status

[calm coral]

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

@ornate cape Has your question been resolved?

[calm coral]

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[blazing coyote]

this is example 5

This can be restated as if C is a simple closed path that doesn't pass though the origin. or encloses the origin, then $\int_{C} F \cdot r=0$

Wondering what the contrapositive would look like

if $\int_{C} F \cdot r \neq 0$ , then C passes via the origin or C is not a simple path, or C is not a closed path?

[potent lotus]

What a wonderful world !

[tidal grotto]

the original question includes the path possibly enclosing the origin too

[potent lotus]

What a wonderful world !

[blazing coyote]

What would the contrapositve be though

If $\int_{C} F \cdot r \neq 0, then \text{ case (i) } C doesn't enclose the origin ;case 2: then C passes via the origin or C is not a simple path, or C is not a closed path

Does this work?

[nocturne heron]

u can use green theorem ?

[blazing coyote]

If $\int_{C} F \cdot r \neq 0, then \text{ case (i) } C doesn't enclose the origin ;case 2: then C passes via the origin or C is not a simple path, or C is not a closed path

[blazing coyote]

If (\int_{C} \mathbf{F} \cdot d\mathbf{r} \neq 0), then:

\begin{itemize}
\item Case (i): (C) does not enclose the origin.
\item Case (ii): (C) passes through the origin, or (C) is not a simple path, or (C) is not a closed path.
\end{itemize}

[potent lotus]

What a wonderful world !

[blazing coyote]

oop

[nocturne heron]

$\int_C F \cdot dr = \int \int_D (\pdv{Q}{x}-\pdv{P}{y}) dxdy$

[blazing coyote]

This is meant to be the contrapositive

[blazing coyote]

yea, I know

[potent lotus]

tm

[nocturne heron]

in your case, F is closed

$\pdv{Q}{x}=\pdv{P}{y}$

[potent lotus]

tm

[nocturne heron]

so the difference is 0

[blazing coyote]

no, I get that

but Trying to determine what the contrapositive is first

[nocturne heron]

oh k

[blazing coyote]

Is it right

?

[nocturne heron]

solving with contrapositive ?

[blazing coyote]

Want to see if that helps

wait

If (\int_{C} \mathbf{F} \cdot d\mathbf{r} \neq 0), then:

\begin{itemize}
\item Case (i): (C) does not enclose the origin.

\item Case (ii): (C) passes through the origin, or (C) is not a simple path, or (C) is not a closed path.
\end{itemize}

no my bad

[potent lotus]

What a wonderful world !

[blazing coyote]

Is this right

[nocturne heron]

its the contrapositive ?

[blazing coyote]

I'm asking

[nocturne heron]

i would just say, If ( \int_C \mathbf{F} \cdot d\mathbf{r} \neq 0 ), then ( C ) passes through or encloses the origin

[potent lotus]

tm

[nocturne heron]

why u took different cases

[blazing coyote]

oh yeah

fair

it either passes though the origin, or encloses the origin or is not a simple path or is not a closed path

This will be much harder go show , IMO

[nocturne heron]

if it passes thought the origin the integral diverges

wait

[blazing coyote]

wait, I 'm confused

[nocturne heron]

(\int_C F\cdot dr \neq 0) implies C either:

1. Is a closed path enclosing origin (with net winding), OR
2. Passes through origin (making integral undefined)

[potent lotus]

tm

[nocturne heron]

now it’s good

[blazing coyote]

how

Can't it not be a simple curve

I think I'll do this tomorrow

gn

.close

[calm coral]

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[blazing coyote]

Thanks!

[nocturne heron]

no, only the winding matter

not the curve simplicity

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[blazing coyote]

So I have $\int_{0}^{1} \int_{0}^{x^2} \frac{y}{x^5+1}dy dx$

[finite oasis]

Don't forget the d's

Also y in the numerator

[potent lotus]

What a wonderful world !

[blazing coyote]

so $\frac{4}{2(x^5+1_}$

[potent lotus]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

which i elemenatry tyo integrate

don't know why I chose to do this

thanks

.close

[calm coral]

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[calm coral]

@honest flume Has your question been resolved?

[calm coral]

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[calm coral]

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[eternal fulcrum]

yo

i have a big algebra 1 EOC coming up and i dont realy remember any of it

could someone hop in a call that i can screenshare it?

[calm coral]

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[limber lagoon]

Draw an angle with the given measure in standard position 1. 260 degrees 2. -750 degrees

[limber lagoon]

Hellooo

[golden hearth]

Can you use a protractor or no

[limber lagoon]

Yeah ig

I js need rhe draw it

I js got back from spring break n forgot how to do all of this 💀

Chat gtp messed up the drawings

[limber lagoon]

He

Yes*

Dude

<@&286206848099549185>

[meager sparrow]

For one you will have to draw a straight horizontal line then another line will make the angle, for the first one you need to add some more to the protractor since it stops at 180, the second angle you need to make a few 360 rotations before you arrive at how much it actually is, the minus is just for the direction of the rotation(+ve counterclockwise, -ve clockwise)

Try your best and send your attempt, if it’s correct then congrats, if not I’ll tell you where you went wrong

[tranquil wasp]

!nogpt

[calm coral]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[limber lagoon]

Alr

@meager sparrow

? Like this

[meager sparrow]

Isnt that perfect

Yup you’re correct

[limber lagoon]

Oh ok

[meager sparrow]

You used the protractor right?

[limber lagoon]

Would the -750 have 2 loops

And 1 incomplete loop

[limber lagoon]

No

I don’t have one

[meager sparrow]

Close, 3 complete loops

[limber lagoon]

I just eye balled it

[limber lagoon]

How come

360 360

30

[meager sparrow]

Not bad, although should be even closer to the negative y axis

[limber lagoon]

Ah ok

[meager sparrow]

Since that would mark exactly 270

[meager sparrow]

Actually you’re correct

[limber lagoon]

Alr on the test Ikk use a protractor

[meager sparrow]

Yeah 2

[limber lagoon]

O damn

Alr lemme try that out

The circle goes the other way right

[meager sparrow]

Alright

Yup

[limber lagoon]

Thanks

[meager sparrow]

Anytime

[limber lagoon]

Like thúc

This?

[meager sparrow]

No

[limber lagoon]

O

What is wrong w it

[meager sparrow]

It goes round and round that’s correct

[limber lagoon]

Oh

[meager sparrow]

How much left after that?

[limber lagoon]

So I never take my pen off?

[meager sparrow]

30 right?

[limber lagoon]

Yeah

[meager sparrow]

No no nobody would bat an eye for that dw

[limber lagoon]

Oh

[meager sparrow]

Since it’s negative go the other direction

[limber lagoon]

Right or left?

[meager sparrow]

You (conventionally) start from the positive x axis going in the direction of the positive y axis

since you have a negative sign you should go toward the direction of negative y axis

Got it?

[limber lagoon]

Something like this?

[meager sparrow]

Not quite

[limber lagoon]

O

[meager sparrow]

You correctly went with direction

[limber lagoon]

Ah

[meager sparrow]

But

It should be from the right

The right half

Always

[limber lagoon]

So flip it around?

[meager sparrow]

No just translate it to the right

[limber lagoon]

Oh ok

Like this?

[meager sparrow]

Hooray

[limber lagoon]

Yay

Thank you so much

I was stuck on thuế two problem for awhile

[meager sparrow]

Although since im very precise guy, this is more like 45°, you should make a bit closer to the x axis than this

[limber lagoon]

Yeah

[meager sparrow]

[limber lagoon]

Ill use a protractor on the test

[meager sparrow]

Anytime

Btw

[limber lagoon]

?

[meager sparrow]

You know how a clock works?

[limber lagoon]

Yes

[meager sparrow]

Yeah

So

Positive angle is from 3:00 then go backwards in time

If you have a negative sign

Then from 3:00 and go with the clock

[limber lagoon]

Ohh wait that’s actually very useful

Imma take a Ss of that

[meager sparrow]

That’s what they mean when they say counterclockwise (because it’s really counter-clock-wise)

[limber lagoon]

Ahh

Wow ok thanks very useful

[meager sparrow]

And this is clockwise

[limber lagoon]

Yeye

Tsym

[meager sparrow]

Anytime

Np

You can close with .close

Asidr from that i should sleep

[limber lagoon]

Alr goodnight

.close

[calm coral]

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[meager sparrow]

Gn

[calm coral]

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[gleaming frost]

can anyone explain to me this question

[granite dust]

find the probability of each difference that you get when you roll 2 dice. then add up those probabilities to get the expected value

[gleaming frost]

what does "of each difference" mean?

I don't understand how I should do it

[granite dust]

say you roll a 6 and a 6. the difference is 6-6=0

6-5=1, etc

so there are 5 possible differences

1, 2, 3, 4, 5

[gleaming frost]

wait
why 5 not 6

[gleaming frost]

it can be 0 too

as you write
6-6=0

[granite dust]

oh ya mb

ye 6 differences

[gleaming frost]

oh okay thank you

.close

[calm coral]

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[gleaming frost]

.reopen

[calm coral]

✅

[gleaming frost]

aaa
sorry but I have other question
should I calculate it like first dice - second dice (and count the nonnegative difference) or like greatest dive - smallest dice

[granite dust]

doesn't matter cuz its nonnegative difference. So if x and y are the values of the dice, then difference = |x-y|

so what you wanna find is what's the probability that the difference is 0? is 1? 2?, etc.

[gleaming frost]

oh then there are 6 expected values and I should calculate the probability for each one that would be 1?

because the 36 option would be nonnegative?

or I have misunderstanding?

[granite dust]

there are 6 expected values yes, but out of the 36 pairs, count which pairs have each difference

[gleaming frost]

0 = 6 pairs
1 = 10 pairs
2 = 8 pairs
3 = 6 pairs
4 = 8 pairs
5 = 2 pairs

[granite dust]

yes but 4 = 4 pairs actually

so the expected value is 6/36 + 10/36 + ...

[gleaming frost]

yeah I understand then the final result will be 1? 36/36

[granite dust]

wait no

before u sum each fraction, u need to multiply each by its respective difference

so 0x6/36 + 1x10/36 + ...

[calm coral]

@gleaming frost Has your question been resolved?

[gleaming frost]

Why this step?

[granite dust]

the formula for expected value is = each possible outcome x its probability.

[gleaming frost]

Oh okay

[granite dust]

ye

[gleaming frost]

Thank you

.colse

.close

[calm coral]

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[wheat wyvern]

I tried using the power to change everything, but that didn't work and im confused 😓

[swift laurel]

what you have to do is plug in x^6 in the place of x in the series

nothing else should change

[wheat wyvern]

that's it?

[swift laurel]

yes

[wheat wyvern]

I can just treat it like function notation?

[swift laurel]

it is a function composition, yes

[wheat wyvern]

thank you

.close

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[patent phoenix]

how do i do this question

[patent phoenix]

why is this wrong

[thorny steeple]

I need help olcing the solution

[patent phoenix]

huh

[thorny steeple]

the general solution

[patent phoenix]

are you tryna differentiate that?

also open a new channel lol

[thorny steeple]

thats a differential equation

[patent phoenix]

oh

[thorny steeple]

and I want to solve the general solution

[patent phoenix]

srry idk multivariable calc

[patent phoenix]

!helpers

[calm coral]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[patent phoenix]

<@&286206848099549185>

[thorny steeple]

ok thanks ca

[patent phoenix]

sorry

[thorny steeple]

do I go to another channel

[patent phoenix]

wait

[shy swallow]

yes make your own channel

this will only clutter stuff up

anywaus

i think you should find the decay constant first and foremost

[keen flare]

im confused on what you're trying to find

[patent phoenix]

i did

[keen flare]

like you found n

[patent phoenix]

d i do it correctly

[patent phoenix]

yea

thats the constant right

[keen flare]

but why did you relate it to 1/2

[patent phoenix]

because we're finding the halflife

[shy swallow]

yeah i was gonna ask

[keen flare]

but n is the half life...

[patent phoenix]

HUH

[shy swallow]

no 💀 thats not how that works

[patent phoenix]

omg

what

whaaaaaaaaaaaaaat

[shy swallow]

$N(t)=N_0 e^{-\lambda t}$

[keen flare]

you have 1000(1/2)^(t/n)

[potent lotus]

parabolicinsanity

[shy swallow]

the base is always e

💀

you have to find lambda for the decay constant what

[patent phoenix]

😭

[keen flare]

i mean their way technically works

[patent phoenix]

i forgot all of this

wait so like

if i

set base to 1/2

then the lambda

[keen flare]

in terms of base of 1/2

[patent phoenix]

is

[keen flare]

not base e

[shy swallow]

it could work but dont do it

its WRONG

[patent phoenix]

what why

whaaaaaat

[shy swallow]

The e is there because its a solution from

[patent phoenix]

yea so the n IS the halflife?

[shy swallow]

the differential equation for decay

[keen flare]

in this case it would be

[shy swallow]

$\dv{N}{t}\propto N$

[potent lotus]

parabolicinsanity

[shy swallow]

just dont do it that way because its exponential decay so we always pick e as the base

i'm telling you changing bases will change your decay constanf

[patent phoenix]

but i got a negative answer

[shy swallow]

keep it e and re do

[patent phoenix]

okay

can i learn the proper way pls

[keen flare]

do it in terms of e

[shy swallow]

the equation i gave

its pretty easy, just N(100)=100g

[patent phoenix]

so the lambda is negative because it's decay?

[shy swallow]

yes

[patent phoenix]

if it's growt then its posite

okay

[shy swallow]

im just a little surprised because i have never seen anyone use ½ as base for decay

[patent phoenix]

omg

i got it

[patent phoenix]

lowkey

thats what we were taught

😭

like whenever half life

[shy swallow]

weird

[patent phoenix]

do base = 1/2

[shy swallow]

oh wait i think i know what

[patent phoenix]

o rlly

[shy swallow]

thats the nth halfife

what

[patent phoenix]

huhh

what doe sthat mean

[shy swallow]

why would you need it

[patent phoenix]

idk because halflife so i hoguht we do that

[shy swallow]

$N(n)=N_0\left(\frac{1}{2}\right)^n$

this is for the nth half life 😭

[potent lotus]

parabolicinsanity

[patent phoenix]

bruh

so like

i CAN solve it like thqat

but it's not t/n

it's nt?

[shy swallow]

no you just did it incorrectly

and you wont find

half life that way

[patent phoenix]

oh

[shy swallow]

only no of particles

[patent phoenix]

😭

what

hos is so confusing

[shy swallow]

for half life set $N(t)=\frac{N_0}{2}$

[potent lotus]

parabolicinsanity

[patent phoenix]

and then

use

the

e^lambdat

?

[shy swallow]

yes

[patent phoenix]

thaks

btw your pfp is very goated

[shy swallow]

you too ||death devil!!1!||

dont

open that if you dont want spoilers

[patent phoenix]

🔥

i stopped reading like

[shy swallow]

anyways

[patent phoenix]

1/4th of part 2

[shy swallow]

first solve for N(100)

[patent phoenix]

what

isnt t given

[shy swallow]

im still going

[shy swallow]

lambda isnt

i siggest simplifying to this

[patent phoenix]

caught up?

[patent phoenix]

ok ay wait

wait

i got lamba

ldmba

[patent phoenix]

huh

[shy swallow]

yes then

nvm you already found lambda

for half life

[patent phoenix]

wait

wait

[shy swallow]

just set N(t) =½N_0

[patent phoenix]

thats the lambda for HALF LIFE??

is it not for the function

[shy swallow]

no you use that to find the half life

[patent phoenix]

bruh

wha

[shy swallow]

lambda and half life are related

[patent phoenix]

what

[shy swallow]

i'll show how

[patent phoenix]

this is insane

oka

[shy swallow]

$N(t)=\frac{N_0}{2}=N_0e^{-\lambda t}$

[potent lotus]

parabolicinsanity

[shy swallow]

then divide N_0 away givjng you

$\frac{1}{2}=e^{-\lambda t}$

[potent lotus]

parabolicinsanity

[patent phoenix]

oh

[shy swallow]

then take the natural log

[patent phoenix]

thats how you did it

oh ok

But like

[shy swallow]

$\ln\left(\frac{1}{2}\right)=-\lambda t$

[potent lotus]

parabolicinsanity

[patent phoenix]

if i did it this way would this lamda work for all values

[shy swallow]

all values of?

[patent phoenix]

like

the way you solved for it

kambda would be half life rigt

[shy swallow]

its related to half life

[patent phoenix]

but like if i did it this way, would this lamdba be usable to find N(t) where t is any number

[shy swallow]

uhm obviously

you know lambda

and you know N_0

[patent phoenix]

lowkey

[shy swallow]

so for a tims t

[patent phoenix]

nm

nvm

[shy swallow]

you can find N(t)

[patent phoenix]

okayy

[shy swallow]

have yiu found the half life now?

[patent phoenix]

yep

tyty

so like

[shy swallow]

so then time for 300g

should be simple

[patent phoenix]

for any function then if i have to come up with the equation myself it would always be N(t) = N_0(e)^-lambda(t)?

[patent phoenix]

ya i used the lambnda

[shy swallow]

for decay yes

becauase its the solution to the differential equation for decay

so you have to use it

[patent phoenix]

uhhh

would exponential growth make a differnce

[shy swallow]

only makes the minus a plus

hmmm

[patent phoenix]

okyayy

[patent phoenix]

wait so like when do we use this and when do we use e

😭

[shy swallow]

ugh that looks disgusting but perhaps it could be used

[patent phoenix]

Bri

[shy swallow]

just that e as a base is thd gold standard in a way

[patent phoenix]

We were taught that since Gr 11

[patent phoenix]

ngl

I like that way better

But like

Does it always work

[shy swallow]

its for decay you have to use it for decay

because its the solution to the ODE for decay

[patent phoenix]

What’s ode..

[shy swallow]

thats why i've been insisting

[shy swallow]

ordinary differential equation

the rate od decay proportional to no. of undecayed atoms yada yada

[patent phoenix]

this

Is so confusing

omg

[shy swallow]

$-\dv{N}{t}=\lambda N$

[potent lotus]

parabolicinsanity

[shy swallow]

sorry its just what it literally is, i dont have much more to say

its true for decay and growth so idk up to you but e is the gold standard

[patent phoenix]

oh okay

ill use e from now on then

[shy swallow]

i guess you've not gone through it, which is weird so just rememver that sticking to e is the proper way

[patent phoenix]

okayyy

thanks for teaching me

[shy swallow]

hope that helped

[patent phoenix]

🙇

[shy swallow]

you can .close if you have no more problems

and

happy mathing

[patent phoenix]

lmfao

you too

.close

[calm coral]

Channel closed

Closed by @patent phoenix

Use .reopen if this was a mistake.