help-38

[elder pond]

I did the inequality and it gave me +2-a<x<a+2

[burnt mulch]

I was talking about the quadratic inequality here

[elder pond]

oh yeah

how do you know that

like

did you substitute the x with something?

i did it with 5 and -2

[burnt mulch]

I phrased it a bit sloppily

But the idea is to find an interval of all x that is some distance from 2 such that the entire solution set is included

If we go from 2 to 5, that’s 3 units away

If we go from -2 to 2, that’s 4 units away

So we can only include the entire solution -2<x<5 if we’re at least 4 units away from x=2

[elder pond]

but can you do that using the inequality of |x-2|<a please

I think that I will understand it better in that way

like +2-a<x<a+2

then I replace it

and solution one would be

+2-a<5<a+2

then I have something like -a<3<a

Second solution is

-a<-4<a ( using -2)

if I take the -a<-4 it would give me a>4

but it doesn't include that number

so I would be more likely to say that a is greater than four but it doesn't include the four

but what can I do to have the correct solution

[burnt mulch]

Uhhhh ig you can just say the Union of the two solutions is the solution to |x-2|<a

🤷‍♂️

[elder pond]

but is it legal to say that the union doesn't exclude the 4?

[burnt mulch]

I mean a=4 does achieve that condition

So you need to include it

[elder pond]

and how can you know that it achieves that

I mean

you just have an A

or the |x-2| gives you some kind of data that I'm not catching?

[elder pond]

oh now I get it

tysm man!!!1

[burnt mulch]

Anything else?

[elder pond]

yeah

how can you know that a is outside of the interval (-2,5)?

[burnt mulch]

???

[elder pond]

yeah because the two conditions for the quadratic being negative is that x should be in the interval (-2, 5)

but you have that a>4 in your equation

[burnt mulch]

you're conflating x with |x-2|

(I think)

(I'm honestly not sure what you're doing)

[elder pond]

aren't they the same x?

[burnt mulch]

yes

but a refers to a bound on |x-2|

not x

[elder pond]

I just saw this video

But I still don't understand why it includes the 4 aksdhjasdhasjlfhslhfjlas

sorry

the answer cannot be greater or equal than three for the same reason

and I would think that bc of the a>4 the 4 would not be included

but it may be that because of the interval (3, infinite) it is included?

I normally would think that if I have that two answers

the interval would be something like (3, infinite) except 4

ykwim?

@burnt mulch

[burnt mulch]

oh we can do a number line thing ig

I'll assume you understand where this came from

[elder pond]

yup

I do

[burnt mulch]

As said earlier, we need the union of these two to be the same as (-a+2, a+2)

If you have the union of some number of intervals of the form (a,b),

• The left endpoint of the union is the leftmost (or smallest) of the left endpoints of the intervals
• The right endpoint of the union is the rightmost (or largest) of the left endpoints of the intervals

So this gives us $$\begin{cases} -a+2 \leq -2 \ a+2 \geq 5 \end{cases}$$

[solid kiln]

Civil Service Pigeon

[burnt mulch]

solving these give a__>4 and a>__3

meaning that they're only both true if a__>__4

[elder pond]

but in that case why don't you just use the a is equal or greater than zero

like, that already includes the 4

they are both sufficient ig but

the most sufficient would be a greater or equal than 3 for me

[burnt mulch]

if a=3 then you have -1<x<5

in which case that's not necessary

you can have x = -1.5

wait

yk what "necessary" means here right

[burnt mulch]

and why that means this

[elder pond]

yup

but the first question is sufficient

[burnt mulch]

wait are we talking abt the same question ...

I was talking about (ii)

[elder pond]

Yeah

[burnt mulch]

whre it talks about |x-2|<a being necessary

[elder pond]

ii

[burnt mulch]

nvm then

[elder pond]

sorry

my bad

I thought that the ii was necessary

haahfahafhsdfh

well now I think that I get it

[burnt mulch]

._. lol

[elder pond]

if I plug the 4 in +2-4<x<+4+2

I would have that

-2<x<6

and that includes everything but it left something in

that is the interval of (5, 6)

sorry

[5, 6)

[burnt mulch]

eh imma just start over

wait no

that is right

._.

[elder pond]

what is right

[burnt mulch]

ii. |x-2|<a must be true for -2<x<5, so a__>__4

[elder pond]

what I'm saying

[elder pond]

yeah yeah

[burnt mulch]

iii. 2-a<x<2+a implies -2<x<5, rinse and repeat

[elder pond]

Although the [5,6) interval is there, it is still correct

right

because it's necessary

and needs to include everything

[burnt mulch]

yeah yeah

[elder pond]

despite it includes something that do not belong there

right????

[burnt mulch]

yeah

[elder pond]

okaaaay now I get it

And the third one?

[burnt mulch]

.

[elder pond]

7 and 8 are sufficient

i g

but

the most sufficient is

8

Am I right?

[burnt mulch]

they prob want the complete range of values of a that work

[elder pond]

oh so that is not the answer

[burnt mulch]

$2-a \geq-2, 2+a \leq 5$

[solid kiln]

Civil Service Pigeon

[burnt mulch]

uhhh I'm too lazy to do this

also 2-a<2+a as well so yeah

[elder pond]

thats the same thing tho

right?

as the previous one

[burnt mulch]

no....

the answer is 8 if that's what you want to know

there's the setup

take it as you will

[elder pond]

a<=( I dont know how to type the two of them sorry)\

a<=4

and

and a<=3

but why xd

[burnt mulch]

(2-a,2+a) needs to be contained within (-2,5)

[elder pond]

yeah

7 can be the answer in that case

[burnt mulch]

so 2-a is to the right of -2, 2+a is to the left of 5

[elder pond]

because it is correct

but

8 is MORE correct

than 7

[burnt mulch]

Again, they likely want the set of all a.

[elder pond]

DAMN

now I get it

goddamnnnn

tysm for the patience

hahaahah

now I cleared all my whys

i fkin love you man

fr

u the best

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[gray zinc]

for | z + 14 - i | <= | z - 3i |
what would be the right one
z-(-14-i) , z-(3i)
or
z-(-14+i) , z-(3i)

[gray zinc]

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[gritty notch]

is h(x) going up or down

[gritty notch]

also, in order to find if f(x) is an odd function do i need to prove that:
-f(x) = f(-x)
or
-f(-x) = f(x)?

[minor rover]

There the same

But usually the top one is written

[gritty notch]

?

[minor rover]

Usually f(-x) = -f(x)

And h is increasing

[gritty notch]

thats what i thought, but you havent accounted for when x = -1.5

[minor rover]

Why it said 2.5 on your graph

[gritty notch]

(red) vertical asymptote?

[minor rover]

So what's wrong with x being -1.5

[gritty notch]

@minor rover

[minor rover]

But it is still increasing

It just starts at 0

[gritty notch]

no its starts from -1.5

and ends at -1.5

[minor rover]

The value of h is increasing

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[wispy dew]

this is my current graph

im trying to make it look like this:

how can i move that "s" shape up but still cap it at 1?

i just want it to be more prominent

the 2 functions i have tried are tanh and erf

[shrewd ridge]

you just do

you move it up and move it right

and then divide by the cap it has

so by 2

[wispy dew]

oh shit

u smart

thanks

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[dull thorn]

? For any x WE have that equality

You need to know like sin 2x = 2sin x cos x

And it's done

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[wraith hinge]

whats the difference from the magnitude of the velocity and its displacement

[true sierra]

In what context?

[wraith hinge]

sorry

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[tender siren]

Hey, I was wondering how to simply this. It's giving me -8/|A| but I'm not sure if this is the right result.

[ionic pendant]

do you know the size of A?

[tender siren]

Yes, my mistake. A is a 3x3 matrix.

[ionic pendant]

we can work through it one part at a time

[tender siren]

Sounds good

[ionic pendant]

what is the relationship between the determinant of a matrix and the determinant of the transpose?

[tender siren]

They're equal

[ionic pendant]

yes

so then we can say the determinant is the same as here: [ |(-\frac 12 A)^{-1}| ]

[solid kiln]

pnoןɔ

[tender siren]

Agreed

[ionic pendant]

then, what is the determinant of the inverse of a matrix?

[tender siren]

I don't know LaTeX but det (A^-1) = 1/det(A)

[ionic pendant]

yes

so we can say then that this is also the same determinant: [ \frac 1{|-\frac 12 A|}]

[solid kiln]

pnoןɔ

[tender siren]

Yeah

Then you take out the -1/2*3 because it's a 3x3 matrix right?

(-1/2)*3

to the power of

sorry

[ionic pendant]

yes, then we have that this is the same: [ \frac 1{(-1/2)^3 |A|} ]

[solid kiln]

pnoןɔ

[tender siren]

Yes

[ionic pendant]

and then we just simplify the fraction

[tender siren]

so 1/(-1/8)*|A| which equals -8/|A|?

[ionic pendant]

yes

[tender siren]

Woah can't believe I got that right lol

Thank you very much!

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[wraith hinge]

So uuuh

[wraith hinge]

How exactly am i supposed to do this when there are only two eignvalues?

How did this example get two vectors for one eigenvalue?

[wooden plover]

what equations did you get for your eigenvectors ?

[wraith hinge]

For the first image? i didn't. they told me the eigenvalues were 12 and 3

[wooden plover]

eigenvectors

[wraith hinge]

huh

[wooden plover]

i know they're saying the eigenvalues are 12 and 3

[wraith hinge]

$Ax = \lambda x$ ?

[solid kiln]

Book Reader

[wooden plover]

ok yeah

like you tried finding the eigenvectors for lambda=3 and lambda=12 right ?

how did you attempt that ?

[wraith hinge]

I only tried for 12

#3 i deemed impossible

[wooden plover]

yeah what do you have for 3 ?

even if you find it impossible

[wraith hinge]

5 2 4
2 8 -2
4 -2 5

1 0 1
0 1 -1/2
0 0 0

eigenvector for 3 is

-1
1/2
1

?

I watched this video
https://www.youtube.com/watch?v=ChZG3uPm5uo

and apparently the second vector is k where $(A- \lambda I)k = x$ where x is the first vector
-1
1/2
1 ?

[solid kiln]

Book Reader

[wooden plover]

,w eigenvalues {{8,2,4},{2,11,-2},{4,-2,8}}

[solid kiln]

Results provided by WolframAlpha

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Requested by _aplatypus

[wraith hinge]

its two 12s?

[wooden plover]

ah so 12 is the repeated one

yea

[wraith hinge]

pain

[wooden plover]

[-4 2 4]
[2 -1 -2]
[4 -2 4]

that's A-12I

so yeah all the rows are multiples of one another

so there's really only one equation for your eigenvectors here

2x -y -2z = 0

w/ only one equation you'd expect 2 linearly independent solutions

[wraith hinge]

because nullity = 2?

[wooden plover]

well the nullity of that matrix is 2 yes

[wraith hinge]

so

x_1 = x_2 (1
1/2
0)

• x_3( 1
  0

or something like that?

[wooden plover]

that first vector doesn't work

you prolly meant (1 2 0)

@wraith hinge

[wraith hinge]

uuuh

no?

2 -1 -2 = 0
is the same as
1 -1/2 -1 = 0

1
1/2
0

1
0
1

[wooden plover]

(1 1/2 0) doesn't satisfy the equation :/

like just plug it in

[wraith hinge]

idk then, thats how I've solved all my problems

[wooden plover]

well here's one way

we have 2x-y-2z = 0

we know we're gonna have 2 free variables

so let's just isolate one of the vars (which will then depend on the other 2)

so for example we get z = x - y/2

so all the solutions to this equation look like (x, y, x-y/2) [where x, y can be any real number]

these are all your eigenvectors for 12

now if you want to extract a basis from that, well you can seperate the x and y parts

(x, y, x-y/2) = x*(1, 0, 1) + y*(0, 1, -1/2)

i.e. all your eigenvectors can be generated from (1,0,1) and (0,1,-1/2)

@wraith hinge

[wraith hinge]

i see

so any two vectors from that would work?

[wooden plover]

well you want to take linearly independent ones

but as long as you respect this condition sure

[wraith hinge]

okay

[wooden plover]

the whole thing I did is one way to get the linearly independent vectors decently fast

[wraith hinge]

so

1 0 1 is a eigenvector
and 0 1 -1/2

[wooden plover]

yes

[wraith hinge]

okay

ty

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[knotty locust]

What's the area of the current unshaded rectangle

The 9ft x 12ft one

Okay

Now it's saying, we're going to add a strip of a certain length to all sides of this rectangle

and the strip is a certain length such that after adding it, the area of the new rectangle is 180 ft

instead of 108

How does adding this strip affect the side lengths of the beginning rectangle?

it would increase the area yes but I was asking how it would affect the side legnths

like say you add a strip of length x

then the side length of the rectangle is no longer 9 and 12

it's what and what?

I'm asking for a specific length

not a term

the rectangle has base length 12 and side length 9

you add a strip of width x around it

you get a new rectangle

what is the base length and side length of the new rectangle

Close but no

you're moving onto a different step already but you still haven't finished this one

the strip is being added to all sides of the rectangle

I already told you 12+x and 9+x was wrong

No

the length in green is 12+x

do you see what you're missing

Why would it not make sense

you're very clearly not measuring the left side of the strip

which has length x...

so another x makes perfect sense

that's one side length

and what is the other

So your rectangle is 9 by 12
you add this strip of uniform length x,
and now it is 9+2x by 12+2x

the area was 9*12

now it is (9+2x)*(12+2x)

But also, they say this new area is 180

So

(9+2x)*(12+2x)=180

Solve for x

Solve the problem

Literally how

you didn't solve it yet

how does it seem off?

Solve for x

Okay

?

[tired shoal]

and do you know how to solve a quadratic

[knotty locust]

a+bx+cx^2

Okay

Does it make any sense to add a strip of length -12

No

It doesn't

negative lengths don't exist

so the answer is what

yes

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[tribal vapor]

required to evaluate this expression. I know that 1/2 bcomes the exponent to 9 but how do I manage +1 ? or how do I convert it into a relevant log form ?

[lean spear]

log base b of b?

=1

So you bring the exponent to the 9

So u have log3

And the 1

1 can be log10

Then u know what to do

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[tribal vapor]

owhhh okay I got it, thank you very much

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[astral moat]

Hi

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[wraith hinge]

I've tried x= 70Cos(135), 70Sine(135) and y=500Cos(135), 500Sin (135). Not sure if on the right track.

[wraith hinge]

*x=70Cos(45), 70Sin(45)

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[surreal bridge]

Um

I'm not the best at physics or anything but what r u struggling with

[wraith hinge]

I'm a little lost onn what their asking for

like which angle

[surreal bridge]

I think I shld answer this cos I'm not the best at this either
Let someone who has actually properly studied this to answet

Thanks

[wraith hinge]

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[drifting rose]

can someone explain to me part c?

[drifting rose]

I watched a video that solved this and I am confused when he wrote the equation for c he wrote 1+x+x^2+...+x^(2n)

why is it to the power of 2n?

[bright junco]

because your final power to end whatever series you're working with will be even

if thats not the case, then the function isn't necessarily always positive. in fact, we positively know it isn't since taking a limit at -infinity shows that u get as a result -infinity, meaning that your function has to be negative somewhere

you need your final term to be an even power, otherwise the statement simply isn't true

makes sense?

[drifting rose]

so that equation he wrote is basically saying, this statement will always be positive so the end value will always has to be even

[bright junco]

well we're looking at it from a more intuitive sidepoint

[drifting rose]

yea that makes sense, I think I confused myself with the last number describing the equation itself which is silly

[bright junco]

we dont want to say "oh its always positive, so the final power has to be even!". what we're doing is "well i notice that when the last power is even, the series i have will always be positive"

[drifting rose]

oh I see

[bright junco]

our conclusion is that its always positive, so we then connect it with the fact that it has to do with the even power at the end

what you said is also true, but its like you are using your conclusion as a given and then ending up at the fact that the last power is even

does that make sense?

[drifting rose]

yes

[bright junco]

of course its a small difference, but a crucial one for mathematics and the logic of thinking

[drifting rose]

so its more of a "from observation this makes that"

I get it

[bright junco]

yup exactly

[drifting rose]

thank you for the help bill

[bright junco]

no worries

[drifting rose]

.close

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[wraith hinge]

If
∫vdv = ∫ads
Then limits have to be same?

[wraith hinge]

V velocity a acceleration

[trim jolt]

@wraith hinge Has your question been resolved?

[wraith hinge]

<@&286206848099549185>

[wild imp]

They're integrals over different variables so no

[wraith hinge]

Real thanks

And another doubt

In this question
For first one they used the same
∫vdv = ∫adx
( Acceleration is in terms of displacement)
So they used the fact that at displacement 8 velocity is 4
And they did
∫vdv( limits from. 0 to 4) = ∫c/(x+4)^2 dx ( limits 0 to 8)

But but but
Velocity at 8m is 4m/s
This is fine
But what's logic behind integrating?

[wraith hinge]

@wild imp this is the actual doubt
Ik the solution but logic of process isn't clear to me

[wraith hinge]

Acceleration is in form of displacement but it's still acceleration
So why to integrate that

I didn't find any logical explanation other then "it's the formula just learn it"

[vernal warren]

ur confused on how we get this ∫vdv = ∫adx?

@wraith hinge

[wraith hinge]

Nope
I am confused WHY and HOW we using it
When displacement is 8m so velocity is 4
So we integrate lhs and rhs with limits 0-4 and 0-8 respectively
But why?

[vernal warren]

a = v (dv/dx) is how acceleration can be defined , rearrange
adx = vdv
dx and dv terms are change in x and v
definite integral is basically anti derivative so to "remove the rate of change" we just integrate

idk how to word it

[wraith hinge]

Can you hear the voice note?

[wraith hinge]

Never mind no voice note option

[vernal warren]

like velocity is rate of change in position, so its definite integral will give us the displacement of the moving object

[wraith hinge]

Exactly that's my point
∫vdt = s
But that is integration of ∫vdv represent?

[wraith hinge]

@vernal warren

[vernal warren]

Velocity probably

That's just change in velocity integral

So it should be velocity

[wraith hinge]

Oh ok

[vernal warren]

Or acceletation times distance

Integral is equal to
V²/2 which is equal to a x

[wraith hinge]

But when it's given velocity at 8m is 4
Why we integrate lhs and rhs by 4 and 8
How do we know they are gonna be same?

[vernal warren]

A = dv/dt

Multiply and divide x

A = dv/dt x dx/dx

A = dv/dx (dx/dt)

A = v dv/dx

adx = vdv

Derived from definition of a

[wraith hinge]

Yeah that i got
But how do u know limiting it to 4 and 8 will be same?

[vernal warren]

U can solve it without limit and then use v = 4 and a= 8

Both are same

solve the indefinite integral
adx = vdv
and then use v and a

U would get the lhs in terms of x , using the formula given find value of xand use that

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@wraith hinge Has your question been resolved?

[wraith hinge]

Oh damn
Thankss a lot man

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[wraith hinge]

also need help answering my physics activities.. Thank you

[rare herald]

a:

b:

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@wraith hinge Has your question been resolved?

[wraith hinge]

so I need to differentiate first?

[wraith hinge]

what happened here? can you please explain?

@rare herald

oh the formula for ave velocity in letter a

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[fair meteor]

how do we evaluate a limit like this

[pseudo grotto]

what's happening in the num and what's happening in the denom

[fair meteor]

num makes it go positive and negative

denom makes it

denom goes to infinite

[pseudo grotto]

yep

[fair meteor]

oh its just 0

im stupid

okay ty LOL

[pseudo grotto]

yep

[fair meteor]

.close

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[pseudo grotto]

yo

[fair meteor]

sleep deprived be like

[pseudo grotto]

stop saying you're stupid while im typing yep

[fair meteor]

admin help im being verbally abused

[pseudo grotto]

cause it makes it seem like i replied to "im stupid" with yep

[fair meteor]

😭

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[lime vigil]

how do I find the image of this function I don't get it

[lime vigil]

I found the vertex (3/2, -5/4) and I don't know what to do from here

?

[plush flare]

when you plot this

the parabloa does not have y cordinates lower than the vertex

so the range is everything above -5/4

[tall kernel]

U don't have to plot it to see that

[plush flare]

yeah

[tall kernel]

It's an upward parabola since the leading coefficient is positive

[tall kernel]

Plug in points for x

And see what values you get as y

And then approximate the graph when you have 5 points

Preferabble vertex

And 2 left and 2 right

[lime vigil]

they got all that yapping in the schoolbook answer when I could just look at the vertex, figure out it's the lowest point and so it's that point and all the numbers above it?

[tall kernel]

Well

The vertex isn't always the lowest point

It can be the highest

If the "a" value of the leading term (x squared) is negative

[lime vigil]

Ik, in that case it'd be all the numbers below that point including the point right

[tall kernel]

Yep

But there are

Horizontal and diagonal parabolas

[lime vigil]

leave that to future me to suffer with, I haven't met those yet

but thanks for the warning

[tall kernel]

lmk if u got the graph^

[lime vigil]

you mean for example do x=1 to get 1^2-3*1+1, and so on until I get 5 points?

[tall kernel]

well try to get values near the vertex

your vertex was (3/2, -5/4)

1.5

so get values like 3

or 0

[lime vigil]

I'll be back in 10 minutes and try it

[plush flare]

not so sure what the point of that is 🤔

[lime vigil]

I was studying for 50 minutes till now I need a break

[plush flare]

i mean like getting the 5 points

[lime vigil]

oh I think that's to draw the parabola?

[plush flare]

makes sense then

but if you just need the range then all you need is the vertex and the coefficient of a

[lime vigil]

is this true for any upward/downward parabola?

[plush flare]

yes when x is to the second power

[lime vigil]

I see, still it would be useful to know how to draw such a parabola right?

[plush flare]

yes obv need to know how parabolas look like

[lime vigil]

my first thought for drawing it out was to use the quadratic formula to find x1,2 but the discriminant in this case is 5, and square root of 5 doesn't really give a whole number

[plush flare]

are you trying to draw that function?

[plush flare]

this one

[lime vigil]

yeah, I wanna know how to do it in-case I encounter another one like that

[plush flare]

you can do -b/2a to find the x cordinate of the vertex first

and plug that x into the original equation to get the y cordinate of the vertex

you know the y intercept is 0,1 and you can use the quadratic formula or factor to find the x intercepts

connect them together and then you should be good

[lime vigil]

I think a better way to find y cordinate of the vertex is -D/4a

D being the discriminant

[lime vigil]

wym the y intercept is 0,1

[plush flare]

(0,1) cordinate

[lime vigil]

where do you get that information from

[plush flare]

you can see from the equation

+c

[lime vigil]

lmao top secret

[lime vigil]

I don't really understand why c being 1 means that y intercept is 0,1

[plush flare]

https://www.youtube.com/watch?v=ijNMeheYiOo check this

[trim jolt]

@lime vigil Has your question been resolved?

[lime vigil]

but in this case I can't use a binomial formula or factor it

and using the quadratic formula will lead me to x1,2 = (3+-√5)/2

I figured that because I got (0,1) all I needed to find was the second x for the same y, that means I already have x1 for y=1, and using x2 = -b/a - x1 I got x2 = 3, so x1 = 0 and x2 = 3 for y=1

which got me this graph

and that's correct right @plush flare

[plush flare]

this would be correct, those are the solutions

[plush flare]

not so sure I understand this

it is symmetric so the distance from the x cordinate of the vertex to the y intercept would be 1.5 and if you move 1.5 right from the vertex you would get 3 if that is what you mean

[lime vigil]

so at first I had gotten (0,1) which gave me one side of the parabola, to draw out the parabola I figured that all I have to do is find the second x on the same y

[plush flare]

that works as well. it would be more accurate if you had the x intercepts, but this is a good rough sketch of how it would look

[lime vigil]

I wouldn't be able to accurately pinpoint where x1,2 = (3+-√5)/2 are by hand anyways

[plush flare]

true

[lime vigil]

you helped me deepen my understanding of parabolas, thank you

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[rain shell]

O centered circle with diameter AB, diameter EF such that A and E don't overlap, tangent at B intersects AE, AF at H and K relatively, P and K are midpoints of HB and BK relatively, find the configuration of diameter EF such that the area of APQ can be the smallest

[rain shell]

so I tried finding the area and it is $\frac{1}{2} \cdot AB \cdot (BP + BQ)$

[solid kiln]

cffex

[rain shell]

then got stuck there

I think it's related to the AM GM somehow but idk

so if I'm in the correct direction $BP + BQ \geq 2 \cdot \sqrt{BP \cdot BQ}$

[solid kiln]

cffex

[rain shell]

then find the equivalence of BP * BQ through similar triangles

thats what I thought in mind

[trim jolt]

@rain shell Has your question been resolved?

[rain shell]

<@&286206848099549185>

[snow saffron]

Hey, I am a bit confused by the "configuration of diameter EF" part, could u pls explain more clear?

[rain shell]

I meant find the position of the diameter EF

[snow saffron]

hmm

[rain shell]

so like rotate around the circle, at which rotation would have the triangle's area be the smallest

[snow saffron]

Triangle APQ?

[rain shell]

yes

I suspect a perpendicular diameter to AB

[snow saffron]

Im sorry I dont think I understand how EF affects the area of APQ at all

[rain shell]

the tangent of B intersects AE, AF at H and K

P is the mid point of HB

and Q is the mid point of BK

[snow saffron]

Ahh I see

[snow saffron]

Yeah that could work

Try drawing it out

I suspect there could be a smaller one tho

[rain shell]

you get an isosceles or possibly equilateral

oops

wrong points name

[snow saffron]

which tool are you using?

[rain shell]

Geogebra

[snow saffron]

Alr

Lets try to look for a pattern here

Since we can only manipulate the position of EF, and the Ar(APQ) is depended on EF and AB(cuz everything else indirectly follows EF)

[rain shell]

yea

[rain shell]

here is the formula for it

[snow saffron]

What if you draw EF really close to AB, almost overlapping but not

[rain shell]

the area would converge to infinity

[snow saffron]

@rain shell is there a way you can share this drawing to me like a link or smth? i wanna open it in Geogebra myself

[rain shell]

wait a minute

[snow saffron]

Converge to infinity is really close to zero

Does that count as smallest area

[rain shell]

https://www.geogebra.org/geometry/adhe8zjt

the area would converge to infinity as E approaches A

so no smallest area here

[snow saffron]

wdym

what if E is point right next to A

[snow saffron]

Ohh i get it

[rain shell]

the closer E is to A, the larger the area of APQ

[snow saffron]

its approaching infinty

[rain shell]

yes

[snow saffron]

So lets do the opposite

whats the farthest

I feel like perpendicular is the farthest we can go

[rain shell]

yeah I tested

it is

it AQP becomes an isosceles triangle at that point

which we can prove

we just need a reason why at that point the area is the smallest

[snow saffron]

Did you prove the isosceles triangle using similarity

[rain shell]

nope

[snow saffron]

or congruency

then?

[rain shell]

congruency

[snow saffron]

oh

[rain shell]

actually congruency and then a few more steps

[snow saffron]

Did you just measure the angles?

[rain shell]

no

[snow saffron]

Wait we wanna prove that the area is smallest when its perpendicular, right?

[rain shell]

we want to prove at what rotation EF makes the smallest area

so we cant just jump to perpendicular yet

[snow saffron]

hmm

We did do that empirical but unless we find a rigor way to prove it..

[rain shell]

yea

[snow saffron]

1/2 * ab*bp+bq

[rain shell]

yes

$\frac{1}{2} \cdot AB \cdot (BP + BQ)$

[solid kiln]

cffex

[rain shell]

fancy text

[snow saffron]

lol

So, when the angle EOB increases, angle Q of the triangle decreases

meaning BQ increases

Hence area increases

we just gotta prove that value of Angle AQP is least when EOB = 90 deg

[rain shell]

actually

the more EOB increase, the closer it is that PQA approach a right angle

so the more EOB decrease, angle Q of the triangle decrease

[snow saffron]

wait no not EOB

I meant E`OB

[rain shell]

oh its F

[snow saffron]

ah

[rain shell]

i overwrote the geogebra link

[snow saffron]

are we allowed to measure any angle using the measure feature in Geogebra

[rain shell]

no we cant measure stuff

[snow saffron]

oh

Are we allowed to construct stuff?

[rain shell]

yes we can draw more stuff

@snow saffron

[snow saffron]

Ok

So

[rain shell]

lol take ur time

[snow saffron]

ignore the 90.5 just think its 90

Assume that AB perpendicular to EF

[rain shell]

uh huh

[snow saffron]

now, construct a ray CD || AB

hence, BCD is 90 deg

[rain shell]

yes

[snow saffron]

and so is QCD

[rain shell]

yes

[snow saffron]

Now, using trigonometry, we can prove that the hypotenuse of Triangle DCQ increases when AOE increases

but we arent done yet cuz it also decreases when the angle decreases

we can prove that the hypo of triangle ABP also inc/dec depending on angle

and using sin and cos, we can find that the least value would be 90 deg i.e 1 and 0

[rain shell]

how is the hyp of DCQ related to the area

and the hyp of ABP?

well

.close

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[snow saffron]

Sorry my internet went out

@rain shell

Hypo of DCQ is part of side of APQ

According to herons formula side of triangle affects area

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[half fjord]

i have done some work in this

but not able to get answer

[vagrant prism]

well x must include 2^3 * 3 as its factors

y must include 2^? * 3^2 where ? is either 1 or 2

z must include 5^2 * 2^? where ? is either 1 or 2 too

that's all i got in my head so you have at least 4 triplets

not sure what degrees of freedom you can have for x, if any

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[wraith hinge]

One hundred reservations are taken for a cruise on a ship that has only 95 cabins. If experience shows that only 90% of the people making such reservations show up. What is the probability that everyone who shows up has a cabin?

[wraith hinge]

so i used binomial distribution

$\binom{n}{x} \times (0.9)^x \times (0.1)^{n-x}$

[solid kiln]

Derivative

[wraith hinge]

with a sum i would assume

from 0 to 95

[vagrant prism]

hmm

[wraith hinge]

$\sum_{x=0}^{95} \binom{100}{x} \times (0.9)^x \times (0.1)^{100-x}$

[solid kiln]

Derivative

[wraith hinge]

because it says the probability that everyone that shows up has a cabin

so it can't be from 96 to 100 because then that would be the probability of overbooking

in total 100 people have booked

chances of showing up is 90%

[vestal saffron]

Yes that's right assuming you mean each individual person has a 90% chance

[wraith hinge]

yes

i will put that answer

i think it is right because it says that 90% of the people show up

and the average is np = 90

so on average 90 people will show up, but sometimes more sometimes less

now, what i will do, is estimate this answer, cuz i cant use a binomial table since n is too large

i will estimate using normal distribution

right now, i get 0.9664 (96.64%) chance

which is pretty reasonable

.close

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[ancient shadow]

how ? 💀

[vestal saffron]

Couldn't you multiply numerator and denominator by what's in the numerator already to simplify

That gets rid of the square root up top and you use difference of squares in the denominator

[ancient shadow]

yeah i did that

but now am lost

[wraith hinge]

Send what you got doing this

I can’t be bothered to do it but I’ll help past that stage

Ping me with response

[ancient shadow]

@wraith hinge i got this

[wraith hinge]

Hm

I mean

If you expand the bracket on the top and then split the fractions the (x+1) terms will become a bit nicer

Although I’m not sure how beneficial that will be for dealing with the 2X part

[ancient shadow]

uhh but the 2x

yes

is there a scope for beta function

[wraith hinge]

Do it and see if there’s a nice substitute that could work

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[frosty heron]

@fallow lodge

But how do we know that we have to take 25x out

[fallow lodge]

We can spot that the equations for V and B are very similar, and that by moving 25x across to the LHS, the RHS will match

Basically our goal is to build an equation where the RHS is the same as the RHS for V. So that then we get that V = LHS

[frosty heron]

Oh

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[civic thorn]

Why is a function not differentiable at a "corner" or cusp?

[marble wharf]

what should the derivative at that point be?

[civic thorn]

0?

[marble wharf]

if you come from the left you get a different slope than from the right

[civic thorn]

oh yeah i guess so

ok thanks

.close

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[fossil cobalt]

how do i find the integral of this: sqrt(x^4 +2x^2+1)

[fossil cobalt]

i dont khow how to start

[fallow lodge]

Try factorising the thing inside the square root. If it's hard to see, try the substitution y = x^2 to help you see it.

[kindred pier]

Find a way to put it into the form $\sqrt{1+u^2}$

[fossil cobalt]

so x* ( x^2+2) +1?

[solid kiln]

SWR

[fossil cobalt]

i do nto know how

[opaque dirge]

seems like square of a sum

[fossil cobalt]

can u help me work it out

i do not want the answer as.much as i wanna know how to integerte it

[opaque dirge]

yes i will not give the answer

hmm

[fossil cobalt]

i did this

[opaque dirge]

i dont think that'll help us here

Do you know how to write a polynomial like (ax+b)²+c ?

I can show how to do it in another example so you can use it in this problem, is it fine?

[fossil cobalt]

completeling the square?

i know how yes

[opaque dirge]

ok then try doing that with x^4+2x^2+1

[fossil cobalt]

wait no

[opaque dirge]

mhm

wait no

[fossil cobalt]

it will be

(x^2 + 1 )^2

[opaque dirge]

that is indeed correct

[fossil cobalt]

and the +1 will cancel with -1

no i do u substitution?

would take work?

[opaque dirge]

i dont see the need to do that here

the question was square root of (x^4+2x^2+1)

[fossil cobalt]

how would i integrate (x^2 + 1 )^2

[opaque dirge]

square root of square of (x^2+1)

[fossil cobalt]

oh

i forgot to do that

then i integrate x^2+1

ty

[opaque dirge]

np ^_^

[fossil cobalt]

completing the squares looks like a useful method

but when do i know when to use it

[opaque dirge]

there are some specific equations like x²+2x+1 or x²+4x+4

[fossil cobalt]

so when i see a polynomial

but sometime ppl use long division

it confuses me

[opaque dirge]

this wont work if its not a square of a sum

(ax+b)²+c

the one that was given was a square of a sum (c=0)

completing the squares is really helpful when finding the reverse functions

In some cases like this in integrating too

[fossil cobalt]

i dont understand this

can u please give an exmaple

[opaque dirge]

lets say x²+4x+4

its just (x+2)²

but in the case of x²+4x+6

its (x+2)²+2

this is the sum of a square

What im trying to say is

we would've taken another road if it was something like squareroot((x+1)²+2) since we cant get this out of the square root

[fossil cobalt]

thanks alot

[opaque dirge]

i hope it helped

no pun intended

[fossil cobalt]

sorry but
can i write √(1+4x^2) a 1+2x ?

i cant imagine am struggling with algebra

[opaque dirge]

no you cant

[fossil cobalt]

why

and if not how could i simplify it

[opaque dirge]

since (1+2x)²=1+4x+4x²

[fossil cobalt]

oh yea

damn am struggling with algebra

:)!)!)!)!)

[opaque dirge]

i know you'll fix the problem, i feel like you only need more practice

[fossil cobalt]

my test is tmr

and its 12 am for me

starting to lose hope

but ill push forward

thank you

[opaque dirge]

same timezone huh

you are welcome

[fossil cobalt]

sorry another question 😭

[opaque dirge]

go for it

[fossil cobalt]

√(2+4x+4x²)

i wanna integrate that

using completing of squares?

i do (2x+2)^2

i think yes

[opaque dirge]

thats incorrect

[fossil cobalt]

why

[opaque dirge]

(2x+2)²=4x²+4x+4

(2x+2)²-2 was right

[fossil cobalt]

2

[fossil cobalt]

where did the 2 here go

oh 2-4

= -2

[opaque dirge]

mhm

[fossil cobalt]

now how do i integrate that 💀

√((2x+2)²-2)

integral of that

[opaque dirge]

4(x+1)²-2

maybe we can do an u subtitution to get rid of that -2

[fossil cobalt]

wont work

i was integrating this

[opaque dirge]

can you even take that integral?

like, is it even possible

[fossil cobalt]

no idea

[opaque dirge]

if that x² at the beginning was an x, then yes we would've been able to take the integral

[long basin]

Ts a trig sub

Hate to be the bearer of bad news

[fossil cobalt]

nah i quit

[long basin]

You're gonna have to let x = 1/2 sec(x)

[opaque dirge]

oh i see

[fossil cobalt]

i hate trig subs so much

thanks tho

[long basin]

Don't worry

Everybody hates them

Every time I have to do a trig sub my will to live decreases

[opaque dirge]

always forget trig sub exists ngl

[long basin]

Peter griffin no likey

[opaque dirge]

understandable have a nice day sir

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[paper willow]

Let ( f ) : ( \left(-\frac{1}{2}, +\infty\right) \to \mathbb{R} ):
[
f(x) =
\begin{cases}
\frac{\ln(2x+1) + \cos(x^2) - 2x -1}{x} & \text{if } x \neq 0, \
0 & \text{if } x = 0.
\end{cases}
]
Find, if it exists, ( f'(0) ).

[solid kiln]

milanesa de pollo

[limpid dawn]

You first need to see if f is continuous at x = 0

You may use L'Hopital quickly for this one to check

[paper willow]

alright

.close

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[queen dawn]

Is this the correct set up for the outward flux?

[ionic pendant]

the integrand is correct, but your bounds are not

[queen dawn]

oh

would it be y=x (top) and y=0 bottom for the y bounds?

?

[ionic pendant]

that works

[queen dawn]

okay thank you

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[vale abyss]

I am confused on why my N doesn’t work

[vale abyss]

epsilon is .001

Oh it should be 1/1000

But it still doesn’t work

I’m assuming I can’t make it 1/N

I changed it to 35/N and it works

.close

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[kindred pier]

I have no idea what this is saying. Why do I need axiom of choice to say cartesian product would be non-empty, and what does this H function have to do with it?

[limpid dawn]

the almighty has a doubt ⚔️

[kindred pier]

fix it

I'll be disappearing soon, so may ask again later.

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@kindred pier Has your question been resolved?

[kindred pier]

Someone smart please help me

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@kindred pier Has your question been resolved?

[crude cipher]

In regular mathematics terms, H is an I-indexed family of non-empty sets.
The cartesian product of (H_i)_{i \in I} is the set of all (h_i)_{i \in I} for which h_i \in H_i.
So saying that for any such choice of H the cartesian product is non-empty is essentially saying that every family of non-empty sets has a choice function, which is probably the most common formulation of AC.

[kindred pier]

So I need to AC to say Cartesian product of non empty sets is non empty? What's wrong with $\exists (x\in X) \exists (y\in Y)\to (x,y)\in X\times Y$?

[solid kiln]

SWR

[crude cipher]

That only proves that the cartesian product of two nonempty sets is nonempty

By induction you may also prove that the cartesian product of any finite family of nonempty sets is nonempty

The statement equivalent to AC talks about all families though

[kindred pier]

But isn't that what the first sentence is saying in my image? "The Cartesian product of non empty sets is always non empty". They're making it sound like I need AC to be allowed to say that

[crude cipher]

cartesian product here refers to the general version (of arbitrary families of sets), not the binary product

I.e. ∏, not ×

[kindred pier]

I don't think i follow.
What's the first one?

[crude cipher]

Second sentence here, though I'm sure the book must include a similar definition earlier in the text if they're talking about the cartesian product formulation of AC

[kindred pier]

I'll have to review the book but I did not see that definition

Oh here we go. I never read this section

Okay that probably answers all of my doubts