help-23

[frigid osprey]

isnt it already done?

[silent forum]

We can prove this in several ways

[frigid osprey]

The easiest

[silent forum]

Altitude nope

Ok denote the small angles that are formed by the radii and the sides

QAB=QBC=c

And similarly for the other 4

Those being a and b

[frigid osprey]

Why

if c is a lenght

[silent forum]

Just so we can manipulate them

[frigid osprey]

and QAB a triangle

[silent forum]

We dont need the length

[frigid osprey]

So what c mean

if its not a lenght

[silent forum]

c is the measure

Of angle QAB

[frigid osprey]

wdym measure

[silent forum]

How many degrees/radians

It has

[frigid osprey]

oh okay

[silent forum]

We dont know so we're using a variable

[frigid osprey]

lets call it x

as i do have c already

[silent forum]

That's a bit messy

[frigid osprey]

these small angles are the same on each one?

[silent forum]

Lets use c for the measure and discard it for the length

As we wont use that

[frigid osprey]

oki

[silent forum]

Ye cuz the triangle is isosceles

So they come in pairs

[frigid osprey]

okay

[silent forum]

Of equal anglrs

[frigid osprey]

so c is equal in A both sides

[silent forum]

Two will be c, two will be a, and two will be b

[frigid osprey]

and in B both sides

oh

okay

i got you

[silent forum]

Aight

Now we know that they sum up to 180

So 2(a+b+c)=180

[silent forum]

As the angles in a triangle sum up to 180

[frigid osprey]

oop`s

[silent forum]

And we can prove thid as well

[frigid osprey]

Nah i know this dw

[frigid osprey]

like this?

[silent forum]

Yes

You got it

Now recall that we wanna show

That angle AQB is twice ACB

We know ACB

But what about AQB

Can you find it?

[frigid osprey]

Ok I see AQB

[silent forum]

Yes

But what is its measure?

In terms of a b c

[frigid osprey]

Um

2radii and c

Oh you mean the angles

2c and the other that I dont know

[silent forum]

Yep

But recall

That we know they add up to 180

So knowing 2 angles in a triangle really tells us all the angles

[frigid osprey]

180 -2c?

[silent forum]

Yes

But also

[frigid osprey]

Hm

[silent forum]

We know this

[frigid osprey]

Yeah

[silent forum]

So

Are we done?

[frigid osprey]

So each angle

Is 30

No

[silent forum]

Nop

a b and c are arbitrary

Be careful here

[frigid osprey]

Yeah

[silent forum]

As we are looking at ANY triangle

[frigid osprey]

Ye ye

[silent forum]

Ok so

What is AQB

In terms of b and a now

As ACB is in terms of b and a

[frigid osprey]

the angle QAB is th sum of 2a+2b

[silent forum]

Yup

[frigid osprey]

Yay

[silent forum]

But its AQB

Not QAB

[frigid osprey]

wdym

[silent forum]

There is a difference

Angle QAB has A as a vertex

[frigid osprey]

Ohhhh

ok i thought it was Q the vertex

[silent forum]

Yeah

Q is

[frigid osprey]

Okay

[silent forum]

Now back to our problem

[frigid osprey]

I meant the one with q vertex

oki

[silent forum]

Draw an altitude

To any one of the sides

[frigid osprey]

From where

To where

[silent forum]

It doesn't matter

Try to figure it out

:))

What point

Is very useful

Here

[frigid osprey]

Q

[silent forum]

Yes

So drop from it

To any side

Whichever you like

[frigid osprey]

okay

Like this

[silent forum]

Yup

[frigid osprey]

Okay now this line is the sin of this new triangle

this red line/AQ

[silent forum]

Not quite

[frigid osprey]

?

[silent forum]

Well

We're looking for the sin

True

But is that the angle we should be taking the sin of?

[frigid osprey]

I dont really know

xd

[silent forum]

Let's denote the point where our perpendicular touches AB with D

[frigid osprey]

Whats D

[silent forum]

Simply a point

Red line

Intersected with AB

We're giving it a name

So we can talk about ut

[frigid osprey]

this one

[silent forum]

Yes

Now, I'll give you some time to think about it

As you're really close

And i dont want to spoil the whole solution to you

And take away the satisfaction of finishing it and seeing it all connect

So take ur time

And pin me if u need help

Alright?

[frigid osprey]

Okay

[silent forum]

You'll need to find sin of ACB in terms of things we know here

Like the sides

Or the radius

[frigid osprey]

Okay

@silent forum

So we can do the same with the other angles

and get the 3 sides

[silent forum]

Careful

If your finding cos

Then it wont be that

[frigid osprey]

I dont get u

[silent forum]

But yeah once you find one sin the other 3 are found exactly the same

I mean

[frigid osprey]

okay

[silent forum]

Cos c is DB/R

Where R is the radius

[frigid osprey]

Okay

[silent forum]

But dont be fixated on these angles we've used variables for

[frigid osprey]

Oh you are right is B

[silent forum]

You need to find sinA

sinB

sinC

[frigid osprey]

Okay

[silent forum]

How do we find sinC

In triangle AQB

Thats a big hint

Btw

[frigid osprey]

What

How could you do that

[silent forum]

Magic of math

C is related

To an angle

[frigid osprey]

AB/BC is the sin

[silent forum]

In AQB

[frigid osprey]

AQB i guess

[silent forum]

Nay, we dont know ABC is a roght triangle

[silent forum]

How?

[frigid osprey]

Its connected with a line xD

[silent forum]

:))))))

Look up

We talked about this

And even proved

Some property

In circles

Involving angles

[frigid osprey]

180degree

2a+2b+2c

[silent forum]

Thats the sum

Of all the angles

Now we want a relationship between C (angle ACB) and some angle in triangle AQB

As some motivation for why exactly we want this, note that it's in triangle AQB that we've split into right triangles which are very handy for finding sin

[frigid osprey]

So

the angle of C

is

the same as 2c

because

C is split

in 2

so

we got them right there?

[silent forum]

Hmm

I dont quite get

What you mean

[frigid osprey]

2 times c

is C

[silent forum]

Try to get a habit of using 3 verices

To denote an angle

[frigid osprey]

well ACB

[silent forum]

As thats unambiguous

Aight

[frigid osprey]

the angle c 2 times is ACB ?

[silent forum]

Nop

Not c

[frigid osprey]

then it just can be Q

well AQB

[silent forum]

Yes

Because Q is the center

Of the circle

[frigid osprey]

Alright

[silent forum]

Ok

Now you have

All you need

Just put it together

And you're done

[frigid osprey]

Okay let me try

but i cant get the sin here

As i dont know the bottom yellow line

[silent forum]

Good

You're going in the right direction

We do know AB

As a side

[frigid osprey]

Do we?

Oh

okay

[silent forum]

Yeah

That we can use

No problem

Otherwise we dont have enough info to express the sin

So just write AD or DB in terms of AB

[frigid osprey]

why To D?

[silent forum]

Wdym?

[frigid osprey]

I mean

why not Sin AQB = AB/QB

[silent forum]

I see

You have a misunderstanding

Of the sin

We need a right triangle

[frigid osprey]

Oh

Oh

[silent forum]

As its defined as the opposite side over the hypotenuse

[frigid osprey]

I see

okay now i get u

[silent forum]

Aight

Let me know what answer you get

[frigid osprey]

okay give me some time

[silent forum]

Btw theres a slicker proof

But ill show it to you after

[frigid osprey]

Ok i cant get past here

[silent forum]

Yes

Now express those in terms of the sides

Radius

And angles of our triangle ABC

[frigid osprey]

So sin of C = 2radius /c

[silent forum]

Whats C

You mean AB?

Be careful with notation

As its a bit ambiguous

[frigid osprey]

Yep

[silent forum]

Aight

Thats the answer

Poggers

Now if you take a look at that

[frigid osprey]

ohhh

god

thanks xD

[silent forum]

You can write 2R as sin(C)/AB

And do the same for all the angles&sides

[frigid osprey]

aha

[silent forum]

Thats the sine theorem

[frigid osprey]

SinC/AB?

[silent forum]

Glad to help

[frigid osprey]

isnt it inverted?

theorem isnt AB/SinC

?

[silent forum]

Oh yeah

You miswrote

[frigid osprey]

probably xD

[silent forum]

Its AB/2R

For the sine

[frigid osprey]

Why?

[silent forum]

Look at your previous photo

[frigid osprey]

Alright

im not good at seing the sins

i can see the sin and cos just if the angle is on left corner

if not idk how

[silent forum]

Thats just cuz u didnt have enough practice

If theres smthing i want you to take away from all this

Is that math really requires effort and practice

And its not you thats the problem

[frigid osprey]

i noticed hahaha

[silent forum]

Yeah just be confident

And it will work out

Btw for the second slicker proof

If you want

I can show it to you

We already have all the stuff we need

[frigid osprey]

Alright

Go ahead

[silent forum]

Kay

So

First off

Draw the diameter from B

And connect it to A

[silent forum]

With another segment that is

[frigid osprey]

Okay

Now?

[silent forum]

Show me the drawing :))

To make sure you wont be confused

[frigid osprey]

like this?

[silent forum]

Yes

Now recall that cool property

With the double angle

At the center

[frigid osprey]

Where

[silent forum]

If you denote the intersection of the diameter with the circle with T

The other intersection besides B

[frigid osprey]

okay

[silent forum]

We have angle BQT=180

As its a straight line

[frigid osprey]

Q was the middle?

[silent forum]

And its an angle with its veryex at Q

Yup

So an angle that also rests on B and T

But its vertex on the circle

Is 90 degrees

This is a very useful particular case

Any angle with its veryex on the circle that rests on the diameter is 90 degress

So we have our right angle

Also

Again from our property back

Any two angles that rest on the same chord

Are equal

So ACB is equal to ATB

[frigid osprey]

Why? It doesnt make 180 it makes 360 then

[silent forum]

Careful :)

Its the angle at the center

Thats doubled

[frigid osprey]

Aha

[silent forum]

Ok so now sin ACB =sin C

Will be just sin ATB

Which we can easily find as this angle is in a right triangle

And it is just AB/2R

[frigid osprey]

Oh so acb should be the same as atb

Alright

That one was easierrr

[silent forum]

Yup

I like this one more also

[frigid osprey]

I do too lol

[silent forum]

Kay, hope you got more interest in math from this

And dont be afraid of it lol

[frigid osprey]

I still dont get why is ACB the same as AQB

[silent forum]

Math is hard, but thats why its cool

[frigid osprey]

Im not, just that what i learn in school is usually boring

[silent forum]

You mean same as ATB?

[silent forum]

It is, that is very true

School math does a disgrace

To the real thing

[silent forum]

ATB is ACB

Thats cuz

Well, let me introduce some vocabulary so that its easier for me to convey this

[frigid osprey]

No no, in the first one

[silent forum]

Aight

In the first one

AQB is twice

ACB

They are not equal

[silent forum]

Well i myself am not ashamed to admit that math does scare me

[frigid osprey]

I know, I like programming (that involves math) and I really enjoy it there but in school is Boeing af

[silent forum]

Its more of an awe feeling

[frigid osprey]

What? I mean AQB not DQB

[silent forum]

Mm

Im confused now as to what exactly your question is

But im glad you're asking questions

[frigid osprey]

I mean, I dont get why ACB is Same (or double) as AQB

[silent forum]

We proved it b4

Long ago

Lol

[frigid osprey]

I didnt understood it

[silent forum]

Here somewhere

Well you showed that the angle at the center

Is 2(b+a)

And the angle with its vertex on the circle

Was b+a

So its double

You'll have to go thru this all again carefully

And redo the proof

[silent forum]

Aight ill make a dinner for myself, but let me know if after this smth is not clear

[frigid osprey]

So its because its 2a+2b?

[silent forum]

Yes

We expressed one angle as a+b

And the other as 2(a+b)

And these a and b were just placeholders

[frigid osprey]

Aha

Ahh

I see

Okay

Got it

[silent forum]

Epic!

[frigid osprey]

Thanks you!

[silent forum]

Glad to help!

.close

You shoudl write this btw

[frigid osprey]

enjoy your meal

Yeah I know

See u :)

.close

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[stone osprey]

How to measure mass

[inner tapir]

use a scale

[safe radish]

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[brittle crane]

Can someone help me with this question?

[lean otter]

Fermat’s little theorem

[brittle crane]

thank you

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@brittle crane Has your question been resolved?

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[flint coyote]

can\ someone help please

[elfin yew]

$$5x+2 = 180 - (4x-3) - (x^2 -1)

I don't know how to do the LaTeX

[flint coyote]

oh

do you know the equation i could use at least?

[elfin yew]

5x+2 = 180 - (4x-3) - (x^2 -1)

[flint coyote]

thanks

[elfin yew]

Because the sum of a triangle is 180 degrees

[flint coyote]

yes

<@&286206848099549185>

can someone walk me through the process

[elfin yew]

The sum of the angles of a triangle is 180 degrees, correct?

[flint coyote]

yes

[elfin yew]

And in this image a + b = 180, right?

[flint coyote]

yes

[elfin yew]

So this means that ∠DAB + ∠DBC = 180

Right?

[flint coyote]

yes

[elfin yew]

So ∠DBC = ∠DBA - 180

[flint coyote]

yea

[elfin yew]

Now,

∠DBA = 5x+2

Therefore, ∠DBC = 180 - (5x+2)

[flint coyote]

oh i understand

[elfin yew]

So now you have the 3 angles in terms of x

[flint coyote]

yes

[elfin yew]

Now add all of those angles, set them equal to 180, and solve for x

[flint coyote]

ok

ill do that

ty

.close

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[wind anvil]

How do I use solve it

Let me show you how much far I have reached

[inland ivy]

Multiplying exponents of the same base add exponents

Dividing exponents of the same base subtract exponents

[wind anvil]

Ik those rules but

Now what

Should I do

@inland ivy

Have I done smth wrong

??

<@&286206848099549185>

[safe radish]

@wind anvil Has your question been resolved?

[wind anvil]

No

It hasn't

[still abyss]

yeah now you can expand

[wind anvil]

Can I do like this

??

[still abyss]

that seems wrong wait

[wind anvil]

Hm

Ok

I'll wait

[flat frigate]

HELLOBELLO
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[still abyss]

@wind anvil

[wind anvil]

Ok

[still abyss]

now u can use the exponential identity $x^{a} * x^{b} = x^ {a+b}$

[flat frigate]

HELLOBELLO

[still abyss]

it solves quite nicely

[wind anvil]

Thanks

[still abyss]

ur welcome

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[lean otter]

How

[flint ledge]

what?

[lean otter]

Nvm lol I got it

Thanks

.close

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[cyan marlin]

How is this found?

[inland ivy]

How is wot found?

[cyan marlin]

Oh my bad

I was playing with this simulation and i am unable to understand how 230° is found here.

@inland ivy hope that helps

Got it

.close

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[ruby plume]

<@&286206848099549185>

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[ruby plume]

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[rain inlet]

what can't you understand?

@fast spoke

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@fast spoke Has your question been resolved?

[pulsar condor]

I've also helped you b4 w/ this.

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[fast spoke]

😄 yea I just remembered lol

[shy glade]

.close

L

[fast spoke]

.close

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[foggy loom]

Does anyone know what im doing wrong here?

[proper crypt]

@foggy loom You did it right

I think.

Wait

[foggy loom]

ok

[proper crypt]

You forgot the dx.

From start to finish

[foggy loom]

oh yea

but thats just adding a dtheta at the end its not gonna change the integration numerically, is it?

[proper crypt]

No

When you plugged in sec(theta), dx is still dx.

You have to find what dx is.

x = sec(theta)
so
dx = sec(theta)tan(theta)d(theta)

Because the derivative of sec(theta) is sec(theta)tan(theta).

You have to plug that in.

[foggy loom]

ah ok wait one sec let me do that

yea that seems more right now

i forgot to change dx to dtheta

@proper crypt thanks

[proper crypt]

np!

[foggy loom]

.close

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[whole ridge]

Very confused by this bit of text. It states that an ordered field has to meet 2 conditions. The first condition is then described and called closure but I thought a closed field was one in which the root of all polynomials within the field also exists within the field. This messes up the definition of the Real numbers because as far as I'm aware, the Real numbers are an ordered field but not a closed field. So is closure used to mean something different here, or am I messing up one of my definitions?

[lean otter]

Personally I’ve never seen “algebraically closed” be called “closed”

But why does it even matter, the property is clearly laid out in 1), doesn’t matter what else you associate with the word closure

And nothing they wrote is unusual anyway

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[proper crypt]

@candid depot First, in order to find the gradient/slope of the function y = 2x^2 + c, what do you need to do?

[candid depot]

find the derivative?

so 4x+'c = -8?

[proper crypt]

yes

but

c is a constant, a number

[candid depot]

oh

right

[proper crypt]

so when you take its derivative, it becomes 0

[candid depot]

so just 4x then

[proper crypt]

yup

4x

[candid depot]

so then x is -2?

[proper crypt]

what is x?

[candid depot]

-2?

[proper crypt]

yeah, but what else?

the slope is -8 WHERE?

[candid depot]

idk

[proper crypt]

at what x?

it's given in the question

the gradient at the point (p, 5)

so what's our x?

[candid depot]

-2?

uH

[proper crypt]

Look,

y` = 4x

we know that at (p, 5), the gradient is -8

[candid depot]

yeah

[proper crypt]

so what do we plug in, other than y` = -8?

we plug in x = p

because the gradient at the point (p, 5), meaning at x = p, is -8

so 4p = -8

[candid depot]

oh i see

[proper crypt]

makes sense?

okay.

[candid depot]

yeah

[proper crypt]

so what does p equal?

don't overthink this

what does it equal?

you've said this already

[candid depot]

-2?

[proper crypt]

yeah

p = -2, great.

now let's find c.

Do you have an idea on how to do this?

Our point now becomes (-2, 5)

[candid depot]

and then what do i do

plug it into the derivative?

[proper crypt]

No, you already know both the slope and the point

there's nothing else to do with the derivative.

Let's look at the function instead

the point (-2, 5) is on the function, correct?

[candid depot]

yeah

[proper crypt]

so that means?

[candid depot]

you sub -2 into the function?

[proper crypt]

yeah, but what else?

the x is -2

what is the y then?

[candid depot]

5

[proper crypt]

good

then plug those in

and you'll get an equation with only c

to sum up this part:
since (-2, 5) is on the function, then we know it must satisfy the equation
y = 2x^2 + c
and therefore
5 = 2*(-2)^2 + c

[candid depot]

so this?

[proper crypt]

yup

[candid depot]

oh

tysm!

[proper crypt]

np!

[candid depot]

.close

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[dry basin]

how i proof that the f(x) is bounded in R>0

[lean otter]

What is f?

@dry basin

[dry basin]

f is a function continuosus and the function are logarithms

[pulsar condor]

$f(x)\leq g(x)$ for some g

[flat frigate]

Mosh

[pulsar condor]

is bounded above.

[lean otter]

For some constant function g?

[lean otter]

That doesn’t sound too promising

[dry basin]

the problem say proof that g(ax)=g(x) for all

$x\in$R_+

and that implies that g is bounded

[flat frigate]

elinder
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lean otter]

I have no idea what you are asking anymore

Can you define all the letters you are using and type the question in full, or post the problem as it was originally stated, in full

[dry basin]

let me translate the problem

and g(x)=f(x)-logx

[lean otter]

I still don’t really understand the question

[dry basin]

the question is how to proof that g is bounded

[safe radish]

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[north niche]

this is whole thing

i need help with d and e

[safe radish]

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[lean otter]

how many unique ways are there to arrange the letters in the word PRIOR?

[lean otter]

i know i have to use the permutations fomula

and im not sure if 5!/(5-4)! is right

[ember bough]

(5-4)! is just 1

[lean otter]

how do i use the formula when the r's repeat?

[ember bough]

you divide by (number of repeated letters)!
if only 1 letter is repeated

[lean otter]

so would it be 5!/(5-2)!?

[ember bough]

no

i mean

divide by 2!

since there are 2 r's

[lean otter]

what do i divide with 2?

i dont really get it

[ember bough]

i mean
5!/2!

since there are 2! ways to arrange the repeated R's

[full girder]

Let me intervene, I'm sorry if it was a disruption.

Let's say, pyre, you have two As. In how many ways you can arrange them?

[lean otter]

only one

[full girder]

Nice.

Now if I say

A1, A2

Then?

[lean otter]

2

[full girder]

Exactly

Now what we did here is permutations

See when the As are same,

The sortings are divided by exactly the number of repetitions

[lean otter]

okay like 2/2=1

[full girder]

Exactly

We can do this for let's say 5

A1 a2 a3 a4 a5

How many ways you can arrange them?

[lean otter]

5!/(5-5)!

5!

[full girder]

Perfect

If there was 3 As, and 2 Bs

Then?

If you can do this, you can do your problem

Because same principle

[lean otter]

i dont really know but would it be like 3/3 and 2/2 which is like a and b so like 2

[full girder]

Nooo

Let me lead you

In total there are 5

So, you can arrange them in 5! Ways

Now

There are repetitions

When you see there is an element repeated 3 times, you see the original count comes down by a factor of 3!

But wait, there's also B which is repeating

2 times

So in total, the 5! Should come down by a factor of 3! × 2!

Tell me if you don't understand this

[lean otter]

i dont understand sorry

im looking at this and i dont understand how it would correlate

[full girder]

Oh ok no problem, let me take the A1 A2 A3 A4 A5 one. You said you can arrange them in 5! Right?

[lean otter]

yeah

[full girder]

If all the As are same how many ways?

[lean otter]

1 but i kinda just thinking about if theres only 1 then only one combination not formula stuff

how would i enter it in the formula if it was all the As the same?

[full girder]

You can show that in formula as well?

5!/5!

5 arrangements, 5 repetitions

If it's like A A A B C

You do it like

5!/(3! × 1! × 1!)

[lean otter]

does n or r get affected when theres repetitions of letters?

would n stay the same in PRIOR like 5

[full girder]

The thing is the usual formula of permutations is a bit skewed for this type of arrangement

[lean otter]

i dont know how the number of selected changes when theres repetitions

[full girder]

The NpR and NcR is not the basic stuff for permutations

Npr assumes all elements are distinct

And Ncr as well

[lean otter]

what would be the formula for this?

[full girder]

Ok let me write a latex equation just bear with me until the result prints here

$\frac{n!}{a! b! c!.... }$ where, n = total number of things you want to arrange, (a, b, c...) = The specific number of appearences of all elements.

[flat frigate]

rikusp2002

[full girder]

Like, in the 3A, 2B case

a = number of appearences of A = 3

b = number of appearences of B = 2

Total = n = 5

So

$\frac{5!}{3!2!}$

[flat frigate]

rikusp2002

[full girder]

Let's tackle your original problem in the same way

PRIOR

[lean otter]

5!/1!1!1!2!?

[full girder]

Perfect

Niceeee

[lean otter]

yaay

[full girder]

You got it

🔥

[lean otter]

how do i add the factorials?

multiply

[full girder]

Yes multiply

[lean otter]

like regular numbers

[full girder]

3! = 6

2! = 2

So answer is 5!/2!

That is 120/2 = 60

[lean otter]

thank you!!

[full girder]

🥳 you're welcome

.close

[lean otter]

.close

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[lean otter]

Can someone tell how the line xy=0 gives this graph, I had this question
"The orthocenter of the triangle formed by the lines xy = 0 and x + y = 1 is"

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

np, seems no one is there to help

.close

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[heavy glade]

what am i doing wrong for the average velocity

[cinder viper]

@heavy glade avg velocity = (final velocity - initial velocity) / time taken

[heavy glade]

i did

it's velocity at 0 - velocity at 3/ 3

[cinder viper]

velocity at 3 - velocity at 0 / 3

so (16/3 - 0)/(3-0) = (16/3)/3 = 16/9

[heavy glade]

wait

velocity at zero is -2.3

[heavy glade]

say's this is also wrong

[cinder viper]

i thought ur answer was correct

velocity at 0 is -2/3

so you have [(16/3) - (-2/3)] / 3

[heavy glade]

ye that's what i had

which becomes 6/3 = 2

it says it was 7/3

i dont get how it is that

[cinder viper]

[(16/3) + (2/3)]/3 = [(16 + 2)/3]/3 = (18/3)/3 = 6/3 = 2

[safe radish]

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[celest lily]

Hi ! Can you explain me pls that why is the remainder (r) smaller than m-1 ? I understand that r has to be smaller than m, because it is the remainder... But why m-1 ??

[native olive]

It's between 0 and m-1 inclusive. So it's one of {0, 1, ..., m-1}

[celest lily]

ooohhh, okay I understand now, thank you !! 🙂

[native olive]

No problem!

[celest lily]

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[kind dove]

how to create correct proof of transitivity ?

$(\forall a, b, c)[aRb \wedge bRc \implies aRc]\
a-b = 4 \wedge b-c = 4 \implies a-c = 4$

[flat frigate]

Michal

[kind dove]

$b = a-4 \wedge a-4-c = 4 \implies a-c = 4$

[flat frigate]

Michal

[kind dove]

but from $a - 4 -c - 4 = 4$ we get $a-c = 8$ and this is contradiction with $a-c = 4$

[flat frigate]

Michal

[kind dove]

is this correct ?

[quasi bison]

overcomplicated

a - b = 4 and b - c = 4 imply a - c = 8, therefore the relation is not transitive.
that's it.

or even more bluntly: give a counterexample

(100, 96) and (96, 92) are in R, but (100, 92) is not

[kind dove]

Ok thanks

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[bright finch]

find x and y:
(x+y)^2 + (y-2)^2 = 0

[split ether]

Generally $a^2 + b^2 = 0 \iff a = b = 0$

[flat frigate]

Touch Our Beans

[split ether]

We can use this fact to argue that here x + y = 0 and y - 2 = 0

So y = 2

Therefore x + 2 = 0 and x = -2

So the only solutions are x = -2 and y = 2

[split ether]

If you also want the proof of this, I can show you

[bright finch]

ye this is very easy

thank you

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[grim light]

Name the property that is shown in each expression.

[grim light]

Hi

Name the property that is shown in each expression.

[void flame]

what part are you confused by

[grim light]

I need the name of the property of each expression

and idk

[void flame]

what do you notice about each expression

what seems to be the common thing

[grim light]

with each expression i need to know

who its the name

[void flame]

forget about the name for now

how would you describe the property being used

like what's happening

[grim light]

idk anything bro

i say you im so bad in this

[void flame]

what property allows you to say 8*2 = 2*8

[grim light]

sorry

i resolve that already

i need help with other its most easy

[safe radish]

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[potent crane]

Hi, I have a problem I can't seem to understand about splitting a sum

[lean otter]

send your question and someone'll get to it

[potent crane]

Yup I'm trying to properly write my problem 😉

So,
There is 2830 I need to split between 11 people for 7 nights trip BUT
8 persons stay 7 nights, 3 persons stay 2 nights
That would be 2830 / 7n ~= 404.29 (Don't worry about rounding numbers please)
404.29 * 2n = 808.58
808.59 / 11p ~= 73.51
So 73.51 for the 3

For the 8 rest:
(404.29 * 5n / 8p) + 73.51 ~= 326.19

But I can't do it on my splitting app:

Any idea why?

Sorry if anything is wrongly explained, feel free to ask more info

[safe radish]

@potent crane Has your question been resolved?

[safe radish]

@potent crane Has your question been resolved?

[safe radish]

@potent crane Has your question been resolved?

[safe radish]

@potent crane Has your question been resolved?

[marsh walrus]

ooo

this looks like a fun application problem

so is the question like

@potent crane so for 5 nights, there's 8 people

then for 2 nights, there's 3 additional people, for a total of 11?

[potent crane]

That's right

[marsh walrus]

and the total cost of booking this place is 2830?

for all 7 nights

[potent crane]

Yep

[marsh walrus]

okay, so here's what i would do

2830/7

this is just the raw cost of the place per night

,calc 2830/7

[flat frigate]

Result:

404.28571428571

[marsh walrus]

lets round 404.3

now you by nights, like this

for a night where there are only 8 people, each person pays (2830/7)/8

,calc 2830/(7*8)

[flat frigate]

Result:

50.535714285714

[marsh walrus]

there are 5 nights like this

on a night when there are more people, each person pays less

,calc 2830/(7*11)

[flat frigate]

Result:

36.753246753247

[marsh walrus]

so with 8 people in the house, each person should pay $50.54 per night

when there are more people, each persons share is less

so 11 people in the house, each person pays $36.75 per night

then all you need to do is tabulate for each type of person

make sense so far? @potent crane

[potent crane]

Wait

Ok yeah I see

[marsh walrus]

so its like each day, you take the cost of staying in the place for that day, 2830/7, and you divide it up among everyone on the table

[potent crane]

That's 50.4*5 + 36.75*2 for the 8 then ?

[marsh walrus]

yea

and just 36.75 * 2 for the 3

which if you think thats fair or not is up to you

but this is i think the sensible math way to do it

[potent crane]

Yeah yeah that's fair

[marsh walrus]

this is straightforward too if a person decides to cancel or come a day early too

you just recalculate that day, ez pz

or say you still have to pay

up to you

[potent crane]

Haha of course ;)

The cost per night for the 8th is cool thanks!

Do you know how it would work in the splitting app?

Because 7 parts of 2830 for the 8 and 2 for the 3 doesn't add up

[marsh walrus]

do you have to use the app?

id just use a calculator tbh

oh wait

lmc

[potent crane]

Nah it's not necessary it's just for other to see fast

[marsh walrus]

,calc 50.5485 + 36.7582 + 36.7532

[flat frigate]

Result:

2830.1

[marsh walrus]

adds up to me

idk about the app

[potent crane]

Ah I was talking about the app

[marsh walrus]

it should be 50.54, times 8, times 5

plus 36.75, times 11, times 3

[potent crane]

The app works like this:
2830
87 parts (of 2830)
32 parts
So that's 62 parts reparted

I thought it would be good

But it's not

[marsh walrus]

just use it for two different bills

2830/7 * 5 is your first bill

split that by 8

send that to the 8 people

do another one for 11 and 2 and send that to everyone

[potent crane]

Yea that's a better idea

Well thanks a lot!

[marsh walrus]

np

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[torpid fiber]

Hey I have a equation that I can't solve, can someone help?

[worthy hemlock]

My first suggestion is to multiply everything by the common denominator to get rid of all the fractions

[torpid fiber]

I already did that, at the end I reach that:

5ax + 52a^2 = 2x^2 -23a^2

I don't know what to do from here

[worthy hemlock]

I think you might have messed up some math on the left side

[torpid fiber]

I managed to do it, thanks anyways!

.close

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[lean chasm]

why is $$e^{\frac{\ln^2 x}{\ln x}} = x$$

[flat frigate]

illuminator3

[fair dragon]

You can write $ln^2x$ as $(ln(x))^2$

[flat frigate]

DarQ

[fair dragon]

Do you see it now?

[lean chasm]

yeah

thanks