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[tardy lagoon]

calculate the differences betweeen the terms

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@frozen sierra Has your question been resolved?

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[cloud lake]

Hello all, I am trying to convert my context-free grammar, which i obtained from a DFA, to Chomsky normal form. The rules that i need to use are:

Step 1: Remove the start symbol from RHS

Step 2 : Remove ε from RHS

Step 3 : Remove unit rules

Step 4 : Shorten the RHS, add variables.

This was my attempt, would anyone be able to help me out to see if its correct or if im doing anything wrong.

·• S -> 0A | ε
• A -> 0B
• B -> 0A | 1C
• C -> 1D
• D -> 1C | ε
Step 1 : Remove the start symbol from RHS
• There is no start symbol on the right-hand side, so no need for change.
Step 2 : Remove ε from RHS – Remove D -> ε
• S -> 0A | ε
• A -> 0B
• B -> 0A | 1C
• C -> 1D | 1
• D -> 1C
Step 3 : Remove unit rules
• There are no unit rules that need to be removed.

Step 4 : Shorten the RHS, add variables.
• S -> EA | ε
• A -> EB
• B -> EA | FC
• C -> FD | 1
• D -> FC
• E - > 0
• F -> 1

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@cloud lake Has your question been resolved?

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@cloud lake Has your question been resolved?

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[graceful rose]

i dont really get how they factored this? whats the process because its not very clear to me

[tardy lagoon]

it's easy if you let $y=\frac{\lambda}{1+\mu}$ first

[clever fjord]

WhereWolf(ping if needed)

[tardy lagoon]

you get an ordinary polynomial

[graceful rose]

yea normally i think of it as (x+y)^2 =x^2+2xy+y^2 but that doesnt really work in this i dont think

[tardy lagoon]

wait you don't know how to factor?

[graceful rose]

wait hold on i do 😭

[tardy lagoon]

cross method

[graceful rose]

ok wait a sec

oh i see it now, just writing it out

[cold urchin]

[ \wrb{\df{\la^2}{2(1+\mu)^2} - \df \la{1+\mu} - 4 = \df12\p{ \df \la{1+\mu} -4} \p{\df \la{1+\mu}+2}} ]

the 1/2 doesnt matter because thats equal to 0.

[clever fjord]

Pure

[graceful rose]

wait so with y = lamda/1+mu

then we'd have 1/2y^2-y-4 right?
1/2y^2+y-2y-4
1/2y(y+2)-2(y+2)
(1/2y-2)(y+2)?

[tardy lagoon]

that works

[cold urchin]

you can factor out 1/2 from the first bracket.

[graceful rose]

even though it gives something slightly different than whats written

[graceful rose]

true, makes it easier

[cold urchin]

It's the same thing.

[graceful rose]

yea

okay thanks alot both of you, i would not have been able to see that otherwise even though it seems clear now

♥️

[cold urchin]

[graceful rose]

.close

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[tardy lagoon]

both works

I don't think there's a "best" way

they are the same thing written diferently

yep

?

what do you mean by "this"

[low locust]

if the exponent is just a number then you can just use the power rule

so you dont need to do this

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@heady badger Has your question been resolved?

[tardy lagoon]

when you have exponent taking log almost always works

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[thorny otter]

Why did the index here not shift ?

[thorny otter]

of why did the index not change starting position after the differential

I understand why its beneficial here to not change the starting index due to how the equation works out but is there something im missing as to why after differentiation the the index didnt change

Like here

Found an answer online that it's due to the method of frobenius we can start always from x=0

.close

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[normal sundial]

how did they factor out voc

[normal sundial]

.close

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[coral saddle]

Good day,

[lament walrus]

hi

[coral saddle]

I have the following question about probability

[lament walrus]

go ahead

!ask

[coral saddle]

i seem to not be able to get the answer to 😛

It is about a game.

i am currently farming. i require seeds to do so.

i have a 50% chance of keeping the seed with every action performed.

My question is, how many actions can i on avarage perform with 1 seed.

i have tried 2 things, which both end up with the same answer. roughly.

1 action takes 99 second. and generates 5 fruits.

so in 1 hour, 3600/99 i can perform 36,36 actions

With 50% of keeping/losing a seed.

That means i lose 18,18 seeds per hour.

and i gain 181,18 fruits per hour

Meaning i get 10 fruits per lost seed.

Meaning 1 seed would get me 2 actions.

The next thing i tried was.

Lets say i start with 1000 seeds. and i run 1000 actions requiring that 1 seed per action

i will be left with 500 seeds.

And i run that over and over

until i am left with 1< seed.

Which makes me come to 1998 seeds used

Before i get to lower then 1 seed.

Which is again 1,998 actions per seed.

[keen nova]

What i understood:

1. One action consumes one seed.
2. Every action has 50% of chance of not consuming the seed.
3. Each action gives 5 fruits.
4. Each fruits gives 1 seed.

Did i get it right?

[coral saddle]

1,2,3 Yes

4 no

fruits dont give seeds

Fruits is the end product

[keen nova]

Ok.

So the number of seed never grow.

[coral saddle]

No.

Only the 50% chance of saving the seed with every action

[keen nova]

So there are two cases starting with 1 seed?

Case 1:
I have 1 seed.
I have 0 fruits

... action perfomed ...
... the seed was not saved ...

I have 0 seeds
I have 5 fruits

.

Case 2:
I have 1 seed.
I have 0 fruits

... action perfomed ...
... the seed was saved ...

I have 1 seed.
I have 5 fruits

[coral saddle]

Yes

[keen nova]

In the case 1: only one action was perfomed.

In the case 2: at least one action was perfomed.

[coral saddle]

Yes

Well case 2

only 1 action aswell

[keen nova]

But it will do another one.

[coral saddle]

Yes

[keen nova]

It still has one seed.

[coral saddle]

yes

[keen nova]

Then the question is:
What is the avarage number of actions perfomed starting with one seed?

Ok. Now i get the problem.

[coral saddle]

oke thankyo

[keen nova]

Let's go.

Let's say that we knew the avarage number and it was some number n.

Let's also state the following: at every action we either have 0 or 1 seed.

[keen nova]

Proof: we started with 1 seed and we don't gain new seeds, we can only keep it or lose it.
The number stays the same (1) or decreases (0).

Do you follow?

[coral saddle]

Yes

[keen nova]

So each step of the thing looks like:

1 seed.
[Number] fruits.

Let's say that we are at some step.

What are the two possibilities from now?

About the number of seeds.

There are two cases: we lose the only seed we have or we keep it.

[coral saddle]

Yes

50% of keeping it 50% of losing it

[keen nova]

Exactly.

[keen nova]

How this affects the total number of actions?

[coral saddle]

Yes

[keen nova]

We are at the state:

X actions done so far
1 seed left

We can do one more action and lose the seed going to the state:

X + 1 actions done so far.
0 seeds left.

Or we can do one more action and not lose the seed going to the state:

X + 1 actions done so far
1 seed left.

Do you follow?

[coral saddle]

Yes

[keen nova]

Remember:
the avarage number of actions = N.

[coral saddle]

yes

[keen nova]

If we start the process, then how many, on avarage actions until it ends?

[coral saddle]

Yes

[keen nova]

N. I just write it.

It's N.

But there is other way of computing it.

50% of the time we lose the seed => 1 action was perfomed in total.

50% of the time we keep the seed => 1 action was perfomed and we get back to the previous state.

In avarage we take N actions.
In avarage: 50% of the time we take 1 action and the other 50% we take N actions.

Makes sense?

1. You, in avarage, makes N actions.

2. If you lose the seed, you do only 1 more action.

3. If you keep it you get back to 1)

. You lose
the seed
|
N = 0.5 × 1 + 0.5 × N
| |
| |
| you keep
| the seed
|
|
|
Avarage
Number
of actions

Do you follow?

[coral saddle]

Yes

it still makes sense

[keen nova]

Now you just have to solve for N.

N = 0.5 × 1 + 0.5 × N

N = 0.5 + 0.5 N.

N - 0.5 N = 0.5

0.5 N = 0.5

N = 1

In avarage: you will perfom only one action.

[coral saddle]

The answer is not making sense to me

[keen nova]

So i didn't explain the equation so well.

[coral saddle]

you are saying with 1 seed

on avarage i perform 1 action ?

While i have a 50% chance of saving that 1 seed ?

[keen nova]

Yes.

[keen nova]

Exactly.

Some times you will do 3.

Some times will do none.

[coral saddle]

none is not an option

1 seed will always perform 1 action

[keen nova]

Yes.

I see the moment i wrote.

[keen nova]

Always at least one action.

[coral saddle]

Yes but on avarage 1 action per 1 seed

is an impossible asnwer

[keen nova]

Now i see.

The N i calculated.

That equation.

The N does not represent the avarage number of actions in total.

The N does represent the avarage number of actions left.

In avarage you will have 1 action left.

If you always do at least 1 action.

[coral saddle]

So the answer is 2 actions on avarage per 1 seed ?

[keen nova]

Yes.

[coral saddle]

Oke that works out to the same math i have done

[keen nova]

Yes? The problem seems more complicated than what i think.

[coral saddle]

Yea, i couldnt figure it out

i did 2 ways of math

and i came to 2 actions per seed on avarage

But i wasnt sure if that was the right answer

[keen nova]

1 seed.
0 actions perfomed.

Then i do one action.

Case 1 :
0 seed.
1 action pefomed.

Case 2:
1 seed.
1 action perfomed.

First perspectitive: in avarage i will do N actions.

Second perspectitive:
50% of time i will do 1 action (lose the seed)
50% of time i will do N action (don't lose the seed).

N = 50% × 1 + 50% × N

50% of time i do 1 action.
50% of time i get back to the start (having 1 seed).

[coral saddle]

Yes

[keen nova]

Solving the equation gives N = 1.

It's 1.

[coral saddle]

But its 1, on avarage after the first action?

[keen nova]

I don't think so.

I think in avarage you do only one action.

[coral saddle]

But thats not possible.

if i use 1 seed per action

That means i dont save 50%, but i save 0%

[keen nova]

Exactly.

50% of the time you will lose the seed.

Half of the time you do one action.

25% (0.5 × 0.5) of time you'll do 2 actions.

[coral saddle]

Yes

So it cant be 1 action per seed on avarage.

[keen nova]

Red : do 1 action
Green : do 2 actions
Light blue: do 3 actions
Dark blue: do 4 actions.

[coral saddle]

Yes i know what you mean

50% chance for 1 action

25% chance for 2

etc etc etc

[keen nova]

You only perfom 2 actions 25% of time.

[coral saddle]

12,5% 3 times

6,25% 4 times

[keen nova]

And perfom 1 action 50% of time.

[coral saddle]

etc

[keen nova]

1 actions have the biggest frequency.

[coral saddle]

1 actions is 100%

with a 50% chance of saving a seed

1 action is 100%

with 1 seed

second action is 50%

3'rd is 25

i think that is where we are making the mistake

[keen nova]

What i am saying:

Pick one seed and do actions until having 0 seeds.

You will certanly do at least one action.

[coral saddle]

yes

50% chance for a second action

25% chance for a third action

[keen nova]

Yes.

[coral saddle]

So

On avarage how many actions, can i do with just 1 seed

It cant be 1 action

[keen nova]

What is the probability of getting exactly one action?

Wait.

1 seed.
0 actions.

... action perfomed ...

50% of time:
0 seeds.
1 action

50% of time we get exactly one action.

[coral saddle]

No

We always get atleast 1 action

so 1 action is 100%

[keen nova]

I saiyng exactly.

Exactly.

[keen nova]

What is the probability the total number of actions is equal to 1?

50%.

What is the probability the total number of actions is equal to 2?

25%.

[coral saddle]

my man, im going to thank you for your time. this is going no where.

.close

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[hushed flower]

help please, apparently the answer is 3

[hollow lintel]

@hushed flower What do you want help with?

[odd edge]

@hushed flower Has your question been resolved?

[hushed flower]

like whats the graph, idk how to get that

[hollow lintel]

The horizontal axis (usually the x-axis) is the passage of time, so the further right you go the more time has passed

The vertical axis (usually the y-axis) is how fast the balloon is moving

It starts at rest, then it gradually increases in velocity until it is forced to come to a stop

[hushed flower]

is it a decreasing acceleration?

[hollow lintel]

I honestly don't know what the difference is between 2 and 3, and I probably would have picked 2 here (or 1? acceleration should be constant?)

[hollow lintel]

the acceleration is increasing downward

so that depends on what you mean by 'decreasing'

it is speeding up, so it is accelerating

[hushed flower]

how tho? Doesn't air resistance increase

[hollow lintel]

Are you doing air resistance with this problem?

[hushed flower]

yeaa

its written in the question

[hollow lintel]

sorry, the question is hard to read :P

I thought it said no air resistance

[hushed flower]

oh sorry my bad

[hollow lintel]

gimme a second to re-evaluate 😭

I am on my phone

[hushed flower]

A large balloon filled with air is dropped from the ceiling of a school lab to the floor about 10 m below. Which of the following graphs would best represent the variation of its velocity (v) with time (t) if the balloon comes down vertically

Assume that the air resistance on the ball is directly proportional to its velocity

[hollow lintel]

you can immediately know that 2 and 3 are the only possible options because the balloon will be accelerating with time

this is more of a thought problem than a math problem, but you could do math to double check your answer

it looks like two has a perfect smooth curve like a parabola and 3 is more of a tapered slope, not perfectly curved?

[hushed flower]

3rd is increasing acceleration curve + constant acceleration line

[hollow lintel]

these are velocity curves

i think it says v at the top of the graph, right?

[hushed flower]

yea they are

straight line in vt is basically constant acceleration right, thats what i meant

[hollow lintel]

yes, it should be

the acceleration should be starting off very strong, then go to 0 as it reaches terminal velocity

[hushed flower]

apparently the balloon will move with a constant velocity in the end, i just dont get why

[hollow lintel]

because at some point the pressure of air against the balloon and gravity will be equal

the faster it falls, the harder air presses against it

you know how when you try to move your hand very quickly through a pool it is much harder than moving it slowly? kind of similar here

[hushed flower]

how do we explain this using f = ma?

[hollow lintel]

I haven't done air resistance in a long time but I am thinking that you would calculate the air resistance force and the force of gravity and add them together

Should cancel out and get 0

[hushed flower]

also tbh i don't get the first part either

F = ma
mg - resistance = ma

resistance increases
so F decreases

since m is constant a should also decrease

but why does it increase? what am i doing wrong here

[hollow lintel]

🤔 what is your actual calculation for resistance?

acceleration should be decreasing yes

I don't know if I have all the information you're working with on the problem

[hushed flower]

yea you do

thats all it says

[hollow lintel]

what are you using as 'resistance' here?

[hushed flower]

how is it 3 then 💀

[hushed flower]

air resistance

[hollow lintel]

how is it calculated? or are you calculating it?

[hollow lintel]

because this is velocity

it also is only falling 10m

so we would not reach terminal velocity maybe I don't know the math here

[hushed flower]

well yes but its decreasing acceleration right? isnt that the curve in 4?

[hollow lintel]

sorry maybe I am being confusing

i think 4 would be more correct with terminal velocity but this is going from 10m to the ground

[hushed flower]

no i mean yea the velocity increases but shouldn't in be the way its shown in (4)

[hollow lintel]

I think you should go back and ask if this balloon reaches terminal velocity

Because I am not sure of the calculation off the top of my head and this is only going from 10m to the ground

[hollow lintel]

by 'straight line' you mean something like '1' or a straight horizontal line?

because 1 would be constant acceleration

[hushed flower]

y = mx line sorry

that type of line without curves

[hollow lintel]

something like the graph for 1 would be constant acceleration

something like the second portion of 5 would be 0 acceleration and constant velocity

I think this question is just kind of confusing

But you may be supposed to calculate when it would hit terminal velocity and then use that in your decision-making

I will let someone else handle this because I have not done air resistance in a few semesters, sorry I did not see it at first in the question 😓

[odd edge]

@hushed flower Has your question been resolved?

[odd edge]

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[mystic saffron]

Where are you stuck?

[snow minnow]

how to solve this problem

i'm lost

[mystic saffron]

Can you color on the graph where you think it’s decreasing?

[snow minnow]

(-4,16) to (0,-4) and (4,6) to (6,4)

i think?

[mystic saffron]

Yeah but we don’t need the y value

We just care that the y value is decreasing between those x values

So now what is it

[snow minnow]

-4 to 0 and 4 to 6?

[mystic saffron]

That’s the answer you just need to write it in interval notation

[snow minnow]

so [-4,0] and [4,6]?

[mystic saffron]

Yea

It’s not increasing or decreasing at -4, 0, 4, or 6 though

The tangent line has a slope of 0 at those points so you don’t include them

Do you know how to represent that in interval notation?

[snow minnow]

no?

[mystic saffron]

[-4,0] means from -4 to 0 including -4 and 0
(-4,0) means from -4 to 0 not including -4 and 0

You get why you don’t include those values of x right?

[snow minnow]

but doesnt the interval start at -4 and end at 0?

[mystic saffron]

It’s only decreasing for these x values

The interval starts at -4 but doesn’t include -4

The green tangent line has a slope of 0

That’s why you don’t include that point

[snow minnow]

so would it be (4,0) then?

[mystic saffron]

Do you understand the difference between these 2?

[mystic saffron]

*(-4,0) yes

As well as (4,6)

[snow minnow]

top one doesn't includes the 0 and 3, bottom does

[mystic saffron]

If you understand everything then feel free to .close

[snow minnow]

.close

[odd edge]

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[stray schooner]

idk how to start

[stray schooner]

<@&286206848099549185>

ik im supposed to wait 15 mins but i asked like 5 days ago

[mystic saffron]

What part are you stuck on

[stray schooner]

idk which formula to use

or how im supposed to begin

[stable robin]

Start by drawing a diagram. Label the square side length and give it a variable x

[mystic saffron]

I must be not understanding the question because everything I’m coming up with doesn’t equal any of the answers

Also you will need 5 squares in total correct?

[stray schooner]

alr did it

got the equation x(11-2x)(8.5-2x)

[stable robin]

Can you show it here

Show everything you’ve done so far

[stray schooner]

https://i.imgur.com/ekeqmYC.png

[stable robin]

So that’s volume

[stray schooner]

yeah

how do i find the max possible volume tho

[stable robin]

Now how do you find the maximum volume for a given x

Take the derivative of V

[stray schooner]

and the length of th4 esides of the squares

alr

[stable robin]

The reason why derivative is useful here: the derivative will show you where V increases and decreases; if V increases before some point then decreases after it, then V has a (local) maximum at that point

[stray schooner]

alr what next

[odd edge]

@stray schooner Has your question been resolved?

[odd edge]

@stray schooner Has your question been resolved?

[odd edge]

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[normal sundial]

how do i know which vector is the height, length and width?

[normal sundial]

or doesnt it matter?

.close

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[crisp flume]

how should i show that this set of matrices creates vector space?

[crisp flume]

or defines? dunno the right terminology

[low locust]

is a vector space

well hopefully you know that the set of all 2x2 matrices is a vector space

so you just have to show that this is a subspace

for which you hopefully have a list of things you need to check

somewhere in your notes

[crisp flume]

that it is closed under addition and scalar multiplication?

[weary pelican]

yes, and one or two final conditions often overlooked?

[crisp flume]

associativity?

[low locust]

no, that is inherited

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[candid turret]

i dont really know how to do this one

i know the equation an = ar^n-1

so sqrt(6) * sqrt(6) would be the square root of 36 which evaluates to just 6^n-1

[weary pelican]

uh

it's not (ar)^n-1

it's a r^n-1

[candid turret]

oh probably

let me see again

i see okay

so sqrt(6) * (sqrt(6))^n-1

would be an

and then finding a5 would just be plugging that 5 in for n

36sqrt(6)

nice thanks

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[west spindle]

How do i find the area of this triangle? Unsure about what steps i need to take for this specific question.

"
The triangle △ABC is right-angled with right angle at corner C and angle α at corner A.
Calculate the area of the triangle, given that c=|AB|=5, and that tan α=7/11
"

[west spindle]

I understand that im supposed to use the area formula for a triangle, 1/2ab

[odd edge]

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[west spindle]

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[quartz pier]

im done 1) and 2)

how do u do 3)

[echo ginkgo]

Your 2) doesn't work

You didn't check the vectors you got from gram schmidt were even orthogonal

You typo'd your dot product, I think that's why it didn't work out

@quartz pier

[quartz pier]

oh

one sec

let me review

[echo ginkgo]

As for 3, well there's the long way of solving the system (1,1,-1).v=0 and (2,1,1).v=0

[quartz pier]

oh it should have been 2 + 1 - 1

not 2 + 1 + 1

[echo ginkgo]

And there's the short way by screwing around with gram schmidt again

I.e take a vector not in U and continue the gram schmidt from q2 with that vector

This will get you a non zero vector orthogonal to both basis vectors of U

[quartz pier]

@echo ginkgo

is this correct

,rotate

[echo ginkgo]

Yeah looks good now

[quartz pier]

i still dont understand how to get 3)

even after reading what u said

[echo ginkgo]

Well U^perp is the set of vectors orthogonal to all vectors of U

The first way just exploits the fact that 'if you're orthogonal to a basis of U, then you're orthogonal to all vectors of U"

So you only need to look for vectors which are orthogonal to both u1 and u2 (u1 and u2 the basis vectors of u)

The second way works by finding an orthogonal basis of R^3 which contains u1 and u2

That 3rd vector u3 you'll get after gram schmidt will be orthogonal to both u1 and u2

And you'd expect Uperp to have dimension 1

So that Uperp = span {u3}

[quartz pier]

how would I find an orthogonal basis which contains u1 and u2

[echo ginkgo]

1st thing, orthogonal basis of what? be precise
Now if you take any vector v not in U, {u1, u2, v} is a basis of R^3

[quartz pier]

So I need u1 * x = 0 and u2 * x = 0 right?

for U^T

right

[echo ginkgo]

Sure

[quartz pier]

oh then can I do

1/sqrt(3) 1/sqrt(3) -1/sqrt(3) | 0
(2 sqrt(2))/3sqrt(3) 1/(3 sqrt(6)) 5/(3 sqrt(6)) | 0

and then convert to rref

[echo ginkgo]

Yeah

[quartz pier]

and then ill get the basis for U^T

alr I get it now i think

and for 4)

am I doing it correvt

[echo ginkgo]

Wait i didn't even check your normalization for 2)

Your second basis vector doesn't have norm 1

[quartz pier]

here

why

[echo ginkgo]

Idk what you did but the norm of [4/3 1/3 5/3] is certainly not sqrt(4+1+1)

[quartz pier]

oh fuck

i took it from

[2, 1, 1]

my bad

so it would be

1/(sqrt(16/9 + 1/9 + 25/9)) * [4/3 1/3 5/3]

[echo ginkgo]

Still need the sqrt tho

Alright

It's not beautiful but it's correct at least

Now you can simplify it a bit

1/sqrt(42) * [4 1 5]

[quartz pier]

ah

so it would be

(4/sqrt(42) + 2/sqrt(42) + 15/sqrt(42)/(1) * f_2

[echo ginkgo]

Yea

[quartz pier]

so it would be 0f_1 + sqrt(42)/2 f_2 right

[echo ginkgo]

Indeed

[quartz pier]

but its not working

[echo ginkgo]

Wym?

[quartz pier]

so x = [1, 2, 3]

and 0 + sqrt(42)/2 * 1/sqrt(42) neq 1

[echo ginkgo]

Why should the projection equal the vector inputted ?

There's no reason to in general

[quartz pier]

oh

alr

[echo ginkgo]

The only case where that happens is if x is in U already

[quartz pier]

tysm

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[onyx wadi]

how log2 is 0,301?

[signal shard]

log(2) = x
10^ x = 2

this x happens to be 0.301

[onyx wadi]

i cant understand

[signal shard]

what don't you understand exactly

,w 10^0.301

[clever fjord]

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[signal shard]

approximately 2 because I left off a few decimals

[onyx wadi]

since 10 to the power of x = 2 gives 0.301

[signal shard]

yes

x = .301 in that case

[onyx wadi]

how

[signal shard]

approx

what exactly are you confused about

[onyx wadi]

how is this calculated?

[signal shard]

back in the day it was calculated using a table of log values

[onyx wadi]

the math doesn't add up, where did this 0.301 come from, what calculation?

[signal shard]

which were determined from experiments

[onyx wadi]

so log2 is always 0.301 unless the question gives a different value?

[signal shard]

ni

no

log(2) is always approx 0.301

[onyx wadi]

that what i said

[signal shard]

when would it give a different value

2 is always = 2 yk

log(2)'s value never changes

[onyx wadi]

ohh

so it will be always 0,301?

[signal shard]

yes

it is a number

like any other

[onyx wadi]

ok, but there's no calculation for that right? like, there just is

[signal shard]

they are calculated using algorithms but there is no exact value

because they are usually to infinite decimal places

its like calculating the digits of pi

the exact value of pi is pi

the exact value of log(2) is log(2)

exact value of sqrt(2) is sqrt(2)

you see the idea

these were all calculated using algorithms/experiments

[onyx wadi]

now i see

Given log 3 = 0.477 and log 5 = 0.699,
what is the value?
a) log 2

the answer is 0.301?

[signal shard]

yes

but I think they didn't want you to use a calculator here

<@&268886789983436800>

[onyx wadi]

they just want to know if I know?

[signal shard]

@onyx wadi no you need to use what's given in order to find it

[spice fiber]

They want you to use properties of logarithm to find it

[signal shard]

yes

[onyx wadi]

can you use the properties in this question? log2

[signal shard]

yes

[onyx wadi]

how

[signal shard]

you must use the property $log(\frac{a}{b}) = log(a) - log(b)$

[clever fjord]

WOWWA

[spice fiber]

Try to relate 2 5 and 3 with the propertied

Hint: you know the log for another value

Besides those which are given

[onyx wadi]

I understand, but when I have log2 I can answer that it is approximately 0.301 or not

[signal shard]

yes

[onyx wadi]

thank you very much that's all

.close

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[pearl patio]

how are they related?

[pearl patio]

how can you use that?

[spice fiber]

1-log 5=log 2

Hence the hint

Log (10/5)

[pearl patio]

it's not

[spice fiber]

It is

[clever fjord]

Result:

0.69314718055995

[boreal crag]

,calc log_10(2)

[clever fjord]

The following error occured while calculating:
Error: Undefined function log_10

[boreal crag]

:(

[pearl patio]

,calc 1-log 5

[clever fjord]

The following error occured while calculating:
Error: Unexpected type of argument in function multiplyScalar (expected: number or Complex or BigNumber or Fraction or Unit or string or boolean, actual: function, index: 0)

[pearl patio]

,w 1-log 5

[clever fjord]

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[pearl patio]

,w log 2

[clever fjord]

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[boreal crag]

that's log base e

[spice fiber]

Thats in ln

[boreal crag]

,w log_10(2)

[clever fjord]

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[spice fiber]

Log 5+log 2 =log 10

Log 5+log 2=1

Log 2 =1-log 5

[pearl patio]

sure yeah

i thought it had to be linear combinations of log(3) and log(5)

thanks~

[summer cradle]

@pearl patio

[boreal crag]

The solution to 10^x = 2 is found by finding values of x that make this very close to 2

[spice fiber]

P sure its impossible to get log 2 with just log 3 and log 5

[pearl patio]

[boreal crag]

10^(0.3) is 10^(3/10)

which is the tenth root of 10^3

which in turn you can calculate by finding which numbers multiplied by themselves 10 times are really close to 1000

[boreal crag]

U mean without 1-?

[spice fiber]

Yeah

With just that info

[boreal crag]

what info

[spice fiber]

Value of log 3

Value of log 5

And properties of log

What

[brittle beacon]

Ignore me

Had a moment lol

[spice fiber]

Since for division theres has to be a factor

[boreal crag]

So not allowed to use log(10)=1?

like you did there

[spice fiber]

Yeah

Why would the q give log 3

[brittle beacon]

Possibly for another part of the question

[cold urchin]

what even is the question

[boreal crag]

To trick people that don't know what they're doing into doing log(5)-log(3) probably

[odd edge]

@pearl patio Has your question been resolved?

[pearl patio]

what question

oh yeah hahah

[odd edge]

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[runic venture]

if x^siny = y^cosx find dx/dy(pi/4,pi/4)... i've been trying this problem for a while

can i rewrite the problem as sinyln(x)=cosxln(y) then implicity differentiate?

like - > cosy ln(x) + siny 1/x dx/dy = -sinx lny dx/dy + cosx 1/y

then solve for dx / dy?

i eventually get it to the form like

dx/dy = (cosx 1/y - cosy lnx)/(siny 1/x + sin x ln y)

but i dont think i am getting it correct

[runic venture]

is this the correct path?

<@&286206848099549185>

[odd edge]

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[runic venture]

<@&286206848099549185>

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@runic venture Has your question been resolved?

[runic venture]

<@&286206848099549185>

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[gusty totem]

How can i calculate the vector of

[gusty totem]

The angle CBA

[tardy lagoon]

angle is not a vector

[gusty totem]

How can I calculate it

[mystic saffron]

it can dictate the direction of a particular vector, though

[gusty totem]

To the calculate à vector

[mystic saffron]

!original

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[mystic saffron]

Missing context. Crucial context in fact.

[gusty totem]

Ba. Ci

[mystic saffron]

[gusty totem]

How can I calcute ba . ci

The vector

Squalar product

[gusty totem]

For that, i have to do

Ba × ci × cos of (?)

@mystic saffron

[mystic saffron]

You refuse to provide context

[gusty totem]

What context?

[mystic saffron]

I have no idea what Ba and ci are meant to he

[mystic saffron]

the original question

[gusty totem]

Ba and ci are vectors

[mystic saffron]

as it was sent you

[gusty totem]

I have the drawing

[gusty totem]

I need to find the sqalar product of the vector ba and ci

We have a triangle with a 90 degre angle

[half fulcrum]

you mean dot product?

@gusty totem

[gusty totem]

Yes

[half fulcrum]

what is vector BA equal to?

BA=< ?, ?>

and what about CI?

CI= < ? , ?>

[gusty totem]

Ba = 8

[half fulcrum]

ok

what is the length of CI?

[gusty totem]

5

[half fulcrum]

ok

the dot product of two vectors is P * Q=|P|*|Q|cos(theta)

do you know the angle between BA and CI?

BA * CI = |BA|*|CI|cos(theta)

[gusty totem]

Yes

[half fulcrum]

ok

[gusty totem]

But I don't know the angle value

[half fulcrum]

are you sure?

this is a right triangle

you can find the angle by looking at it

[gusty totem]

How

[odd edge]

@gusty totem Has your question been resolved?

[odd edge]

@gusty totem Has your question been resolved?

[gusty totem]

<@&286206848099549185>

[mystic saffron]

Yes?

[odd edge]

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[odd edge]

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[timber notch]

If a² + b² = a³ and value of a is 64. Find the square root of b

[mystic saffron]

4096+b^2=262144

@timber notch

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[faint briar]

How do I manually calculate for the excel function "NORMINV"? I got a pen and paper test tomorrow and I can't find how to solve the NORMINV manually or in a calculator.

[faint briar]

It is the formula we used in Service Time in waiting line simulation model.

[south plume]

inverse of a normal distribution?

[faint briar]

Yes, if that function is the inverse of a normal distribution

Not sure about it cause my prof didn't mentioned anything about it

[south plume]

you can try setting up an integral and solving it by hand but that's not easy

[faint briar]

We can use calculators, so it's not that really manually calculated

[echo ginkgo]

what calculator do you have then ?

[faint briar]

it's a casio fx-350ms

[echo ginkgo]

seems like you can't for this model

I don't see an equivalent function in the manuals

@faint briar

[rain pendant]

So I need some intuitive help guys,

Suppose you got a string of arbitrary length using the characters A B C D,such that they appear in the same frequency, 25% each

Would the permutations and combinations of A B C D also have the same frequency?

[faint briar]

I saw a "INV" but I'm not sure if that's the function I should use

[echo ginkgo]

nah that's for regression stuff I think

[faint briar]

Hmm I see, thanks though, I'll try to figure it out

[echo ginkgo]

I hope they'll give you normal tables or something in the exam

[odd edge]

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[civic folio]

I cannot find a calculator online to chekc my answer so can someone check this real quick

[civic folio]

,rotate

[gritty holly]

u have to subract not multiply

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[civic folio]

.close

[odd edge]

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[sleek compass]

Let $n$ be the degree of the polynomial $P(x)$. Prove that the equation $P(P(P(P(P(x)))) = x$ has at most $n$ roots.

[clever fjord]

kenshin

[low locust]

!original

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[sleek compass]

Let n be the degree of the polynomial P(x). Prove that the equation P(P(P(P(P(x)))) = x has at most n roots.

Пусть n - степень многочлена P(x). Докажите, что уравнение P(P(P(P(x)))) = x имеет не более n корней.

for russian

[low locust]

well ok but its just completely false

[sleek compass]

no, you're wrong

<@&286206848099549185>

[sleek compass]

but if it's really false prove it

[low locust]

well take P(x)=x^2

x^32=x has far more than 2 roots

[sleek compass]

stop. we can't take any polinomial

[toxic monolith]

Пусть P(x) – многочлен степени n > 1 с целыми коэффициентами, k – произвольное натуральное число. Рассмотрим многочлен
Qk(x) = P(P(...P(P(x))...)) (P применён k раз). Докажите, что существует не более n целых чисел t, при которых Qk(t) = t. <- that is orignal text,

read it carefully

and then discuss

Q is a superposition of polynomials

to make it clear

[sleek compass]

спасибо, это, думаю, поможет

[toxic monolith]

нет проблем )

[odd edge]

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[crystal zinc]

Hi there, I have the answer to part (b) to this which is $^nC_r \frac{365!}{365^n \cdot (365+r-n-1)!}$

[clever fjord]

Xetrov

[crystal zinc]

Although I have derived the answer, I am not too sure that this is the correct answer, even though the answers says it is.

Currently, I am looking at this as, choose r spots in n people $^nC_r$ to have the same birthday and then everyone else gets a different birthday which is where the factorials come from. Then I divide it by $|\sigma|$ to get the probability

[clever fjord]

Xetrov

[crystal zinc]

The issue is I haven't incorporated the fact that we have assumed that all of the other birthdays are unique. Surely in this sense, this is flawed?

And how would I go about correcting this if I am correct in saying that such an assumption has been made

[hasty dome]

It doesn't say that the others can't share birthdays.

It just says that r people share a birthday.

n - r could also share a birthday, or two could.

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[spare kindle]

Hello

[spare kindle]

the function f:[0,2]>R, f(x)=2x-x² is considered. Find the values of the real parameter m, so the equation line y=mx divides the area of the subgraph f into two sets of equal areas

<@&286206848099549185>

<@&286206848099549185>

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[normal sundial]

how can i solve this? Im stuck

[normal sundial]

@toxic monolith ?

.close

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[raven plaza]

im confused

[torpid owl]

check the derivative and 2nd derivative

[raven plaza]

ok so for second derivative i am getting

something insanely long and crazy though

what do i do with the second derivative

[torpid owl]

see here: https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-6b/a/concavity-review

[raven plaza]

ok thanks ill look at that

.close

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[high vigil]

I am struggling to understand circle theorems and i'd like someone to help me with how this is supposed to work

[high vigil]

.close

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[valid cove]

can someone help me on this question?

[valid cove]

i know that the t statistic is -4.108, p value is 0.0261, and point estimate here for fitted value y hat is 2.15

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<@&286206848099549185>

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@torpid owl can you help

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[elder vault]

Here’s my answer

[elder vault]

I’m trying to prove the equality

[elder vault]

Is it acceptable

[late flint]

if it leads to the pythagorean identity then I suppose it can be taken, but I would do it like

take the ^2 out of both sin and cos

then use sin(2x) = 2sin(x)cos(x) -> sin(2x)/2 = sin(x)cos(x)

so sin^2(2x) / 4

$\frac{sin^2(2x)}{4}=\frac{1}{4}(\frac{1-cos(2\cdot2x)}{2})$

[clever fjord]

Triaxyz

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[half thistle]

can someone exolain this

[lament forge]

Pretty sure its C

might be wrong though

[tardy lagoon]

Two triangles are similar if the proportion of two sides are equal and the included angles are equal

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lament forge]

Oh shit my bad

technically im not a helper so 🧐

[half thistle]

k but how can u determine if it aa or sas or sss

also how u know which side is proportinoal to the other side

@tardy lagoon

[tardy lagoon]

AA means two angles are the same

[half thistle]

ok

[tardy lagoon]

You onyl got one angle so you onyl need to check sas

check the proportions of the corresponding sides

[half thistle]

ik but how do u know which one goes to which

[tardy lagoon]

For example for option C, you check if ML/GF=MK/GE

Option C says that ∆MLK~∆GFE

So the side ML corresponde to GF

MK corresponde to GE

[half thistle]

wait is it just smallest to smallest like 20 goes with 8

[pearl patio]

yes u can try that way

[half thistle]

is it the correct way to do it tho

[tardy lagoon]

That works too

[pearl patio]

or you can try reflecting MKL and rotating to align with the smaller triangle

[half thistle]

so the answer would be c

[tardy lagoon]

It's important to know the correspondence though

[pearl patio]

what is c

[pearl patio]

yeah

[half thistle]

because mk corresponsds with ge

[half thistle]

ok

[pearl patio]

mhm and angle G congruent to angle m

[half thistle]

do u know if u have to put ~ this in front to let it now it a similarity not a congruence like the threom

[pearl patio]

they just did

[half thistle]

oh shoot mb

sry but one last question

[eager nymph]

can u help me with my question

[pearl patio]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[pearl patio]

there are two triangles that share an angle

can you identify those two triangles?

[half thistle]

nvm i think i got it

it just proprtions

thks alot for your help tho

[pearl patio]

hmm sure

but you should still show the triangles are similar before applying proportions

the easiest one being AA similarity

cuz you know angle G is shared between the smaller and bigger triangle so it's same on both

then you can show angle E is congruent to angle U

[half thistle]

mb for not showing the whole prblem

wht abt this one

[mystic saffron]

Hi guys

I have a mathematics question.

<@&286206848099549185>

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[tardy lagoon]

!status

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What step are you on?
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[zinc glacier]

whats your question

think about it this way, if you shuffled a deck of cards and then spread them out in a row, what is the probability the third card in that row is an ace?

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[tardy lagoon]

yo

do you need help

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[restive dew]

I need help with a confidence interval problem

[restive dew]

i want to do one for car

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@restive dew Has your question been resolved?

[restive dew]

.close

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[frozen sail]

hello

[wooden python]

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
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[frozen sail]

so i undertsand where i got my p

but how did they get D

[lofty spade]

by multiplying with A

[frozen sail]

what is A

[lofty spade]

that should be given in the question

it should be a matrix

[lofty spade]

ys

yes

so they got D by multiplying together P, A, and P^-1

[odd edge]

@frozen sail Has your question been resolved?

[frozen sail]

can you look over my work its getting wrong

[viral bone]

bottom left should be 0

in the product of the matrices

1 x (-1) + 0 x 0 + 1 x 1

which is -1 + 1 = 0

it kinda sucks that it ruins the rest of the exercise a lil bit

[frozen sail]

I'm still wrong

[viral bone]

no its 0 😩

you put a 1 there haha

oh that was a 2 there

[frozen sail]

aaahhhh i seee noww

[viral bone]

ye ye

1x1 haha

dang gotta fix the whole thing over again 😭

matrices are such a pain

i shouldve just sent it like this tbf

[frozen sail]

,rotate

[frozen sail]

Idk what's u0

[viral bone]

u0?

[frozen sail]

meant whats wrong

[viral bone]

what was the initial question

of the exercise

[viral bone]

looks right to me (depending on what you are meant to do)

[viral bone]

i mean you can just turn that matrix you got into ^^ in just a few steps, i think its just a possible solution no?

was P^-1 given?

[frozen sail]

wait whats p^(-1)

no it wasnt

[viral bone]

did you only get A?

[frozen sail]

a is given