help-42

[pure kayak]

when (x+h)=0, y=2

here you see thats when x=0

so then h=0

[marsh blade]

i dont understand really

can u pls explain

oh u mean to say there is no horizontal

shift

[pure kayak]

a^(-x) =1 when x=0 right?

[marsh blade]

what about this

[marsh blade]

yes

[pure kayak]

so a^[-(x+h)]=1 when x+h=0

[marsh blade]

yes

[pure kayak]

since we are translated up one

were looking for when a^[-(x+h)]+1=2 for x+h=0

we see the point (0,2) is there

so x=0

0+h=0, h=0

[marsh blade]

let me try in this

so we ar shifted 3

so -a^(x-h) +3

x-h = -1?

h = 1

right?

[pure kayak]

no, when x+h=0, -a^(x+h)+3=-1+3=2

we see there that x=-2 [from the point (-2,2) ] so h=2

[marsh blade]

i really dont understand now

can u please explain again

why did u choose point -2,2

why that one?

why not 0,-1

[pure kayak]

i didnt choose it, that just happened to be the y value when x+h=0

we want x+h=0 because then there is no dependence on what a is

-a^0+3=-1+3=2

the point (-2,2) is on the graph, so x must =-2 here

x+h=0 then becomes -2+h=0

so h=2

[red umbra]

think about it like this you have the upper limit of the graph 2

so your function is 2-something

because it goes downwards as x is growing

f(x)=2-a^(x-h)

[pure kayak]

the upper is 3, small nitpick (but keep it going)

[red umbra]

my bad

)))

but that x-h is annoying

u cant really do much with it

how do u get rid of it

you equal it to 0

so now you have 3-a^0

which is 2

and when you plug x=0

you have 3-a^(-h)

and this has to be equal to -1

wait im now confused

[pure kayak]

yeah you went a bit off

to use (0,-1) you need to know a

you do know that for y=2, x=-2

use that

[red umbra]

oh yeah my bad

basically you calculate f(h)=3-a^(h-h)

and see what x gives you that relation

and since its an injective function

[pure kayak]

we had x+h not x-h here but i see the vision

[marsh blade]

hm

[red umbra]

f(x)=f(h) <=>x=h

and 2=f(-2)

h=-2

[red umbra]

how do you knoow

like am i missing something

?

oh

its if it is concave or convex ?

[pure kayak]

f(h)=-a^(2h)+3
isnt too helpful

is if its written as f(x)=-a^(x-h)+3
then f(h)=-a^0+3 we would know f(h)=2 so h=-2 as you said

then we arrive at f(x)=-a^(x+2)+3 which is fine

your ideas fine it just required a tweak to the setup

[red umbra]

if f(x)=3-a^(x+h) then you can just plug in f(-h) right?

[pure kayak]

that you can

then h=2

nice idea

[red umbra]

but where do you get the a^(x+h) or a^(x-h)

never in my life did i do these types of problems?

[pure kayak]

you can choose it arbitrarily

[red umbra]

like i dont know when its + or when its -

oh

[pure kayak]

youll just get opposite signs for h when you solve in each case

[red umbra]

doesnt that matter tho?

[pure kayak]

nope

[red umbra]

like what is the point in finding h then

[pure kayak]

to know the shift of course

if you used x+h we found h=2 so x+2

[red umbra]

or do they give you f(x)

[pure kayak]

if we used x-h we found h=-2 so again, x+2

either way we found a left shift of 2

[red umbra]

oh yeah

[pure kayak]

theyre working from the parent function a^x

[red umbra]

didnt even think of that)))

[calm coral]

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[violet flicker]

how can we integrate this

[unique jackal]

use a substitution

[shy swallow]

sub for x^2

[unique jackal]

there’s a fairly natural one to decide on

[shy swallow]

its pretty much the power rule from then on

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[winter iron]

hi i think this is a really basic quesiton idk if it goes here but question (c) is kind of confusing me
this is what i did but i have no idea if it makes sense; i dont really understand fields and like how to prove stuff in them
i couldnt even prove (-1)*(-1)= 1,

[civic dirge]

If you want to prove something isn't a field, you just need to give a counter-example

Taking your logic, but turning it into a counter-example, it should be true that:
3 + (2×4) = 3+2 × 3+4

But clearly that's not true

@winter iron

[winter iron]

i see

bc idk what is like

field axioms vs like normal maths

[civic dirge]

Normal math is fields, but you don't really get into the details

[winter iron]

i see

[civic dirge]

That is, addition/multiplication on the real numbers are a field

[winter iron]

is it bc real numbered infinite

[civic dirge]

Make sure to take a look at the field axioms. They're all pretty natural, and you should realize you've been using them forever

[winter iron]

i see

[civic dirge]

Just words like "identity" and "inverse" get mixed in

[winter iron]

how would u prove (-1)(-1)=1

[civic dirge]

I'm not one to give answers directly. However, you'll need these:
1 is the multiplicative identity. That is,
1•a = a.

-1 is the additive inverse of 1. That is,
1 + (-1) = 0.

I will assume it's already been proven that 0•a = 0.

[winter iron]

i’ve seen the proof but i like don’t understand it

wait not that i don’t understand

more like i don’t really know how to think that way

in an exam for example

it felt really unintuitive

[civic dirge]

Okay, so if you've seen it, I'll finish my proof.
Take 1 + (-1) = 0

Multiply by -1:
(-1)1 + (-1)(-1) = 0(-1)

-1 + (-1)(-1) = 0

(-1)(-1) = 1

Notice how, once I pulled up all the definitions of every object, the proof is immediate

That can be the hard part for many people. Knowing that 1 is the multiplicative identity, and -1 is the additive inverse to it

[winter iron]

so i should probably like set out what i know from the axioms

and go from there

[civic dirge]

Writing out the definitions of things and unraveling them is like 80% of higher mathematics haha

[winter iron]

thank u 😊

[civic dirge]

I hope I can figure out the other 20% soon

[winter iron]

Lol

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[final brook]

check weather these Relations R1 and R2 in Real numbers re transitive or not.

R1 = { (a,b): a ≤ b }
R2 = { (a,b): a ≤ b^3}

[final brook]

m stuck tht if both belongs to real numbers.... then why we cant take same examples for both questions.

i proved tht R1 is transitive

but for R2 the same example is not wroking

[wild badge]

the second relationship is nonlinear

[final brook]

yeah i also heard this can explain what is this?

[wild badge]

because its cubic it exhibits different properties than the standard order of two numbers

it has an exponential growth rate but that can work to its disadvantage in terms of transitivity, consider the pairs (1,1) and (1,0.5)

[final brook]

hmmm so i have to remember this...

idk why it seems simple but still confusing me

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[steady sky]

i worked out angle x to be 42.4

how do i solve 2.3? are they vertically opp angles?

[steady sky]

also will the outer angle of AP be 90 degrees or 180 - x ?

It looksa like a square

[calm coral]

@steady sky Has your question been resolved?

[trim galleon]

No they arent vertically opposite angles

do you know whats angle APR

[steady sky]

yes, 42,4

so to find APB = 180 - (?)

[calm coral]

@steady sky Has your question been resolved?

[versed sluice]

180 - x should work

[steady sky]

So I don't need to figure out the angle? Just 180 - x?

[versed sluice]

Yes

[stone root]

Use the "fact" that AFPR and BQCB are squares

[steady sky]

They have 90 degrees on the inside or the lengths of the sides?

[stone root]

I think that problem is posed so you guess that are squares

[stone root]

both

[steady sky]

I see, so area will be 1/2(9)(17)sin(P)

but im still confused for that angle

i worked out QPR as 42.4

[stone root]

Can't you see APB as a substraction of angles?

APB = 360 - ...

I can't give you more hints

You should get it straightforwardly

[steady sky]

360 - 90 - 90 - 42-4 = 137.6

[primal bear]

!status

[calm coral]

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

[steady sky]

6 i think

[unkempt drift]

if that's true type .close

[steady sky]

.close

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[tall moon]

is there a reasonable way to find the ammount of factors of 300!

[velvet osprey]

yeah factorize it into primes

there is legendre's formula for the exponent of a prime p in n!

and like maybe you need to know how many primes there are up to 300, 300^(1/2) and 300^(1/3) or so.

[tall moon]

i meant like is there a more elegant way than that

[velvet osprey]

no

[unkempt drift]

not really

[tall moon]

ok

ty

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[crystal breach]

i dont understand 5

[primal bear]

!status

[calm coral]

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

[crystal breach]

1

like what do i do

[upper prairie]

I suppose by corresponding point they mean the same y coordinate

[crystal breach]

uh so

what exactly am i meant to do

[upper prairie]

Its a goofily worded problem

But I suppose u need to find the same y coordinate

[crystal breach]

hm

[upper prairie]

Now can u do it

[crystal breach]

what do we sub what in

[upper prairie]

f(10)=25

g(?) = 25?

[crystal breach]

wait wait what

why is f(x)=g(5x)?

and isnt 25 the y coordinate not x value

[fast heath]

f(x) = y

[crystal breach]

i dont rlly understand the quesiton like why are we subbing it in to find the same point

and how do we know what to sub it into

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[violet magnet]

Im trying to solve this problem to find x:

I have broken the problem down into this sector, so i need to find the angle at the bottom, but my brain just wont work:

[velvet osprey]

you want to find that arc length?

[violet magnet]

ultimately, yeah

[velvet osprey]

so r and h are known right

[violet magnet]

yep

[velvet osprey]

you can express BC in terms of those and cos(θ) via the sine rule and also find the length of chord AB once as 2(r+h)sin(θ/2) and in another way via triangle ABC and the law of cosines again

though this sounds like it might get pretty disgusting.

[near helm]

May I say something?

I see another method of solving it

[violet magnet]

oh, it doesnt actually matter that the top part is curved at all, because BC isnt

idk why i was being thrown off so much by that

[near helm]

Nice

[violet magnet]

thanks all, should be fine from here 🙂

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[solid nymph]

Need notes of Logarithms

[solid nymph]

Sources pls

[edgy leaf]

search "logarithms cheat sheet" on google

find any source that suits u

[solid nymph]

Got any u made?

Or any fellow classmate of yours?

Pls dm if you got any

[dull wagon]

what's wrong with online ones

[calm coral]

@solid nymph Has your question been resolved?

[solid nymph]

They aren't efficient from my pov

So if you got anything related to log which you learnt during your time then pls send

[dull wagon]

what about them isn't efficient

[solid nymph]

It just doesn't seem right and handwritten notes are better than them

Ykw nvm

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[river pike]

hi again, lowkey for someone to guide my through this one

[river pike]

i know that e^x is like the biggest one possible

[trim galleon]

try applying l'hopital's rule

[river pike]

will do i know that one well

[short salmon]

this is also just sort of proof by heuristics

e^x in the numerator and denominator grow at the same rate

so it should behave the same as $\frac{5e^x}{e^x}$

[potent lotus]

00100000

[short salmon]

which, of course, is 5 as $x\rightarrow\infty$

[potent lotus]

00100000

[river pike]

what about the -1 and -6

what makes us allowed to like ignore them

[short salmon]

consider what happens as e^x grows very large

it's lower bounded by 2^x

[river pike]

2^x?

[short salmon]

try x=100, and you get an absurdly large number. the constant factors of -1 and -6 aren't going to really make much of a difference

[river pike]

lemme try that

[short salmon]

to see why this is truly true, you'll need to look toward the epsilon delta definition of a limit, but I assume if this is for your standard calculus class, such a level of rigor won't be necessary

[river pike]

ya it does give me 5 when i put it in

[desert lichen]

For larges values of x, adding constants to e^x doesn't cause a big difference

[short salmon]

:^)

[river pike]

ya no hahahaha

[river pike]

is this a strategy i can implimit

in all hr problems

[fickle musk]

Divide the top and bottom by e^x maybe

[short salmon]

pretty much

[river pike]

cool

def helpful

[fickle musk]

Then you can do lim x->∞ e^-x

[short salmon]

oh, fascinating

[river pike]

HAHAHAHA

ya calculus mostly is just to sort students

[short salmon]

well, understandable honestly. math does have a toxic "intellectual" culture, so I do get it

[river pike]

make sure its either the ones are are naturally good at school or the ones that work hard enough to stay

[short salmon]

as someone who self-idenitifies as very dumb at math 💀

[river pike]

this is so real

my prof instantly starting having no hope for me early semester

[fickle musk]

Then you get 5

[short salmon]

oh. well he's a mean one fr

[river pike]

then i got a tutor

and got like 73% at my midterm which is descent according to me

[short salmon]

that doesn't seem bad at all for a college calc class. usually, averages tend to be quite low

at least at my school

I mean, the average on my ODE exam was like, less than 50% 💀

[river pike]

OUF

[calm coral]

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

[river pike]

could i do something similar here

it looks like the previous one

[short salmon]

in a sense I suppose...

[fickle musk]

Divide through by e^x then evaluate or you can go straight ahead

[river pike]

oki

[fickle musk]

Look at the size of the denominator as x->∞

[river pike]

would it be 0?

[fickle musk]

Yes, I suppose.

[river pike]

what about the scond one since they are the same

[fickle musk]

,w lim 1/(e^x - 2)

[river pike]

whats up with that

[potent lotus]

Results provided by WolframAlpha

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Requested by ishigamisenku_

[river pike]

can you show me how to divide by e^x

I struggle with that

[fickle musk]

Okay

$\frac{\frac{5}{e^x}}{{\frac{e^x-2}{e^x}}$

I'll just write this instead then

[river pike]

oki

[river pike]

i understand the second step but not the 3rd

[fickle musk]

Splitting the single fraction into two

[river pike]

ya

[fickle musk]

$\frac{a}{b} + \frac{c}{b} = \frac{a+c}{b}$

[potent lotus]

David

[river pike]

ah gotcha

ok lets do the next step

i see e^x on e^x

cancels

[fickle musk]

Yeah

Also lim x->∞ c/x = 0 if c is a number

[river pike]

i know theres an algebra rule to sorta break the fraction

a/b/c/d

[fickle musk]

Well you just need to evaluate lim 5/e^x and 2/e^x

[river pike]

ummm

[fickle musk]

Plug in and that's all.

[river pike]

so 5/e^inf

and -2/e^inf

[fickle musk]

Yes

[river pike]

and those results on a fraction

[fickle musk]

What do you think this equals

[river pike]

wouldnt that lead me to hospitals

0/0

[fickle musk]

No

[fickle musk]

.

[river pike]

so ya both give 0 no

thats why i said 0/0

[fickle musk]

Both give zero

But there's a 1 in the denominator

[river pike]

wait where

[fickle musk]

_ _

[river pike]

-1 2/e^x

[fickle musk]

1-2/e^x

[river pike]

so 1-0?

[fickle musk]

Yes

[river pike]

so 1 would be my other limit

[fickle musk]

In the denominator

[river pike]

so its 0/1

[fickle musk]

Yes

For the first one

[river pike]

so what does that mean for our context

[fickle musk]

0/1=0 for the first one

[river pike]

i already had that one

[fickle musk]

The limit is 0

[river pike]

cause small/big= 0

[fickle musk]

The second one is shorter
Just use the fact that ce^-x = c/e^x

[river pike]

I have no clue for this one

[fickle musk]

Start off with replacing the xs with -∞

[river pike]

oki

[fickle musk]

I meant c times e^x

[river pike]

oh my bad

il rewrite it

[fickle musk]

Yeah and the x gets replaced with the value (-∞) not e

[river pike]

oki done

[fickle musk]

[river pike]

its the c thats putting me off

but its looks like small/big again

[fickle musk]

So we go from ce^-∞ to c/e^∞

[fickle musk]

Yes

It is

[river pike]

so 0

[fickle musk]

Yes

[fickle musk]

Now time to do the actual limit

[river pike]

oh ok

[fickle musk]

re write e^x on the denominator

[river pike]

ok

[fickle musk]

Remember the minus sign on the ∞

and x on the left

[river pike]

wait

because of the -2

[river pike]

where is that from again

[fickle musk]

5/(0-2)

The e^x at the bottom and the -∞

[river pike]

oh shit

thats a -

they didnt code that well

[fickle musk]

Remember when you switch to the denominator the sign is dropped

[fickle musk]

_ _

[river pike]

yes

but when does it switch

like when it goes from * to /

[fickle musk]

When you have 1*x^-n you get 1/x^n

[river pike]

gotcha

i removed the - from the denom

[fickle musk]

[river pike]

so we have c/e^inf

[fickle musk]

Yes

[river pike]

so

whats up

[fickle musk]

That's zero

[fickle musk]

What do you think that e^x will be

[river pike]

inf?

i cant say why

but i have a feeling

[fickle musk]

No it's 0

[river pike]

BUT WHY IS IT ALWAYS 0

WHEN WILL IT NOT BE 0

[upper sparrow]

(Careful of how you phrase this here)

[fickle musk]

Yeah

I mean the e^x is 0

Because e^x = e^-∞ = 1/e^∞

[river pike]

yes cause 1/x^n

[fickle musk]

Yeah then replace it with 0 in the denominator

And continue with it

[river pike]

isnt that 1/0 again

[fickle musk]

No

[river pike]

oh its 1

[fickle musk]

5/(0-2)

[river pike]

5/-2

[fickle musk]

e^x in this case ->0

Yes

-5/2

[river pike]

YAY

thank you so much, il go start another one!

appreciate it

[fickle musk]

Good luck 👍🏼

!done

[calm coral]

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[river pike]

.close

[calm coral]

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[unborn stone]

how do i do this

[wanton wave]

They give you the formulas for the parametric equations at the top

You just have to plug in the values for x1 and x2, or y1 and y2

@unborn stone do you see what I mean?

[unborn stone]

what about the t

what i plug in for it

[wanton wave]

nothing, just leave it as t

The idea is, you could plug in a number for t and the equations would give you an x and y of a point on the line

But here, they're just asking you for the equations

[unborn stone]

i got x=3 and y=-1

but its wrong

[wanton wave]

I think you dropped the t

For x, if you plug in x1 and x2, it'll look like

x = x1 + (x2-x1)t

[unborn stone]

i added in and still doesnt work

[wanton wave]

no hang on

[wanton wave]

what is x1 and x2?

[unborn stone]

x1=-2 and x2=3

[wanton wave]

so

x = -2 + (3 - -2)t

do you agree?

then just simplify the 3- -2 part

[unborn stone]

5

[wanton wave]

so

x = -2 + 5t

[smoky wolf]

hey i dont mean to interrupt but you could find the value of t for that particular line segment and substitute it back in the equation?

[unborn stone]

ok so y =3-8t

[unborn stone]

i only need x and y

[smoky wolf]

did you just subsitute

?

does that work atlesat

[wanton wave]

not 3

[smoky wolf]

x=-2+5t

[wanton wave]

I'm not really sure what you mean

[unborn stone]

i wrote -1-8t and it doesnt work

[smoky wolf]

i think i was wrong now that i think about it

[unborn stone]

ok nvm

[wanton wave]

not -1 either, what's y1?

[smoky wolf]

what about the x part

did that work

[unborn stone]

i wrote t wrong

[unborn stone]

yeah

.close

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[ripe jacinth]

is this even allowed? lower bound is greater than upper bound??

[drifting seal]

nah bro it’s illegal

[smoky wolf]

why not

[calm coral]

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[strange loom]

Hey anyone this message may concern,

I have a PSMT for Specialist Maths but I am stuck with find the solution and I don't know where to start to find out for the solution and I would like assistance to get started.

Kind Regards,
Clixqlofy

[strange loom]

I will be back in 30

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[calm coral]

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[strange loom]

Not yet

[strange loom]

Not yet

.reopen

/.reopen

[calm coral]

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as this action is irreversible, and this channel may abruptly lock.

[strange loom]

.close

[calm coral]

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[wintry trench]

How do I do this I’m completly lost

[wintry trench]

Don’t know where to begin

[harsh siren]

do you know the margin of error formula

[wintry trench]

This?

[harsh siren]

whats U and L

[harsh siren]

^

[wintry trench]

Upper and lower

[harsh siren]

of what

[wintry trench]

Of the problem??

Idk 😓

[harsh siren]

that doesnt make sense lmao

also no

[harsh siren]

does this ring a bell?

[wintry trench]

I currently being a messenger for my cousin

And she said I did get a similar question and gonna try chat got again

I don’t think she needs help anymore

[minor gust]

🙏 😭

[wintry trench]

Ty though

.close

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[median cloak]

It is Bullshirt

Bullshirt

Horseshirt

P(1) is not 6/16

Stop

Stop putting Bullshirt content

Hello

Bull

Bull!

[shadow path]

4-3, 3-4, 3-2, 2-3,2-1, 1-2

[calm coral]

@median cloak Has your question been resolved?

[median cloak]

What

[median cloak]

Did you see 4 on the spinning wheel?

Hello

Excuse me

<@&286206848099549185>

[bleak gust]

even if there was no four, you can still use striker's format

you can find all possible answers and get the probable mean or smth

i'm not sure if its called that but

since there are a minimum of values, you can find all of them relitivily quickly

[exotic falcon]

the correct grammar would be bullshit

[bleak gust]

😭

[median cloak]

What

Do you see 4 on the spinning wheel

[rapid cliff]

1: 1-0, 2-1, 3-2
2: 2-0, 3-1
3: 3-0

P(0) = 0
P(1) = 3/6 = 1/2
P(2) = 2/6 = 1/3
P(3) = 1/6

[bleak gust]

yeah so do it with your values

ty shuba

[exotic falcon]

horseshirt

[rapid cliff]

how

[exotic falcon]

idk I just wanted to say it

[shadow path]

whoops

mybad

[rapid cliff]

lmao

[shadow path]

let me present a better answer as inspiration from my reading

[median cloak]

The denominator is 16

Excuse me

[bleak gust]

wait lol i just realized its a video

that makes it easier than

[median cloak]

And P(3) = 2/16 not 1/6

[bleak gust]

yeah khan academy is correct

are you confused about the answer?

[median cloak]

Yes

[bleak gust]

which part?

[median cloak]

The probability they made

[bleak gust]

in general or in a specific column?

[rapid cliff]

ah, ignore my last. I didn't read properly XD

[shadow path]

6 ways to have a difference of 1/ total possibilities = 4*4 for two spins

[bleak gust]

^^

[median cloak]

All of them. P(1)=n(1)/n(s)

[bleak gust]

yup!

the combinations of 1 are:

1-0
2-1
3-2

then you need to multiply that by two cause you can spin them twice

so if i spin 0-1 and a 1-0 it will still be the same

[rapid cliff]

ways to get 0: 0-0, 1-1, 2-2, 3-3
ways to get 1: 1-0, 2-1, 3-2
ways to get 2: 2-0, 3-1
ways to get 3: 3-0

P(0) = 4/10
P(1) = 3/10
P(2) = 2/10
P(3) = 1/10

[median cloak]

Why

Why

Why I have to multiply them by 2

Hello

Excuse me

[bleak gust]

so if i span the wheel and i got 3 and 1 for example,

wouldn't it still hold the same value if i span and got 1 and 3?

[median cloak]

They are the same thing

[bleak gust]

exactly

but you need to count for the two possibilities

[median cloak]

Why

[median cloak]

Why I have to

[open meadow]

Ik this will be out of the context, but your username is hard and diff respectively

[bleak gust]

otherwise the values will be wrong

[median cloak]

Why

[bleak gust]

and because khan is only looking at positive values

[median cloak]

Why I have to consider one thing as two

[bleak gust]

i want to try to explain this clearly give me a second

[median cloak]

What do you mean hard and diff respectively

[bleak gust]

in an example, if i said that you need to roll a 1 and a five from a dice, there are two ways to do so:
one way is if the first roll gives me a 1 and the second gives me a 5
and the second way is if the first roll gives me a 5 and the second gives me a 1
both fufill the "roll a 1 and 5" so i need to count both of them

However, some questions in probability tell you to exclude the other way though, but i would assume from the answer that Khan academy wants you to consider the two ways

[median cloak]

Why

The sums are both 6

[bleak gust]

could you elaborate your question a little bit more?

[bleak gust]

but the order is different

sometimes you need to count for that

[median cloak]

Why order matters

[bleak gust]

sometimes it doesn't

but it wants you to count for all possible ways you can spin the differences

[median cloak]

When? Could you give me an example

[bleak gust]

ALL ways means that you need to count for order

[bleak gust]

sure!

[median cloak]

Why

Why all ways means order matters

[bleak gust]

wouldn't you say that 3 then 2 is a valid roll?

then what about 2 then 3?

both are valid rolls

and if we discount one, it would affect the probability

[median cloak]

Sure but they are the same

[bleak gust]

yup but discounting one affects the probabilityh

they have the same outcome, but have different orders

i think that if you want a better explaination, you should see what khan academy needs to say about the question

[lyric ravine]

Uhm thats called vocabulary...

[calm coral]

@median cloak Has your question been resolved?

[jolly pilot]

You mean spelling

Just make a table to organize all the outcomes, something like all columns marking the first roll, and all rows marking the outcomes of the 2nd roll. Then you can count how many possibilities there are, and what differences you get and count those

[calm coral]

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[floral vessel]

ok so i was practicing some derivative questions and i got stuck on this specific question [ Basically i needed to find the derivative of f(x) and in this pic, purple one is my answer and blue one is the desmos calculated one ] As you can see, my function graph does not exactly overlap with the desmos one. where did i go wrong?

[floral vessel]

the curved purple part hid the rest of the blue graph btw

[sweet pike]

The derivative of arcsecx is $1/|x|root(1-x^2)$

[potent lotus]

Aria

[floral vessel]

yeah i used that

and then chain rule

[sweet pike]

But here the x is 1/1-2x^2

[floral vessel]

wait

[sweet pike]

So you have to put that instead of x

[floral vessel]

it's x²-1 inside root

[sweet pike]

It won't be that

[floral vessel]

not 1-x²

[sweet pike]

it's 1-x^2

[floral vessel]

i'm pretty sure it was the other way around

[sweet pike]

but the thing is, it's not arcsecx, it's arcsec(1/1-2x^2)

[floral vessel]

yeah and for that we use chain rule afterwards

[sweet pike]

uhh lemme check cause I think it's 1-x^2

[floral vessel]

yeah sure

[floral vessel]

it think this one was for arcsin and arccos

[sweet pike]

Arccosecx is x^2-1

[floral vessel]

it's the same for arccsc and arcsec, just the diff of a negative sign

[sweet pike]

No that's root(1-x^2) for arcsin

[floral vessel]

yeah

but i'm talking about the terms

[sweet pike]

Oh yeah mb

it would be x^2-1

But still it won't work

[floral vessel]

i can show my work how i got the purple one if that helps

[sweet pike]

No no I got it

The thing is you only derivated arcsecx

[floral vessel]

the whole purpie function here

is my answer

[floral vessel]

well i used chain rule tho

[sweet pike]

Wait I'll solve it and send you the pic of what I'm saying cause that's a lot of writing 😭

[floral vessel]

i'll send my work as well

then you can see what's wrong there

that works?

[sweet pike]

okayy

[floral vessel]

ok wait a bit

this one

@sweet pike

[sweet pike]

oh

[floral vessel]

well you made a mistake there

i did that before as well

[sweet pike]

yeahh

[floral vessel]

any mistakes on mine?

[sweet pike]

I'll check yours just a min

[floral vessel]

ok

[sweet pike]

it looks good to me

[floral vessel]

then why is my graph m=nt matching?

is desmos broken or something

<@&286206848099549185>

[sweet pike]

How would it be a straight line

[floral vessel]

that's the prob

[sweet pike]

your answer's correct

[floral vessel]

but i hovered over it and it wasn't straight

[sweet pike]

Do you know the solution?

[floral vessel]

nope

it was an exercise for the reader in a book

[sweet pike]

oh okay okay

[floral vessel]

well i'm stuck now

[sweet pike]

Oh I got it

So the problem is in the equation

Wait I'll send you the ss

[floral vessel]

umm well it's the same as before

[sweet pike]

Ohh I thought it was only just the straight line lol

[native fox]

what is the problem?

[calm coral]

@floral vessel Has your question been resolved?

[calm coral]

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[sleek dust]

Can someone help me with this am I simplifying it correctly?

[azure willow]

why is it ^5 here?

[sleek dust]

Is it supposed to be ^9?

[azure willow]

no, ^6

[sleek dust]

Oh

[azure willow]

it means multiplying (x^2+1)^3 by itself once

[sleek dust]

And for the numerator do i multiply 8x^2 * 3(x^2)^2

[azure willow]

i think you should start by factorizing (x^2+1) first

from numerator and denominator

[sleek dust]

Like this?

This.

?

.close

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[verbal stag]

Can someone see for me this question part C, the answer is x^2/4 instead of my answer (x/4)^2. why is it not like (x/4)^2? because isn’t x^2/4 a vertical dilation instead of a horizontal?

[soft remnant]

(x/4)^2 is not the same as x^2/4

i think that is the flaw here

it would be technically correct if it was (x/2)^2

[verbal stag]

Yeah but wouldn’t x^2/4 be a vertical dilation though?

[calm coral]

@verbal stag Has your question been resolved?

[calm coral]

@verbal stag Has your question been resolved?

[verbal stag]

.close

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[unborn relic]

I'm stuck again. by the way, im in year 8 so I don't know complex ways if you know what i mean

[fringe swan]

use a² - b² = (a+b)(a-b)

[unborn relic]

Yes, i have already tried that

but im not sure what to do after that

[fringe swan]

then whats the issue

u will see a lot of stuff cancelling out

[unborn relic]

yes, ive cancelled a lot out, but as since it is a big range, i kept the elipses, and therefore I dont know which ones are left as i wont do all of them from 2 to 2010

[fringe swan]

the first term ever

in the product

(1 - 1/2²)

[fringe swan]

if u apply this formula on it

u obtain 1/2 × 3/2

now go for second

(1 - 1/3²) is 2/3 × 4/3

so 3/2 and 2/3 cancel out

u have 1/2 × 4/3

third factor (1 - 1/4²) = 3/4 × 5/4

3/4 and 4/3 go disappear

u remain with 5/4 which will disappear with the following factors

that 1/2 at the start will stay

[unborn relic]

oh, ok, thanks, i get it

[fringe swan]

and the last factor is (1 - 1/2010²)

2009/2010 × 2011/2010

2009/2010 will be cancelled out js like any other factor

but 2011/2010 wont

cause the product ends there

so ur rlly left with 1/2 × 2011/2010

[unborn relic]

ok thanks

.close

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[spice barn]

hello, I would like help with understanding bernoulli's trial, i understood nothing of it despite trying so hard

[balmy grove]

It's also called a binominal trial where there are exactly two designated outcomes. Often in "success" or "failure".

Is there something in particular you do not understand about it?

[spice barn]

yes, particularly the formulas, i dont understand what i should expect while learning about it, like..how can i exercise ? and i feel like i didnt understand it at all

[balmy grove]

I'll need you to point at any formula first, post them here and we can take it one step at a time from there.

[spice barn]

alright give me a second

this one right here

[silver tapir]

what do u not understand

[spice barn]

i dont understand what binomial trial is about

at all, when i feel i understood i end up in front of other exemples and i feel even more lost

[silver tapir]

I doubt u need that

[spice barn]

?

[silver tapir]

it's for remembering the coffecients when u expand a binomial

[spice barn]

yes, and i dont understand what is a binomial, i keep seeing more and more formulas with more and more exemples, i just need someone to please help me understand what it is about

the whole binomial thing

please :(

[silver tapir]

i don't understand what do u need help with it is pretty straightforward and easy

[balmy grove]

You know how factorials work, right?

[spice barn]

that's the thing, what is bernoulli's trial? (binomial) are there formulas? am i just lost and misunderstood the whole thing ?

[spice barn]

yes

[silver tapir]

do yk the general formula for expanding an expansion?

general equation

[spice barn]

yes i do

[silver tapir]

basically binomial is when u have two terms

like (x+y)^n

x and u are two terms

y*

i thought u were asking for pascals triangle sry

[silver tapir]

do u need the derivation or what

[spice barn]

oh uh no i was just lost about what was binomial about

[balmy grove]

They're actually talking about the binominal coefficient from the Pascal's triangle. lol

Look at the formula above.

[spice barn]

oh man im just getting more and more lost

[silver tapir]

sorry i jumped in mid convo without reading older messages

[spice barn]

i think its called binomial ditribution

[balmy grove]

n refers to the amount of possibilities, k refers times you pick from the set

[spice barn]

or something

[silver tapir]

choose k things out of n

[balmy grove]

n ≥ k ≥ 0

That's what the coefficient does.

[spice barn]

ohhhhh!!! alright

thank you so much i feel stupid

[balmy grove]

The rest of the variables from the trial comes from the actual experiment.

[spice barn]

alright so its normal i dont have values, i need the experiment for it to work

[balmy grove]

Actually, I'm misreading a little.

Looks like they refer to the chance of failure and chance of success.

Like I said earlier, designated success and failure.

Coin flips selects either side for success. For dice, they select one or more sides for success.

[spice barn]

so like... i just need to know the formula ? thats it ? i felt like i had to understand something else but now that you explained it i feel like its just meant to be applied

[balmy grove]

Yeah, just having this formula will help calculating the probability of certain statements with binary outcomes.

Usually for "yes or no" questions in simple terms.

Formulas aren't named formulas for nothing.

Most of the time spent is inserting the values(or calculating the values beforehand).

It's not exactly a problem, but a tool.

[spice barn]

thats exactly what i got wrong !! thx

really that helped out a lot

[balmy grove]

You're welcome. Feel free to close this channel via .close.

[spice barn]

alright

.close

[calm coral]

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[blissful carbon]

wroing one

dis one

ok

so basically

my teacher said to do these u do the opposite of what u would do in differentiation

but for this one i would use quotient rule

how do i do opposite quotient rule

[dusky flax]

i don't think theres one

[random escarp]

u can type it as:

e^2x / e^x + 1/e^x

[potent smelt]

You can simplify this into two fractions

[dusky flax]

youd have better luck writing it as e^x + e^-x

[blissful carbon]

oh ok

thank you

.close

[calm coral]

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[calm coral]

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[ionic harness]

I'm not really sure how to approach this. Why would the index be finite for either of these when the order of G seems to be infinite?

[ionic harness]

I'm not entirely clear on fundimental domains in this context either. Like I know finding the fundimental domain might give the order of H_1, though I'm not entierly sure how to approach doing this for an arbitrary equivalence relation

[thick zinc]

what is the meaning of m(2,0) and n(4,3) and 2m+4n,3n

and also of '+' in G=(Z^2, +)

[ionic harness]

its just an additive binary operation. My confusion comes from how the order of this group could be finite when computing the index

my thought towards m(2,0) is when the first component of the paris is the same and the second component of the pairs are the same with opposite polarity

[stoic oyster]

that's not a super good argument isn't it ? [Z : 2Z] = 2 and Z is infinite

[ionic harness]

I was referencing Legranges thm on accident

[stoic oyster]

not sure what you're trying to say here

what do the cosets look like in this case ?

if you had to draw the coset containing (1, 1) for example what would it be ?

[ionic harness]

something like {(a,a) , a in Z}?

[stoic oyster]

no I'm not talking about the subgroup generated by (1, 1)

@ionic harness

(1, 1) + H1, what is it ? how does it look ? that's my question

[ionic harness]

{(2m+1,1); m in Z}?

[stoic oyster]

yea right

if you had to visualize it on the xy-plane, what would it be ?

[ionic harness]

all of the odd integer x values along the line y=1

[stoic oyster]

yeah right, a horizontal line going through (1,1) but only half the points essentially

now let's say we're looking at all the cosets (0,0)+H1, (0,1)+H1, (0, 2)+H1, ..., are these distinct or not ?

[ionic harness]

yes since they would lie along parallel lines in the xy plane. would this be a fundamental domain then?

[stoic oyster]

not exactly the fundamental domain

but it tells you the number of cosets is infinite at least

so the index of G in H1 is infinite

now yeah there's more stuff in G/H1

first of all you can pick the negative y's

so ...(0, -2), (0, -1), (0,0), (0,1), (0, 2)... all give rise to distinct cosets

[ionic harness]

the hint telling me to think about fundamental domains of ~_H is really throwing me off cuz this doesn't seem to be that related other than looking at distinct cosets

[stoic oyster]

that's essentially what fundamental domain means from what I can gather from wikipedia yes

I didn't know this name before tbh

[stoic oyster]

and there's also the fact that these cosets only cover the even x-coordinate points

so you also have ...(1,-2), (1, -1), (1, 0), (1, 1), (1, 2).... to get the odd x-coordinate ones

all these guys would form a fundamental domain

[ionic harness]

so the index is infinite because there are infinitely many distinct cosets that partition G in the form of (x,y) + H_1

[stoic oyster]

yes

[ionic harness]

ok cool so for the second one the index appears to be finite since it looks something like this as n scales

(2m,0)
(2m+4,3)
(2m+8,6)

so using the same idea as before it kinda seems that we can only add (1,1), (1,2), (0,1), (0,2) to make distinct cosets?

since i think the only way to make a distinct coset is for 2m+4n to be odd and 3n to be 0,1,or 2

[stoic oyster]

well the index is finite yeah

I'm not super sure how you get your conditions

[ionic harness]

im trying to find the way to make distinct cosets. Since 2m+4n is always even i figured it being odd would make a distinct coset

[stoic oyster]

it's a bit more complicated than that

the problem with the fact that the index is finite is that you really have to find all the cosets exhaustively, you need to give the exact number in the end

you can't just be like "ok i found infinite cosets, even if I haven't found them all, done" like the previous q

so just saying I found some distinct cosets doesn't cut it

that's what it looks like when you draw H2

a nice grid

the lines are here to represent the (2,0) and (4,3) from the def of H2

each integer point in Z^2 has to end up in some parallelogram

and by some amount of translations by (2,0) and (4,3) you can bring every point in Z^2 inside this parallelogram

i.e. writing every point p = q + h where q is in the pink thing and h in H2

in other words, the number of cosets is essentially the number of integer points inside the pink parallelogram

[ionic harness]

So any ordered pair of integers inside the parallelogram would generate a distinct coset?

Yeah

[stoic oyster]

with a bit of precautions

you have to be careful with the boundaries

you can't accept all the boundaries cause you'd be counting some cosets twice

[ionic harness]

It can’t be on the boundaries cuz then it wouldn’t be distinct

[stoic oyster]

well you have to keep some, not all

1 boundary for each "direction" (2, 0) and (4, 3) really

for example bottom and left

[ionic harness]

So my cosets would be (2,1) + h, (3,1) + h, (3,2) + h, and h

[stoic oyster]

it's like for Z and 2Z, the cosets are [0] and [1], you don't have [2] cause that's the same as [0] (that's the equivalent of rejecting one boundary)

but rejecting both would be the equivalent of removing [0] also

you're missing cosets now

[ionic harness]

Would (4,2) + h be distinct? Cuz isn’t it just 2*(2,1)?

[stoic oyster]

yes

the only equivalences there are are translations by (2,0) and (4,3) that's it

[ionic harness]

Ok

[stoic oyster]

so 6 points really

[ionic harness]

So my index would be 6 then since there only exists 6 distinct cosets?

[stoic oyster]

yea

[ionic harness]

Ok thanks a ton I was having a lot of trouble conceptualizing these kinds of problems

.close

[calm coral]

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[calm coral]

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[wraith nest]

,tex \displaystyle{\displaylines{\frac{\sqrt{x+a}+(a+1)\sqrt{x-a}}{a}=\frac{(2-p) +\sqrt{p(p+2a)}}{\sqrt{p+2a}-\sqrt{p}}}}

[potent lotus]

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

[wraith nest]

I am trying to simplify this problem, the answer is x=p+a, but idk how to take this equation there.

What I have tried is just rationalizing the rhs side but don't know how to go further

[primal bear]

i dont know if rationalisaing is the best call here

[fast heath]

this is straight up hell

😭

[wraith nest]

:(

[buoyant moat]

Have you tried just multiplying both sides' denominator and seeing what that gets you

[fast heath]

yeah try that

[wraith nest]

The denominator gets cleaned up after rationalization

Only 2 gets left on rhs

[buoyant moat]

what do you have after rationalisation

[wraith nest]

Just a second

@digital oak

[wraith nest]

My friend who has solved this equation suggested that, he is not currently available so asking here

If I proceed to square I may get rid of roots but it would be absolute hell and this is supposed to be solved under 5 mins

[primal bear]

fair enough

this is hell

i cant believe youd be asked this on a test

if youre doing it just for curiosity or something let wolffram handle it for you

[wraith nest]

It's in the sample paper, the test will be on 30th and a similar question may be asked so yeah

[primal bear]

😭

[wraith nest]

Can i post another question here only?

[primal bear]

i dont think so

[wraith nest]

Ohk

[digital oak]

I think I got it

Just how would I help you, post answer?

[wraith nest]

Probably telling me what to do next or some hint to reach it

[digital oak]

Rationalize the right part

The left part can stay

[wraith nest]

@digital oak

I already did that I think

[digital oak]

Hold up let me look

Um I think u did wrong the first step

[wraith nest]

Ah yes

Mb lemme redo

[calm coral]

@wraith nest Has your question been resolved?

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[wraith nest]

Got it thanks again

[calm coral]

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[fast heath]

for part a

what are all the upward forces

[simple musk]

let me do a diagram

[fast heath]

just tell me what forces can you see

theres 2 scales