help-39

[still dawn]

lol

[tame smelt]

Can't I do all this without the absolute value?

Oh hi there

[quick bramble]

Nah, it's actually the most compact way to cover every case.

[tame smelt]

[quick bramble]

You could do without it, but you'd be checking a lot more cases.

[tame smelt]

🗿

Okay okay

[quick bramble]

Every absolute value inequality is actually two inequalities.

[tame smelt]

So can you tell me all the cases that are available now

How much are they

[quick bramble]

|x| < 1 says both that -1 < -x, and x < 1.

So everywhere we've used an absolute value multiply the number of conditions by 2.

[tame smelt]

That's a lot lmfao

4?

[quick bramble]

Looks like 8 or 16, not sure tbh.

[tame smelt]

OH MY GOD

so this is without and absolute value or with?

[quick bramble]

Without.

[tame smelt]

Ah.

So 4 with

2

[quick bramble]

|x| ≤ |y| is 4 statements.

[tame smelt]

Ah ok

[quick bramble]

|y| ≤ |x| is another 4.

[tame smelt]

Conclusion:

I am screwed

[quick bramble]

So checking |x|-|y| is 8 statements if you do it the hard way.

No, we are basically done.

[tame smelt]

Oh?

[quick bramble]

Do the two cases I said for you to do.

We want |x|-|y| < 1.

[tame smelt]

Yes

[quick bramble]

So get there with the two cases: |x| ≤ |y| and |y| ≤ |x|.

You are allowed to use |x| ≤ 1 and |y| ≤ 1.

[tame smelt]

Oh so we can't just say x is positive and y is negative

We have to try both

[quick bramble]

We already have.

That's how we have x + y = |x|-|y|.

[tame smelt]

Hmmm

[quick bramble]

One being positive and the other negative doesn't tell us which one has a bigger absolute value.

Which is why we need to check both |x| ≤ |y| and |y| ≤ |x|.

I say we're done because we have a single substitution left to do, but I want you to do it.

[tame smelt]

And that

Is why you confuse me

Lmfao

Ok I'm going to school now

I'll check it out okw

Omw

[quick bramble]

k

gl

[tame smelt]

Thanks I'm gonna need it

[quick bramble]

Case 1: |x| ≤ |y|.

So |x|-|y| < 0 and we're done.

Case 2: |y| ≤ |x|.

Here, you can just say that |x| - |y| < |x|.

And we use |x| ≤ 1.

Done.

So we have x + y < 1.

Contradiction achieved.

[still dawn]

@tame smelt leaving?

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[midnight haven]

Hello, all done? can someone help me out?

[still dawn]

see how to get help

[still dawn]

via https://discord.com/channels/268882317391429632/488120190538743810

[pearl pond]

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[midnight haven]

.ask

.ask Factorizing

@still dawn channel 0 is occupied

[midnight haven]

Factorizing and sketching f(x)

.rotate

How would we factorize this?

and find the roots

[merry stirrup]

,rotate

[midnight haven]

cheers

[merry stirrup]

Do you know how to factor cubic polynomial

[midnight haven]

I know how to factor where there isn’t a 3rd number 🙂

I don’t know what it’s called

[pulsar lark]

Third degree polynomial

[midnight haven]

https://youtu.be/rnAKWPM-YBc

[pulsar lark]

Or cubic function

[midnight haven]

so this should explain how to factorize my question??

[merry stirrup]

Find a value of x where you get a reminder 0@midnight haven

[merry stirrup]

Yes

You could watch it

[midnight haven]

I will, thank you

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[midnight haven]

.ask

.ask Transformation Graph

[pearl pond]

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[midnight haven]

how would i find coordinates

more clear view

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[balmy ember]

Hi please help me with this problem
Show that this sequence is convergent:
$$Sn=1+1/1!+1/2!+1/3!+....+1/n!$$
Thanks

[jolly parrot]

Zyssalone

[balmy ember]

I tried (Sn+1) - (Sn)

but couldnt prove if its increasing or decreasing

[dark aurora]

look at Vn= Sn + 1/n and prove its decreasing

[balmy ember]

okay ill try

[dark aurora]

you will have adjacent sequencies

so S(n) is convergent

[balmy ember]

OHhhh okay

i proved that Vn is a decreasing sequence

and convergent since its understated by 1

[dark aurora]

and you have that S(n) is an increasing sequence as well, with S(n) - V(n) = 1/n converging towards 0 right?

[balmy ember]

yeah

owww

OKay

i get it

wow

wich means they are adjacent sequencies

Thanks man

[dark aurora]

np

[balmy ember]

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[sterile helm]

Hello could anyone help me set this problem up I’m a bit confused

[sharp quest]

what have you tried?

[sterile helm]

I’ve tried
C(50,5) multiplied by C(40,1)

[sharp quest]

ok, so that's the number of outcomes from the lottery drawing

Now, what's the probability my lottery ticket matches those numbers exactly?

[sterile helm]

it’s 1/ 184,750,400?

[sharp quest]

yep

i think, if that's what that product comes out too

[sterile helm]

okay I think I may have made an error when I first tried to solve it

Thanks

[sharp quest]

,w (50 choose 5)(40 choose 1)

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Zybikron#5611

[sterile helm]

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[wanton roost]

help

<@&286206848099549185>

[pearl pond]

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[wanton roost]

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[frozen chasm]

The pairs i got was (1,1), (1,2), (2,1), (1,3), and (3,1)

is it 5?

[ionic quiver]

2,2

it is missing

[frozen chasm]

is that all that's missing?

(1,1), (1,2), (2,1), (2,2),(1,3), and (3,1)

6?

[ionic quiver]

yes rest is fine

[frozen chasm]

ok, thank you

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[spare citrus]

Help please

[pearl pond]

Please don't occupy multiple help channels.

[midnight haven]

Hopefully that helps for the first one

For the second one, follow the same steps but leave the variable a in your answer

Hopefully that helps

[spare citrus]

Thanks bruda

[pearl pond]

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[wind fox]

trying to find the curve length of $\sqrt{2}ti+e^tj+e^{-t}k$

[jolly parrot]

InfinityMango

[wind fox]

got to $\int{\sqrt{2+e^{2t}+-e^{-2t}}$

but dont know where to go from there

[jolly parrot]

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

[shell oxide]

I’d use calculator for curve length integration problems

[wind fox]

prof dosent let us use calculators

[shell oxide]

Alright, fair enough

Curve length is the sum of infinitely many Infinitely small distances $\int{ds}$

[jolly parrot]

❤ Pauli Excluder ❤

[shell oxide]

Tiny distances can be expressed in tiny Pythagorean theorems

$\int{\sqrt{dx^2 + dy^2 + dz^2}}$

[jolly parrot]

❤ Pauli Excluder ❤

[shell oxide]

We can divide the inside the sqrt by dt^2 and multiply the outside the sqrt by fr

[wind fox]

yeah, thats how I got to the integral originally

[shell oxide]

$\int{\sqrt{dx^2/dt^2 + dy^2 /dt^2 + dz^2 /dt^2} dt}$

[jolly parrot]

❤ Pauli Excluder ❤

[wind fox]

wait, (e^t)^2 is e^2t right

[shell oxide]

U r right

Ohhh we DID get the same thing

I thought I was leading to somewhere productive

My bad

Also for your k component in trout integral the square of the derivative of e^(-t) is e^(-2t), this term is not negative

[wind fox]

oh

god

that makes everything so much easier

[shell oxide]

Wait it does? Can u explain to me?

[wind fox]

was trying to figure out how to integrate that last one

because you can rewrite it as 2cosh(t)

[shell oxide]

Whoop, that’s outside the scope of my knowledge lol

[wind fox]

just hyperbolyc trig identities

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[somber yacht]

What does the L symbol mean?

[hollow axle]

well considering the developpement of e^x, seems to be the factorial

[somber yacht]

alright thank you

[hollow axle]

np

[pearl pond]

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[teal saffron]

can someone help me how i would solve this

[teal saffron]

im trying to solve it and i get the answer

-5n + 17

[lavish hare]

there’s no equal sign… how are you supposed to solve?

do you mean simplify?

[teal saffron]

idk i got a whole worksheet of these equations like this

[teal saffron]

is that what its called?

[lavish hare]

maybe

if you simplify you just group together like terms

[teal saffron]

its like this

[lavish hare]

yeah it’s simplifying

To solve it you just need to know to properties

[teal saffron]

wait can u circle which one is the property

and like terms

[lavish hare]

1. a(b+c)=ab + ac (Distributive property)
2. ab+ ac= a(b+c) (like terms)

@teal saffron

[teal saffron]

wait so a(b+c) is same as ab + ac

[lavish hare]

yeah

you can distribute a to b and c

[teal saffron]

what does it mean by distributing

it looks like im multiplying them

[lavish hare]

yeah

it’s called distributive because you multiply both terms

6(x+2) = 6x + 6*2 = 6x +12

[teal saffron]

so if i wanted to solve this it would be 3n -2n + 10

[lavish hare]

careful with the signs

-2(5) = -10

[teal saffron]

so it would be 3n - 2n - 10

[lavish hare]

yup

[teal saffron]

alright thats pretty easy then nvm

[lavish hare]

what did you get as your final answer?

[teal saffron]

i was multiplying the 3n and 2 and then i multiplied 2 and 5 i think

for before

[teal saffron]

this 3n - 2n - 10 right

[lavish hare]

you have two n terms

remember property 2

like terms

[lavish hare]

1. a(b+c)=ab + ac (Distributive property)
2. ab+ ac= a(b+c) (like terms)

[teal saffron]

so i have to solve 3n - 2n

but i dont know what n is

[lavish hare]

it’s okay

start with the formula

ab + ac = a(b+c)

Let a=n, b=3, c=-2

then just plug it into both sides

[teal saffron]

so n * 3 + n * -2 = n(3 - 2)

[lavish hare]

yeah

[teal saffron]

how does it help me to solve what 3n - 2n is without knowing n value?

[lavish hare]

n(3-2) = n(1) = n

so (3n-2(n+5)) = n-10

and that’s the simplest form you can get it in

@teal saffron

[teal saffron]

its still confusing me

theres so many brackets and not knowing the n variable to solve it seems impossible

[teal saffron]

oh wait this is what im supposed to get as the final answer?

[lavish hare]

yeah

n-10

[teal saffron]

so the n-10 is how i would solve 3n - 2(n + 5)

so i just subtracted 3-2 to get 1n

[lavish hare]

no it’s how you would simplify

[teal saffron]

what is the difference

[lavish hare]

solving means getting n=…

like n+1 = 3

n =2

that’s solving for n

but if you just re arrange the terms thats just simplifying

You can simplify an expression or you can solve an equation

@teal saffron you wanna try another q?

[teal saffron]

let me do the rest of my worksheet first

[lavish hare]

okay

[teal saffron]

its getting late and i need to finish this

[lavish hare]

Imma go offline then

cya

[teal saffron]

so is just literally make it the simplest form

[teal saffron]

thanks for all the help good explanations

[pearl pond]

@teal saffron Has your question been resolved?

[pearl pond]

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[neon pier]

hello, can someone tell me why my answer is wrong?

[copper kestrel]

how did you arrive at your answer?

[neon pier]

so i multiplied top and bottom by sqrt7

[grim patrol]

lol

no

how about the 2

did u cancel the 2

[neon pier]

yes?

[grim patrol]

yea

u cant be doing that

[neon pier]

was that wrong

why

[grim patrol]

yes

very wrong

like

imagine

$\frac{1+1}{1}$

[grim patrol]

its not the same as

$\frac{1}{0}$

[grim patrol]

u can only cancel multiplication

[neon pier]

wdym

[grim patrol]

of non zero numbers

not addition

[neon pier]

interesting

so is it

[grim patrol]

yes

completely wrong

yes

[neon pier]

2sqrt7 over -14

[grim patrol]

er

idk

[neon pier]

same bro

[grim patrol]

can u write it step by step

D:

like

whats the denom

can u write em both

[neon pier]

the denominator is 2-sqrt7

wdym write them botm

both

[grim patrol]

like

dont write them separately

i cant understand what u r doing

is so confusion

[neon pier]

oh

wait

is that right

can there be a negative in the bottom?

[grim patrol]

no like

how did u get there

what step did u take

D:

[neon pier]

oh

one sec

was i supposed to multiply by the conjugate

[midnight haven]

Hopefully this helps

[midnight haven]

Correct

To figure out what you multiply it by, you get the denominator and change the sign of the root

[neon pier]

oh yeah

i did that later on but kept on messing up on the denominator

thank you!!

[midnight haven]

All good :)

[neon pier]

thank u both

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[pale token]

Can someone please explain what the question wants from me, i get that its a coordinate, but im not sure how to use the equation i found

[pale token]

ive found the equation of bm, and now im a little confused as to how i find out P

heres the previous part

wait nevermind i think i got it

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[maiden compass]

Hey

[fiery vortex]

heyyyy

[maiden compass]

how do i solve this over Z5

😦

[fiery vortex]

welp sadly can’t help but still, heyyyy

[maiden compass]

heyyyyy broooo ❤️

[fiery vortex]

easiest way

just use graphing calculator

[maiden compass]

by gauss ofc

my lecturer told us, that we can solve it regular and then convert all the numbers to z5

[fiery vortex]

plug in values into GC if u are allowed to use one

[maiden compass]

GC?

oh graphic calc

nope

is it legal just to add rows without multiplying them with a scalar?

i mean, is R2+R3 in R3, legal?

[pearl pond]

@maiden compass Has your question been resolved?

[maiden compass]

<@&286206848099549185>

[edgy heart]

Hi

[pearl pond]

@maiden compass Has your question been resolved?

[chrome patio]

so far so good. IMHO, it's better to write the augmented matrix [A | b] with the vertical bar in the middle

imagine adding two equations

[maiden compass]

already solved

i wonder if i did it correct

[chrome patio]

,align
a &= b \tag{$R_1$} \
c &= d \tag{$R_2$} \
a + kc &= b + kd \tag{$R_1 + kR_2$}

[jolly parrot]

vin100

[chrome patio]

i only looked at the solutions. LGTM

[pearl pond]

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[midnight haven]

8

[shrewd basin]

What have you tried?

[midnight haven]

i wanted to put 8sina in cos but i dont know how

[shrewd basin]

Put 12cos^2 in sin^2 instead

[midnight haven]

so what is it its 12sin²2(pi/2-x) ?

[shrewd basin]

No use $$\sin^2{x} + \cos^2{x} = 1$$

[jolly parrot]

Pure

[shrewd basin]

Ah what

[midnight haven]

so 12cos²x+sin²x=1

then its 1-12cos²x=sin²x

[shrewd basin]

No $$12(1-\sin^2{a}) = 5 +8\sin{a}$$

[jolly parrot]

Pure

[midnight haven]

how u got 12(1-sin²a)

[shrewd basin]

Because cos^2(x) + sin^2(x) = 1

And so cos^2(x) = 1-sin^2(x)

[midnight haven]

and then

[midnight haven]

how u got 12(1-sin²a) after this

[shrewd basin]

Replace cos^2(a) with 1-sin^2a

[midnight haven]

ok

@shrewd basin so its 12-12sin²x=5+8sinx in the end ?

[shrewd basin]

Ye

[midnight haven]

ok isee

[midnight haven]

wait its 12-12sin²x or 12-sin²x

[shrewd basin]

12-12sin^2x

[midnight haven]

@shrewd basinand if i have sin²b-cos2b+1=0 i can still use cos²b-sin²b=1 ?

@shrewd basin hey are u there

[shrewd basin]

I’m not sure what you’re asking

[midnight haven]

@shrewd basin i have sin²b-cos2b+1=0

i can do cos²x+sin²x=1 or no

[midnight haven]

like this

@shrewd basin now u understand

@shrewd basin i want to change the sin²b in cos²b

i can or no ?

i did it but i didn't get the right answer

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

@shrewd basin hey

.close

[pearl pond]

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[junior pawn]

$\int \frac{ln(x)}{1+x^2}$

[jolly parrot]

Max.cc

[junior pawn]

the u-sub is u = 1/x

but even by knowing that idk how to deal with it

[pearl pond]

@junior pawn Has your question been resolved?

[severe quartz]

Thank you

I legit spent 10 min raging before i realized

[junior pawn]

?

[severe quartz]

That it was out of my scope🗿

[junior pawn]

wait me too lol

but normally if they give use the u-sub it must mean that is possible

[severe quartz]

I checked from two websites and plus all of my methods fail

[junior pawn]

keep in mind that the webisites don't use the u-sub

[severe quartz]

They do

[junior pawn]

https://www.integral-calculator.com/

the mine use complex number

[pearl pond]

@junior pawn Has your question been resolved?

[pearl pond]

@junior pawn Has your question been resolved?

[pearl pond]

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[topaz pumice]

I need help

[arctic pasture]

We all do

[midnight haven]

send the question

[topaz pumice]

I'm in Algebra

and hold on

<@&286206848099549185>

I have to go

so can you guys do this for me?

sorry

[midnight haven]

what questions tho

[tacit mulch]

• When asking for help, do not insist on getting just the answer; we are here to help you learn, not do the work for you. Likewise, if you are providing help to others, try your best to explain and elaborate instead of simply giving away the answer.

[topaz pumice]

hey I'm back

what I meant was leave some instructions

sorry

I forgot to say that

or some hints

[pearl pond]

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[fringe solar]

#help-39

[fringe solar]

3x + 6 * 2

[shell oxide]

What is your question?

[autumn tinsel]

3x + 6*2? what

Can you send a picture of the question

[pearl pond]

@fringe solar Has your question been resolved?

[fringe solar]

no

i finished it is ok

[pearl pond]

@fringe solar Has your question been resolved?

[candid spire]

@fringe solar type .close

[pearl pond]

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[pearl pond]

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[fringe solar]

.close

2x+6*3

[pearl pond]

@fringe solar Has your question been resolved?

[pearl pond]

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[midnight haven]

can someone help me

[midnight haven]

also is this correct?

[supple epoch]

yeah

[midnight haven]

Can you help with the first one

[supple epoch]

Isn't it similar?

[midnight haven]

Yeah, but I don't know how to find x because BA doesn't have a given value

[supple epoch]

What's tan x in the first one?

Can you tell which is the hypotenuse in the first triangle?

[midnight haven]

yeah BA

because the right angle

at the top is opposite of the hypotenuse

[supple epoch]

And definition of tan theta

[midnight haven]

wym

like tan(52)

[supple epoch]

umm

yeah tan 52

what will you write it as?

AC/BA?

[midnight haven]

tan(52) = AC/BA

but what is BA?

[supple epoch]

No AC/BA is incorrect

[midnight haven]

oh

[supple epoch]

tan of an angle is opposite side/adjacent side

[midnight haven]

yeah

[supple epoch]

Here adjacent side is the side other than the hypotenuse

[midnight haven]

isnt that opposite/adjacent

ohhhh

adjacent cant be hypotenuse?

[supple epoch]

there are two adjacent sides

hypotenuse is one of them

[midnight haven]

BA and BC

yeah so just use BC since given value then

[supple epoch]

But in the definition of tan, the adjacent side used is not hypotenuse

[midnight haven]

ah ok

I get it

ty

@supple epoch

[supple epoch]

Correct

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[iron shore]

i can't choose between a and c

[iron shore]

the unknown concavity at 0<x<1 throws me off

[tacit mulch]

How did you rule out B and D?

Notice the question is: which of the choices could be false

Three of the choices are definitely true. One of them is not decidable with the given information

@iron shore

[iron shore]

Oh

Hahaha

I didn’t read the question

D

[old marsh]

yea those collegeboard q's are pretty tricky sometimes

[iron shore]

I just assumed it would be asking true

.close

[pearl pond]

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[pearl pond]

@zenith basin Has your question been resolved?

[pearl pond]

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[static zephyr]

uh, hello ?

[static zephyr]

actually nvm

.close

[pearl pond]

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[pale token]

can someone help me with the next step?

[buoyant panther]

u know where to start?

[pale token]

i found out that y=9 so now i have (4,9) as a point, but what do i do to get the equation

[fiery vortex]

differentiate

[pale token]

i differenciated it, then subbed in x

[fiery vortex]

to find eqn of tangent

[pale token]

and got my y

[fiery vortex]

yes

[pale token]

now what

[fiery vortex]

no

[pale token]

im a little lost sorry

[fiery vortex]

when u sub in x, that’s not y

that’s dy/dx

[pale token]

oh yea your right

[fiery vortex]

so 9 is ur gradient

[pale token]

so i can sub into

[fiery vortex]

a tangent is a straight line yes?

[pale token]

y=mx+c

[fiery vortex]

and u found ur gradient

yep

[pale token]

thanks 👍

.close

[pearl pond]

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[midnight haven]

Ayeee that's my man skate right there, teaching people and tings 😎👍🏻

[pale token]

too good man, too good

[fiery vortex]

LEXXXX

[midnight haven]

Yeee

[fiery vortex]

[midnight haven]

SKATEE

[pale token]

.reopen

[pearl pond]

✅

[pale token]

wait wait

i got y=9(4)+c i dont remeber how i get y so then i can find c

do i sub y=0?

or do i sub x into the original and get y

[fiery vortex]

u have y=9x+c as the eqn for the tangent

now look at ur original eqn

when x=4 what is the y coordinate

for the original equation

because we’re finding the tangent at that point

[pale token]

yea i got y from subbing in x into the orginal

[fiery vortex]

what

[pale token]

hold on

[fiery vortex]

I thought u did this

[pale token]

ye

but after

[fiery vortex]

and found the y value

wait so the 9 is which one

the original or the tangent

[pale token]

,rotate

[pale token]

like this

?

[fiery vortex]

okay yeah

that’s dy/ex

dx

so now

I want you to sub in x=4 into the original eqn

not the tangent eqn

,w 4-16/2

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Springsskateboard#8145

[pale token]

yea y=-4 ?

[fiery vortex]

yep

so the coordinate that lies on the tangent

is (4,-4)

[pale token]

yea

[fiery vortex]

remember u had y=9x+c

now to find c

plug in (4,-4)

[pale token]

ye ye thats what i got, thanks

[fiery vortex]

nice

np

[pale token]

.close

[pearl pond]

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[magic gust]

Hello, does anyone speak Spanish?

i need help with a question

[fiery vortex]

so senor

si

okay I don’t

mine is very basic

but u can google translate

[pearl pond]

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[magic gust]

ok

[magic gust]

Can you help me with some exercises?

[timber void]

send your question

[pearl pond]

@magic gust Has your question been resolved?

[pearl pond]

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[graceful wing]

simplify cos^2(x)+sen(x)=0

[graceful wing]

i did 1-sen+sen=0 but -sen+sen is 0 and 1 is not equal that 0

[sharp quest]

sen?

[graceful wing]

sen=sin

[timber void]

Sen x?

Oh

[sharp quest]

cos^2(x) = 1 - sin^2(x)

[graceful wing]

and where u get 1-sin^2

[sharp quest]

pythagorean identity

cos^2(x) + sin^2(x) = 1

[graceful wing]

yes

[sharp quest]

your substitution is wrong

[graceful wing]

i did cos^2= 1-sin

[sharp quest]

which isn't true

[graceful wing]

why

[cursive fulcrum]

It should be 1- sin^2(x)

[sharp quest]

^^

[graceful wing]

ohh yea

the ^2

i didnt remember

thanks

i know how to do now

thanks 🙂

.close

[pearl pond]

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[midnight haven]

sen is sin, I'm just Spanish

[buoyant panther]

you're missing 1/2

and

then do integration by parts

once again

[midnight haven]

Where is 1/2?

[buoyant panther]

u = cos(t) dv = e^(2t)
du = -sen(t) v = 1/2e^(2t)

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[unborn field]

Situation: I have a deck of 36 cards that contains 4 cards with the number 6 and 4 aces.

Task: Whats the probability that player A will receive EXACTLY 2 6's and 2 aces when 9 cards are dealt to each of the 4 players?

Question: I am wondering my approach does not work:

(4 * 3 * 4 * 3 * 28 * 27 * 26 * 25 * 24)/(36 * 35 * 34 * 33 * 32 * 31 * 30 * 29 * 28)

the 4 * 3 * 4 * 3 should be 2 aces and 2 6s. 28 regular cards remain, which I then try to select.

The solution solves the task using a different approach, but I would like to find out how to solve the problem with a calculation similar to the one above.

Thanks for any help

[real jungle]

Here you are calculating only 1 possibility

I think that's it

Because you have many ways to deal the cards

[unborn field]

can you elaborate on how Id incorporate that fact into the calculation then?

[real jungle]

You have to use binomial coefficients here

Because the order in which the cards are dealt isn't relevant here

So you have to take into account all the possibilities of dealing the cards

Idk how to say it in english

[unborn field]

I know what you mean

[real jungle]

Like the coefficients

[unborn field]

the official solution uses the binomial coefficients

but I have been wondering whether there is a way to solve it "my way" with a calculation similar to the one above

[real jungle]

I'd say your solution is logical but you have to multiply it by the number of possibilities

But i'm not sure of that

Try it

[unborn field]

well if I divide the solution by my result I am getting 756,x

and idk how I should arrive at 756,x possibilities

[real jungle]

Well maybe you can't

You have to do it with binomials coefficients

That's all i can explain in english

[unborn field]

alright thanks nonetheless

[pearl pond]

@unborn field Has your question been resolved?

[vestal tapir]

your approach is putting two aces, then two sixes, then the rest of the cards

it's just immediately obvious that you;re missing all the hands where the aces are not at the start together

if i'm not mistaken your answer is divided by 756

not choosing 2 spots out of 9 for the aces, then 2 out of 7 for the sixes

[pearl pond]

@unborn field Has your question been resolved?

[unborn field]

youre most likely right, but how can I calculate that the number of possibilities is 756?

[vestal tapir]

by choosing 2 spots out of 9 for the aces, then 2 out of 7 for the sixes

[unborn field]

ah

its 126*6

[vestal tapir]

it's 36*21

ah yeah, 126*6 also makes sense

either or

[unborn field]

but wait a second

[unborn field]

2 options out of 9 would give 36 possibilities right

if I do that twice I get much more than 756 dont I?

ah nvm

2 out of 7

the second time

not 9

sec

hmm Okay I think I got it

So I assume we dont need to consider the order of the 2 6s and 2 aces because they are treated as the "same" right? And we dont need to do anything with the rest because that simply follows once we have picked the 4 places for the relevant cards?

[vestal tapir]

nice work then

yes for the second part, the positions of non special cards follow automatically

which order do you mean

first 6 and second 6 are "the same"

[unborn field]

well when we use binomial coefficients we divide by the amount of permutations that are possible with the number of places we picked

thats what I mean with "order"

[cyan heath]

.close

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[hoary lily]

What are the first 10 integer inputs for which you get integer outputs?

[pearl pond]

@hoary lily Has your question been resolved?

[pearl pond]

@hoary lily Has your question been resolved?

[hoary lily]

<@&286206848099549185>

[pearl pond]

@hoary lily Has your question been resolved?

[pearl pond]

@hoary lily Has your question been resolved?

[hoary lily]

hmmm

[hoary lily]

The first 3 examples of integer inputs that lead to integer outputs that I could find

are x=1, x=8, and x=49

that was just after manual searching

I'm wondering if there's a way to analyze the function to create a system that can solve for ALL integer outputs, or at the very least, the first 10

[sharp void]

really weird problem

because if you assume

$\frac{x(x+1)}{2} = n^2$ for some $n \in \mathbb{Z}^+$

[jolly parrot]

omeganebula

[sharp void]

then it goes back to $n^2 = \frac{m(m-1)}{2}$ for some $m \in \mathbb{Z}^+$

[jolly parrot]

omeganebula

[sharp void]

for x to be an integer

then we'll have to again impose conditions so that n is an integer

[hoary lily]

so is there no way to do it?

[sharp void]

idk

you could try writing a code for it

then oeis :)

[hoary lily]

well yeah, but that'd be kinda trivial

[sharp void]

could give you some insight

[hoary lily]

ill look through it

ill close the channel in an hour if I can't find anything

[sharp void]

Aight

[hoary lily]

https://oeis.org/search?q=0%2C1%2C8%2C49%2C288%2C1681%2C9800%2C57121%2C332928%2C1940449&language=english&go=Search

wtf is this

I mean yeah

it gives me the solutions

but I can't understand the reason why at all

there's way too many references on this page

[cinder flower]

i think you can solve this with some pell equation theory

[hoary lily]

How so?

[cinder flower]

i'm lazy (and also very fuzzy on the details lol) so let me try to find a reference

but i think i saw a video on a similar problem a while ago

[cinder flower]

https://en.wikipedia.org/wiki/Square_triangular_number how's this?

[hoary lily]

That’ll do for now.

.close

[pearl pond]

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[lavish hare]

(New to linear algebra), what exactly is the point of transposing a matrix

[lavish hare]

It was only briefly mentioned in my book

[plush bramble]

lots of things. what answers are you looking for

[lavish hare]

just an example or two

it can be more advanced I just wanna note it down for the future

[tall tangle]

depressed cube

[lavish hare]

I can’t find anything about it online

if you could can you send a link?

[tall tangle]

lol im just referencing user riemann

😭

[lavish hare]

[cinder flower]

having trouble trying to explain this is in a simple way lol but something nice is how the column and row spaces of a matrix and its transpose are related

[lavish hare]

it’s all good

if you have the time could you explain it normally

I’d save it for later

[cinder flower]

i was thinking of fredholm's theorem

a more friendly thing might be

you can define a symmetric matrix to be a matrix that's equal to its transpose

and symmetric matrices come up a lot and have nice properties

[lavish hare]

would you cover this theorem in a regular linear algebra class?

[cinder flower]

i've seen it done in a first linear algebra class, but i think usually not

[lavish hare]

Okay I’ll try to get to it

also

are symmetric matrices just square matrices

or am I getting it mixed up

[cinder flower]

nope, they're matrices that are equal to their transpose 🙃

[lavish hare]

okay 👍

thanks for all the help I’ll look into it in the future

[cinder flower]

yep ^-^

[lavish hare]

.close

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[odd trench]

How would you find the equation? Any help would be appreciated

[odd trench]

I may be flawed in my understanding, but from what I know would you divide x by -1? (for the dialation) If this is the case the graph would not be possible. I do understand that in the question, the graph would be a reflection in the y-aixis, and hence the negative is placed on the outside of the surd, but for every other question it seems , as a I previously stated that you would divide the x value by the dialation factor? That is the part i do not get.

[tight storm]

Sort of

To move up two units, you want all the y values to increase by 2

So new_y = y + 2

y + 2 = sqrt(x) + 2

so you've sorted up two units

to dilate horizontally you want to inverse the x so new_x = -x

so your final equation should be y = sqrt(-x) + 2

the dilation doesn't affect the vertical translation so it's fine to do it like this 😄

@odd trench

[odd trench]

I still do not understand the square root of negative x, at least from how I am understanding it that would not be possible? Could you please elaborate?

[tight storm]

If x is -5

The square root of negative x is the square root of 5, so you still have real values

It flips the graph horizontally, which is a dilation of -1

[odd trench]

Ooh I do understand what you are saying. But the answer in my textbook puts the negative of the outside of the root? Is this an error?

[pearl pond]

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[runic glacier]

Construct a relation on the set {1,2,3,4} that is reflexive, antisymmetric, and not transitive.

[runic glacier]

here's the relation i constructed

R = {(1,1), (2,2), (3,3), (4,4), (4,3), (4,2), (3,1)}

pretty sure (4,2) isnt right

but i'm not too sure

[vestal tapir]

well it's optional to have

both are correct
{(1,1), (2,2), (3,3), (4,4), (4,3), (4,2), (3,1)}
{(1,1), (2,2), (3,3), (4,4), (4,3), (3,1)}

[severe quartz]

If (4,2) present and (2,2) present, with (4,2) being present wouldn't that mean it is transitive?

[vestal tapir]

you got me, I don't know the answer to that

[runic glacier]

so i think 4,2 should be there

alr i’ll just go with this

[vestal tapir]

no that's not enough

the fact that {4,1} is absent makes it not transitive

[severe quartz]

What does (4,1) have to do with this 🤔

[vestal tapir]

(4,3), (3,1) requires {4,1}

[severe quartz]

Yea but there are things which still make it transitive

Ohhhh

[vestal tapir]

no

other things don't fix the lack of {4,1} it stays not transitive

[severe quartz]

Yes i get it

[runic glacier]

wait i dont

if we got (4,4) and (4,2)

we also got (4,2) do we not?

[severe quartz]

There is still a pair which makes it not transitive

[severe quartz]

^

[runic glacier]

is this circular reasoning?

[tight storm]

Just because there is an example of transitivity doesn't mean it's transitive

[runic glacier]

hmmmm

[tight storm]

If 4 ~ 4 and 4 ~ 2 then 4 ~ 2

That's not a very helpful statement - you don't need to know anything about transitivity for that

If A and B then B

[runic glacier]

maybe i dont know the definition of a transitive relation well enough

[tight storm]

Well the overall answer to this is no

I believe

is there a question we are still trying to solve?

[runic glacier]

it was this one

[tight storm]

i think this is correct then

[runic glacier]

so what does it mean for a relation to be transitive

for all elements in the relation

and not one example?

[tight storm]

yes

[runic glacier]

alr thanks appreciate the help

.close

[pearl pond]

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[runic glacier]

.reopen

[pearl pond]

✅

[runic glacier]

i guess it was more of a definition thing i just didn’t get that part right

.close

[pearl pond]

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[pearl pond]

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[sinful dew]

I don't even know where to start with this.

Consider Gaussian Elimintation carried out with pivoting by columns instead of rows, leading to a factorization AQ = LU where Q is a permutation matrix.

Show that if A is nonsingular, a factorization always exists. Show that if A is singular, such a factorization may not always exist

[pearl pond]

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[sinful dew]

<@&286206848099549185>

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[pearl pond]

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[wintry nacelle]

could someone help with part b ?

ive got that a+15d = 2(a+20d)

and that a + 15 d = 10

but i dont have the other vlaue

[rocky jewel]

a+15d = 0.5(a+20d)

[rocky jewel]

a + 15 d = 10
a+20d=20
5d=10

[wintry nacelle]

why wouldnt it be 2(a+20d) since its twice as big not half as small ?

[supple spindle]

a + 15d is the smaller one

[wintry nacelle]

ohhhhhh

[rocky jewel]

x is twice b, x=2b

[wintry nacelle]

thank you that makes sense now

what about for this bit ?

[rocky jewel]

d=2, a=-20

did u get this?

[wintry nacelle]

yeah i got that from the simultaneous equations

[rocky jewel]

,sum=n(2a+(n-1)d)/2

do uk this

n=number of terms

[wintry nacelle]

ohh

yeah

[rocky jewel]

if yes, sum=n(2a+(n-1)d)/2=0
n=0 or -2a=d(n-1), 40=2(n-1), n=21

ans is either k=0 or 21

[wintry nacelle]

do u mean n/2(2a+(n-1)d) formula?

im aware of that one

and n/2(a+l)

[rocky jewel]

yh

[wintry nacelle]

im kinda confused where you go the 40, 21 from

[pearl pond]

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[pearl pond]

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[midnight haven]

are these correct?

[pearl pond]

Please don't occupy multiple help channels.

[midnight haven]

1, 2, 4, 5, 6,

[pearl pond]

@midnight haven Has your question been resolved?

[agile ledge]

Yeah it's right

[pearl pond]

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[midnight haven]

@light helm sorry to bother but, at the answer do i have to add the + C?

[light helm]

for definite integrals no

(and by no, don't do it at all at the end)

you "could" have a "C" in the antiderivative, but they'd just end up cancelling

<@&268886789983436800>

$$[x + C]\bigg|_a^b$$
having the $C$ there is fine but unnecessary

[jolly parrot]

ℝamonov

[light helm]

$$x \big|_a^b \wthonk b-a + C$$
would be wrong

[jolly parrot]

ℝamonov

[vital estuary]

Omg i love that

[pearl pond]

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