help-33

[echo bough]

if they look alike

if someonelse's work didnt make sense they are probably doing unnnecessary stuff

[woeful shadow]

wouldnt it just go on forever

[echo bough]

because it should take one line

[woeful shadow]

for the right hand side

[echo bough]

wdym?

[woeful shadow]

you said r = n - r

[echo bough]

i didnt say that

[woeful shadow]

oh

the other person did

[echo bough]

yeah its an abusive notation

try to do what i said

write nC(n-r)

[woeful shadow]

what nCr = nCn-r ?

[echo bough]

like write what its equal to

i feel like you are having trouble with notation and variables no?

[woeful shadow]

I am

this is very new to me

is nCn-r just

n!/(n-(n-r))!(n-r)!

[still temple]

your parentheses are off, but yes

[woeful shadow]

how is this meant to help me

[still temple]

yes

simplify now

expand the - in the brackets

[woeful shadow]

huh

[still temple]

distribute the negative sign inside the parentheses

using the fact a(b+c)= ab + ac

[woeful shadow]

im stumped

[still temple]

how would you simplify something such as $-(x+1)$

[woeful shadow]

ohh like

(n-(n-r))! = (n-n+r)!?

[still temple]

yes

n-n = ?

[woeful shadow]

0

[still temple]

ok and your expression now is?

[woeful shadow]

n!/r!(n-r)!

oh

it’s the same

[still temple]

yes

yes

[woeful shadow]

ah

is that it?

[still temple]

yes

[woeful shadow]

okay

now how do i do b

[echo bough]

see, wasnt that hard

[woeful shadow]

do i substitute nCr as whatever the fraction is

[echo bough]

you should probably write the left hand side of b

express n-1C(r-1)

times n/r

[still temple]

I feel like starting from nCr is even simpler

[echo bough]

and do some algebra

whichever you like best

its honestly the same

[still temple]

just remember the definition of the factorial: [
n!=n\cd(n-1)\dots2\cd1
]

[woeful shadow]

how do i deal with the n/r

[still temple]

what do you have right now

[woeful shadow]

nothing

[echo bough]

express the left hand side

as you did for a with nC(n-r)

express it as a global fraction

[echo bough]

but expand the n-1Cr-1

aswell

[still temple]

write out $\binom{n-1}{r-1}$ fully

[echo bough]

inside the fraction

[woeful shadow]

ahhh

i think i did this very wrong

hello?

[still temple]

how did you get that ((n-1)(r-1))! in the fraction

[woeful shadow]

im not sure

:/

this is so confusing

[still temple]

so in nCr you would have (n-r)!(r)! in the denominator. How would you transform that in the (n-1)C(r-1) case

[woeful shadow]

uhh

((n-1)-r)!(r-1)!

[still temple]

the other r?

[woeful shadow]

((n-1)-(r-1))!(r-1)!

[still temple]

yes great

simplify the first one

((n-1)-(r-1))!

what do you get

[still temple]

okay great

wait

your numerator is incorrect

[woeful shadow]

is it meant to be n-1

[still temple]

yes

now do you have any idea on what to do

[woeful shadow]

no clue

[still temple]

let's start with the numerator

[woeful shadow]

wait does n/r stay normal

or do i have to change that too

[still temple]

you have n*(n-1)! Correct?

[woeful shadow]

yes

isnt that just n!

OH

[still temple]

yes

[woeful shadow]

AHH

I understand now

thanks a lot for your help dude

i appreciate it

have a good day

.close

[marsh citrus]

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[marsh citrus]

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[slate stump]

Hello, I have a question regarding Ramey's theorem.
Given an infinite set X={x1, x2....}
and A a group of finite sub-sets of X.
I need to prove the following:
If for every coloring of X with K colors there's a monochromatic set in A, then there exists n such that for each coloring using K colors of {x1, .... , xn} there's a monochromatic set in A.

[marsh citrus]

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[marsh citrus]

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[still temple]

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of
the opposite plate, 1.61 cm distant from the first, in a time interval of $1.44\cd 10^{-6} \tsx{s}$. The objective is to find the magnitude of the electric field

[still temple]

so what I did is like calculate [
E = \p{9.0\cd 10^9} \f e{\p{1.6\cd 10^{-2}}^2}
]
Which leads me to think there is more to this question than what i wrote

[still temple]

So first question: why is what i wrote incorrect

Second question: What would be a good nudge in the right direction for me

[deft apex]

uniform electric field is a big enough hint

[still temple]

hmm elaborate

[deft apex]

any charged thingy in that uniform field experiences a constant force

[still temple]

yeah

[still temple]

but why is this incorrect in this casE?

[deft apex]

which means the acceleration is constant

[deft apex]

honestly im not sure I even req what this is rn, I can pay more attention in a moment just in a tft game

[still temple]

its like

[
E = k\f q{r^2}
]

[main idol]

You need an equation for force and electric field

[still temple]

wait wdym

[still temple]

isnt this all we need?

[main idol]

There's no acceleration in that equation

[deft apex]

no

[still temple]

okay i understand why what your describing would be correct

but at the same time

im not entirely sure what is invalidating my attempt above either

[fathom ridge]

bzzzzz

That's incorrect because it's for a point charge

[deft apex]

ok so I believe you are describing the non uniform electric field around a charge of e at a distance of 1.61cm, this does not fit the scenario laid out by the question , I think it's better to consider what you can deduce from a constant acceleration , a given distance and time. then move from there with other relationships you know?

[deft apex]

what this guy said

[fathom ridge]

A plate has a different formula for electric field

I think it's $\frac{\sigma}{2\epsilon_0}$ where $\sigma$ is its surface charge density

[elfin berry]

jan Nejon

[fathom ridge]

I don't remember though, derive it for yourself

[still temple]

oh this is a plate

[still temple]

yeah i think i get u

[
F = ma \implies a = \f Fm \implies a = \f{E}{e\cd m} ]
now like [
x-\underbrace{x_0}{=0} = \underbrace{v_0t}{=0} + \f12 at^2 \implies x = \f12 \f E{e\cd m}t^2 \implies E = \f{2xem}{t^2}
]

[still temple]

i did this @fathom ridge @deft apex

would this be fine

[deft apex]

yes the bottom bit was the thing I was getting at

[deft apex]

although shouldn't the charge be on the numerator of this?

[still temple]

oh shit right

[deft apex]

I think after that is ammended it would be gud?

[still temple]

[
F = ma \implies a = \f Fm \implies a = \f{E\cd e}{ m} ]
now like [
x-\underbrace{x_0}{=0} = \underbrace{v_0t}{=0} + \f12 at^2 \implies x = \f12 \f {E\cd e}{ m}t^2 \implies E = \f{2xm}{et^2}
]

[fathom ridge]

Sure I suppose

Looks good

[deft apex]

think its gud?

dimensions line up also <?>

[fathom ridge]

well yeah id expect that

[still temple]

i get like 165

i hope thats fine

[fathom ridge]

,w calc 21.61(1/100)(1.67)(10^(-27))/((1.6)10^(-19)(1.44*10^(-6))^2)

[elfin berry]

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[still temple]

yea i approximate protons as 1.7

lmao

[fathom ridge]

ah I see

[deft apex]

https://tenor.com/view/lizard-dancing-poggers-lizard-dance-poggers-gif-18527737

[fathom ridge]

You could do this without a calculator

The numbers seem to line up well enough

[still temple]

wait @fathom ridge @deft apex

one more q

like

im asked to calculate teh speed of teh proton when it hit the plate

so i can use $v = v_0 + at$

[still temple]

question is teh acceleration

like

[deft apex]

tis constant

[fathom ridge]

eE/m

[deft apex]

u already worked it out previously

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[still temple]

Why the tangent plane is not horizontally inclined rather than vertical ?

[still temple]

The book says that since the gradient vector at P is perpendicular to the tangent vector it is normal to define the tangent plane to the level surface F(x,y,z)=k at P as the plane that passes through P and has normal vector grad F(x_0, y_0, z_0)

But there are two vector that are perpendicular to the tangent vector : grad F(x_0, y_0, z_0) but also the cross product of grad F(x_0, y_0, z_0) and r'(t_0)

So why choose grad F over the cross product of the two vectors as the normal ?

it seems perpendicularity is not enough of a criteria

Oh it is inclined

it embraces the surface right ?

[marsh citrus]

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[prisma trellis]

Someone please check my work

[wild quiver]

Hi genious people

[marsh citrus]

@prisma trellis Has your question been resolved?

[marsh citrus]

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[quaint sky]

does this seem correct mathematically?

[quaint sky]

ping me when you look at it

<@&286206848099549185>

[lone heart]

No

Because you have units in mm then made them in m

You should convert the mm to m first

Then apply the process again

[quaint sky]

but [mm]^2 is m^2

[lone heart]

No it's not

It's mm^2

mm is millimeters, m is meter

You have milimeters squared

And you just changed the units the meters

[quaint sky]

OH

so here is the sample calculation tha tmy ta did, i thoguht i did the same process

[lone heart]

But notice how the units are mm but Pa are N/m^2

So you have to convert still

[quaint sky]

oh i multiply by 1.0 e-6?

to get to m^2

[lone heart]

Yeah

[quaint sky]

ok let me redo

give me a second

Just to make sure, i did this part correct

for the area

[lone heart]

Not quite

If you wanted to convert from mm^2 to m^2, find the area first, then multiply by 1e-6

[lone heart]

Yeah that's it

[lone heart]

If you're using excel to do the calcluations, you should use it to find the area so you can use a more precise number

Because 0.00006 is very truncated

[quaint sky]

Ok let me try

[lone heart]

,w (pi* 81)/4 * 1E-6

[elfin berry]

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[lone heart]

See

It's actually 0.000063617...

[quaint sky]

oh i just used 3.14

i think that should be fine

i used your calculation, i got this

[lone heart]

Yeah that's fine now, a lot more accurate than 907000, which was what you got before

[quaint sky]

should i use 0.000063617 ?

[lone heart]

You can, or your value of 0.000063685, it only differs by 200 Pa

[quaint sky]

alright!

ok so in the next step, i need to find gamma. i asked my friend where gamma came from and she said it was in the lab manual but tbh i cant find it. do you think it is calculated or found? ill send my lab manula.

it just states that it is cp/cv but does not give us an actual value, which is where i am confused

i think it is calculated

[quaint sky]

OH ok

ty

do yoy know where 5.02e-5 is coming from. i know 50.16 is the force, do you think 5.03 is the area?

[lone heart]

Yeah could be

[quaint sky]

so the pressure i just found is the initial pressure?

[lone heart]

I'm not sure, it could be, you might be better off asking your classmates

[quaint sky]

wait let me do the calculation based on the sample calc and ill lyk

okay so i used the TA data

here is where im confused

i got this

but by 10^-11

not 10^-5

[lone heart]

That 0.008 is probably in m already

[quaint sky]

Oh, i see

[lone heart]

Like the ta converted the mm to m so they didn't need to multiply by 1e-6

[quaint sky]

so yes it is my initial pressure!

let me try to calculate initial volume and ill show you

this is what i did

Remember this is my data

idk if i did the correct si unit

[lone heart]

Volume should be m^3

But looks right

[quaint sky]

why m^3?

wouldnt it be m^2

because (m)(m)=m^2?

[lone heart]

Volume is cubic meters

[lone heart]

You're forgetting it was mm^2 converted to m^2

Then times by the height which is m

So m^2 * m = m^3

[quaint sky]

so this?

[lone heart]

Yes

[quaint sky]

Can you check this

[lone heart]

Isn't Pi supposed to be the pressure you calculated?

[quaint sky]

im referencing off my ta work

ill send it

im up to here

[lone heart]

Okay then yeah this is good

[quaint sky]

since its a constant, is there an si unit?

if it does, maybe joules

which is the si unit of pascal

[lone heart]

It would be N/m, because pressure is N/m^2 and volume is m^3 so you get N * m^3/m^2 = N/m

[quaint sky]

or its just joules

[lone heart]

But you did pressure times volume, the units for pressure is N/m^2 and the units for volume is m^3

So [N/m^2][m^3] = N/m

[quaint sky]

hence, joules?

because n/m^2 = pascal

and[N/m^2][m^3] = N/m

meaning Pa/m^3= N/m

which = joules?

[lone heart]

Joules is N * m

Not N/m

[quaint sky]

Oh okay!

did i do this right so far?

idk the si unit for this, i assume m because [N/m]/[N/m^2]=m

also i dont know how to get vf alone without the exponent

[lone heart]

It's going to be a volume so it'll end up being m^3

[lone heart]

Logarithms

[quaint sky]

how?

isnt it this N/m]/[N/m^2]=m or did i math it wrong?

[lone heart]

I made a mistake, it's [N/m^2][m^3] = N * m

Because N * m^3/m^2 = N * m

So it was joules

[quaint sky]

so Joules/ Pascal = m^3

[lone heart]

Yes

[quaint sky]

give me a second to do log

soemthing like thus?

[lone heart]

You can actually just take the 1.4th root

[quaint sky]

what do you mean?

[lone heart]

1.4 = 7/5

[quaint sky]

where are you getting 7/5?

[lone heart]

So you have $V_f^{7/5} = 1.04 \times 10^{-8}$

[elfin berry]

CaptainNova22

[lone heart]

Convertingg 1.4 to a fraction

[quaint sky]

oh okay!

ok but how do i get rid of 7/5

[lone heart]

Raise both sides by the reciprocal

Because 7/5 * 5/7 = 1

So when you raise V^(7/5) to 5/7 power, you get V^1 then you have to raise the other side by 5/7 power too

[quaint sky]

please check

[lone heart]

,w (1.04E-8)^(5/7)

[elfin berry]

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[lone heart]

Should have been to the 10^-6

Not 10^-12

[quaint sky]

oh okay!

in m^3?

[lone heart]

Yeah because it's volume

[quaint sky]

ok let me try temperature and ill lyk

[lone heart]

You can just dm me, I gotta go, I'll look at it when I can

[quaint sky]

okay! im kind of confused with it tbh. ill dm it

[marsh citrus]

@quaint sky Has your question been resolved?

[marsh citrus]

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[wicked wharf]

$\lim_{x \to \frac{\pi}{2}} \left( \cos(x) - \frac{2}{x} \right)$

[marsh citrus]

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[wicked wharf]

.close

.open

\lim_{{x \to \frac{\pi}{2}^+}} \frac{-2}{\cos(x)}

.close

[marsh citrus]

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[unique breach]

Can I have help on this q please

[unique breach]

it’s mathematical induction

my main problem is getting confused when factoring

[marsh citrus]

@unique breach Has your question been resolved?

[unique breach]

<@&286206848099549185>

Pls 😭

[marsh citrus]

@unique breach Has your question been resolved?

[unique breach]

<@&286206848099549185>

[quaint elm]

hmm

[main idol]

what does that ... mean

[crystal grove]

?

Troll?

Definitely a troll

@weary ibex i think theres a troll here

[teal grove]

sup

ya'll need help?

what

@unique breach

what?

you need help?

nvm

close this channel ig

[unique breach]

Oh it’s in the question

I’ll add the actual question

[unique breach]

I’m a troll?? 😭😭

[weary ibex]

Dealt with, but please ping @Moderators instead of individual mods another time.

[unique breach]

This is the actual question

I did up to the plug in k part

[weary ibex]

No, they're gone now.

[unique breach]

Then I add k+1 and then it all goes downhill

[unique breach]

Oh okay thanks

[crystal grove]

Alright mb I’ll ping the role next time thank you

[main idol]

put this sum over a common denominator

[unique breach]

So multiply 3k+1 on both sides of the fraction?

[main idol]

no

there are two fractions in that sum, so i don't know which one you're referring to

[unique breach]

I mean multiply 3k+1 by 1 / (3k+1)(3k+4)

[main idol]

does that get you a common denominator?

or maybe you call it greatest common factor
https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:gcf/a/greatest-common-factor-of-monomials

[unique breach]

Uhh I haven’t tried that yet but I’ll work it out

Umm it’s like completely different from how my teacher did it

Like that bottom portion

[main idol]

,w expand k(3k+4) + 1

[elfin berry]

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[main idol]

,w factor 3k^2 + 4k + 1

[elfin berry]

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[main idol]

yes that's right

[unique breach]

Ohh yk I think I kind of got it ?

it’s really just the factoring that gets me confused

But thanks for helping

[main idol]

when you plug in (-1) to 3k^2 + 4k + 1, you get 0 so (k + 1) is one of the factors

[unique breach]

Wait why would u plug in -1

[marsh citrus]

@unique breach Has your question been resolved?

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[final skiff]

!status

[marsh citrus]

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

[terse folio]

i dont know how to get B

or b prime

[final skiff]

Can you explain more? I do not know what you mean by b prime and the base area is basically given

[terse folio]

have you taken calc?

[final skiff]

Yes

[terse folio]

b prime

???

[final skiff]

I see you mean the derivative of b

[terse folio]

yeah

so i take the deriv of

\the equation

for the area

in order to find the rate of h

because im given the volume rate

but i dont know how to find the rate of b prime

because you need it from product rule

[final skiff]

Send what you got

[terse folio]

v prime = 1/3 b prime * h + h prime * b

thats the deriv

and i can plug in for v prime, and h and b

but not b prime

and h prime is what im finding

so i think you need to find b in terms of jh

h

but i dont know how

[terse folio]

can you help

[final skiff]

Yeah I think you will want to use similar triangles since you are given the base and height of the pyramid, and you are given the height of the water, so you should be able to find current b with that

[terse folio]

how do i find b prime tho

[final skiff]

Try taking the derivative again after replacing B in the volume formula

cuz now it will be in terms of h

[terse folio]

ok and then when using triangle to find b

i only have one side and no angle

so...

am i missing something

[final skiff]

Yeah you need to setup a relation between the triangle from the original pyramid, and the triangle from the water pyramid

[terse folio]

ahh

thanks

[terse folio]

ok i got h prime being 6 but thats wrong

i tried plugging in b before taking the deriv

[terse folio]

yoo

[final skiff]

how did you get this

[terse folio]

i plugged in

[final skiff]

what did you plug in for B in the original equation

[terse folio]

2

[final skiff]

I think you forgot B is the area

2 is the side length

[terse folio]

so i plug in 4?

[final skiff]

yeah

[terse folio]

so the asnwer would b 3

thats wrong tho

[final skiff]

I am not getting 3

[terse folio]

lemme try plugging it all in again one sec

2.5?

wait

9/4

so 2.25

idk what im doing wrong

i have

dv/dt = 1/3 4 dh/dt

then plugin 4

[terse folio]

what did you get

[final skiff]

I double checked and I did it wrong I also got 3 after rechecking let me review everything

[terse folio]

ok thanks

[terse folio]

did you end up figuring it out?

[final skiff]

I think I was wrong about being able to plug in b before taking the derivative (but idk how you would find b' then)

[terse folio]

ok lol

yah idk

[final skiff]

Ok I realized, you have to find a general formula for the base in terms of h, then substitue B with that (remember to square base length) then take the derivative @terse folio

[marsh citrus]

@terse folio Has your question been resolved?

[terse folio]

ok what would the general formula be

i dont really understand that

[final skiff]

Use same method you did to find that base length = 2, but instead of plugging in 3 for the water pyramid height, keep it as h

[terse folio]

so 1.5h

[final skiff]

yeah

actually no

I got (2/3)h

[terse folio]

oh yeh

i got 1

[final skiff]

You plugged it into V=1/3Bh to get V=1/3(2/3)h^2 right?

(I am not getting 1)

[terse folio]

yah

[final skiff]

Then you took derivative?

[terse folio]

yaqh and got dv/dt = 2/3 (2(3)) h prime

and then you get 12/3

which is 4

then divide both sides by 4

i think its 1

[final skiff]

believe your 2/3 in here should be 2/9 because 1/3*2/3 is 2/9.

[terse folio]

where are you getting 1/3

[final skiff]

This simplifies to 2/9h^2 then taking derivative of that is 2/9(2)h(h')

so from the volume formula

[terse folio]

ohhh

well i checked the answer as 1

and it was right

but i see what ur saying

[final skiff]

lol ok then idk

[terse folio]

cause 1/3 times 2/3 is not 2/3

idek

just glad its over

thanks for helping

[final skiff]

np

[terse folio]

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[indigo quarry]

Need help on #7 calculating M

5.13 is that not just 12 so my N is 1 and for 5.14 if it’s quartered is that not just 4?

I got 1052.52 for 5.13 and 1052.39 for 5.14

[mystic ledge]

yes you are correct

but for 5.14 if you round that number you get 1052.4

but anyways 5.13% is better

[marsh citrus]

@indigo quarry Has your question been resolved?

[indigo quarry]

Number 9 also

I’m getting 262.5

<@&286206848099549185>

i have 0 idea why i keep getting 262 on 9

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[marsh citrus]

@scenic pumice Has your question been resolved?

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[still temple]

chain rule!

[
\dv[y]x = \dv[y]u \cdot \dv[u]x
]

[still temple]

they applied the fundamental theorem of calculus

[
\dv x \int_a^x \m ft \dd t= \m fx
]

[marsh citrus]

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[thorn marsh]

can someone guide me to how they derive the second function from the first one

where did the x go

and then where did the -1 go.. lol

[hard gull]

l'hopital rule

[thorn marsh]

omfg

dammit ok

tysm

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[granite geyser]

I have a doubt if E>0 THEN why we took |an-l| <E/2

[tacit fjord]

its an implicit sort of substitution

Your usual quantification is "forall positive epsilon" right?

[granite geyser]

Yes

[tacit fjord]

If you multiply or divide this epsilon by 2, it will live in the same set

the positive reals

So the statement can be rewritten "forall e/2 > 0, blah blah blah"

[granite geyser]

Okay they just write it to get epsilon in end so no special meaning

Thanks

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[granite geyser]

Why we let h=Min.(l-E,...)

And same for k = Max.(l+E...)

Other half of question is this

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[brisk badger]

,rotate

[brisk badger]

how can i solve this integral?

[marsh citrus]

@brisk badger Has your question been resolved?

[brisk badger]

<@&286206848099549185>

[blissful terrace]

Use trignometric identities and break them seperately

2sin(a)cos(b) = sin(a+b) + sin(a-b)

then you can do by parts

[brisk badger]

ohhh

thanks

one more thing

Five soldiers A, B, C, D and E volunteer to perform an important military task if their following conditions
are satisfied:
i. Either A or B or both must go.
ii. Either C or E but not both must go.
iii. Either both A and C go or neither goes.
iv. If B goes, then A and C must also go.
Suggest a single gate implementation for each of the above conditions (one gate for each condition).

i. OR
ii. XOR
iii. ?(AND)
iv. ?
I don't understand the last 2

[sleek lake]

3 is XNOR

4 doesn't sound like a gate

maybe you can use multiple gates as long as they are the same

i think it's a mistake in question

[marsh citrus]

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[whole jetty]

An investor has made a profit of €1,500 after investing a total of €40,000 in three different companies. These profits are broken down as follows: the amount invested in company A has brought him a 2% profit, the amount invested in company B, 5%, and the amount invested in company C, a 7%. The money invested in company B has been the same as in the other two companies together. What was the amount invested in each of the three companies?

[marsh citrus]

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[graceful nacelle]

What is meant by "express you answers exactly"? Aren't these numbers irrational since they are factors of pi?

[lone heart]

It means don't approximate the answers

[graceful nacelle]

How do I answer the question without approximating it?

[lone heart]

Like instead of sqrt(2), you put 1.414 instead

[graceful nacelle]

So how do I answer this?

[lone heart]

Unit circle

[graceful nacelle]

Ohh

So, I know 5pi/4 is in the 3rd quadrant because it is between pi and 3pi/2

and cos is the coordinate x

and x is negative in quadrant 3

and

since it is a factor of pi/4 or 45deg I know that the absolute value of cos is the same as the absolute value of sin

which means both answers will be either -sqrt2/2 or sqrt2/2

the cos will be negative and sin will be negative

So, this would be correct?

[lone heart]

Yes

[graceful nacelle]

It was correct thanks

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[brisk badger]

thanks

[sleek lake]

did you find a solution that uses one type of gate?

or you just mean thanks for answering

[brisk badger]

i can use a 3 input AND gate

[sleek lake]

i don't see how it helps?

[brisk badger]

I mean a single AND gate having 3 inputs can suffice here right?
since, there is no other way round

[sleek lake]

that's not what it says

whatever, i think there's no solution

bad question

[brisk badger]

lol
i'll ask the professor

[sleek lake]

the condition is something like (~B or A) and (~B or C)

the closest you could do is build a huge thing from NAND or some other universal gate

[brisk badger]

aight

[marsh citrus]

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[still temple]

what does this limit become, how do i evaluate its value

[desert socket]

you can divide numerator and denominator by x

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[fringe trellis]

How did u get 6log8(2) as 8log8(8)
Can u prove that?

U gotta use this property and u will get the answer of all three logs

I will show u how to do one

Take 6log8(2)
If u look carefully,As per the property n=6
So take 6 to thr power of 2
U will get log8(2)⁶ which is log8(64)
And we know that log8(64) is 2

I hope u can do the rest now

[marsh citrus]

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[main idol]

what does then mean here?

[marsh citrus]

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[neon lark]

The segment MN is the diameter of the circle, and the point K is outside the circle. The circumference of sections MK and NK intersect at points A and B, respectively.

Area AKB / Area MNK = 1/4
Find the angle ANK

[neon lark]

<@&286206848099549185>

[main idol]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[marsh citrus]

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).

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[neon lark]

<@&286206848099549185>

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@neon lark Has your question been resolved?

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@neon lark Has your question been resolved?

[runic parcel]

How is that a question

Its a picture wow

[marsh citrus]

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[narrow gate]

How could I find x?

I know I have to make the bases the same but I'm struggling to see how to do that here with the 64

[narrow gate]

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[umbral sparrow]

demostration.

"if a cannot divide with 3b^3-b^2+5b, then a cannot divide b"

i did by absurdism. (p n not q) --> F (that means, a cannot divide 3b^3-b^2+5b, then a can divide b"

could that be good? and with a good procedure?

[marsh citrus]

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@umbral sparrow Has your question been resolved?

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[long cape]

is this on the right track or wrong way to solve?

[summer trench]

i dont know what u are doing

what we have

[long cape]

I bring -3/2 to the right hand side to solve for x or is that incorrect?

[summer trench]

different conditions for this

[long cape]

x is just a/2?

[long cape]

so is RREF the incorrect way to solve this?

[lone heart]

No RREF is a good way to solve it

[long cape]

oh right

0 + 0 = something

is no solution

0 + 0 = 0
is infinite solutions

so Row 2 tells me there is no solution

technically this is fine?

even this tells me the answer too, right?

this is the giveaway @lone heart ?

[lone heart]

Yes

[long cape]

but it's asking "under what conditions" will it have each type of solution

[lone heart]

And what conditions will it be infinite solutions?

[long cape]

those are the two conditions on a and b where we get infinite solutions

[lone heart]

There's only one condition

2a = b is the same as a = b/2, you just rewrote it

[long cape]

but "on and and b"

that doesn't mean for each?

a =
b =

[lone heart]

No because it's the same equation, just written

If b = 4 then a = 2 using the bottom one, and if a = 2 then b = 4 using the top one

[long cape]

I'm not sure how to write this

no solution when we have anything but 0

[lone heart]

Yeah so just saying $-2a + b \neq 0$ is enough

[elfin berry]

CaptainNova22

[long cape]

OK

[lone heart]

Because if it equals 0, then it's automatically no solutions

[long cape]

wouldn't this help to find the conditions for a and b to give a unique solution?

[long cape]

or is it not helpful

[lone heart]

No because there isn't a unique solution

That row is 0 0 | -2a - b

So -2a - b can only either be equal to 0 or not equal to 0

[long cape]

and that's because y is a free variable?

[lone heart]

Because the 0 0 | part doesn't have any values to make it unique

[lone heart]

Yes, if -2a + b = 0

[long cape]

so this is the correct way to solve this?

One solution: DNE
Inf solutions: 2a = b
No solution: 2a ≠ b

[lone heart]

Yeah that's fine

[long cape]

OK ty

.close

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[long cape]

is there any reason to make a matrix with this question?

or is this "parametric table" kinda the same idea as a matrix?

"parametric equation" = "parametric table" = "matrix"?
similar ideas going on here?

or are they not the same thing?
"parametric equation" ≠ "matrix"?

[lone heart]

No this is not the parametric solution form the problem is referring to

[long cape]

just as an example tho

[long cape]

how about this one @lone heart ?

[lone heart]

That's just the matrix equivalent of your system of equations

Not parametric solution form

[long cape]

hmmm

I thought it means x_1 + 3x_2 - x_3 = -4

[lone heart]

Parametric solution form looks like this

[lone heart]

That's just one of the equations

You need to row reduce that system

To find the free variables

[long cape]

weird, when I Google "parametric equation" I get this:

[lone heart]

Then create the parametric solution form

[lone heart]

Not specific enough

[long cape]

it looks different for systems of equations, compared to linear functions?

[lone heart]

You need to search something like "parametric equation form matrix"

And specify that it's a matrix solution

[lone heart]

Because there are free variables

[long cape]

the only variable that is not free is x_1

x_1 = -4 - 3x_2 + x_3

x_2 = t
x_3 = u

[lone heart]

No

[long cape]

really?

[lone heart]

What happened to the 3 and -1?

x_1 depends on x_2 and x_3

[long cape]

that is better now?

[lone heart]

Also, you should be consistent on the right side

[lone heart]

If you are using t and u as representation of x_2 and x_3 then you need to use that in the x_1 equation

[long cape]

yeah I forgot to write the line for the augmented matrix

oh right

lemme rewrite that

ty

[lone heart]

x_1 = -4 - 3t + u
x_2 = t
x_3 = u

[long cape]

do I need to specify what t= and u= like that too?

or they are free variables, so they are just t and u

[lone heart]

What is this?

[long cape]

oops

yeah you are right

I can't do that

[lone heart]

You don't need that for parametric solution form

You want to do something like this

[long cape]

not sure about this..

x1, x2, x3 better?

[lone heart]

The only one that is correct is the last vector

[long cape]

ooof

better?

[lone heart]

All you did was change t to x_2 and u to x_3, and added a subscript 1, there was no change

Take this part, line up all the terms

So all the t terms are in a column, and u terms

[long cape]

why are we doing the answer like this?

[long cape]

the answer should be like this

I already found it

this is what is in the answer key so I'm confused..

[lone heart]

It's the same, most of the time, people write it the way in the image I sent

Because that's how most online sites take the answers

[lone heart]

And it's the exact same as your answer, they just used r and s instead of t and u

[long cape]

so what is this called, if it's not parametric form?

clearly these are different ways of writing the same thing

and each can't be called the same thing, because they look so different, so it would be good to know the correct terminology for each

[lone heart]

They're both parametric solutions to a matrix, one is in equation form, the other vector form

[long cape]

it's like perpendicular and orthogonal.. why we making up new terms for no reason

when they both mean the same thing

[lone heart]

Because synonyms

[long cape]

so the first is called "parametric solution in equation form"?
and the second is called "parametric solution in vector form"?

if they ask for parametric form how do you know what they want?

[lone heart]

Yeah that's what I would refer it as

[lone heart]

Either is fine

The second is just taking it one more step and making it into vector form

[long cape]

so these are equal?

[lone heart]

Yes

[long cape]

wait I forgot x_1

[lone heart]

As mentioned

They're both parametric solutions to a matrix, one is in equation form, the other vector form

[lone heart]

No, x_1 is not a free variable

That is the answer

Parametric form shows what variables are dependent and independent

[lone heart]

Like this, you can see that x_1 depends on t and u, and x_2 only depends on t and x_3 only depends on u

[long cape]

so they could be in any order, all of this is equal to one another?

[lone heart]

Normally, by convention, or at least how I've seen it done frequently is the constant vector is first, then the free variable vectors are in order numerically/alphabetically

But yes those are all the same because addition is commutative

[long cape]

just to be extra clear I labeled each above

interesting, ty

and each of these for the second example are considered vectors

whereas for the first they are just considered variables

(yet the first is also vector?)

the variables are just not written in vector format, but underneath of the hood they are vectors too

is this true for ALL math? whenever you see a variable, technically it's considered a vector too?

regardless of the context

[lone heart]

That I don't know

[long cape]

https://www.reddit.com/r/econometrics/comments/1aasgq/the_difference_between_vectors_and_variables/

Each variable is a vector of n observations. But we usually just call them variables.

n = real number, I'm assuming?

not complex

https://tenor.com/view/whatif-matrix-make-gif-723526056553365657

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[wheat haven]

Can someone please explain to me why this works?

[dry prawn]

It doesn't

Can you show more of the page?

[wheat haven]

This is the entire problem. My professor said that I got the problem correct, but now that I am going back to study for the midterm, I feel I did do something wrong

[dry prawn]

Yeah that last equality is miswritten

[wheat haven]

Is it supposed to be theta/n?

[dry prawn]

It should be $\theta \frac{n}{n+1}$

[elfin berry]

Steakanator

[dry prawn]

That's why they compute E[(n+1)/n*Y_n]

[wheat haven]

This is what I am reworking with

[dry prawn]

Yeah that last equality is better

[wheat haven]

Alright, I am going to try to rework it and see if I still end up with the result that the professor marked correct. If I do not, I will come back.

Thank you!

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[dry prawn]

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[odd swan]

what is wrong about this one

[marsh citrus]

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[cloud dawn]

So it's a P=E/t problem where I need to find t, so t=E/P. E is 14,055 Joules, P is 14 horsepower

Honestly I feel like I did this right (considering I got it to work out to seconds) but also this is a huge block of unit replacement and conversion so I wouldnt mind someone going over what I did and making sure I did this right

also I did 30.48^2 because I knew to get rid of two feets and m^2 I would need to convert two instances of feet twice

(hopefully that logic is right)

[marsh citrus]

@cloud dawn Has your question been resolved?

[marsh citrus]

@cloud dawn Has your question been resolved?

[cloud dawn]

ok so I did get this one wrong by a landslide

[cloud dawn]

ah I see what I did

Man

Okay so basically my brain forgot that its 550ft-lb per 1hp, and that ftlb is a unit of energy so I could just cancel out the ft-lb and J lol

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[eager ibex]

Is this correct?

[stoic saddle]

appears so.

[eager ibex]

okay thanks

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[tulip folio]

Need stats help

[tulip folio]

How do we know that A and B are independent events here?

[marsh citrus]

@tulip folio Has your question been resolved?

[whole sleet]

@tulip folio Worth making sure you know what all of the above terms mean. ie, write all of the algebraic facts A,B,C,D satisfy

For example, A and C are mutually exclusive. That means
P(AC) = 0
P(A U C) = P(A) + P(C)

[tulip folio]

I know those

Why is A intersection B independent tho?

Is it just assumed because we r not told that they aren't independent

[whole sleet]

Why do you think they are?

I see that A and B' are independent, so:
P(AB') = P(A)P(B')

But that is the same as:
P(A) - P(AB) = P(A)(1 - P(B))

Oh lol, yes okay A and B are independent then

@tulip folio

Have any questions about that? I did quite a bit there

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[violet relic]

Hints please

[waxen dust]

Well

Seeing as you only have 8 values to choose from you can try plugging them in

Or you can solve the inequality

And seeing the largest z which is the solution to the inequality

[small ravine]

you have three cases to consider

[waxen dust]

4 actually

[small ravine]

When z is between -2 and 2,
When z is less than -2,

and

when z is greater than 2,

[small ravine]

how?

[waxen dust]

What

[small ravine]

:

[waxen dust]

Go on

[small ravine]

When z is between -2 and 2, both |z + 2| and |z - 2| will be positive.
When z is less than -2, both |z + 2| and |z - 2| will be negative but since we're dealing with absolute values, we still have

|z + 2| + |z - 2| = -(z + 2) - (z - 2) = -2z - 4
When z is greater than 2, |z + 2| will be positive, and |z - 2| will be positive

then

Now, you have three cases to consider:

2z ≤ 6 => z ≤ 3
-2z - 4 ≤ 6 => -2z ≤ 10 => z ≥ -5
2z ≤ 6 => z ≤ 3

i guess it'll solce in this way..... but not sure

solve*

[stoic saddle]

is z not supposed to be complx?

complex*

[small ravine]

hmm yeh

[stoic saddle]

anyway

!nosols

[marsh citrus]

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.

[small ravine]

alright i'll avoid next time sorry

-:)

[waxen dust]

What

[small ravine]

so we'll consider both real and img parts

[waxen dust]

The union of that is everything

[small ravine]

hmm

[violet relic]

Nevermind i have got a shortest method

[waxen dust]

This is not the solution

[violet relic]

|z+2+z-2|<=6
2|z|<=6
|Z|<=3

Some people are very stick to the server moderation as high

Just ignore them

[small ravine]

alright what's weong with that solution

wrong*

[waxen dust]

If we find the union, we get $z \in \mathbb{R}$

[elfin berry]

USS-Enterprise

[small ravine]

okhhh got it

[waxen dust]

Try z=8

[small ravine]

what?

[waxen dust]

Abs of 10 plus abs of 6 is 16

16 isn't less or equal to 6

Your solution says we can pick any z

I picked z=8

z=8 is not a solution