help-0

[summer dirge]

I guess the easiest way to check if there is a solution is to directly solve for a and b

and if you run into a problem like b = sqrt(-4), then there would be no (real) solution to the problem

but the easier answer is that

if the problem is given by a teacher and you're being asked to solve it, there's probably a solution

[real glen]

okay

thank you

:D

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[summer dirge]

have a great day!

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[stone summit]

hello

[stone summit]

i need to figure out how to determine the percentage value and actual value between two numbers, and i dont know/dont remember what it's called or how to do it

for example, i have the lower limit of this problem as 0.40, and the upper limit as 1.00, and an in-between value of 0.55

how do i figure out the percentage value of 0.55 if 0.40 and 1.00 are my lower and upper bounds respectively?

[shadow sinew]

formula to find percentage

(value - lower) / (upper - lower) * 100%

[stone summit]

thank u

[shadow sinew]

np

[stone summit]

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[rapid sparrow]

if someone could write this out and fit it into here that would be great

[teal orbit]

,rccw

[teal orbit]

i would...uh...change that name of yours.

[teal orbit]

do you know how logs work

[willow wadi]

wym

[teal orbit]

just to let you know, $f\qty(\frac{1}{25})=\log_5(\frac{1}{25})$

[ocean seal]

ooaa

[willow wadi]

oh this one already occupied my bad

[teal orbit]

shit

thought you were an alt or sumn

lol

[willow wadi]

nah I thought this was mine for a min mb😭

[vale wigeon]

!noans @rapid sparrow

[lone heart]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[teal orbit]

what are these usernames today

jesus

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@rapid sparrow Has your question been resolved?

[toxic wharf]

i'll give you a brief summary for logs, since i think it should help. Log or logarithms is a function that (sort of) finds an exponent. The format for logarithms is like

m = logb x

You can also think of this function like b^m = x. Answering this question is basically saying "to what power of b need to be to get x?" So as an example from your question, to what power of 5 equals 1/25?

[rapid sparrow]

forget it dawg im too drunk rn

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[toxic wharf]

😭 get sober soon man

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[desert siren]

I need help with my assignment

[vale wigeon]

gonna have to post your questions here one at a time

and show all of your current progress on each one (and if on some questions there is none, then say it)

[meager nova]

fax

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@desert siren Has your question been resolved?

[limpid turret]

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[thick vigil]

Hi

[thick vigil]

It's about arcs and central angles of circles

This is my solution for number 1.

Can someone help me with these?

[true field]

your answer for number 1 is correct

[thick vigil]

Thank youu

Can you help me with number 2?

Idk how to answer it

There's no given values or idk where to get it

[slim kite]

I think that red angles represent 90°

[true field]

i think the two angle markings are showing symmetry of 90 degrees yea

i mean it sure looks like 90 deg

[slim kite]

More like, they r equal and since they sum to 180, they're 90

[slim kite]

Yea

[thick vigil]

Ohh okay okay it makes sense cuz mYZ is equals to 180 and it looks like half of it

[true field]

number 3 the process would be take the measure of rotation from I to K, and then add the measure from K around to H

and then number 4 if you are observant you can see a congruent angle to the one you are trying to find

[thick vigil]

What do you mean when angles are congruent?

@true field

[slim kite]

Congruent means same angles

He means JKI = FKG = 50°

[thick vigil]

Okay but how can we tell if two angles are congruent?

[true field]

do you know supplementary angles theorem

[thick vigil]

Not yet

[thick vigil]

But the word supplementary is familiar

[true field]

supplementary means the two angles add to 180

[thick vigil]

Okay

[true field]

basically we know that JKI + IKG = 180 and we know that GKF + FKJ = 180 and because JKI and FKG are directly across from each other we know they are the same angle

[thick vigil]

I see

[thick vigil]

Can you look if these solutions are correct?

[true field]

they all look correct to me

[thick vigil]

Okayyy

Thank you so much!!!

@true field and @slim kite

[true field]

np

[thick vigil]

[slim kite]

Np

[thick vigil]

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[alpine sable]

question c

[alpine sable]

where did i do wrong

final answer is 4 unit/s

[coral flower]

after dy/dx , try to solve again but this time writing each step (u would find ur mistake , hopefully)

[alpine sable]

ooookay

[alpine sable]

so my dy/dx was correct?

OH

I HAVE TO USE DX/DY INSTEAD OF DY/DX?

😭

[coral flower]

huh?

i mean u could if u want to

[alpine sable]

oh

were u thinking of another thing?

[coral flower]

i was just thinking of the way u did , u just made a mistake

[coral flower]

dy/dx = dy/dt * dt/dx
using ur values of everything
6 * dt/dx = (18-12)/4
send the 6 to the other side
dt/dx = 6/4 * 1/6
taking reciprocal on both sides
dx/dt = 4

if u write like all steps out , u would understand better

[alpine sable]

ohhhh

okay thanks

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[golden nacelle]

i don't get what it wants me to do

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@golden nacelle Has your question been resolved?

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@golden nacelle Has your question been resolved?

[void nymph]

so do you know basic combinatorics

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[wooden hinge]

Ok why when shifting the graph of y=(3x+2)^2+5 to y=x^2 using my teachers working out, why do I get a translation of up 5 units when it’s obviously down 5?

[void nymph]

what does 'shifting one graph to another' mena

[wooden hinge]

Like what translation would occur to change the graph of y=(3x+2)^2 + 5 to y=x^2

[void nymph]

huh.

okay so

what transformations in x and y?

[wooden hinge]

Yeah

[void nymph]

so

your thing appear okay

the thing to note here is that

y=y+5 doesnt shift the graph up

it shifts it down

[lime bobcat]

It depends on in which direction you are describing the transformations.

If it is (3x+2)²+5 to x² or x² to (3x+2)²+5.

[wooden hinge]

The first option

[void nymph]

$y+C=f(x) \ \implies y=f(x)-C$, clearly shifting the graph down by $C$

[ocean seal]

Percy

[void nymph]

i think that's it lol

[wooden hinge]

Hmm

[lime bobcat]

So clearly (3x+2)²+5 is 5 units "above" x²

So when transforming it goes down 5 units

[wooden hinge]

I got that but I thought that would mean it would be -5 on the translation

[lime bobcat]

If y=(3x+2)²+5 and y'=(x')². Then y'=y-5.

[void nymph]

well it doesnt

hope that helps

[wooden hinge]

I’m getting mixed messages lol

[wooden hinge]

This is what I thought it would be

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[lilac elk]

Hints plz

[lavish cave]

difference of two squares

[void nymph]

!occupied.

[lethal dune]

Guys i think its meant to be easy but i dont understand😭

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[lethal dune]

oh

Sory

[pulsar aspen]

What is the factorization of $x^2 - y^2$?

[ocean seal]

Xwtek

[lilac elk]

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[steady thorn]

just wanna check is the answer just r 1/10

[lavish cave]

again what's your working?

[gritty dune]

hmm yeah 1/10 seems to be incorrect

[plucky jasper]

"again"

[steady thorn]

uh the sum infinite of the first series is

a/1-r = 10 which could also be a=10-10r

[plucky jasper]

bro is traumatised

[lavish cave]

no I answered another question from them

that's all

[steady thorn]

yes this is true

[lavish cave]

if I was fed up I wouldn't go back for this one

[plucky jasper]

fairs

[steady thorn]

uh then the second one is ar/1-r^2=1

yo what am i doing

[lavish cave]

ar = 1 - r^2 then you can sub in a = (10 - 10r) into ar = 1 - r^2

[steady thorn]

lemme redo my working a bit neater rq

right so i got the 9r^2 -10 r+1

then do i just solve the quad

[lavish cave]

I actually have another idea for you:

$\frac{a_2 + a_4 + a_6 + \cdots}{a_1 + a_3 + a_5 + \cdots} = \frac{1}{9}$

[ocean seal]

south

[lavish cave]

yes

[steady thorn]

how does this work

[lavish cave]

1 + x = 10

[steady thorn]

is x the sum

i dont really get that fraction thingy

i got the answer of 1/9

[lavish cave]

$(a_ 1 + a_2 + a_3 + a_4 + \cdots) - (a_2 + a_4 + a_6 + \cdots)$ is?

[ocean seal]

south

[lavish cave]

besides from being equal to 10 - 1 = 9 of course

[steady thorn]

uh a/1-r - ar/1-r^2

hmm thats interesting but howd u like

think of that thing

like imagine u in a test like how u gonna know that

[lavish cave]

practice

you already figured out $a_2 = r \cdot a_1$ so why stop there

[ocean seal]

south

[lavish cave]

$a_4 = r \cdot a_3, a_6 = r \cdot a_5$ and so on

[ocean seal]

south

[lavish cave]

$\frac{ra_1 + ra_3 + ra_5 + \cdots}{a_1 + a_3 + a_5 + \cdots}$

[ocean seal]

south

[steady thorn]

ohhh

very cool method

yeah alr i think thats all the questions i got

thanks for the help cro

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[warped topaz]

Part (b)

I know we need to get the range of x(t) and y(t)

Thats the length and width which will give us the area

[void nymph]

i mean

[warped topaz]

And the range of y(t) is obviously [-1, 1]

[void nymph]

im ready to sit for another 5 ninutes

[warped topaz]

But idk how to get the range of x(t)

[earnest slate]

same

[void nymph]

close the old one though

[earnest slate]

what different methods do you know

to find the range

[void nymph]

like do you have no ideas at all or just dont like the ones you have

[warped topaz]

I know how to get the range of certain functions but not this one

[warped topaz]

I feel like I suggested plenty of ideas lol

Im 1 step away from solving it

[sour mica]

what are the bounds of a range? Like a name from what are they called?

[earnest slate]

try completing the square

[void nymph]

yeah try completing the square if you're comfortable with that

if not, go with what facter some letters is suggesting

[warped topaz]

The max and min?

[sour mica]

thank you for describing my name

[sour mica]

any hints you picking here?

[void nymph]

think we were going for 'global extrema' but yes

[warped topaz]

I get the max and mins of x(t) = 0?

[void nymph]

no the extrema of the function itself?

[earnest slate]

get the max and mins of x(t)

[warped topaz]

Not equal to 0?

[earnest slate]

no

that would give you roots

[sour mica]

if its already equal to zero, its gonna be constant

[warped topaz]

Ah ok

[earnest slate]

try expressing the expression in terms you know the range if

[sour mica]

min and max exist coz the value is not constant

[earnest slate]

for eg sinx

[void nymph]

If you're fine with it, just complete the square.

It's the best approach in general

Finding the extrema will eat time

[warped topaz]

Alright I got the answer 😄

Thanks everyone!

❤️

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[sharp bronze]

I need help minimizing the following DFA
I have used the partition method and I have reached P3 containing {s8},{s6,s7},{s3},{s4,s5},{s1],{s2} and I wanted to make sure that was the correct answer
Thank you

[alpine sable]

hi guys

what is the answer for a+b on square

[wanton jasper]

!occupied

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@sharp bronze Has your question been resolved?

[sharp bronze]

<@&286206848099549185>

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@sharp bronze Has your question been resolved?

[young dirge]

anyone interested in afrobeats here text me

i will play snippets of my songs tell me if it interest you

[lavish cave]

post in #chill

don't disturb people asking their maths questions

[surreal fog]

I tend to think that's correct

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[twilit elbow]

can anyone explain me this question

[earnest slate]

how would you write AB in terms of radius

[prime lynx]

um not hard

im new here

so um

[twilit elbow]

??

[prime lynx]

the distance between op and pr is the sam eright

because the diameter

extending the argument

you can prove that the sides of the square are equal

or necessarily prove that it IS a square

idk if thats the evntual aim

let me know if it helped!

[twilit elbow]

can you do it for me plz?

[prime lynx]

(english is not my first language so please excuse and grammar mistakes!)

[prime lynx]

so

[twilit elbow]

np

[prime lynx]

it goes like this

OP = OR

(radius)

[twilit elbow]

yep

[prime lynx]

as b and c are equidistant from each other

PR = BC

so necessarily

BC =AD

similarly you can do it vertically too

AB = CD

now, adding both equations:

AB+BC = AD+CD

simple

@twilit elbow

[twilit elbow]

ok

thank you very much

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[worn bane]

i forgot how i did that...anyone help me understand this

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@worn bane Has your question been resolved?

[prime lynx]

you have split the term form powers of 6 to powers of 3

using the difference of squares

then you have split it again using the difference of cubes and substituted the values

into the equation

[worn bane]

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[ruby lion]

I have big problem

[ruby lion]

Going from spherical to cartesian coordinates and the other way around

$z = r \cos(\phi)$ and $r = \sqrt{x^2+y^2+z^2}$

$\frac{dz}{dr} = \cos(\phi)$ and $\frac{dr}{dz} = \frac{2z}{2 \sqrt{x^2+y^2+z^2}} = \frac{z}{r} = \cos(\phi)$

So $\frac{dr}{dz} = \frac{dz}{dr}$ ? something must be wrong

[ocean seal]

Mr. Rotor

[ruby lion]

I know it's dr/dz that's wrong

Wait I think I understand now what I got wrong

I think I know what's wrong but it's weird like

ok wait let me do a quick schematic

I think one is essentially the partial derivative

while the other one isn't

[north canopy]

well iirc ur dr/dz should be partial since u have multiple variable

[ocean seal]

Mr. Rotor

[north canopy]

but u have the base formulas for cartesian to spherical, so where does it block for u?

[ruby lion]

I'm trying to understand how to get the partial derivatives

Or like, when I am getting one or the other

[north canopy]

uh not sure i understand what ur asking

how to get partial? u do the partial derivative and treat other variables as constants

[ruby lion]

$r = \frac{z}{\cos(\phi)}$ and $r = \sqrt{x^2+y^2+z^2}$

[ocean seal]

Mr. Rotor

[ruby lion]

I'm just confused in how to make sure I get the partial

[north canopy]

well lets say u have z = 6x^4 + 2y^3 and i ask u to find partial x partial x, what would u get

[ruby lion]

you get 24x^3

but now let's say you have y=f(x)

and you rewrite z taking in that in count

[north canopy]

yes, then u would need to do implicit, bc u can rewrite z as 6x^4 + 2f(x)^3

[ruby lion]

Ok I think I know how to phrase it

Let's say I have f(x,y) = x + y
and now I use the fact that y and x are linked by x = y and so f(x,y) = 2x

Isn't there an issue

cause like, f(x,y) = 2x = 2y

but depending on how i write It down I won't get the same partial derivatives no ?

if I write it down x + y , 2x or 2y, I won't get the same partial

Ok screw it let's get to the point

I'm trying to redo the divergence in spherical and I've made a mistake, I think it's on partial derivatives

tell me if the reasoning is good

wait a sec

[north canopy]

my fault i was tryna help someone

[north canopy]

seems valid

[ruby lion]

I'm going to copy paste something big

[north canopy]

u will get the same partial depending on what variable u use

bc if u do partial f partial x on x+y, u will get 1+f'(x), which will result to 2

[ruby lion]

I think the moment i use x = y in the formula of f(x,y), I'm essentially breaking the independency, so I just won't get the proper partial derivatives

[north canopy]

u will, just need to convert if they are related

[ruby lion]

Fk, how do you go back to line with latex

😭

[north canopy]

,rccw

[north canopy]

here is what u said

[ruby lion]

$\text{div}(\vec{F}) = \frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} + \frac{\partial f_z}{\partial z}$

$\vec{F} = f_x \vec{e_x} + f_y \vec{e_y} + f_z \vec{e_z} = f_r \vec{e_r} + f_{\phi} \vec{e_{\phi}} + f_{\theta}\vec{e_{\theta}}$

USING :

$\vec{e_r} = \sin(\phi) \cos(\theta) \vec{e_x} + \sin(\phi) \sin(\theta) \vec{e_y} + \cos(\phi) \vec{e_z}$

$\vec{e_{\phi}} = \cos(\phi)\cos(\theta) \vec{e_x} + \cos(\phi)\sin(\theta) \vec{e_y} - \sin(\phi) \vec{e_z}$

$\vec{e_{\theta}} = - \sin(\theta) \vec{e_x} + \cos(\theta) \vec{e_y}$

WE GET :

$f_x = \sin(\phi) \cos(\theta) f_r + \cos(\phi)\cos(\theta) f_{\phi} - \sin(\theta) f_{\theta}$

$f_y = \sin(\phi) \sin(\theta) f_r + \cos(\phi)\sin(\theta) f_{\phi} + \cos(\theta) f_{\theta}$

$f_z = \cos(\phi) f_r - \sin(\phi) f_{\phi}$

AND SO :

$\text{div}(\vec{F}) = \frac{\partial \left(\sin(\phi) \cos(\theta) f_r + \cos(\phi)\cos(\theta) f_{\phi} - \sin(\theta) f_{\theta}\right)}{\partial x} + \frac{\partial \left(\sin(\phi) \sin(\theta) f_r + \cos(\phi)\sin(\theta) f_{\phi} + \cos(\theta) f_{\theta}\right)}{\partial y} + \frac{\partial \left(\cos(\phi) f_r - \sin(\phi) f_{\phi}\right)}{\partial z}$

BY DEVELOPPING WE GET :

$\frac{\partial r}{\partial x} \left( \sin(\phi)\cos(\theta)\frac{\partial f_r}{\partial r} + \cos(\phi) \cos(\theta) \frac{\partial f_{\phi}}{\partial r} - \sin(\theta)\frac{\partial f_{\theta}}{\partial r} \right) + \frac{\partial \phi}{\partial x} \left( \sin(\phi)\cos(\theta)\frac{\partial f_r}{\partial \phi} + \cos(\phi) \cos(\theta) \frac{\partial f_{\phi}}{\partial \phi} - \sin(\theta)\frac{\partial f_{\theta}}{\partial \phi} + \cos(\phi)\cos(\theta) f_r - \sin(\phi)\cos(\theta) f_{\phi} \right) + \frac{\partial \theta}{\partial x} \left( \sin(\phi)\cos(\theta)\frac{\partial f_r}{\partial \theta} + \cos(\phi) \cos(\theta) \frac{\partial f_{\phi}}{\partial \theta} - \sin(\theta)\frac{\partial f_{\theta}}{\partial \theta} - \sin(\phi)\sin(\theta)f_r - \cos(\phi)\sin(\theta)f_{\phi} - \cos(\theta) f_{\theta} \right) + \frac{\partial r}{\partial y} \left( \sin(\phi)\sin(\theta) \frac{\partial f_r}{\partial r} + \cos(\phi)\sin(\theta) \frac{\partial f_{\phi}}{\partial r} + \cos(\theta) \frac{\partial f_{\theta}}{\partial r} \right) + \frac{\partial \phi}{\partial y} \left( \sin(\phi)\sin(\theta) \frac{\partial f_r}{\partial \phi} + \cos(\phi)\sin(\theta) \frac{\partial f_{\phi}}{\partial \phi} + \cos(\theta) \frac{\partial f_{\theta}}{\partial \phi} + \cos(\phi)\sin(\theta)f_r - \sin(\phi)\sin(\theta)f_{\phi} \right) + \frac{\partial \theta}{\partial y} \left( \sin(\phi)\sin(\theta) \frac{\partial f_r}{\partial \theta} + \cos(\phi)\sin(\theta) \frac{\partial f_{\phi}}{\partial \theta} + \cos(\theta) \frac{\partial f_{\theta}}{\partial \theta} + \sin(\phi)\cos(\theta)f_r + \cos(\phi)\cos(\theta)f_{\phi} - \sin(\theta)f_{\theta} \right)$

[ocean seal]

Mr. Rotor

[ruby lion]

I think one question would be :
Is this true :

[north canopy]

i havent learned div so i cant tell, i can learn it in a bit tho but i got stuff to do rn so i gtg

[ruby lion]

np dw

$\frac{\partial f(r,\phi,\theta)}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial f}{\partial r} + \frac{\partial \phi}{\partial x} \frac{\partial f}{\partial \phi} + \frac{\partial \theta}{\partial x} \frac{\partial f}{\partial \theta}$

[ocean seal]

Mr. Rotor

[ruby lion]

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[polar chasm]

How do find the angle of a triangle with an adjacent side of 40/3 and opposite side of (80(11)^(1/2))/3

[hard harness]

Tan^-1 (Opposite/Adjacent) = Angle

[polar chasm]

I haven’t learned tan^-1

[hard harness]

Don’t forget to put it on Degree mode

[hard harness]

Tan(angle) = O/A

[polar chasm]

How do u put it to the power of -1?

[hard harness]

Depends on the calculator

But there should be a built-on function

[polar chasm]

What I mean by that is

[hard harness]

Tan^-1 is also Arctan if that helps

[polar chasm]

Oh

Thx

[hard harness]

Yw

[polar chasm]

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[coarse elm]

Can someone helpp

gr 12 calc

the answer key says that the answer is: length 190 m, width approximately 63m

idk how they got that bc i got different numbers so can someone do it from the start and show me the way they did it

[buoyant saddle]

show your work

[buoyant saddle]

where is your perimeter from?

[coarse elm]

From the box that I drew? idkk

because at first i thought why cant i just do the box for the 2400, but the fence is connected to the other boxes so i wasnt sure

ig i could have done 10w + 6h

but last time i did that someone told me not to

[buoyant saddle]

i mean the fence should be separating each compartment

i don’t see where 2 and 6 come from

especially 2

[coarse elm]

the perimeter represents the fence

[buoyant saddle]

where did you get 2 from?

[buoyant saddle]

6 you got from the 6 vertical fence pieces

[coarse elm]

the red is the 2

[buoyant saddle]

no

[coarse elm]

2 red lines

[buoyant saddle]

it’s 5 separate pens you can’t treat the top and bottom as one for a single pen

it’s spans 5 pens

there are 10

[coarse elm]

oh

ok

well if i do that

then i get that w = 38

but if i plug in w as 38 then i get height 63 which is right

[buoyant saddle]

so 2400 = hw and the total perimeter is P = 6h + 10w and
h >= 10, w >= 10 then we want P to be a minimum so P = 6h + 24000/h

[coarse elm]

oh the answer multuplied 38 by 5 which gives 190

[buoyant saddle]

👍🏻

[coarse elm]

thats why at the start i didnt count them as each seperate one

ugh i dont get why the answer key says its 190 then

if in the equation we cut each one by pieace, shouldnt the answer key say the length is 38

[buoyant saddle]

how did you even get your answers?

[buoyant saddle]

did you do this

[coarse elm]

ya

its fine

thanksss

.close

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[buoyant saddle]

,w sqrt(4000)

[ocean seal]

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[buoyant saddle]

,w 2400/sqrt(4000)

[ocean seal]

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[buoyant saddle]

hmm i see

,w 12000/sqrt(4000)

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[lethal dune]

I am stuck

[lone heart]

@lethal dune Has your question been resolved?

[lethal dune]

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[lusty dove]

That's... hard

(I was writing it for half an hour)

[kind trellis]

what is it and what is ur question

[lusty dove]

I was was trying to summarise all math symbols (very usefull ik lol)

and...

I started making s#&* up, and this abomination

[forest marsh]

Missing the primorial

[kind trellis]

Yea there are a lot of missing symbols, but if all the symbols is your goal, you can rewrite 1 as pi * 1 / pi

for example

(unless thats too easy)

[lusty dove]

I was just messing around and now im trying to get an answer lol

[kind trellis]

an answer? For what, there is no question...

[lusty dove]

answer i.e. any simplification, anything, literally anything

[tiny fable]

there's a lotttt of simplification in there

integral of e^(-x^2) from 0 to infinity is sqrt(pi)/2 for example

[kind trellis]

So you just put a lot of different things in 1 equation, and you want to see what it is equal to?

[lusty dove]

basically - yeah

[tiny fable]

i don't think that's gonna give you anything pretty as an answer

[kind trellis]

I think it diverges

[lusty dove]

I'm trying to get to know math in absurd ways (alan becker did well also lmao)

(and also studying in more "normal" ways for polish final exam of highschool [matura])

[kind trellis]

If you really want a more intuitive understanding of mathematics, I'd suggest reading up on analysis

[lusty dove]

analysis are cool

we are learning them in school in this month

thanks for help tho :)

[kind trellis]

np :)

[lusty dove]

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[tacit marten]

I’m trying to derive the formula of the sum of triangular numbers. I can’t think of any way to go from here.

[slim kite]

Try taking 4 terms

So, $T_{n-1} + T_n + T_{n+1} + T_{n+2}$

[tacit marten]

Was it intentional to write T_{n-2} at the end or did you mean T_{n+2}?

[slim kite]

N+2

[ocean seal]

Suika

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@tacit marten Has your question been resolved?

[tacit marten]

I looked through previous exercises and found a sequence of the sum of odd perfect squares. That should help me derive this formula. Thanks for the help.

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[eternal sinew]

I’m learning linear and angular velocity and this is what I did so far. My thought process was to find the radians because 1 revolution can be converted to radians?? But idk if im doing it right because im not sure what to do from here

[wet linden]

do u know dimensaionl anay;sis

like converting units

[eternal sinew]

Yes

[wet linden]

do that!!!

remeber

[eternal sinew]

Convert what?

[wet linden]

what is the radius

in this problem

[eternal sinew]

1 ft

[wet linden]

how many radii are in a revolution

[eternal sinew]

2pi?

[wet linden]

yes

so if we recall

a radian is just a radius

right

[eternal sinew]

Sorry I don’t follow

[wet linden]

an angle in radians

is how many radii the arc length of that angle is

think about this

if we have a radian angle of pi

the arc length of that angle is equal to pi

which is pi amount of radiii

heres a graphic to explain

[eternal sinew]

What do you mean by this?

[wet linden]

i meant arc length for a angle in radians

is just going how many radii

so an angle with a measurment of 1 radian

is equal to 1 radius of the circle

basiclaly if take the raidus of the circle and wrap it around the circle

it will give me the measurment of one radian

sorry for the lenghty explanation, just think it helps to understand for the problem

[eternal sinew]

No yeah I want to understand this concept I really appreciate you explaining it but I am still having trouble comprehending lol

[wet linden]

so take pi/2 on the unit circle for example

basically if i wrap 1.57 radii around the unit circle

it will reach pi/2

[eternal sinew]

Oh

How does that relate to the question?

[wet linden]

for units

b/c 2pi for 1 rev

is an angle

but we can a raidus

for 1ft

just shows why units can cancel the way they do lol

[eternal sinew]

What do you mean it’s an angle?

[wet linden]

1 revoltuons is

2pi

[eternal sinew]

1 revolution is 360° which is why I converted it to 2pi in radians

[wet linden]

2pi is the meausremtn of the angle

[eternal sinew]

Yes

Sorry mb

[wet linden]

ye like u can do what u were doing

which is the equations for angular stuff

but i dont like do use them b/c they just get kinda messy

[eternal sinew]

So I need to convert from radians to revolutions is that right

[wet linden]

first do time

u see how seconds cancel

and we are left with just minutes

on the denom, which is what we want

now we need to get rev on the numerator

im just going to show the work this time

b/c its hard to explain with words

do u see how the units cancel out

and we are left with rev/min

which is rpm

[eternal sinew]

Why is 1 rad = 1 ft

[wet linden]

radius=1ft

[eternal sinew]

Why is radius the same as radian

Am I asking you to repeat what you were trying to say before

[wet linden]

if the radisu was 2

not 1

it would be this

[eternal sinew]

So radians is the same as radius??

[wet linden]

ye

not the ssame

radians isnt same as radius

[eternal sinew]

Same value?

I know radius is a length and radians is an angle measurement

[wet linden]

in 1 radian

there is 2ft of radius

the arc length is 2 ft

1 ft for this example

[eternal sinew]

Huh

[wet linden]

not 2ft sorry 1ft

sorry im not doing a very good job explaining

[eternal sinew]

So radians are proportional to radius

[wet linden]

not exaclty

s is arc legnth

your radius is constant

but if u inc theta

then your arc length increases

but if u find an arc legnth of pi

then there is pi radii that could wrap around that angle

it all sounds confusing, sorry

they are related, but they are not proprtional... proprtional implies that inc/dec one directly affects the other

[eternal sinew]

If r = 3ft then what is theta?

[wet linden]

1 rad

[eternal sinew]

When does the rad change?

[wet linden]

it doesn’t when you dimensional analysis

the radius will always be 1 rad lol

Sorry it’s all very confusing

I would recommend just sticking to the way you learned in class

[eternal sinew]

Wdym the radius is 1 rad

Radius isn’t measured in rads

[eternal sinew]

Could you simplify it for me

[wet linden]

Basically

we know they’re are 2pi rad in one rev

and in 1 rad for an angle it is equal to the amount of radii there is

[eternal sinew]

?

[wet linden]

The arc length is equal to theta x radius right

s=rtheta

If theta is 1

then s=r

[eternal sinew]

I think I get it?

[wet linden]

I would just stick to how u learned

But if u wanna try this on another problem

It’s super fast one you get the hang of it

[eternal sinew]

Idek what this is to be honest

[wet linden]

This method

[eternal sinew]

I don’t know what this method is

I don’t understand

[wet linden]

Sorry :(

I would just stick to how you learned in class

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[eternal sinew]

.reopen

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✅

[eternal sinew]

Is this right?

[wet linden]

11000x60

/2

is not 22000

[eternal sinew]

Oh

Oops

33000/pi

[wet linden]

ye!!!!!

https://cdn.discordapp.com/attachments/1035210635870474271/1262557893656445049/BF22DEC9-AFD7-49A7-AE54-2A386EBE5122.gif

https://cdn.discordapp.com/attachments/1035210635870474271/1262557553859235890/togif-4.gif

[eternal sinew]

That’s so cute omg

Tyy!

[tepid badge]

LOL

[eternal sinew]

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[static star]

Find the errors with this solution

[vale wigeon]

ok, have you found any lines you think are erroneous or even just sus?

[static star]

Yeah i have a few ideas

line 5 is (4 choose 2) * 100

the 100 is because two of the digits are already 7

but maybe it should also include cases with more than 7

actually i think thats wrong

yeah i guess im a bit confused because i know how to do the problem but not how to explain why this is wrong

but definitely line 5 is sus

you are basically subtracting 4 digit numbers including 0 start with 2+ 7s which should be

2 7s) 6 * 9 * 9
3 7s) 4* 9
4 7s) 1

= 523

[vale wigeon]

you sure abt this

oh my bad i cant read

[static star]

yeah was just restating it

[static star]

do you have any tips to start?

[vale wigeon]

i think the problem with the proof is a deeper one

namely that we're not really counting the 5-digit numbers with at least 2 sevens, but rather a different kind of object

namely: five-digit numbers with at least 2 sevens in which 2 of the sevens are red

[static star]

yeah i think that is correct

but the question asks to explain what incorrect assumptions the proof has made

just to clarify

[placid zinc]

Ann is bang-on, that step is not correct

[static star]

which step?

[placid zinc]

It's not exactly a single step, but a combination of steps that don't mesh.
Namely, you can't place two 7's, then "place a 7 later lol" and expect to avoid double counting

[static star]

I see

[placid zinc]

I will put the first two 7's from step 1 in ().

(77)711
and
7(77)11

Are different objects, according to our counting method. That's wrong, since they're obviously the same number.

[static star]

just to make sure im understanding

are you saying that in step 1 you place two 7s, then in step 5 you are placing two more 7s which double counts

[placid zinc]

I have read and understood the question correctly

[static star]

im still a bit confused on your explanation

[static star]

would it be accurate to say there are 10,000 4 and 5 digit numbers with at least 2 7s?

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[coarse elm]

Im confused because I dont even understand the wording

can someone draw the visual for me

i dont understand how its supposed to look like

whats inscribed

[tight pier]

inside

[coarse elm]

oh LOL

[tight pier]

like draw a half circle

and inside a rectangle

and you want to adjust width/height such that area is max

[verbal blaze]

https://www.desmos.com/geometry/qhk4rxbztn

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[timber coral]

Can anyone help me with part d?

[cedar rune]

i think your suppose to have made a graph

by that point

[timber coral]

I have

And the tangent

[cedar rune]

what part are you confused about just not sure how to find the values of x?

[timber coral]

What does the equation=7 for range 1≤x≤7 mean

We already have the values for x

[cedar rune]

on the graph

[timber coral]

Sorry for y the values for y

[cedar rune]

where does the function = 7?

[timber coral]

On 3.88

[cedar rune]

if thats it

[timber coral]

I calculated it and made the point on the curve

[cedar rune]

thats the solution

atleast one of them

ur just finding what values of it satsify the equation

so it equals 7

[timber coral]

I dont think so

[timber coral]

Look if you see it has asked us to draw the graph and given the values for x and y

[cedar rune]

1 sec ill be back

someone is helping me

[timber coral]

Then it has asked to find the missing value k which is y completing the table

[cedar rune]

lets say y = x^2 / 6 + 12/x

than you would draw a horizontal line over y=7

and wherever the function intersects with the line are the solutions

thats how id do it atleast

assuming its all graphically

also

this is a new curve

im pretty sure

i dont think its the one from the tables values

forget i said that

no actually im prettu sure it is a new graph

idk man

hope that helped

[timber coral]

So you're saying I draw a horizontal line and take out 2 solutions?

Basically

[timber coral]

No its the same curve from the table's values unless I managed to screw it up

[cedar rune]

well assuming what u have graphed is the curve

wherver the y values = 7 are your solutions

[timber coral]

Sorry I did screw it up

This is the correct graph hopefully I didn't draw the last point right

I don't think the graphs long enough

shid now i will have to draw it over again

I can get the answer online from graphing tool

[cedar rune]

def ideal if its allowed

[timber coral]

Its homework

The thing is to practice drawing quadratic graphs

[timber coral]

This is also how you would solve for (e) part ii as well yeh?

[cedar rune]

yea

but

[cedar rune]

its not a constant number this time

its x/4

[timber coral]

Heres the question

[timber coral]

Yeah

[cedar rune]

you can rearrange the inequality i believe

put 6 on right side

x/4 on left

graph the curbe

than horizontal line at 6

and those should be ur solutions for the range

[timber coral]

Wait its an inequality

[cedar rune]

yea? lol

[timber coral]

But I thought you just take 2 values as the solutions

[cedar rune]

lessthan or equal to

[cedar rune]

wym

if u rearrange it all u do is make sure ur x value is lerss than or equal to 6

[timber coral]

Like this

I can actually plot this one

[cedar rune]

plotting always works

there isnt 1 right way to solve this i dont think

[cedar rune]

this is right

im like 95% sure

[timber coral]

But its y≤x/4

Does it require shading the region under the line

[cedar rune]

I honestly dont know but 2.05 is definitely ur first solution

the other one is starred

looks like 5.6ish

and thats ur final answer im p sure

[timber coral]

But theres an inequality its ≤x/4 not equal to

Or does it mean the same thing

[cedar rune]

It doesnt make a difference here

[timber coral]

I can't even find the past paper to see the answer its November 1999/2/10 4024

[timber coral]

Are you sure

[cedar rune]

positive

ur graph looks good

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@timber coral Has your question been resolved?

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[oak temple]

Good evening, I was wondering if I could get some pointers on this question

,rotate

[oak temple]

This is what I’ve done and I assume that’s not correct

I also assume the answer isnt just 49/50000

<@&286206848099549185>

[cedar rune]

do you know bayes theorum

[oak temple]

Never heard of it

[cedar rune]

p sure if u do some research on it u will be able to solve this pretty easily

[oak temple]

I’ve just looked it up to see if it’s on my course and apparently you don’t 100% need it, it’s just an accessory

[oak temple]

Is it possible to do it without?

[cedar rune]

i mean ofc its possible

you just gotta think carefully about the probabilities

the theourm just makes it so u dont have to think very much

[oak temple]

Oh I see

I’ll look into it now thank you

.close

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[lost terrace]

Let $X$ be an infinite set. Is $$|{(A,B) : \text{$A,B\subseteq X$ and $A \cap B = \emptyset$}}| = |{(A,B) : \text{$A,B\subseteq X$ and $A \cap B \neq \emptyset$}}|$$ true?

[ocean seal]

Eduude

[forest marsh]

Are you sure of the question ?

[lost terrace]

Why?

[forest marsh]

Didn't you write the same thing twice or am crazy

Oh mb

Its a not equal

[lost terrace]

No. The first set is A and B have empty intersection and in the second one they don't have empty intersection

[forest marsh]

Ye

[lost terrace]

I'm guessing that this is indeed true and that the cardinality of both sets is 2^|X| but I'm not sure

[lone heart]

@lost terrace Has your question been resolved?

[twin nimbus]

If we use the map of (A,B) -> A ∪ B, then we can find that this map is surjective on P(X) for both sets. So each set is no smaller than 2^|X|

On the other hand, I'm pretty sure there is a handy proof that |{(A,B) : A, B ⊆ X}| = |P(X)| but I can't recall it off hand, (a similar technique is used to prove |N| = |Q|)

[lost terrace]

.close

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[fleet hawk]

How do i solve this? Not looking for answer but help learning to solve

[small lance]

Hint: ABD is similar to ACD

I believe you can find the solution with this hint, since you’re hikaru

[fleet hawk]

is the hypetenuse of abd 4?

still unsure of how to solve lol

[blazing tiger]

show the triangles are similar

[fleet hawk]

the hypotenus of bcd will be twice the size?

[blazing tiger]

that's not clear

[fleet hawk]

ok then im confused

[blazing tiger]

show triABD is similar to triDBC

[fleet hawk]

how do i do that

[blazing tiger]

similar just means the have the same angles in order between the vertexs

[fleet hawk]

oh ok so a=c and b and d are the same for both triangels?

<@&286206848099549185>

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[hallow summit]

Who can help me solve this?

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[alpine sable]

i was wondering if i could get help with stats

[alpine sable]

Like the steps and what to plug

[cedar rune]

send your question

[alpine sable]

Im trying to understand this

[cedar rune]

what are you confused about

[alpine sable]

Step threee

And hiw to plug it into a calc

Yk calculator

[cedar rune]

do you know what the confidence interval formul is

the one you sent above

thats what that is

[alpine sable]

Yea

[cedar rune]

whats ur question im confused

u just dont know how to plug it into a calc?

[alpine sable]

Yea and which is whic

Is jt t twst

Z test

I have an 83 ti

[cedar rune]

i gotchu

maybe

[cedar rune]

is this your calculator

[alpine sable]

Yea mine has a sticker on the back lowkey

So maybe

Close enough

[cedar rune]

so what do you not know how to plug in

[alpine sable]

Yep

[cedar rune]

No like

what part do you not know how to

[alpine sable]

Step theee i dont k oe ehetr 2.4 and 3.06 came from

Im pretty shre a kut 2.262 and 10

Mb phone screen cracked a lit

[cedar rune]

the 2.4 is the mean of the differences

3.06 is the standard deviation of those 10

idk if those numbers are calculated right

but thats what they stand for

[alpine sable]

Oh fa show

I got the 3.06 and 2.4

Now how do i plug it into the calc which test or interval fo i use

[cedar rune]

i think you just plug it in normally

i could be wrong

[alpine sable]

But i have to find rhe numbers at the bottom

.209 and 4.591

[cedar rune]

thats probably from the + and - of the equation above

so plug the equation into the calc and those will probably be the numbers

[alpine sable]

But i dont know which equation to use

T test

Z test

2 samp t test

2 samp t int

[cedar rune]

is the equation not written down in the 3rd step box

am i buggin

[alpine sable]

No

[alpine sable]

I jusr know i have to use one of these

[cedar rune]

is this not what your suppose to plugin

i very well could be wrong

but like im pretty sure thats it

[alpine sable]

Yea but i dont know which test to plug it in

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[static star]

stuck on integrating this

[cedar rune]

that can be rewritten as d/dx (xsinxy) =xe^x

[static star]

yeah

but u need to integrate both sides right

[cedar rune]

yea

[static star]

to find the solutions

[cedar rune]

integrating left side just gets rid of the d/dx

than you integrate xe^x

[static star]

i thought u need to like move the dx on the right side before u can integrate

[cedar rune]

wdym

[static star]

nvm

yeah so then u integrate the right side

[cedar rune]

after that you solve for y

and your done

[static star]

how do you handle the abs value

xe^x−e^x

i got

• C

but that i did assuming no abs value sign

[cedar rune]

i dont think thats right

i think its

e^x (x - 1) + C

[static star]

how did u do it

integration by parts?

[cedar rune]

ibp

yea

[static star]

u=x, dv=e^x ?

[cedar rune]

yea

[static star]

xe^x−∫e^x dx

[lone heart]

@static star Has your question been resolved?

[lone heart]

@static star Has your question been resolved?

[static star]

<@&286206848099549185>

[granite crest]

do we have to prove lhs=rhs by integration?

[static star]

Find all the solutions to this differential equation.

[granite crest]

okay wait

general or particular?'

[static star]

doesnt specify :/

[granite crest]

cool

okay so listen

just divide both the sides by xsinx

as you need dy/dx form

then it will be a linear differential equation

[static star]

yeah

[static star]

i got to this step

[granite crest]

okay hollon let me solve further

[static star]

im not sure how to handle abs value

im pretty sure the answer is supposed to look relatively simple

[granite crest]

yeah one sec

[static star]

got this tho

[granite crest]

what was your integrating factor?

[static star]

abs value xsinx