help-23

[timid mason]

(ping me please)

[solar hazel]

no yes

asymptotes are a “behavior as x or y goes to infinity or -infinity” thing

[timid mason]

the thing is, if the functions would be unrestricted, f(x)/g(x) would approach 0

[solar hazel]

wait i may not have read well enough

yea i did not read well enough

my answer is now yes

[timid mason]

This is what they would look like if they werent restricted (the formula for the functions are a guess lol)

[timid mason]

oh, gotcha

so when writing down the asymptotes, one should ignore restrictions to the domain?

[solar hazel]

well it’s more like

asymptotes on the boundary of the domain are still asymptotes

[timid mason]

wait

"hol up"

This would be with the restrictions

and the y=0 asymptote-

i just relaized i wrote the question wrong-

istg

i meant y=0...

not x=0

[solar hazel]

i chose to read it as y=0 based on the rest of the question anyway

[stoic dune]

I see two vertical asymptotes

I don't see enough information for a horizontal asymptote

[timid mason]

then you understood me even when i wrote it wrong 😭

[timid mason]

the horizontal asymptote is there if the functions werent limited

but with the restrictions

well, thats the thing

i dont know

[timid mason]

so (by your first answer) y=0 should not be counted as one

yes?

[solar hazel]

if the domain is restricted to (0,56) or whatever, an asymptote at y=0 (what i meant with at the boundary of the domain) is an asymptote

[timid mason]

[the 0,56 is an arbitrary number, It just appeard that that's the argument where f(x) and g(x) intersect]

[solar hazel]

yea i know

[timid mason]

so in that case the answer to the task "list all asymptotes of f(x)/g(x)" would be:

1. x=0
2. x=a [ where g(a)=0 ]
3. y=0

Correct?

[solar hazel]

i don’t know how you could say there are any horizontal asymptotes on a function with domain (0,56)

(again i know the (0,56) is pretty arbitrary)

[timid mason]

im so confused rn im so sorry 😭

[solar hazel]

horizontal asymptotes are a “behavior as x goes to infinity” thing

[stoic dune]

You need to see what the function is doing as x goes to infinity

[timid mason]

does that imply that functions whose arguments are restricted dont have horizontal asymptotes?

[stoic dune]

Instead, you're guessing what the function might do.

[timid mason]

i think i get it...

so in order to say if a function has a horizontal asymptote, we need to see what it does when its approaching infinity, not make an assumption, yea?

[solar hazel]

sure, sometimes you are expected to make assumptions about graph behavior though

[timid mason]

gotcha

and what about oblique asymptotes? is it the same story?

[solar hazel]

yea

[timid mason]

Tysm <3

so sorry if that was tiresome to explain

[solar hazel]

[timid mason]

.close

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[dapper ledge]

Compute the integral over D:
Double integral (x+y) dx dy
y = -x + 6
y = -x + 3
y =2x
y =x

[safe radish]

Please don't occupy multiple help channels.

[mellow idol]

Oops sorry

[dapper ledge]

get outta here

[mellow idol]

U beat me to it

Sorry

[dapper ledge]

unless u wanna help

😛

Compute the integral over D:
Double integral (x+y) dx dy
y = -x + 6
y = -x + 3
y =2x
y =x

<@&286206848099549185>

[compact wraith]

What have you tried?

Also pretty sure you need an domain for x otherwise your double integral is unbounded

[dapper ledge]

wdym

i got the intersects

x=2 y =4 x=1 y =2 x=3 y =3 x =3/2 y =3/2

im not too sure what ot do now

shud i change to polar?

can u help plz @stable estuary

@compact wraith

<@&286206848099549185>

[vale robin]

why polar

its not a circular region

[dapper ledge]

oh i c

dy y=x to -x+6
dx x=2 to 3

then another one
y=2x to -x+3 dy
then dx x=1 to 1.5

there should be two right

is that corret?

@vale robin

<@&286206848099549185>

[wooden forum]

what about x=1.5~x=2?

[dapper ledge]

how

[wooden forum]

Your overall Integrating range has x=1~x=3 right?

[dapper ledge]

yea

x =1 x =3

[wooden forum]

And this talks aboht x=1~x=1.5 and x=2~x=3

[dapper ledge]

those r the intersects

[wooden forum]

There is also the area between x=1.5~x=2

Did you plot the overall shape of D?

[dapper ledge]

yes

https://gyazo.com/bcbfd5a4ede57abca3f2d30856d58770

ah yes

you ar eright

[wooden forum]

then what should be the interval of x & y for x=1.5~x=2?

[dapper ledge]

i mnot sure

[wooden forum]

what is the upper y boundary for x=1.5~x=2?

[dapper ledge]

4

y=4 y = 1.5

[wooden forum]

I mean

in function of x

[dapper ledge]

y=2x and then -x + 6

[wooden forum]

there is no y=-x+6 for boundary in x=1.5~x=2

[wooden forum]

here

[dapper ledge]

ohhhh

y=2x

y=x

[wooden forum]

yes

now you can integrate

[dapper ledge]

but i need 3 intgrals right

3 double

[wooden forum]

this two

so overall 3 integrals

[dapper ledge]

then another one

[wooden forum]

wdym?

[dapper ledge]

like 3 integrals

one on left one in mid then another on right

right

[wooden forum]

yes

you already found the left and right one

[dapper ledge]

double integral from
dy from y= -x + 3 to y = 2x
dx from 1 to 3/2

then
dy from y = x to y = 2x
dx from 1.5 to 2

then
dy from y = x to y = -x + 6
dx from 3/2 to 3

right

[wooden forum]

yes

[dapper ledge]

danke

[wooden forum]

np

[dapper ledge]

dy must be in terms of x

right

[wooden forum]

wdym?

interval of integral?

[dapper ledge]

dx 1 to 3/2 dy -x+3 to y = x

dx 3/2 to 2 dy y=x to y = 2x

dx 2 to 3

dy y = x to y = -x+6

right

[wooden forum]

yes

[dapper ledge]

then just add them all together right

[wooden forum]

yes

[dapper ledge]

ty

[safe radish]

@dapper ledge Has your question been resolved?

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[white gate]

yo, i cant wrap my head around this, can anyone explain why the exponents apart from the largest ones are ignored whilst the largest ones are chosen?

[thin bridge]

smaller powers are insignificant for large values of x

consider dividing the numerator and denominator by the highest power of x in the denominator

[white gate]

and what will happen if the numerator and the denominator had different values of x? (x^3/x^2)

[frozen marlin]

so like just x^3/x^2?

that simplifies to x, and as x tends to inf, it turns to inf too

[thin bridge]

show me what you have after dividing by what i mentioned

[white gate]

lim x-> inf (x^3/x^2) = 1/1 = 1?

[thin bridge]

no

[white gate]

1/0?

[thin bridge]

no

[white gate]

oohohhhhh

[thin bridge]

you're not using your original questions,
and how are you getting 1/1 or 1/0

[white gate]

dividing the numerator and denominator by the highest power of x in the denom?

[frozen marlin]

x^3/x^2 can be simplified to x, right?

[white gate]

so 0/1?

[frozen marlin]

not here

no!

[white gate]

im confused

[frozen marlin]

this is for more complex equations

[frozen marlin]

but in the equation x^3/x^2, it can be simplified easily

[white gate]

so uh

[thin bridge]

forget about limits

for a sec

[white gate]

okoke

[thin bridge]

if you were asked to just simplify
$$\frac{x^3}{x^2}$$
what would you do?

[flat frigate]

ℝαμΩℕωⅤ

[white gate]

oh damn

yeah its x

[thin bridge]

yes

[white gate]

didnt give much thought into it

what about (x^3+1/x^2-2)?

if lim x approaches infinite

[thin bridge]

do you mean
$$\frac{x^3+1}{x^2-2}$$

[flat frigate]

ℝαμΩℕωⅤ

[white gate]

ah yes this

[thin bridge]

consider dividing the numerator and denominator by the highest power of x in the denominator

[white gate]

since x^2 in the num is equal to 0, then 0/1?

[thin bridge]

no

[white gate]

damn

[thin bridge]

wdym by x^2 in the num is equal to 0

[white gate]

since in the numerator, (x^3 + 1), x^2 isnt present

i just assumed 0*x^2 = 0

since x^2 in the denom is 1

i filled in 0/1

[thin bridge]

uh huh, but that adds no new info

there's no need to introduce/mention that at all

and doesn't affect the overall limit

don't overthink the division

[white gate]

im stuck on this, could you explain to me how you do it?

[thin bridge]

consider dividing the numerator and denominator by the highest power of x in the denominator

don't worry about introducing additional terms to divide

just divide directly

the highest power of x in the denominator is x^2

[white gate]

yes

so i use 2 to divide?

[thin bridge]

wdym

you're butchering my words

you divide by numerator and denominator by x^2

[white gate]

sorry man i wasnt taught this before

[thin bridge]

and then dividing the numerator and denominator by $x^2$:
$$\frac{\blue{\frac{x^3}{x^2}} + \red{\frac{1}{x^2}}}{\blue{\frac{x^2}{x^2}} - \red{\frac{2}{x^2}}}$$

[white gate]

ohhhhhhh

[flat frigate]

ℝαμΩℕωⅤ

[thin bridge]

simplify the blue parts

the red parts → 0 as x→inf

[white gate]

x for num and 1 for den

[thin bridge]

this is the reason why you can ignore all the lower degree terms

[white gate]

so im left with x/1?

equals to x

right?

[thin bridge]

sort of

[white gate]

then what is left is x = 0 as x-> inf

[thin bridge]

no

[white gate]

damn

[thin bridge]

involving the limit you'll have
$$\lim_{x\to \infty} \frac{\blue{\frac{x^3}{x^2}} + \red{\frac{1}{x^2}}}{\blue{\frac{x^2}{x^2}} - \red{\frac{2}{x^2}}} = \lim_{x\to \infty} x$$

[flat frigate]

ℝαμΩℕωⅤ

[thin bridge]

there is no x = 0 here

x→inf as x→inf

[white gate]

nono not on the lim

let me try the bot rq

.
$$\lim{x\to\infty} x = 0$$

damn

[thin bridge]

no

[flat frigate]

Potus

[thin bridge]

as x gets infinitely large
x doesn't get anywhere close to 0

x gets large means x gets large

[white gate]

oh damn okok i think i understand now

[thin bridge]

as far away from 0 as you could possibly be

[white gate]

i just mixed up my brain with 1/x

anyways thanks a lot man, real sorry if you got frustrated trying to explain limits approaching infinity to me 🙏 🙏

no wonder i got a D in calc 1

[frozen marlin]

is fine

calc is weird and very unintuitive

just learn the concepts behind stuff and ur gonna ace it

[white gate]

got it 👍 thanks

.close

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[dapper ledge]

Find the volume of a sphere of radius 2 from which a central cylinder of radius 1 has been removed

[dapper ledge]

<@&286206848099549185>

[frozen marlin]

please don't ping helpers before 15 mins have passed

[dapper ledge]

ok so

soz

[frozen marlin]

np
i'd help but am horrible at polar

:(

[dapper ledge]

well not polar coordinates

but double integrals

Find the volume of a sphere of radius 2 from which a central cylinder of radius 1 has been removed

[frozen marlin]

no idea what those are either

[dapper ledge]

i dont understand

[drowsy karma]

sketch the graph, to observe the situation,

[dapper ledge]

i dont understand...

do i get th evolume of sphere

[drowsy karma]

wait

[dapper ledge]

then i - it from the volume of the cylinder

[drowsy karma]

if you talked about polar coodiantes, then youhave to know it

yes that is other way

[dapper ledge]

how do i do other way

[drowsy karma]

form volume of sphere you can subtrat volume of cylidner, but

cylinder inside the sphere

[dapper ledge]

no i meant using double integrals

[drowsy karma]

is not onyl cylinder

[royal kiln]

you really should have said if you are expected to use cylindrical or spherical coordinates

[dapper ledge]

there is none

thats just the question

Find the volume of a sphere of radius 2 from which a central cylinder of radius 1 has been removed

which is easiest

[drowsy karma]

i precisely wrote it for you, general method

[dapper ledge]

oh that is it?

so square root(2^2 - 1)

• r dr

then just a normal integral?

[royal kiln]

$dV=ρ^2 \sin{φ}dρdθdφ$

[flat frigate]

Melvin Eugene Punymier

[lean otter]

You're doing great kiddo.

[dapper ledge]

so i am correct?

[drowsy karma]

it must be a double integral, or triple, but double is ok here, R = 2, and r = 1, in your case

[dapper ledge]

how did u find that equation?

[drowsy karma]

i sketched the graph of sphere and cylidner inside sphere, in my mind, and then i made it projection on plane z = 0

then i get two circles, one inside

R =2, r = 1

[dapper ledge]

ic but is that equation online

[drowsy karma]

fo rvolume ?

[dapper ledge]

yea

[royal kiln]

here's a neat related fact for later:
https://www.sfu.ca/~adebened/funstuff/sphere_cyl.pdf

[drowsy karma]

that is general formula with help of the double integral:

[dapper ledge]

i dont understand

[drowsy karma]

in your case g(x,y) - the equation of upper part of sphere

h(x,y) = 0, hence i multiplied by 2

[dapper ledge]

i dont get it

[drowsy karma]

have you ever calculated some doublle integrals ?

[dapper ledge]

yea but not like this

can i use double integrals for volume of sphere then double integrals for volume of cylinde

then take away the 2?

[drowsy karma]

the volume of the area bounded by the sphere, i.e. the volume of the sphere, you know that there is no need for integrals, but what you have to subtract consists of the cylinder and something else between the cylinder and the sphere

[dapper ledge]

i c

what about

a triple integral then

[royal kiln]

$V = 2\int_{0}^{\sqrt{3}} \int_{1}^{\sqrt{2^2 - z^2}} \int_{0}^{2\pi} r d\theta dr dz$

[drowsy karma]

no, it is not easier

[royal kiln]

I like it...

[drowsy karma]

sure

[royal kiln]

it's in cylindrical coordinates

^

[dapper ledge]

is that right

[flat frigate]

Melvin Eugene Punymier

[royal kiln]

yeah

[dapper ledge]

howd u find dat

[royal kiln]

based off this

https://www.sfu.ca/~adebened/funstuff/sphere_cyl.pdf

[dapper ledge]

i c but what about normal double integral tho

[royal kiln]

I just had to draw a little picture to get the values right:

[dapper ledge]

I c but what about joannes metho

[royal kiln]

The "normal double integtal" is just this with the inner integral already evaluated

iirc

[dapper ledge]

can i c

[royal kiln]

Yeah go for it

The other form may have been in a different order than this

And yes: this envelope IS the murder weapon

[dapper ledge]

i c

but what about double

[royal kiln]

This is a double integral now

You want to do the one that Joanna introduced?

[dapper ledge]

ya

[royal kiln]

oh ok-ok: It's the same as this, but z integration is already done

[drowsy karma]

i have calculated yoru task in two ways, some details i omit:

the secodn way is: volume of the sphere minus volume of region consisted of cylinder and region ebtween sphere and cylinder

[dapper ledge]

square root (4-r^2) * dr * d0
dr is from 1 to 2
and d0 is from 0 to 2pi

right

[drowsy karma]

i do not know how to call it nicely in english 🙂 forgive me ))

[dapper ledge]

and then solbve?

[drowsy karma]

but math is correct

[dapper ledge]

square root (4-r^2) * dr * d0
dr is from 1 to 2
and d0 is from 0 to 2pi

this right set up?

[drowsy karma]

first my integral shows calculation of the volume of the region, located outside cylinder but inside sphere

[dapper ledge]

i c bu is that right set up

[drowsy karma]

yes, both are

[royal kiln]

Hurray!

[drowsy karma]

second maybe, is easier for you to understand

[dapper ledge]

square root (4-r^2) * dr * d0
dr is from 1 to 2
and d0 is from 0 to 2pi

[drowsy karma]

do not forget about r = jacobian

[dapper ledge]

is this right?

[royal kiln]

There's the rest of the triple btw

[dapper ledge]

ic danke

but is this right

square root (4-r^2) r dr * d0
dr is from 1 to 2
and d0 is from 0 to 2pi

[drowsy karma]

now ok

[dapper ledge]

theres no 2* thi

[drowsy karma]

you need to multiply by two, your set up

[royal kiln]

the 2 is so that the integration only needs to happen on half of the sphere. The total volume is double the volume of the hemisphere

[dapper ledge]

bruh

[drowsy karma]

smiles that i have known )

[dapper ledge]

i did it wrong

so first

square root ( 4-r^2)

rdr

dr is from 1 to 2

right

[royal kiln]

I responded to the wrong avatar but I was telling @dapper ledge

[dapper ledge]

@drowsy karma dat wrong>

melvin

did u try ths

square root ( 4-r^2)
rdr
dr is from 1 to 2

[royal kiln]

I mean, I just did it and I got the same answer as @drowsy karma

[dapper ledge]

square root ( 4-r^2)
rdr
dr is from 1 to 2

u did dat?

[royal kiln]

I'm doing z^2

not r^2

I'm sure the way she does it is logical

[drowsy karma]

make a photo of your exercise book and show it here

[dapper ledge]

square root ( 4-r^2)
rdr
dr is from 1 to 2

right

then i got -1 d0 from 0 to 2pi

[royal kiln]

hers is "r dr d\theta"

[dapper ledge]

yea

[royal kiln]

she integrates with respect to r, then theta

[dapper ledge]

i did

i got -1

[royal kiln]

I was doing \theta, r, z

[dapper ledge]

i got 4pi

thats wrong

[drowsy karma]

look once again

[dapper ledge]

nvm i did it danke

ans is

21.73

right

[drowsy karma]

great

)

[dapper ledge]

dank

ty

[drowsy karma]

yw

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[median abyss]

,(5^3 \times \binom{5}{3} \times 3^2)

[median abyss]

.close

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[hard crest]

#latex-testing or #bots plz

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[rare vale]

Is arctan(infinity) always pi/2?

[spice grove]

yes, though mostly you'd use x -> infty and not just simply arctan(infinity)

Many people do add npi and shit. But if you're looking at the inverse function of tan(x) then it needs to be bijective so you only want pi/2 and nothing else.

It's usually context dependent whether you're looking at the function or not. But in general, it is only the principal value.

[neat patrol]

its not bijective

it is injective, but not surjective

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[haughty nymph]

Im actually stuck on the first part

I beleive I am supposed to factor the top side of the function

but I dont know hoow

Nevermind I got it lol

.close

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[native slate]

Hi All, I'm trying to calculate the probability to win a prize draw.

[native slate]

Its been many years since i've done probability and statistics so i'm finding myself a little stuck

I'm struggling to calculate the total number of chances there in the draw.
The draw open to all account holders of a bank.
The chances to win are based on the balance in the account.
You get 1 chance for the first 1000 in the account, and an additional chance for every 200 in the account.
We know:

• the total amount in all accounts: 3,000,000,000, and
• the total amount of clients: 1,250,000

My first step is to try and figure out what kind of client distribution we have over the 3bln balance. For that

To me it seems logical to have an exponential distribution: more people have less, and less people have more.

Looking back on the past wins, the exponential distribution looks correct

Number of Wins Frequency Sum of Frequency 3 3 9 2 38 76 1 1394 1394 Grand Total 1435 1479

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[native slate]

No

[lean otter]

ok

post a question or else close the channel @native slate

[native slate]

How can I close it

[lean otter]

.close

[native slate]

.close

[safe radish]

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[lean otter]

cool

#❓how-to-get-help

[safe radish]

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[hushed mantle]

i have no idea how to do the first one

is -b the gradient or the y intercept?

[dense egret]

M is the middle of the line segment of AB with (3,6) so the the length between M and A is sqrt(3^2+6^2) (pythagoras). the length is sqrt(45). the length of AB has to be 2*sqrt(45). so the length between M and B has also to be sqrt(45). with that you can write sqrt(45) = sqrt(6^2+(x-3)^2) and solve for x. x is 6. so A will be (6, 0). so B has to be B(12,0). With that the gradient has to be b =2 so -b = -2. c = 12

[safe radish]

@hushed mantle Has your question been resolved?

[hushed mantle]

i dont rly get it

can you explain it bit by bit

[dense egret]

so to find the gradient of b and the intercept of the slope c you have to know where A and B is. You can just draw a triangle in the left bottom corner. then you can mirror it to the right. To receive A you can calculate the length of OM which is sqrt(45). But OM has also to equal AM. so AM is also sqrt(45). With the height of 6 of M and the length AM you can calculate the coordinate of A: sqrt(45) = sqrt(6^2+(x-3)^2). the solution is 6. because the question states that M is in the Middle. the lengths of BM and AM have to equal (in sum) the length BA. To receive the height of B you can write sqrt(45) = sqrt((x-6)^2+3^2). the solution is 12. because A and B are intercepts their other coordinate has to be 0. so A(6,0) and B(0,12). With that you can draw a line between A and B and recognise that the gradient has to be -2. because B is also the intercept of the slope, c = the y-coordinate of B, so c = 12. use always a sketch to simply problems

[hushed mantle]

the square root part confuses me

like where is the square root coming from

[dense egret]

you know c^2 = a^2 + b^2? pythagoras theorem?

if you know the length of 2 sides of a triangle with one corner being 90 degrees you can calculate the length of the third side with that formula

because you know 3 being the length on x and 6 being the height on y you can state: AM^2 = 3^2 + 6^2. by using the square root you receive AM = sqrt(45)

[hushed mantle]

ohh ok

ok thanks

is there a way to find gradient without drawing a graph?

oh wait nvm

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[random lynx]

I'm gonna sound dumb asf rn but why does it have to be done the first way and can't be done the 2nd way

[lime dust]

You made a mistake

(x+5sqrt(x))^2 is not x^2+25x

It’s x^2+25x+10xsqrt(x)

That’s why you don’t solve this way, because you don’t get rid off the sqrt

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[safe radish]

@last phoenix Has your question been resolved?

[last phoenix]

/close

.close

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[last onyx]

help

[last onyx]

how he calculate this?

<@&268886789983436800>

<@&286206848099549185>

[warped roost]

please do not ping moderators for math help

[spiral shuttle]

Yes

[last onyx]

ok sorry for that

[thin bridge]

factorial definition,
pascals triangle

[last onyx]

what?

[thin bridge]

have you done anything with combinations, binomial coefficient before?

[last onyx]

this?

[thin bridge]

yes

[last onyx]

I'm very smart!!!

.close

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[lone jay]

when a question says

[lone jay]

K'(L)

Like here

Does this mean the implicit differention in part b) is oging to be

dK/dL for every time I do a derivative of a K

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[lavish cosmos]

7= square root 9−x ^2 + square root 49−x ^2
what is x

[safe radish]

@lavish cosmos Has your question been resolved?

[lavish cosmos]

x>

X?

[safe radish]

@lavish cosmos Has your question been resolved?

[safe radish]

@lavish cosmos Has your question been resolved?

[stray ice]

@lavish cosmos has it been solved or?

[safe radish]

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[lean otter]

need help finding this derivative its question number 13

[lean otter]

i got the basic idea but i think i fucked up somewhere in my algebra

cause im getting a quadratic equation for f^1(x)

which doesnt make any sense

[pastel abyss]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[mellow cobalt]

Why do you use the limit?

[lean otter]

cause thats how u get the slope of the tangent line

[mellow cobalt]

There is another way to do that

[lean otter]

i have to use the definition of the derivative

for this specific question

[pastel abyss]

this is correct

when u're differentiating f/g the denominator of the dervative will be g²

[lean otter]

then when i plug in x to find the slope i get a slope of -2 at the point (3,3)

[pastel abyss]

so it makes sense that u got a quadratic

yes

[lean otter]

OH

I was looking at the wrong question answer on my txetbook

thx for help

[pastel abyss]

np

[safe radish]

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[safe radish]

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[loud bronze]

hey can someone help me with picard theorem

the x at the top is what i got as the general solution before

and f(x,t) is next to b)

how do ik if t^2+3tx+x^2 is continuous

[safe radish]

@loud bronze Has your question been resolved?

[loud bronze]

<@&286206848099549185>

[lean otter]

@loud bronze

[loud bronze]

ah okay

sorry how do yk 3tx is continuous

[loud bronze]

.close

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[upper moth]

Given two 32 bit RNGs that have been seeded by an NIST compliant entropy source, is it acceptable to rely on the generated numbers being unique? What are the chances of a collision occurring?
I'm thinking the birthday paradox could come in to play here and a collision could end up being fairly likely

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[short sparrow]

Let $A,B \in \mathbb{R}^{n\times n}$ be symmetric matrices.\
\
(a) Show that every local extreme point $\overline{x}$ of $f(x) := \langle Ax,x\rangle$ under contraint $g(x) := \langle Bx,x \rangle - 1 = 0$ is a generalised eigenvector of $A$

[flat frigate]

Levens

[short sparrow]

i never really merged linear algebra and analysis together like that so i dont really know what to do here

<@&286206848099549185>

[safe radish]

@short sparrow Has your question been resolved?

[obsidian oracle]

maybe try starting to characterize local extreme points of f

(as in compute df)

[short sparrow]

how do i do that? since we're not given an explicit A

[obsidian oracle]

you can compute f(x+h)-f(x)

[short sparrow]

so <A(x+h),(x+h)> - <Ax,x>?

[obsidian oracle]

yes

[short sparrow]

we're probably using the standard inner product since they didnt specify right

[obsidian oracle]

yes <x,y> = sum(xiyi)

[short sparrow]

but i dont see how that gets us any further

apparently the derivative is Ax? but i dont get why

[safe radish]

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[short sparrow]

lol idk why it closed on me..

[short sparrow]

i dont understand how i can find the derivative of <Ax,x>

[peak estuary]

just write out what <Ax,x> is

in sum notation or with ... or however you like

if you want do it for n=2 or n=3 first

to see whats going on

after that you can just differentiate with respect to each variable

[short sparrow]

so $\sum_{i,j} A_{i,j}x_j x_i$ right?

[flat frigate]

Levens

[peak estuary]

yes

[short sparrow]

okay i got (A^T + A)x is that right

what now?

right and since A^T = A we have 2Ax

<@&286206848099549185>

[safe radish]

@short sparrow Has your question been resolved?

[short sparrow]

what if B was invertible? what could we say about f(x) and its extreme points?

maaan..

.close

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[lean otter]

I'm getting

-63/65

but that's not the answer

apparently

I can show steps

but i mostly need someone

to

verify

[faint zealot]

your cos(a) should be -3/5

[lean otter]

how come?

a^2+b^2=c^2

-5^2+b^2=13^2

b^2=13^2-(-5)^2

[faint zealot]

looking at the graph of cosx between pi and 3pi/2, it should always be negative though, so we need the negative root of b

[lean otter]

wait wrong

(-4)^2+b^2=5^2

5^2-(-4)^2=b^2

b^2=9

b=3?

[faint zealot]

we could have b = 3, but b = -3 is also an option

looking at the graph we see it has to be negative (do you see why?) so we have to pick -3

[lean otter]

I'm not sure if on an actual assessment I'll be allowed to use one of those calculators so

mathematically speaking

why is that

[faint zealot]

you dont need the calculator, its just to look at

[lean otter]

oh wait a minute

square root is either

[faint zealot]

knowing the shape of the cosine function is just a standard thing you'll be expected to know

[lean otter]

+/-

your right

[faint zealot]

yeah exactly

[lean otter]

but how would you verify

?

[faint zealot]

you mean verify the final answer or verify whether we want the positive or negative root?

[lean otter]

yes

positive/negative

root

[faint zealot]

so we're told pi < a < 3pi/2

[lean otter]

yup

oh i see

thought that info was useless

[faint zealot]

and looking at the graph of cos(x) between those points, its always negative, so our root should be negative too

[faint zealot]

its always a good idea to think about why the information's there, and what changes if it wasn't. If we weren't told, we wouldnt be able to pick between 3 and -3

[lean otter]

hm weird

oh okay i see

pi is -1

so -1 to 0

wait no

well then

technically

none of those answers

are valid

because a must be

-1 to 0

[faint zealot]

true, but remember we were only looking at the numerator there. Really we're picking between 3/5 and -3/5

[lean otter]

okay

2nd option

would have to be it

[faint zealot]

yeah, do you feel fully satisfied it can't be 3/5?

[safe radish]

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[lean otter]

I thought I understood this but whenever I convert I don't get a exact number so IDK what it wants me to put it as

[safe radish]

@lean otter Has your question been resolved?

[left gyro]

,rccw

[left gyro]

you keep subtracting 2π from 9 radians until the resulting angle is between 0 and 2π
then what you get is... ||9 - 2π because thats literally what you did||

[lean otter]

Woah that's way simpler than what I was making it thank you

[left gyro]

np

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[lean otter]

how would phase shift

[lean otter]

affect the period

say you had

a phase shift

of 2 units right

and a period of 8

[spring glacier]

Phase shift shouldnt affect the period

[warm stirrup]

phase shift does not affect period

[spring glacier]

Because f periodic when f(x)=f(x-a), for some a
Then the phase shift will hold
f(x-b)=f(x-a-b), which still has period a, if we let g(x)=f(x-b), then
g(x)=g(x-a), which has the same period of a

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[rose thorn]

an urn contains 6 white balls and 9 black balls. 4 balls are drawn randomly twice in succession without replacement. what is the probability that the first 4 balls are white and the next 4 balls are black

[rose thorn]

i used the bayes theorem to solve

[inner current]

what is the doubt here

[rose thorn]

how to solve the question @inner current

cause my answer was wrong

i'll tell you my approach

for first 4 balls we have

[inner current]

go on

[rose thorn]

(6/15 x 5/14 x 4/13 x 3/12)/(9/15 x 8/14 x 7/13 x 6/12) + (9/15 x 8/14 x 7/13 x 6/12) + (9/14 x 8/13 x 6/15 x 5/12...)

for all cases i.e. 4w 0b, 3w 1b, 2w 2b, 1w 3b, 0w 4b

[rose thorn]

got this from bayes theorem

the denominator cancels in the end

as there is no replacement, we'll do the similar thing for next set of balls, but the only possible cases are 4b 0w, 3b 1w, 2b 2w as 4 white balls are aready drawn and only 2 white balls remain in the urn

so for the next set of balls we have

(9x8x7x6/11x10x9x8)/all possible cases

then we multiply both to get the answer

so what's wrong here

@inner current

[safe radish]

@rose thorn Has your question been resolved?

[delicate bobcat]

You have a sequential series of events occurring

4 black then 4 white

Or

B, B, B, B, W, W, W, W

You can answer this problem like flipping a coin multiple times

Just multiply the probabilities together

[rose thorn]

that's what i did

look at the expln i gave

[delicate bobcat]

Ok well let's double check the arithmetic then

[rose thorn]

i did that 3 times ngl

even used a calculator after that

still wrong

[delicate bobcat]

Hmm well that should be the correct method

Is your question marked wrong on computer?

Or do you have a key you're looking at

[rose thorn]

no it was an mcq

answer wasn't matching

[delicate bobcat]

Do you have the choices

[rose thorn]

eh wait

i do

[delicate bobcat]

That's a different question

[rose thorn]

it's same just the colours are mismatched

[delicate bobcat]

Oh yea my bad

[rose thorn]

@delicate bobcat

[delicate bobcat]

Let me take a look im gonna check by hand real quick

I got 3/715

[wispy portal]

yeah i just solved and agree with you

[delicate bobcat]

Nice

What you have to do is write out that long product of fractions

Then observe that a lot of them reduce

Then observe that a lot of the numerators cancel with a lot of the denominators

And when you cancel everything on top and bottom you end up with

3 / [(13)(11)(5) ] = 3/715

I'm not sure where the source of your error is but i would restart and write all those fractions in a row and carefully cancel and reduce line by line

@rose thorn

[wispy portal]

yeah thats what i did lol

[delicate bobcat]

Nice perfect demonstration

[rose thorn]

okay got it

thanks

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@rose thorn Has your question been resolved?

[safe radish]

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[deep haven]

having trouble how to do word problem like these...

[pine spoke]

Essentially this tells you, every 6000 years the amount will decrease to half of it

[deep haven]

see i yunderstand that and i can solve it like that

but when it says y years

thats where im stumped

cuz u basicallu need a formula for that

and you can't do it without a formula

[pine spoke]

Well you can always come up with one

[deep haven]

i dont know how to

[pine spoke]

As we know, every 6000 years the amount halves

We can define a function

f(x+6000)=1/2 x

Or similar things

Does that give you a good approach?

[deep haven]

i think so

[pine spoke]

Have a shot at it then

[deep haven]

okay

il ping u when i figure something out

@pine spoke is it like this

100(1/2)^w/6000

100 is the initial value

1/2 is like the decay factor

and we are dividing by 6000 cuz it happens every 6000

[pine spoke]

Something like that

[deep haven]

ok this question i dont get at all

[pine spoke]

It is a function respect to time and substance

[pine spoke]

Uh

I don’t think we worry about the concrete part

It’s just a similar question

But this time you are looking for a fractional answer

[deep haven]

how do i start

the process

to solve it

is it something like

50000(1/2)^n/25000

[pine spoke]

Where did 50000 come from?

[deep haven]

since te half life

is

25000

im guessing

the full life

is 50000

[pine spoke]

Woa

Nope

Not like that bud

[deep haven]

no?

oh

[pine spoke]

25000 is the time taken to halve it

50000 years later it’s going to be a quarter

Not gone

[deep haven]

oka

but for all these problems i use the same structure right

like

a(r)^n

something like that

[pine spoke]

Rather than using a constant, try using a percentage or a variable

[deep haven]

for what

the initial value?

[pine spoke]

Like 25000 years later, X amount of radioactive plutonium become x/2

[deep haven]

oh

[pine spoke]

Because you are looking for a fractional answer

[deep haven]

yeah

[pine spoke]

This way you will save yourself lots of time

[deep haven]

is the answer .997

[pine spoke]

I haven’t calculated it

But it looks fine

[deep haven]

okay

[pine spoke]

For a million it’s quite easy as it’s just 40 of 25000 years

So it halves itself for 40 times

[deep haven]

alr alr

tysm

[pine spoke]

Your welcome

You get the hang of this type of questions once you know how to make functions about them

[deep haven]

okay

[pine spoke]

Other wise You will break your head when you have some gruesome numbers

Like 42195

💀💀💀

[deep haven]

lmfao

alr ty

.close

[safe radish]

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[rough matrix]

Masses m1 = 10.0 kg and m2 = 5.00 kg are connected by a light string
that passes over a frictionless pulley in the figure to the right. If m1,
initially held at rest on the table, travels 1.0 m in 1.2 s across the table, determine the friction force between it and the table.

[rough matrix]

I keep getting 38.58N

[light shoal]

show work?

also pls show the figure

[rough matrix]

Figure

Work

<@&286206848099549185>

@light shoal

Or <@&286206848099549185>

[light shoal]

sry, i can't really read what you did

[rough matrix]

Okay so acceleration is 5/6

Right?

M/s?

Or .83

I have an assessment tomorrow

On this

Thas y a little nervous

I rewrote it

Am i on right track @light shoal or <@&286206848099549185>

Im tryna find t on right sude

Side

[light shoal]

these are ok so far

now you need to write down what Fg is

and what Ff is

[rough matrix]

Im getting a rn

Is this good for a?

Iv been doing this but getting wrong answer

Can u jus do whole question and see how ur answer compares to mine? @light shoal

[light shoal]

the velocity is not constant, right?

so what you wrote is only the average velocity

during the first 1.2 seconds

[rough matrix]

Wdym

[light shoal]

well initially nothing is moving

then the objects start accelerating

[rough matrix]

Velocity average is .83

But since starting at rest it dont matter

Cause vi is 0

[light shoal]

ok..

[rough matrix]

Does it?

[light shoal]

yea if the acceleration is constant then velocity will increase linearly

is the acceleration constant here?

[rough matrix]

I think

Yes it is

[light shoal]

ok

i think you're correct, the only external forces here are gravity and kinetic friction

[rough matrix]

Wait how is friction positive

Answer key says positive

[light shoal]

well it depends on which way your friction force arrow points

it's positive to the left, or negative to the right

[rough matrix]

Isnt it other way?

Around

[light shoal]

friction force is in the opposite direction of motion, which way is the motion?

[rough matrix]

Right

[light shoal]

so friction is pointing left

[rough matrix]

So negative

[light shoal]

so if you draw the friction force arrow to the left then it's positive in that direction

if you draw it to the right then it's negative in that direction

[rough matrix]

I made my signs - for left and + for right

And got friction to be +

How is it possible

[light shoal]

ma = T - Ff
a is in the rightward direction, so:
positive T is pointing to the right
negative Ff is pointing to the right
positive Ff is pointing to the left

if you had written ma = T + Ff then negative Ff would mean left

[rough matrix]

I wrote ma=T-Ff

[light shoal]

yes

so see above

[rough matrix]

Is that not what i did

Like m,a and T are positive

And ff negative

[rough matrix]

How @light shoal

@light shoal or <@&286206848099549185>

[wooden forum]

I think he set both T and Ff to be positive to the direction m1 is moving

so if Ff becomes negative, then friction is acting left

which is opposite direction of which m1 is moving

[rough matrix]

Isnt ma=T-Ff The same as ma=T+Ff?

It doesnt look the same tho

[wooden forum]

It's just on how you would decide the axis of Ff

[rough matrix]

Wdym

Like left negative and right positive

[wooden forum]

Ill try elaborate it

[rough matrix]

Ok

I don’t understand how it doesn’t wrk out

[wooden forum]

I think you can ignore what I said before

In my opinion ma=T+Ff form is to express Ff in vector; you dont know the direction, so we just wrote as T+Ff to write as sum of two forces. T&Ff's negativeness will decide which direction their forces are

In other side, ma=T-Ff form is from the idea that we already know the direction of forces, and the fact that we already know the direction of net force. So we only think about size of them, not considering direction

[rough matrix]

So how are they not the same

Im so confused why it doesnt work out

[wooden forum]

It's not same as equation, but can used to solve same problem

[rough matrix]

If i fo ma=t-ff i get positive

Then negative for other

Is that how it supposed to be?

[wooden forum]

Let's say $|F_f|$ is size of friction

[rough matrix]

Ok

[flat frigate]

Dri111

[rough matrix]

Arent both ways considering direction

How come direction different

@wooden forum

[wooden forum]

If we write $ma=T-F_f$, where $F_f$ is positive, then our formula can be written as $ma=T-|F_f|$ right?

[flat frigate]

Dri111

[wooden forum]

because Ff is positive

[rough matrix]

How?

[wooden forum]

if real number a is positive, then |a|=a

[rough matrix]

Ok true

Alr

So in that we saying friction is positive?

[wooden forum]

if you write ma=T-Ff, yes

Remember, Ff is the number representing the force. You need to interpret what the value and negativity of Ff in equation according to your setup

[rough matrix]

Ok

So other way is bttr

[wooden forum]

if $ma=T+F_f$ where $F_f$ is negative, then $ma=T+F_f=T-|F_f|$

[flat frigate]

Dri111

[wooden forum]

So same as this

[rough matrix]

So for this equation u hace to kinda look at ur diagram and make ff negative according to your arrow whereas the other way will give exact value and direction

[wooden forum]

T-Ff can also give you directions, but overall you are right

It's just that you already included direction with minus on Ff

[rough matrix]

How does it jus go away

[rough matrix]

Wdym by this

[wooden forum]

For your question, the direction of friction, tension, and net force is obvious; so ma=T-Ff is from thinking of size

but in some question directions are not obvious; this is the case Inwas saying

[left gyro]

amazing seeing cell lab in the wild

[wooden forum]

you can find discord server if you want

[rough matrix]

So the other way will be good for all questions

[wooden forum]

It would be general method, I think

[left gyro]

not good enough to really do much but thanks

[wooden forum]

Cell lab server welcomes new people, so feel free to visit anytime

[safe radish]

@rough matrix Has your question been resolved?

[safe radish]

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[crude lynx]

Topic: TI-84 Help
How can I input the red into my TI-84, specifically the y variable?

[crude lynx]

<@&286206848099549185>

[left gyro]

the y variable is called a "dummy variable" - it doesnt affect the calculation

,,\int_{-\frac{5\pi}6}^0(-2\sin y-2\sin y)\dd{y}+\int_0^\pi(2\sin y+2\sin y)\dd{y}

[flat frigate]

matt07734

[left gyro]

is the same as

,,\int_{-\frac{5\pi}6}^0(-2\sin x-2\sin x)\dd{x}+\int_0^\pi(2\sin x+2\sin x)\dd{x}

[flat frigate]

matt07734

[left gyro]

since both integrals are the same, you have to enter it in with x instead

[crude lynx]

aah I see

ty matt

[left gyro]

np

[crude lynx]

.close

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[safe radish]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[trail oriole]

The radius of each circle is 1 cm, how do we find the area of the rectangle?