help-13

[narrow scarab]

true that

[cedar kiln]

@midnight bramble Has your question been resolved?

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[ornate rune]

What is the probability of all the dice getting a value of less than 4 in all 30 rolls

[solid juniper]

what’s the probability if you just roll 1 die?

[ornate rune]

1/2

[solid juniper]

right

[mental trail]

and all the dices are independent

[solid juniper]

so what’s the probability of that happening 30 times in a row?

[mental trail]

so what happens when you repeat a "bernouilli experiment" independently 30 times?

[ornate rune]

Idk

[mental trail]

think about what happens when you repeat a simple experiment independently over and over

the probability that a fair coin lands on heads is 1/2

what's the probability it lands on heads twice in a row

[ornate rune]

Well I suppose it depends I guess

[ornate rune]

Still 1/2?

[vagrant elbow]

What are all possible outcomes for two coins being tossed?

[ornate rune]

HH, HT, TH, TT

[mental trail]

ok

so what's the proportion of outcomes where we land on heads twice

[ornate rune]

1/4?

Ah so that's why it's being multiplied

[mental trail]

have you not seen P(A and B) = P(A)P(B) for independent events?

[ornate rune]

Wdym

[mental trail]

you've never seen definition of independence?

for events in probabilities

[ornate rune]

Not that I remember

[mental trail]

oof

[ornate rune]

Well how do you know if a operation to be used is multiplication or addition

[mental trail]

independence: two events A,B are independent if one happening doesn't affect the other

effect on probability: P(A and B) = P(A)P(B) (multiplication)

incompatible: two events A,B are incompatible if it's impossible for both to happen together

effect on probability: P(A or B) = P(A) + P(B) (addition)

[ornate rune]

I see thanks

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[tidal blaze]

Is it possible to do this equation without having to spend 6 hours ?

[normal cipher]

sub x = 1+r i guess

then you get 4 degree polynomial which you can guess roots and do long division

[tidal blaze]

Wdym by that?

[normal cipher]

let x = 1+r

[tidal blaze]

Alright

Oh

[hidden hound]

$ \text{Let x = (1+r)} \ 850 = \frac{15}{x} + \frac{15}{x^2}+ \frac{15}{x^3} + \frac{1015}{x^4}$

that font kinda funky so im not sure if i got the numbers right

[tidal blaze]

Yeah you did

[hidden hound]

but yeah thats a visual representation for you ^

[normal cipher]

fr

thats right

[tidal blaze]

Its my colleagues handwriting lol

[hidden hound]

$\text{Let x = (1+r)} \ 850 = \frac{15}{x} + \frac{15}{x^2}+ \frac{15}{x^3} + \frac{1015}{x^4}$

[tidal blaze]

So what can I do next?

[wraith dagger]

pixel

[tidal blaze]

Just solve for x?

[hidden hound]

yeah

then sub back into x = 1 + r

and solve for r

[normal cipher]

$850x^4 -15x^3 - 15x^2 - 15x - 1015 = 0$

[wraith dagger]

Death.cv

[tidal blaze]

Smart mfs

[normal cipher]

first simplify

$170x^4 -3x^3 - 3x^2 - 3x - 203 = 0$

[wraith dagger]

Death.cv

[normal cipher]

then guess roots

wait that might be hard for this qn ngl

[tidal blaze]

U did times x^4

correct?

[hidden hound]

yeah

[normal cipher]

yes

[tidal blaze]

alright just had to make sure

[hidden hound]

honestly lol idk how you'd do it for this funky ahh equation

ugly ass equation i cant lie

[normal cipher]

meh

bad af equation

nah im stumped

you could try using quartic formula but that'd take hours

[hidden hound]

(long division)

thats how id do it

but would take likeeeeee

15m?

if yk what ur doing

[normal cipher]

yeah

[tidal blaze]

What about horner's

Narrow it down to quadratic

then solve x1/x2?

[hidden hound]

honestly idk what that is

but

if it works then it works

go for it

whatever you know

and check ur answer

[tidal blaze]

i barely remember it but it was like scheme for when u had roots of 3 in x things

[normal cipher]

idk what horner's is tbh

go for it ig

[tidal blaze]

nvm i remembered it wrong

it's just scheme for dividing polynomials

[hidden hound]

long division is calling your name

[normal cipher]

fr

[tidal blaze]

how do i use it here?

[normal cipher]

hmmmmmmmmmmmmmmmmmmmm

good question

[hidden hound]

what 'x' value do you sub in

to give you RHS = LHS which means 0 = 0

actually

would long division work here

hmm

🗣️ 🔥 just graph that shit

[tidal blaze]

this shit paralyzed my college class

i mean i got the answer

but professor said no external help like excel all that stuff

i just did =IRR

[normal cipher]

welp

desmos time 🙂

[hidden hound]

College class?

😭 i aint built for that

im in year 11

[tidal blaze]

its on 1 and -1

stale

[normal cipher]

hmm

tf

alr then those are your two solutions 🙂

[tidal blaze]

it doesnt make sense though

when u put 1 or -1 in

it wont equal 0

oh wait its not fully -1 and 1

its -1,041 something

jesus christ

now

it makes sense

only one right answer since rates cant be -

[normal cipher]

ah

[tidal blaze]

1+r = 1,0581

r=0,0581

r=5,8%

[normal cipher]

i guess

was this some compound intrest qn

[tidal blaze]

yea sorta

[normal cipher]

ah

gon do amc 10 tmw wish luck 🙏

[tidal blaze]

i'll forget in 5 mins but right now wishing luck

ty for help @hidden hound @normal cipher

[normal cipher]

ty

np

[tidal blaze]

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[idle crystal]

!status

[cedar kiln]

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

[crimson sedge]

4

[idle crystal]

whats your answer?

[cedar kiln]

@crimson sedge Has your question been resolved?

[mild spear]

Again?

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[tawdry lodge]

Hi, i got given this homework on matrices, can someone help with this one question that i just dont understand

[cedar kiln]

@tawdry lodge Has your question been resolved?

[tawdry lodge]

<@&286206848099549185>

[dusky forge]

Hey how can i help u

[tawdry lodge]

Hi, my teacher gave me this homework, i understand part a) but i just havent been taught how to do part b, do you know what to do?

[dusky forge]

Write the line equation

y=2x+1 in terms of coordinates, like a point
(x,y).

Apply the transformation matrix by substituting
x and y into it.

Calculate the resulting coordinates after the transformation.

Use these new coordinates to determine the equation of the transformed line.

[tawdry lodge]

thank you so much i will try this now

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[grim umbra]

Demonstrate that any n>=1, 3^(n)+2 is not divisible with 13
Can anyone help me with this problem that is solved through leapt step induction?

[cedar kiln]

@grim umbra Has your question been resolved?

[royal kindle]

Ion know about induction but it can be proven using binomial theorem

Or congruence modulo maybe

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[wanton shadow]

Hello I dont get where i'm going wrong here

[median lantern]

p should be >= -9/50

thats what you wrote down on paper

[wanton shadow]

No the negative sign is there its just cropped out of the thing for some reason

you can see a little speck of it

Ahh i got it

its supposed to be greater than or less than

not greater than or equal to

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[proud sable]

if I have an answer of 1134 and an uncertainty of 21,5 (1134 ± 21,5 ) , how should I write it with significative figures?

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@proud sable Has your question been resolved?

[proud sable]

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[green arrow]

Can anyone help me find the Y for B?

[dreamy void]

Can you tell me what AD is, I cannot read it

[green arrow]

y=1_2x - 3

the 1_2 is baso 0.5 ig

[dreamy void]

ok

Is ABCD a rectangle?

[green arrow]

yeah

[dreamy void]

Ok good

That means

BA is perpendicular to 0.5x-3

So its slope will be -1/0.5 = -2

y = -2x + c

[green arrow]

yeah right i got that but then what

[dreamy void]

It goes through the point (6,0)

[green arrow]

uh huh

[dreamy void]

0 = -2(6) + c

solve for c, which is your y-intercept

[green arrow]

ohhh

alright

so it owuld be

3 right?

the y is 3?

[dreamy void]

explain.

[green arrow]

i mean

i do -2 times 6 right

[dreamy void]

yup

[green arrow]

then uh

hold on this part gets me everytine

so like

after that

i would move the c

then like

[dreamy void]

What is -2 times 6

[green arrow]

-3

[dreamy void]

dont play no games

[green arrow]

then uh

yea

if thats -c

[dreamy void]

what is -2 times 6

[green arrow]

OH FUCK

IM STUPID

WAIT

-12..

i thoguht

it was

whatever

yea

thennn

-c = -12

then

c = 12..?

is that it

[dreamy void]

yes

[green arrow]

alr thx

[cedar kiln]

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@sick vessel Has your question been resolved?

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[queen kindle]

sorry i hope you can read my working out, how do i make the values negative like in the question

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[thorn idol]

So i got this equition

To get a i did 3 - 1

so i get 2a = 4, a = 2

but the answer book sayus a = -1/2

does someone knows why

[steel fractal]

ummm…

can you explain how you got 1?

[thorn idol]

So i basically put them all in the formule y=ax^2+bx+c

I did the equitions (-1,3) - (1,1)

so i got

1. 1= a+b+c , 0 = a +b -1

3. 3 = a -b+c, 0 = a - b - 3

so i got

[steel fractal]

what about c?

[thorn idol]

i thought the y would replace the c

[steel fractal]

no no!

you see, c is a seperate entity itself

[thorn idol]

ok so if i want to do (-1,3) - (1,1)

i should have 0 = a + b -1c?

[steel fractal]

0 = a + b + c - 1

[thorn idol]

aha

wait im going to work this out on my paper

[steel fractal]

alright!

[thorn idol]

ok

so i got

0 = 2a + 2c - 4

[steel fractal]

yes

taking the 4 that aide

[thorn idol]

2a + 2c = 4

[steel fractal]

divide by 2

[thorn idol]

a + c = 2

[steel fractal]

yes, call it equation 4

now, substitute the value of b is equation 2

what do you get?

[thorn idol]

i dont understand soryr

so i put the b in the formule -5=25a-5b+c?

[steel fractal]

Yes

[steel fractal]

umm, that’s b, not x

[thorn idol]

true

-5 = 25a +5 + c

0 = 25a + 10 + c

[steel fractal]

yes

Call that equation 5

[thorn idol]

alright

[steel fractal]

Solve 4 and 5

subtract them and find a

[thorn idol]

man you have giga brain

i got a= -1/2

[steel fractal]

haha!

exactly!

did it help?

[thorn idol]

wait

0 = 24a - 12

24a = 12

a = 1/2

[steel fractal]

Nope…

Its 24a + 12 = 0

[thorn idol]

wait i got 0 = a + c -2

0 = 25a + 10 + c

[steel fractal]

Yea…

[thorn idol]

your right i see it

i did 25 -1 and i did -2 -10

but its 25 - 1 and 10 - - 2

thank you very much

i will keep praticing this for today xD

[steel fractal]

Youre always welcome!

[thorn idol]

😄

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[pseudo merlin]

excuse me how would i do part A

[pseudo merlin]

ohh energy released = total binding energy of the two nuclei that make up u-235 - binding energy of u-235

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[hot crag]

yeah don't do that

[woeful spruce]

.close

[woeful spruce]

I couldnt finish writing

Can I still continue?

In this channel

[hot crag]

i'd suggest you don't

[woeful spruce]

Ok

[hot crag]

#help-25

[woeful spruce]

Done

[hot crag]

seen it

[woeful spruce]

Sorry for creating the channel

I mistakenly pressed enter without completing

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[heavy crypt]

Why is this wrong

[cedar kiln]

@heavy crypt Has your question been resolved?

[frank solar]

Because it's not true that $\int f(x) g(x) dx = \int f(x) dx \cdot \int g(x) dx$ (see second row)

[wraith dagger]

cristorenzo99

[heavy crypt]

right

lol thx

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[fallen hemlock]

Hey i really dont get what variables im supposed to use idk if im placing them wrong ill show my diagram

[cedar kiln]

@fallen hemlock Has your question been resolved?

[fallen hemlock]

<@&286206848099549185>

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[fallen hemlock]

<@&286206848099549185>

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[empty bobcat]

Factor the following polynomial and find the zeros : 2x^3+3x^2-18x-27

[dusk goblet]

factor by grouping

factor the first two terms then factor the last two terms

and there will be a common factor

[empty bobcat]

?

soorry factorings not my strongest point in math

ohh so take out x and x(2x^2+3x^2?

[dusk goblet]

$(2x^3+3x^2)+(-18x-27)$

[wraith dagger]

knief

[dusk goblet]

take the gcf of each

[empty bobcat]

x and -9

[dusk goblet]

well x^2 for the first one

but yes

$x^2(2x+3) + (-9)(2x+3)$

[wraith dagger]

knief

[dusk goblet]

do you notice anything we can factor

recall that $ab + cb = (a+c)(b)$

[wraith dagger]

knief

[empty bobcat]

12x^2?

[dusk goblet]

what?

[empty bobcat]

got this far now im confused

[dusk goblet]

2x+3 is common to both x^2 and -9

so we can factor it out

[dusk goblet]

2x+3 serves as our b here

[empty bobcat]

ohohh wait

(x^2-9)(2x+3)?

[dusk goblet]

yep

and you can factor the x^2-9

[empty bobcat]

(x-9)(x-9)?

or +

(x-9)(x+9)?

[dusk goblet]

no

difference of two squares

[dusk goblet]

this is x^2-81

[empty bobcat]

would it be (x-9)(x-9) then?

[dusk goblet]

you just said that

and i said no

[empty bobcat]

i used diffren signs thought \that would cahnge sorry

(x-3)(x+3)

sorry

[dusk goblet]

yes

so your final answer is..

[empty bobcat]

(2x+3)(x-3)(x+3)

how would we find the zeros?

[dusk goblet]

great

[dusk goblet]

set it equal to zero

for a product to be zero at least one of the factors has to be zero

hence set each factor equal to zero

[empty bobcat]

???

[dusk goblet]

what’s confusing

[empty bobcat]

setting each factor to zero

[dusk goblet]

^

suppose p(x) = (x-a)(x-b)(x-c)…

if p(x) = 0

then we know at least one of (x-a), (x-b), (x-c), … is zero

since for a product of numbers to be zero we must have one or more of the numbers being zero

so we solve for all possible ways in which the product is zero by setting each factor equal to zero, since if that factor is zero, then the product is zero

you can’t multiply two or more nonzero numbers and get zero

you can have for instance (0)(3)(1)(8) = 0

or

(0)(2)(0)(5)(1) = 0

so long as there is one factor that is zero, the product will be zero

[empty bobcat]

got it it makse sense now, I have to go also thanks for your help and time 🙂

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[cedar kiln]

@woeful spruce Has your question been resolved?

[woeful spruce]

<@&286206848099549185>

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✅

[woeful spruce]

😭

<@&286206848099549185>

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[crimson sedge]

Someone help me how to do this please

I see the bottom part but i would like a real person

[dire geode]

what confuses you

[crimson sedge]

The whole thing

[dire geode]

which part first

stock?

[crimson sedge]

Ive haven’t done problems like this in a while. I just dont really understand the math process

[dire geode]

do you know the meaning/formulas of all the words in the question

[crimson sedge]

Dont know what dividend means

[crimson sedge]

I just struggle with percents and the order to find out the answer

[crimson sedge]

Actually i think i got it

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[storm seal]

I think it's 1, because the n > 1 applies to the Fibonacci sequence, and not for the theorem I'm try to prove

Is this right?

[kindred gale]

why not try it

when you try to write down proofs you will see what are the components needed in your proof and show that they follow this relationship

[cedar kiln]

@storm seal Has your question been resolved?

[storm seal]

how am i supposed to know if it's the base case

[kindred gale]

you first the the inductive case supposing you have the base case, once you have the induction done, turn back and support it with base cases nedded

[storm seal]

Nevermind I got it

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[slow palm]

how do i reduce

this

[cedar kiln]

@slow palm Has your question been resolved?

[fair geyser]

hints?

too hard to guess

[cedar kiln]

@slow palm Has your question been resolved?

[slow palm]

do you wanna join vs so i can explain?

[slow palm]

or should i show

or try to explain

[fair geyser]

sorry im too shy

[slow palm]

its alr

[fair geyser]

just try to explain in words, i'll guess the rest

[slow palm]

this is an easier example ig

so here

x1 on the left and x1 on the right

can be merged

[fair geyser]

sure

[slow palm]

because they are identical

but the x1 in the middle cant

[slow palm]

you get this right?

[fair geyser]

why do they even have names

i can't merge nodes that are identical except the names?

[slow palm]

the thing in the lecture we have a chinease man who doesnt speak german so he doesnt explain well

[slow palm]

the rules are either merge or delte

[slow palm]

see in the the x1 on the right and on the left both have the property that 0 goes to 1 and 1 goes to 0

[fair geyser]

i don't know what i'm looking at, but it looks pretty optimal

[cedar kiln]

@slow palm Has your question been resolved?

[cedar kiln]

@slow palm Has your question been resolved?

[cedar kiln]

@slow palm Has your question been resolved?

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[cedar kiln]

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[sonic vine]

hello and good day. as you can see this question gave me a migrane and i have no idea how to solve this. if anyone can help, it will be much appreciated

[tranquil gulch]

In triangle ACN, can you calculate CN?

[sonic vine]

like by using TOA?

[tranquil gulch]

no?

[sonic vine]

oh wait i read that wrong-

[tranquil gulch]

You don't know any angles apart from NAC, right?

[sonic vine]

yeah

[tranquil gulch]

Is there another method you can use?

[sonic vine]

do u.. do u use the phytogoreas theorum to find CN??

[tranquil gulch]

Yes.

Pythagorean theorem

[sonic vine]

okok hold on

got 116.6

[tranquil gulch]

,w sqrt(100²+60²)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by bluegaming

[tranquil gulch]

yes

[sonic vine]

OH OK COOL

[tranquil gulch]

Now, in triangle BCN, can you calculate angle BCN?

[sonic vine]

angle BCN.... how do u.. do that..

[tranquil gulch]

SOH CAH TOA

or SOS CAS TOA, or whatever you've been taught

[sonic vine]

OH THE THE MEASURE IS THERE I JUST SAW IT

okok

[tranquil gulch]

👍

[sonic vine]

13.8

degrees

ok thats one question down

[tranquil gulch]

,w arctan(28.8/sqrt(100²+60²)) in degrees

[wraith dagger]

Results provided by WolframAlpha

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

Requested by bluegaming

[tranquil gulch]

Have you tried the other question yourself yet?

[tranquil gulch]

Rounding up to 13.9, this time, because of the 7.

[sonic vine]

....i'm too braindead..

[sonic vine]

O okok

[crimson sedge]

Hello

[sonic vine]

hello

o wait i know what to do now uhm woops

[crimson sedge]

i need help what is (142.134-132.153)

[cedar kiln]

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[empty locust]

in this question, can i write the differential equation as
q" + Aq' = c- bt?

where A = R/L
c = V/L
b = 1/CL
i am not sure about the integral part

[long swan]

yeah this seems like a really convoluted way to write things

i mean why not just write the entire thing in terms of q

L q'' + R q' + 1/C q = V_s

that's how i'd put it

[empty locust]

okay so the question then asks us this

so basically we try to solve the differential equation. we first solve the homogenous equation and then the particular solution right

[long swan]

yeah

just remember that V_s is a constant

[empty locust]

wait so why is it 1/c q ?

its a definite integral right

[long swan]

yeah you're right

the integral actually evaluates to:

q(t)-q(0)

mb

we can't assume q(0) = 0

so final answer:

L q'' + R q' + 1/C (q - q(0)) = 0

does that make more sense?

[empty locust]

yeaaa

so we treat q(0) as a constant?

[long swan]

yes

so really this is just a second order DE

if you can solve:
a y'' + b y' + c y = d

then you can solve this

and the conditions on the constants will result from solving the quadratic equation

let me know if i'm going too fast

don't want to lose you

[empty locust]

so if we substitute e^xt then

ax^2 + bax + c-d = 0

and then x = -ba +- sqrt(b^2a^2-4a(c-d)) right

[long swan]

not quite

[empty locust]

but we dontk now what q(0) is so how can i know how the discriminant behaves

[long swan]

you need to solve the homo equation first

let's back up for a second

[empty locust]

okay yeaaa

okay

so the homo is

ay'' + b y' + c y=0

right

[long swan]

no.

hint: where did a go?

looks good

[empty locust]

yeaa typo

[long swan]

now you make the substitution y = e^(kt)

[empty locust]

so according to the condition we know it should be a complex solution so in form of cosx and sin x right

[long swan]

let's take this one step at a time

[empty locust]

shouldnt this be L q'' + R q' + 1/C (q - q(0)) = V??

[long swan]

yeah mb

you'll have to rearrange to get into the 'standard form'

you're right though, we need the value of q(0)

[empty locust]

okayy so our homogenous equation is

Lq" + Rq' + 1/c q = 0

q" + R/lq' + 1/clq = 0

substituting e^kt

k^2 + R/l k + 1/cL = 0

K = -R/l +- sqrt(R^2/L^2 - 4/cL) so according to the condition K will be complex so can be written as trig form

now trying to find the particular solution

we have

Lq" + Rq' + 1/c q = V - q(0)/c right

i am not sure how to solve that? do we substitute At or somethibg?

[long swan]

not quite

the homogeneous equation looks good

but remember we have 1/C (q - q(0))

so you'll need to bring that factor of 1/C to the other side

it's very easy to make algebraic mistakes here, be careful

[empty locust]

yeaaa sorry

fixed it

[long swan]

what's your favorite letter?

trust me, this is important.

[empty locust]

alpha. should i take V - q(0)/c as alpha?

[long swan]

yes

[empty locust]

okayy. but i kind of forgot how to solve this kind of differential. i know if its in the form of ay" + by' +y = R(X) then our particular solution takes the form of R(X)

but here R is a constant

how do we deal with that? is that equilant to a polynomial?

1 degree or something?

[long swan]

yes exactly

the particular solution is a polynomial

i'll give you another hint

it's a constant polynomial

[empty locust]

so just alpha ?

[long swan]

almost

do the substitution q(t) = beta and try to solve

where beta is just a number

[empty locust]

ohhh

so beta = c alpha

[long swan]

yes

[empty locust]

and we can just leave q(0) the way it is right. and the general solution would just be sum of all of this?

[long swan]

yes

i think they want you to assume q(0) is 0 because otherwise there's no way to actually express it in terms of the constants given

but leave q(0) in your solution for now and see what you get

[empty locust]

got it

thankyouuu so much!!!

you were very helpful!

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[empty locust]

@long swan i am so sorry for pinging you

[empty locust]

but

i tried the question again

so like

k = -R/l +- sqrt(R^2/L^2 - 4/CL)

so simplifying the discriminant i get

(CR^2 - 4L)/C

but we dont know if C is positive or negative

if C is negative then we will still get real solution

[long swan]

let me latex this

$k = \frac{-R \pm \sqrt{R^2 - \frac{4L}{C}}}{2L}$

[wraith dagger]

Mr. Gamer 🇵🇸

[long swan]

and the condition given was $4L > CR^2$

[wraith dagger]

Mr. Gamer 🇵🇸

[long swan]

yeah you're right

[empty locust]

should i just assume C is positive

[long swan]

yeah

that's a fair assumption

[empty locust]

thankyouu

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[rare solstice]

What does this formula is used for?

P/100 = SV/KV -1

[rare solstice]

SV - Discount price
KV - Original price

i mean sell price

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[buoyant latch]

Hi guys I’m in need of assistance

[buoyant latch]

How does one find the particular solution

[raven shard]

undetermined coefficients

[buoyant latch]

What does that mean

[cedar kiln]

@buoyant latch Has your question been resolved?

[raven shard]

short for “method of undetermined coefficients”
here it means assuming (from the forcing term Fsin(gamma t)) that the particular solution is of form Acos(gamma t) + Bsin(gamma t)

(where no additional scaling by t is necessary due to the non-resonant condition gamma neq omega)

[buoyant latch]

Ah

Like this?

[raven shard]

looks about right yeah

[buoyant latch]

Thx

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@eager hemlock Has your question been resolved?

[frank solar]

If I remember correctly, I studied two types of differentiability: Gateaux-differenciability and Frechet-differenciability. With the latter one, it is true that f is continous in (x_0, y_0)

[eager hemlock]

Hmmm i haven't learned that deep😭

[frank solar]

But I studied those things at university, not high school

[eager hemlock]

Sorry maybe your university is good or your teacher is good or you're smarter or smth else idk

I just haven't learned that deep, that's a fact, idk why you can't accept that, it's kinda weird tbh

[eager hemlock]

I just want to ask a simple question here and then I got questioned by haven't learning xxx, lol, this is funny

[cedar kiln]

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[last shoal]

Would a sequence like 8,8,8,8,8... be an arithmetic sequence or a geometric one??

[void sage]

It would be both

[hot crag]

oh yeah actually

good point lmfao

arithmetic because you're adding 0 each time, and geometric because you're multiplying by 1

although this isn't a very typical example is it

[last shoal]

its just a random one

[normal cipher]

both

the common difference is 0 and common ratio is 1

[last shoal]

thankss

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[pearl cloud]

why is this wrong? supposed to be -(1/2) * (1/(x^2+3)^2) + C

[slender ginkgo]

use

power rule

after 1/u^3

[pearl cloud]

so u^-3

why is my method wrong

[slender ginkgo]

u cant just cahnge

the variable of the integration

[pearl cloud]

oh

[slender ginkgo]

${\int u^{-3} du = \frac{u^{-3+1}}{-3 + 1} + C = \frac{u^-2}{-2} + C = -\frac{1}{2} \cdot \frac{1}{u^2} + C = -\frac{1}{2}\cdot\frac{1}{(x^2 + 3)^2} + C}$

[solar agate]

f(x) = x^3

G(x) = | f(|x|) |

Can someone type this out in form of

g(x) = { }
{ }

like when we have

|f(x)|

it means
{ f(x) when f(x) > or equals to 0 }
{ -f(x) when f(x) < or equals to 0 ]

but the absolute value here is for both f(x) and x
please help

[wraith dagger]

k

[pearl cloud]

i was taught this way, idk

[slender ginkgo]

u can do u sub

but if u sub u = x^2 + 3 back

it will be this instead

${\int \frac{1}{(x^2+1)^3} du }$

[wraith dagger]

k

[slender ginkgo]

now u obviously cant integrate f(x) with du

especially when du and dx are linked

[pearl cloud]

so i integrate before i switch variables

always?

[slender ginkgo]

yes

u integrate while keeping the variable matching

[pearl cloud]

okay then

thank you

[cedar kiln]

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[pseudo merlin]

excusem e

[pseudo merlin]

how do di do this

[split pike]

Draw FBD

and show the forces being applied with resp angles

[pseudo merlin]

this is what i bave

@split pike

[nova snow]

BUM CHICKEN

[pseudo merlin]

HI

im crying

my spec exam is on monday

[peak minnow]

probably something to do with resolving the forces into x and y components?

[pseudo merlin]

oh

how..

[peak minnow]

Fx = F sin theta
Fy = F cos theta

[pseudo merlin]

um

wait......

[peak minnow]

isnt this angle wrong tho?

[pseudo merlin]

oh..

[nova snow]

Lol

[pseudo merlin]

wait how

[peak minnow]

F1 and F2 is 150
150+60 ins not equal to 150

ok nvm , i just cant read it

[pseudo merlin]

wut

[peak minnow]

nothing

[nova snow]

Can’t imagine having anymore exams

[pseudo merlin]

i only have 2.

meth and spec

[nova snow]

I finished the HSC

[pseudo merlin]

ugh whatever

[nova snow]

I’m free

forever

[nova snow]

You should be happy for me!

[pseudo merlin]

OK FIEN IM HAPPY I GUESS..

but.. u will be here less often

[nova snow]

yes probably now that I no longer need help!

[pseudo merlin]

yes u will

in uni

ill see u next year

when im doing 12 spec and 12 methdos

[nova snow]

ok buddy

[pseudo merlin]

😈😈

we have grown water beam.

[nova snow]

I actually ate in my physics hsc tho

[pseudo merlin]

i found the y components now wat

[pseudo merlin]

OMG i hate physics

i had my science exam and the physics questions were BS

[nova snow]

I love it

[peak minnow]

ok maybe that may not work , i was thinking something else
ig wait for more help ? ;d

[peak minnow]

wait maybe it works
resolve into x and y components

and that should give 2 equations , solve for F2 and F3

[pseudo merlin]

what

[peak minnow]

if its in equilibrium
Fx (total) = 0
Fy (total) = 0

[pseudo merlin]

how do i get the components of 400N

[peak minnow]

400N is already y component , and it would have no x component

[pseudo merlin]

why

ohb

[pseudo merlin]

ok i have this now

[peak minnow]

yea thats for equilibrium of y components

now do same for x components

[cedar kiln]

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[zinc crypt]

i dont understand why my answer is divergent?

what makes it divergent

[mortal grove]

$\frac{x^2}{x^3+4} \leq \frac{x^2}{x^3} = \frac{1}{x}$

[wraith dagger]

doaby

[mortal grove]

the integral of 1/x diverges from 1 to infinity, so by comparison this integral should too

oh bad argument

[hot crag]

??

yeah lol

[mortal grove]

I always flip the two lol

[hot crag]

i mean you can just do a u-sub?

oh that's exactly what you did lmao

[broken mist]

Not sure where your limit disappeared to but limit of ln(x^2+4) as x->infty is infty

[mortal grove]

i think they want an intuition

[broken mist]

infty - ln(4) is still infty

[zinc crypt]

so if there is a infinity inside my ln it means divergenet

[hot crag]

yeah

[broken mist]

Depends on which infinity

[zinc crypt]

oh just like 1/ infinity is 0

[broken mist]

ln(x) approaches infty as x approaches infty

[hot crag]

well
sure

1/inf isn't a thing--inf isn't a number

[broken mist]

But zero as x approaches negative infty

[hot crag]

1/n approaches 0 as n approaches inf, yes

[zinc crypt]

same thing lol

I get what your saying

[hot crag]

no, those are different actually

[broken mist]

Pedantry aside it's close enough to the same thing to understand

[hot crag]

*shrugs*

[broken mist]

I say that in support of pedantry, I'm not taking a stab at you lol

While they're not the same thing, the intuition that they are doesn't tend to lead to incorrect reasoning at this level

[zinc crypt]

oh ok thank you everyone.

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[zinc crypt]

why is it divergent I dont understand if 1/0 is undefined

[hot crag]

because as x->0, 1/x^8 -> inf

[broken mist]

$\lim_{t\to0} -\frac{1}{7t^7} \to \infty$

If you want it in latex

[wraith dagger]

m. frost

[broken mist]

As t gets smaller, 7t^7 gets smaller too

It's indeterminate at 0 of course because dividing by zero is heresy against Brahmagupta

But as 7t^7 gets smaller, it divides 1 by smaller and smaller numbers, which results in bigger numbers

I guess it would be -infty not infty, but still diverges

[crimson sedge]

help

maximum = translation + amplitude right?

minimum= transaltion - amplitude

amplitude here is B

and translation is a?

[zinc crypt]

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[warped garnet]

Where did my solution go wrong?

[cedar kiln]

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[keen hound]

heresy against Brahmagupta

Holy shit I'm wheezing

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[lament obsidian]

Hello

[lament obsidian]

Can anyone please look into the example on the bottom right ?

why we got backwards to find the determinant of the interchanged row we didn't say -36

it should be -36 right

[cedar kiln]

@lament obsidian Has your question been resolved?

[pastel vault]

nah, the determinant of an upper triangular matrix is just the product of the diagonal entries

to see why, expand the determinant along the first column

(1, 0, 0, 0)^t

reduce to a 3x3 matrix and you multiply by 1

reduce to a 2x2 matrix and you multiply by 3

so 1 * 3 * 21 * 4/7 in total

[lament obsidian]

okay that's good for the last operation

then we can backward

until the very first step

we did interchange

then det(A)=-det(B)

so -36

[pastel vault]

yeah like they didn't swap any rows right or multiply a row by a constant

oh they swapped a row you implied

right, at the very beginning they did ahh

[pastel vault]

cool

[lament obsidian]

yep

[lament obsidian]

okay must be a mistake then

Thanks

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[pastel vault]

yeah it happens that people forget

I did not realise until you pointed it out lol

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[elfin jolt]

can someone help me with the graphing of this parametric curve: x=t^2-4, y=t^3-4t. i have done the domain,limits,derviatvie,double derivative,table of signs, im just stuck at where to start when graphing.

[pastel vault]

what's the domain?

the idea is to sub in t = 0, then plot the (x, y) you get
now you repeat this with t = 1, t = 2 and so on

and then join up all the points with a continuous curve

[elfin jolt]

R, as t belongs to (-infinity,+infinity)

[pastel vault]

well actually you should realise that you have y = tx

[elfin jolt]

wait lemme show u my work so far

[pastel vault]

it's not a straight line cause of the t

but should save you calculations

[cedar kiln]

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@elfin jolt Has your question been resolved?

[fringe oasis]

im sorry, im not far enough into math to be able to help you, but your handwriting is gorgeous

[cedar kiln]

@elfin jolt Has your question been resolved?

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@elfin jolt Has your question been resolved?

[keen hound]

Hi, do you still need help

Also notice that y = t(t²-4)

and x = t²-4

so we have y = tx

and x + 4 = t²

So ±√(x+4) = t

So we have |y| = x√(x+4)

Or y = ±x√(x+4)

@elfin jolt

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[shell radish]

if we're given a curve like the one on the left, and we're told to revolve it around the y axis: are we meant to produce the solid formed by the area under it like on the left? or the solid from the area above the curve on the right?

[shell radish]

because ive seen two different sources seemingly interpret the question in either way, and this makes me paranoid for the test i have to take coming up

[crimson delta]

well I would interpret it as the option on the right

the part between the axis and the function

[vivid radish]

Like this

[cedar kiln]

@shell radish Has your question been resolved?

[shell radish]

so for something like this, the only reason its shaded under the graph and not over it is because the line x=2 is there, which implies that they want us to do (2)^2-x(y)^2?

whereas without x=2, then we would integrate the area over the graph (relative to this view)?

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[agile onyx]

how do i go about doing this? some pointers would be much appreciated

[twilit escarp]

La question est bizarrement posée

[crimson sedge]

Vous pouvez utilisez long division?

Je pense

Mais je ne suis pas certain.

vous pouvez également utiliser la preuve par induction

[twilit escarp]

Ouais une recurrence mais dans le cas où n n'est pas définie faudrait trouver vous même le rang où ça marche bien

[crimson sedge]

Oui

[twilit escarp]

(Induction proof = raisonnement par recurrence =D )

[crimson sedge]

Ahhhh okay lol

I translated it directly :(

sorry im not a native

[twilit escarp]

Thats still impressive !

[crimson sedge]

Merci beaucoup

I actually think of the sentence in English in my head and then jot that down in french

Anyways

Do you think that inductive proofs could work here? @twilit escarp

[twilit escarp]

I do

[crimson sedge]

gotcha

[twilit escarp]

But theres probably a more direct one, that i actually don't see

[agile onyx]

i've been at this for quite a bit and seeing that this exam is only one hour im trying to find a direct way to go about it

(this is not my exam to be clear! this is from like 2021)

[kindred ruin]

cant this be done by applying binomial theorem ??

[crimson sedge]

,w evaluate n(1)^(n + 1) - (n + 1)(1)^n + 1

[wraith dagger]

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[crimson sedge]

guys