help-0

[vale wigeon]

show! your! work!

[dusky chasm]

Can i somehow type my working out

i dont discord on my phone

[vale wigeon]

sure if you can make it readable that way...

[dusky chasm]

ok um actually ill write in symolab and snip it

[vale wigeon]

ok => and -> are inappropriate there

but also

[dusky chasm]

those are like transitions

[vale wigeon]

notice your density isn't equal to x over the entire interval

it's only equal to x on [0,1)

and on [1,2) it is not x but (2-x)

so what you should've had was: $$\mu = \int_0^1 x \cdot x \dd{x} + \int_1^2 x \cdot (2-x) \dd{x}$$

[ocean seal]

Ann (glomed)

[dusky chasm]

is there a reason cause my teacher never taught me that

[vale wigeon]

$\mu = \int_0^2 x f(x) \dd{x}$

[ocean seal]

Ann (glomed)

[vale wigeon]

do you agree or disagree w/ this

[dusky chasm]

yes i agree

[vale wigeon]

$\int_0^2 x f(x) \dd{x} = \int_0^1 x f(x) \dd{x} + \int_1^2 x f(x) \dd{x}$

[ocean seal]

Ann (glomed)

[vale wigeon]

and this?

[dusky chasm]

yes

[vale wigeon]

from 0 to 1, f(x) = x

so the first integral rewrites as $\int_0^1 x \cdot x \dd{x}$

[ocean seal]

Ann (glomed)

[vale wigeon]

agree or disagree?

[dusky chasm]

agree

[vale wigeon]

and from 1 to 2, f(x) = 2-x

so the second integral rewrites as $\int_1^2 x (2-x) \dd{x}$

[ocean seal]

Ann (glomed)

[vale wigeon]

agree or disagree?

[dusky chasm]

agree

[vale wigeon]

yeah ok so like

[vale wigeon]

do you now understand this

[dusky chasm]

it gives me o ne

cause 1/3 plus 2/3 = 1

ur working out makes sense but i dont understand which equation you looked at to get the [0,2]

[dusky chasm]

can you futher explain more as to why that is

sorry im kinda slow

[vale wigeon]

your density function is positive on that interval

its right there

idk how else to explain it except by pointing at it with increasing vigor

[dusky chasm]

ok

thanks

[lone heart]

@dusky chasm Has your question been resolved?

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[stray panther]

whats the radius of an inscribed circle given a square side length 20 m

[stray panther]

do we js half 20?

[echo socket]

Yes

[stray panther]

oh okayy

and for an inscribed circle what do we do?

i js want the rule because then ill know how to solve

[wind cloak]

Drawing a diagram should make it obvious why you're halving the side

[stray panther]

yea but for an inscribed circle do we js add radical 2??

so lets say we have 20 as the side length

then we half it to find for an inscribed circle

we get 10

thwn for the circumscribed would it js be

10 radical 2

wait i may have miced them up

sorry

[wind cloak]

Yes it's the other way round

[vale wigeon]

this kind of attitude is not welcomed here

[stray panther]

what??

i was just saying how all i would need is the rule because i know how to solve it for the inscribed

i didnt mean any rude way

[vale wigeon]

learning rules while specifically throwing away reasoning and understanding is not the way to learn math

[stray panther]

i dont think u get what i mean

im saying

like

i know what everything is so if i could get the rule from someone that would be good so then i could practice this question on my own and see if i get it which i know i would

i didnt mean anything rude haha

[wind cloak]

You've got it down

[stray panther]

i think i get it now

[wind cloak]

But as Ann said

Do you understand why there's a radical 2

[stray panther]

yea yea

[wind cloak]

Alright

[stray panther]

thats just the way i prefer when i learn math i like to have the rule and then do it by myself

thanks so much for the help anyways

have a nice day

i will close this

.close

[lone heart]

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[alpine sable]

why isn't the ans to b (5,7)

in the answer key it says (5,-1)

if the original y was -4 then the transformation would look like: -4*-1= 4 + 3 = 7

[modern sedge]

I get (5, 7) too

[lone heart]

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[willow patio]

$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

[ocean seal]

jpsz

[willow patio]

It looks like you just went ahead and kept the sqrt part @alpine sable, correct?

so $(k+2)^2-4(3k)(4)$

[ocean seal]

jpsz

[lone heart]

@alpine sable Has your question been resolved?

[willow patio]

Hmmm, I see your point

Sure, but I got the same answer as you. I'm just as confused as you

I got $22 \pm 4\sqrt{30}$ as well.

[ocean seal]

jpsz

[willow patio]

Is there any chance that those answers are to a different problem?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[barren elbow]

from the mark scheme

why can z not be 0

[minor needle]

because then 3 = 0

which is false

[subtle birch]

try putting z=0 and check if the value is zero

[minor needle]

(aka z = 0 isn't a solution)

[barren elbow]

ahh

thx

.close

[lone heart]

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[lone heart]

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[empty meadow]

Quick question, if there term isn’t x^n and like the greens i highlighted does that change anything while doing the ratio test

[clever nimbus]

you would do the ratio test as usual

if the term is something like 2n, you would still substitute n+1 and it would become 2(n+1) = 2n + 2

for example, in part b), x^2n would be x^(2n + 2)

[empty meadow]

But while doing the ratio test normally don’t we just omit the x^n term?

[clever nimbus]

wdym by that

[empty meadow]

Let me show

The videos i watched do it like this

I know the second one isn’t the ratio test

But they just use an

[clever nimbus]

an/an+1 means you take the equation and you divide it by the equation where n is replaced with n+1

[empty meadow]

Yes i know

But why she is not including the (x-2)^n part?

[clever nimbus]

what do you mean

[empty meadow]

Whoop

Nvm

Got it thanks

[lone heart]

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[safe tartan]

How would you change the bounds if you let u=e^x ? Thanks

[safe tartan]

.rotate

Umm

Six

Soz

[ebon sparrow]

,rotate

[reef grove]

do you know this integral

$\int \frac{1}{1+x^2}dx$

[ocean seal]

tobi.

[safe tartan]

Tan

Identity right

[wild trail]

let e^x = t

[reef grove]

arctan

[safe tartan]

Oh right

Yep

I thought to use that you would need to let u be e^x and change the bounds

I may be wrong though.

[reef grove]

you have to substitute

[safe tartan]

Could you elaborate pls

[reef grove]

you want your integral in this form

[safe tartan]

Yep

[reef grove]

because you know what that is equal to

[fossil latch]

Basically you referred to as if you ghave to substitute u=e^x
Yes, ultimately

Also damn, they teach inverse trigonometric function in korean highschools now???

[safe tartan]

…

[reef grove]

do you know how to substitute

[safe tartan]

I know that you change dx in terms of the new one and that you need to change the bounds

But idk how to change the bounds

[reef grove]

you dont have to change bounds you can first solve the Indefinite integral

but I can show you what changing bounds mean wait a sec

[fossil latch]

u=e^x -> plug in the lower and upper limit for x

[reef grove]

yeah

[safe tartan]

So e^x = 0 and ln 3?

[fossil latch]

sorry, I was unclear
Substitute x to 0 and 1/2 ln3 and then find the values of u

[safe tartan]

Oh so it becomes 1 and whatever that it

[fossil latch]

Sqrt 3, yes

[safe tartan]

Ah that makes sense

I’ll get back to you I’ll just try this now

How do you do 1/1+u^2 integral again

Oh wait that’s the arctan

Ok I think I got it thanks guys!

.close

[lone heart]

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[alpine sable]

I was trying to solve this problem

I was able to show that P(1) holds true

I was trying to show that P(k+1) holds true whenever P(k) holds true

Idk if I made a mistake somewhere, but my expression doesn’t cancel out properly

[ocean hawk]

show your work

[alpine sable]

One sec

It’s uploading

It’s really messy so please let me know if I need to write it down properly

[ocean hawk]

I think you're overcomplicating things

[clever nimbus]

induction?

[ocean hawk]

actually you've made a mistake it seems

[alpine sable]

Is there a simpler way?

[ocean hawk]

looks like you're summing the terms, not multiplying

[alpine sable]

Where?

Oh ya

[ocean hawk]

it's very straightforward once you fix that, you'll see

[alpine sable]

Sorry I thought it was summation 😭

Thank you so much

.close

[lone heart]

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[alpine sable]

Here I was thinking I missed the plus sign so many times 💀

[ocean hawk]

no worries

[lone heart]

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[vale wigeon]

progress?

[junior gorge]

That’s the question

[vale wigeon]

yes, i see the question just fine.

i'm asking you how much progress you have made.

[junior gorge]

I solved it but I think there’s something wrong

[vale wigeon]

show your work.

then we can look at it and tell you if there is anything wrong.

you might also want to tell us why you think there is something wrong.

[vale wigeon]

you say you solved the problem. does this mean you have an answer?

because i don't see any answer

this looks unfinished

[junior gorge]

It wants me to do this x^2+y^2-2xy

[vale wigeon]

what's "it"?

also you have not answered my yes/no question.

do you or do you not have an answer?

[junior gorge]

Not really but the way it was solved doesn’t make sense to me, I just wanna make sure it’s correct

[vale wigeon]

... what you are saying is confusing me.

so you wrote this, but also now say it doesn't make sense to you.

[junior gorge]

Let’s start from here

Is this mathematically correct?

[vale wigeon]

as opposed to what other notion of correctness?

that statement is correct, yes.

[junior gorge]

How?

[vale wigeon]

as in "what's going on??" or "i think it should be false"?

[junior gorge]

So this is also correct

[vale wigeon]

yeah sure is

do you think otherwise?

[junior gorge]

I think it shouldn’t be right

[vale wigeon]

why do you think it shouldn't be right?

[junior gorge]

Because of the minus

[vale wigeon]

yeah and?

a + (-b) + c = a + c + (-b)

[junior gorge]

Wait wait wait

[vale wigeon]

x^2 + (-2xy) + y^2 = x^2 + y^2 + (-2xy)

x + 1/x - 2 = x - 2 + 1/x

[junior gorge]

Oh I see it now

I appreciate it lol

.close

[lone heart]

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[lilac jolt]

$\frac{x-3}{2} = \frac{x+1}{3}$

[ocean seal]

Sterling

[lilac jolt]

I got bored and stumbled upon this equation that I answered on my 1st semester exam

[stone oxide]

okay

[lilac jolt]

I forgot how to solve it

[stone oxide]

thats fine

[vale wigeon]

do you know how to solve linear equations in general

and/or do you know how to deal with fractions in equations in general

[stone oxide]

the first step for anything like this is to find the LCD

and multiply both sides by it

[lilac jolt]

6

[stone oxide]

so that there are no fractions

yea

multiply

[lilac jolt]

what do I do with the denominator 2 and 3?

[lilac jolt]

nope

[stone oxide]

i see

so, a fundamental idea to solving these is that

[vale wigeon]

$6 \cdot \frac{x-3}{2} = 6 \cdot \frac{x+1}{3}$

[ocean seal]

Ann (glomed)

[stone oxide]

whateverr you do to on eside u do to the other

[vale wigeon]

if you do not know immediately what to do, then you should only do what you set out to do, and write down the result

and look at the equation from there

[stone oxide]

soin order to remove the denominators

we multiply both sides by 6

that way, 6 and 2 cancel

and 6 and 3 cancel

does that make sense?

also thanks ann for the visual

[lilac jolt]

How do they cancel

6 divided by 3?

[stone oxide]

yes

lcd*

does that make sense sterling?

least common denominator

but ig lcm works too

both work

lets focus on the problem

ig lcm is a better way to put it tho

but @lilac jolt

making sense?

[lilac jolt]

ye

[stone oxide]

good

so what does that leave us with

[lilac jolt]

I got an idea and tried solving it but still got the wrong answer

sorry for the late reply

[stone oxide]

its fine

yea

what does that leave us with ster

[lilac jolt]

so back in multiplying 6 in both sides

[stone oxide]

ye

[lilac jolt]

I got $6x-18+3=6x+6+2$

[ocean seal]

Sterling

[stone oxide]

not quite

[lilac jolt]

$=6x-15=6x+8$

[ocean seal]

Sterling

[stone oxide]

no

[lilac jolt]

how

[stone oxide]

so

her

e

yea

[lilac jolt]

where did 1/3 come from

[stone oxide]

$6 \cdot \frac{x-3}{2} = 6 \cdot \frac{x+1}{3}$

[ocean seal]

Fishy

[stone oxide]

see

the left becomes:

3*(x-3)

and the right:

2*(x+1

)

[lilac jolt]

I get it

[stone oxide]

same

[lilac jolt]

Add a dollar sign at the end

[stone oxide]

i just copied from what ann did

[ocean seal]

Idris20h

[stone oxide]

put parentheses around x-3 and x+1

[ocean seal]

Idris20h

[stone oxide]

yea

[lilac jolt]

it's pretty easy to learn the bot. just understand one input from a user and you're good to go

[stone oxide]

yea

[lilac jolt]

Okay I got the final answer

$4=4$

[ocean seal]

Sterling

[stone oxide]

no

not quite

because we are solving for x

[lilac jolt]

x=11

I substituted 11 to all the x's sorry

To check if it's correct

can I get another problem like this and try to solve it

[stone oxide]

okay

here

(x+4)/2 = (x-3)/6

might be a decimal

[ocean seal]

Idris20h

[stone oxide]

okay

but i think your problem is solved

bye

[lilac jolt]

$x=-\frac{7}{2}$

[ocean seal]

Sterling

[lilac jolt]

?

no nvm

wait

[lone heart]

@lilac jolt Has your question been resolved?

[lone heart]

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[proud mortar]

Does anyone can teach me this one ? I think I went something wrong....

[gray isle]

multiple things went wrong

using alpha as one of the roots,
the other root is alpha/2, not beta/2

you could also use the pair:
alpha, 2alpha which is probably less tedious

also the sum of roots doesn't have x in it

the -b/a is only the -(-(p+1))/1 = p+1

[proud mortar]

So my SOR is alpha + alpha/2. Then, POR is alpha × alpha/2 ?

[gray isle]

yes

[proud mortar]

Okay

[proud mortar]

Any suggestions for finding value of p ?

[gray isle]

show your updated equations/work

[proud mortar]

why duhh... -(-(p+1))/1 = p+1 ?

[gray isle]

what's your b here

[proud mortar]

-(p+1)

[gray isle]

yes, and what's a

[proud mortar]

oh is compare betweem the form a quadratic equation and original equation right ?

[gray isle]

a,b,c in the sum/prod of roots refer to the respective coefficients and constant of the equation in general form

[proud mortar]

why is = p + 1 ? not = -(p+1) ?

[gray isle]

sum if -b/a, not just b/a

[proud mortar]

my origin b is -(p+1) menawhile my sum of roof is p-1. By comparing, why not is -(p+1) = p-1 ?

[gray isle]

no

where's sum of p-1 coming from

b =

-(p+1)
what's a?

[proud mortar]

Oh wait I copy wrong thing sssssh I apologize...

so is p + 1 = -(p+1) ?

[gray isle]

no

[proud mortar]

SOR = b ?

[gray isle]

no

SOR = -b/a

[proud mortar]

then whats the point of alpha that part ?

[gray isle]

those represent the roots themselves

[proud mortar]

both are p + 1 = p + 1 ?

[gray isle]

-b/a simplifies to p+1

[proud mortar]

Yea

[gray isle]

that's all i was saying

as opposed to the px+x that you had

so from that your equation from the sum of roots is
alpha + alpha/2 = p + 1

[proud mortar]

So I should form x² - (sor) x + (por) = 0 to compare the equation given to find the p right?

[gray isle]

yes

in summary
$$\alpha + \frac{\alpha}{2} = p + 1$$
$$\frac{\alpha^2}{2} = 8$$

[ocean seal]

ℝamonov

[proud mortar]

wait i can find the value of alpha right ? after that i can sub into the sum of root there

[gray isle]

yes

[proud mortar]

Should I keep going ? Or went something wrong too ? Because I found that x has many decimal places

[gray isle]

you didn't distribute the -2 on the left side properly

and you also shouldn't turn that > into an =

[proud mortar]

how do i find x ?

[muted fable]

first change mistake u made w the -2

[gray isle]

fix the above mistakes first

[proud mortar]

aight

[muted fable]

and remember when u multiply or divide inequalities, u have to flip the sign

so whats the quadratic u get

[proud mortar]

Is that correct?

[muted fable]

u can graph the equations on desmos to check if theyre correct

also no

[muted fable]

@proud mortar

[gray isle]

horrid notation at the end

not quite sure what your intended solution is

and/or if its even absent

[proud mortar]

Nvm I have verify my ans with geogebra it looks like correctly

[gray isle]

x_1 = 1.5 is a root, that value is greater than 0
but that tells me nothing about the range of values of x
same for the -4

[proud mortar]

Owhh

Should I do the sor/por that thing?

[gray isle]

discriminant more efficient

[proud mortar]

b²-4ac ?

[lone heart]

@proud mortar Has your question been resolved?

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[strange imp]

Need help

[lean plover]

.close

[lone heart]

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[strange imp]

How did they turn the acceleration into v·dv/dx?

[vapid shuttle]

this channel is occupied and now closed, read #❓how-to-get-help to figure out how to post your question

[lone heart]

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[alpine sable]

hello

[lone heart]

Please don't occupy multiple help channels.

[rustic coral]

#help-11

.close

[lone heart]

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[alpine sable]

.reopen

[lone heart]

✅

[alpine sable]

anyone can help me?

[rustic coral]

You already have an open help channel, why are you opening another one?

Anyway, you’re being answered in the initial channel, so there’s no reason for this one to be open

.close

[lone heart]

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[steady basin]

need some help on part a plz

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[steady basin]

1

.close

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[limber depot]

Can anyone explain to me the Nth term, I know its stupid, I get it but 2 hours later I forget. The videos on youtube are great I just cant remember. Please 🙏

[limber depot]

prolly a stupid question at this point xD

[fiery storm]

its basically the pattern a sequence follows

[lone heart]

@limber depot Has your question been resolved?

[lone heart]

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[manic hearth]

i want help please

[lone heart]

Please don't occupy multiple help channels.

[manic hearth]

.close

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[lone heart]

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[elder crane]

struggling to understand what happens here

where does the limit part go

[raven rover]

That limit is 1

[elder crane]

e^0 = 1

1 - 1 = 0

0/0

how do you get 1

[elder crane]

can you explain why

[raven rover]

I don’t know how you get 1 to be honest

Maybe a graph of the function?

I don’t know how you’d do it algebraically

Taylor Series?

[elder crane]

I'm in my first week of calc so haven't gotten there yet

[raven rover]

Ah ok

I think you can use the definition of e to do that limit

[elder crane]

this is just tripping me up though

[raven rover]

Google it, this pops up

[elder crane]

i mean i know the easier method

[raven rover]

https://socratic.org/questions/given-e-h-1-h-how-do-you-find-the-limit-as-h-approaches-0

[elder crane]

i just want to understand how we get there through the limit definition

I don't really understand the answers in that link, I guess I'll just disregard it for now

[raven rover]

I don’t think it’s super important unless your teacher says so

[elder crane]

it's not, at the very least I know it equals 1 which completes the derivative.

.close

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[nocturne iris]

i dont know how to solve

[nocturne iris]

cos(1/x)>0

[deft lagoon]

is there a restriction like [0,2pi]?

[nocturne iris]

no

[deft lagoon]

ok

[nocturne iris]

but x should be negative actually

[deft lagoon]

wait what? why?

[nocturne iris]

yeah just [0, +inf[

[deft lagoon]

uh

Well then

[nocturne iris]

f:R+ ---> R

[deft lagoon]

when is cos=0?

(within 0, 2pi)

[nocturne iris]

when x=pi/2

[deft lagoon]

... and?

3pi/2

but we can just say that x in cos(x)=0 is equal to: x=pi/2+pi*n

where n is all integers

helloo?

[nocturne iris]

yes

[deft lagoon]

ok

so

in this equation, the x in the previous example is now 1/x

therefore 1/x=pi/2+pi*n, and n is all integers

[nocturne iris]

so it's the inverse of that

[deft lagoon]

yep!

[nocturne iris]

and if we want it to be bigger?

it's bigger than the inverse?

[deft lagoon]

what do you mean

[nocturne iris]

x>(1/(pi/2 + npi)

[deft lagoon]

oh

then we need to (kinda) go back to the beginning

[nocturne iris]

yeah because it's periodic so that's no obvious

i mean i can't just say this

[deft lagoon]

yeah

so

cos(x) is bigger than 0 when -pi/2<x<pi/2 and it repeats every pi

kinda

[nocturne iris]

ok so i take the inverse

[deft lagoon]

yeah

[nocturne iris]

so it is

-1/(pi/2+pin)> 1/x >1/(pi/2+pin)?

@deft lagoon

[deft lagoon]

hmm

I think so?

wait

in the middle it should be x, not 1/x i think

[nocturne iris]

why?

[deft lagoon]

well

[nocturne iris]

here you said cos(x)>0

[deft lagoon]

yes

that was an example

[nocturne iris]

but now it's 1/x

[deft lagoon]

so in that example, your thing would work

cause you just substituted things in the equality

[nocturne iris]

yes

[deft lagoon]

bur for your question, you need to do that, but add the extra step which is the 1/x

so you would get -1/(pi/2+pin)> 1/(1/x) >1/(pi/2+pin)

which simplifies in the middle to being
-1/(pi/2+pin)> x >1/(pi/2+pin)

also you reversed the signs of inequality

final answer should be

[ocean seal]

TheWiseDragon

[nocturne iris]

but is it never the inverse with the bounds?

[deft lagoon]

huh

you just did the inverse on both sides of the equation

so then you can ignore the inverse

[nocturne iris]

i mean when i did the inverse and do the inverse once again; i need to do the same with every side?

[deft lagoon]

no

because you were just simplifying 1/(1/x)

so simplification does not require you to do the operations on the other sides

[nocturne iris]

wait i dont understand

if we go from this

-1/(pi/2+pin)> 1/x >1/(pi/2+pin)

then what?

[deft lagoon]

ok

so what you just got was for cos(x)

but what you have is cos(1/x)

so you take the -1/(pi/2+pin)< 1/x <1/(pi/2+pin) and replace the x in there with 1/x

to make ti work with cos(1/x)

[nocturne iris]

so it's not the same x

we should write it diffrently then

[deft lagoon]

it is and isn't

you're basically just taking the same base equation but replacing the variable inside

it would be the same if it was cos(5x) instead

then you'd get -1/(pi/2+pin)< 1/5x <1/(pi/2+pin)

[nocturne iris]

i mean, if i want to change something into something else, i cannot say x=5x, i need to call the first x differently like Y for example because algebra just dont work like that?

[deft lagoon]

yeah

[nocturne iris]

so i prefer to say 1/x = Y

[deft lagoon]

ok

do that then

[nocturne iris]

-1/(pi/2+pin)< Y <1/(pi/2+pin)

[deft lagoon]

yes

wait no

pi/2+pin < Y < pi/2+pin

[nocturne iris]

if Y=1/x i mean

[deft lagoon]

oh yeah then sure

[nocturne iris]

so x=1/Y

[deft lagoon]

yes

[nocturne iris]

and then what?

[deft lagoon]

well that's basically it

you just need to put the equatino together

-1/(pi/2+pin)< x <1/(pi/2+pin)

and voila

[nocturne iris]

i still don't get it

[nocturne iris]

why don't we take the inverse

[deft lagoon]

bruh

[nocturne iris]

if x=1/Y

[deft lagoon]

ok

-1/(pi/2+pin)< 1/y <1/(pi/2+pin)

[nocturne iris]

yeah

[deft lagoon]

(new small y)

so -1/(pi/2+pin)< 1/(1/x) <1/(pi/2+pin)

[nocturne iris]

yes ok

soooo

if we do everything from the start

cos(1/x)>0 is what we need to solve

cos(x)>0 when -(pi/2+pin)> x >(pi/2+pin)

and if we take y=1/x

so we want to solve cos(y)>0

cos(y)>0 when -(pi/2+pin)> y >(pi/2+pin)

so x=1/y=1/(1/x)

[deft lagoon]

yes!

[deft lagoon]

I mean you still need to reverse both sides because of this but yea

[nocturne iris]

and then we have the result

ok

thanks

.close

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[hazy moat]

If you know the volume of a helium balloon, how do you calculate how much weight needs to be attached to the balloon for the balloon to not rise or fall? I have taken physics 11 and precalc 12.

[prime badge]

weight of air of this volume minus weight of balloon

so you're missing one value

how much the shell weighs

also the volume of the thing you;re attaching but it's like tiny

[lone heart]

@hazy moat Has your question been resolved?

[hazy moat]

Ig I should have asked how to calculate the weight of the helium and how quickly it will rise in the air around it

[prime badge]

well you can't, because the rubber weighs it down and you don;t even know how thick it is

[hazy moat]

Not calculating the weight of the balloon, bcs I can do that once I understand the question

[prime badge]

any object in air is pushed up by the air, with force equal to density of air multiplied by object's volume
balloons are "big" so this force is large

if it's larger than weight of the baloon, the net force is upwards

[hazy moat]

Is there an equation for this type of question?

[hazy moat]

.close

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[azure barn]

if the limit approaches 6 from the left side, is the answer "infinity" or "dne/unbounded"???

[hushed locust]

it would be infinity if and only if both directional limits approached infinity

if the directiional limits don't match then the limit just doesn't exist

[azure barn]

the answer's infinity in tht case?

what's the difference between infinity and unbounded? they look the exact same

[hushed locust]

infinite limits and unbounded limits are essentially synonymous although unbounded limits can go to either +∞ or -∞. technically a limit that goes to infinity doesn't exist but it's more useful to say it goes to infinity than just saying it doesn't exist

[azure barn]

unbounded limits r infinite limits?

[hushed locust]

they mean the same thing

[azure barn]

so if i put "∞" or "unbounded / dne", they would both be correct?

[hushed locust]

for the directional limit on the left side both would be correct although usually it's preferred to say it goes to ∞ since that's more descriptive. For the overall limit since the directional limits are different the only thing you can say is that it doesn't exist

[azure barn]

ohh

so it doesn't exist cuz if it approaches js 6, but from both sides, it would be "dne" since the answers r different?

[hushed locust]

yes. a limit can only exist if both directional limits approach the same value

[azure barn]

okok, tysm<33

.close

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[obsidian shell]

Just a quick question, how do you find the gradient between two parabolas, like what formula do i sub the points into

[waxen flame]

Between two parabolas or two points on a parabola?

[obsidian shell]

two parabolas sorry for the bad wording

[waxen flame]

What class is this for?

[obsidian shell]

math

[waxen flame]

What type of math class?

[obsidian shell]

math methods

[steady basin]

to find the gradient of a parabola isnt it just the gradient of the tangent?

[waxen flame]

OP is asking about the gradient between two parabolas which has me confused.

[cursive badger]

^

[steady basin]

yh ive never done this before

[cursive badger]

are you asking for the gradient between the points on two different parametricized parabolas or something?

ie you have two parabolas v(t) and q(s) and you want the gradient of the line of the points from v to q at a given (t,s)?

[waxen flame]

Better yet, can you post an image of the math problem you are working on?

[obsidian shell]

like this the gradient between the red and green one

like where they connect

[cursive badger]

you mean the slope where they connect?

[waxen flame]

Looks like you are working on piecewise functions.

[cursive badger]

let red be r(x), green be g(x), solve for x in r(x)-g(x)=0 (to find the intersection), then plug that x into r' or g' and they should be the same, assuming they both meet and have the same derivative at that given x coord

[obsidian shell]

yeah

[cursive badger]

is there a reason you were confused?

[obsidian shell]

but how do i find the slope at where they connect between the two

[cursive badger]

given they intersect at some x_0

then find r'(x_0) or g'(x_0)

do you know how to take derivatives?

[obsidian shell]

no i dont we are covering derivatives next term

[cursive badger]

ohh ok I see

do you know the (f(x+h)-f(x))/((x+h)-h) version of finding a derivative? as h->0?

[obsidian shell]

yes

[cursive badger]

you can just use that

in this case use x = x_0 the intersection point

[obsidian shell]

oh ok so what values do i sub them for

[cursive badger]

it's a polynomial so it'll cancel out properly

[obsidian shell]

yeah

[cursive badger]

if they share the same derivative at the point then it doesn't matter which of the two you use for f

but x has to be the intersection x, obviously

[obsidian shell]

i just needed to do that and my math teacher said ill probably get a 20 out of 20 in the assessment pice but i have been stuck on what she had meant by it for ages

[lone heart]

@obsidian shell Has your question been resolved?

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[dapper lynx]

Quick question

[dapper lynx]

When dealing with velocity and acceleration and such

We use v for velocity and a for acceleration

What do we use for speed? s?

[rustic coral]

Speed is the magnitude of velocity, so you can represent it by $|v|$

[ocean seal]

Civil Service Pigeon

[dapper lynx]

Fair enough!

Thanks 🙂

.close

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[copper delta]

i believe i got two formulas while doing this earlier but i ended up with a quartic

[ocean seal]

[code{RED}]

[copper delta]

$a^2 = \frac{b^2}{b^2-1}$

[ocean seal]

[code{RED}]

[copper delta]

$a^2 +b^2 = (5-a-b)^2$

[ocean seal]

[code{RED}]

[rustic coral]

what are a and b defined as

[tacit arch]

Gotta read bottom up

[rustic coral]

oh wait new messages

we love slow internet

bonk

[copper delta]

oh my bad

i have a triangle

[tacit arch]

I only see where the last equation comes from, not the previous two

[copper delta]

inverse pythagorean theorem

i saw it in a 3b1b vid

[tacit arch]

Well TIL inverse Pythagorean theorem

[fallen verge]

Why not ask in MODS?

[copper delta]

true

i didn’t know i could ask there

[fallen verge]

Anyway

I think equal areas might work better

[tacit arch]

But you have two equations and two unknowns so you can solve?

[copper delta]

yes but i got a quartic

[fallen verge]

ab/2=(5-a-b)/2

[tacit arch]

Oh right I see that now

[fallen verge]

ab+a+b=5

Use sfft

[copper delta]

i asked this question earlier

$96b^4 -460b^3 +425b^2 +500b-625 = 0$

[ocean seal]

[code{RED}]

[copper delta]

got this

[copper delta]

today i learned?

[fallen verge]

Do we assume integer?

[tacit arch]

,w solve
96b^4 -460b^3 +425b^2 +500b-625 = 0

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by riemann#2529

[fallen verge]

Oof rrt-able

[tacit arch]

Ah yes obviously

[copper delta]

i don’t see a reason to

[tacit arch]

,w factor
96b^4 -460b^3 +425b^2 +500b-625

[ocean seal]

Results provided by WolframAlpha

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

Requested by riemann#2529

[tacit arch]

Ah yes so obvious

[copper delta]

have i made a mistake?

[tacit arch]

Is the polynomial in a any better

[copper delta]

surely it must be the same

because of the symmetry

[fallen verge]

Yeah its symmetric

[tacit arch]

Oh right

[fallen verge]

Tbf you know if it works by rrt, b must have a 5 in the numerator

[copper delta]

i’ll go through the equations again

[fallen verge]

Not "works by rrt", has a rational root

[copper delta]

you can feed wolfram multiple equations right

,w 1=1/a^2 + 1/b^2,a^2 + b^2 = (5-a-b)^2

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by [code{RED}]#8104

[tacit arch]

What's sfft

[fallen verge]

,w sfft

[ocean seal]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by GarlicBredFries#2110

[tacit arch]

This looks promising

[copper delta]

what’s rrt

[fallen verge]

Oops

[copper delta]

oh is there another equation

[fallen verge]

Sfft = simons favorite factoring trick
Rrt = rational root theorem

[copper delta]

oh shit

,w ab=(5-a-b),a^2 + b^2 = (5-a-b)^2

[ocean seal]

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Requested by [code{RED}]#8104

[copper delta]

that makes a whole lot more sense

rather than me pulling equations out of my ass

equal areas

i gotta remember that

thanks guys

even if i don’t understand half the acronyms

.close

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[rotund shoal]

Need some help on where to start

Do I need to use f(2) and f(3) to find f(x)?

[solemn juniper]

you won't be able to find f(x) for general x, you just have to use the values given

[rotund shoal]

Thats what confused me

Without f(x) how can I find f of g(3)

g(3)=2

[solemn juniper]

and f(2) = ?

[rotund shoal]

5

[solemn juniper]

there you go

https://tenor.com/view/shrimp-as-that-clash-royale-hee-hee-hee-haw-gif-25054781

[rotund shoal]

I still dont understand that part

Why are we finding f(2)

How does that help with the f? Are we trying to find the same x value for both f and g?

[median oar]

f takes some value as input, in this case it takes g(3) as input

Well we know g(3) is 2

So then f(g(3)) = f(2)

[rotund shoal]

Hmm thats a little confusing but I see where thats coming from

so g(3) = 2

[median oar]

Yeah functions are a bit wack when you first learn them

[rotund shoal]

are we pretty just replacing the g(3) with 2 within the f

so f(2)

thus giving 5?

[median oar]

This mean wherever we see g(3) we can write 2

And vice versa as well, wherever we see a 2 we can replace it with g(3)

[median oar]

Exactly

[rotund shoal]

Ahh gotcha, that makes sense

[median oar]

The = sign tells us we can replace one with the other at any point in time

[rotund shoal]

Righto, thanks for the help

that clears things up

.close

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[supple laurel]

Weird question

[supple laurel]

I have a number let's say 1.34

is there a way to convert it to 1 only using +-*/

[crisp cargo]

multiply by 100, subtract 133

[neon scarab]

If you are on some programming language, their math library should have some math-round function i guess(?)

[supple laurel]

the thing is

the number varies

[crisp cargo]

multiply by how many decimal places u have

[copper delta]

multiply by 0

add 1

[crisp cargo]

true

[supple laurel]

then I just type 1

it's a stupid question to a poor developed thing

I just had to ask to be sure

[crisp cargo]

what is your goal for the question

[copper delta]

i don’t understand the premise of your question

do you want a function that always outputs 1

[supple laurel]

let me give some context

I am configuring a minecraft plugin where I have to make an operation on how much points you get based on the level you are at

income-progression-equation: baseincome * (1 + 0.5 * joblevel // 5)

baseicome es the default amount of points you get

and joblevel is the level you are at

I need to make so every 5 levels baseicome gets multiplied by 1.05 (or added a 5%)

[copper delta]

so 0.05

otherwise that looks good to me

[supple laurel]

I can't use // to get an int

that's mi issue

I can only use */+-%

[wary stream]

What's // in code mean?

[copper delta]

floor divide

round down

i see

you have % though?

[wary stream]

What programing langauge is that?

[copper delta]

it’s psuedocode for rn

i think

[supple laurel]

some weird limited math

[wary stream]

Well, whatever coding language, you're using, type casting will truncate numbers, like

double d = 10.123
int i = (int) d \\rounds to 10

[supple laurel]

I can't

[supple laurel]

.

im limited by that

3rd grade math operations

[neon scarab]

In your language, what is the result of (10.123 / 1) then?

[supple laurel]

there's no language

[copper delta]

yeah

so

[supple laurel]

it's kinda difficult to explain

dw

[copper delta]

baseincome * (1+0.05 * (joblevel/5 - ((joblevel/5)%1))

or if it were a programming language you could cast to int

but we have established that you can’t do that

[supple laurel]

I wish I could do that

the code is closed source and I really dont want to deobfuscate it

[copper delta]

is my solution wrong?

[supple laurel]

let me check it

IT WORKS

you saved me a lot of time

really appreciate it

[lone heart]

@supple laurel Has your question been resolved?

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[twilit cypress]

I have no idea what I am doing here and how to go about proving q24. Anyone willing to help an idiot see the light?

I just know the definition of floor is "n <= x < n+1", and that is supposed to help me somehow lol

[copper delta]

let x = a + b where a is an integer and 0<b<1

then floor(x) = a

[twilit cypress]

okay, I am with you so far

[copper delta]

and floor(-x) = floor(-a-b) = floor(-a-1+1-b)

the -1+1 will be useful later

since 0<b<1, 0<1-b<1 must also be true

do u want me to show why

i’ll do it anyway

so 0<b<1
x-1
0>-b>-1
-1<-b<0
0<1-b<1

anyways because of this,

floor(-x) = -a-1

[fossil latch]

We can see the -a-1 as an "interger part" of -x and 1-b as the "non-interger" part of -x
It will be natural then floor(-x)= -a-1

[neon scarab]

I think you can just go for this and do the algebra on left-hand-side to try to reach to m-1 as the end-result.

[copper delta]

then floor(x) + floor(-x) = a-a-1 = -1

please stop cluttering my nice proof 😡

[twilit cypress]

sorry for my slow replies, I am just trying to parse my way through this still. Going right to the top, how did b get re-introduced here?

[copper delta]

what do you mean

i defined x = a+b

[twilit cypress]

floor(x) = a makes total sense since its just an integer, but how do we then get from there to floor(-x) = floor(-a-b)?

[fossil latch]

x= a+b

[copper delta]

yes

we haven’t evaluated the floor at that point

[twilit cypress]

oh hole up, this wasnt a step, just an observation, eh

[copper delta]

yes

[fossil latch]

Where a is the "interger part" and b is the "non-interger part" of x

[copper delta]

yes

[fossil latch]

Not to be confused with -x.

[twilit cypress]

Im slow af (forgive me, I am learning lol), but I just parsed my way through to here

[copper delta]

okay

it is fine

you can take your time

[twilit cypress]

so up to that point, everything is making sense to me, but this isnt the final answer right?

since we need to show that its m-1?

[copper delta]

what is m-1

oh

right

that shouldn’t change anything

just add it in

and floor(m-x) = floor(m-a-b) = floor(m-a-1+1-b)

[fossil latch]

Well, if m is an interger, floor (m+x)= m+ floor(x)

This should be obvious if x= a+b, m+x= m+a+b=(m+a) +b

[copper delta]

let them respond

i will sleep as it is 2:11 here now

gn

[fossil latch]

So floor(m-x)= floor(m+(-x))=m+floor(-x)

[twilit cypress]

yeah, sorry, I am just trying to wrap my mind around this. I am lost right about here. I am not sure how this is helping currently

or is that where you are just adding the m back in?

[fossil latch]

You understand why floor(-x) is -a-1?

[twilit cypress]

^ I think so, yeah since 1-b is between 0 and 1, its effectively the fractional part (which when removed leaves us with the floor)

[fossil latch]

floor(m-x)+floor(x)= {m+floor(-x)}+floor(x)

So yeah

[fossil latch]

The reason why floor(m+x)= m+floor(x) is described here

[fossil latch]

And floor(x)+floor(-x)= a+(-a-1)= -1

[twilit cypress]

are you taking a whole different route than code red took, or is this just adding on to what they were saying?

[fossil latch]

I'm taking a different route from here

The description till floor(x)+floor(-x) is the same

He is trying to find the "normal" way, by finding floor(m-x) , whereas I'm trying to modify the equation from floor(x)+floor(-x) to floor(x)+floor(m-x)

Both works fine

[twilit cypress]

so this is basically where i currently am

but I am not really sure how to continue on the left side there. I think I am getting confused between the two different approaches at this point

[fossil latch]

a-a-1 is -1, not 1

[twilit cypress]

oops, you are def right

but still

I am not quite sure how to continue there

[copper delta]

another approach for extra confusion; since m is an integer, floor(m-x) = floor(m)+floor(-x)

[twilit cypress]

^ that is kind of what I was thinking. Like break it up at the start

[fossil latch]

Anyways, I see that you're stuck on {m+floor(-x)}+floor(x) part on the right of the page

So that is m+ floor(-x)+floor(x)

Which is m+ {floor(-x)+floor(x)}

=m-1

[twilit cypress]

is that even correct though, I am not sure if that is actually relevant to what I currently have down lol. I wrote it down to try to wrap my brain around it there

[fossil latch]

Kinda irrelevant

As what we want is floor(m-x)+floor(x)

[twilit cypress]

ok hot damn

[twilit cypress]

I am following this

[fossil latch]

But ultimately, we need to know what floor(-x)+floor(x) is

So it's relevant

The left side is just the process of proving floor(x)+floor(-x)=-1

[fossil latch]

Which we can use that fact to solve this

[fossil latch]

Yes

Anyways the question is solved, it seems

[twilit cypress]

(apologies for the chicken scratch writing here) Is that stull below my line the solution that we were trying to get at?

[fossil latch]

Yes