help-38

[velvet bolt]

so?

whats the fraction

yes]

same thing

now you do?

no

[shrewd ridge]

no

[velvet bolt]

what's 1/3

?

,calc 1/3

[solid kiln]

Result:

0.33333333333333

[velvet bolt]

see?

0.(3)

is

1/3

?

[inland orbit]

This is not really helpful. It makes it look like it terminates

[velvet bolt]

ok

take a piece of paper

and

[inland orbit]

But yes, 1/3 = 0.(3)

[velvet bolt]

divide

1

by

3

ok?

wrong explenation

please revise fraction

s

and division

you can't not know what 1/3 is

,w 1/3

[solid kiln]

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[velvet bolt]

see?

yes

so

1/3 is

0.(3)

do you agree>

?

no you dont

the only thing you can do

is to take a peice of paper

and divide 1 by 3

you will see the 3's go on forever

yes that's what 0.(3) means

yes

yes

what's 10/3

in decimals

decimals means with .

yes

ok

tell me the fricking answer

i dont care how it is

tell me what 10/3 is

divide

yes

10/3

is

3.(3)

1/3 is 0.(3)

right?

why not

?

[inland orbit]

1/3 = 0.(3) exactly

[velvet bolt]

yes

[inland orbit]

It is not an approximation

[velvet bolt]

yes

YES

[inland orbit]

1/3 is rational, because it is a fraction.

[velvet bolt]

0.(3) is rational, because it can be written as 1/3, a fraction

no

[inland orbit]

0.33333333 and 0.3333333333333333 are approximations, but 0.(3) is exactly 1/3

[velvet bolt]

repeating isnt a valid explenation

[lavish hornet]

Oh, this dude is still going on about rational/irrational numbers?

[velvet bolt]

yes...

i dont think you do

[inland orbit]

Rational means it can be expressed as a ratio of integers, a fraction

[velvet bolt]

are you sure you understand?

https://tenor.com/view/oogway-master-oogway-kung-fu-panda-gif-26485559

good

[solid kiln]

Master Oogway

[velvet bolt]

any whole number

can a irrational number times an irrational number equal a rational number?

u sure?

yes

[solid kiln]

Master Oogway

[inland orbit]

Bro are you tryna give him a whole course? lol

[velvet bolt]

lol

cmno

answer

ok

[solid kiln]

Master Oogway

[velvet bolt]

we are multiplying two irrational numbers

it's not 4

nope

yes

yeah

can 2 rational numbers multiply to a irrational number?

u sure?

wrong

good

you got it

[velvet bolt]

so?

i'm glad

can a irrational number times a rational number equal a rational number?

nope

yes

but

there is a very special rational number

when multiplied by any number

will equal the special number itself

do you know what it is?

yes

you should have

0 is rational

np

sup mikkel

[lavish hornet]

The product of two rationals will always be rational: if we have numbers x and y be rational numbers, they can be expressed as ratio of integers a,b,c,d such that x = a/b and y = c/d. If we multiply our numbers x and y we get

x • y = (a/b) • (c/d) = (ac)/(bd)

Both ac and bd are integers, since the product of any two integers will result in an integer.

[lavish hornet]

Sup🧍‍♂️

[velvet bolt]

.close

not !

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[wraith hinge]

I feel really dumb and tbh I’m not the best at math but what do these dots mean?

[edgy cedar]

Which dots

Like the …

Or

•

[wraith hinge]

Both

[trim lichen]

you're asked to calculate a product

but it is too long to write out in full

so instead they give you the first few things being multiplied, so as to indicate what pattern the rest of them follow, and they tell you where it stops

if written out in full, you would have

(1 - 1/2) * (1 - 1/3) * (1 - 1/4) * (1 - 1/5) * (1 - 1/6) * (1 - 1/7) * (1 - 1/8) * (1 - 1/9) * (1 - 1/10)

[wraith hinge]

Oh I got it now thank you!

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[kindred crypt]

$$\sum _{p=1}^{\infty }:\frac{1}{\sqrt{p}}$$

[solid kiln]

ANASTECH

[kindred crypt]

Hello

how can I prove that this sum diverges ?

[wraith hinge]

compare with sum 1/p

[kindred crypt]

wdym ?

[marble wharf]

do you know that the sum 1/p diverges?

[wraith hinge]

Limit comparison test

Yes that

[kindred crypt]

I don't

[wraith hinge]

U can also do integral test here I think

Since the thing is always decreasing and positive and continuous

Well its a p series I suppose

[kindred crypt]

the thing is, I can't use any advanced technique of these to solve the problem, cuz we've only just studied the concept of divergence and convergence at class

[wraith hinge]

How far is ur knowledge

Oh yeah

[kindred crypt]

maybe I should provee that the limit is infinity using the definition

[wraith hinge]

U mean nth term divergence test?

[kindred crypt]

by that i mean for every A>0, i need to prove that the exists a rank p, such that for every n : if n>=p then Un > A

[kindred crypt]

with Un = equals this sum

[wraith hinge]

Oh you mean that

Yeah

[trail ingot]

if you have literally nothing i think proving sum 1/p diverges first would make things easier to work with

[kindred crypt]

i can also prove the divergence of Vn = 2(sqrt(n+1)-sqrt(n)) since I proved that the sum of 1/sqrt(n) is always greater than Vn, in the first part of the exercise

$$2\left(\sqrt{n+1}-\sqrt{n}\right):\le \sum _{p=1}^n:\frac{1}{\sqrt{p}}\le 2\left(\sqrt{n}-\sqrt{n-1}\right):$$

[trail ingot]

oh thats probably how they want you to do it then

is that right?

[solid kiln]

ANASTECH

[kindred crypt]

I think so

do u have an idea how to prove thee divergence of the left side of that inequation

[trail ingot]

yes but

the first inequality looks false?

sum p=1 to n 1/sqrt(p) is smol

[kindred crypt]

my bad

[trail ingot]

try evaluating that at like n= 4, and same with 2(sqrt(n+1) - sqrt(n))

[kindred crypt]

$$2\left(\sqrt{n+1}-\sqrt{n}\right):\le :\frac{1}{\sqrt{n}}\le 2\left(\sqrt{n}-\sqrt{n-1}\right):$$

[solid kiln]

ANASTECH

[kindred crypt]

I meant this, sorry

[trail ingot]

isnt that even worse? 🥲

[kindred crypt]

lol

It is

so the problem now is to prove the divergence of the sum of the left side

[trail ingot]

how does that help?

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[echo fossil]

Would this be written as f(x) = 1x^2 where k = 1 and p = 2?

[echo fossil]

or would it be a(x) = 1x^2

[lament jewel]

er

idt they specified a function?

i think u can use either ig?

[echo fossil]

does y = x^2 look better, given the instructions?

[lament jewel]

um

i guess

[echo fossil]

ok, im guessing it doesnt matter too much aside from identifying k and p

thanks!

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[lament jewel]

yea

lol

np

💕

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[hearty sparrow]

What's up with the 1.5

[inland orbit]

It says the lift accelerates with a constant acceleration of 2ms^-2 until it reaches a velocity of 3ms^-1

That would take 1.5s

[hearty sparrow]

it says the area of that triangle is 27 tho

@inland orbit

[inland orbit]

No, it says when the lift comes to rest, it has moved a distance of 27m

That's the entire trapezoid

[hearty sparrow]

Oh

Shit yea

Jeez im dumb

Sorry

[inland orbit]

Lol nah it's tricky

No problem

[hearty sparrow]

Also can u check a proof I did in latex if thats ok

Proving that if f(x)=4x²+2x, f'(x)=8x+2

[inland orbit]

The result is right but the 2delta at the bottom should be 4delta

[hearty sparrow]

No?

[inland orbit]

$\frac{4\delta^2}{\delta} = 4\delta$

[solid kiln]

tatpoj

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[dim kiln]

just wondering how to know whether or not it converges

ik its something like a telescoping series

but i can't figure it out

[long basin]

Simplify 7k/(7k)!

[naive crest]

the numerator is 7^k * k!

[dim kiln]

wat

[wraith hinge]

Umbra with the straight facts as always 🙏🏻

[dim kiln]

what's there to simplify

[long basin]

The numerator and denominator both have a common factor

Try expanding the factorial to a couple of terms, you're gonna notice soemthing

[dim kiln]

yea i realize you can do

u can factor out 7k

and it becomes

1/(7k-1)!

[glacial karma]

$\frac{7k}{(7k)!}$ = $\frac{1}{!}$

[solid kiln]

Springsskateboard

[glacial karma]

SEYSRY

[dim kiln]

so i guess i find convergence from that

[naive crest]

on a more serious note, Raemelia, do you know the ratio test ?

[dim kiln]

ya

[long basin]

1/!

Ah yes

[naive crest]

it works quite well here

[long basin]

Well you would use the integral test after that

But like

[glacial karma]

😉

[long basin]

1/n! Converges

[dim kiln]

well on the test i gotta show that it converges cuz of __ test

[naive crest]

because the numerator is 7*1 * 7*2 etc
= 7^k * k!

[dim kiln]

so i just gotta

be thorough

[wraith hinge]

Mathematicians hate him..

[dim kiln]

wat

so does it simplify to 7^k / 1

or 1/(k-1)!

[naive crest]

this

[dim kiln]

yea

im talking about the whole thing

if u replace the top with 7^k * k!

it all would simplify to 7^k

[long basin]

I think 7k is the general term for the numerator

The 7 . 14 . 21 just visually shows the pattern

The actual summation is 7k/(7k)!

Otherwise it'll diverge

And then it's just a lame question

So you have to use integral test for 1/(7n-1)!

[dim kiln]

can you integrate factorials

[naive crest]

a_(n+1)/a_n = 7^(k+1)*(k+1)!/(7(k+1))! * (7k)!/(7^k * k!)

and now it simplifies

and you take the limit

[long basin]

Nth term for divergence only diverges if it results in a number

If it results in 0, use another test

[dim kiln]

uh im gonna act like i know and try to learn from that

thanks

.close

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[wraith hinge]

The life of a calculus student

Welcome to the gang

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[quasi sorrel]

when plugging this back I was wondering if the square would affect the result

or am I confusing the order of operations.

because if you were to square the -1 when plugging back it would be positive

[long basin]

I mean jjst make it into one log

And then since the log bases are the same, so are their arguments

But check for extraneous solutions

[split chasm]

you left something out in your work

[quasi sorrel]

what should I include?

[split chasm]

note that it's
2 log_2(x^2) in the question

not log_2(x^2)

[torpid trail]

i saw that too i think the first superscript is pencil

[quasi sorrel]

yeah sorry I wrote that 2 there

it should be 2 Log_2(x)

in the equation

[split chasm]

oh

[long basin]

Well once you do that, you open possibilities for extraneous solutions I think

2log(x) cannot have x be negative, log(x^2) can

[quasi sorrel]

I'll rewrite it as one big log and see

[split chasm]

when checking whether a solution is extraneous, you'd consider the original equation

[long basin]

That's perfectly fine, just check for extraneous solutions

[quasi sorrel]

only the original equation no manipulation or anything?

so making one big log wouldn't be a good idea?

[long basin]

It would

But you have to check for extraneous solutions

Which shouldn't take long to check anyways

[quasi sorrel]

this is what I'm plugging in both solutions to

[long basin]

Oh no no no

[quasi sorrel]

I should have no extraneous solutions yeah?

[split chasm]

when solving, you could apply whatever log laws, but when checking you'd consider the original restrictions

[long basin]

_ _

so x^2 = 2x + 3

[quasi sorrel]

hm so no extraneous solutions?

[long basin]

Well there could be

That's why I said to check

Once you have them

[quasi sorrel]

so why is this wrong?

[long basin]

It ain't, but you're just making useless steps

[split chasm]

2log_2(x) is originally present,
so x must be greater than 0

[quasi sorrel]

I thought that's what you wanted me to do lol

[long basin]

No

[long basin]

If you saw this text

Then no

[quasi sorrel]

either way it makes it easier to visualize. plugging in both solutions they are positive

[long basin]

If $\log_b(a) = \log_b(c)$, then $a = c$

[solid kiln]

Umbraleviathan

[long basin]

That's what I've been trying to get you to do

[quasi sorrel]

how would this help me check for extraneous solutions?

[long basin]

You have to check yourself

Just find possible candidates for x, test all of them manually

[quasi sorrel]

once I have a = c then what

or was that on a different topic not extraneous solutions

just solving for x?

[long basin]

Just x

But then again

[long basin]

_ _

[quasi sorrel]

yep

[long basin]

Once you solve for x, you have to test them all (or at least winnow out the negative ones)

[quasi sorrel]

so my final answer is that there are no extraneous solutions

-1 and 3 are both solutions

[long basin]

My guy you didn't even test them

How can -1 be a solution to the original

[split chasm]

note that also the left side with
2^x = x^2 and
2^x = 2x+3
is wrong

[long basin]

Go back up here

What equation am I trying to get you to set up

[quasi sorrel]

I was just rewriting the logs on both sides

[quasi sorrel]

I already tested it what do you mean

[long basin]

Look

$$2\log_2(x) = \log_2(2x+3)$$
$$\log_2(x^2) = \log_2(2x+3)$$
$$x^2 = 2x+3$$

[split chasm]

that's not how long laws wirk

[solid kiln]

Umbraleviathan

[long basin]

Again

[quasi sorrel]

bro

[long basin]

This comes into play

[quasi sorrel]

what is that doing

I already factored it lmao

[long basin]

Okay

But what are your values for x

When you solve that

[quasi sorrel]

literally look at my work

[long basin]

-1 and 3 right

One of them is extraneous

[quasi sorrel]

how

I already told you my answers are both positive

when plugging back in

You can test it yourself

[long basin]

Is -1 positive

Ready for this

,w calc 2 Log[-1]

[solid kiln]

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[quasi sorrel]

My brother you cannot be serious right now

lmao

it's x ^2

not x

[long basin]

Ain't gonna work for the original

No

The ORIGINAL

[quasi sorrel]

🤦

[long basin]

The original is 2 Log(x)

2 log(x) = log(x^2) only when x > 0

[dim kiln]

wat is happening in here

[long basin]

I can physically show you

[long basin]

If you're not convinced

[dim kiln]

2x + 3 = x^2

x^2 - 2x - 3 = 0

[quasi sorrel]

why does it have to be the original

[dim kiln]

(x-3)(x+1) = 0

[quasi sorrel]

If the equations are the same

[split chasm]

not quite the same

[long basin]

They're not the 100% the same

They share similar features

,w plot log(x^2)

Good program

[split chasm]

the restrictions are based on the original expressions

[solid kiln]

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[long basin]

That's the edited

This is the original:

,w plot 2 log(x)

[solid kiln]

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[long basin]

Note how x cannot be negative in the original

[quasi sorrel]

hm okay

so -1 is extraneous

[long basin]

Yes

[quasi sorrel]

and 3 is not

[long basin]

Mmhm

You have to test solutions in your ORIGINAL, like the raw shit your teacher gives you

The nasty

Uh

Stupid equation they gave you

Even if it's dumb looking

Gotta plug it in

[split chasm]

in case you missed my response, what you did on the left wasn't a valid application of log laws

[quasi sorrel]

how come?

[long basin]

a^b ≠ b^a unless a, b = 1

[quasi sorrel]

That's not what I was writing

I was literally rewriting the log on the left side

into another equation

[split chasm]

you inappropriately used x when you shouldn't have

[long basin]

You wrote 2^x = x^2

Iirc

[quasi sorrel]

yeah x is just another value

I should have used a or something

[split chasm]

yes

[long basin]

Oh that's not the same x

Bruhg

That's hella confusing

[quasi sorrel]

lmao

[split chasm]

do NoT use the same variable to represent different things

[quasi sorrel]

sorry I didn't make my work user friendly lol

help friendly

well anyway thanks for the insight

how do you close the channel

[long basin]

I gotchu

.close

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[strong hare]

i’m so confused

[long basin]

Alright alright alright so there's this painful thing called implicit differentiation

Absolute ass to do

Try that

[strong hare]

ok i tried but

idk how to find the derivative of sqrt(x+y)

i think it’s a chain rule

[long basin]

Chain rule

The fun stuff

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[strong hare]

.close

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[manic ibex]

I asked 4 servers, none could solve this, i need help it's serious

[silent canyon]

We can see that AB=AD=BC=CD=√200

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[manic ibex]

Someone serious?

[wraith hinge]

Why is @silent canyon not serious

[manic ibex]

I definitely dont know the area of a square formula

[manic ibex]

.close

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[chilly sundial]

Help

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[velvet bolt]

l'hopital

[stark bison]

Or the definition of a derivative

Actually no wait I think there's something simpler

Just rewrite it like this

[raw gate]

we can't use that yet

[raw gate]

this works thanks

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[stoic saddle]

so im pretty sure c and d are required for sure

although I think b is implied through d isnt it

so it doesnt need it

and then for a, not sure

[past comet]

Can’t you just read rolles theorem?

Says the conditions

[wraith hinge]

[past comet]

Continious does not imply differentiable if that is what you are saying

[wraith hinge]

Rolle's theorem, in analysis, special case of the mean-value theorem of differential calculus. Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

That's all you need

[stoic saddle]

ok

makes sense

what about for this one

[past comet]

Easiest is figure out which ones are false

Probably

[stoic saddle]

b is true , c is false , d is false

[past comet]

p can’t be even?

Can you say the first few prime numbers?

[stoic saddle]

if its even then 2 is a factor so its not prime right

oh

kek

can you translate the notation in the first q for me pls?

for a.

well I guess it doesnt matter its the correct answer but kind of want to know what its saying

[granite cove]

The | means devides

[stoic saddle]

ty

ok last one

[wraith hinge]

Very interesting homework you got there

It's like going from three completely different topics

[stoic saddle]

its a pre reading week thingy

do this and we cover it over a week

[past comet]

What are you confused about?

This is just like Rolle’s theorem

Read what it is

And you will know the answer

[granite cove]

Hint: It's not b

[stoic saddle]

alr so

got them all, so for this one

is there any tricks I can use

or do I just have to straight up derive 3 times

and pull my hair out by the end of it

[trim jolt]

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[vital plank]

hey

[vital plank]

quick question, is it possible to do partial fraction decomposition if the denominator is a sum?

[edgy cedar]

[vital plank]

like specifically I have it as x^2 + 1

[wraith hinge]

It's pure HIMSELF?!

Maybe that's ur answer

Be a wolf holding a sword with its mouth

[trim lichen]

x^2 + 1 is unfactorable unless you're willing to involve complex numbers.

[wraith hinge]

Also you need to factor it somehow

[vital plank]

yeah but in general does it work with sums or no

[wraith hinge]

They just answered your question..

[vital plank]

without complex..

[edgy cedar]

Depends what you mean by sum

x^2 +2x +1 is a sum

[trim lichen]

if you want to do partial fraction decomposition you need your denominator to be factorized

[vital plank]

ok, so if no complex factorization then no pfd

deal

in part two how are we to do the last finishing move, I don't get it

[trim lichen]

$\int \frac{\dd{x}}{1+x^2} = \arctan(x) + C$

...bot is down for some reason

$123$

[vital plank]

I can still read don't worry

we haven't taken integrals with inverse trig functions but let me do it by hand hang on

how did we get arctan x, I got ((1+x^2)^2)/4 + c (assuming the numerator is x)

[trim lichen]

the numerator isn't x. it's 1.

[vital plank]

oh

I was just doing it now, then yeah it's arctan x

but how does integrating help with part 2

[trim lichen]

well you will have that $\frac{\log(x)}{2x} \leq \frac{x \log(x)}{1+x^2} \leq \frac{\log(x)}{x}$ on $[1,2]$

[vital plank]

I tried rechecking what I did and I'm not sure if it's just me but the change in the green squares looks bad, I tried it with numbers

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@vital plank Has your question been resolved?

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[lusty raptor]

is there a name for this function

[stark bison]

Not sure but you can definitely simplify the what's inside

[lusty raptor]

how?

ill run it trough mahway

[stark bison]

You can get r/x inside the sqrt

[lusty raptor]

how

this is what mathway gives

which is already like 100% better

[stark bison]

r/x * sqrt(1 - x^2/r^2) = sqrt(r^2/x^2 - 1) (as long as r and x are of the same signs)

[lusty raptor]

they are

[stark bison]

Then just sqrt(r^2/x^2 - 1)

[lusty raptor]

r^2/(x^2-1) 0r (r^2/x^2)-1

[stark bison]

The latter one

[lusty raptor]

k thanks for the help

.close

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[wraith hinge]

This is a sequence not a series yes?

im not sure about the name in english but i think sequence

yes sequence

<@&286206848099549185>

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@wraith hinge Has your question been resolved?

[lusty raptor]

infininity?

oh

well we know that lim n->inf (1/2^2*n) =1

so your sequence must be >=1

if we only accout for the first three terms (1+1/2)x(1+1/4)x(1+1/8)=2.1

so your seq. >=2.1

and it most definitely increases for the next couple terms so seq.>2.1

but it decreases too rapidly to be 3

so 3>seq>2.1

thats a good enough approximation imo

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[finite shuttle]

Hi, I got feedback on my work saying "lack of determination of the domain of the function", what did they mean by this?

[finite shuttle]

another example of the same mistake

how do i determine it? the calculations are correct aren't they?

so in 1a only x1 would be considered

so you'd say something along the lines of x∈((-1+sqrt(29))/2, +infinity) as the final answer?

im right there with you bud i'm doing this in polish xd

i understand the domain is missing for this to be a fully correct answer, would adding x∈((-1+sqrt(29))/2, +infinity) at the bottom of 1a make it correct?

hmm

isn't this determinable without doing all of the calculations?

i see, i struggle to see how the x1 and x2 are needed for any of thtis though

like what did we discover by finding them

gotcha

and the mistake here is putting -3 innstead of -2 right?

because the line above it literally sttates x=> -2

so it'd be x∈<-2,1)

[finite shuttle]

in the top right it also says there's "not enough written about the conditions"

i guess we can also say x cannot be 0? 🤔

true i think i wanted to make it a nice number like 3 but that's a bit illegal

unless i did it to both sides or something

crap that's right

feels like i figured out the hard part which is the calculations but something really basic like the domains are holding me back

im not really sure when we're supposed to do something like -1<x<+1 or x∈(-1,1, u (+1,infinity)

or if it's even the same thing

gotcha

thanks for the help

think its a lil clearer now

.close

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[zinc dome]

Hi, I was trying to convince myself that 100* (x-a) / x will always be superior to 1, so I began with x - a < x and I multiplied only the left hand side with 100, to obtain 100*(x-a) > x, with this I was convinced that my numerator will always be superior to the denominator, and hence the result will always be superior to 1

[zinc dome]

What do you tguys think?

[stoic iron]

What values are x and a allowed to take?

[zinc dome]

I forgot a 100 in the numerator

they are integers

positive

natural numbers basically

The only exception is when x and a are the same I guess

and 1 <= a < x, with the exception of x = a which will give us zero

[stoic iron]

100x - 100a > x

99x > 100a

See if that's always true

[zinc dome]

you mean with random numbers right?

[stoic iron]

Well if you suspect it is true, checking some random numbers won't be enough

[zinc dome]

Well, x = a won'T work but this was the exception to my argument

I think it's true, it's not some thing for a class It's just a personnal challenge i gave myself

yes it seems to work with the exception of x = a

anyway I wouldn'T kno whow to prove such a thing, I just need to convince myself it's true

I mean, I could try but I don't have enough time for this

I was just curious to see if you agreed with my reasoning

[stoic iron]

a=100, x=101 doesn't work

[zinc dome]

wow

how is that possible?

[stoic iron]

Why do you think it wouldn't be possible

[zinc dome]

wait a min

yes you're absolutely right, wow my world was shatered

how did you find it?

[stoic iron]

There's infinitely many times it doesn't work

a=100, x = anything bigger than 100

Plus loads more for bigger values of a

If you mess around with 100*(x-a)/x <= 1 (the opposite of what you thought was true) it will simplify to a constraint that isn't impossible

[zinc dome]

ok wait a min let me see something

yes now it starts to make sense. I think I understand why it did that now

thanks this was a humbling experience

man i thought I had something bro

lol

oh well

this also answers my question which I was having about this.

i think my error was wanting to beleive i was right

this blinded me

[stoic iron]

Good life lesson

Try and prove yourself wrong sometimes

[zinc dome]

yes thanks agian my friend,!

!close

/close

[stoic iron]

🙂

.close

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[zinc dome]

btw I have a last question concerning this

[zinc dome]

after I had x-a < x set up, was it true to say (x-a) * 100 > x ?

I mean, this seems to be true here

wait forget it

no it's no always true

not*

i think this was my error here

yes I think this was where I blundered I assumed it would be true but I did not think about cases where this would be false

this isn't a vlaid operaton also afaik

valid operation* unless you,Re willing to take risks

it seemd so true though.. sometimes life be so deceiving lol

.close

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[alpine fern]

I believe the point of inflection is 5/24? Why am I getting it wrong?

[delicate bobcat]

show your work

[alpine fern]

$24x^{2}-10x$ this is the first derivative

[solid kiln]

cookie2

[alpine fern]

$48x-10$ this is the second derivative

[solid kiln]

cookie2

[split chasm]

5/24 is only the x-coord of the point of inflection
and not the actual point

[alpine fern]

FUCK

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[limber herald]

I got no idea how to work this

[limber herald]

Anyone knows how?

.close

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[rotund nova]

im trying to evaluate this and turn to polar coordinates and did i set this up right because i feel like rcostheta isnt supposed to be there

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[rotund nova]

<@&286206848099549185>

[rotund nova]

thanks guys

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[wraith hinge]

I need help coming up with a rival hyptoehsis for my research question

[trim jolt]

Please don't occupy multiple help channels.

[wraith hinge]

if anyone can help me ill love u forever

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[uneven parcel]

can someone tell me the transformations

[uneven parcel]

in this equation

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[uneven parcel]

<@&286206848099549185>

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[wraith hinge]

There are 7 students in grade 11 and 15 students in grade 12 that are part of the robotics club. A committee consisting of a president, vice president, media rep, and accountant is being chosen.

In how many ways could the panel be chosen if the president and vice president must be from different grades?

[turbid spruce]

so the president and vice president must be in different grades, meaning that the president can be an 11th grader and the vice president is a 12th grader or vice versa, and the rest can be simply anyone

so there will be two cases: 1. the president is 11th grader (and consequently, the vice president is 12th grader), and 2. the president is 12th grader

then use the multiplication rule for each case

[wraith hinge]

So it would be P(7 and 15)?

@turbid spruce

[turbid spruce]

So it would be P(7 and 15)?
what

[wraith hinge]

you said to use the multiplication rule

[turbid spruce]

hmm
try to do it first

[wraith hinge]

ok

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[static wharf]

I need some help with this cuz I'm falling behind in class

[static wharf]

How do I make pressure a number?

How do I use newtons and a base area to calculate the answer

[static wharf]

Where does the force come from

[tall schooner]

and always convert the units to newtons and meters

[static wharf]

What units

Sorry I'm very stupid

[tall schooner]

its basically weight that is expressed in newtons

[static wharf]

How do i convert them

[tall schooner]

a unit is a measure of things. Newtons (N) are a measure of forces

Meters are measure of space

use this table

[static wharf]

So how can I get the pressure from newtons and the area

[tall schooner]

you will usually need only to know what are kilometres, centimetres and decimetres

[tall schooner]

u plug the numbers into the firmula that I sent

lets do this as an example

we have an area of $0.09 m^2$ and a weight (force) of 180 N

[static wharf]

so i divide 180 by 0.09squared

[solid kiln]

ta

[tall schooner]

here you dont need to convert anything

[static wharf]

okay

[tall schooner]

yes

so $P=\frac{180N}{0.09m^2}$

[solid kiln]

ta

[static wharf]

so its always n/a=p

f

so how do i write out the answer

[tall schooner]

$P=2000 Pa$

[solid kiln]

ta

[static wharf]

Pa is pressure area?

what is it

[tall schooner]

Pascals (Pa) are also a unit.

[static wharf]

?

ohh

[tall schooner]

when you do physics you need to keep track of units

[static wharf]

okok thanks a lot

ill try the rest myself and ill come back if i need to

thanks

[tall schooner]

np

[static wharf]

wait how do i divide them 180 by 0.09m squared

and how do i umm unsquare 0.09

[tall schooner]

you do not need to unsquare it

because for the formula you need area

and area is a squared number

[static wharf]

an area is always squared?

[tall schooner]

units for area are always squared

[static wharf]

got it

but how do i divide them

[tall schooner]

so if you were to measure area of your room youd measure width and length and multiply them to get area

[tall schooner]

like regular numbers

[static wharf]

what do i do with the square

[tall schooner]

nothing. just forget about it and calculate 180/0.09

[static wharf]

so the square is just irrelevent?

[tall schooner]

in physics when you calculate something you treat units separately from numbers.

[tall schooner]

yes

[static wharf]

ok got it

thanks again

[tall schooner]

np

[static wharf]

for the second question, do i mutiply 75, 25 and 50? because the number i get will be really big am i doing this correctly

42,750^3

i know im doing something wrong

[tall schooner]

well

as you know for area you only need width and length right

[static wharf]

🪦

[tall schooner]

i can see a drawing on the right that shows that suitcase

[static wharf]

sry im really bad at maths

[tall schooner]

its ok

it would be easier for me to explain if you send the drawing of the suitcase on the right

[static wharf]

the width is 25 and the length is 75

[tall schooner]

yeah

so now multiply it

[static wharf]

so i multiply them both and i get the are

[tall schooner]

to get area

but remember you treat numbers and units separately

[static wharf]

1875^2?

am i doing somthing wrong

[tall schooner]

so you calculate numbers like this: $75\cdot 25=1875\newline$
and units like this: $cm\cdot cm=cm^2\newline$
so the answer would be $1875cm^2$

[solid kiln]

ta

[static wharf]

oooohh

[tall schooner]

as I said earlier you need to convert units to metres. here you have $75 cm = 0.75 m$ and $25 cm = 0.25m$

[solid kiln]

ta

[tall schooner]

so the area is different now

its actually $0.25m\cdot 0.75m=1.875m^2$

[solid kiln]

ta

[static wharf]

so we have to always do it in metres

[tall schooner]

yes exactly

same with newtons

you may need to know these:
1 kN = 1000 N
1 km = 1000 m
1 cm = 0.01 m
1 dm = 0.1 m
1 mm = 0.001 m

but it's only in physics. if you're in maths class units don't change usually

[static wharf]

so if i want the its maximum weight what do i need to do

i only have the area so what can i do

[tall schooner]

in a) you have to calculate force
you can use a hint left in brackets that 1 kg = 10 N.

suitcase's mass is at maximum which is 20 kg. you need to convert 20 kg to newtons (N)

[static wharf]

so its just 20 time 10?

[tall schooner]

yes

[static wharf]

i thought i was doing something wrong cuz it seemed too easy

[tall schooner]

yeah it's easy

[static wharf]

whats the difference between force and pressure

[tall schooner]

force is expressed in Newtons and is used to describe how much mass is being accelerated. Pressure is expressed in Pascals and is used to describe how much force is applied to some area. Calculation of pressure is needed to ensure that for example oil pipes don't burst when oil runs through them. Calculation of force is used to see how much something can push something else and it is used in mechanical engineering

[static wharf]

i was getting somewhere but i lost my train of thought

[tall schooner]

so are you understanding all the other points

[static wharf]

i think i get it now

so force is like pushing a pencil away from you and pressure is how much force youre putting in that that push?

or pull

[tall schooner]

well pretty much yes

you just forgot about area, so when you push the pencil force defines how quickly the pencil accelerates and how much energy you spend to make it move. while pressure describes how much force will be applied to a part of pencil under your finger.

[static wharf]

im not keeping up

[tall schooner]

imagine pressure like trying to poke a balloon with a bat and a needle. the force applied to both bat and needle is about the same while area of application is a lot smaller for needle, which allows it to penetrate balloon easier because more force is being applied to a smaller area.

[static wharf]

so the smaller the area the more the more force is being put into it?

[tall schooner]

force stays the same. pressure increases

[static wharf]

i think i kinda get it know but still dont know how to put it into words

[tall schooner]

look at the formula: $P=\frac{F}{A}$ when A decreases P increases. when A increases P decreases.

[solid kiln]

ta

[static wharf]

so instead of force its pressure

[tall schooner]

say F = 10, $A_1=10,A_2=100\newline$ then $\frac{10}{A_1}=1Pa$ while $\frac{10}{A_2}=0.1Pa$

[solid kiln]

ta

[tall schooner]

yes

[static wharf]

so the amount of force used to poke the balloon with a bat and the needle stays the same while the pressure varies depending on what we use to poke the balloon?

[tall schooner]

yes exactly

[static wharf]

so what would the balloon, the bat and the needle be in the second question?

[tall schooner]

the side with the smallest area will apply the biggest pressure. opposite for the biggest area.

the bottom of a suitcase would be a needle and it's wider side would be a bat and a balloon would be ground

[static wharf]

ooohh

thank you so much

[tall schooner]

np

[static wharf]

i understand now

but igg so byebye

thank you

[trim jolt]

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[wraith hinge]

Let A be a set containing exactly n elements for a positive integer n, and let R be a total ordering on A.

Prove that A contains a largest element. [Hint: Use strong induction. For the inductive step, choose two
nonempty proper subsets of A whose union is A, and compare the largest elements of each.]

[dusty hound]

Did you manage to make the hint work ?

[wraith hinge]

not really

im not really sure how partitions of A inducted could prove A having a largest element

[dusty hound]

You have a set of n+1 elements. By the induction hypothesis, you know that a set of n or fewer elements has a maximum

But n+1 = (n) + (1), i.e. you can easily cut A into two sets that both have a maximum

[wraith hinge]

oh i see

so if i were to prove the proposition true for all positive integers would i use 1 and 2 for base cases

and how would i write the base cases to be arbitrary

[dusty hound]

First time writing a proof ?

At least by induction

[wraith hinge]

ive done induction before this is just my first time doing relations

[dusty hound]

Let A = {a1, ..., an+1}

Then {a1, ..., an} and {an+1} both have a maximum right ?

[wraith hinge]

i guess

if R is a total ordering

but isnt that circular logic

[dusty hound]

It's the strong induction hypothesis

We're doing the inheritance part, as the property is obvious for n = 1

[wraith hinge]

yeah

[dusty hound]

So if their maximums are m and m', then one of them is bigger (since the ordering is total) and therefore is an upper bound for both sets.

[wraith hinge]

oh im pretty sure that we have to use a different definition for maximum

We say x ∈ R is a largest element (or
greatest element) of A with respect to R if r R x for all r ∈ R.

largest*

[dusty hound]

It works anyways by transitivity

[wraith hinge]

so the largest element is pretty much the element that satisfies r R x for all r

[dusty hound]

I purposefully said it in a way that didn't entirely say outright that it's the maximum of A

[wraith hinge]

oh

im kinda confused by the syntax of the definition of largest

We say x ∈ R is a largest element (or
greatest element) of A with respect to R if r R x for all r ∈ R.

how would r be an element of R if the relation R contains ordered pairs and r is a singular element

[dusty hound]

usually when R is an order, we write it as <= or occasionally >= for obvious reasons
Reminder: R is a subset of A², and we only write a R b to mean (a, b) in R. That's why it's like a <= b

[wraith hinge]

yea but if r is an element of R doesn't that mean that r is an ordered pair

so how can r R x

[dusty hound]

should be elements of A yes

x, r in A rather than R

[wraith hinge]

oh so is it just a typo

[dusty hound]

yes it is

[wraith hinge]

for the base case in the question above would i only need n = 1

because i could form any number of cases by just adding 1

this is what i have so far

[dusty hound]

I'd say it's fair to just say a1 is the max directly because it's very obvious

unless you want to prove it in detail

[wraith hinge]

isn't the base case adequate proof for p(x) being true for n=1

[dusty hound]

your proof works. I wouldn't detail it as much, but I'm no longer at the level where you'd detail such trivial things

[wraith hinge]

eh im just being safe

also i have no idea how to proceed from here lol

is n=1 all i need to move onto the inductive step?

also how would i even do the inductive step

[dusty hound]

the thing is if you detail n = 1 you might want to do so for n = 2 as well as it's needed for the induction step

[wraith hinge]

i did n = 2 but i got stuck lol

if n= 2 then A = {a1,a2}

so a total ordering on A would be {(a1,a1),(a1,a2),(a2,a2)}

but that doesnt even have a largest element

[dusty hound]

if a1 is the max, we're done. Otherwise since R is total and we don't have a1 >= a2 that means a2 > a1, so a2 is the maximum

that's n = 2

[dusty hound]

you don't want max(R), you want max(A)

[wraith hinge]

but isn't x the largest element only if r R x for all r in A

[dusty hound]

yes

[wraith hinge]

man i have no idea what im doing

these relations are confusing af

[dusty hound]

I frankly don't think you ever need to state what R is

[wraith hinge]

oh

but how would i demonstrate that A has a largest element if there isn't an element in A such that r R x for all r

[dusty hound]

also you don't actually know what R is because you don't know what order you're dealing with

so that's another reason why you shouldn't state what R is: you don't even know

[dusty hound]

there is

[wraith hinge]

oh

oops

ok i think i proved the base cases

if A has a largest element for n = 1 and n = 2 then i can suppose that A has a largest element for n and n + 1

right

so how would i use that to prove the inductive step

[dusty hound]

it is obvious for n = 1
that was n = 2
That's how I'd do them, and for you probably detail a little bit more. I don't know what you did but probably too much and maybe wrong

[wraith hinge]

maybe i could restrict the relation

[dusty hound]

it's what we did at the start

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@wraith hinge Has your question been resolved?

[amber patio]

Can anyone explain why its C and not A? Im rlly confused

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[fickle helm]

This is just a question about proving s=sqrt(2) is the sup of the set S where x^2<2

I’m a bit confused as to why they used epsilon as the min of 2-s^2/2(2s+1) and 1/2

Why is the 1/2 necessary?

[zinc ginkgo]

what happens if s=0?

[fickle helm]

2-s^2/2(2s+1) would equal to 1

However we set epsilon to be less than 1

But I still don’t see why they chose 1/2?

[zinc ginkgo]

It's a somewhat convenient number less than 1