help-39

[orchid tapir]

how?

n is infinity

[hard kite]

how would it be 1?

[solemn linden]

exactly 1/n is 1 over someting very big

[orchid tapir]

0+1

[hard kite]

$\frac{1}{a + b} \neq \frac{1}{a} + \frac{1}{b}$

[jolly parrot]

kaue

[orchid tapir]

so you have to divide everything by n first?

[hard kite]

if you want to it also works, but its pretty clear that 1/(n + 1) goes to 0 as n approaches infinity

[orchid tapir]

ahh okay

thanks

appreciate the help

.close

[pearl pond]

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[sharp smelt]

Let ${e_1,e_2,\dots, e_n}$ form a basis of $Ker(T)$ . We now extend this to a basis of $U$. ${e_1,e_2,\dots, e_n,f_1,\dots, f_m}$ Let $v \in U$. We then have $T(v) =\sum_{i=1}^{m} T(a_if_i)$

this is what I'vd done thus far

I'm lost beyond this

[jolly parrot]

A dense set

[sharp smelt]

I could construct a basis of $V$, but I don't see how that'd help

[jolly parrot]

A dense set

[pine jay]

No basis needed

[sharp smelt]

Let ${e_1,e_2,\dots, e_n}$ form a basis of $Ker(T)$ . We now extend this to a basis of $U$. ${e_1,e_2,\dots, e_n,f_1,\dots, f_m}$ Let $v \in U$. We then have $T(v) =\sum_{i=1}^{m} T(a_if_i)$. We now extend this basis to one of $V$; The basis would now be ${T(e_1),\dots,T(e_n), g_1,\dots, g_k}$. At most, all these $k$ vectors map to $0$. Thus $dim(null(S))\leq k$
\ we also know $dim(null(T))= n$. We thus have $dim(null(S)) + dim(null(T)) \leq k+n$.But dim (null(ST)) is atmost $n$
\
$n \leq k+n$
We thus have
\
\
$dim(Null(ST)) \leq dim(null(S))+ dim(nulll(T))$But dim (null(ST)) is atmost $n+m$

I feel I'm close to the answer

But dim (null(ST)) is atmost $n+m$

[jolly parrot]

A dense set

[sharp smelt]

ah

that's it

no?

[pine jay]

You’re free to go that route if you feel that way but do note that choosing a basis is not needed

[sharp smelt]

So I pressume you feel the rank nullity theorm, can be used here?

[pine jay]

Yeah it’s really straightforward in some sense (not in an easy way maybe)

But if you like this route we can go through this one

And see if it works

[sharp smelt]

Can we use my method first

[pine jay]

Yeah sure

[sharp smelt]

Okay, so this is my proof

[pine jay]

Ah okay, let me take a look

So when you then extend it to V what happens there?

You’ve got a bunch of zero vectors in it

[sharp smelt]

yes

I just realised, there are a few mistakes in my proof

[pine jay]

And whose to say V doesn’t have less of a dimension?

Take your time and see if u can patch those up

[sharp smelt]

Yeah

sorry

[pine jay]

No worries at all

[sharp smelt]

Let ${e_1,e_2,\dots, e_n}$ form a basis of $Ker(T)$ . We now extend this to a basis of $U$. ${e_1,e_2,\dots, e_n,f_1,\dots, f_m}$ Let $v \in U$. We then have $T(v) =\sum_{i=1}^{m} T(a_if_i)$. We now extend this basis to one of $V$; The basis would now be ${T(e_1),\dots,T(e_n), g_1,\dots, g_k}$. At most, all these $k$ vectors map to $0$. Thus $dim(null(S))\leq k$
\ we also know $dim(null(T))= n$. We thus have $dim(null(S)) + dim(null(T)) \leq k+n$.But dim (null(ST)) is atmost $n$,as the pre-image of S is the outputs of T.
\
$n \leq k+n$
We thus have
\
\
$dim(Null(ST)) \leq dim(null(S))+ dim(nulll(T))$

[jolly parrot]

A dense set

[sharp smelt]

There we go

[pine jay]

Hm the issues I addressed seems to still be there?

1. your extended base to V contains zero vectors

2. what if V has lower dimension than U or even Ker T?

[sharp smelt]

Well shit, It should have been f_1s not es

As for the second one, cases, I suppose>

[pine jay]

I see

[sharp smelt]

Ok, what's the rank nullity proof

[pine jay]

So for 2) don’t really mention extending it

[sharp smelt]

this si going to be messy

[pine jay]

Ah okay well, I liked your idea in any case

You might be able to revisit it

With the idea from the rank nullity proof of this

Okay so for starters let me try and guide you towards it. The first few steps are only the maybe nontrivial ones

Consider the equation STv = 0, that is v is an element of KerST in this scenario.

Notice how we can view this equation as S(Tv) = 0

Where Tv is a vector in Im T, but more specifically the image of T restricted to Ker ST

Take your time thinking through why that is the case

[sharp smelt]

hmm

That does make sense

yes

[pine jay]

Great, so $\text{Im}, T|_{\text{Ker},ST} \subseteq \text{Ker}, S$

[jolly parrot]

Aslan

[pine jay]

Would u agree?

[sharp smelt]

yes

[pine jay]

Now as an aside thing, what does the rank nullity say about T restricted to KerST?

[sharp smelt]

dim(T) = dim(ker(ST))+ dim(S)?

[pine jay]

What is dim of T and S supposed to mean?

[sharp smelt]

dim(Im(T))

and dim(ImS)

[pine jay]

I see, well not quite; like okay say you were given some map P

With domain idk K

what is the formulation of the rank nullity theorem on this map?

[sharp smelt]

$dim(K) =dim(Im(P))+ dim(Ker(P))$

[jolly parrot]

A dense set

[pine jay]

Right

[pine jay]

So for this

P is the map T restricted to Ker ST

what would K (the domain) be in such case ?

[sharp smelt]

Ker(ST)

[pine jay]

Exactly!

Now restate the above but with those instead

[sharp smelt]

so $dim(Ker(ST)) = dim(ker(T)) + dim(Im(S))$

[jolly parrot]

A dense set

[pine jay]

Uh no

Remember we’re only replacing P with T |_KerST

And K with KerST

[sharp smelt]

$dim(Ker(ST)) ) =dim( im(T|{ker(ST)})+ dim(Ker(T|{ker(ST)})$

[pine jay]

Right!

[jolly parrot]

A dense set

[pine jay]

Now what did we know about the image of T restricted to Ker ST from before again?

[sharp smelt]

It's a subset of $S$

[jolly parrot]

A dense set

[pine jay]

Indeed!

Or well

No

Ker of S

But yeah you understood the idea I think

[pine jay]

So this

[sharp smelt]

oops

yeah

[pine jay]

But yeah no worries

So what does that mean dimension wise?

If we have a subspace

[sharp smelt]

dim(ker(S)) is more

[pine jay]

Indeed!

So now your rank nullify equality has turned into an inequality

And you’re almost done

Just write that inequality up

And you might see what can be done as a last step

[sharp smelt]

$dim(ker(S)) \leq dim( im(T|{ker(ST)})+ dim(Ker(T|{ker(ST)})$

[pine jay]

Oh the other way

[sharp smelt]

we now expres $dim(ker(S))$ in terms of it nullity rank

[jolly parrot]

A dense set

A dense set

[sharp smelt]

which gives us the desired result

[pine jay]

Oh no I mean that was correct before but I meant the other way as in with respect to Ker ST

Or wait no

That wasn’t correct either way that way u wrote it before

[pine jay]

U had this right

Now u know dim Im T|KerST <= dim Ker S

[sharp smelt]

Yes

[pine jay]

So don’t change the LHS and only make it an equality to the right

[sharp smelt]

what

how

[pine jay]

Sorry I’m wording it poorly

[sharp smelt]

There's a lecture I want to attend now, didn't expect this to take so long

[pine jay]

I’m talking about how u now have that dim Ker ST <= dim Ker S + dim Ker T|kerST

well I’m only going slowly that’s why

want to make sure you’re understanding the proof is all

[sharp smelt]

ah

[pine jay]

Same lmao

[sharp smelt]

ah, it follows directly from here

got it

Thanks!

[pine jay]

Np!

[sharp smelt]

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[fallow bane]

Not sure when to go from here

[velvet tiger]

what is the question

[surreal relic]

Hello again @fallow bane

[fallow bane]

It’s in pink

[fallow bane]

Lmaooo hiii

[surreal relic]

Let a+ b = x

Treat it as a quadratic and factorize

[quiet tendon]

right but that's not an equation or something to "solve", do you want to find the roots?

[surreal relic]

See where you can get from there

@fallow bane

[fallow bane]

Oh sry I need help factoring it

[quiet tendon]

wasn't it already factored to begin with?

oh wait

[fallow bane]

Like a form of substitution?

[quiet tendon]

yeah you can further factorize that lmao

[quiet tendon]

yeah just mkaes it easier to work with

then resubstitute back

[surreal relic]

Yup

[quiet tendon]

so if you let x = a + b, now you're just trying to factor x^2 - 3x - 10

[fallow bane]

omg I never thought you could do it like that ngl

[quiet tendon]

and once you do that in terms of x, just replace x with a + b

[fallow bane]

how do you guys see that and just recognize what to do ;-;

[surreal relic]

You don't have to

It just makes it easier

I like my math easier

[quiet tendon]

parentheses look like the variables of a quadratic, two variables is annoying to deal with so just wrap them up in a new one with a substitution

is how i think about it

[surreal relic]

@fallow bane you got it?

[fallow bane]

Doing it rn

[surreal relic]

Just put x back in as a+b

[fallow bane]

Ohh

Ok I got the answer ty

[surreal relic]

Great job again

Lol

Please close the chat

If you done

@fallow bane

[fallow bane]

Is this correct

@surreal relic

[surreal relic]

Perfect!

👏

[fallow bane]

I just gotta remember this somehow lol but tysm

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[brisk light]

How do I do a without multiplying the expansion of both of them indivudlaly

[pulsar lark]

Hmm binomial expansion?

[brisk light]

I did but if I do it on both individually then I gotta multiply like 10 terms with each other

[little sundial]

binomial theorem?

[brisk light]

yeah I did it on each bracket, but then what

[little sundial]

wait does "term independent of x" mean the constant lmao

[brisk light]

yea

[austere quail]

If you multiply the constants you should get the right answer

You should only have two of them, one from each bracket.

[hot stone]

I'm guessing the cancelations from having the x in the bottom make this much harder than it sounds at first

[brisk light]

yeah sometimes they cancel

but idk if there's another way than ut manually multiplying every term

[hot stone]

You can figure out which terms would interact that way

Then use binomial expansion to find just those terms

And multiply them together

[brisk light]

oh ok thx

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[sacred hollow]

Help

Messed up somewhere

[limber glacier]

,rccw

[sacred hollow]

Ok found my mistake TwT, it was 37-b, I wrote the other way around

[crimson mango]

the problem is: From $1908$ to $1940$, a house could be mail-ordered from the Sears catalog. Shown here is a floor plan for the Shelburne No. $1$ model which was sold during the $1920$s. The dimensions of each room are given in feet and inches, and adjacent walls meet at right angles. In square feet, what is the area of the dining room of the Shelburne No. $1$ model? Express your answer to the nearest square foot.

[jolly parrot]

𝕯𝖆𝖗𝖐_𝖝.

[sacred hollow]

Can someone give me the answer to this please, I got 78.5

<@&286206848099549185>

[pearl pond]

@sacred hollow Has your question been resolved?

[zenith cipher]

Seems correct

But instead of calculating like that you could use sine rule

[sacred hollow]

Like?

[pearl pond]

@sacred hollow Has your question been resolved?

[zenith cipher]

I'm sorry if my handwriting is bad.
You can ask me any step you don't understand.

[pearl pond]

@sacred hollow Has your question been resolved?

[sacred hollow]

Thanks!

.close

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[tawny pewter]

anyone know how to set this up and do it?

[zenith cipher]

!status

[pearl pond]

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

[zenith cipher]

Can you describe your steps?

The bottom is square, assume length to be l.
So area of bottom is l^2.
Assume height to be h.
So volume of the container would be V=(h)(l2)...(1)

Now let's create the cost function:
It'd be
40(l^2) + 30(4hl)..(2)
Put the value of h from (1) into (2)
You'll get the cost function.
After that to minimise it. Make it's first derivative wrt l zero and find l.
After that put l back into eq(2) to get minimum cost.

[pearl pond]

@tawny pewter Has your question been resolved?

[tawny pewter]

@zenith cipher

It still says its not right

H=5.04
L=7.56

[zenith cipher]

Mine is same as well.

[zenith cipher]

I'm not sure why it's not accepting.

[pearl pond]

@tawny pewter Has your question been resolved?

[primal robin]

it looks like they want the exact answer and not the decimal form (?

[pearl pond]

@tawny pewter Has your question been resolved?

[pearl pond]

@tawny pewter Has your question been resolved?

[pearl pond]

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[long iron]

Euler phi gives you how many numbers are co prime from 1 to n. But what gives you the set of numbers?

[main oxide]

I guess simplest way is you can check gcd(k,n)=1 for 1<=k<=n to determine if they're in the set

we can probably get them in better ways, like remove multiples of distinct prime factors of n from {1,2,...,n} kind of like the sieve of eratosthenes I suppose

probably better/faster algorithms out there, depends on what you need it for I guess

[long iron]

is there a not a name for that set?

[summer imp]

Z/Zn with multiplication is essentially that set

Since any element mod n has to be coprime to n in order to have a multiplicative inverse

[long iron]

thank you

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[wind meadow]

I dont get how this relates to this equation

[wind meadow]

why r theta cos theta instead of just r cos theta

and how do you know to use that

I do understand getting the equation in terms of polar coordinates which gives you rcos theta, rsin theta but idk why you use r of theta in this equation

[tender gate]

have you perhaps used ai to get that equation?

[wind meadow]

... yes... XD

class resources are limited

[tender gate]

yeah that might be why you are running into an issue

try finding the equation of the raduis of the curvature

and doing the derivative yourself

as for the r(theta) i got no clue

maybe that the radius changes depening on the angle

which makes some sense

[wind meadow]

o i see

I worked it out with r cos theta r sin theta and i get 1/r but i dont really see how that relates to the r prime and stuff

I did notice the magnitude equation on the bottom cubed which is the same as the curvature equation

[tender gate]

r' is the derivative no?

[wind meadow]

oh yeah

Im kinda stuck once I get to 1/r

[pearl pond]

@wind meadow Has your question been resolved?

[pearl pond]

@wind meadow Has your question been resolved?

[pearl pond]

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[lone sorrel]

Did I do 11 right nobody came back😭?

[unborn abyss]

your answer should look like $y = \psq x + \bsq$

[jolly parrot]

hayley is stateside!!

[lone sorrel]

Huh

[unborn abyss]

like this one

[lone sorrel]

Oh ok let me write that

[lone sorrel]

Did it!

[unborn abyss]

on the right, you mean -1/2 x

[lone sorrel]

Ohhhhhhhh so I flip it at the end

[unborn abyss]

no

you should have m = -1/2

[lone sorrel]

Ik flip the -2/1 to -1/2

[unborn abyss]

er, well, you shouldn't have -2/1 in the first place

[lone sorrel]

Wait

[unborn abyss]

i'm not sure where that came from

[lone sorrel]

Wdym?

I divided

4 & 2

[unborn abyss]

you divided what?

sure, that's how you got the "+2" part

you did $\frac{-x+4}{2}$, right?

[jolly parrot]

hayley is stateside!!

[lone sorrel]

Yes

[unborn abyss]

well, that's

$\frac{-x}2 + \frac42$

[jolly parrot]

hayley is stateside!!

[unborn abyss]

$\frac{-1x}2 + \frac42$

[jolly parrot]

hayley is stateside!!

[unborn abyss]

$-\frac12x + \frac42$

[jolly parrot]

hayley is stateside!!

[lone sorrel]

.

So how do I get the slope I don’t understand

[unborn abyss]

the same way you got it on the left 🤨 with the -6/2

just this time it's -1/2

[lone sorrel]

Ohh ok

Someone did the one on the left with me 🙂

[unborn abyss]

ok, well a lot of this is doing the same steps but with different numbers

[lone sorrel]

So now I got y=-1/2x+2

[unborn abyss]

yep great

[lone sorrel]

Ok

I’m try the next one and see if I can get it right

And put it here

[lone sorrel]

Finished

[unborn abyss]

[lone sorrel]

WAIT FR?!? I got it right!?

[lone sorrel]

If I got -x and -3 would it be 1/3 or -1/-3?

[unborn abyss]

i can't tell what you're asking

but yes you got it right

(you wouldn't normally say y = -3 though, probably c = -3)

[lone sorrel]

Need a pic of what I’m talking about?

[unborn abyss]

why don't you just do the next one

[lone sorrel]

Okay!

What do I do here?

[unborn abyss]

i like how you said okay and then didn't do the next ones lol

[lone sorrel]

I only did 2 of them lol

You want to see em?

[unborn abyss]

well, for these you can use the y-intercept

so set x = 0 and see what y is

and then compare that to the graph

[lone sorrel]

Ok I’m try em

[unborn abyss]

if you want to get a feel for how these numbers influence the graph, you can use this https://www.desmos.com/calculator/0wlqvzs5tw

[lone sorrel]

Ok

I don’t really get it

How do I make it ax+by=c form?

I just got-5/3 for counting rise over run

And the 1 from the y-intercept

[cunning comet]

if -5/3 is the slope (if x increases by 1 then y decreases by 5/3), and 1 the y-intercept then you have y = -5/3 x +1

[pearl pond]

@lone sorrel Has your question been resolved?

[lone sorrel]

Ok

But I have to put it in ax+by=c (standard form)

[cunning comet]

then move x on the other side of the equation.

[lone sorrel]

How do I set it up?

[cunning comet]

sorry, i dont understand? are you really asking for y = -5/3 x +1 is equivalent to y + 5/3 x = 1?

[lone sorrel]

.

I don’t know which answer is correct and how to figure it out

[cunning comet]

yes, you are really asking for this.

[lone sorrel]

👍

I am

[cunning comet]

you should know such techniques:

from y = -5/3 x + 1 start with

1. add 5/3x on both sides
2. multiply both sides by 3.

[lone sorrel]

Wait is it B?

.

5x+3y=3

Ty

[pearl pond]

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[hybrid oriole]

Hello, I had this question on my exam and I tried replicating it as best as possible:

I’m trying to find the lengths of AB, AC and CD
The equation to AB is: 3x+2y+c=0
AB and CD are parallel, AC is perpendicular to AB and CD
Coordinates of D are (21,0)
Also, I know that the right triangle on the bottom left is 48cm^2 (I don’t know if this has anything to do with finding the lengths of the segments)

Is there even a way to solve this?

[summer imp]

One way is to find the points A, B and C and then use the distance formulas between them to get distances.

A and B are the intersections of 3x+2y+c = 0 with the x and y axes respectively.

Using this point, and the fact that AB and AC are perpendicular, you can find the slope of AC and then its equation (since you have the point A the is common to both).

You can also find CD by considering the fact it is parallel to AB (same slope) and you know a point on it (D).

Now you can solve for the intersection of AC and CD to get the point C, and you have all 4 points.

[hybrid oriole]

I know I can find the equation of CD which is: y = -1.5x + 31.5
But I don’t understand what you mean in the second and third paragraph, how do I find the equation of AC if I don’t know the coordinates of point A?

[summer imp]

Yeah mb I though that the equation for AB was all there

[hybrid oriole]

I was stuck on this finding point A for the whole exam 😭

[summer imp]

You can still find it though using the area. You can find the intersections of AB with the axes in terms of the constant c, and then that gives you the legs of the triangle, so you can get the area in terms of c, and solve for c since you know the area is 48

[hybrid oriole]

I’m ngl, I do NOT understand this

[summer imp]

The equation of AB is 3x + 2y - c = 0, we don't know c.
The intersection with the x axis is at the point (x,0), which yields 3x - c = 0, or x = c/3.
The intersection with the y axis is at the point (0, y), which yields 2y - c = 0 , or y = c/2.

But the length of the legs of the triangles is then exactly c/3 and c/2, so the area of the triangle is c^2 / 12 = 48. You can solve this to get c = 24.
This gives you points B(8, 0) and A(0, 12).
Then you have the equation AB completely, so you can find the equation of a line perpendicular to it going through A(0,12), and then compute the intersection of that line with the line CD to get the point C.

[pearl pond]

@hybrid oriole Has your question been resolved?

[hybrid oriole]

Where do you get the 12 in c^2 / 12?

[summer imp]

Area of the triangle is ab/2

a = c/3 and b = c/2, so ab/2 = c/3 * c/2 / 2 = c^2 / (3 * 2 * 2) = c^2 / 12

[hybrid oriole]

Ok so

I sort of get it

For (x,0) x = c/3

And (0, y) y = c/2

Since the area of the triangle is 48cm^2

And the base of the triangle is y and the height is x

(xy)/2 = ((c/3 x c/2))/2 = c/12?

And c/12 is also equal to 48

Wait

c^2*

c^2 /12 = 48

c^2 =576

Square root both sides, c = 24

I get it

Thanks

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[fast sluice]

idk how to find that equation right there

Or the answer

From the problem

[hybrid oriole]

Wait, c should = -24, right?

Oh sorry I didn’t see someone else typing

[lofty aspen]

wait… what do you mean by water level?

from the bottom or the top?

[fast sluice]

..

aight guess ill end myself

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[willow ridge]

One million dollers to whoever can tell me where -1127/27 comes from?

[haughty minnow]

Yep, cant tell you

[sharp vigil]

plug in lambda = 7/3 to the characteristic polynomial

[willow ridge]

lmao

[cinder flower]

i can

[willow ridge]

Who knew parenthesis were important 🥲

Shhh

[cinder flower]

give me 1 million dollars

[willow ridge]

its okay guys

its late

[pure rapids]

$1000000 when 💀

[willow ridge]

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[rich jolt]

got this for a solution of a differential equation

[rich jolt]

now how do i simplify it ot this form and find m and n

[main oxide]

depends on what you mean by simplify, this is just an expression so you can't really solve for anything here

there are a handful of things you could do to manipulate it but idk if that'd really get you what you want either

[rich jolt]

gotta convert to this

maybe i made a mistake then while solving

hmm

[main oxide]

for what it's worth, you could write this thing as, $$\frac{1}{\ln m - \ln n} = \frac{1}{\ln(m/n)} = \log_{m/n}(e)$$

[jolly parrot]

Merosity

[rich jolt]

yup did

i found the value of the constant wrong

[main oxide]

gotcha

[rich jolt]

🤦🏻‍♂️

[main oxide]

well if you want to back up to the original differential equation we can see

or what's the original problem

[pearl pond]

@rich jolt Has your question been resolved?

[rich jolt]

@main oxide i did it correct but i made a mistake here:

made algebra mistake here

[main oxide]

I stepped away for a bit, but you ended up getting it you said? good stuff

[pearl pond]

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[hollow hamlet]

can someone explain what im missing? why isnt its just 10C6 x 6C4?

[pearl pond]

@hollow hamlet Has your question been resolved?

[pearl pond]

@hollow hamlet Has your question been resolved?

[cosmic garnet]

not 6C4

named roles are distinguishable

so 6P4 (permutation)

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[tulip wharf]

hello could i have some help with my math homework just struggling on the last two questions

[tulip wharf]

a b are calculated and correct not sure about b and c

i mean c and d

[cosmic garnet]

what happened in c

what is this

and how

[tulip wharf]

sorry i didn't know what i was doing

i accidentally wrote that

[cosmic garnet]

solve again

[tulip wharf]

but idk how to

like what do i sub in

[cosmic garnet]

same way u did a

[tulip wharf]

but do u have to make t equal zero

wait nvm

[cosmic garnet]

48 = -2t^2 + 18t + 20

what happens if you add -48 on both sides

[tulip wharf]

-28

[cosmic garnet]

so -2t^2 + 18t - 28 = 0

solve it

[tulip wharf]

OH WAIT

GOT IT

ty

so do u sub in fourty eight bc ur tryna find the time when the heights the same

[cosmic garnet]

not "sub"

you equate the given height function to 48

you are trying to find t when h = 48

and you have function of h

so equate both

[tulip wharf]

how do you do d

[cosmic garnet]

think along the lines of b and c

the flare was at h=48 when t =2 and t=7

so it was above that height for 7-2=5 seconds

[tulip wharf]

oh yeah

so i do just do that

and that's correct

[cosmic garnet]

do what

[tulip wharf]

js figure out that it's gonna be higher in between both of the

thme

[cosmic garnet]

and how will you do that

[tulip wharf]

and then js figure out the difference

so u just 7-2=5 so then it will be above the lowest visible height at 56m

[cosmic garnet]

what

no

why 7-2

those times are for when h=48

[tulip wharf]

oh yeah

do i js sub one in and see if it works

i subbed in six

and it led me to 56

is the lowest visible height just the maximum height

[cosmic garnet]

equate the height function to 56

you will get 2 values of t

these two values of t are when height is 56 (first it passes 56 at t=3, then goes up and comes down to 56 again when t=6)

so 6-3

[tulip wharf]

ohhh got it

thank you

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[wheat stream]

Ik how you do the steps im just not sure how you get infinite from the left side and minus infinite from the right side it’s very confusing

[snow sail]

you can do sign analysis

just like in algebra

find regions where the expression is positive or negative

do you know how to do that?

[wheat stream]

Umm can you explain it rq cos I learnt math in another language

[snow sail]

you know its not defined at x=-2

but everywhere else is good

[wheat stream]

Yea

[snow sail]

we know a negative divided by a negative is positive

[wheat stream]

Ofc

[snow sail]

and -4 is always negative

so, where the denominator is positive

expression in negative

[wheat stream]

So when u replace x for example by -2.1 its becomes negative didvided by negative so you get +infinite?

[snow sail]

yes

because we know expression is positive there

[wheat stream]

Well that makes more sense, and we also got infinite because it’s a number over 0?

[snow sail]

sometimes

other times no

say

[wheat stream]

Oh, how did you know that it’s infinite and not 0

[snow sail]

(x) / { x (x-2) }

this has a removable discontinuity at x=0

the limit exists there

but at x=2...

[wheat stream]

2/0”

[snow sail]

because its a non removable reciprocal discontinuity

[wheat stream]

Oh

I think I get it now thanks

[snow sail]

[pearl pond]

@wheat stream Has your question been resolved?

[wheat stream]

Can I ask another question

In this question why do we divide by e^-9x , I know that we need to divide by the biggest exponent

Like for this question we used e^4x cos it’s the largest

[wheat stream]

Why’d we take the smallest exponent in this one

[pearl pond]

@wheat stream Has your question been resolved?

[primal robin]

Hello, that's because the limit is x-> - inf (not + inf). In this case, e^x approaches zero (not e^-x), so you want the exponents to be "positive"

[wheat stream]

Oh thanks

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[gilded jacinth]

I have to differentiate this equation. I know that e^x differentiated is e^x, but do i ignore the fact that x is also raised in ^3?

[autumn fossil]

Nope

you can't ignore it

try applying chain rule

[gilded jacinth]

Is the function split between e^x and x^3?

[autumn fossil]

Yes, it's a composition of e^x and x^3

[gilded jacinth]

Thank you

[gilded jacinth]

Is it possible to reduce my result?

[autumn fossil]

hmm are you sure it's correct?

Have you appleied the product rule?

it's also a product of e^x^3 and (sin(x) + cos(x))

[gilded jacinth]

I did not

[autumn fossil]

Alright, try re-doing it with the product rule then

[gilded jacinth]

Now im confused on what i should do

Do i first do the product rule?

[autumn fossil]

Yes

because the outter-most operation is product

the function is a product of e^x^3 and (sin(x) + cos(x))

so the derivative will be

(sin(x) + cos(x)) * d/dx e^x^3
+
e^x^3 * d/dx (sin(x) + cos(x))

[gilded jacinth]

So like this?

[autumn fossil]

yes

now if you want to simplify it, you could factor out e^x^3

actually

idk if that simplfies it

it probably cant be written too niceyl

,w differentiate e^x^3 (sin(x) + cos(x))

[jolly parrot]

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[autumn fossil]

okay, yeah, they factored it out here

[gilded jacinth]

Man thats a really ugly equation

[autumn fossil]

ye

[gilded jacinth]

Im not sure how i would do that

Where do they get the two 1's from

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[midnight haven]

this is not intuitive to me at all

[north orbit]

find the total possible events when one ball is taken out

whichis

$3 \choose 1$

[jolly parrot]

mint

[midnight haven]

but this would only work if all 3 balls are different

[north orbit]

then the number of ways to 1 white ball coming out

Let E be the event of 1 white ball coming out, then P(E) should be

[midnight haven]

what if twoballs are white then we can't use this operator right

[midnight haven]

.

[north orbit]

intially there are three possible setups

either both balls are white

one ball is white, other is black

both balls are black

then we add a white ball

[midnight haven]

yeah

[north orbit]

so depending on the initial setup

we have the following cases

1. if both were intially white, we now have three white balls

2. if one was white one was black, we have two white one black

3. if both were black, then we have one white two black

so

[midnight haven]

yes

[north orbit]

we can assume each intial configuration has the probability of $\frac{1}{3}$

[jolly parrot]

mint

[north orbit]

case 1:

probabilty of drawing a white ball would be 1 since it's 3/3

case 2:

[midnight haven]

yes

[north orbit]

probablity of drawing a white ball would be 2/3

case 3:

1/3 agaiin

so overall

[midnight haven]

this is what i did

but what after that

[north orbit]

then we calculate the overall probabilty which is

$P(white) = \frac{1}{3} \times 1 + \frac{1}{3} \times \frac{2}{3] + \frac{1}{3} \times \frac{1}{3}$

$P(white) = \frac{1}{3} \times 1 + \frac{1}{3} \times \frac{2}{3] + \frac{1}{3} \times \frac{1}{3}$

FUCK

ugh give me a minute

[midnight haven]

so you add them right ?

[north orbit]

yea yea

[midnight haven]

then ?

[north orbit]

$P(\text{white}) = \frac{1}{3} + \frac{2}{9} + \frac{1}{9} = \frac{3 + 2 + 1}{9} = \frac{6}{9} = \frac{2}{3}$

[jolly parrot]

mint

[north orbit]

then you get 2/3

which is the answer

[midnight haven]

I see I got it wrong here I didn't take in account the probability of one configuration

the factor of 1/3 I mean

[north orbit]

what i was saying before was simply that total possible events when one ball is taken out is 3 choose 1, number of ways to 1 white ball coming out is 2 choose 1
then P(E) = 2 choose 1/3 choose 1 which is also 2/3

but the other solution is simpler to udnerstand!

[midnight haven]

I see

yeah

I just finished cal 2 and some linear aljebra

it felt easy

why does this feel a little tougher

[north orbit]

you almost had it

you just made a silly msitake

[midnight haven]

yeah i guess ill work on more problems

thanks

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[worthy grail]

this is kind of a chemistry/ math question

[worthy grail]

a photon falls from the 4th EL in a hydrogen atom to the 2nd EL

I need to find the joules released using the formula Delta E= -2.179* 10*-18 ( 1/4^2 - 1/2^2)

[midnight haven]

energy of photon

is -Rhc(z^2/n^2)

in nth bohr orbit

[worthy grail]

what about the formula I gave?

[midnight haven]

do you need to find energy in electron volts ?

[worthy grail]

in joules

it’d be something like this i think

[midnight haven]

this should be right

i guess

[worthy grail]

i have the first step but things go wrong after that

(1/4^2 - 1/2^2) is -3/16 i think

[midnight haven]

the energy released cant be negative

[worthy grail]

oh is the example incorrect?

[midnight haven]

i guess

yes

[worthy grail]

then it should be energy absorbed?

[midnight haven]

this is correct

[worthy grail]

it’s the algebra that comes after though that has me confused

[midnight haven]

yeah

that is just calculation

[midnight haven]

2.179*10^-18 ( 1/16 - 1/4)(-1)

calculate this

it would be basically 2.179*10^-18 (3/16)

yeah

thats pretty much it

[worthy grail]

4.085625e-19

which is like

[midnight haven]

nah dont use a calculator

[worthy grail]

4.085625* 10^19

[midnight haven]

its basically (2.18)(3)/(16)

multiplied by a factor of 10^18

[worthy grail]

what about the negative? before 3/16

[midnight haven]

yes its -3/16 but you already have negative sign on the outside

check it

[midnight haven]

here

so -(-3/16) = 3/16

[worthy grail]

oh i see

[midnight haven]

yes

yep

[worthy grail]

this would be wrong though

it would be *10^18

[midnight haven]

yeah

[worthy grail]

okay

thanks!

[midnight haven]

close

[worthy grail]

.closw

.close

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[vestal gust]

I still need help with limits and asymptotes a bit

[vestal gust]

I have gotten better at the algebra portion of it but i struggle a lot with very complex problems

these are the topics my curriculum requires you to know for limits

[pearl pond]

@vestal gust Has your question been resolved?

[pearl pond]

@vestal gust Has your question been resolved?

[pearl pond]

@vestal gust Has your question been resolved?

[pearl pond]

@vestal gust Has your question been resolved?

[pearl pond]

@vestal gust Has your question been resolved?

[pearl pond]

@vestal gust Has your question been resolved?

[coarse harbor]

What exactly do you want to discuss about limits?

Definition, methods of calculation, existence of limits, something else?

[vestal gust]

calculation of limits and limits at infinity

and asymptotes

i know that if i have a limit with a radical i need to take the conjugate

or factor if it's a polynomial expression

but i still seem to slip up a lot when the structure of the equation gets more difficult

[coarse harbor]

Ok so first let's recall that limits have the following properties:

6 and 7 might be an overkill if you know properties of continuous functions

[vestal gust]

adding this to my formula sheet notes because it's very helpful

i will upload a few examples i have worked on

heres the problems i am asked from my course

these are the exam level questions

now my instinct

for each of these

the first one is to take the left hand and right hand limit for both values in the numerator

the second and third ones are to multiply it by the conjugate root(x^2 - x) +x

the fourth one is to apply the sum rule

and the fifth one would be to test the limit from both sides of the absolute value in the denominator

however i feel like i'm stilll missing something fundamental in terms of how i'm thinking about this

limits approaching infinity are still slightly confusing

[coarse harbor]

Hold on, the sum property doesn't work here because we'll get -inf + inf which indicates that we must be careful

Those fractions should be added up and possibly simplified

[vestal gust]

hm

[coarse harbor]

Your instincts are on point, however you should be careful when applying limit properties, just like in 4

For example when solving $\lim_{x \rightarrow 0} \frac{x}{x}$ one could say that it's $\frac{0}{0}$ but that's nonsensical. The quotient rule only works if the denominator has non-zero limit, and even then, if the limit is infinity, then the rule is only applicable if the numerator limit is finite

[jolly parrot]

EQUENOS

[vestal gust]

hm

i'm still confused a bit

couldn't i take the derivative of the top and the bottom

and get a limit equal to 1

or is that not the right thinking here

i feel super jumbled up about what to do

[coarse harbor]

Yes, that's L'hopital's rule

[vestal gust]

0/0 in my understanding

[coarse harbor]

Also in case of x/x you could just cancel out x and get 1

[vestal gust]

isn't necessarily indicative of the absence of a limit

it's just that you need to take the derivative of the top and bottom

[coarse harbor]

Yes

Or use Taylor expansion and see if anything cancels out nicely

Although I'm not sure if you're allowed to use Taylor series

[vestal gust]

we have a unit on Taylor series

but that's in about

3 weeks

We're in week 2 of the course

[coarse harbor]

oh okay

[vestal gust]

in a minute i'm gonna upload some of the equations ive worked on

i still think limits at infinity is the most confusing to me

i understand in principle

like if we have the denominator going to infinity we're getting infinitely smaller

but i find i mess up on those problems a lot more

[coarse harbor]

Yes, but what if the numerator is also going to infinity?

In this case L'hopital also works

Also conceptually limits at infinity and limits at zero are the same thing. If you're solving a limit with x -> +inf, you can substitute x=1/y and solve for y -> 0+

[vestal gust]

can you elaborate a bit on this

[coarse harbor]

It's just a side note, nothing important

But the idea is

If y --> 0+, then 1/y --> +inf

[coarse harbor]

I'd rather elaborate on this instead

Have a look at these three limits:
$$\lim_{x\rightarrow +\infty} \frac{x}{x^2} \quad \lim_{x\rightarrow +\infty}\frac{x^2}{x^2} \quad \lim_{x\rightarrow +\infty}\frac{x^3}{x^2}$$

[jolly parrot]

EQUENOS

[vestal gust]

Here are some I’ve worked on

Trying to find the limit at infinity problems

[vestal gust]

I’m going to attempt these and post here

[coarse harbor]

okie

[coarse harbor]

In 35 you missed a sign

You expanded 2 + x^2 - (2 - x^2) as 0 but it's 2 x^2

In 36 you probably should've got rid of |...| instead of substituting 0.01

Also you made a typo and missed a + in the second |...|

And another note is that if a function of x is 0 at some point it doesn't mean that it appeoaches zero at another point

[vestal gust]

Explain? I didn’t catch it maybe

[coarse harbor]

General rule: never say that inf/inf = 1, that's not true. Otherwise all three limits would be equal to 1, which is not the case.
In the second and third limits you should've continue using L'hopital rule until you arrive at a/b, infty/c or c/infty, where a, b, c are finite

[vestal gust]

Ah I multiplied them incorrectly

[vestal gust]

Hm

So then the answer for 3, would be 3/2

[coarse harbor]

The first L'hopital is miscalculated

(x^3)' = 3 x^2, not 3 x

Reminder: $(x^n)' = n x^{n-1}$

[jolly parrot]

EQUENOS

[vestal gust]

Oh yeah

[coarse harbor]

Another remark is that you can cancel out things in numerator and denominator when taking limits. For example in all three limits you could start with cancelling out x

[vestal gust]

Dumb mistake on my part

Question

In theory if I’m able to reduce a problem to 1/x would it be. 0

Like for example

Let’s take the first one

Let’s say I do l’hopital’s rule and get 1/ 2x

Infinity is going to weigh the problem down

Plugging in* infinity

So would it just go to 0

[coarse harbor]

yes

[vestal gust]

So the first one should just be. 0

[coarse harbor]

Yes, the first one is zero

[vestal gust]

Hm.

Let me see if I can retry these knowing my mistakes

I realize I didn’t put the original problem back for number 1

[coarse harbor]

nice

That's correct

So as you can see, even though in all three problems both numerator and denominator approach infinity, we get three completely different answers

That's because the "speed" of going to infinity matters.

In the first example, numerator grows "slower" than denominator, so we get zero

However, in the third example numerator grows faster and we get infinity

[vestal gust]

I’m going to attempt a few more to post in here so I can really pinpoint my issues

[vestal gust]

@coarse harbor i find i have a lot of trouble working with radicals

or more trouble

do you have any helpful formula sheets for this

[coarse harbor]

Depends on which sort of problems you have

[golden birch]

L'Hopital

doesnt work in every single case

use taylor if you cant

[coarse harbor]

We already discussed Taylor, it turns out they can't use it

[golden birch]

yeah usually they wont make the question that hard

[coarse harbor]

But I also mentioned that L'hopital doesn't always work

[coarse harbor]

yep

[golden birch]

sometimes l hopital ends you up in a loop

especially if theres trigs

sin to cos then -sin and you cant get the part out of the way

[vestal gust]

I don’t know Taylor’s series yet

I will learn it

But not in this unit

[vestal gust]

Usually ones where I need to eliminate the radical. I know to multiply by the conjugate but what if I can’t?

[coarse harbor]

Please give an example

Usually roots are either obvious or solved with conjugate multiplier

[vestal gust]

Here’s an example

@coarse harbor

[coarse harbor]

Ah, I see

1 sec

So I'd just use Taylor series sum up these fractions and use properties of logarithms
Or actually, since we arrive at a 0/0 situation I'd use L'hopital

[coarse harbor]

Are you sure that's what your teacher expects you to solve without knowing Taylor series?

[proper sapphire]

Sure, this is very easy if you use L'hopital's formula.

[coarse harbor]

It takes unnecessarily much space if you use L'hopital

[proper sapphire]

if then?

is there other method without L's formula?

[coarse harbor]

Yes, the one that involves Taylor series

[proper sapphire]

maybe...

you are right.

[coarse harbor]

This one took like 2 lines to figure out that the limit is -1/2

[coarse harbor]

This one is 1 page of differentiation

[vestal gust]

The thing is we haven’t learned it yet

I know we will learn it in about 2 weeks

But I don’t know anything about it yet

[coarse harbor]

yes yes, I remember

Well, then your only option is to apply L'hopital 2 times

(express the limit as a single fraction; the numerator and denominator will both approach 0, so you apply l'hopital)

After that you'll once again have a fraction that approaches 0/0 so you apply l'hopital again

[tawdry plaza]

Hi

[vestal gust]

so after playing around with some equations myself

I think i realize that if i get any number over x

it's automatically a 0

and i can disqualify it from my final answer

if the limit is approaching infinity

right?

[coarse harbor]

Not sure what do you mean

If c≠0 and x --> 0, then lim c/x doesn't exist

However, 0/x --> 0 when x --> 0

[coarse harbor]

Oh you were talking about x --> +inf situation

Then yes, c/x --> 0

But you can't always remove it from the answer

You can only do that according to these rules: #help-39 message

Otherwise you'd never prove things like lim (1 + 1/x)^x = e

[vestal gust]

Hm

I’m still figuring out these factoring rules

Did a lot of practice with factoring out from under the radical

But I feel like there’s some stuff that still isn’t clear yet

Here’s one I got wrong

Last one I tried

Answer is 3/4 but I got 0

I multiplied by the conjugate

Ended up with 3x in the numerator, and then x in the denominator times the conjugate

Canceled x’s to get 3/sqrt(4x^2 + 3x) + 2x

Factored x^2 out of the root to get 3/ sqrt(4 + 3/x) + 2x

That simplifies to 3/ sqrt(4) + 2x

Which then gives you

3/ 2 + 2x

I missed something along the way because the answer should be 3/4

[verbal whale]

The correct answer for this is 0, indeed

,w lim (√(4x^2+3x)−2*x)/x as x \to +\infty

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by zuanon.alberto

[verbal whale]

So who wrote the exercise either put the wrong answer or maybe miswrote the function

[vestal gust]

Okay so I’m not crazy

What the fuck

Tbh I’ve been using chat gpt to generate extra problems for me to solve

So maybe there’s that

[verbal whale]

Ahn definitely yes then

[vestal gust]

Question if I’m factoring out the root

Like the value under the radical in the denominator

Why does everything else in the entire fraction not need to be divided by x^2

[verbal whale]

What do you mean? If you're working with the denominator why would you change the rest of the function?🤔

[coarse harbor]

Sorry, what happened here, exactly?

So you were trying to simplify $\frac{3}{\sqrt{4x^2+3x}+2x}$

[jolly parrot]

EQUENOS

[coarse harbor]

I think you forgot to factor out x from 2x as well

But anyways, the answer is obviously zero

Because the denominator is unbounded

[vestal gust]

That’s what I got too

I think I still have some confusion about factoring out from under the root

I understand why we factor out x^2

But why do we not do this for every value in the equation

[coarse harbor]

When factoring something out, we should factor it from each term of the expression:
$$\sqrt{4x^2+3x}+2x = x\left(\sqrt{4+\frac{3}{x}}+2\right)$$

[jolly parrot]

EQUENOS

[woeful oxide]

https://www.youtube.com/live/uWqq1wLOPqs?si=KuerNNu9jNuWnI1h

sinister is on live

[vestal gust]

So out of each term of the original expression

Not out of the fraction we made

[coarse harbor]

I meant that after factoring something out we should end up with an expression that's equal to the original expression

That's the general rule

For example $\sqrt{4x^2+3x}+2x \neq x\left(\sqrt{4+\frac{3}{x}}+2x\right)$

[jolly parrot]

EQUENOS

[vestal gust]

Sure but I mean

Okay here’s where my confusion is

We multiply by the conjugate

On top we get 3x

On the bottom we have this expression

And then we factor out of the denominator expression

We do we not apply the division by x^2 to the numerator as well

[coarse harbor]

Oh, I think you're talking about an equivalent approach

We can:

1. Just factor out x in the denominator
2. Divide both numerator and denominator by x

Both approaches achieve the same result, essentially

$\frac{3}{\sqrt{4x^2+3x}+2x} = \frac{3/x}{(\sqrt{4x^2+3x}+2x)/x} = \frac{3/x}{\sqrt{4+3/x}+2}$

[jolly parrot]

EQUENOS

[vestal gust]

But it would be fine if I just factored x^2 out from the root?

[coarse harbor]

Oh yes, if you mean this: $\frac{3}{\sqrt{4x^2+3x}+2x} = \frac{3}{x\sqrt{4+3/x}+2x}$

[jolly parrot]

EQUENOS

[vestal gust]

yeah

i think that works

new somewhat related question

we are working on derivation now

here's an example

I skipped out on 2,8,11,and 12

i'm unsure how to work with these values that are arbitrary for this kind of problem

[coarse harbor]

Just like with non-arbitrary values

For example in 2. y' = 1/2, so the slope of the tangent line is 1/2

And that line passes though (a, a/2)

Hence the equation of that line is y = 1/2 (x - a) + a/2, so y = x/2

[vestal gust]

is that really it?

what about 11 though

that seems less clear

[coarse harbor]

Same stuff

Calculate f'(x0), that will be the slope of the tangent line at x0

After that the answer is f'(x0)(x - x0) + f(x0)

(Here I assume that you know how to write the equation of a line with a given slope passing through a given point)

[vestal gust]

I know the formula for a line yeah

But