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[verbal turret]

should I do ((7.17/7)x100)-100 or 100-((7/7.17)x100)

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[lone heart]

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[covert epoch]

Question b is what I'm trying to do.
I've been able to deduce that whenever two variables(e.g y and z = 0, then x will equal -2 and 2)

[covert epoch]

And I've marked all those points on a graph

[mellow grail]

a hollow sphere

how u might ask? well in 3 dimensions the equation for a sphere is x^2+y^2+z^2 = r^2

[covert epoch]

yh

[mellow grail]

if x^2+y^2+z^2>4 it means x^2+y^2+z^2>2^2

which means u have to remove a sphere of radius 2 units on the graph

[covert epoch]

a centre of

0,0,0

cos theres no

[mellow grail]

mm not so sure about that

[covert epoch]

oh

equation of a sphere is (x-a)^ +(y-b)^2 + (z-c)^2

so since theres no number after the x y and z

i was thinking it would be a centre of 0,0,0

[mellow grail]

oh right

[median oar]

I’d think it’s centred at the origin

[mellow grail]

yeah that does make sense

[covert epoch]

and how do we know its hollow in this case

[mellow grail]

now z <= 100 so z^2 =< 10000

[median oar]

Cos of the strict inequality

[mellow grail]

u mean z<100 right

[covert epoch]

well its less than or equal to

but yh

[median oar]

No the sphere

[mellow grail]

its not visible..

[median oar]

It’s hollow cos everything inside the sphere doesn’t satisfy the first inequality

[covert epoch]

the 4 part?

[median oar]

Actually that has nothing to do with strictness of inequality

[mellow grail]

but we need to also make sure that its not the entire 3dimensional whatever right

[median oar]

Wat

[mellow grail]

we need an upper bound as well

[covert epoch]

what plane?

[median oar]

Well yes

But concerning just the sphere part

[mellow grail]

ohhhh

I see

[median oar]

Yeah I wasn’t talking about the 100 yet

[covert epoch]

its either referring to inside the sphere or the surface and the entire space outside the sphere right?

[mellow grail]

its always a sphere with a sphere of r=2 at the centre removed

we dont rlly care about size anyways

[median oar]

Everything inside the sphere doesnt satisfy the inequality

[covert epoch]

ahh ur right

ive just used the centre of the sphere to test that out

i can use 0,0,0 to substitute into x y and z right?

cos 0 ^2+ 0 ^2+ 0^2 is not greater than 4

[median oar]

Yup

So the origin is not included

[covert epoch]

what about the surface🤔

[mellow grail]

less go i was right 🙂
(this is i think the 1st time i have done a problem related to 3 dimensions)

[covert epoch]

what point could i use for that

[mellow grail]

wont be included

[covert epoch]

ig

[mellow grail]

z is the x axis perpendicular upwards to x and y right (in some places its switched, like blender)

[covert epoch]

it would only be included if it was less than or equal to?

[mellow grail]

yes

[covert epoch]

🤝🏿

thanks for the help guys

[mellow grail]

np

[covert epoch]

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[ocean seal]

1shero1

[median oar]

So you go from x=0 to x=r

Now instead of x you have rsin θ

So you’ve got rsin θ = 0 to rsin θ = r

That’s the 2nd one

[median oar]

In here

Yeah rsin θ = r

That’s x = r

Then you go oh sin θ = 1

θ = …?

So your new bounds are to integrate from 0 to pi/2

[median oar]

Look at this

x = r

x = r sin θ

r sin θ = r

sin θ = 1

Yep

[lone heart]

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[south crystal]

well i actually wanted to post the 8th question

but i dont get how to write and make calculations with the 2^- 's either maybe the logic behind them is the same as the question below?

[mellow grail]

can u translate the question?

[south crystal]

"what is the result of the following calculation"

[mellow grail]

ah i see

[mellow grail]

let the fraction to be found = x

i am not gonna tex it, cus idk how

[south crystal]

does the ... mean 2/2/2/2... infinitely

[mellow grail]

yeah

the pattern repeats

[south crystal]

but then wouldnt thay give me some 1's?

[mellow grail]

wdym by 1s?

[south crystal]

but i should know how many there are to add them?

[south crystal]

literally 1

[mellow grail]

yeah see thats the trick

[south crystal]

isnt that gonna be 2/2 so thats one "1"

[mellow grail]

its not 2/2

[south crystal]

then again above that suppose another 2/2

[mellow grail]

its 2+.../2

so u put 2+.../2 in ....

[mellow grail]

its not 2/2

[mellow grail]

notice how the pattern is formed

[south crystal]

does it mean we dont know the number on the dots?

[south crystal]

yes

[mellow grail]

yes we dont know

but it still converges

[south crystal]

hmm

should the ... be 2?

because of the pattern

[south crystal]

.

[mellow grail]

eh..hold up lemme write it out

[south crystal]

or does it like go as the others

go

2/2 but with a +2 in front

then again 2/2 ...

[mellow grail]

ok lets just for a minute

[south crystal]

(rewriting the .../2 thats already written

[mellow grail]

think that there is a finite number of ...

[south crystal]

hm

[mellow grail]

say 5

so we would have:
$$2+\frac{2+\frac{2+\frac{2+\frac{2+\frac{2}{2}}{2}}{2}}{2}}{2}$$

[ocean seal]

itzkraken.

[south crystal]

yeah okay

[south crystal]

but how do i calculate this

[mellow grail]

before doing that, if we had ... = infinity instead of 5

this would go on forever

[mellow grail]

so if u let this = x

[mellow grail]

can u write this interms of x?

[lone heart]

@south crystal Has your question been resolved?

[south crystal]

lemme see

$$x+\frac{x+\frac{x+\frac{x+\frac{x+\frac{x}{x}}{x}}{x}}{x}}{x}$$

[ocean seal]

ajax4074

[south crystal]

like this?

[lone heart]

@south crystal Has your question been resolved?

[south crystal]

is that x+1 then

hm since our x=2

should be 3 i suppose

[flat lava]

Its 4

[south crystal]

hm its not right

[south crystal]

yeah how is that?

[south crystal]

(i said this to myself)

[flat lava]

The fraction part can be looked as

2/2+the infinite part/4

And the nxt term will be
2/4+the infinite part/8

And this thing will be equal to
2+1+1/2+1/4+1/8+1/16+1/32....

Wich is equal to 4

[south crystal]

okay

thanks to both of you

.close

[lone heart]

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[lean wolf]

Just cross checking this. Thanks

[lone heart]

Please don't occupy multiple help channels.

[upbeat hornet]

You can ask this in your other help channel if you’re not working on the original problem

[lean wolf]

oh im sorry i didnt know that

how do i like end that chat?

im not sure how this works sorry

[upbeat hornet]

you mean close this channel, or end the conversation in your first one?

[lean wolf]

end the convo in the first one, coz i got tht..thank you for that

[upbeat hornet]

how about you end this one and go back to the other chat? you can ask multiple questions in one help channel

[lean wolf]

okok

[upbeat hornet]

.close

[lone heart]

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[true sandal]

can someone explain to me how they got this sketch

[true sandal]

I understand these other two

[true sandal]

but not this

can you guys teach me how to get that graph

is it just y = -x

[lone heart]

@true sandal Has your question been resolved?

[true sandal]

nope

[lone heart]

@true sandal Has your question been resolved?

[vale wigeon]

it's y ≥ -x, which is all of the stuff above the red line y=-x

[lone heart]

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[tidal stream]

Hey! How do we find the value of p which gives us the maximum area of a rectangle under the curves g(x)= e^x and h(x) = e^-2(x-p)

[echo socket]

Any progress so far?

[lone heart]

@tidal stream Has your question been resolved?

[tidal stream]

nope!

<@&286206848099549185>

[echo socket]

Let's assume that the point A has coordinates (x, 0)

Then the length of AD would be e^x, right?

[tidal stream]

yeah

[echo socket]

Note that BC has to have the same length

Okay wait before that let's introduce a variable denoting the length of AB

Let us call it a, doesn't matter

[tidal stream]

mhm

[echo socket]

Then B has coordinates (x + a, 0)

Meaning the length of BC, using a similar argument, should be e^-2(x + a - p)

But, like already mentioned, we should have AD = BC, so here comes our first equation: e^x = e^-2(x + a- p)

Before moving on, we can simplify that

Taking ln of both sides yields x = -2x - 2a + 2p and x = 2(p - a)/3

[tidal stream]

yeah mb got it!

so...

after this....

[echo socket]

So now let's find out what would the area be in terms of our variables

S = AD * AB = a * e^x

Okay maybe it could have been better to write a in terms of x, that yields a = p - 3x/2

Meaning S = (p - 3x/2)e^x

Btw there is no more information given I assume, is that right?

[tidal stream]

and now diff and make =0 and sub back into s

[tidal stream]

yeah

[echo socket]

Could you show how the problem is originally stated?

[tidal stream]

Umm we had our test this morning and i couldn't do it and was thinking about it....we havent gotten our papers back!

but yeah they gave us the 2 functions and a similar diagaram and asked us to find the value of p which gives us the max are of the rectangel

[echo socket]

Okay, I guess we need to assume p is some fixed number and solve for it only after we get the expression for S in terms of p

So it's actually finding the maximum of the maximum area

The derivative of S with respect to x becomes -3/2 * e^x + (p - 3x/2)e^x

We set it equal to 0 and divide both sides by e^x since it can't be zero

We get -3/2 + p - 3x/2 = 0 and solving for x yields x = 2p/3 - 1

And yes, like you said, we now plug in back into S

[tidal stream]

s = p here

[echo socket]

What's that

Ah you already subbed in?

[tidal stream]

the value u get once u sub it back

[echo socket]

Yeah that's correct

Which is like really weird as that expression doesn't have a bound

[tidal stream]

mhm

can we sqrt that and make it equal to a side?

coz it needs to be a square right?

[echo socket]

Are you sure it wasn't something like "so that the maximum area of such rectangle is minimal"?

[echo socket]

We aren't dealing with a square though

[tidal stream]

yeah but isnt it assumed that the max area of a rectangle only occurs if its a square right?

[echo socket]

Alright, let's check if what we got is a square

The width was a, which turned out to be p - 3x/2

Oh, which simplifies to 3/2 actually

Interesting, the width of the largest rectangle always has to be 3/2 regardless of p

And the height was e^x, which isn't a constant (it's dependent on p in the maximal rectangle case)

So, no, what we are working with is a rectangle

[tidal stream]

hmm okay...

[tidal stream]

yeah im sure

@echo socket

[echo socket]

Either way the question doesn't seem to make sense, I would like to take a look at how it's stated

[tidal stream]

yeah....

im pretty sure thats what the question said tho....

@echo socket

Anyway...youve been great help!

Is it okay if i dm you more doubts and stuff?

[echo socket]

Sure

[lone heart]

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[strange fractal]

Hi

[strange fractal]

Can someone tell me how i got the x intercepts wrong

[tidal stream]

it should be +- 3

[west girder]

Subtract

[strange fractal]

Ohhhhhhhh

Ok thnx

.close

[lone heart]

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[west girder]

No that isn’t it

(x+4)^2+9=0 is not the same as (x+4)^2=9

It’s (x+4)^2=-9

@strange fractal

[strange fractal]

Oh

And u cant square root a negative

So theres

No x intercepts

[west girder]

No real x intercepts

[strange fractal]

Ohhh ok thank u

[tidal stream]

yeah mb!

[strange fractal]

Wait wdym by real

[tidal stream]

dww

[west girder]

You’ll learn later

[tidal stream]

you can get non real sols

[strange fractal]

Are u serious am i gonna be learning about fake numbers soon

[tidal stream]

yeah imaginary numbers!

Pretty fascinating imo!

[strange fractal]

NOOOOOOOOOOOOOOOOOOOOIIIOOOOO

[tidal stream]

haha!

Just make sure then the eqn of the quadratic is in vertex form you realise where ther vertex is and the shape of the graph

then u'll easily catch it when u dont have x intercepts!

[strange fractal]

Thanks 🙏🙏🙏

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[alpine sable]

how to read the RHS after the equals sign? i understand the integrand from negative infinity up to y. f(x) is the probability density function. what is meant by dx

[celest terrace]

derivative of x

[alpine sable]

can you help me to understand what is meant when there is both a derivative and the integrand symbol in one equation

is it meaning like the integrand of f(x) in that space times the derivative of x?

[lone heart]

@alpine sable Has your question been resolved?

[vale crag]

easy answer : it's just part of the notation of the integral : \int f(x) dx means integrating f with respect to the variable x (i.e. x is the variable of integration)

hard answer https://en.wikipedia.org/wiki/Differential_form

@alpine sable

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[cyan grove]

hi :)
i need help with this problem, specifically 2.3
i got 720 degrees for 2.1, and for 2.2: p = 1/2; q = 30 deg

i dont know where to start with 2.3 :(

[vale wigeon]

channel busy please move

[vale wigeon]

h(x)/k(x) ≤ 0 precisely when one of the following holds:

• h(x) and k(x) are both positive;
• h(x) and k(x) are both negative;
• h(x) is 0, but k(x) isn't.

[cyan grove]

is my understanding correct that it will be for all the values green except for those 2 points then?

[vale wigeon]

fuck ok sorry hold on

i accideally misread my own ≤ as ≥

for ≤ you want either h and k to be of opposite signs, or h(x)=0 and k(x)≠0

some of the green bits you highlighted don't work for that, such as just to the left of -180

[cyan grove]

oh i see okay

should i just exclude that whole section between the 2 x intercepts

the first pink area up to -180

[lone heart]

@cyan grove Has your question been resolved?

[lone heart]

@cyan grove Has your question been resolved?

[lone heart]

@cyan grove Has your question been resolved?

[lone heart]

@cyan grove Has your question been resolved?

[carmine reef]

Yes

Can you figure out which endpoint you're supposed to include and which one you aren't

[lone heart]

@cyan grove Has your question been resolved?

[modern topaz]

ah

[lone heart]

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[teal plaza]

hey guys what does this statement exactly mean
"Give the expressions of the joint probability distributions fX1,X2(x,y) and FX1,X2 (x, y) in terms of γ1 and γ2 and θ. Deduce the cumulative distribution function of S, FS(s)."

[teal plaza]

where y1 (gamma) and y2 (gamma) are parameters of X1 and X2 and theta is the parameter of copulas
like clayton or frank copula has Cx
S is X1+X2
Cθ(x,y)=xy+θxy(1−x)(1−y), θ∈[−1,1]andx,y∈[0,1]

[lone heart]

@teal plaza Has your question been resolved?

[teal plaza]

.close'

.close

[lone heart]

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[obtuse folio]

I dont see what I'm doing wrong in my work?

[lone heart]

@obtuse folio Has your question been resolved?

[obtuse folio]

I literally followed this example^ :((((

I dont know what I'm doing wrong

ok whatever idk anymore

im gonna sleep

.close

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[daring rapids]

How do you determine whether this is no n linear or not

[echo socket]

Can you recall the definition of linearity?

[daring rapids]

That if you graph it it will look like a line and will go up by a set amount?

[echo socket]

f is linear iff additivity and homogeneity are satisfied, that is:
f(x + y) = f(x) + f(y)
f(ax) = af(x)

[daring rapids]

Ah ok

[echo socket]

Can you check whether f(x) = 9x/5 + 32 satisfies any of these two?

[daring rapids]

Sorry give me a minute

Ykw ill figure it out thanks

[lone heart]

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[brazen berry]

Lin Alg Question:

[brazen berry]

can someone explain how c is false and d is true?

I originally had C as true and here is the explanation i had:
Since tx is in U, then scalar multiples of tx are also in u. For values of t not equal to 0, we can multiply by 1/t to get x alone, and for t equal to 0, we get the zero vector which is part of the subspace by definition

[tardy stag]

okay, you remembered that 0 is a real number but I don't think you really thought about what you were saying

i think you proved the converse, that if x is in U then tx is in U

which is true

[ruby current]

simply let t = 0 and x be something not in U

you basically already got there

[thorn monolith]

For d , linear combinations also belong to that space

[brazen berry]

wouldnt that just give 0 vector?

[echo socket]

Yeah, and it's clear that 0 is in any subspace

[ruby current]

use my substitutions and read the statement

tell me what it translates to

[echo socket]

So when t = 0, the statement becomes "If 0 is in U, then x is in U"

Which is false assuming U is not V (can't remember whether subspaces are supposed to be proper or improper subsets)

[ruby current]

the “for all” could be confusing

[tardy stag]

improper, but ofc it doesn't matter here

[ruby current]

subspaces not necessarily proper

[echo socket]

Alright

[brazen berry]

if 0 is in U then x is in U

oh, wait

but what if its just a random x value

cuz 0 is always in the set

[ruby current]

!!

[brazen berry]

ayyyyy

[teal plaza]

which topic is this related to and what grade?

[ruby current]

that’s why i chose x such that x is not in U for a counterexample

[brazen berry]

this is linear algebra 2 in my university

[teal plaza]

thanks thanks

[brazen berry]

for d, would i use the additive inverse to prove it?

im not 100% sure how that would work

[ruby current]

for d, see this

consider the sum and difference of the given elements in U

[brazen berry]

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

therefore x is in U

am i on the right track there?

and if x is in U then -x is in U

(x+y) + (-x)

does that work?

to prove y is in U

[ruby current]

that’s even better than the one i suggested

yes, that works

[brazen berry]

cool! thank you! i understand now!

much appreciated from all of you

.close

[lone heart]

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[teal plaza]

can anyone here help me out with copulas and how to find joint cdf of 2 continous random variable?

[thorn monolith]

Cdf is continous distributive function?

[teal plaza]

cumulative

[lone heart]

@teal plaza Has your question been resolved?

[teal plaza]

<@&286206848099549185> please

humble request lmao

[lone heart]

@teal plaza Has your question been resolved?

[pale kestrel]

try an advanced channel and a more specific question

[teal plaza]

i did put : (

[verbal epoch]

what problem

[teal plaza]

i have a copula dependence function

[verbal epoch]

send it

[teal plaza]

for 2 exponential random variabe

i want t find the joint cdf

both exponential variables are not independent

[verbal epoch]

You are above grade 9?

In school

[teal plaza]

i have written in #advanced-probability

[teal plaza]

lmao

[verbal epoch]

If you are I can't help lol

Oof

[teal plaza]

are you?

[verbal epoch]

I saw it

[teal plaza]

which grade are you in?

[verbal epoch]

I have not graduated

[teal plaza]

youre still in school?

[verbal epoch]

Yes

[teal plaza]

no hate or rudeness. But sadly this is post grad level stats

so you might not be able to help me

but thanks for your time

[verbal epoch]

Oh

Its ok

[pale kestrel]

latexing your question would also help in that channel

but yeah, the activity is not as high

and ur just gonna have to wait for someone who knows. . .

[teal plaza]

what does latexing mean?

[lone heart]

@teal plaza Has your question been resolved?

[lone heart]

@teal plaza Has your question been resolved?

[pale kestrel]

writing in latex. @ocean seal

[lone heart]

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[vapid shuttle]

I'm trying to solve $$\cos{(x)}\frac{dy}{dx}+\sin{(x)}y=1$$ I don't see what is going wrong. I rewrite as $$\frac{dy}{dx}+\tan{(x)}y=\sec{(x)}$$ and then choose my integrating factor to be $$u=e^{\int{\tan{(x)}}}=\ln{(|\sec{(x)}|)}$$ But then when I multiply this factor back through I get $$\frac{dy}{dx}\ln{(|\sec{(x)}|)}+\tan{(x)}\ln{(|\sec{(x)}|)}y=\sec{(x)}\ln{(|\sec{(x)}|)}$$ Which then I should be able to rewrite the left hand side as $$(\ln{(|\sec{(x)}|)}y)'$$ but it doesn't shake out when I differentiate that, it is missing a term and not equal to he left hand side. Could anyone point out what I am missing?

[ocean seal]

austinu

[vale crag]

integral of tan(x) is ln(|sec(x)|)

you forgot the exponential for the integrating factor

@vapid shuttle

[vapid shuttle]

ah crap

yup that is what it is

ty!

.close

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[brave dagger]

Not sure how to solve these questions... the hint at the back for Q15 suggests to use the fact that 5^2 is congruent to 2^3 (mod 17)

but idk what to do after

[ocean seal]

toby____

[alpine sable]

hint: $5^{2m+1} = 5\cdot (5^2)^m$

[ocean seal]

toby____

[lone heart]

@brave dagger Has your question been resolved?

[brave dagger]

hey Toby thanks a lot for the hint

I tried it out, but I'm still stuck : ((((

so here's what im getting

$5^{2m+1}-2^{3m+1} \cong 3 \cdot 2^{3m} (mod 17)$

[ocean seal]

statisticalcat

[alpine sable]

theres a 3 in front of the 5^(2m+1)

try writing this in terms of 2^(3m)

[brave dagger]

Nope, I tried doing that rn

im getting everything but the thing to be proved

if $17|(3 \cdot 5^{2m+1} + 2^{3m+1})$ then the $2^{3m+1}$ factor next to (mod 17) must be negative....

[ocean seal]

statisticalcat

[alpine sable]

no?

can you show your work? @brave dagger

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[lone heart]

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[mortal trellis]

just look at the first two equations

(did you already post this a few weeks ago? I remember seeing it)

[fickle oyster]

i just checked

i didnt

this is the first time of me asking this question

[fickle oyster]

also yeah

wow

it looked so complicated i didnt see that the first two equations being the same giving different answers

.close

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[alpine sable]

How would I get an equation for this?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[rigid prawn]

soo you just gotta think of a function that is defined for x>3 and has limf(x)=-inf

right?

[alpine sable]

Do you still need help?

yepp

[alpine sable]

Yea

do you know what Graph A is saying?

or is asking you to draw

as X gets bigger Y gets smaller

what function could do that?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[proper vine]

how do I know the equation based on the table of values that it provides

or whatever you call this sort of chart

[tardy stag]

i mean it'll be guessing regardless

[proper vine]

can i tell what sort of

like

is it x

or x^2

or 1/2^x

is there a way to know

[coral thorn]

you can’t guess

there’s an infinite number of functions to guess from

so you can’t know the exact function

[tardy stag]

there's no way to know for sure
the exercise here is to come up with a "simple" function that matches these points

[coral thorn]

i mean you can guess

[tardy stag]

so yes, built out of multiplication, addition, exponentiation, maybe some squares or square roots, etc

since x is going up by 1 each time, it may be helpful to examine what happens to the output on each line, compared to the previous output

[proper vine]

ok

thats gonna take a while

thanks

.close

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[tender anchor]

This is a graph of f'

i need to find x values the inflection point(s)

i thought it was 10 but i was wrong and this graph is kinda wonky its confusing me

[nimble fern]

hmmm handmaded graph, this is too difficult for me lol

[coral thorn]

why’d you hand draw the slope function

use desmos

[tender anchor]

i didnt!!!

[alpine sable]

anyone help with a-level math?

[tender anchor]

its just like this lol

[lone heart]

Please read #❓how-to-get-help

[coral thorn]

bruh

[tender anchor]

its hella small too i had to zoom in

[coral thorn]

send a screenshot of the original question

[nimble fern]

that's the graph for f', not f

[tender anchor]

yeah

[coral thorn]

do you know what an inflection point is?

[tender anchor]

yes

when it changes concavity and f'' = 0

[coral thorn]

when is f’’ = 0?

[tender anchor]

4,10,14?

[coral thorn]

why 10?

[tender anchor]

bc it looks slightly cubic and the slope appears to be 0

idk

it looks like its where the concavity changes

wait nvm no its not bc f' is decreasing from (4,14) which means f'' is concave down that whole time so its not actually changing

its only changing at 4 & 14

but thats kinda throwing me off it looks cubic? does that mean anything

@coral thorn

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[tender anchor]

oh max is -4 min is 1

.close

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[lone heart]

@crisp heron Has your question been resolved?

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[lyric narwhal]

How do you factor integers in quardratic fields such as Q(√-2)? I know that, in the case of Q(√-2), every integer has a unique factorization but how do you acutally work this out? Does Q(√-2) and Q simply have the same primes because the primes in Q can't be further factorised in Q(√-2)?

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[grand karma]

Do we know of any set of numbers which are 'sparser' than the primes (I suppose I mean the percentage of numbers in the set under n grows slower than the primes at the limit, sorry for the math gibberish) where some two of the numbers can add to make any positive even number?

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@grand karma Has your question been resolved?

[waxen flame]

Non-even primes?

[plain flame]

thats a bit of an unclear question hahah but i guess you could take the set of prime pairs

or twin primes

that is, primes that only differ by 2

for example 11 and 13

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[hot bluff]

determine convergence/divergence, evaluate if it converges.

[hot bluff]

so i used l'hospital's here, but how would i find this limit by hand?

intuition tells me this converges to 0, but im trying to prove it.

[slow hound]

what class is this for

real analysis or calc

[slow hound]

if it's real analysis, you'll have to take a bit more care

[hot bluff]

this is just calc

[slow hound]

so just use what i said above

[hot bluff]

that makes sense, basically power series rule of convergence right?

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[raven dirge]

I got lost at integrating, which is at the bottom. Did i write my integral correctly in the first place?

[mortal trellis]

dont upload random files

[raven dirge]

okay ill get a photo one sec

[mortal trellis]

you didnt plug in your bounds correctly

[raven dirge]

I tried both ways so i think thas why its switched in the photo but either way I got confused at that point. Whichever term you decide to subtract, I didn't undersand what to do with the x's i guess

or how to simplify that expression

[mortal trellis]

upper bound minus lower bound

what do you get if you plug y=x^2 into the antiderivative

and what for y=sqrt x?

[mortal trellis]

I should have asked in the other order cause sqrtx is the upper bound

so after you switch around, yes

[raven dirge]

okay, ive switched them but im still confused on the simplifying to then integrate 1 and 0. Here is my revised expression

[mortal trellis]

you plugged the bounds in. why are they still there

now its just a bit of power rule

[lone heart]

@raven dirge Has your question been resolved?

[raven dirge]

this looks really wrong

[lone heart]

@raven dirge Has your question been resolved?

[raven dirge]

I integrated wrong at x^2/2, it should be x^3/3 but otherwise I still don't know how to do the rest of this problem

I went on symbolab and it said 3/4, but I didn't understand the solution

[raven dirge]

is anyone able to help :(

[twin nimbus]

@raven dirge your setup looks fine, I didn't follow the integration along carefully, but it's polynomial so I don't doubt you did it correctly. However, I do have one tip!

Because you have symmetry about the line y=x you can do the integral from x^2 to x instead of x^2 to sqrt(x) and double the result. This should make your integral simpler to compute

It also gives you a way of verifying your work

\begin{align*}
\int 3x \sqrt{x} \dd{x} &= 3 \int x^{3/2} \dd{x} \
&= 3 \cdot (2/5) x^{5/2} + C \
&= (6/5) x^{5/2} + C
\end{align*}

[ocean seal]

omnipotententity

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[steel canyon]

Hello so I didn’t have enough time for my test so I randomly picked the last answer

[steel canyon]

Give me one second

The question was

Was this right

I think it was wrong

It shoulda been the last one

.close

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[nimble marlin]

i have a physics problem

[nimble marlin]

but im not sure if this is where to post but heres the problem

[ebon sparrow]

There's a physics server in #old-network

@nimble marlin

[nimble marlin]

ah okay thanks

.close

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[raw nacelle]

for these 2

[lone heart]

@raw nacelle Has your question been resolved?

[raw nacelle]

<@&286206848099549185>

id appreciate some help

thx

[tacit arch]

What work is there

You just have some numbers

[raw nacelle]

well not necessarily my work, just my answers

[tacit arch]

Well show your work

[raw nacelle]

my work is on my notebook but i use pc discord

i can type it out

if ud like

[vapid shuttle]

That works

or a picture

I used to take pictures on my phone and email them to myself to get them on my pc to send here

now I just upload them from phone discord, or type my work

[raw nacelle]

lol i'll just type em out, thanks for the recommendation tho

-.1 < | x^2 - 16 | < .1

[ocean seal]

austinu

[raw nacelle]

close

[ocean seal]

zoumaldo.

[raw nacelle]

that gave me the decimal, which is in my answer up there

and then I took the absolute value of the difference for my 2nd answer

first picture btw

[ocean seal]

zoumaldo.

[raw nacelle]

solved to get

[ocean seal]

zoumaldo.

[raw nacelle]

^ my answer for the first blank in the 2nd pic

then subtracted c

[raw nacelle]

was wondering if this was sufficient?

[lone heart]

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[raw nacelle]

wtever

.close

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[safe cape]

hey guys

[safe cape]

you know how when you differenciate cos, it becomes -sin

[tardy stag]

✅

[safe cape]

is there such thing as differenciating -cos

[tardy stag]

well if you differentiate -cos it becomes sin

[safe cape]

and does it result in sin

[twin nimbus]

Yes, it becomes sin

[safe cape]

so like

the - rules apply

as usual

[twin nimbus]

Yup

[safe cape]

oh ok ty

.close

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[safe cape]

.reopne

.reopen

yes

[lone heart]

✅

[twin nimbus]

d/dx(-f(x)) = - d/dx(f(x))

Due to linearity of the derivative

[safe cape]

oh

ok ty

.close

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[chilly idol]

i need help on all these questions

3. not sure where to begin

4. also not sure where to begin

5. I know how to do polynomial division and i have attempted this problem, but im confused on how to do this problem when you have a gap in two exponents

[median oar]

do you know that you can uniquely define a quadratic with 3 distinct points?

[chilly idol]

no?

[median oar]

how many variables does y = ax^2 + bx + c have?

[chilly idol]

2

[median oar]

variables as in "free parameters"

we can't change y and x those can't be different

[chilly idol]

ok, so a, b, and c

[median oar]

if we changed them we wouldn't have a quadratic anymore

right

so a quadratic is defined by what a, b and c are

[chilly idol]

yup

[median oar]

so we need 3 "pieces of information"

[chilly idol]

and these three pieces of information are given in the question

[median oar]

suppose we are told 3 distinct points

[median oar]

$(x_1, y_1), (x_2, y_2), (x_3, y_3)$

[ocean seal]

frosst

[median oar]

and we are told these 3 points lie on the quadratic

[chilly idol]

yup

[median oar]

then we can construct such a system

$y_1 = ax_1^2+bx_1+c\
y_2 = ax_2^2+bx_2+c\
y_3 = ax_3^2+bx_3+c$

[ocean seal]

frosst

[chilly idol]

so, by making this system, we could combine all three of these to make the quadratic equation

[median oar]

yes

[dry shard]

The second option would be to use Lagrage Interpolation

[median oar]

but actually we are going to do something different to this

because if we look at our points

they are special!

they are not just any 3 points on the quadratic

what do you notice about the 3 points given?

[median oar]

this was just to show you that a quadratic is uniquely defined by 3 distinct points

[chilly idol]

they have a negative symbol?

ok

[median oar]

no, what do you see in the first 2 points

(-5, 0) and (2, 0)

[chilly idol]

the x intercept

[median oar]

they are the x intercepts!!

[chilly idol]

ok

[median oar]

we know another form of the quadratic besides this

$y = a(x-b)(x-c)$

[ocean seal]

frosst

[median oar]

where b and c are the roots of the equation

[chilly idol]

vertex form?

[median oar]

not quite

[chilly idol]

ok

[median oar]

we know from the points that -5 and 2 are roots right?

[chilly idol]

yup

[median oar]

(also note these a, b, c's are not the same ones as the one before)

[median oar]

then from here, we can get

$y=a(x+5)(x-2)$

[ocean seal]

frosst

[chilly idol]

and now we plug in the third point

[median oar]

then we just need this quadratic to also pass through the last point

yep

[chilly idol]

ok

let me write this down

[median oar]

so this is the fundamental concept for that question

[chilly idol]

ok

[median oar]

a few ways to find the a b c's but they all use the fact we need 3 pieces of information

[chilly idol]

ok

[median oar]

ok

now

Q4

do you know what a cubic is?

[chilly idol]

i got 1/-18 = a for Q#

Q3

btw

[median oar]

,w solve for a, -1 = a(4+5)(4-2)

[ocean seal]

Results provided by WolframAlpha

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Requested by frosst#0

[median oar]

looks good

[chilly idol]

yup, a cubic is a parent function of y=x^3

ok

[median oar]

if i told you to make a quadratic with roots √2 and -3

how would you do it

[chilly idol]

use the previous formula

[median oar]

(also notice that a quadratic needs 3 pieces of information, a cubic needs 4 pieces of information)

[chilly idol]

ok

[median oar]

this means we will get a family of cubics that satisfy the given condition

because the conditions dont constrain us enough to have only 1 solution

[chilly idol]

ok

[median oar]

which formula

[chilly idol]

y=a(x-h)(x-k)

[median oar]

do you think you could extend this to cubics?

[chilly idol]

yes

[median oar]

what would it look like?

[chilly idol]

by putting another x-

[median oar]

show me

[chilly idol]

y=a(x-h)(x-k)(x-q)

h, k, and q are the roots

[median oar]

ok now make this cubic have roots 2, √7 and -√7

[chilly idol]

ok

so we plug it in

ok

i have put the roots in

[median oar]

what does it look like now

[chilly idol]

y=a(x-2)(x+sqrt(7))(x-sqrt(7))

[median oar]

there you go

that's the answer

[chilly idol]

ok

wait thats it

[median oar]

you do need a non-zero a

that's the degenerate solution

[chilly idol]

ok

so we need to find a

[median oar]

and let's see what you've done for Q5

[chilly idol]

ok

[median oar]

nah you just need some a that's not 0

and it's fine

[chilly idol]

ok

so ive pretty much got the answer

[median oar]

i would write (in an exam) $a\in\mathbb{R}\setminus{0}$

[ocean seal]

frosst

[chilly idol]

for Q5, I know how to do polynomial division but I dont know how. to do it when you have a gap in exponents

[median oar]

let's see your work

send us a picture

[chilly idol]

so in Q5

it has 2x^4-x^2

ok ill send a picture

here it is

[median oar]

,rotate

[chilly idol]

im sorry its slanted

[median oar]

well

consider that $2x^4-x^2+3x+5=2x^4+0x^3-x^2+3x+5$

[ocean seal]

frosst

[chilly idol]

i wasnt really sure what to do with that -x^2 and 2x^3

[median oar]

try putting in 0x^3 between x^4 and x^2

[chilly idol]

ok

give me a minute or so to do it

does 0x^3 + 2x^3 = 2x^3?

nvm

@median oar ok i think i got it

i got 2x^3+2x^2+x+4. remainder is 9

[median oar]

Then there you go

[chilly idol]

ok

next one is Q6

for this one are you supposed to solve for x

[rigid prawn]

find one solution

[chilly idol]

ok

i dont get it

[rigid prawn]

x^3+5x^2-2x-24=0

there is an "obvious" solution

when you are given something like this start with checking 0->1->-1->2->-2

[chilly idol]

ok

[chilly idol]

so you should plug in numbers for x

?

[rigid prawn]

yes

[chilly idol]

ok

[rigid prawn]

if the number you plug in results to 0 then that number is a solution

[chilly idol]

yup

ok let. me do it

i might need a couple of minutes

[rigid prawn]

sure but it shouldnt be too hard to do it with you head

obviously 0 doesnt work for example

[chilly idol]

yeah

1 doesnt work

-1 also wouldnt work

bingo

2 works

so x=2

x=2 is the solution?

[rigid prawn]

x=2 is one solution

might have more

[chilly idol]

ok

[rigid prawn]

when a polynomial has a solution x=r

then it can ALWAYS be written as (x-r)(something)

[chilly idol]

ok

[rigid prawn]

so how do you proceed?

[chilly idol]

not sure

[chilly idol]

your saying (x-2)(something)

?

[rigid prawn]

yes in this case x=2 is a solution

[chilly idol]

ok

[rigid prawn]

we got x^3+5x^2-2x-24=0 so (x-2)(something)=0

[chilly idol]

oh

polynomial division

[rigid prawn]

if you were able to find that (something) then you can find it s solutions

yeah

[chilly idol]

x^3+5x^2-2x-24 divided by x-2?

ok

im gonna do it real fast

[rigid prawn]

funny thing is i dont remember how the division works...i ve always just done it by myself

[chilly idol]

ok i got it

[rigid prawn]

whats the something?

[chilly idol]

ya i get that feeling

x^2+7x+12

[rigid prawn]

i can show you if you want

[chilly idol]

im a little bit short on time

almost bed time

[rigid prawn]

correct

[chilly idol]

so now that i have that whats next

[rigid prawn]

so what s this things solutions?

[chilly idol]

dont know

[rigid prawn]

well it s just ax^2+bx+c

[chilly idol]

oh

sooooooooo

the a, b, and c is our real solutions?

yay

[rigid prawn]

lmao no

[chilly idol]

oh

[rigid prawn]

to find the solutions of this you gotta find the discriminant

[chilly idol]

whats discriminant?

[chilly idol]

ok

basically

the x value

or quadratic equation?

[alpine sable]

isnt b^2 - 4ac the discriminant and not the whole formula?

[rigid prawn]

yeah but students normally find the discrim first then use it for the whole thing

[alpine sable]

oh thats interesting to know ! 👍

[rigid prawn]

bing you ve seen this before right?

i mean you ve been taught how to solve ax^2+bx+c=0

[chilly idol]

yes

[rigid prawn]

well then you can solve x^2+7x+12=0

[rigid prawn]

with this

[chilly idol]

using quadratic equation?

ok

ok

i got x = -7 + or - sqrt(1) divided by 2

well i could simplify the sqrt

[rigid prawn]

well somthing tells me that if you plug in -7 or -1/2 none of them will work

[chilly idol]

ok

[rigid prawn]

do it again or show your work if you cant work it out

[chilly idol]

ok

ill send a picture

lower left is my work for quadratic equation

[rigid prawn]

yeah this is correct...so the solutions are?

[chilly idol]

-3.5 plus or minus 1

=x

[rigid prawn]

bruh

[chilly idol]

oh wait

thats not what your asking for

[rigid prawn]

it is

[chilly idol]

oh

[rigid prawn]

do the +sign first

[chilly idol]

ok

[rigid prawn]

the solutions are (-7+1)/2 and (-7-1)/2

[chilly idol]

ok

oh

[rigid prawn]

it s -3.5+-1/2

[chilly idol]

ya

[rigid prawn]

you forgot to divide 1 by 2

so the solutions are?

[chilly idol]

wait i did

[chilly idol]

let me do

i got -4 and -3

[rigid prawn]

nice

[chilly idol]

so is it correct?

[rigid prawn]

yeah

[chilly idol]

ok

[rigid prawn]

so what are the solutions of x^3+5x^2-2x-24=0

[chilly idol]

the real solutions are -4, -3, and 2

[rigid prawn]

those are also all the solutions

[chilly idol]

ok

[rigid prawn]

a polynomial of degree 3 has 3 solutions

we found all 3 so all of them are real

[chilly idol]

ok

alr im going to stop here

i need to go to bed

[rigid prawn]

alright

[chilly idol]

ill figure out the last question on my own or ill come back