help-23

[glass carbon]

I help you, and you still complain ... lmao

[polar knot]

i see ty

[polar knot]

i wasnt i just said

it was funny that

i thought u were writing a paragraph or somehting

so i was waiting and u just typed 3 letters xd

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[hazy totem]

hey how would i go about solving this, ive tried plugging in -2x+4 for y in the upper equation but im still stuck, any tips?

[worthy hemlock]

What more did you do?

[hazy totem]

went into a quadratic equation

uh

[worthy hemlock]

Can you post your work?

[hazy totem]

(-6 +- sqrt(36-4(5)(-35))/2

ill send ss

am i on the right track though?

[worthy hemlock]

Can't tell, that format is horrible to read

[hazy totem]

true

ok my phone died

i cant send ss

am i supposed to go through quad formula tho

thats all i need

[worthy hemlock]

Looks like it

[hazy totem]

alr

if i have better format and stil lcant figure it out ill repost

ty for the lead

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[warm rose]

I just need help getting started because idk where to start

[glass carbon]

to find it set y' to 0

then solve

[warm rose]

Yeyey

I didnt think of that

Thanks

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[barren falcon]

Can someone help me with this?

[barren falcon]

This is what I did

<@&286206848099549185>

[safe radish]

@barren falcon Has your question been resolved?

[barren falcon]

<@&286206848099549185>

[barren falcon]

no

<@&286206848099549185>

<@&286206848099549185>

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@barren falcon Has your question been resolved?

[barren falcon]

Ok i solved it finally haha

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[devout harbor]

This section is on trigonometric substitution

but I have absolutely no idea where to even start on this problem

all of my problems previously have had a clear x^2 or Sqrt

I'm completely lost on this one

[lean thorn]

have you done integration by parts yet?

[devout harbor]

yes

that is the only thing I could think of here

but idk why I would be doing that on a section that has pretty much exclusively been trig

I'll try that real quick

[lean thorn]

oh no wait my bad

first try a u sub with u = x^3

it's easier than the IBP

[devout harbor]

ok word

and that will lead into trig sub somehow?

[lean thorn]

uhhhh not really

it doesn't have the trig sub form

This is the form for trig sub

you'd need a sqrt in there

[devout harbor]

right that's what Ithought

but how could I possibly sub an x in here??

[lean thorn]

so this will be u-sub then an IBP

[devout harbor]

ugh

[lean thorn]

it's not too bad 🙂

[devout harbor]

thank you

yeah should be fine

[lean thorn]

well rather than x

let u = x^3

[devout harbor]

right

[lean thorn]

now find the derivative of both sides

[devout harbor]

and then du = 3x^2

honestly I think I would prefer parts first

[lean thorn]

don't forget the dx 😉

it's important here

[devout harbor]

just cause it's hard to go up powers

[lean thorn]

no I promise this way is better

I mean do both

get practice in

but this is faster

because if you start with IBP, then you have IBP --> u-sub --> IBP

this skips a step

[devout harbor]

oof

so with u = x^3

how do I get that x^5 out of there?

what does x^5 change to

[lean thorn]

you'll see 🙂

keep going!

do du = 3x^2 dx

now divide both sides by 3x^2

then you get dx = 1/3x^2 du

now sub that back into the original equation for dx

and you'll see something magical happen with the x^5 🙂

[devout harbor]

holy shit

can I swear here

how could you possibly think of something so elegant

x^5 is turned into X^3

then it's just ucos(u)/3

then in parts you have a 1 for your u'

[lean thorn]

DING DING DING

now do IBP

just playing with it lol

[devout harbor]

okay I got it right

thank you!!

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[tribal crater]

bro check if shit right

[plucky elk]

looks right

[tribal crater]

bro there actually shit load of mistakes

funny typos

bunch of terms missing f(y)

[plucky elk]

ah yea haha

just write $d\mu = f(y) dy$ and boom no more mistakes

[flat frigate]

riemann (eric tao for honorable)

[tribal crater]

ye br0

[plucky elk]

it's just these 3 then?

you have the f(y) in the other two integrals

[tribal crater]

ye

[plucky elk]

yea woulda been dumb to write variance = infinity - mean^2

[safe radish]

@tribal crater Has your question been resolved?

[tribal crater]

yee bro

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[pliant moon]

How would I use exponent laws to solve for x^3

[green lichen]

hellooo

[pliant moon]

I already know the third term is x^3

[green lichen]

you’ve learnt binomial theorem right

[pliant moon]

But I want to figure out how to do this with exponent laws

Yes I have

[green lichen]

you can make use of general term

[upbeat ridge]

Pascal's triangle might help ya

[green lichen]

I wouldn’t recommend this because it’s not very efficient

[green lichen]

if you’re working with higher powers

[green lichen]

yes

but replace a and b

[pliant moon]

that's what I have

[green lichen]

what is a and b here?

[pliant moon]

x and 2

[green lichen]

yes

[pliant moon]

so what's teh next step I should take

subbed both of them in

[green lichen]

what do you have now

[pliant moon]

I'll type it out one sec

that's what I have now

[green lichen]

yes

now you have (5Cr)(2)^r (x)^5-r

in this case what’s the power of x

for b

[pliant moon]

for b there is no power of x

[green lichen]

no for part b

[pliant moon]

oh

[green lichen]

in your question hahaha

[pliant moon]

x^3

Mb

[green lichen]

yes

the power is 3 right

[pliant moon]

yes

[green lichen]

so here 5-r = 3

and you can find r

[pliant moon]

why wouldn't we use the other r?

[green lichen]

other r?

on 2?

[pliant moon]

yeah

[green lichen]

why would we use that though

[pliant moon]

Would we only use it if we had another x?

[green lichen]

we know that the power of x is suppose to be 3 right

[pliant moon]

yeah we do

[green lichen]

and we have x^5-r

to make x^5-r = x^3

5-r = 3

we don’t touch the r in numbers

when we do this

only focus on the r on x

[pliant moon]

ok that makes sense

thanks for the help!

I'lll keep this open for 1 more minute since I'm doing practice with the biniomial theorm

[green lichen]

okay!

you’re welcome!

[pliant moon]

for this question would the r be 7?

also I just realized it looks kinda sus but

I'm using revision village NOT doing a test

[green lichen]

yes seems right

but here they want the coefficient

so plug r=7 back inside

[pliant moon]

ok thanks

[green lichen]

you’re welcome!

[pliant moon]

.close

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[pliant moon]

So my question is for this question

How would I find the indpendent value of x

just wondering how to expand my terms using expoenent rules

I have

this so far

[nova creek]

You want to find the value of r that makes all x's cancel

[pliant moon]

yeah

but with exponent rules

so like

Making an equation with the exponents to equal 0

[nova creek]

What have you tried to that end

[pliant moon]

Idk how to start the equation

But I know that r should be 2

what I started with

is 0 = 7-r +r

But I know that won't work

[nova creek]

Try combining all the xs into one exponent

Remember (a/b)^c = (a^c)/(b^c) and (ab)^c = (a^c)(b^c)

Along with the other exponent rules

[pliant moon]

combining all the xs?

[nova creek]

Yeah

Like if I had x^r * x², I could combine them to give x^(r + 2)

[pliant moon]

oh so do you mean 0 = -x -2x +3x

or something liek that

Because of the powers

oh

[nova creek]

No

[pliant moon]

Sorry just slightly confused

[nova creek]

Just use exponent rules to combine all the exponents

a^b * a^c = a^(b + c)

a^b / a^c = a^(b - c)

[pliant moon]

Wait are you saying combine all the exponents to equal 0

Because I'm trying to find r while pretending I don't know what r is

[nova creek]

Well, the whole point of combining things is to find r

But if you already know r

Then you can just plug it in

[pliant moon]

yeah but trying to learn how to get r without just looking at the question

like if I got a bigger expoenent it would be harder to look at it

[nova creek]

Then yeah, just combine all the x's

[pliant moon]

Can you pls give an example

like ik x^3/x^2 = x

like what I understand from what you're saying is multiply 1/x*1/2x*x^3/3

[nova creek]

$\left(\frac{1}{x^a}\right) * x^b * x^c = x^{b + c - a}$

[flat frigate]

Gamer Dio

[nova creek]

Don't forget the other exponents, like r and 7 - r

You can't just ignore them

[pliant moon]

oh wait so are you saying do

1/x * (7-r)2 * 3(r)

Wait what do I do with x^3

[nova creek]

What happened to the x's

(1/2x)^(7 - r) is not (7 - r)2

And (x^3/3)^r is not 3(r)

[pliant moon]

wait I just want to use powers

Right

for exponent laws

[pliant moon]

so couldn't I do 0 = 7 - r - 1 +3r or something like that

[nova creek]

No

Let's simplify the equation

We only care about x, so let's remove the other numbers

[pliant moon]

ok

[nova creek]

$\left(\frac{1}{x}\right)\left(\frac{1}{x}\right)^{7-r}\left(x^3\right)^r$

[flat frigate]

Gamer Dio

[pliant moon]

right

[nova creek]

Can you use exponent rules to combine those into one exponent?

[pliant moon]

you can combine 1/x with the other one and make it (1/x)^8-r

idk how you could put the x^3 on the top unless you could just multiply it out

[nova creek]

Remember (a/b)^c = (a^c)/(b^c)

I'm particular, (1/a)^b = (1^b)/(a^b) = 1/(a^b)

[pliant moon]

Still confused on what I did wrong

unless it's (1/x^2)

[nova creek]

No

If (1/a)^b = 1/(a^b), how can we simplify (1/x)^(8 - r)

[pliant moon]

(1/x^7-r)

[nova creek]

How did 8 - r become 7 - r

[pliant moon]

7 terms sorry that was my fault

Bc it was7 terms I made a typo somewhere

oh wait you combined the -1/x

sorry

[nova creek]

So what is (1/x)^(8 - r)?

[pliant moon]

(1/x^(8-r))

[nova creek]

Yes

And how can we combine 1/(x^(8-r)) with (x^3)^r

[pliant moon]

wait but it was 1/2

Just a little confused how we got to 1/x

[nova creek]

We're ignoring the numbers, remember

[pliant moon]

oh right

mb

[nova creek]

The only parts that matter for finding r is the x's

[pliant moon]

so now we have 1/(x^5)

[nova creek]

Why x^5

Where'd 5 come from

[pliant moon]

When we cancelled

It would be (x^3)^r/1/(x^(8-r))

[nova creek]

It's 1/(x^(8 - r)) * (x^3)^r, not (x^3)^r/1/(x^(8 - r))

And we can't combine anything without first writing (x^3)^r as one exponent

[pliant moon]

so x^3r?

[nova creek]

Yes

$\left(\frac{1}{x^{8 - r}}\right)(x^{3r})$

[flat frigate]

Gamer Dio

[pliant moon]

ok so what's the next step then

[nova creek]

Combine them

[nova creek]

.

[pliant moon]

Just did that

x^3r/x^8-r

[nova creek]

And then?

[pliant moon]

not sure what to do after?

[nova creek]

.

[pliant moon]

x^(3r-{8-r})

[nova creek]

Yes

Simplify the exponent

Combine like terms

[pliant moon]

Ok I have x^(4r-8)

Oh and then x^4(r-2) right?

and then r =2?

[nova creek]

Yeah

Though you don't really have to factor it

[safe radish]

@pliant moon Has your question been resolved?

[pliant moon]

oh

ok well ty

[safe radish]

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[frozen plover]

would this be the correct way to write the domain? not to include 0 or 2 but all other numbers

[frozen plover]

for number 17

[severe laurel]

sure

you could also write R - {0, 2} or R \ {0, 2}

[frozen plover]

alr cewl

just wanted to check teacher likes it the union way

.close

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[warm lintel]

can someone help with this? "What is U∅? Prove your answer."

[sonic sierra]

Do you have a photo of the question?

[warm lintel]

@sonic sierra yes, sorry just saw this

[sonic sierra]

I guess the union of an empty set would just be an empty set

[warm lintel]

that's what i was thinking, but im not sure how to prove it. maybe just an empty set has no members so the union of an empty set would also have no members?

[sonic sierra]

Sounds right to me

[tame charm]

hm yeah how do you prove that

contradiction?

[warm lintel]

im not sure. idk how complicated or simple im supposed to make it

Uo is the set of all objects that are elements of o. But, there are no elements of o since o is an empty set. Therefore, Uo=o. ?

where is o is the null symbol

[safe radish]

@warm lintel Has your question been resolved?

[tame charm]

wait I thought $\bigcup A$ was the union of all the sets inside $A$

[flat frigate]

a.b.s._.0.

[warm lintel]

oh no it's just UO

[warm lintel]

could i also possibly get help with 7? not sure where to start or what it means

we did the first half of the proof in class and now i think he wants us to do the second half

[safe radish]

@warm lintel Has your question been resolved?

[warm lintel]

i just need help with the first part

[safe radish]

@warm lintel Has your question been resolved?

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[marble oasis]

What do I do 😭

[safe radish]

@marble oasis Has your question been resolved?

[thick sparrow]

You have a point, one more thing you need is a slope, which can be obtained via taking derivative.

Once you obtain the slope, you can apply point-slope form to get your equation. 🙂

[marble oasis]

x/4 is the derivative

but idk what to do with x/4

[marble oasis]

<@&286206848099549185>

how is a=2 given

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@marble oasis Has your question been resolved?

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[golden lance]

Hello, everyone! I need some help. In the problem I am asked to solve the ABSOLUTE DEVIATION and COEFFICIENT OF VARIANT. I tried to solve it but there seems to be a problem on my CV. I dunno if I am doing it right. Below are the photos of my solution.

PS: I based my solution on the blue table, since that’s our lesson last time, and we are asked to solve for the deviation and cv based on that.

[safe radish]

@golden lance Has your question been resolved?

[golden lance]

<@&286206848099549185>

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@golden lance Has your question been resolved?

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@golden lance Has your question been resolved?

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[viscid grail]

need help solving
|x^2+xy-2y^2=0
|xy+x-y=1

[viscid grail]

there's 9999999999999 ways I've been taught to solve systems of equations

I need just a little guidance

like what do I do first

[spice grove]

xy + x = y +1
x(y+1) = (y+1)

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

[viscid grail]

hmm ok

ohh I see it

ty @spice grove

.close

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[astral vine]

This is my question.

The distance between two parallel lines is 3x+4y+c1 = 0 and 3x+4y+c2 = 0 is 4 where c2>c1>0. The minimum distance between the point (2, 3) and the line 3x+4y+c1 = 0 is 6. Find the value of c1+c2.

I get the answer in an inequality but it's supposed to be a number.

[astral vine]

am i at least headed in the right direction with this? i hope the handwriting is readable lol

[untold canopy]

do you know

the formula for distance from point to line?

[astral vine]

yes

[untold canopy]

so apply that here

[astral vine]

yeah i did that

wait

distance > 6

[untold canopy]

what is |ax_1 + by_1 + c|

[astral vine]

that gives c1 > 12

[untold canopy]

no the distance is not > 6

it equals 6 right?

[astral vine]

nope

minimum distance > 6

[untold canopy]

"The minimum distance between the point (2, 3) and the line 3x+4y+c1 = 0 is 6"

[astral vine]

distance > 6 then, right?

[untold canopy]

no it says is 6

which means equals

[astral vine]

it says the minimum distance is 6 though
which means it can be larger

can i take it as being equal to 6?

[untold canopy]

No

Just

Focus on the distance formula

And make it equal to 6

[spice grove]

Minimum distance of a line from a point given that the point is not on the line, would be a perpendicular from that point on the line. Hence that is what the formula is for, the perpendicular's length.

And in the question, it's mentioned that the perpendicular is actually 6.

[astral vine]

it says that the minimum distance is 6, reading that i assume it can be larger than 6 too but cant be smaller than 6

can i take it as being equal to 6

[spice grove]

Read the above two messages i wrote

you HAVE to take it to be 6, short answer

[untold canopy]

Just trust us, that’s not what it means

[astral vine]

ahhhhhhhhh

shit i get it

[untold canopy]

Good

So now

[astral vine]

thank you both :>

[untold canopy]

Back to the formula

[astral vine]

yeah

wait im gonna get the answer, i'll take a minute

[untold canopy]

Do you know how to sub in the values

[astral vine]

yes ofc

[untold canopy]

Sure

[astral vine]

so c1 = 12 right?

and |c2-c1| = 20?

[spice grove]

How

[astral vine]

did i do it wrong?

[spice grove]

I'm asking for your steps, that is all

[astral vine]

ah

okay

the formula is |ax1+by1+c1|/(a^2 + b^2)^(1/2)

[spice grove]

There's an absolute value on the top.

[astral vine]

a, b = 2, 3
x1, y1 = 3, 4

[astral vine]

ahh sorry

yes

[spice grove]

a not 2, b is not 3

[astral vine]

what

[spice grove]

x1,y,1 aren't 3,4

You got them mixed up

[astral vine]

ah

changed those two

[spice grove]

a is 3, b is 4

[astral vine]

yeah, i wrote the numbers from my notebook, probably got them mxied up

sorry

i can handle it from here, thanks again :>

i should close this now right?

[spice grove]

yes

[astral vine]

.close

[safe radish]

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[marble yarrow]

Help

[marble yarrow]

if anyone is available

[split ether]

Rewrite 18 as 2 * 3^2

[idle parrot]

Use the property $\log (ab) = \log(a) + \log(b)$

[marble yarrow]

Just the c problem

I solved for a and b

@idle parrot

[idle parrot]

You can rewrite 1/5 as 10/2

Use the property $\log (\frac ab) = \log(a) - \log(b)$

[marble yarrow]

Genius

Thanks

[idle parrot]

Sure np

[marble yarrow]

@idle parrot sorry but i cant seem to figure it out lol

Im missing on something

[idle parrot]

Hmm well when we write log without any base it is standard that it is in base 10

So log(10)

[marble yarrow]

Ye

[idle parrot]

Is $\log_{10}10$

[idle parrot]

And what do we know

[foggy salmon]

huge

[marble yarrow]

1

[idle parrot]

Yes!

[marble yarrow]

Imagine i was trying to figure it with log2 and log 3

lul

1 - log(2)

[idle parrot]

[marble yarrow]

Vielen Dank

[idle parrot]

What dat?

[foggy salmon]

dat miku

[marble yarrow]

boi

how do i write 2/10

Mathematically

From 1/5

@idle parrot

sorry lol

[idle parrot]

Umm what?

[marble yarrow]

I mean

I have

log(1/5)

how do i make it 10/2

2/10

[idle parrot]

You multiply and divide by 2

[marble yarrow]

I mean

Huh?

[idle parrot]

$\frac 15 \cdot \frac 22$

[marble yarrow]

I just write it like this

Should i write something at the end of the log

Like / *2

Or smh

[idle parrot]

We’re multiplying by 1

2/2=1

So u can write log(1/5) = log(2/10)

[safe radish]

@marble yarrow Has your question been resolved?

[safe radish]

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[craggy trout]

Hi

[craggy trout]

[ f(x)=\frac{1}{x-6} ]

[ g(x)=\frac{7}{x} + 6]

[ f(g(x))=\frac{1}{\frac{7}{x}} ]

[ g(f(x))=\frac{7}{\frac{1}{x-6}} + 6]

[flat frigate]

dopediscorduser

[craggy trout]

How would I go about reducing these composite functions?

[safe radish]

@craggy trout Has your question been resolved?

[craggy trout]

<@&286206848099549185>

[safe radish]

@craggy trout Has your question been resolved?

[craggy trout]

.close

[safe radish]

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[loud ridge]

im on the right track right

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[signal igloo]

my friend sent me this and now im confused too

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[patent vault]

how do i do this

Mark 6 points on a plane so that any 3 of them form vertices of an isosceles triangle.

[unique bison]

Are you sure it's possible?

[patent vault]

idk is it?

[steep lily]

Yes it is

I'm trying to think of a way to hint at it without giving away the answer

Try the same problem with 4 and 5 points?

[patent vault]

sure?

[steep lily]

And keep in mind that the more symmetry you have the easier it is to add another point

[patent vault]

for four points its just a square i think

not sure about five

@steep lily

[steep lily]

An isosceles triangle has 3 sides the same

[patent vault]

actually with 5 points its a pentagon

[steep lily]

So the 5 points should be placed so that a lot of the distance-

[patent vault]

no only 2 need to be the same

[steep lily]

Yes

-s are the same

Yes sorry typo

[patent vault]

ah no worries

but a hexagon doesnt seem to work for six points

[steep lily]

Can you find another way to get 4 points keeping in mind you want lots of distances the same?

Not all the same just a lot of similar ones, like a square has the sides and the diagonals and a pentagon has the sides and the... kinda diagonals

[patent vault]

wdym?

[steep lily]

In a square and pentagon, there are only two possible distances between points

Which is why you can form isosceles triangles because if you have two sides the third has to be the same length as one of the previous two

[patent vault]

right

[steep lily]

The 6 point case won't be quite as nice, but same idea of not many options

[patent vault]

a hexagon has three, so thats a problem i think

[steep lily]

It doesn't have to be a problem, the actual solution also has three, but that's not the point

The point is that you want a small number of distances and a lot of symmetry

[patent vault]

i cant think of any other shape besides a regular hexagon

[steep lily]

Can you add a fourth point to an equilateral triangle to get the triangles?

Can you do something similar for a square? (Allow degenerate triangles)

(i.e. sides that lie on top of one another, triangles aren't possible because of angles but you can get the distance idea)

[patent vault]

still not sure i understand

sorry

[steep lily]

Let me try give a different hint

I'm not sure I can without just giving the answer

Play around with adding points?

[patent vault]

ive already tried 4 and 5 points

[patent vault]

what does this mean exactly

if i add a fourth point

its not a triangle

[steep lily]

I mean so that if you pick any three of those four, it's isosceles

So really that point and any two of the equilateral one

[patent vault]

a a square works for that shape

[steep lily]

Can you get it so that three of them form an equilateral triangle

Not as a square

[patent vault]

i dont know of any other shape besides a square

[steep lily]

Can you find a point that is the same distance from all three points of an equilateral triangle?

[patent vault]

that would be in the center i believe

[steep lily]

Can you see that those four points work?

As in any three of them form an isosceles triangle

[patent vault]

they dont work tho

wait

i gotta draw this

[steep lily]

The outside three are equilateral so work by assumption

[patent vault]

yeah you are right

[steep lily]

If you pick the inside and any two on the outside, the distance to the inside one is the same

[patent vault]

what about for 5?

points

[steep lily]

Do you mean adding one to a square?

Or going from 5 to 6

[patent vault]

we just did 4 points

what would the shape be for 5

[steep lily]

In the case of a square the center works for the distances, but unfortunately the angles mean you can't get triangles

[patent vault]

hows this for 6

oh no that doesnt work

not sure

doesnt work either

im stuckkkkk

[steep lily]

A regular pentagon works for 5, add a point to those 5 to get 6

[patent vault]

hm

aha is it this

[safe radish]

@patent vault Has your question been resolved?

[patent vault]

@steep lily ?

[steep lily]

Yep

[patent vault]

cool!

.close

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[dusty bone]

i don't understand part b, why cant y have more hyphens than x?

[safe radish]

@dusty bone Has your question been resolved?

[safe radish]

@dusty bone Has your question been resolved?

[safe radish]

@dusty bone Has your question been resolved?

[fair beacon]

the idea is that each time you do the inductive clause you are adding a hyphen either to both the left and right, or just the left

so if the original string satisfies (b), the new string also satisfies (b) and you are meant to prove that using induction

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[main carbon]

Calculus help

[foggy salmon]

um so

we know the tangent will have 12 gradient

and passes through 2,6

[main carbon]

What is a gradient

I’ve never heard that term

[foggy salmon]

its like

y=mx +c

m is the gradient

[main carbon]

Oh the slope?

[foggy salmon]

um ye

[main carbon]

Okay go on please

[foggy salmon]

ye this

so uve to the the line that does that

[main carbon]

How do I do that tho

[foggy salmon]

https://m.youtube.com/watch?v=rNQ36DK6aBk

[main carbon]

I tried using 12 for the slope but I don’t think that’s right

[foggy salmon]

what did u get/do

[main carbon]

I got -12

Idk if that makes a difference

And then when I used the info on the video, I got -12x+78

[foggy salmon]

wait

nono

12 is the gradient

2,6 is the point

this graph is that of f'

u cant pick 2 points from the f' graph

[main carbon]

How do you know 12 is the gradient

[foggy salmon]

the

gradient at the point is the derivative at the point

its kinda the defn of the derivative

so the derivative of f at 2 is

f' at 2

aka 12

[main carbon]

Okay that makes sense I think

So the equation would be 12x-66 right?

[foggy salmon]

um

how did u get that

[main carbon]

From the video u sent me

The gradient is 12 so I used that for the m

And then the point (2,6)

Y-y1=m(x-x1)

Y-6=12(x-6)

And got 12x-66

[foggy salmon]

um

x1 is 2

[main carbon]

Oh my god

I’ve been making slip ups all day

So I got 6x-11

[foggy salmon]

i think

its still wrong

[main carbon]

I-

[foggy salmon]

also for lines u gotta add

y=

[foggy salmon]

[main carbon]

So it’s not y=6x-11?

[foggy salmon]

no

[main carbon]

Did u get something different

[foggy salmon]

Y-6=12(x-2)

its this right

[main carbon]

Ugh

I replaced y1 with 2 also

[foggy salmon]

lol

it happens

❤️

[main carbon]

Okay so THIS TIME 12x-18??

[foggy salmon]

yes

rmb to add

y=

[main carbon]

Okay thank you so much lol

[foggy salmon]

❤️

[main carbon]

.close

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[oblique trench]

?

[whole acorn]

is that the whole question?

[oblique trench]

from what i know yeah

is it possible to solve?

[thin bridge]

you have a function

[whole acorn]

that's just a piecewise function

[severe laurel]

collatz

[thin bridge]

there is no find "n"

[oblique trench]

oh

how would u solve?

[thin bridge]

it's like asking given
f(x) = x^2
find x

makes no sense

[oblique trench]

oh so not enough info

what info would u need to solve?

[severe laurel]

if you mean to find n such that repeatedly applying f to it never ends up at 1 thats going to take a while

[thin bridge]

have the actual question?

[oblique trench]

nope

[foggy salmon]

given x, find x

[whole acorn]

where did you get the question @oblique trench

[oblique trench]

so i would need conditions like n>2?

[oblique trench]

oh some friends challenging other to solve

and i ve got no idea so i just tried it

might be trolling each other

[severe laurel]

they are

[oblique trench]

so thats why ive come here

[thin bridge]

i mean seeing this, we all know this is supposed to be collatz

at least have the proper problem statement

[oblique trench]

that doesnt sound like high school math

so would i need a condiiton for n to solve?

[oblique trench]

like this

[thin bridge]

at least have the proper problem statement

[oblique trench]

i dont follow

eg?

[thin bridge]

it's like asking given
f(x) = x^2
find x
makes no sense

[oblique trench]

right

what would the proper problem statement be then

[severe laurel]

what properties do you want your n to have

[oblique trench]

like n>2?

[thin bridge]

its better for you to just look up collatz conjecture

[severe laurel]

you'd want to involve the function f in your properties too

[oblique trench]

right so f(x) >2

or no

[severe laurel]

sure, "find n such that f(n) > 2" is a valid problem statement

[oblique trench]

right gotcha

[severe laurel]

its probably not the problem your friends are trying to solve, though

[oblique trench]

yeah ok

[quasi bison]

your friends are definitely trolling you.

[oblique trench]

so i know its a troll then

loll yeah thank youse

thanks

.close

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[silk bough]

Help

[silk bough]

1 b and d

For 1b the relative minimum is (0,-1) innit? And maximum 1,2

[quasi bison]

yes (except you want to be talking about absolute min/max here as per the problem)

[silk bough]

What about d?

The Lines discontiniou so no answers?

[quasi bison]

the function has neither a maximum nor a minimum but that is not due to discontinuity alone

you can see how if x approaches 0 from below then f(x) gets closer and closer to 1 but never gets there

[silk bough]

Oh i see

[silk bough]

Ok wait wym

[quasi bison]

was remarking on your use of "relative minimum" as opposed to "absolute minimum"

just a correction on terminology nothing else

[silk bough]

Oh oke

[silk bough]

Still dont get this one

[quasi bison]

look at the part of the graph immediately to the left of x=0, that's the one i'm talking about

[silk bough]

What i see is that the 1 has been approached there

[quasi bison]

there's a hollow circle at 1

which means the graph of f includes all points going up to (0,1) except (0,1)

idk how else to say it really

[silk bough]

Oh the tiny white hole?

[quasi bison]

yes that one

[silk bough]

What about 2nd question?

What do i do? Find the critical points?

[quasi bison]

yes, but not just that

you verify that your function is differentiable on your interval (which here it will be in all cases afaict), then take its derivative, find where it equals zero, and that's your critical points

then evaluate the function at all critical points within the interval and at endpoints of the interval

get a list of values

the smallest and largest among those will be your absolute min and max

@limpid shell please open your own channel, even if you have been in this one before. also do not ping individual users

[limpid shell]

How to open that ...sorry I didn't knew it !

[silk bough]

Ok so firstly, i take the derivative? Then set it to zero?

[quasi bison]

yes that is what i said to do

[silk bough]

i got the derivative 6x^2+6x-12=0

for 2 a

[quasi bison]

well have you solved the equation 6x^2+6x-12=0 for x?

[silk bough]

I tried to but ion know how

6x^2=12-6x ?

[quasi bison]

this is a quadratic equation

do you know how to solve quadratic equations?

[silk bough]

Like that? @quasi bison

[quasi bison]

this does not really answer my question

like, sure, that's a valid algebraic move, but do you know how to solve quadratic equations?

[silk bough]

Nah

[quasi bison]

ok so you DON'T know how to solve quadratic equations

this means you need to review how to do that, pronto.

stop doing calculus and get yourself an algebra review.

[silk bough]

😅

[safe radish]

@silk bough Has your question been resolved?

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[somber spire]

Hi, I am trying to understand how to find an inverse of f in this case. Any help would be appreciated 🥰

[quasi bison]

can you show the entire problem? is f a polynomial over Z_3 or over some other ring?

[somber spire]

I am learning about NRTU cryptosystem. It gives me this example

[quasi bison]

ah uh. hm.

might not be my area of expertise, but...

,w expand (x^3+x^2-1)(x^4+x-1)

[flat frigate]

Results provided by WolframAlpha

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

Requested by Ann#0413

[quasi bison]

hm...

maybe they're talking about this expression being pointwise congruent to 1 mod 3? on Z, or something?

[somber spire]

Here is the whole description

I think I am working in a quotient of the polynomial ring

[safe radish]

@somber spire Has your question been resolved?

[peak estuary]

dont make us download random files to help you

[safe radish]

@somber spire Has your question been resolved?

[safe radish]

@somber spire Has your question been resolved?

[safe radish]

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[sharp moon]

if 'a' is in...
slot 1: 25^9
slot 2: 26 * 25^8
slot 3: 26^2 * 25^7
...
slot 10: 26^9
by this logic, their answer seems wrong

also im not sure how they got the (9 choose 2) part of this

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[storm bone]

2x^3-x^2-10x+1=0 help

[compact ferry]

$x^3 - \frac{x^2}{2} -5x + \frac{1}{2} = 0$

[flat frigate]

Gijs

[compact ferry]

i would just find a trivial solution and then use long division

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[outer swallow]

guys how do i find the apothem of a quadrangular pyramid with only sides
or atleast find the height
the side is 12cm

[outer swallow]

help

[nova creek]

Are the triangle faces equilateral?

[outer swallow]

yes

wait

let me check

i think they are

if they are then its easy

let me check the results

im so fuckign dumb

.close

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[outer swallow]

i read the question wrong

ffs

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[pallid dune]

I need to find whatever angle. I've been trying to use law of cosine but it doesn't work. Any help?

[nova creek]

Wdym whatever angle

[signal glade]

Wah you tryna find

[thin bridge]

you can start with cos law to find the third side

and then cos law again to get the other angles

[pallid dune]

I tried that but I got 22 I think? Which doesnt make much sense

[thin bridge]

show work

[safe radish]

@pallid dune Has your question been resolved?

[pallid dune]

Sorry I'm not getting notifications

[thin bridge]

you played around to much and made too many unnecessary manipulations and in the process messed up your signs

a^2 was already by itself, why bother with shuffling everything around

[pallid dune]

So the direction I'm going to is correct I just messed up the signs yeah?

[thin bridge]

direction, no

it looks like you had no idea where you were going after plugging stuff into the cosine law

a^2 was already by itself, why bother with shuffling everything around

to get the third side a, simply sqrt both sides

[pallid dune]

Oh okay thanks bud!

.close

[safe radish]

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[tired fjord]

Can someone explain line 3 of the proof, what is this method of factoring that whole chunk out? I don't really understand the process

[split ether]

They factored f(x + h)f(x) out of the denominator, and then switched signs in the nominator, so it resulted in the -1 there

[tired fjord]

Oh I see it now, thanks! Also sidequestion: Shouldn't the factor have limit notation surrounding it? I suppose this is the product rule of limits, but isn't it incorrect for the the lim not to be written?

[split ether]

The limit is applied to the product of the fractions

And yes the product rule of limits was applied, because the limits of both of the fractions exist

[tired fjord]

Ah that clarifies it. Thank you Bean ❤️

.close

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[hollow trench]

i need help

[hollow trench]

any1 here

how u do this

[split ether]

It says that y = 3, so every point should have a y-coordinate of 3

[hollow trench]

oh just that?

[split ether]

Yes, so it's 3 in each gap

[hollow trench]

thx fam

can I stay in this chat

just in case i need more help?

this a bit confusing, any idea

[split ether]

Since y-coordinate of any of the points will still be 3, you can just choose 3 random x values and put them

So (x1, 3), (x2, 3), (x3, 3)

For whatever x1, x2 and x3 you picked

[hollow trench]

@split ether I love u bro fr

??

[split ether]

This time you have to pick 3 values for x but this time the y-coordinates of each should be -4 times that value

So if I, for example, took x = 2, then y = -4 * 2 = -8

So (2, -8) will be a point

[hollow trench]

it didnt work

whats wrong

[split ether]

I didn't say that it's -4 for all of the points

Alright look for the first point you chose x = -4

So y = -4 * (-4) = 16

So the point should be (-4, 16) instead

[hollow trench]

gimme an example for the next point, what numbers do i use?

im struggling bro, explain

[formal wren]

Now it's the same as previously, but for y (the previous one was for y). So if you take x = 2:
y = -5*x
y = -5*2
y = -10

A(2, -10)

[hollow trench]

ok got it

whys the slope wrong?

[plucky elk]

slope should be negative here

rise is measured right y value minus left y value

[safe radish]

@hollow trench Has your question been resolved?

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[hollow trench]

thx for all the help

[safe radish]

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[spark yoke]

In a triangle ABC state weather the following equation is true or false:

$\sin^2 A + \sin^2 B = \sin^2 A + 2 \sin B \sin C \cos A$

[spark yoke]

I have to use one of these formula but idk how

[compact ferry]

you can simplify the equation

[spark yoke]

How tho

[compact ferry]

$ \sin B = 2 \sin C \cos A$

[plucky elk]

subtract a term from both sides

[compact ferry]

looks pretty tough tho i wont lie

[spark yoke]

Ohh shit

I just looked

I copied wrong question

My b

$\sin^2 C + \sin^2 B = \sin^2 A + 2 \sin B \sin C \cos A$

[spark yoke]

I’m really sorry

[compact ferry]

no worries haha

it could be false right?

[spark yoke]

Yes but we need to prove that

[compact ferry]

fair enough

and A, B, C represent the angles in a triangle right?

[spark yoke]

Yes

[compact ferry]

interesting

well i dont think it will hold to be honest

[spark yoke]

Well most of my classmates say it is true

[spark yoke]

Okay

[compact ferry]

wait actually i take that back

[spark yoke]

But we don’t need to make our decision based on them tho

I need to understand why and how

[compact ferry]

A, B, and C are all angles between 0 and 180 degrees right

[spark yoke]

Yes and A+B+C =180

[spark yoke]

Ohhkay I think we might get something out of this right?

We can write A= 180-(B+C)