help-28

[wicked edge]

equals this?

[cinder vale]

is correct

yep

they are the same thing

amazing isn't it

[wicked edge]

wtf wait if i plug something into x lets say

these two will equal different things wont they

[cinder vale]

I can set the constants such that they will be the same

[wicked edge]

oh shit...

[cinder vale]

C is whatever you want it to be

[wicked edge]

the constants in the natural log you mean

thats why theyre equivalent

[cinder vale]

not the natural log, just C

[wicked edge]

C is just there, we dont care about it

[cinder vale]

yes C is just there, but I think given our situation, we do care about it

go ahead and plot this in desmos

and this too, but with the C as a D instead

if they ask you to add sliders for both of them, add them

then I want you to move the slider of D so that both functions are right on top of each other in the x-y graph. i.e: they are equal

from there I hope it is clear why those two answers are equivalent

and are therefore correct

[wicked edge]

hot damn

i pulled 2 in for x and then solved

thinking if i got the same answer they'd be equivalent

that failed, but that graphing it yeah that makes sense.

[cinder vale]

righto, pretty cool isn't it

[wicked edge]

because the graphs are the same its just the constant

didnt learn this shit in calc 1

lmao is this common?

[cinder vale]

yes it is

two integrated functions are equivalent iff they only differ by a constant

that's a definition/theorem

[wicked edge]

the thing causing this weird discrepensy is the natural log correct?

or rather the values in it

[cinder vale]

yes I suppose you could see it that way

[wicked edge]

i assume this never happens without natural log?

[cinder vale]

try using log base 10 instead

instead of the natural log

I bet they are still equivalent

[wicked edge]

base 10?

[cinder vale]

in fact, have a look at the process at which I showed both answers are equivalent.
What properties did I use?
I used a log law and that's it

the log law I used works irrespective of base

[wicked edge]

i see

thank you very much

[cinder vale]

no worries mate

now here's my question for you if you don't mind me asking

where are you from?

[wicked edge]

im in kansas city if thats what youre asking, doing calc 2 but thats a question my calc 1 friend gave me

we both ended up with different answers due to how we started, but both were actually correct as you showed

[cinder vale]

nice

I'm from australia

[wicked edge]

that friend is australian

[cinder vale]

lmao

[wicked edge]

interesting

you have a good day, i need to do something

[cinder vale]

maybe that person could be me

:illuminati:

righto cya around

make sure to type .close if you're all g

[wicked edge]

.close

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[cinder vale]

nice

good luck

[wicked edge]

tyty

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[gaunt heart]

"we consider the in the set E ....1)show that.....2)show that......(notice that ..)
3) is ]0,2] included in E? justify"

[gaunt heart]

how do i start the second question

<@&286206848099549185>

[abstract vapor]

The remark in problem 2 factors the expression as x(1-y)+y(1-x)

It is key to note the behavior of this equation

In particular, if you fix an x value and increase y by one, the first term subtracts x and the second adds (1-x)

Which shows that it is decreasing in x, and similarly is decreasing in y

What is the maximum value this expression can take?

[full forum]

@gaunt heart Has your question been resolved?

[gaunt heart]

well, 1?

because N*

[gaunt heart]

what expression

x+y/xy?

how can x+y/xy be x(1-y)+y(1-x)

<@&286206848099549185>

[silk vine]

Assume (x, y) = (1, 1)

For this the fraction is 2

For any values for x and/or y larger than that the fraction gets smaller

x and y cannot get smaller than 1 so for every (x, y) the fraction is smaller than or equal to 2

It must be larger than 0 cause the enumerator must be larger than 0 and so must the denominator

So the fraction is between 0 and 2 for every (x, y)

Making E a subset of that interval

@gaunt heart

[gaunt heart]

so it can be (1,1) and (2,2)

@silk vine

[silk vine]

Or (3, 1)

Any combination of natural numbers

[gaunt heart]

wait not 3

they said prove it's between 0 and 2

[gaunt heart]

.

[silk vine]

The fraction

x and y can both be 3

[gaunt heart]

im really confused whats the relation between the remark and the fraction

[silk vine]

It shows that x+y-2xy<=0

Meaning x+y<=2xy

[gaunt heart]

i dont see the symbol of inferior there

[silk vine]

Well that’s deduced inductively

For (1,1) it’s 0

For any consecutive it decreases

[gaunt heart]

for (2,2) it's negative

o isee it now

[silk vine]

Exactly

[gaunt heart]

what do we say after deducing that x+y-2xy<=0

[silk vine]

Add 2xy on both sides

[gaunt heart]

x+y<=2xy

[silk vine]

Divide by xy (we can do that cause xy>0)

[gaunt heart]

(x+y/xy)<=2

[silk vine]

With parentheses

(x+y)/xy<=2

[gaunt heart]

why xy>0

[silk vine]

Addition before division, that’s why they are important

[gaunt heart]

oh cause they are from the set of positive integers mb

[silk vine]

Cause x>0 and y>0

[gaunt heart]

alright

(x+y)/xy<=2 then?

we prove it's inferior or equal to 2

and it can't go lower than 1

meaning we solved number 2

[silk vine]

Can’t to or below 0

Cause it’s a positive divided by a positive

It could go below 1

For (3, 3)

[gaunt heart]

0 isnt allowed

in that set

N*

[silk vine]

The fraction

Are we talking about proving the lower bound?

[gaunt heart]

proving ]0;2]

[silk vine]

Or are you stating that we have proven the upper bound for all (x, y)

[gaunt heart]

we already proved the upper bond here

[silk vine]

Yes

[gaunt heart]

and the lower bond i said since N* doesnt have 0 then ]0;2]

[silk vine]

Yeah I suppose that works

[gaunt heart]

for (3;3) the result isnt integer

[silk vine]

I think you may be misunderstanding the set definition

x and y are positive integers

[gaunt heart]

N*={1,2,3,4,5,6,7,8,..}

[silk vine]

But not all elements of E are

In fact, only two of the elements are positive integers

[gaunt heart]

i thought E was supposed to be integer

[silk vine]

No

[gaunt heart]

that means i was wrong

[silk vine]

E is a set of rational numbers

[gaunt heart]

where should i correct myself after knowing this?

[silk vine]

Only here

You need to state that a fraction can only be 0 of the enumerator is 0

Which can’t be the case here

[gaunt heart]

because x and y are positive integers

right now we proved 0 isnt in the set E

right?

[silk vine]

Yep

[gaunt heart]

how do we prove the rest of all negative numbers arent in set E

[silk vine]

And it can’t be negative for the same reason

For a negative number one must be negative

Can’t be

[gaunt heart]

okay thats the third thing we proved

all them three (upper bon/no 0/lower bond) mean E=]0;2]

[silk vine]

Not quite

E is a subset of that interval

But they’re not the same

[gaunt heart]

wait E in the question isnt the same E in the exercice?

[silk vine]

The answer to 3 is no

That’s why they’re not equal

[gaunt heart]

so thats the answer of the question

that yes E is a subset of that interval

[silk vine]

Yes

But the interval isn’t a subset of 3

[gaunt heart]

and which concerns 3 you say its no because they're not equal, who arent esual?

[silk vine]

There is an irrational number between 0 and 2

[gaunt heart]

what 3 it's supposed to be E, 3 is the number of the question

[silk vine]

pi/3 for example

No irrational number can be included in E

So not all numbers in the interval can be included in E

[gaunt heart]

i see

[silk vine]

In fact, there are infinitely many more numbers in the interval than in E

But that’s a different topic

[gaunt heart]

i understood, by the way did you confuse 3 with E here?

[gaunt heart]

.

[silk vine]

Yes

[gaunt heart]

okay then we're done

THANK YOUUU

[silk vine]

No worries

[gaunt heart]

@silk vine on the first question if i say that for every (x,y) in (N*)^2 , we have E the set of rationals

isnt it wrong because the set of rationals is R

[silk vine]

R is the set of reals

[gaunt heart]

oh okay

[gaunt heart]

what about Q

[silk vine]

That should be the rationals

[gaunt heart]

so i cant say E is the set of rationals
and therefore sqrt(2) is irrational meaning it isnt in E

[silk vine]

E is a subset of the rationals

sqrt(2) isn’t in the rationals

So it can’t be in E

[gaunt heart]

oooo subset

okay

[gaunt heart]

@silk vine problem

how do i prove x and y are positive

[silk vine]

Well it’s given

In the set definition

[gaunt heart]

N* not N*+

my bad my bad N has only positives i forgor

[silk vine]

So it’s resolved?

[gaunt heart]

yes

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[cyan trench]

Hi, I am working on a related rate practice.

[cyan trench]

Question:

I am confused as to what I am doing wrong.

Here are my steps:

1. Write down rate you want ✖️
2. Write rate(s) you have ✔️
3. Write a basic equation
4. Take derivative of both sides
5. Plug in and solve

I feel like I have to get rid of x somehow.

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@cyan trench Has your question been resolved?

[cyan trench]

<@&286206848099549185>

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@cyan trench Has your question been resolved?

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@cyan trench Has your question been resolved?

[cyan trench]

<@&286206848099549185>

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@cyan trench Has your question been resolved?

[formal scarab]

you can express x in terms of h using pythagoras

[cyan trench]

Thanks!

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[fierce roost]

How would I start 33? And what number am I trying to find? I know it’s synthetic division but idk where to go from here

[empty sapphire]

can you show the full instructions? It got cut off

[fierce roost]

Yea sorry

[empty sapphire]

,rccw

[spiral vigil]

,rccw

[empty sapphire]

I win

[fierce roost]

Lol

[empty sapphire]

Ok yeah so you'll do polynomial division of P(x)/(x-r) and show yo have no remainder

Then do it again with the result from the first division and show again that you get no remainder

so let's start off with the first part

[fierce roost]

Ok

[empty sapphire]

have you done the divison of $\frac{x^4 + 2x^2 + 8x + 5}{x+1}$?

[glossy valve]

MellowDramaLlama

[fierce roost]

Not yet because I wasn’t sure to start but I’ll do that now

[empty sapphire]

kk sounds good

the instructions just wrote a more verbose way of "divide by r = -1 two separate times"

[fierce roost]

Do I put this into equation form? Or do I continue dividing

[empty sapphire]

nice work! So that shows the first round of division.

If you want to be verbose then write out what this polynomial is and then do the synthetic division again but with the new polynomial

[fierce roost]

If I do that, what do I divide by? Is it just -1 again?

[empty sapphire]

but you can just continue on with the next iteration:

-1 | 1 -1 3 5
   | 
   -----------

[empty sapphire]

yep exactly

[fierce roost]

Okay

So after I divide, do I convert it into ax^2+bx+c?

Or just leave it as my final answer?

[empty sapphire]

What you're trying to show is that both iterations of division produced no remainder

[fierce roost]

Oh wait yeah so I’d have to solve it

[empty sapphire]

to me, that would show sufficient proof that you have a root of degree 2

sorry what do you mean solve it?

[fierce roost]

By plugging in the final polynomial=0

Or is that not what it’s asking? I apologize this unit is confusing me

[empty sapphire]

The thing that it asked you do do was to show that r = -1 is a double root. You do that by doing synthetic division 2x and showing that both divisions have a remainder of 0 (aka that's a root).

once you show that, that's sufficient enough to show that r = -1 is a double root 🙂

[fierce roost]

Alr, thanks! Would you be willing to help me with a similar problem?

[empty sapphire]

Do you want me to tell you how I'd write the answer?

[fierce roost]

Sure

[empty sapphire]

yeah sure

[fierce roost]

Would you write it out like “it’s a double root factor because after dividing twice the remainder is 0”?

[empty sapphire]

First, we will show that x^4 + 2x^2 + 8x + 5 has a single root of r = -1. Using sythetic division, we see that: 
-1 | 1  0  2  8   5
   |   -1  1 -3  -5
   ----------------
     1 -1  3  5 | 0
Since our remainder is 0, r = -1 is a root of x^4 + 2x^2 + 8x + 5. 

After our division, we get the polynomial x^3 - x^2 + 3x + 5. We want to show that, once again, r = -1 is a root of this polynomial. We can do synthetic division again. 
  -1 | 1  -1  3   5
     |    -1  2  -5
     ----------------
       1  -2  5 | 0
Since our remainder is 0, r = -1 is a root of x^3 - x^2 + 3x + 5. 

This implies that r = -1 is a double root for x^4 + 2x^2 + 8x + 5. 

Okay so it's a little verbose but I wanted to show the logic of why it's our answer

[fierce roost]

Mhm

I think our teacher is looking for something more simple but this helps

[empty sapphire]

Yeah probably lol

just showing the synthetic division with a remainder of 0 is sufficient

[fierce roost]

This is the other one I needed help with, 23

[empty sapphire]

,rccw

[fierce roost]

Once again, I’m not really sure how to set this up/what answers I’m looking for?

[empty sapphire]

ah ok. So it's just asking for any polynomial where this is true. The most simple and minimal polynomial we'll have to create is a polynomial of degree 4.

If we have roots a, b, c, d, then we can have a polynomial that is of the following form: (x-a)(x-b)(x-c)(x-d)

It's as simple as that

So just plug in your numbers where the letters are

then expand out the polynomial if you wish

[fierce roost]

Okay

[empty sapphire]

yep looks good

[fierce roost]

How would we go from there?

[empty sapphire]

I would expand it out

if you need to

you might be able to just leave it in that form, depends on your teacher

but that's literally the whole problem

[fierce roost]

Hm

Ok I think I got@it

[empty sapphire]

where the problem gets more difficult is when you have complex roots or non integer roots.

For example, if you had a polynomial with a root at 1/3, you can't do (x - 1/3) since that would cause non integral coefficients. But you can do, say (3x - 1), then you're good to go

[fierce roost]

Okay

[empty sapphire]

I would double check your factorization

$(x-2)(x+3) = x^2 + x - 6$. You have $x^2 - x - 6$

[glossy valve]

MellowDramaLlama

[fierce roost]

I redid it but got the same answer

[empty sapphire]

you should get x^4-7x^2-6x

[fierce roost]

Hm

[empty sapphire]

oh wait

hmmm

my math goofed up too lol

one second

[fierce roost]

I’m doing it one more time but starting from beginning

[empty sapphire]

lol ok

sorry I think Im tired

[fierce roost]

Yeah same

It’s a bit too late for me

[empty sapphire]

there we go lol

[fierce roost]

Ooooh I’ve been doing something stupid and dumb

[empty sapphire]

lol no worries

it's the algebra that will always get you

ALWAYS

seriously, it's the algebra that trips you up, even in your highest math courses

[fierce roost]

alright I figured it out

I did it again and got the correct answer

Thank you, I think imma turn in

You were very helpful

[empty sapphire]

yeah get some sleep

no problem, have a good one

[fierce roost]

good night

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[prime galleon]

ok

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[agile thunder]

Oops

[prime galleon]

LOL

[agile thunder]

Ok do it in #help-27

That one is open

Same for the other 4

U can figure it out xD

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[vale path]

the question was

as x approaches 2

what does f(x) equal?

i put in 2

am i wrong?

[tough marsh]

No

But why? (I know the answer, just want to test your reasoning)

Sorry, I answered a wrong question

[vale path]

because the left side and right side are both approaching the same y value (2)

so im right?

[tough marsh]

Yes

[vale path]

ok tyy

[tough marsh]

I previously answered yes because I misread your question as does f(x) approaches 2 as x approaches 2

[vale path]

its fine lol

i appreciate ur help

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[haughty glade]

What did I do wrong gor my inverse why is it wrong

[rocky vale]

,rotate

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[haughty glade]

<@&286206848099549185>

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[fiery vigil]

How can an object transfer heat if the object does not possess a discrete quantity of heat?

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@fiery vigil Has your question been resolved?

[fiery vigil]

<@&286206848099549185>

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[fiery vigil]

.close

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[hardy snow]

can someone help me wrap my head around this problem

[hardy snow]

it's exploding my brain

it's a counting problem

each green dot has 17 lines

orange has 19

pink has 18

this is too hard to imagine in my head

[ebon glacier]

When you draw a line from red to yellow f.e. No other red lines make an intersection with it right?

Nor do the yellows

Wait no nvm

[hardy snow]

there should be intersections

[ebon glacier]

Yeah right

[hardy snow]

😭

[ebon glacier]

Shit ok

[hardy snow]

wellll

i guess I can calculate all the possible paths

10*9*8

what

ok so there are 720 paths right

idk how that helps

[ebon glacier]

It doesnt

[hardy snow]

well

each green point

has 72 paths

esiy

wait

nvm idk

what about

the amount of intersections if you drew points from green to orange to pink

from green to orange there is 10 green and 8 orange points

so 80 paths

the top green point would intersect with the bottom points

[ebon glacier]

But red quantifies that

Wait

What if we find lines

From red to green

Then we multiply bu the other color

[hardy snow]

im just trying to simplify it rn to like one case

what

idk

80 paths how many intersections

[ebon glacier]

*the other color #

[hardy snow]

how many times does the top green point intersect with the 2nd top green point when going to any orange point

[ebon glacier]

Uhhh

[hardy snow]

so ther are 8 orange points

i'd say

uhh

7+6+5+4+3+2+1

[ebon glacier]

Wait

[hardy snow]

that's the answer

[ebon glacier]

Green red orange

[hardy snow]

it would intersect 28 times

[ebon glacier]

I forgot right

So green to red line count is what

[hardy snow]

wehat

what

[ebon glacier]

80

What if

You do this problem

For green red

And orange = 1

Then generalize

Make an assumption

[ebon glacier]

Prove by whatever

[hardy snow]

ok so

[ebon glacier]

Wait

[hardy snow]

10 green 9 red

90 lines

[ebon glacier]

Yeah

So set orange = 1

Find intersection count

[hardy snow]

so

[ebon glacier]

And tru to come up with a formula

[hardy snow]

8+7+6+5+4+3+2+1

*9

[ebon glacier]

Hmm ok

[hardy snow]

so

[ebon glacier]

Wait really?

[hardy snow]

9*(8+7+6+5+4+3+2+1)

so it would be....................

no

this is intersections with top green point

with all the bottom points i think

ok so then

wiat

isn't it just

90*(8+7+6+5+4+3+2+1)

ok but green to orange intersections are

80 points

8 orange

so

80*(7+6+5+4+3+2+1)

wait

ughhg

72*(8+7+6+5+4+3+2+1)

[ebon glacier]

Brain not braining

[hardy snow]

idk

bruh this is so weird

[ebon glacier]

Ill ask my teacher

Abt this

[hardy snow]

i can't ask my teacher cuz they don't want to help 🤩

I should probably ask for help again

[ebon glacier]

Is this discrete maths? Given that its counting exercise

Or comb?

[hardy snow]

IB Math HL 1-2

[ebon glacier]

Hm

How the fuck am I not solving IB math

Ok wait

Ill do it later

[hardy snow]

ok I'm just gonna ask again

!close

ok i don't know the command

.close

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[tawdry acorn]

Any tips on how to tackle this?

[tawdry acorn]

@me when you respond please

[torn jolt]

what have you tried

limit....to what?

[tawdry acorn]

I haven’t tried anything idk where to start

That’s the kinda tips I’m looking for honestly 😂

[torn jolt]

we dont even know what your limit is approaching to to know how to help you

[tawdry acorn]

That’s all the problem gives me

So I’m assuming it’s N-> infinity

Or n rather

[torn jolt]

if that's all it gives you you can probably assume it's to infty

this is a hidden limit of e

if it is approaching infinity

[tawdry acorn]

Yes I’m assuming it’s approaching infinity

How do u show that?

[torn jolt]

Wait

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[tawdry acorn]

If there is any assistance please @ me so I receive a notification

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[obtuse plank]

ok so I know how to do it

but

so when i divide (x-1) by the first equation I get

kx - 1 = 4

[onyx glen]

you mean the other way around?

[obtuse plank]

so kx = 5

[onyx glen]

hold on

kx - 1 isn't the remainder

it's got degree 1

you can and should subtract away k(x-1)

[obtuse plank]

wait

why x -1

[onyx glen]

that's... your divisor?

[obtuse plank]

no ik but why are you multiplying k by x - 1

no cause look what i did was this

[onyx glen]

ok show me all your work them

then*

[obtuse plank]

srry i cant figure out how to upload a pitcure

gimme a sec

ok finally

so is this rong

wrojg

how do i rotate this pitcure

[inland moth]

,rccw

[obtuse plank]

so it says the k is 5 right but isn't kx = 5?

or am i being stupid

[inland moth]

i mean you still have 1 x left

why dont you divide it

after subtracting the 3x^2, you still have a (k+3)x

you need to divide (k+3)x by x, not just 3

@obtuse plank

the remainder shouldnt have any x left over

[obtuse plank]

wait so what you're suggesting it

is

wait what

[inland moth]

do you see that 3x + kx -4

[obtuse plank]

yes

[inland moth]

factor the x out

[obtuse plank]

ok x(k + 3) - 4

[inland moth]

yes

now divide the x

[obtuse plank]

so ill have to divide the 4 aswell

bc what u do to one variable u do to the other aswell no?

[onyx glen]

no,

[inland moth]

no

[obtuse plank]

OH wiat nvm

[inland moth]

(k+3)x - 4 is just the same as ax-4

what do you do with that

[onyx glen]

you're looking for a multiple of (x-1) to subtract away from this polynomial to eliminate the x^1 term.

[obtuse plank]

ok

[obtuse plank]

ok so i got that, but what would i have to multiply (x-1) to get 3x + kx

i multiplied 3 first

left with kx + 3

[onyx glen]

k+3.

[obtuse plank]

wait what about the x

[onyx glen]

you had made a misstep earlier but it was not fatal

and i tried to do a course correction

work out kx+3x-4 - (k+3)(x-1) and see

[obtuse plank]

ah it makes sense now

but the question is

how did u know to multiply it by (k+3)

[inland moth]

do you know how to do long division

[obtuse plank]

lol

i think so

[inland moth]

ok sorry thats kinda bad way to ask it

you divide the first term by first term yes?

so (k+3)x divided by x

you get k+3

now multiply (k+3) by x-1

[obtuse plank]

no so when u divided by x right

like srry if this is a dumb way to put it but where

OH

WAIT

LMAO

oh i get it now

hahahah

alr i think i get it now

tysm for ur patience guys

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[narrow ermine]

Surely they're both countable no?

I mean N and NxN have the same cardinality so I can only imagine that N and NxNxN as well.

And the set of primes is also countable

So by definition you can take the bijection that links them to N and make a bijection from N^3 to P

[vast lily]

Can anyone solve this question of btech

[spice orchid]

!occupied

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[narrow ermine]

Your idea that the set N^3 is bigger than N is flawed, for the same reason that the set of even numbers has the same cardinality as N even though it feels half as big.

Yes but no

Again the set of even numbers is a subset of N

A proper subset at that

And yet there's a bijection between them

[spice orchid]

Intuition about "size" of sets breaks down a bit once they're infinite

[mighty ridge]

Since P minus an element p of P is also countable, there is a bijection from N^3 to P-{p}, so strictly injective in P

Oops

I meant countable

My phone miscorrected sorry

Azyra was right about how you can just inject twice
If you inject N^3 in N and N in P, compound and you injected N^3 in P

Making it strictly injective is then just injecting P in P-{p}

It's a strict injection in P

For the same reason n -> n+1 is a strict injection from N to N

Map 1 to the first prime (2)
2 to the 2nd prime (3)
3 to the third prime (5)
...

No primes don't have a known pattern

The only way is to list them

The obvious bijection is mapping k to the kth prime

And to make it strictly injective you can just skip the first prime for example

[spice orchid]

A function doesn't need to have a specific formula like it seems you're trying to make it have, like melo said just declaring the function to be "send k to the kth prime" is a fine description of a function

[mighty ridge]

Not possible, if you could find a formula, you would have partially solved two problems of the millenium

The main difficulty being nobody found a pattern yet

A bijection is an injection

Which is also surjective

[spice orchid]

Well every prime is going to get mapped to with this map

If you want a strict injection then just map k to the k+1th prime

[mighty ridge]

We answered 3 times about how to make it strict

2 is first prime

Not 2nd

[spice orchid]

Lol right we should've spotted that here

[mighty ridge]

You're writing the bijection

Yes

Nothing maps to 2

Otherwise it would be bijective

And you said you didn't want a bijection

[spice orchid]

You wanted a function that wasn't surjective

The codomain is just the set in which your function can take values, not required to hit all of them

That's exactly the property that defines a surjective function

There are standard bijections from N² to N that you can look up

You can just do them twice, N² x N -> N x N -> N

Nah you're totally fine

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@main pasture Has your question been resolved?

[spice orchid]

Yeah I think that's fine

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[next basalt]

x^(3)-2x-4
How do i factorise this?

[next basalt]

Since the 4 doesn't have an x in it

WIth this? https://gyazo.com/660155ab913684641f4e9d3a61c7df7d

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[rapid rain]

I suggest you find an obvious root

find a such that a^3 - 2a-4 = 0

then you can factor x^3 - 2x - 4 = (x-a)(...)

[next basalt]

But how did you know that

(x-a)(...)

Is it because factorising always is in the form (x-a) or (x+a) multiplied with (...) ?

@rapid rain

How do you find the obvious root?

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[tired drift]

i need help with this

[quick cairn]

U need to find the derivative first

Then find the critical points by solving f'(x)=0

[tired drift]

idk where to put what

[tepid root]

Do you know how to find the derivative of that function

[tired drift]

yes

[tepid root]

What is it

[tired drift]

6x^2-48x+72

[tepid root]

When that f'(x) equals 0 it means that there is a horizontal tangent in f(x) there

So its a minimum or maximum

[tired drift]

minimum

?

idk

[tepid root]

It wasnt a question lol it could be one of those two

[tired drift]

idk what to do tbh

[tepid root]

To know if its a minimum or maximum you have to find the second derivative

[tired drift]

12x-48

[tepid root]

Step 1 : find f'(x) [you did it correctly]
Step 2: find values for f'(x)=0 [do that]

[tired drift]

is the first derviative the local minimum/maximum

@tepid root

[tepid root]

Look at this

The red curve is f(x) and the blue parabola is f'(x)

(These are examples not the ones in your exercise)

[pseudo cape]

f'(x) = 0 ==> x is a possible local minimum/maximum

[tepid root]

Can you see the extreme values in the graph?

[tired drift]

can u just walk me thru it for this

[tepid root]

Equal this to 0 and solve

For x

[tired drift]

6/2

6,2*

i got 7 mins left 😂

would i just put the 2 points in the boxes

bruh

alright thanks mathematics discord server

useless ass server

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[swift fulcrum]

Hello

[swift fulcrum]

Not sure how to do the second part

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[dense bridge]

The blue circle is the correct answer from the answer key. I'm tryna figure out why it is right, the answer key gives a unclear explanation

[dense bridge]

Ik that for p(e and f) to be independent it must be equal to p(e)*p(f)

I'm just tryna find out how to figure out what p(e) and p(f) is

Like what their individual values are

<@&286206848099549185>

[barren sage]

hi Guo

the superscript C means compliment of the set right?

[dense bridge]

Ye

[barren sage]

events E and F are independent if P(E n F) = P(E) x P(F)

[dense bridge]

Yep

[barren sage]

P(E^c n F) is the intersection of the compliment of E with F

so it is F without E, p(Ec n F) = P(F) - P(E n F)

and P(E n Fc) = P(E) - P(E n F)

[dense bridge]

Right...

[barren sage]

working steps in paper

[dense bridge]

Oh I see

Ty

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[barren sage]

what

[dense bridge]

How to get the question

[barren sage]

get what question?

like answer it?

[dense bridge]

Finish the question

Ye

[barren sage]

ok one min

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[barren sage]

look i wanted to reach some way where we could deduce the result straight away but i think here you just substitute and deduce per choice

[dense bridge]

Yep

[barren sage]

so i guess going on from here it is just substitution, find P(E) and P(F) and get their intersection

[dense bridge]

Ye, I just solved it

[barren sage]

alright, sorry if i didn't help much i wanted to reach an elegant way yknow

[dense bridge]

👍

[barren sage]

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@hearty lake Has your question been resolved?

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[gilded creek]

not sure how to start this one

[plush egret]

you know $f''(x) = 0$ looks a lot like a separable equation

[glossy valve]

jan Niku

[plush egret]

maybe it helps to let $f''(x) = g'(x)$

[glossy valve]

jan Niku

[plush egret]

where $g(x)$ is the first derivative of $f$

[glossy valve]

jan Niku

[plush egret]

then you are solving $\dv{g}{x} =0$

[glossy valve]

jan Niku

[plush egret]

does that help?

[gilded creek]

what do you mean seperable equation

[plush egret]

separable differential equations means you can manipulate it to get all the f's on one side and all the x's on the other

then you can integrate both sides

[bitter beacon]

f"=0 implies that f' is constant

[plush egret]

ye

alternatively g'=0 implies g is constant

if a first order equation is less spooky

[plush egret]

convention is to start here, and then write like $$\int \dd g = \int 0 \dd x$$

[glossy valve]

jan Niku

[plush egret]

did we loose you @gilded creek

[gilded creek]

yea kinda

[plush egret]

okay we can go way further back

if you see g''=0, you should think what kind of functions you differentiate twice and get 0

you know how if you differentiate a constant, it becomes 0?

[gilded creek]

so that means g'(x)is just a constant

and f(x) is just an x+c

right?

[plush egret]

well you want to represent all the solutions

you remember how if you have a solution to a differential equation, and you find another one, then you can add them to get a new solution?

if you dont thats fine, but you should know it

this means that if you can think of a solution, you just add it to the list, then add them all up at the end

[gilded creek]

i did just learn differential equations earlier today

so this is news to me

[plush egret]

thats fine

this is a pretty important property so its good to know early

so i can think of an easy solution

say f is a constant

then f' = 0

and if you differentiate it again, it will still be 0

so f(x) = C for any constant C should work

[plush egret]

you mention x

does 2x work? how about 10x?

what happens if you differentiate 10x twice

[gilded creek]

oh yeah its still 0

[plush egret]

yea

so in general, lets call it Ax+C seems like a good solution

its a combination of the C solution and your x solution, made a bit more general

so you have your general solution, $$f(x) = Ax+C$$

[glossy valve]

jan Niku

[plush egret]

two unknown constants, but thats okay, because you have 2 more pieces of information

$f'(-4) = -2$ and $f(-4)=-2$

[glossy valve]

jan Niku

[gilded creek]

ohh i see

[plush egret]

how do you feel about moving forward from here?

[gilded creek]

so f'(x)=A

right?

[plush egret]

yup

so whats f'(-4)

[gilded creek]

-4

[plush egret]

well, -2

but i mean, what equation can we make

with respect to the unknown numbers A and C

[plush egret]

you say this

[gilded creek]

wait

so A has to be 1/2

[plush egret]

no

[gilded creek]

fuck

[plush egret]

f'(x) = A right?

[gilded creek]

oh does a just = -2

[plush egret]

but theres no x here

yea

[gilded creek]

bc theres no x input

[plush egret]

f' is just A

for all x

[gilded creek]

oh the f' (-4) was confusing me

[plush egret]

so that its f'(-4) is actually superfluous

[gilded creek]

so then it would be -2(-4)+C=-2

and C=-10

i got it right!

ty!

[plush egret]

np

[gilded creek]

how did i use this, still not sure what it means exactly

is it just when i incorporated the A=-2 into my f(x)

[plush egret]

i think at this point you should think of it like

its a tool that helps you when you are guessing answers

it tells you that you need to make sure that you find all of them

but theres no confusion if you can think of two answers

you just add them together

good exammples like the one you just did

or maybe f'' = -f

(cosine and sine both work here...)

[gilded creek]

is there a name for this rule

[plush egret]

its called the principle of superposition

[gilded creek]

ok!

thank you so much!

[plush egret]

np

[gilded creek]

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[tawdry grove]

i was a bit lost on how to take the derivative of ln(xy)

[tawdry grove]

maybe something with this?

[formal scarab]

you could split it into lnx + lny if that helps

[tawdry grove]

well

that jus gives me

1/x and 1/y

i think the y is wrong

uh

y prime?

[formal scarab]

y'/y yup

[tawdry grove]

?

could someone explain why it is not a

[formal scarab]

$\dv x(\sec2x)=\sec2x\tan2x\cdot2$ by chain rule

[glossy valve]

chlamydia

[tawdry grove]

what what

[formal scarab]

the 2x inside

[tawdry grove]

where does the extra 2 come from at the end

thats not on teh inside tho

thats in teh argument

sec2x is together

[formal scarab]

what if you were to diff sec(x^2)

[tawdry grove]

wellll then yes i would multiply by 2x

[formal scarab]

so what separates that from sec2x

you still need to diff the inside

[tawdry grove]

alright

say that iw as meant to take the product rule of this

which 2 terms would i pick

and if i picked fx and gx. would i just leave the 4 on the outside

or would i have to do something with it ?

[formal scarab]

4 is just a constant, so you just multiply the end result by 4

of d/dx(f(x)g(x))

[tawdry grove]

but then i would not be taking the derivative of it

[formal scarab]

you don't need to because it's a constant

even if you did so, its derivative is 0

so the derivative cancels out anyway

[tawdry grove]

well then it becomes 0 times the rest which is all 0

[tawdry grove]

im confused

the drerivative of a constant is 0

and if they are beign multiplied then it is all 0

[formal scarab]

$h'(x)=\dv x(4)(f(x)g(x))+4\dv x(f(x)g(x))$

[glossy valve]

chlamydia

[formal scarab]

$=0\cdot(f(x)g(x))+4\dv x(f(x)g(x))$

so you're still left with diff of f(x)g(x)

what's wrong

[tawdry grove]

welll......

[formal scarab]

[tawdry grove]

idk im at a loss

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[pearl pumice]

A straight line is parallel with another line given by 2x+6y = 5. The line goes through the point (3,4). Find the equation for the line?

[pearl pumice]

<@&286206848099549185>

[wintry whale]

so you know 2 parallel lines have the same slope

so i would begin figuring out the slope of the first line

[pearl pumice]

Yeah

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[pearl pumice]

Okok

Wait I’m in class brb

[main hinge]

Put everything but 6y on the right

then divide both sides by 6

boom, general form

$y=-2x/6+5/6$

[glossy valve]

sol

[main hinge]

so the slope is -2/6

or -1/3

[wintry whale]

then you need to figure out the y value at x=0

you know that at x=3 y=4

and y=-1/3x+a (you need to find a)

so you have 4 = -1/3 * 3 + a

and solve for a

then your equation is y=-1/3x+a (with your a value you just solved for)

[kind frigate]

look, 2x+6y = 5 <=> y = 1/3 * x + 5/6. so equation of paralel line is y=1/3 * x + a. If line goes thru (3,4), it means that in that point y=4=1/3*x+a = 1/3 * 3 + a = 3+a. If 4=3+a, whats a=?

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[austere rose]

Express sin(301) in terms of trigonometric functions of a single argument in the range [0, pi/4].

I have no idea how to do this because you can't substract 360 because then that wouldn't be in the range

[onyx glen]

you can as an intermediate step

sin(301°) = sin(-59°)

think about what to do from here

[austere rose]

I did that and then said -sin(x) = sin(-x) so it would work as -sin(59)

but its still not in the range

[onyx glen]

you can instead do sin(x) = -cos(x+90°)

[austere rose]

Oh right, I didn't think of that, thank you very much, that's very helpful

Have a nice day

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[plain rune]

$x*s=23/23

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[onyx glen]

what is this supposed to mean?

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[amber bone]

Hints please

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@amber bone Has your question been resolved?

[night bough]

System with 3 unknown no ?

[amber bone]

yes

[night bough]

I not sure but I think it's b because I have found y=ay+b

But I think I made mistakes

[amber bone]

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[torn jolt]

$y=-2x/6+5/6$

[glossy valve]

Madi💖

[torn jolt]

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[long bolt]

A,B,C are 3 partners in a business. If twice the investment of A is equal to thrice the capital of B and the capital of B is 4 times the capital of C, then out of a total profit of $29,700, the share of B is:

14000
18000
10800
14800

[long bolt]

i wrote the answer as 10,800 but it was a guess

<@&286206848099549185>

why does no one help with this question

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[slate night]

For any x c C, x o x is defined as: (x+i)(x+i) - i

[slate night]

prove that x1 o x2 o ... o xn = (x1 + i)(x2 + i) ... (xn+i)-i