help-28

[patent grotto]

cuz im confused how the discriminant can be all 3

i thought a parabola can only have one

and now we can have all 3

[rocky saddle]

,rccw

[patent grotto]

i know how to complete the work

i just dont understand how the nature of roots can be all 3

[rocky saddle]

a rational means that a number that can be written in the form of a fraction

like 5 is rational cuz 5=5/1

or 2/3 is rational

[patent grotto]

nono for a i meant

ignore the bottom part

[normal tree]

you pick d to change the nature of the roots

[patent grotto]

so basically anyone quadratic equation can have all 3 roots

[normal tree]

that's not what it says

[patent grotto]

and it just depends on the value of a variable

like im asking in general

[normal tree]

a quadratic equation has at most 2 roots

[patent grotto]

this is just an example

[patent grotto]

i mean all 3 nature of roots

mb

[normal tree]

there are quadratic equations with two real roots

one repeated root

[patent grotto]

yeah

[normal tree]

and no real roots

yes

there are examples of each

x^2 - 4 = 0 has 2 real roots

[patent grotto]

so basically that question is just dependent on d

and depending on d

[normal tree]

x^2 = 0 has one real root

x^2 + 4 = 0 has no real roots

[patent grotto]

it classifies in one of the nature of roots

hmm alright

[patent grotto]

what i said here is correct right?

or am i wrong

[normal tree]

it depends on all of the coefficients

[patent grotto]

okok understood

thanks

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[sly sail]

I'm not understanding what is being said here- specifically, the final expression of $(n+1)^3 - 1$. Why is it expanded as such?

[glossy valve]

Kiameimon | Welt Rene

[onyx glen]

you have n equalities above the line

[sly sail]

yep... but how does it factorise to $3[1^2+....+n^2] + 3[[1+....n] + n$?

[glossy valve]

Kiameimon | Welt Rene

[restive viper]

you obtain it by summing all n equalities above the line

[tulip marlin]

for future reference this technique is "telescoping"

I think ann wanted to hint rather than give the answer but oh well

[restive viper]

welp sorry then

[sly sail]

a h..........

kinda like a shorthand notation, icic

thanks shuwuri and ann and con

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[onyx glen]

i was going to say "and we add them together to produce the one at the bottom"

but i got distracted

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[untold delta]

what the heck does this plot mean

[vast fossil]

The colour at a point represents the value of whatever is being measured

And on the right you have the list of meanings of those colours

E.g. if you want to know what happens when the rostrum length is 20 and elytra length is 10, you would look where the (20, 10) points would go in Cartesian coordinates and check the colour present there

[untold delta]

why is that purple (5, 10) so weird tho

[vast fossil]

(5, 10] means that it's between 5 and 10 btw (excluding 5 and including 10)

What's so weird about it?

[untold delta]

the shape was kinda weird but i think im understanding it now

i was confused with why it like that when 15+ is neatly done

what does level mean tho

[vast fossil]

Approximate scale of whatever is being measured

It shows you what colour indicates what

E.g. yellow here corresponds to (40, 45], meaning the value would be from 40 to 45 (excluding 40 and including 45 again)

[untold delta]

Okay cheers I think im getting there

thanks for the help!

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[grave elm]

Is $(\mathbb{R}, +)$ same as $(\mathbb{R}^+, \times)$ up to isomorphism?

[glossy valve]

MathIsAlwaysRight

MathIsAlwaysRight

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@grave elm Has your question been resolved?

[cloud shore]

No

[grave elm]

thanks, why is that?

[onyx glen]

do you mean to ask if the groups (R, +) and (R^+, *) are isomorphic

if so, they are

[grave elm]

yeah, thanks. Does the fact that they're groups matter?

[grave elm]

Oh I think I understand it now

Is it isomorphic because there exists an inverse homomorphism to the f I found?

[grave elm]

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[grave raft]

I already know that this limit is equal to 0, but I don't know how to get to this answer

[sharp flame]

Maybe try a substitution

$$\frac 1x = t$$
$$\lim_{t \to \infty} te^{-t}$$
$$\lim_{t \to \infty} \frac{t}{e^t}$$

[glossy valve]

NEONPerseus

[sharp flame]

The exponential grows faster than the linear

So you have it approaching 0

[grave raft]

thank you very much, the substitution helped

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[deep fern]

How do you find the domain of this function.

[cloud moon]

@deep fern what have you tried

[deft tulip]

find out where various parts are undefined (0 denominator, log can only take > 0 input, same with square root), then it's all values except the values \ intervals that are excluded

[deep fern]

X^2 +2x not equal to 0 . And (-X^2 +X+6) positive

[deep fern]

I did that and got the answer but it's incomplete

[cloud moon]

good you are only missing number under square root should be ≥ 0

[deft tulip]

log can result in negative output

so you have to make sure you only allow values that give a positive value for the log

[deep fern]

Oh right thx

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[dry rover]

Trying to find out the volume of a solid of revolution around the y-axis for the function $f(x)=x+2$ in the interval $x\in[0, 2]$. Correct answer is $\frac{8\pi}{3}$. Could someone check my work? Cheers

[glossy valve]

Dex

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@dry rover Has your question been resolved?

[sly cove]

You're supposed to square the function before integrating, your mistake is that you multiplied by y instead

And added a 1/2 for some reason

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[ebon grove]

whats the explanation for negative numbers? this solution just ignores them

[onyx glen]

n ∈ N

[ebon grove]

oh-

mb

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[wanton mesa]

isn't h pi/12?

becuase the difference is pi/12

why did they put h as pi/6 ?

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[sharp rover]

To combine [(x-3)-5i] [(x-3)+5i] would that be (x-3)²-5i²?

[twilit leaf]

Yup

Be sure to simplify (5i)^2

[sharp rover]

How would I do that?

[twilit leaf]

Split it

What 5^2?

[sharp rover]

25

?

[onyx glen]

encouragement of hypoparenthetosis

[twilit leaf]

And what is i^2

[sharp rover]

Uhhh

-1?

[twilit leaf]

Im sorry ^^'

Yes

So (25)(-1)=?

[sharp rover]

-25??

[twilit leaf]

Yes

[sharp rover]

Oooooh okok

[twilit leaf]

so then you plug it into (x-3)^2-[(5i)^2]

Replace the (5i)^2

[sharp rover]

Hmmm hold on I wanna try to do it and could u just tell me if it's correct pls?

[twilit leaf]

Sure

[sharp rover]

Hmmm

So for this (x-3)²-5i²
It would become (x-3)²-25?

[twilit leaf]

[sharp rover]

Or positive 25 since like there's already a negative sign?

Thonk

[twilit leaf]

-(-25)=25 yes

[sharp rover]

Ooooh okay

So (x-3)²+(25)?

[twilit leaf]

+25

[sharp rover]

Oh

Yea looks better lol

Okay hold on

x²-6x+34?

[twilit leaf]

Looks good

[sharp rover]

Whew

Okay tysm!

I'll be closing it now ^^ Have a good one

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[glad whale]

surely i made a mistake here but im not sure what

as the graph is not what i expect it look like

[drifting dune]

what did you expect it to look like

[glad whale]

some type of u looking line going like

like idk how to explain but the problems im doing the teacher said all of them look some certain way

[drifting dune]

check the -4

[light sonnet]

Parentheses matter

Fyi

[drifting dune]

you flipped some signs

[glad whale]

ah

[light sonnet]

Like how did you insert a + ?

The problem is (x + 1)(x - 1)(x + 3) but the work at the bottom, you did x + 1 + x - 1 + x + 3

[glad whale]

so if i did f(-2)=(-2+1)(-2-1)(-2+3)

im gonna do -2+1 and get -1

etc do each parentheses first then combine?

[light sonnet]

It's not combine as in addition

It's multiplication

[glad whale]

oh im an idiiotott

so t would be like

waitttt

how would a set up even look like

f(-2)=

[light sonnet]

f(-2)=(-2+1)(-2-1)(-2+3)
Literally just simplify that

[glad whale]

oooooooooooooooooohmygod

(-1)(-3)(1)

i cant believe im doing this stuff like this

[light sonnet]

And this equals?

[glad whale]

3

so is there even a scenario where i can just do what i did earlier

i guess without the paranthess

[light sonnet]

No

[glad whale]

Nice ok

will be sure not to do tht for function concepts

.close thnx

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[nimble crescent]

i need help

[nimble crescent]

i need to factor this

[drifting dune]

!status

[full forum]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[nimble crescent]

1

[wide sundial]

you can do by inspection

1, 5 factors the constant 5

one of them needs to be negative for -5

3 times 1 of the factors added with 1 times the other factor gives you -2

[nimble crescent]

okay i think i see that

[wide sundial]

so we want to make -2, ∓ 3*5 ± 1 or ∓ 5 ± 3*1 = -2

∓ and ± mean the opposite, so if we choose + for ∓ then ± represents -

if we choose - for ∓ then ± represents +

(thats how we can notate only 1 of them is negative and the other is positive)

[nimble crescent]

okay okay

[wide sundial]

∓ 3*5 ± 1 = -2
or
∓ 5 ± 3*1 = -2
from here we can just inspect which one gives the answer

clearly 3*5 is too far away from 2

-5 + 3 = -2
5 - 3 = 2
so we want -5 and + 3*1

those are the 2 factors

so we write the bracket as
()()
(3p)(p)
(3p - 5)(p + 1)

keeping in mind the position of the 3 and the 1

they have to be in different brackets so they multiply when you expand it

it's kinda hard to explain these by inspection methods of factorising tbh

[nimble crescent]

omg

thank you

it says its right

[wide sundial]

i know it says it's right

the question is if you understand the method

and can you apply it to another question

of course i know how to factorise these, i've factored countless quadratics, im trying to show you my method of doing it

[nimble crescent]

ye i think

il try the next one

[wide sundial]

👍

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[hot mango]

would
(7/(16+x))/inf * (0n^(n)+1)
work?

bc at any value of n it ends up being *1

[atomic blade]

Factor out -1/16 from f(x)

[hot mango]

and if that expresion is added infinitely then the inf on the bottom should cancle

oh

divergent

[atomic blade]

Wait no, not -1/16

Just factor out 1/16

[hot mango]

if that one

hymonical

[atomic blade]

And no it's not divergent

[hot mango]

o

[atomic blade]

It's not harmonic

[hot mango]

I thought 1/x

ok

[atomic blade]

1/x is harmonic

But factoring out 1/16 doesn't make it 1/x

[hot mango]

7/x

[atomic blade]

No

[hot mango]

lol

7/1+(x/16) ?

[atomic blade]

Yeah

[hot mango]

mk

what abt it

[atomic blade]

That should be easy because you know that $\frac{1}{1 -x} = \sum_{n=0}^{\infty} x^n$. You have $7\cdot \frac{1}{1-\parens{-\frac{x}{16}}}$.

[glossy valve]

Umbraleviathan

[atomic blade]

So building the series shouldn't be too difficult if you understand composite functions

[hot mango]

I should study series more

I dont

but this looks like an easy concept

so ill go learn it

[atomic blade]

Well we can start but taking out the 7 because it's a constant factor

1/(1-x) -> 1/(1-(-x/16)), the x got "replaced" with -x/16

So really your series is $7\sum_{n=0}^{\infty}\parens{-\frac{x}{16}}^n$

[glossy valve]

Umbraleviathan

[hot mango]

mk

[atomic blade]

But notice how like

[atomic blade]

When the argument gets replaced with "something", the series contains that "something"

And then $\sum Cx = C\sum x$ if C is a constant

[glossy valve]

Umbraleviathan

[atomic blade]

So like if you have 5/(1-2x)

Shouldn't be too bad to do

[hot mango]

mk

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[hard temple]

From this question, I understand how you would get x-y=3, but not 2y=x

[hard temple]

solving 70y-4y=40x-7x is 64y=33x

[astral hollow]

70-4 isnt 64 its 66

[hard temple]

oh yea so 66x=33y, then how would you get 2y=x?

oh

wait simplifying it

shit

[astral hollow]

no 66y=33x ,not the other way around

[hard temple]

understood

How would you solve this question?

[shut thorn]

write some equations

[hard temple]

how would you write the equations? Im struggling in that

x would be the speed of car A and y would be the speed of car B

[shut thorn]

you know kinematics?

what is distance in terms of time and speed

[hard temple]

the formula?

speed is distance/time

distance would be both of the cars distance?

[shut thorn]

ues

yes

[hard temple]

while time is 8 hrs for the 1st eq?

[shut thorn]

the wat will be distance?

[hard temple]

160?

[shut thorn]

$\text{distance} = \text{speed} \times \text{time}$

[glossy valve]

bettim

[shut thorn]

right?

[hard temple]

yea

[shut thorn]

so now we have two cars at different speeds, thus it is linear equations in two variables, so we need atleast two equations to solve it eh?

[hard temple]

distance would be 8x of car 1 and 8y of car 2

[shut thorn]

let that be aside

let me take the first case

so

they move in the same direction

[hard temple]

yes

[shut thorn]

to solve it better

we may have to convert hours to minutes

[hard temple]

that would be complicated wouldn't it?

nvm its better

so 8 hrs=480 mins

[shut thorn]

yes let that be

now let me take the first case

let car X with speed X start from A and let car Y with speed Y start from B

now which car is faster?

[hard temple]

A?

[shut thorn]

yes

you know why right?

[hard temple]

not really

[shut thorn]

well

suppose if B was faster than A

A cant reach B after 8 hours right?

in fact it will catch it right?

now you know why?'

[hard temple]

ohh ok understood yea

[shut thorn]

now

going on with the first case

what is the relative velocity of our faster car?

[hard temple]

im not sure

[shut thorn]

consider you are in side car A

which is goin 80 mph

and car B is goin 70 mph

what will be your speed comparing to car B

?

[hard temple]

10

[shut thorn]

how much faster are you going than the other car

yes

so can i say that as $x-y$?

[glossy valve]

bettim

[shut thorn]

considering X is the speed of the faster car

[hard temple]

yes

[shut thorn]

now

$time = \frac{distance}{speed}$

[glossy valve]

bettim

[shut thorn]

take the first case

[hard temple]

ok

[shut thorn]

speed is x - y now

time is 8 hours

which is 480 mins?

[hard temple]

distance is 8x+8y?

yes

[shut thorn]

we converted to minutes remember?

[hard temple]

yea ok

[shut thorn]

and distance is given 80 kms

[hard temple]

alr

[shut thorn]

so what is the equation now/

?

[hard temple]

480=80/x-y?

[shut thorn]

yes

so what is it now

$x-y = \frac{1}{6}$?

[glossy valve]

bettim

[hard temple]

yup

[shut thorn]

now can you do it for the other case?

[hard temple]

I'll try, I need to go somewhere rn
Can I leave this channel open?

[shut thorn]

i dont know

it will close automatically tho

[hard temple]

alr well I'll do the next case myself, thanks for helping me

[shut thorn]

okay

[hard temple]

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[lucid marten]

Just wanna ask if the work I've done here is correct

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[hot herald]

what's the original question

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[undone pendant]

Why is this answer 1500 and not 7637

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[thorn fjord]

Does anyone know the formula for 2nd order total differentials of 3 variables?

[thorn fjord]

This is the formula for 2 variables (x,y) btw

[elfin stream]

That looks like a whole square for sm reason...

[thorn fjord]

Yeah you can rewrite it as (.....)^2

[thorn fjord]

But its easier like this imo

[elfin stream]

ig

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@thorn fjord Has your question been resolved?

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@thorn fjord Has your question been resolved?

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[wind prism]

-2,0 2,-0 -5,0 X intercepts

Then what i should do next is make a table like this?

X | Y
-5 0
-3 ?
-2 0
-1 ?
2, 0

[torn jolt]

A is the point (-5,2). B is the point (7,-2). C is the point (-2,5) a) find the gradient of the line perpendicular to the line AB
b) find the equation of the line perpendicular to line AB and passing through point C

A is the point (-5,2). B is the point (7,-2). C is the point (-2,5)
b) find the equation of the line perpendicular to line AB and passing through point C

[wind prism]

?

[ember shadow]

!help

[full forum]

Please read #❓how-to-get-help

[ember shadow]

Just plug in the x values

[wind prism]

Yep but the table looks good?

f(-3)
f(-2)
F(2)

[ember shadow]

What do you mean "good"

[wind prism]

like co rrect table

[ember shadow]

Yes looks like it

[wind prism]

Nice ok

[glad whale]

@wind prism Ur table looks wrong

where is 0

[torn jolt]

!15m

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[wind prism]

how

dont u only use 3 points in between them

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[sly sail]

Pls work bot

[sly sail]

.... I've been stuck on this logic question forever (2 darn hours....) it's to prove that the above expression is equivalent to (P and Q) -> R... I managed to derive that expression, but there's a whole lot of jumbo entailing behind it 💀

.close

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[normal tiger]

honestly a bit embarrassed to even show my train of thought for this problem... but can someone point me in the right direction / let me know if im already headed in the right direction with this problem?

[ebon grove]

are you sure your angular velocity is 4pi rads/hr?

1 rev is not 2 miles

its 2pi miles

[normal tiger]

Wait a second

Okay that’s a good place to start

[ebon grove]

ok what are they taking theta, is it the angle with the horizontal or vertical?

I'm more comfortable with taking theta with the horizontal

this is what I would start with

from here you have to write the relationship between theta and x which I'm sure you can do, then differentiate it wrt to t to get your final result

got it?

[normal tiger]

Ehhh

[ebon grove]

which part did you not get?

[normal tiger]

Where does the angular velocity come into this

[ebon grove]

Do you know the definition of angular velocity?

$\omega = \frac {d\theta} {dt}$

[glossy valve]

roxyit

[ebon grove]

when you differentiate that relation, you will get it in terms of angular velocity and normal velocity

[normal tiger]

I think I need to sleep on it and try again in the morning

Especially this help too

Thanks

[ebon grove]

alr gn

[normal tiger]

This would be the cos(theta) = x/2

[full forum]

@normal tiger Has your question been resolved?

[ebon grove]

yup

now just differentiate and dont forget the chain rule

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[normal tiger]

nice

[drifting dune]

nice

[normal tiger]

I’m guess I’m confused about the walking speed and angular velocity part

[atomic blade]

When t = 0, is x the diameter?

[normal tiger]

well yes

[atomic blade]

Well what you know is that $x^2 = 2 - 2\cos(\theta)$

[glossy valve]

Umbraleviathan

[atomic blade]

You can start by identifying that

[normal tiger]

how do i know this

[atomic blade]

Look at the diagram.

You have a triangle with sides 1, 1, x

[normal tiger]

okay

so thats the law of cosines part

[atomic blade]

And then linear velocity is just rω, and you know rω = 4 mph. You know that r = 1, so ω = 4 rad/hour

Wsll no that doesn't seem right

Well atcually that's correct

Okay so we know that $\theta = 180 - 2φ$. We know that $\dv{φ}{t} = 4$

[glossy valve]

Umbraleviathan

[atomic blade]

Because of this

This means that:

$$x^2 = 2 - 2\cos(π - 2φ)$$

[atomic blade]

Read from here @normal tiger

[normal tiger]

i am

im a bit lost

[atomic blade]

Which part is confusing to you?

φ is the angle Amy and Bill each make as they walk. Since they have the same linear velocity, they're going to make the same ref angle

That's why θ = π - 2φ

[normal tiger]

that makes sense

[glossy valve]

Umbraleviathan

[atomic blade]

Had to fix it I had it in degrees lol

Also cos(π-2φ) = - cos(2φ)

So that's a nice simplification

[atomic blade]

So which part is confusing you?

[normal tiger]

so this would become:
$$x^2 = 2 +2\cos( 2φ)$$

?

[glossy valve]

tripp

[atomic blade]

Yeah

And then differentiate both sides with respect to t

[normal tiger]

the part im confused about is having the second variable

can i represent φ in terms of x

if i do arcsin(sqrt(4-x^2)/2)

[atomic blade]

Where is the sqrt(4-x^2)/2 coming from

[normal tiger]

i used Pythagorean theorem and rewrote the y of the triangle in terms of x

but maybe thats a mistake

[atomic blade]

Yeah no, idk how you got x/2 and so on. This angle changes

[atomic blade]

Just start by differentiating both sides with respect to t

And see where that goes

We can worry about the pythag stuff later

[normal tiger]

2x dx/dt

-4sin(2φ) dφ/dt

[atomic blade]

And we know that dφ/dt is 4

Hmm

[normal tiger]

okay

so we can sub that

so -16sin(2φ)

[atomic blade]

How to determine sin(2φ)...

Well that's the same as 2sin(φ)cos(φ)

Oh damn this question is harder than I thought

[normal tiger]

hold on

[normal tiger]

can you run me through how you did this derivative

[atomic blade]

Up here

4 rad/hour is dφ/dt

Although now that I think about it

[normal tiger]

thats the angular velocity?

[atomic blade]

Yeah

[normal tiger]

okay

[atomic blade]

But I've also took note because if they're moving at the same pace, then their distances will remain parallel to the diameter

So I've also noticed that x/2 = cos(φ)

Ah that's what it is.

[normal tiger]

which is where i got the x/2 from

[atomic blade]

Okay okay I see I see now

[normal tiger]

i think i was approaching it from another angle

[atomic blade]

I mean you used a diff theta than the one they gave

That diff theta is my phi so

[normal tiger]

yeah

[atomic blade]

Alright alright I gotcha I gotcha

[normal tiger]

thats why i said cant i rewrite the phi in terms of x

[atomic blade]

I mean you technically could

cos(2φ) = 2sin(φ)cos(φ)

[normal tiger]

would be harder differentiating with the arcsin?

[atomic blade]

So we can replace each trig thing

You found that sin(phi) = sqrt(4-x^2)/2

And cos(phi) = x/2

[normal tiger]

oh

wow

[atomic blade]

But then now I'm thinking about the x on the numerator of cos(phi)...

[normal tiger]

wait

so when i sub the values into -32sin(φ)cos(φ), what exactly is that result going to tell me

like not the numerical value but what does that number represent

[atomic blade]

That's gonna tell you dx/dt

But now I'm having doubts whether or not I did it correctly...

[atomic blade]

Unless I'm overseeing something hm

Hm so

$x^2 = 2+2\cos(2φ)$

[glossy valve]

Umbraleviathan

[atomic blade]

Lemme look at something

I think???

Oh no it works out just fine

I'm dumb

I forgot that x^2 -> 2x dx/dt

And it'll all simplify down to what Amy has

[normal tiger]

what

[atomic blade]

This implies:

$$2x \dv{x}{t} = -4\sin(2φ) \dv{φ}{t}$$

[glossy valve]

Umbraleviathan

[atomic blade]

That you get right

We know that dphi/dt is 4, and expanding sin(2phi):

[normal tiger]

i got for the right side: -8x sqrt(4-x^2)

[atomic blade]

Yeah the RHS will simplify down to that

The LHS is still 2x dx/dt

That's what I forgot lol

So just divide both sides by 2x

[normal tiger]

oh

duh

thank you

[atomic blade]

Np

[normal tiger]

i definitely would not have solved this on my own

[atomic blade]

Yeah it's a bit tricky

[normal tiger]

crazy ass problem

.close

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[spark anvil]

If x^2+5x-5=0
(x+1)(x+2)(x+3)(x+4)?

[hot herald]

!status

[full forum]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[spark anvil]

2

[keen shoal]

show your working

[spark anvil]

I rewrote the identities
(X+1)(x+2)= x^2+3x+2
(X+3)(x+4)= x^2+7x+12

aw

[hot herald]

you didn't expand (x+1)(x+2) correctly

[spark anvil]

And

Yeah

Edited

Now

I think I understood smth

[hot herald]

also you should consider doing multiplication in a certain order/way
that allows you to easily apply the given condition

[spark anvil]

(x+2)(x+3)= x^2+5x+6
And we know
X^2+5x -5 = 0
Then that is equal to 11

So we got 11(x^2+5x+4)
Which is 11(9) = 99

just didn't try to think

Thx

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[stark geyser]

hey is this correct?

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[old crane]

How to find the sides of an isosceles right-angled triangle by knowing the base (which is : 0,849) ?

[old crane]

Easy problem, plz help

;)

[light compass]

hi

why don't you draw a line from the apex to the middle of the base

[old crane]

my logiciel can't

but i think there is an maths equation

[light compass]

there is

btw it's "program" not "logiciel"

do you know about the 45-45-90 triangle

[old crane]

oh okk im french lol :)

no i don t

[light compass]

same, techincally

right so

[old crane]

ah d'accord

[light compass]

do you want me to speak in french if its easier?

[old crane]

ping moi quand tu me dis qqc

yeah of course

[light compass]

bien sur

mais

je ne suis pas supposee de faire le probleme pour toi

[old crane]

ah d'accord

j'ai pas bien compris mdrr

[light compass]

non c'est correcte

[old crane]

il faut faire de la trigo ?

[light compass]

pas technicallement

il faut utiliser une formule

dans des triangles isoceles avec une angle de 90 et deux angles de 45 il y a une relation speciale entre le cotes

[old crane]

ah

[light compass]

est ce que tu sais c'est quoi?

[old crane]

base = coté x √2

c'est ça ?

[light compass]

oui

[old crane]

donc coté = base / √2

[light compass]

oui!

[old crane]

donc 0;849 / √2

= 0;6003336572

merci beaucoup

;)

[midnight dew]

*√2

pas ÷

[light compass]

pas de probleme!

[light compass]

nono

[old crane]

si pour trouver la base

[light compass]

Jaunedoeuff a raison

c'est 0.6

[old crane]

je peux te mettre une note ou qqc ?

[light compass]

(environ)

non ca fonctionne pas comme ca je pense

[old crane]

okk ça marche

[light compass]

mais il faut fermer ca avec ".close"

a plus tard

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[fiery valve]

how do i do this qn

[limber flicker]

setup an inequality

we want when h(x) >= 1

[full forum]

@fiery valve Has your question been resolved?

[fiery valve]

@limber flicker is this wrong?

[limber flicker]

I would say this is close

also keep in mind you want greater than or equal to ,

and also remember what solutions and equation has when a square root is involved

[fiery valve]

oh yeah

but why greater than or equal to tho?

ah

okay i got it

wait

[limber flicker]

I'm pretty sure you can solve it for when it's equal to but you have to give extra reasoning about why they are the points of the domain where h(x) is >= 1 and <= 2

blabber on about some circle nonsense or something

[fiery valve]

yeah i dont think i know how to do that its just kinda making ie more complicated let me just try greater than or equal to

[limber flicker]

also maybe a note on that h(0) is just the radius of the circle would provide the upper value of 2, and h(x)<=2

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[dense edge]

How would I answer b?

[drifting dune]

so true

hi

[dense edge]

It’s a new question tho

[drifting dune]

why did you make another channel

[dense edge]

I answered a already

[drifting dune]

so use the same channel

D:

[dense edge]

Do I need to use that answer for b?

[drifting dune]

no

[dense edge]

OK

[drifting dune]

use the formula

[dense edge]

Linearization formula is like point slope formula? Very similar?

[keen shoal]

f’(a)(x-a)+f(a)

f’(4)(x-4)+f(4)

f’(4)(5-4)+f(4)

[devout valley]

[some people have suggested you're "supposed" to use a new channel for each "new" question tbf]

[drifting dune]

i didn't know that

[devout valley]

[Well personally, I don't care either way and prefer using the same one, but that's probably why I think ]

[full forum]

@dense edge Has your question been resolved?

[viral jasper]

I can see the merit in new channels. Keeps the question you want answered at the first comment

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[dense edge]

.reopen

[full forum]

✅

[dense edge]

f’(4)(x-4)+f(4)

So how would you solve for this question? After you make it here

[keen shoal]

what’s the value they’ve given you to approximate

[dense edge]

f(5) I think?

[keen shoal]

yes so your x is 5

[keen shoal]

yes

now solve that

you know the derivative

and you can calculate f(4)

[dense edge]

So 10/3 is the final answer?

Pretty close.. I guess that’s the point

3.333 is the approximate answer

3.316 is the exact answer

I can always check afterwards to see if they are close?

[viral jasper]

That's the idea of approximation

Wait until you get to Taylor series

[dense edge]

What do you use this for? In real world

I don’t think we have learned that yet

Approximation is required for Taylor?

[viral jasper]

Local linear approximation is useful for quick approximation

[viral jasper]

Other way around, Taylor series give you infinitely precise approximation.

[dense edge]

Oh

[brave blaze]

engineer use it to say dumb shit like sin(x) = x when x approaches 0

[viral jasper]

And cos(x)=1-x^2

[brave blaze]

Its useful for linear interpolation too

when dealing with too much data

[viral jasper]

Or was it (x^2)/2? I forget

[brave blaze]

i forgot

[viral jasper]

,w plot cos(x), 1-(x^2)/2

[glossy valve]

Results provided by WolframAlpha

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Requested by SWR#8588

[viral jasper]

It's /2

,w plot cos(x), 1-(x^2)

[glossy valve]

Results provided by WolframAlpha

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

Requested by SWR#8588

[viral jasper]

,w plot cos(x), 1-(x^2)/4

[glossy valve]

Results provided by WolframAlpha

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Requested by SWR#8588

[dense edge]

Oh OK

.close

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[tawny latch]

can anyone help with this

[tawny latch]

.close

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[cosmic shoal]

so for divergence is it just phi,cos(phi)zr ?

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[wet minnow]

Having trouble with understanding what closed means.

For 1b.

[rapid rain]

do you know what an open set is?

[wet minnow]

I don't

[rapid rain]

Ok

I'll remind you of the definition of an open set:

Do you know definition of a neighborhood maybe?

[wet minnow]

No haha

For a bit of context I'm taking linear algebra right now

We just started learning about general vector spaces

[viral jasper]

I think just the linear algebra definition of closed is needed here

[wet minnow]

Would you mind giving the definition?

Having trouble understanding it from the book

[viral jasper]

By definition, V is closed if any linear combination of elements of V is also in V

[wet minnow]

So if the given ten axioms work, then there it's closed?

By ten axioms

I mean this

[viral jasper]

If I recall, yes. That may be definition of a vector subspace tho

I forget the distinction

[wet minnow]

This was the vector space axioms I believe

subspace is next section

[viral jasper]

But if you want to prove closed, just show that any linear combination remains in V

[rapid rain]

Yes that's a necessary/sufficient condition w linear algebra

[wet minnow]

Would you mind showing me how you would prove it's closed?

this is the explanation I'm currently working with

[rapid rain]

Let's do it on another example

[wet minnow]

ok

[rapid rain]

Let's say $E = \bR$, the set of all real numbers, viewed with the regular addition and scalar multiplication

[glossy valve]

rafilou2003

[rapid rain]

We want to show that E is closed under those operations

Let's show that E is closed under addition :

Let $a,b\in E$. Then $a+b\in \bR = E$. Thus E is closed under addition

[glossy valve]

rafilou2003

[rapid rain]

Now, for scalar multiplication :

Let $a\in E, \lambda \in \bR$. Then $\lambda a \in E$. Thus E is closed under scalar multiplication

[glossy valve]

rafilou2003

[wet minnow]

Ok to summarize so far so I make sure I'm understanding correctly

E is equal to R which is the set of all real numbers, then we move on to prove closure with addition and scalar multiplication. We do this by showing a + b is a set of all real numbers which shows E is closed under R. Then we do the same with scalar multiplication since we receive a real value which shows again it is closed under scalar multiplication.

so since both operations have a set of all real numbers, it shows E is closed under R

Is this logic correct?

I think I understand it better now

[rapid rain]

the wording on the "a+b" part is a little wrong, but you got the gist

a+b is not a set of real numbers, we want to show it is part of E

since $a+b\in \bR$ and $\bR$ is the same thing as $E$ here, $a+b\in E$

[glossy valve]

rafilou2003

[wet minnow]

Ok I think I get it

Thank you so much

!

[rapid rain]

you're welcome

[wet minnow]

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[old tendon]

To do distance vector algorithm, i know that i need to make a chart like i attached in this message, but i want to make sure i get the algorithm correctly. So, the output will need to be a chart that has all the values of the shortest path, but, as im making it, the values will initially be wrong for some of them, since they wont know the existence of other nodes until later. So when do i know that i can update a value?

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[placid jackal]

yo

[placid jackal]

how does logging 5^x make it just x

[tepid cedar]

.

[placid jackal]

i dont understnad the math behind it

[drifting dune]

wait what

log 5^x = x?

[torn jolt]

what is your log base

[tepid cedar]

The expression "log base 5 of x" (denoted as "log5(x)") is the inverse function of "5 raised to the power of x" (denoted as "5^x"). In other words, if you take the logarithm of a number x with base 5, you are essentially asking the question "to what power do I need to raise 5 to obtain x?"

Mathematically, the relationship between "log5(x)" and "5^x" is defined as follows:

log5(x) = y if and only if 5^y = x

This means that "log base 5 of x" gives you the exponent y that you need to raise 5 to in order to obtain x. So, if you have an equation in the form "5^x = y", you can take the logarithm of both sides of the equation with base 5 to isolate x:

5^x = y
log5(5^x) = log5(y) (taking logarithm of both sides with base 5)
x = log5(y)

This is how "logging 5^x" results in just "x". It helps you solve for the exponent x in the exponential equation involving 5 raised to the power of x.

[drifting dune]

chat gpt?

[scenic wren]

log_a(b)=c means
a equals b if taken to the power ofc
in other words
a^c=b

[placid jackal]

how does the log become just x

[torn jolt]

[tepid cedar]

y

[drifting dune]

what is the log

[torn jolt]

just no

[placid jackal]

im so condfused

[torn jolt]

log is kinda the inverse operation of exponents

[placid jackal]

i know how to do questions

and the laws

i just dont know why they work

[torn jolt]

Think of it like this

[placid jackal]

gotch

gotcha

[placid jackal]

?

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[simple totem]

this is kind of a copy paste from [#groups-rings-fields](/guild/268882317391429632/channel/496784958430380033/) but i want to know if this proof is correct. the statement i'm proving is:\
let $R$ be an integral domain, $F$ a subfield, and suppose $\dim_FR$ is finite. show $R$ is a field.\
my proof:\
\
let $a\in R$ be nonzero. let $p(x) \in F[x]$ such that $p(a) = 0$ (we've shown this exists already). \
we claim we can construct $p$ such that $p(0) \ne 0$ as well. if $p(0) = 0$, then $p(x) = x^n\cdot q(x)$ for some $q\in F[x]$ such that $q(0)\ne 0$, so we can instead just consider $q$. \
now, consider that $q(0) \in F$, and consider the factorization in R[x] $$ q(x) = (x-a)\cdot \tilde{q}(x)$$
then $q(0) = -a\cdot \tilde{q}(0) $, then $$a\cdot - \frac{\tilde{q}(0)}{q(0)} = 1 \implies a^{-1} = -\frac{\tilde{q}(x)}{q(0)}$$

[nova basin]

the existence of f is because a must be algebraic over F because [R : F] is finite right ?

also then pls call them P and Q like polynomials instead of f and g
Conventions are preferable

[simple totem]

yes

also got rid of the c cause i didn't really need it

sorry that's the last edit

[nova basin]

I don't remember if the factorization by (x-a) holds on integral domains but I think
If so I think this should be good

[simple totem]

oh good point

i'll double check

i've been working with fields too much

[nova basin]

I've only really ever worked with fields

[glossy valve]

maximo

[nova basin]

the trick of looking at a polynomial because a is algebraic is pretty neat, I must say

[simple totem]

credit goes to parrot tea

i did not know where to even go with this at the start, i kept playing around with the basis to no avail

i believe you can in fact factor the (x-a)

thanks for checking it out mateo

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[tawdry ginkgo]

if anyone can help me solve this

[grave elm]

collatz...

[ivory cairn]

🙄

[grave elm]

What's the question btw?

I can see just a definition of function f

@tawdry ginkgo do you have a question or is this just a troll?

[void nova]

What do you mean by "solve"??🤔

[tawdry ginkgo]

All I got

this was the entire problem

thats why im lost

[grave elm]

who gave you this question/problem

[ivory cairn]

then it's not a problem, it's a statement of a function.
Whoever gave it to you is trolling you because that's the function for the unsolved Collatz Conjecture.

[tawdry ginkgo]

Friend did

Did not know she was trolling

Mb guys

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[elder cape]

Should I not be t[-1, 1] ?

<@&286206848099549185>

[storm saffron]

<@&286206848099549185>

[polar valve]

whats the difference?

[elder cape]

Well what r the steps to finding the general solution?

[polar valve]

just multiplication from am matrix with a vector.

[elder cape]

Ok I did that

I get [x1 + x2
0 ]

[polar valve]

and this should be the nullvector.

so x1+x2=0

[elder cape]

Yeah

Which gives x1 = -x2

and we know that x2 is free

so why is the solution not [-x2
x2]

= t[-1
1] ?

[polar valve]

and i asked you what si the difference between this two solutions?

[elder cape]

The first component of each vector is different

[polar valve]

no, as x2 is free and in R. the solution is a vector where the second element has the opposite sign from the first element. both solution express this exactly. they are the same.

[elder cape]

ok

ty

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[signal solar]

why are these not equal?

[signal solar]

i thought that:

$sqrt(x^{2}+y^{2}) = (x^{2}+y^{2})^{1/2}$

[glossy valve]

nino

[signal solar]

so wouldn't distributing that power lead to x+y

[stuck vortex]

No it wouldn't

[signal solar]

why not

[stuck vortex]

The powers are not divided to individual terms in addition function

It only happens in multiplication or division

@signal solar

[signal solar]

ohh so its only if it was

(x^2 times y^2)

right?

not plus cuz itd be one quantity

[stuck vortex]

Well that's one way to understand it

It's cause of the distributive property of addition

Btw what grade are you in?

@signal solar

[signal solar]

im learning calc rn and im a senior its just a random question i had cuz i knew it was a misconception

but i wanted to get to the bottom of it

[stuck vortex]

I see, did I answer your question?

[signal solar]

yep

ima close this now

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[mystic cedar]

was wondering how the Rout is 5-y
shouldn;t it be 3-sqrty
right - left?

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