help-23

[safe radish]

@frank musk Has your question been resolved?

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[neat owl]

A plant is 6.2 cm tall. The height of the plant increases by 11% each week. Find how tall the plant will be after two weeks.

[safe radish]

@neat owl Has your question been resolved?

[neat owl]

<@&286206848099549185>

[arctic condor]

Show what you have tried

[neat owl]

I did 11x2+22 so 22% of 6.2

[arctic condor]

If you take 22% of 'something', will that 'something' decrease or increase?

Let the OP answer

[worthy oxide]

What i think, that this problem has to be solved with a different way

no wait

I have some concnerns

[worthy oxide]

lol

[arctic condor]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[arctic condor]

Stop please

[worthy oxide]

umm...

i apologize

will that really be 22%?

[arctic condor]

Of course not, the point is to make the OP understand why

[worthy oxide]

oooo

lemme delete everything

lmao

done

do it Mr. OP

I am out

[arctic condor]

@neat owl

[worthy oxide]

bro is playing games

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@neat owl Has your question been resolved?

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[urban vale]

how do I solve this?

[elfin yew]

do you know what makes a function continuous at a given point

[urban vale]

nope

I know how to find the limits and stuff tho

[warm vault]

Your goal is making

m(-2) + b = (-2)^2 - 7

and

5 = 2m + b

[urban vale]

ok

why is it -2 and not posive 2 tho

oh I see

[elfin yew]

yeah so basically at the point has to be the same value at the point for it to be continuous correct

[urban vale]

ok

but then what after 5 = 2m + b

oh I see

-2m + b = -3

[warm vault]

yea

try to solve these two equations

[urban vale]

I forgot how to do that

I know that b =1 and m =2

pretty sure thats correct

thanks Cosms6do

.close

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[proud zealot]

How to solve this?

[proud zealot]

Bruh -_-

[uneven jacinth]

graph

[proud zealot]

Hiw

How

[uneven jacinth]

get a graphing paper

and then draw the two axes

then draw the points of each curve, preferably labelling the two functions

[proud zealot]

How do u graph y=x^2 -4

[uneven jacinth]

interesting

draw two axes

the x-axis and the y-axis

[proud zealot]

Eh?

[proud zealot]

yes

[uneven jacinth]

step 2, put some values of x and get some values of y

for example

in the function x^2-4

if you let x = 0

you have y = -4

right?

[proud zealot]

Hm

Yes

[uneven jacinth]

one point

then try another value for x

[proud zealot]

So wdym by another value?

[uneven jacinth]

you choose

[proud zealot]

U put in 1 = x

[uneven jacinth]

yes

[proud zealot]

Oh.

[uneven jacinth]

i did 0

so thats how i got the first point

you try x = 1

x^2-4 --> 1^2 - 4 = -3

we have a new point

@proud zealot choose a point

[proud zealot]

2

[uneven jacinth]

what do you get when x = 2

[proud zealot]

0

[uneven jacinth]

new point

keep going

[proud zealot]

3

...

[uneven jacinth]

you know what we are doing right now?

we are finding points

(0,-4)

[proud zealot]

Eh, yeah.

[uneven jacinth]

(1,-3)

(2,0)

so

keep getting points

and then draw a line

well the thing about quadratics

is that they arent straight

they are curved and form the shape of a parabol

[proud zealot]

So basically, use this to find vertex,

[uneven jacinth]

what

no

we are doing this to find the curve

keep doing it without me

i will go to the toilet

bye bye

[proud zealot]

Ok

<@&286206848099549185>

Please help clarify

.close

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[lean otter]

I dunno how to slove this one , the only thing I can think of is plugging a number into x

[lean otter]

have u done derivatives yet?

oh ummm yes but I still don't really get those

[lean otter]

anyways, yk that the maximum profit is the y value of the vertex, right?

Yes!

we can do this without calculus

do you know the formula for the x value of the vertex?

noo I don't

-b/2a

remember this?

It looks familiar

anyways, once u plug in the a and b value, you'll get the x value of the vertex

then, plug the x value into the P(x) equation to find the y value which is the maximum profit

ooooooh yea okay I get that!!

Kay let me try real quick

okay I got it!! And I double check it , profit per day is 2200 and cans he can sell for max profit is 2000

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[glad zinc]

Hello, I saw a video where there's one part that I can't get through my head, and I wondered if someone else knows how to do it. https://youtu.be/KZ8G4VKoSpQ?t=1322 I don't really understand the way the green part became the blue one.

[elfin yew]

they just found the definite integral or do you have a question on how it was integrated

[glad zinc]

Yes my question is on how it was integrated

I don't understand how mv became mc squared

[safe radish]

@glad zinc Has your question been resolved?

[glad zinc]

@meager igloo

[glad zinc]

Can i get help?

<@&286206848099549185>

[glad zinc]

Yes my question is on how it was integrated
I don't understand how mv became mc squared

.close

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[dusty talon]

implicit differentiation

im blanking on what i can do next

[icy lance]

get all the dy/dx terms on one side and factor it out

[dusty talon]

how should i do that?

[icy lance]

move the x^2 dy/dx to the right, factor our dy/dx

[dusty talon]

lmao

subtract 2x^2y both sides and then divide by -x^2 ?

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[lean otter]

what are linear factors? how are they calculated, do I apply the quadratic equation the same as the zeros of polynomial? although I guess as far as I know about linear factors it only applies if the zero polynomials are zero? for example this one why linear factors is x-2 and x-3?

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[lean otter]

.close

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[stray axle]

so in the previous HW assignment, we had to find how the fibonaci sequence converges to the golden ratio

or, assuming it converges then find its limit

[strange bane]

It's been a while but you could do a comparison test with p-series

since you know fn/f(n-1) approaches phi

[stray axle]

yea, thats something i was thinking. since its |R|<1 then dont we know it converges

or wait

there is no P in this example

[strange bane]

well in this case p>1

I'll do a simpler example first

1/2 + 1/4 + 1/8 ... = sum((1/n)^2) right

[stray axle]

yea that converges

[strange bane]

in this case, the denominator of each successive term is twice that of the previous denominator

[stray axle]

but in this isnt it just sum(Fn)^-1

[strange bane]

now when the denominators are based on the fib seq, we know that dividing fn/f(n-1), the ratio is phi (which is not 2, but still more than 1)

[strange bane]

yeah but the ratio of successive terms approaches a concrete value

which implies we can do a comparison

another argument you could do is say "well fib sequence eventually increases at a rate equal to n^phi"

then you could make the substitution:
sum((1/n)^phi) which converges by p-series test

does that make sense? not the best at explaining stuff

[stray axle]

no thats good, im just trying to understand each step

[strange bane]

alr all good

[stray axle]

confused on how if fib seq increases at a rate equal to n^phi

[strange bane]

eventually it does

[stray axle]

so n^phi is equal to fib seq? hows that

[strange bane]

you can google the fib seq, scroll far to the right, divide successive terms and you'll see the ratio of successive terms approaches phi

[strange bane]

not equal, but it behaves like n^phi for large enough n

[stray axle]

so this is similair enough to the fib seq?

or am i misunderstand it

[strange bane]

in a certain sense yeah

[stray axle]

hot diggity dog! is there a proof or a name for this?

[strange bane]

it's just a property of the Fibonacci sequence for large values of n lol

[stray axle]

🤔 wait how does that work then

ahaha, sorry for the bunch of questions. this is real interesting though

[strange bane]

I was careful to say that fib(n) bahaves like n^phi for large n - this is simply because the ratio of successive terms approaches...

oh I think I messed up

ratio --> geometric series

not p-series

the common ratio of terms in this case --> 1/phi

[stray axle]

that simple?

so because we know Fn = phi, then its in the form of a geometric series and as such it converges because |R|<1

since the original question only asks to determine convergence

am i missing something? that seems like it answers the question

or since we instead say that Fn= (1+sqrt(5))/2

Fn= 2/(1+sqrt(5))

which still works for geometric series

[strange bane]

the ratio of terms --> 2/(1+sqrt(5)) which I'm assuming is 1/phi

so we can compare that with geometric series r = 1/phi

and since |r| < 1, and it can be shown that fn > 1(phi)^x, by basic comparison test, 1/fn converges

[stray axle]

damn alright! thank you, i thought this would be much harder but its just a simple series

[strange bane]

I forgot the difference between p and geo series this shouldn't have taken this long my bad :(

or rather I mixed them up initially

[stray axle]

lmao its alright, my bad that i didnt notice the mistake

.close

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[strange bane]

https://tenor.com/view/predator-hand-shake-arnold-arnold-swarzeneger-gif-9494194

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[slate thorn]

what does cu and cd mean? (for curve sketching)

[slate thorn]

.

nvm

. close

.close

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[heavy terrace]

did i do this wrong?

[river wharf]

@heavy terrace

formula is right but i got more than 8billion

[heavy terrace]

what

[heavy terrace]

can u send ur workings?

[river wharf]

nvm urs right

i did an e lapsus

[heavy terrace]

i see

ig ur an impostor

lmao

(im sorry)

[river wharf]

[safe radish]

@heavy terrace Has your question been resolved?

[heavy terrace]

.close

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[latent imp]

tiny proof check please

[latent imp]

https://i.imgur.com/z5xFK9k.png

If $0v = 0$ then $(1 + (-1))v = 0$ and $v + (-v) = 0$. \newline
Here $(-v)$ is the additive inverse of $v$ and it's existence is given by closure under scalar multiplication.

[flat frigate]

Suro

[river wharf]

suro

you can extend this more

for any vector v

@latent imp

[latent imp]

uhh

the question is giving 0v = 0 and asking me to prove that it can replace the additive inverse in the definition of a vector space

it's a nice proof but you proved the given 😅

oh I got a thumbs up for the proof didn't see, cool thanks

.close

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[river wharf]

that xd

what*

ah

[latent imp]

ye Zybikron gave a thumbs up lol

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[hoary spire]

How would you solve for C

[empty gyro]

C is just a point

[hoary spire]

Sorry I mean side C

like side AB

[empty gyro]

You can use pyhagorean theorem to find CA

[hoary spire]

How would that work? because we dont know what OA is right. does that little c mean something?

[safe radish]

@hoary spire Has your question been resolved?

[trim swan]

@hoary spire can you post the exact problem?

if this is from a homework or book or something

[hoary spire]

No its not its test prep

I dont have the actual problem

I just know I have to solve for the area of the pentagon

[trim swan]

hm

without the values of r and/or c, all you can do is express it in terms of r and/or c

[hoary spire]

maybe i am missing a part of the problem

[trim swan]

It's possible to find the area in terms of those values

but that's all

maybe that's what you're supposed to do?

[hoary spire]

but what is c? I know r is radius - does c stand for anything in particular

[trim swan]

well yeah

this is a regular pentagon right?

it must be, otherwise we don't really know anything

c is the distance from the center to the vertex of the pentagon

[hoary spire]

Yes it is a regular

oh i see

[trim swan]

It is the radius of the circumscribed circle, if you know what that means

like if you drew a circle around the pentagon

it's the radius of that circle

[hoary spire]

Ohh that makes sense

[lethal hill]

hey guys

what is 1+1

[hoary spire]

I think I can solve it from here

[trim swan]

ok great 👍

[hoary spire]

Thank you for clearing that up

[trim swan]

np

[lethal hill]

guys what is 1+1

tat

[hoary spire]

how do i close this?

[trim swan]

with .close

[hoary spire]

.close

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[trim swan]

@lethal hill please don't troll in the help channels

[lethal hill]

i am not

tat

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[rugged dove]

QUESTION MANS

[rugged dove]

im still confused as to how im supposed to draw the triangles in trigenometry

cause like lets take this for example: in a right triangle sea

SA= 13cm and EA = 7cm

how am i supposed to draw the triangle

[devout shale]

Right triangle SEA means that it has vertices (points) labeled S, E and A

[rugged dove]

you see

how

[devout shale]

the fact that it is a right triangle tells you that it has a right angle

like the triangles in those images do aswell

[rugged dove]

yes

but which of those images are the correct one

left? right? both?

[devout shale]

Neither, since neither are your drawing and neither are labeled S E A

[rugged dove]

Bruh

[spiral bane]

there are also multiple correct answers here

since you dont know if SA or ES is the hypotenuse

[rugged dove]

also when i write the S, E, A where do i place them

[rugged dove]

i think 13 is the hypo

[devout shale]

The "E" is in the middle

[rugged dove]

since it is the longest

[rugged dove]

and where does the middle go?

[devout shale]

Try drawing it out....

[rugged dove]

ok

[devout shale]

It obviously has to go in between the other two points

[rugged dove]

ok i cant post my drawing here since i dont have a phone but anyways

i ended up with S as the hypo

[dull sequoia]

Go to google

Search js paint

[rugged dove]

ok

[dull sequoia]

Draw and send us a picture

[rugged dove]

this

@dull sequoia is it correct?

[dull sequoia]

S, E, A are vertices

Not sides

[devout shale]

they are the points (the corners)

not the sides

[rugged dove]

yea

i fucked up with that

[dull sequoia]

Also you don’t know the angle beteeen e and a is 90 deg

[rugged dove]

this?

[dull sequoia]

Yes but you don’t know that angle sea is 90 degrees

[rugged dove]

what?

isnt E 90˚

[dull sequoia]

This is possible as well

[rugged dove]

ohhhhhhhhh

they did not specify wether E was a right triangle or not

[dull sequoia]

That’s right

[rugged dove]

i see isee

thanks for that

[dull sequoia]

That’s also why angle side side is not a valid reason for congruency of triangles

Side angle side is

[rugged dove]

ok

[dull sequoia]

But that means your angle must be between the 2 side lengths

[rugged dove]

ohh

.close

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[prime cobalt]

,w solve e^(x-2) > 1/(x-2)

[flat frigate]

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[prime cobalt]

,w what is W(1)

[flat frigate]

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[obtuse plover]

#bots

[prime cobalt]

hi

is it possible to solve it algebraically

e^(x-2) > 1/(x-2)

[flat frigate]

Stephen

[prime cobalt]

yes

how do you solve that algebraically

[obtuse plover]

Well we would end up with

$e^x (x-2) > e^2$

[flat frigate]

Stephen

[prime cobalt]

wait you fqn just cross multiply?

but we dont know if x-2 is positive or negative

[obtuse plover]

Yea that’s what I’m wondering

[prime cobalt]

we can cross multiply by the square

but idk where to go from there

[grim plover]

you could also try inequating by using factors

[prime cobalt]

how?

[grim plover]

e^2 = e*e

e^x > e
or x-2 > e

[prime cobalt]

wait it works like that?

isnt that only for 0

[grim plover]

well it does work in equations, so why not inequalities

[grim plover]

no?

[prime cobalt]

is @toxic stratus here?

[obtuse plover]

xD calling in the big guns

[prime cobalt]

yeah ahaha shes rly good and helpful

[toxic stratus]

[obtuse plover]

Yea man I don’t have much experience with these things, sry

[prime cobalt]

yay

[obtuse plover]

You have summoned the elements

[prime cobalt]

can we solve e^(x-2) > 1/(x-2) algebraically?

[mortal sandal]

e^(x-2) * (x-2) > 1

that's the form you want

[prime cobalt]

yes

[mortal sandal]

then you can do a substitution

[prime cobalt]

oh

okay

[mortal sandal]

or wait hmm

[prime cobalt]

x-2 = y

[toxic stratus]

you have to worry about x-2 being negative

[mortal sandal]

e^y*y = 1

[toxic stratus]

but you'll have to use lambert W in the end

[mortal sandal]

yeah^

[prime cobalt]

whats that

[toxic stratus]

inverse function of xe^x

[prime cobalt]

oh

[obtuse plover]

That’s this W function here

[mortal sandal]

What were you supposed to do with the answer anyways

Just write it?

[safe radish]

@prime cobalt Has your question been resolved?

[safe radish]

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[undone coyote]

i've attempted this MANY times already but can't figure out the right way to enter the correct answer

[quasi bison]

show the work that led to this

[undone coyote]

ok

[mortal sandal]

where is x coming from?

The message tells you itself: the answer should be a number, not a formula

[safe radish]

@undone coyote Has your question been resolved?

[undone coyote]

it looks like it says the opposite

[mortal sandal]

"your answer isn't a number"

when you entered something that isn't a number

you think that's telling you your answer needs to be less number-like?

[undone coyote]

on the last picture it shows x1 and x2 in a vector matrix

[undone coyote]

oh, I thought the message was telling me "your answer isn't a number' meaning, it should not be a number

[mortal sandal]

nah it wants a number

[undone coyote]

understood

[safe radish]

@undone coyote Has your question been resolved?

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[obtuse plover]

How would I go about this problem? I've attached the formula with which the error term is associated, as well.

[obtuse plover]

wait

do i just isolate E(x,y) in that formula

[safe radish]

@obtuse plover Has your question been resolved?

[lean otter]

yeah, isolate in eq 4.26, then substitute the derivatives etc

its taylor's theorem in 2D for first order approxs

[obtuse plover]

hmm

its saying im wrong, could u check and see if i did something incorrectly?

[lean otter]

evaluate the derivatives at x_0 and y_0

[obtuse plover]

so i should come out with a numerical answer?

hmm

i got 0

which makes sense because it makes the limit evaluation true

but it doesnt feel right

[lean otter]

I dont think it should be zero since f is not linear

can you show your new working?

[obtuse plover]

I didn’t do this right, did I

[lean otter]

,calc -14*1

[flat frigate]

Result:

-14

[lean otter]

-14(x-1)

[obtuse plover]

ohhhh]

thats what u meant

[lean otter]

bruh did i just do that ;-;

[obtuse plover]

LOL

[lean otter]

ya

[obtuse plover]

This is what I end up with

WAITR

[lean otter]

the y's should cancel

since f is linear in y

[lean otter]

working is good now! (though write x_0 there, so you dont get confused)

[obtuse plover]

Ohh, I see, that makes sense. Because I am evaluating at (x_0, y_0) (1,-5) okkk, Thank you for all the help Toby, i will check if the answer is correct now

yayyyy

its right!

ty!

[lean otter]

yw :D

[obtuse plover]

.close

[safe radish]

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[river wharf]

what is that

mind writing it better

@lean otter

[quasi bison]

so somebody else wrote this, not you...?

[river wharf]

so your problem is just a simple calculator problem? and you go on other channels seeking help

[quasi bison]

we can't help you.

we don't know what the problem is.

[torn vessel]

What do we have to do

[quasi bison]

all we see is some poor handwriting that looks vaguely like some kind of expression

[river wharf]

well you can divide 154 with pi and you get 49 over 4

so it's √49/√4

do .close

hey Ann it's been 2 years

[safe radish]

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[safe radish]

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[turbid path]

this problem looks like its missing something but i could be wrong

[compact ferry]

sounds like theyre asking for the derivative

[turbid path]

could you explain, i dont think i even learned this in class..

i know it has something to do with e, but what

would it be like 4.5e^x and then solve x

[safe radish]

@turbid path Has your question been resolved?

[turbid path]

.close

[safe radish]

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[lean otter]

i am confused about Remainder Estimate for the Integral Test

[safe radish]

@lean otter Has your question been resolved?

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[lean otter]

in second last step, i dont get how it is -b = -2?

[lean otter]

b corresponds to -2

how is it b = 2?

[devout shale]

tiny, are you following an answer guide?

[lean otter]

i am watching it through a video

[devout shale]

Okay

so you have 2 equations

and one of the equations says that

negative b equals negative 2

divide both sides by negative one

this will make both sides of the equation positive

so that b equals 2

[lean otter]

i have always done another step in place of this one

[devout shale]

and what is that?

[lean otter]

i am not able to explain

i maybe might come across a question like that

[devout shale]

You could instead do this

[lean otter]

it is -b = -2

[flat frigate]

AustinU

[lean otter]

this one i would do

[flat frigate]

AustinU

[devout shale]

both ways work just fine!

[lean otter]

if you do this then will it be like -b = -2, -b = -1/2?

[devout shale]

No

[lean otter]

okay and this also wont work lsuch as -b = -2 if u cancel both minus signs, this too not right?

.close

[safe radish]

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[autumn rock]

i need help with 5c

[autumn rock]

i dont understand what its asking for

since i dont recognize it as a node

[peak estuary]

the nodes are the subsets of {1,2,3}

draw a few dots and to each one you write one of the subsets as its name

and then put the edges that you found in (a)

[onyx leaf]

Is someone please able to help with this ? Thank your very much, is it by using taylor polynom or not ?

[autumn rock]

thanks i got it @peak estuary

.close

[safe radish]

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[lean otter]

in question b, if i have to find y do i have to make it like this y = a^-1 i

[lean otter]

in few questions that i have come across, they tend to write inverse first

i think i have misunderstood my question

.close

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[lean otter]

Can someone please explain me this?

it's a definition. which part don't you get?

[lean otter]

$$
\|A\|=\max \{\|A \vec{v}\|:\|\vec{v}\|=1\}
$$

[flat frigate]

Mohmoka

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[peak meteor]

I need help with this question..this is my first time encountering double | | (I'm not a native English speaker so I'm sorry that I don't know what they are called)

[peak meteor]

I've asked cymath.com for help and it returned the answer x > 3

But Photomath returned
x € [-∞, -4] U [-2, 1] U [3, +∞]

[buoyant shadow]

the second one makes more sense

[peak meteor]

Ye

But Photomath gives really broad answers and I've already had a situation where it gave me a wrong answer

[pure agate]

We call them Absolute Value signs in English

[peak meteor]

oh so it's same as in serbo-croatian..good to know

[pure agate]

And to solve them, analyze each possible case of + and - for each expression. There will be four cases which will result in the domain found by Photomath.

Then do the same for the next group of absolute value signs. This will result in four inequalities.

[peak meteor]

ah

so each one of these 2 cases has 2 additional cases like this?

[pure agate]

Correct.

The fourth one is incorrect.

It shoud be -1, not +1.

[peak meteor]

oh ye

a typo

ty man

[pure agate]

yw

[peak meteor]

so how do I close the channel now?

[pure agate]

.close

[safe radish]

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[loud ember]

Why abs has 2 solutions for every integers

[misty bay]

it doesn't

[loud ember]

Did i heard lie then

[buoyant shadow]

it doesn't

[misty bay]

well someone might not have said that exactly

you might've just remembered something wrong

[buoyant shadow]

it's mostly true, it's not accurate

[loud ember]

±Every integers isnt?

[peak meteor]

absolute value is a distance of a number from 0 so for example
| - 5 | = 5 & | 5 | = 5

[buoyant shadow]

abs(x) =m has 2 solutions for almost any integer m, if you don't involve complex numbers

[loud ember]

Only for prime numbers because 5 is?

[lean otter]

you are missing the obvious one, i guess

[loud ember]

Isnt ±x = |x|

[peak meteor]

| ±x | = x

[loud ember]

So it has 2 solutions

Because x^2 = -x^2

And $\sqrt{x^2} = \pm x$

[flat frigate]

아리스킨충∪

[loud ember]

= also |x|

[lean otter]

i am not sure what your confusion is? @loud ember

[
\abs x = \map \sgn x \cdot x
]

[loud ember]

Then it is x^2 * -1

I think you meant sin

[lean otter]

no

thats the sign function

[loud ember]

And it would work on x = π/4

[lean otter]

,w plot sgn(x)

[flat frigate]

Results provided by WolframAlpha

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

Requested by ♡Lixera♡#5050

[loud ember]

Or 3 random integers

sgnxx = sgnx^2

So it would be x^2

[safe radish]

@loud ember Has your question been resolved?

[safe radish]

@loud ember Has your question been resolved?

[lean otter]

bro channel name

[fresh copper]

It's korean for ice cream gun

idk actually

[safe radish]

@loud ember Has your question been resolved?

[tropic sable]

e

[river wharf]

breh

[safe radish]

@loud ember Has your question been resolved?

[loud ember]

Is that 5

Its ±5 for abs of 5

[normal charm]

그니까 절대값은 어떤 숫자가 수직선에서 0, 그니까 원점으로 부터 떨어진 거리로, x가 양수이면

$|\pm x|=x$

[flat frigate]

R4f43l1006

[normal charm]

또 실수 중에서도 0은 절대값이 0 하나 밖에 없습니다

그리고 중3 과정에선

$\sqrt{x^2}=|\pm x|$

[flat frigate]

R4f43l1006

[normal charm]

이라 배우는데 그 이유는 좌변과 우변 모두 양수로 나오기 때문인데

루트와 제곱이 만나는 것하고 절대값은 엄연히 다릅니다

[safe radish]

@loud ember Has your question been resolved?

[broken yew]

,,\sqrt{x^2}=\abs{x}

[broken yew]

,,\abs{x}\coloneqq\begin{cases}\phantom{-}x,\quad x\geq0\-x,\quad x<0\end{cases}

[loud ember]

It can be \pm x

[lean otter]

this is resolved right?

[raven fjord]

nerdss

[loud ember]

?

I asked why

[normal charm]

what r u asking abt

[uneven jacinth]

oh you are still doing this?

[normal charm]

ill make it clear the absolute number of a real number is the distance between that number and zero on the number line

so the abs of a number cant be a negative number

[uneven jacinth]

you spelt mahogany wrong

@loud ember

[loud ember]

Isnt it \sqrt{x^2}

In defines

[uneven jacinth]

one second

[normal charm]

@loud ember 절댓값은 수직선에세 임의의 실수가 0으로 부터 떨어진 거리입니다

[loud ember]

Not because 2 solutions for abs?

[uneven jacinth]

네

[loud ember]

감사합니다

[uneven jacinth]

only the positive root

[normal charm]

그래서 |5|=±5가 아니라 |±5|=5인 것 입니다

[uneven jacinth]

네

[normal charm]

다 됐으면 .close 좀

[loud ember]

결과는 1개의 임의 정수인가요?

[normal charm]

네?

[normal charm]

0을 제외한 모든 실수는 절댓값이 같은 실수는 각각 2개 즉 ±x고, 0은 절댓값이 0 하나입니다

@loud ember 됐습니까?

[loud ember]

네

[normal charm]

.close 좀 부탁 드려요

[loud ember]

.치ㅐㄴㄷ

.close

[safe radish]

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[solemn vault]

Hello

[solemn vault]

What does more than mean

Like is it 100-200 or wot??

[whole acorn]

idk, but the person who wrote it probably meant 100-200

it should mean mark > 100 but that should include everything listed below it. Which it clearly doesn't

[solemn vault]

Yes

Isn't the CI always a <= x < b

a,b are the upper and lower limits

.close

[safe radish]

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[safe radish]

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[sleek kettle]

can someone explain this problem to me

[wind folio]

Find the range of x from the given inequality and compare it with given range of x

[safe radish]

@sleek kettle Has your question been resolved?

[sleek kettle]

what's the {x "|" x>7} symbol mean?

[wind folio]

Set of values of x satisfying the given equality is x>7

[sleek kettle]

dang

im still confused

[wind folio]

For all x greater than 7 the given equality holds

Then we have to find the value of m

[sleek kettle]

can you help me find that

[wind folio]

x>18+m/m will be the solution of the given equality

And the question has already given us the solution of equality that is x>7 and we have to find m

So( 18+m)/m=7
18+m=7m
m=3

[sleek kettle]

ohhhh

i get it now

thank you

.close

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[eager niche]

Hello! I need help with a combinatorics problem related to Project Euler problem #806. The text of the Project Euler problem in question is here (https://projecteuler.net/problem=806), but I can reproduce it if needed. In relation to the problem I have determined that there are two recurrence relations and a generating function, all in three variables, related to the question that I need to solve. The problem is I have no idea where to go from here.

[eager niche]

The recurrence relations are $$f(a,b,c) = f(a-1,c,b) + f(c-1,a,b) = f(b,c,a-1) + f(b,a,c-1)$$$$=f(a-2,b,c)+f(a,b-2,c)+f(a,b,c-2)+2f(a-1,b-1,c-1)$$ and the generating function is $$F(x,y,z) =\frac{1+x+z+xy+yz-y^2}{1-x^2-y^2-z^2-2xyz}$$

The main issue is that I need to find a closed form of $f$ so that I can determine $f$ for high values of $a,b,c$ such as $f(50000,32768,17232)$ and permutations of those.

It's also worth noting that $f(a,b,c) = f(c,b,a)$ and that the base case for $f$ is $$f(0,0,0)=1$$$$f(0,b>0,0)=f(odd, even, odd)=f(even, odd, even)=0$$

[flat frigate]

SherlockSage

[eager niche]

If this is too complex a question for these help channels, I'm happy to ask elsewhere

[stray socket]

Try #combinatorial-structures or #combinatorics

[eager niche]

Which channel is under No Access there?

That probably makes no sense to you. I can read combinatorial structures, but not the second one

[stray socket]

Oh apparently it's an archived channel

Whoops lol

[eager niche]

Ahh no worries then

I'll post in the other channel and then close

.close

[safe radish]

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[mint veldt]

ignore the 2

So, the absolute maximum is the highest y value.

Because it's the absolute maximum, this would include the entire domain of f(x)

[hot thistle]

yes

[mint veldt]

and x can't be 0

because of the ln and the restriction

oh! I took the derivative of this

it's 1/x

can I get a hint on this one?

I think I've got the main idea

[hot thistle]

1/x

is it positive or negative on the interval given

[mint veldt]

positive

[hot thistle]

so the function must be increasing over 0<x<=2

[mint veldt]

yeah

I can see that with the graph I wrote as well

[hot thistle]

if it’s always increasing, where must the maximum be

[mint veldt]

so

I feel like I know what you're getting at but I don't want to guess

ughh

Like I want to say infin?

But does that make sense?

[hot thistle]

you have the right idea

but we’re restricted on our domain

[mint veldt]

exactly

[hot thistle]

we want the largest possible value of x

[mint veldt]

that's where I'm losing

oh of x?

I thought the absolute max was a y value

[hot thistle]

it is

but since the function is increasing

a larger value of x means a larger value of f(x)

so we do want the largest value of x

[mint veldt]

well from the restriction, 2 is included

would 2 be the largest x value?

[hot thistle]

yes

in the domain

[neat kiln]

The best thing is the show the graph you drawn 👻

[mint veldt]

yeah

[mint veldt]

gotcha so far

[hot thistle]

that’s it

your largest value is at x=2

so f(2)

[mint veldt]

ohhhhh

I see

now

uhh

f(2) = ln(5*2) right?

[hot thistle]

yes

[neat kiln]

As

Yea

[mint veldt]

dang

see I tried that at first

but the question doesn't mention anything about how many decimal points to round to

so I figured I was missing something

Cause the natural log of 10 is 2.30258....

.close

[safe radish]

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[sullen hemlock]

consider concavity

[proud heath]

there is a website called desmos it may help u if u wanna make sure of ur ans

[lucid charm]

yeah, but I cant really put those points in to check

[lucid charm]

yeah

[sullen hemlock]

if f' is increasing from (-inf,-2) and (0,2) it should be concave up there

[lucid charm]

oh I see

[sullen hemlock]

also

from (-2,0), f' < 0

meaning its concave down

[granite idol]

no, it just means the slope is negative. you look at the 2nd deriv for concavity

closer to the answer

the right-hand part of the graph is not right

[sullen hemlock]

if f' is increasing on an interval, f is concave up on that interval

but yeah that graph isnt entirely correct

[lucid charm]

Looks better i think

[granite idol]

yes

[sullen hemlock]

funny looking graph

[granite idol]

yes which means f'', the second derivative, is positive

[neat kiln]

It's about perspective

[sullen hemlock]

😂

[lucid charm]

so is this it? i believe i am missing something right?

@sullen hemlock

[sullen hemlock]

looks correct

[granite idol]

it looks about right

[lucid charm]

so the points of inflection are the two peaks on -2,2 and 2,2 right?

[granite idol]

there is an inflection point where the 2nd deriv changes from pos to neg (or vice versa)

[lucid charm]

ah the one at the origin

[granite idol]

no... second derivative

look at the text in the problem

[sullen hemlock]

if you just look at the problem, the inflection points are pretty much given

[lucid charm]

so at -4,0

[sullen hemlock]

?

definitely an inflection point at (-1,0)

[lucid charm]

i dont see one

there

graph doesnt go through there

[sullen hemlock]

from (-inf,-1) f'' is negative

and then AT -1, we can see f'' becomes positive

so -1 would be one

(1,0) as well

[granite idol]

inflection points are at x=-1 and x=1

[lucid charm]

so i would label them here?

[granite idol]

yes

[lucid charm]

if you dont mind explaining, I thought inflection points were where the second derivative goes from increasing to decreasing, or vice versa

[granite idol]

they are... re-read the 3rd bullet point

[lucid charm]

ah, so even though tthis is the graph of f, those are the points where f'' has inflection points

[granite idol]

they are inflection points of f. which are found by where f'' is 0

[lucid charm]

I understand, so the relative extreme values are just at the two peaks correct?

[granite idol]

and?

[lucid charm]

the one that goes through the origin?

[granite idol]

yes

[lucid charm]

oh, why would that be one? just need some clarification because i thought that wasnt one since the functin goes to infinity on the bottom half

[granite idol]

it's a local minimum

[lucid charm]

ah I see, so extremes are both maximum and minumum

[granite idol]

yes

looks right to me

[lucid charm]

okay, thanks for the help. see ya!

.close

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[agile plank]

I have 2 questions:
#8: how do you find f6(0)? Do you have to manually take the derivative or is there an easier way? (I tried using the series and I got 0, but I'm not sure if that's how you do it/is the correct answer).
#9c: I'm not sure how to start this, I tried all sorts of things

[agile plank]

Here is my work so far:

[neat kiln]

How did you make the Taylor series?

[agile plank]

Oh. Um

I think I plugged x^2 for sinx then added the cosx taylor series

[neat kiln]

So you have like a formula sheet?

[agile plank]

Not exactly, but the taylor series about 0 for sin x is x - x^3/3! + x^5/5! + ...

and so I plugged in x^2 instead of x to make it sin(x^2)

then I added that series to the cosx series: 1 - x^2/2 + x^4/4! + ...

then I wrote down the first four terms of that

onto my paper

[neat kiln]

I might be completely off but doesn't the 4th term containing 6! denominator indicate that it contains the sixth derivate?

So everything in the sixth term except x^6/6! Is the sixth derivative?

Assuming it's about x=0

[agile plank]

OHhh yeah okay

I didn't think about that

[neat kiln]

I'm not entirely sure :p

We can of course confirm

[agile plank]

Yeah lol

So, then how do you do #9C?

[misty bay]

,w Taylor series of sin(x^2) + cos x around x = 0

[flat frigate]

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[agile plank]

Oh okie thank you lol

I'll fix my work

[misty bay]

your work might be right, not sure

I didn't really check

as for 9c, I assume you can find f(t) = something involving t

and then just integrate normally

[neat kiln]

So f^6(0) is 121??

[agile plank]

I guess? I'll double check that later

[neat kiln]

Why not now XD

[agile plank]

Okay I will

[neat kiln]

,w sixth derivative of sin(x^2)+cos(x)

[flat frigate]

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[misty bay]

,w sixth derivative of sin(x^2)+cos(x) at x = 0

[flat frigate]

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[misty bay]

seems right

[neat kiln]

💀

[agile plank]

okay I guess it's correct

[neat kiln]

How?

[agile plank]

Here

[neat kiln]

Why is it -121?

[misty bay]

wdym

[neat kiln]

Negative

[misty bay]

the series indicates it too

oh why's it negative?

because the term of the series is negative

[neat kiln]

Oh yeah

[misty bay]

remember the Taylor series is just f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + ...

[neat kiln]

Taylor series is pain in ass 💀

[agile plank]

Lol yes

[misty bay]

I cheat with Taylor series when finding weird limits

you can usually just substitute in the first terms of the Taylor series

the formal way to do it is with the algebra of limits and l'hopital's rule

but

[agile plank]

Okay so back to this, why did it say "hint use part b" then?

[misty bay]

because part(b) tells you to find f(t) = something involving t

then part (c) is just integrating that from -1 to -1/2

[agile plank]

I'm dumb, but how is finding the sum of the series, f(t)?

[misty bay]

well they're asking you to find f(x) in terms of x

without the series

[agile plank]

so that's f with respect to t?

I'm just confused on "t" how got into this

[misty bay]

t is just the variable of integration

[agile plank]

oh okay

OH OKAY

WAIT NOW I GET IT

THANK YOU SO MUCH!!! :)

That makes a lot more sense

.close

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[dense yacht]

How is this possible:

If x^2 + y^2 = xy = 4, then I can both reason that (x+y)^3 = 0 and (x + y)^2 = 12 ... how come?

(x+y)^3 = (x + y) (x^2 - xy + y^2) = 0

(x+y)^2 = x^2 + y^2 + 2 xy = 12

[empty gyro]

where the confusion exactly?

[dense yacht]

(x+y) ^ (3/2) = 0 or 12^(3/2)?

[empty gyro]

where did you get that from

wait. Is it (x+y)³ or x³+y³?

[dense yacht]

Oh, hahaha I got it!

SOrry, my bad

[empty gyro]

all good

I missed it at first too

Fun fact: x and y are complex numbers

[dense yacht]

All makes sense now... I'm too tired... I think that's it for me today. Thank you so much!

Thank you! Have a good one!

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[random star]

Hi

[random star]

I'm supposed to give two examples of 3x3 orthogonal matrices.

I already know the definition, which is that a matrix multiplied by its transposed should yield the identity matrix as result.

Now, beside googling a matrix with such property, is there a way I can make one from scratch?

[jade magnet]

one way could be I=I

or just think about what it means for a matrix to be orthogonal

[random star]

like I know a generic definition but I can't write one off my head

this one is from my book

[jade magnet]

it means that the columns of Q are pairwise orthornomal

[random star]

not supposed to use this fact cause its introduced later in the book

and btw that's true ofc

[jade magnet]

idk how else you would do it then?

just do one where you put -1 instead of 1

obviously still orthogonal and still length 1

just in the different corner of your room there are 6

no, 8 i think 💀