help-23

[safe radish]

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[earnest wagon]

The diagonal matrix is the matrix in which all elements other than Principal diagonal's elements are 0.

[earnest wagon]

I wonder if there is a nomenclature for a matrix in which all elements other than Off diagonal's elements are 0?

[granite idol]

I've seen it called a "hollow matrix" I think

[earnest wagon]

Ohh

There is also scalar matrix

A subcategory of diagonal matrix

Is there a similar subcategory of hollow matrix as well?

[granite idol]

by "scalar matrix" do you mean kI where I is the identity matrix and k is some scalar?

[earnest wagon]

Good to know this, I'm just tryna see if there is new knowledge like these

[earnest wagon]

Yes

Identity matrix is the subcategory of scalar matrix

These terms I've studied are related to principal/leading diagonal

Wanted to know , if similar ones can be reciprocated by off diagonal (lower left to upper right)

[granite idol]

I don't know if there are specific names for them

[earnest wagon]

I see

Will close then..

.close

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[wintry mantle]

How do I prove this? every time I try, I get to (1+n)x^n. x is defined as |x|

[empty gyro]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[zinc berry]

There are presumably three coordinates. so \nabla \cdot \vec{x} = 3

Cropping out all the set up does not help

[safe radish]

@wintry mantle Has your question been resolved?

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@wintry mantle Has your question been resolved?

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[harsh pine]

hey

[harsh pine]

[\lim_{x \rightarrow - \infty} \dfrac{|x|}{1-x}]

[flat frigate]

madmike

[harsh pine]

Is there something I can do about this?

the |x| confuses me

idk what to do

should I now do 2 cases?

[fluid token]

l'hop?

[harsh pine]

idk that yet

the original problem is

[flat frigate]

madmike

madmike

[harsh pine]

not sure if what I did there is correct?

.close

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[brisk siren]

Hi, is this integral well made?

[brisk siren]

I have problems when derivating the result (it gives me the same but instead of 4cos(x)^3 it gives me -4cos(x)^3)

[elfin yew]

what is sen?

[safe radish]

@brisk siren Has your question been resolved?

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[tall hare]

did I do this right?

[plucky elk]

,w plot cos(3x) + sqrt(2)/2 = 0 for 0 < x < 2pi

[flat frigate]

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[plucky elk]

,calc pi/12

[flat frigate]

Result:

0.26179938779915

[tall hare]

so no

[plucky elk]

,calc cos(pi/4)

[flat frigate]

Result:

0.70710678118655

[tall hare]

would it not be pi/4

[plucky elk]

not -sqrt(2)/2

,calc -sqrt(2)/2

[flat frigate]

Result:

-0.70710678118655

[tall hare]

i thought even odd identities mean i could use sqrt2/2 as well

so use 3pi/4?

[plucky elk]

cos(-x) = cos(x), not cos(-x) = -cos(x)

[tall hare]

right, so cos(sqrt2/2) = pi/4

cos(- sqrt2 / 2) = cos(sqrt2 / 2)

right

?

[plucky elk]

definitely not

cos(sqrt(2) / 2)) is not what you think it is

,calc cos(pi/4)

[flat frigate]

Result:

0.70710678118655

[plucky elk]

@tall hare read this

[tall hare]

okay, so i should use 3pi/4

right?

[granite idol]

and?

[tall hare]

5pi/4

[granite idol]

,w calc cos(5*pi/4)

[flat frigate]

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[granite idol]

yup

[tall hare]

look good so far?

[granite idol]

yeah, I think so

[tall hare]

heres the rest of my work, pretty sure its right now but just wanted to double check

[granite idol]

,w calc cos(3 * 5*pi / 12)

[flat frigate]

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[granite idol]

I think you're right. I haven't done trig stuff in ~25 years 💀

[tall hare]

@plucky elk all good?

.close

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[safe radish]

Please don't occupy multiple help channels.

[tall hare]

what do i do with the cosx to make this true

[worthy hemlock]

You already have a channel, please do not open more than one

[tall hare]

can u help with this one and ill close the other one

[knotty matrix]

Do they ask you to manipulate one side or can you cross multiply?

[tall hare]

manipulate one side to verify the other is true

[worthy hemlock]

Verifying is manipulating one side only

[knotty matrix]

I was never taught what verifying is

But looks like you already did some of it

[tall hare]

yup

[tall hare]

can u help me since u seem to know about this

[worthy hemlock]

You're taking the wrong approach

[tall hare]

so is my first step good or bad

[worthy hemlock]

First step is fine, writing cot in terms of sin and cos is fine

Next step is not

From there, I suggest finding a common denominator to combine the fractions in the denominator

[tall hare]

can i do this

pulling cos out of denominator instead of whole thing

[worthy hemlock]

Actually yeah, that was a better way to do it

[tall hare]

so its all valid?

[worthy hemlock]

Yes

[tall hare]

cool thanlks

.close

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[high wharf]

Can anyone help me undrstand how to integrate

i understand left hand side but not right

[arctic locust]

What is the antiderivative of 1/x?

[high wharf]

it is lnx

if i remember correctly

dont understand how symbols were assigned sorry

[arctic locust]

Yes

Here think that x is $\theta$

[flat frigate]

black_couscous

[safe radish]

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[dreamy verge]

i feel like my teacher is wrong (this is rounding off)

[dreamy verge]

according to his method, 2.3650 rounds off to 2.36, but how i've always learned it, shouldnt it round off to 2.37?

[mortal sandal]

2.365 you mean?

This is a weird non-standard kind of rounding that's occasionally used in computer science

[dreamy verge]

oh there are 2 systems?

[mortal sandal]

It's only done so that the amount of time something rounds up is balanced with the amount of time something rounds down

[dreamy verge]

ohhhh interesting

[mortal sandal]

There's an old story about a stock system or something that would always round down

[dreamy verge]

so its a valid method right? I was wondering if its just straight up wrong or just a non-conventional method

[earnest wagon]

Is this is the system in which is there is odd you increase the number by 1 and if it's even then you let it be?

[mortal sandal]

always basing it's new number on its old number, basically just doing a very long string of consecutive multiplications

[earnest wagon]

I learned the same system of rounding off in my class of physics

[mortal sandal]

And every time it finished a multiplication it would round down

[dreamy verge]

yep

[mortal sandal]

And there was like, maybe six significant digits at play here

[dreamy verge]

but i always thought it just depends on whether or not next no. >5, so i was confused

[mortal sandal]

So it would round from 927.483922848933 to 927.483

But eventually the errors accumulated and it ended up being way way lower than it should be

So it's an incredibly rare method

It would be more common to just always round down or something

[earnest wagon]

Well that's 927.484 thou

[mortal sandal]

Not if it always rounds down

I'm not even talking about just what happens with .5 cases

It's just a story about how occasionally it's helpful to have your overestimates happen exactly as often as underestimates

Like the joke about two statisticians going hunting, one shoots 5m to the left of the deer, the second one shoots 5m to the right

the first one jumps up and yells "We got him!"

[dreamy verge]

is this one more common in physics? this is a physics class not CS, so not sure why they taught this one

[mortal sandal]

I've never seen it in physics

[earnest wagon]

I was taught the same in physics class of youtube

[dreamy verge]

ic

[earnest wagon]

It's the system in units and dimension of ncert 11th

[dreamy verge]

yep thats what im studying rn lol

well JEE

but same thing

[earnest wagon]

Yh we are taught the same

Everything is alright with this

[dreamy verge]

cool thanks

.close

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[solar plank]

Question: Cooper drove 40m at a bearing of 12 degrees, to where he believed was John's house. However, the correct path to take was 60m at a bearing of 59 degrees. What distance and direction should Cooper now drive to get to John's house. Use right angle trigonometry

I'm struggling which right angles to find and use to calculate this

[quasi bison]

such great distances cooper's driving... the house should be within sight either way lmfao

have you made a diagram?

[solar plank]

yeah

C for coopers start, f for john's fake location and j for John's real location

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[solar plank]

<@&286206848099549185>

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[winter island]

I need to generate in Matlab a random set of real numbers between [0,1], centered in 0.7/0.8 with a strong left tail and close to zero elements near 1. I was thinking of using a Levy distribution, using the command makedist('Stable','alpha',0.5,'beta',1,'gam',1,'delta',0.8); but i can't understand how to 1) shift the tail from right to left 2) center the distribution in 0.7/0.8

[safe radish]

@winter island Has your question been resolved?

[safe radish]

@winter island Has your question been resolved?

[winter island]

Apparently there is no easy way to do this in Matlab. For reference (if anyone is ever interested), I ended up using the skewnorm function of scipy, normalizing the data with the parameters i want, and then importing the result in Matlab

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[normal quartz]

Hi all, I have this graph right, and I need to know how to find the period for the graph. However, the graph does not repeat cycles like I am used to, and has varying minimum/maximum values for each cycle. How would I find the period in this case?

[normal quartz]

I am used to the graphs being consistent and repeatable like above. I don't know where to start.

[safe radish]

@normal quartz Has your question been resolved?

[normal quartz]

<@&286206848099549185> does my question make sense?

[safe radish]

@normal quartz Has your question been resolved?

[normal quartz]

What I am trying to say is that my graph never comes back to its values. For example, at 0:00 it is at 201 and at 10:00 it goes to 200 instead of 201. so there is no consistency, and therefore no cycle.

[normal quartz]

change of question.

.close

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[safe radish]

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[dawn veldt]

Why is the rotation of the camera in the 3d perspective projection equation represented as 3 numbers rather than two? https://en.wikipedia.org/wiki/3D_projection#Mathematical_formula

[pallid nymph]

If theres 3 dimensions shouldnt you need 3 numbers to represent it?

Or 3 angles

I guess

[dawn veldt]

But if it's just rotation, you should just need 2 I thought?

[pallid nymph]

Why is this?

[dawn veldt]

considering spherical coordinates

[pallid nymph]

How many Euler angles are there?

[dawn veldt]

1 represents rotation theta, another is rotation phi, then there is radius

radius is distance from the origin, and with a camera, I don't think you need that.

[pallid nymph]

well the equation here is rotating by the Euler angles

so theres 3 of those

"This representation corresponds to rotating by three Euler angles (more properly, Tait–Bryan angles), using the xyz convention, "

it goes on to do what youre saying i think

[dawn veldt]

I guess I'm just not very familiar with Euler angles, do you know why they need 3 when 2 is viable?

Doesn't that introduce a lot of duplicate points?

[pallid nymph]

i dont think there'll be duplicate points

just looking at the matrcies

its like

rotate by x then y then z

[dawn veldt]

Doesn't the first and third step do the same thing?

Maybe it's easier to calculate?

[safe radish]

@dawn veldt Has your question been resolved?

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[fickle talon]

for a pure 50/50 game, is there a better betting strategy than martingale?

[buoyant shadow]

define better?

all in feels better to me

and not playing is better than all in

in terms of time you spend playing

[lean otter]

You look oddly familiar

Yep, the recursion problem girl.

martingale a strategy

[safe radish]

@fickle talon Has your question been resolved?

[fickle talon]

better odds of winning money

https://youtu.be/zTsRGQj6VT4

[buoyant shadow]

martingale feels too restrictive, my guess is min betting is strictly better

[fickle talon]

what does that mean

[fickle talon]

this makes sense yeah

[buoyant shadow]

i mean like suppose you have $50 and you lose $31 by doubling your bet from $1, what are you even supposed to do then

[fickle talon]

go home

💀

[buoyant shadow]

well then if instead you bet always $1 you have better odds immediately

[fickle talon]

I see

alright thanks

.close

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[slender juniper]

x^3y'+2x^2y=x^2+2

[slender juniper]

what is the order on this?

i have answered the first order

[lean otter]

!15m

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[slender juniper]

okay

<@&286206848099549185>

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[wooden palm]

Need help finding X

[mortal sandal]

Find the centroid in terms of x

at least it's y-coordinate

[safe radish]

@wooden palm Has your question been resolved?

[wooden palm]

how do i go about finding the centroids

35 would be the area for sector 2?

[safe radish]

@wooden palm Has your question been resolved?

[mortal sandal]

yeah

you can find the area of the 2 sectors (sector 1 area in terms of x)

find the centers of the two sectors

And then do a weighted average

[safe radish]

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[fickle trail]

if you can't find this under 10 seconds would you resort to quadratic formula?

i need to find the x-intercepts

i thought maybe 3 and -24 but it needs to multiply to +72

[plucky elk]

10 seconds on a test is probably too little

[fickle trail]

1 min

[plucky elk]

unless that exact problem was on a homework

[fickle trail]

yeah

[native cape]

how do you want to use quadratic formula here

its cubic

[fickle trail]

i changed it to be quadratic

[native cape]

no you didnt

[fickle trail]

the part i'm trying to factor

oh

yeah i guess quadratic needs the entire expression to be quadratic

[native cape]

if you factor it, when the thing you factor zeroes your function is equal to -9, not 0

[fickle trail]

=0

right so I can't use QF on this?

[twilit scroll]

why not just factorise the cubic?

[fickle trail]

ummmm

[native cape]

you can once you find any root of the cubic

[fickle trail]

how?

[native cape]

and the question is for sure made so that at least 1 root of cubic is rational (most probably integer)

[twilit scroll]

find the factors of 9 and then substitute them for x, whichever one equals to 0 it will be (x-(the value of x)

it might’ve been a confusing explanation sorry

but it’s basically the factor theorem

[fickle trail]

why 9? something to do with the constant?

[twilit scroll]

that’s just the way i’ve been taught it

[fickle trail]

so nothing to do with the constant?

[brave wolf]

I actually dont think each of the factors would be 9

0 has that very special property, because 0*anything stays 0

[native cape]

in general you can use rational root theorem or whatever its called

to find all rational roots in finite amount of time

[quasi bison]

rational root theorem

,w 2x^3 - 21x^2 + 36x - 9 = 0

[flat frigate]

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[quasi bison]

nasty

[brave wolf]

damn

[quasi bison]

@fickle trail show the original problem, if you would

[fickle trail]

for part C

i can sketch it but not sure if x-intercepts are super important to get exact

i know the following: local min/max, inflection point, and that it's end behaviour is quadrant 1 and quadrant 3

[quasi bison]

plot just the extrema and inflection points.

(just those, don't sketch the curve through them yet)

[fickle trail]

"extrema" refers to both f'(x)=0 and f''(x)=0?

the slope, and the slope of the slope

critical points

[quasi bison]

no, extrema refers to only the points where f'(x) = 0

well, ok, no

strictly speaking a point can fail to be an extremum despite f' being 0 there

let me rephrase my instruction: i want you to plot the points on the graph at which f' or f'' is zero

you've determined the x coords of those points in parts a and b.

[fickle trail]

the scale is stretched out on the x axis, otherwise the lines would be too close

[quasi bison]

yeah, was about to mention it's wildly not to scale.

[fickle trail]

so now I just draw a line from Q3 to Q1?

[quasi bison]

and those points are kind of fat.

[fickle trail]

lol

[quasi bison]

what's Q3 and what's Q1

[fickle trail]

i will put them on a diet, one sec

Q3 = bottom left
Q1 = top right

i believe it follows trig convention iirc

Q1 - Q4

for the word "quadrant" with cartesian plane graphs

(thank you, Professor Leonard)

generally it is helpful for describing end behaviour

this graph goes towards -Inf in Q3 and +Inf in Q1

[quasi bison]

aeugh.

[fickle trail]

lol wha?

[native cape]

its very nonstandard

[quasi bison]

ok im gonna sketch over your sketch

gimme a moment

[fickle trail]

https://tenor.com/view/laugh-crying-gif-24196106

[fickle trail]

what's a better way to describe end behaviour on a cartesian plan graph, rather than using Quadrants?

[native cape]

just say what happens in inf and -inf

[fickle trail]

yeah but you don't know what part of the graph that is just by saying -Inf and Inf

[native cape]

its a function of x. "what happens with f at infinity" refers toi f(x) for big x

[quasi bison]

this is what i was gonna suggest next

[plucky elk]

avid nobody ever references quadrants when talking about polynomials

[fickle trail]

oh ya i can get the y-intercept easily enough

so it's just the x-intercepts that are a pain with this one

[fickle trail]

Professor Leonard does

[quasi bison]

well i put the y-int in for you anyway

anyway like

end behavior wasnt what i was talking about

these uh. critical points and the way the graph looks around them

is what i was going for

[native cape]

you wont get them precisely unless youre willing to use cubic root formula

[quasi bison]

im taking it on faith that you calculated the y coordinates of all these correctly btw

cbf to check your arithmetic there

[fickle trail]

yes

it's correct

[quasi bison]

the arrangement looks feasible for a cubic

yeah so do you understand the little curve bits i drew near your points

[fickle trail]

so to "sketch" this graph I can just draw a line through these dots using the info given?

x-intercept is juts cherry on top?

[plucky elk]

who that

[fickle trail]

not necessarily required to get full marks?

[fickle trail]

the best YouTuber on planet earth for teaching math

imo

very very good math teacher tho

for real

[native cape]

the question isnt precise

i mean not yours but in the task you sent

[fickle trail]

https://www.ratemyprofessors.com/professor/1158298

[native cape]

but well it says "using the information obtained above". there was no information about zeroes of f

so you shouldnt have to know them

[quasi bison]

this is how i personally would have made the sketch

in order red then green then purple

this is all mouse-drawn so apologies for it not being neat at all plus it was drawn over an existing one

[plucky elk]

alrighty i'll keep that in mind

[fickle trail]

lol ya that y-intercept is making scale pretty bad

[quasi bison]

well where else would i have put it

[fickle trail]

he has some great playlists, all the way up to Calc 3 and beyond

they call it Calc 1, 2, 3 in California school systems

might be different elsewhere

[plucky elk]

yea that's fine for semesters

[tame sphinx]

Bi

[empty gyro]

[empty gyro]

Critical points? Maybe?

[quasi bison]

pretend i have the energy to care about that rn.

[safe radish]

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[orchid horizon]

If i have something like f(x),i have to use substitution,if i want to know when it crosses x. But how do i convert it back

[orchid horizon]

Do i have to use Arc Cos or what do i use

[stray socket]

For the function cos^2(x) + cos(x) - 2?

Well yeah you'd eventually have to use arccos

Because I'm assuming you let u = cos(x)

This means x = arccos(u)

[orchid horizon]

ok ty mate

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[raven coral]

Hello can I get help with this question

[raven coral]

I got the third part of the basis wrong and I'm not sure how

<@&286206848099549185>

[safe radish]

@raven coral Has your question been resolved?

[raven coral]

nvm I see my mistake

.close

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[calm comet]

$$ 4[8-(6-4)^2+1] $$

[flat frigate]

Skyelf

[calm comet]

I know a question this easy shouldn't be this confusing but I'm so confused

$$ 4[8-(6-4)^3+1] $$

[flat frigate]

Skyelf

[calm comet]

.close

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[marble otter]

my guess would be equal to 1 but im not sure if thats correct

im on my last attempt lol

[mortal sandal]

try drawing it out

and maybe consider what x^1/7 < x^1/15 implies

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[faint seal]

so whenever we solve an integral using Feynmann's technique, we use a particular case of that integral to find the constant of integration (after differentiating and then integrating wrt the parameter). But what I dont understand is, why does the constant of integration have to be the same for all values of the parameter?

[floral osprey]

well you get f'(a) = some function, gotta use FTOC to find particular values of f(a)
no?

[faint seal]

no i get that

so whenever you integrate you get a constant of integration

and you need to solve for that using a particular value of the parameter in the original definition of the parametized function

but I dont see why the constant should be the same for all parameters

[floral osprey]

why the +C doesn't change? by definition it's constant

[faint seal]

HMM

hmm

.close

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[floral osprey]

left unsure what the question was or if I answered it

[faint seal]

yeah no i think i was just asking a dumb question

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[left willow]

Prove that a group G of order 24 cannot be simple. [Hint: Consider the action of G on the set of 2-Sylow subgroups of G by conjugation, noting that the kernel of this action is a normal subgroup of G.]

[left willow]

Could someone guide me

<@&286206848099549185>

[topaz tree]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[left willow]

Lol it was 15mins by the time u sent that

I was looking for direction rather than a spoonfed answer

But alr

[young nexus]

the direction did you name in your hint.

[left willow]

Yea but it was not trivial with that hint

.close

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[earnest wagon]

Is identity matrix- both an involutary matrix and orthogonal matrix?

[obtuse jackal]

if by involutary you mean being its own inverse, yes

[earnest wagon]

Yes

Thanks for your answer. I've a follow-up question

[obtuse jackal]

by virtue of being symmetric and orthogonal

[earnest wagon]

Oh, so any matrix which is symmetric and orthogonal is involutary matrix?

[pastel verge]

Yes

[obtuse jackal]

orthogonal: A A^T = I, symmetric: A = A¨T

[pastel verge]

You have AA' = I and A=A'

[earnest wagon]

Ohh

That makes sense, thank you both

Also, is identity matrix the only matrix which is both symmetric and orthogonal?

If not, do you have a counter example?

[pastel verge]

No

[earnest wagon]

Ohh

Thank you

[earnest wagon]

Any counter example?

[pastel verge]

You want a counter? Try finding an example first

[earnest wagon]

Ok

[pastel verge]

Just think about positive and negative signs

[polar cloud]

or you could consider the matrix [0,1][1,0]

[earnest wagon]

That example works. Thanks. Now I'll think one by myself according to what faf said

One example I thought of: [0,-1],[-1,0]

[pastel verge]

Go ahed and think about trigonometric elements in matrix

[earnest wagon]

Ok

[pastel verge]

Done?

[earnest wagon]

Actually what are trigonometric elements

I studied trigonometric matrix, not elements

[pastel verge]

Nah I mean entries

[earnest wagon]

Oh you mean the entries excluding the leading diagonal?

Since they form a triangle

[pastel verge]

Do you think $\begin{bmatrix} \cos x & \sin x\\sin x&-\cos x\end{bmatrix}$ is symmetric?

[flat frigate]

fäf

[earnest wagon]

Yes it's symmetric

[pastel verge]

What do you think we get after doing a matrix multiplication with it's transpose?

[earnest wagon]

Identity matrix

[pastel verge]

Nah find the product

[earnest wagon]

[cos2x,0][0,cos2x]

That's what I found

So it's not orthogonal but only symmetric

[pastel verge]

Really?

Be sure

[earnest wagon]

It's [1,0][0,1]

Sorry for my previous errors

[pastel verge]

Ok

So there are many such matrices for values of x

[earnest wagon]

Yes

Thank you for the help. Now I'll close the channel then

.close

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[pastel verge]

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[little sorrel]

i get a different answer to b

can i get someone to confirm theirs?

[pastel verge]

Show it

[safe radish]

@little sorrel Has your question been resolved?

[little sorrel]

(60(1.006)^240-1))/(1.006-1)

=32025.74

ans= 32217.89

[pastel verge]

Ok Bob let's see where you went wrong

[flat frigate]

fäf

[little sorrel]

yes

[pastel verge]

You forgot that 1.006

,w 32025.74 × 1.006

[little sorrel]

o

[flat frigate]

Results provided by WolframAlpha

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

Requested by yes sam#5028

[pastel verge]

Good enough?

[little sorrel]

yes

could you help me with another question?

[pastel verge]

For sure

[little sorrel]

how to do 9b

[pastel verge]

Wait, I'm just zooming out reading it

[pastel verge]

Not really hard, what did you do?

[little sorrel]

idk how to do geometric series

with the value stopping at ==2

n=22

[pastel verge]

Write it in addition form

Like the first fund goes for 44 years so it's amount at end will be 5000(1.07)⁴⁴

[little sorrel]

i did S22 first

with the 1.007 =r

idk how to do the rest

[pastel verge]

So we go like 5000(1.07)⁴⁴+ 5000(1.07)⁴³+...+5000(1.07)²²

Oh wait it's 7 till this time

[little sorrel]

isnt it till 23?

[pastel verge]

22

[little sorrel]

ye

thought it was the 3% interest one

1-22

then 23-44

[pastel verge]

After this 5000(1.03)²¹+...+5000(1.03)

[pastel verge]

1-21 then 22-44

[little sorrel]

ye

but n is 22

for both

[pastel verge]

Nope

[little sorrel]

how?

he make deposit on 43

[pastel verge]

7% goes for 23 years and 3% goes for 21 years

[little sorrel]

43-21

[pastel verge]

43-21+1

[little sorrel]

-1

[little sorrel]

wouldnt that +1 be -1

[pastel verge]

Well whatever it is, we don't need to do this anyway

[little sorrel]

ok

[pastel verge]

Maybe not

[little sorrel]

so what would i do with the formulas

[pastel verge]

Whatever he funds at 43 goes on for 65-43 = 22 years

[pastel verge]

You get two gp

[little sorrel]

gp?

[pastel verge]

Geometric progression

[little sorrel]

o

[pastel verge]

Cmon do it

[little sorrel]

would the first one be [5000x1.07(1.07)^22-1]/1.07-1

@pastel verge

<@&286206848099549185>

[pastel verge]

No

First factor out (1.07)²²

Then go ahed

[little sorrel]

wdym

[pastel verge]

Factor out (1.07)²² from the first series

[little sorrel]

me no understand

[pastel verge]

Pls Bob try to understand

[little sorrel]

can you just show me

me confused

[pastel verge]

Ok

[pastel verge]

5000(1.07)²²(1 + 1.07 + 1.07² + .... + 1.07²²) now solev the brackets by geometric sum

[little sorrel]

o

5000(1.07)^22 x[1(1.07)^22-1)/1.07-1]

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[plain crest]

radius of cone in terms of slant height and surface area

[plain crest]

so far

oh wait

one second

i got this

[pure agate]

Do you recognize a Quadratic Formula with r being the variable here?

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@plain crest Has your question been resolved?

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[marble crystal]

i just want to know if the A is 2

and b it's +infinite

[thin bridge]

how are you getting 2 for a)

[marble crystal]

my bad it's infinite

but i don't know if it's - or +

[thin bridge]

consider the one sided limits

[safe radish]

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[surreal stratus]

$1 + ln(|x|) ?= 1 + ln(x)$

[flat frigate]

Penca53

[surreal stratus]

why these two should be the same?

[narrow sage]

they're not same

but they are equal if x>0

[surreal stratus]

why can I then simplify the first into the second? It doesn't make any sense

[narrow sage]

does the "?" mean something

[surreal stratus]

for context, I've calculated the derivative of xln(|x|), and the given solution is the one without abs

[surreal stratus]

it's a question, "is this equal to"

[granite idol]

if you call the LHS f(x), then e.g. f(-3) is defined. not so for the RHS

[surreal stratus]

take the derivative of $xln(|x|)$

[flat frigate]

Penca53

[surreal stratus]

I get $1 + ln(|x|)$, the solution is $1 + ln(x)$

[flat frigate]

Penca53

[narrow sage]

well in that way

1+ ln(x) actually gives the same y values as 1+ ln(|x|)

but the given solution does seem incorrect

bcoz according to the solution the derivative will exist only for x>0

so its wrong

[surreal stratus]

it's wolfram

[narrow sage]

wolfram sucks sometimes

[surreal stratus]

the last step is: simplify the expression, and it drops the abs

because then I have to study the sign of the derivative

but it's pretty different to ask $1 + ln(|x|) > 0$ or $1 + ln(x) > 0$

[flat frigate]

Penca53

[surreal stratus]

with the first I get $x > \frac{1}{e}, $x < \frac{-1}{e}$ $ while the second is just $x >\frac{1}{e}$

[flat frigate]

Penca53
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[surreal stratus]

ok, you got the idea

[narrow sage]

yea

[surreal stratus]

so, I assume I keep the abs, and get the final solution to be those two, the positive and negative?

[narrow sage]

yes

[surreal stratus]

ok, thank you very much 😊

.close

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[quasi bison]

make a variable for Anna's current age

incorrect

in the past Mary was as old as Anna is right now, i.e. x

and Anna in turn was half of Mary's age back then, or x/2

[safe radish]

@lean otter Has your question been resolved?

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[lean otter]

is a spiral a polar curve

[lean otter]

and what if my spiral is r = theta

[obtuse jackal]

Would make a spiral yes

It works on desmos btw

[lean otter]

uh im confused about

how to find the length of it

[obtuse jackal]

Pain and suffering ?

[lean otter]

like is it literally intergral sqrt(f(theta) + f(theta)^2)

[lean otter]

pls no

[obtuse jackal]

r = t
x = r cos t = t cos t
y = r sin t = t sin t
dx/dt = cos t - t sin t
dy/dt = sin t + t cos t
dl² = dx² + dy² = (1 + t²) dt²
L = integral of dl = integral of sqrt(1+t²) dt = pain and suffering but doable

sorry but yes

[lean otter]

what the freak is this

[obtuse jackal]

[lean otter]

can we go over it

pls

< cries >

[obtuse jackal]

just went over the proof of the arc length formula, but the polar version

[lean otter]

so what do i do wiuth that

[obtuse jackal]

find how to express the arc length as an integral (fine)
compute the integral (pain and suffering)

https://www.intmath.com/blog/mathematics/length-of-an-archimedean-spiral-6595
I'm happy

I got the formula right

the rest is your problem

[crimson field]

$L = \int_{a}^{b} |\vec{r'}(t)| \dd{t}$

[obtuse jackal]

or WA's problem if you so choose obviously

[crimson field]

If you're familiar with this formula, it can be derivative from here.

[lean otter]

not familiar with that

whats the thing above the r

[crimson field]

How do you normally calculate arc length?

[crimson field]

Derivative symbol like f'(x).

[obtuse jackal]

derivative of a vector function

[flat frigate]

stabulo

[crimson field]

That looks better now. 🙂

[obtuse jackal]

the other way round no ?

[crimson field]

Oh. I think I know what you normally use. Rip.

[obtuse jackal]

which is what I used. Works mighty fine

[crimson field]

It follows much better from the integral formula. 😢

[crimson field]

They might be better able to follow this if they had the integral starting point. 😢

[lean otter]

imm pretty sure i just did what you sent

like wouldnt it be sqrt(theta^2 + 1^2)

[crimson field]

That's the required integrand, yes.

[obtuse jackal]

As I said

[lean otter]

and then you d a u sub?

[obtuse jackal]

I'm curious. Is this better to find the formula ?

cause ofc it's the general one

how would you find |r'(t)| here ?

@crimson field

[crimson field]

It's a bit confused since position vector uses r and we use r in polar. I'll use r to distinguish the two.

[obtuse jackal]

usual

[crimson field]

Set r(θ) = <θ cos(θ), θ sin(θ)> then it would reduce to the desired result. We just consider θ the parameter instead of t now.

[obtuse jackal]

so you do the computation anyways

[lean otter]

do i need trig sub

[obtuse jackal]

right ? It's no better than just doing them directly ?

[obtuse jackal]

according to wolfram yes

[lean otter]

yeah then idk wtf would the trig sub be

[obtuse jackal]

t²+1, what would you like t to be

remember that tan² + 1 = sec²

[lean otter]

thats disgusting

i hate math

i computed it

gross

[obtuse jackal]

yeah right

good old inverse hyperbolic sine coming in

[lean otter]

good old i want to kms

thank you sm

.close

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[lean otter]

$\sum_{n=1}^{\infty} \frac{\left( 6 \right)}{n(n+2)}$

[flat frigate]

ABootLaTributi

[zinc wing]

im having trouble with this radical distribution

oops sorry

[meager chasm]

What have you tried

[lean otter]

I added the first four values of n and used it for the expression

[meager chasm]

Wdym

[safe radish]

@lean otter Has your question been resolved?

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[unique valley]

How have they created a sketch just from this information?

[unique valley]

As in what are these blueish lines around the circle and the line x=y

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[safe radish]

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[unique valley]

.reopen

[safe radish]

✅

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@unique valley Has your question been resolved?

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[safe radish]

@lean otter Has your question been resolved?

[mortal sandal]

i will not

dont ping random users to ask for help

[eternal gyro]

check line 9 if I where u

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@lean otter Has your question been resolved?

[safe radish]

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[fickle trail]

in terms of

[fickle trail]

V=pi r^2 h

that's about as far as I made it with this one

my answer for volume is units^3

area would be units^2

[jade magnet]

A=?

[fickle trail]

A= area

[worthy tendon]

The surface area is basically the areas of the two circles of the top and bottom added with the walls.

[fickle trail]

A= pi r ^2

[worthy tendon]

pi * 2r * h

[fickle trail]

wait, with my A I only found the top area, right?

[worthy tendon]

The top and bottoms, yeah

You need both top and bottom

[fickle trail]

so it should be 2(pi r^2)?

for top and bottom

[worthy tendon]

So A = 2 * pi * r^2 + pi * 2r * h

Yes

[fickle trail]

how about the surface area of the can itself?

is that what this is? pi * 2r * h

[worthy tendon]

A = 2 * pi * r^2 + pi * 2r * h

[worthy tendon]

The can has like the curved wall right?

If you flattened that wall then it would be a rectangle of width pi * 2r (diameter of circle) and height h

[fickle trail]

so 2 * pi * r * h is the surface area for the rectangle of a cylinder?

(first express h as a function of r).

not sure what they mean here

h(r)=?

am I on the right track here?

[jade magnet]

i mean they're correct algebraic manipulations

A is your function that you want to minimize

[jade magnet]

V=pi * r^2 * h is your constraint

[fickle trail]

so how would I manipulate this to be h(r)=

[jade magnet]

divide r^2 * pi

you treat V as a constant throughout

[fickle trail]

like this?

[jade magnet]

no you want to isolate h in your constraint

[jade magnet]

to substitute into your function

[fickle trail]

that's what h is?

[jade magnet]

yes but this is useless

[fickle trail]

where did I go wrong? I thought I did this part

[jade magnet]

do you remember the video you sent

professer leonard or smth

[fickle trail]

I haven't made it there yet heheh

I will watch it now

[jade magnet]

💀

just the 1st 15 min is enough for this problem aswell

[safe radish]

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[vocal rain]

I need help

[vocal rain]

Wut is 23/56×45/78

[hazy elbow]

Just use calculator

0.2369 smth

[vocal rain]

My parents won't allow me to use calculator

[hazy elbow]

Oh okay

use long division

[vocal rain]

U sure

[hazy elbow]

yes

[vocal rain]

Ok

.close

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[torpid fiber]

Hey if I have sin(90-2a) that turns to cos(2a) right?

[hardy lion]

sounds right

[torpid fiber]

thanks

.close

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[sly prairie]

unsure how to approach problem, deals with conditional probability

[quasi bison]

there are 6 questions on the test. can you name, in order, the number of options on each question after the student rules out two options each on the first two questions?

(answer as a comma-separated list of six numbers)

[sly prairie]

yes, but first, hi again

[quasi bison]

it has been a while (not)

[sly prairie]

(3/5, 3/5, 5/5, 5/5, 5/5, 5/5)

[quasi bison]

incorrect.

reread your problem and reread what i am asking you.

[sly prairie]

bruh

[quasi bison]

still incorrect.

reread what i am asking you.

[sly prairie]

3 for first 2 and 0 for last 4

[quasi bison]

you're getting closer, in that your answers are now natural numbers as they're supposed to be, but you're still wrong.

[sly prairie]

idk

can eliminate 2 for the first 2

5-2 = 3?

5 possible answers - 2 possible answers = possible answers for first 2 questions?

[jade magnet]

yes

[sly prairie]

30 possible answers --> 26 possible answers?

hello

[jade magnet]

is just this but dropping the fraction

they asked for possible answers

writing 5/5=1 means there is 1 possible answer

[sly prairie]

that's what im sayin

[jade magnet]

you never said that thoug

[sly prairie]

true

[jade magnet]

so now you know possible answers for each

then do probability for each

if there are 3 possible answers whats the probability to guess the correct one

[sly prairie]

1/3 probability

[jade magnet]

ok do this for each question

[sly prairie]

umm

1/3 for the first two

then 1/5 for the next four

[jade magnet]

yes

[sly prairie]

multiply together = 1/15

for a)

[jade magnet]

well no

[sly prairie]

.

[jade magnet]

its 1/3 twice

and 1/5 4 times

not just 1/3 and 1/5

[sly prairie]

hmm

(1/3 * 1/3 * 1/5 * 1/5 * 1/5 * 1/5)

[jade magnet]

yes

[sly prairie]

= 1/5625

[jade magnet]

,calc 3^2*5^4

[flat frigate]

Result:

5625

[sly prairie]

b) she will answer every question incorrectly

(2/3 * 2/3 * 4/5 * 4/5 * 4/5 * 4/5) ?

[jade magnet]

yes

[sly prairie]

= 1024/5625

c) she will answer at least one question correctly

hmm

1/26?

no way

[jade magnet]

y thats way too low

[sly prairie]

hmm 26 possible answers

[jade magnet]

whats the opposite of getting atleast one correct

[sly prairie]

one wrong

all wrong

[jade magnet]

we calculated that in b)

[sly prairie]

yep yep

[jade magnet]

so c) is the opposite of what we calculated in b)

[sly prairie]

4601/5625

= 0.82%

[jade magnet]

yes

[jade magnet]

82%

[sly prairie]

niceee

that was the last question

thanks for the help big man

[jade magnet]

NP

[sly prairie]

.close

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[wraith prism]

@lean otter hello @thin bridge

[safe radish]

Please don't occupy multiple help channels.

[wraith prism]

The channel is closed

So I'm here again

I didn't know what is the reason behind the mod n^3 it maybe gives positive value

[thin bridge]

`the reason is that
$$|n^3| = \sqrt{n^6}$$

[flat frigate]

ℝamonov

[thin bridge]

which isn't quite the same as n^3

[wraith prism]

Ohh yes it is

Right right

So you mean i have to divide it by root 6

[thin bridge]

and for the limit you're considering where n<0,
|n^3| can be expressed as -n^3

[wraith prism]

Hmm so

[thin bridge]

for $n<0$,\
$$|n^3| = \sqrt{n^6} = -n^3$$
use the forms most convenient for manipulation

[flat frigate]

ℝamonov

[wraith prism]

How it can be -n^3

It should be n^3 i

Complex number no?

[thin bridge]

is n^3 positive or negative when n is negative?

[wraith prism]

Negative

[thin bridge]

and recall the (piecewise) definition of the absolute value

$|\text{this}| = \begin{cases} \text{this}\ &\text{if } \text{this} > 0 \
-\text{this} \ &\text{if\ } \text{this}\leq 0 \end{cases}$

[wraith prism]

So what do you want to do exactly? Divide it with -n^3??

[flat frigate]

ℝamonov

[wraith prism]

Yesssss

[thin bridge]

as those three expressions i've mentioned above are equivalent

[wraith prism]

Yes it works when we use mod

| x|

[thin bridge]

dividing numerator and denominator by |n^3| is the same as
dividing numerator by -n^3 (as |n^3| = -n^3 for the what you're consider)
and dividing the denominator by sqrt(n^6) (as |n^3| = sqrt(n^6))

[wraith prism]

Ahh I'll be back. Got to go urgent work. Yes. I guess this will give me proper answer

I'll use it after 10 minutes

[safe radish]

@wraith prism Has your question been resolved?

[hazy elbow]

well if it is zero then it remains zero no need to make it negative

[thin bridge]

saves ink

avoids explicitly writing the
= 0 if this = 0
case

[hazy elbow]

yeah that

[wraith prism]

So when I divide the numerator and denominator

By dividing it |n^3| it gives -1

But this method is quite unusual to me and I'm not satisfied with it. Sorry

Is there any other idea for a solution?

[gaunt garnet]

First you should see that in the bottom $1 + n^6$ will become $n^6$ because n will be extremely big/small that +1 won't affect it. So you get:

$\lim_{n\to-\infty} \frac{n^3 - n}{\sqrt{n^6}}$

Then get the $n^6$ out of the square root

$\lim_{n\to-\infty} \frac{n^3 - n}{|n^3|}$

At this point you will see that 3 is the dominant power so $-n$ on top won't matter for n values that are extremely big/small

$\lim_{n\to-\infty} \frac{n^3}{|n^3|}$

And you will see that for negative values of n, this will always give -1 because if n is negative, top is negative and bottom is positive. So the answer should be -1.

[flat frigate]

Oğuzhan

[thin bridge]

you could go the roundabout way
and do a substitution like
u = -n

[gaunt garnet]

Yeah it might make it less confusing

Since working with infinity is easier than the negative infinity

[rocky ferry]

imo this is getting a bit complicated. why not bring the numerator into the square root, and then you are able to distribute the limit into the numerator and denominator, expand the numerator, and then use the trick to divide all the terms by the largest variable exponent to find the behaviors at infinity?

[thin bridge]

same issues

requiring you to consider the definitions of the absval

the approach i've mentioned is pretty much the simplest

[thin bridge]

what's unsatisfactory about it