help-38

[neon dirge]

if so then no, because then the rank of the matrices would differ

meaning that they are different linear maps

[languid coyote]

sorry yeah that's one representation where there's k zero columns and n-k non-zero columns

that should be a natural consequence right of the representation with k zero columns

since dim ker T = k

so the other columns are non-zero

[neon dirge]

ah you prob meant to write n-k nonzero before

yeah that's right

[languid coyote]

oh yeah my bad

yes

okay then

[neon dirge]

yup

[languid coyote]

for any general representation of this linear map though

it doesnt necessarily have to have k zero columns

the number of 0 columns it can have

can range from 0 all the way to k inclusive

[neon dirge]

yes

[languid coyote]

okay, and then in general i can think about it like

for every 1 dimension in dim range T

there will be a non-zero column in the matrix

at least

1

column

sliding scale from k zero columns and n-k non-zero columns all the way to n non-zero columns

but in the case where we have n non-zero columns

k of those will be in the span of the other n-k

[neon dirge]

sry I gtg, rlly don't want to cut off here, if you don't mind you can also write it in dms, will be available in a bit :)

[languid coyote]

no worries, this was super helpful i really appreciate the help!!

[neon dirge]

otherwise maybe another wanderer will come along and guide otherwise 🐛

[languid coyote]

sounds good

thanks!!

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[sudden sail]

hey

[sudden sail]

what should N equal for this to not have a solution

the answer is -inf:-20 if that helps

[trim jolt]

@sudden sail Has your question been resolved?

[edgy willow]

try rearranging things to 10^x's on one side and numbers on the other

something like a*b^x will always be between 0 and infinity, it can't reach exactly 0 or a negative

[sudden sail]

done that and got something like thhis

n=25t-20

t being 10^x

or 10^x=20+n/25

[sudden sail]

thats the piece ive been missing then

[edgy willow]

uh it'd be 16*10^x=n+21

[sudden sail]

oh shit im sorry I sent

a

wrong one

the +1

in the end

is actualy

l

10^x+1

or 10x10^x

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[sharp vine]

da fuck do I do in part a

[kindred pier]

Try law of cosines

[sharp vine]

aigh lemme see here

[sharp vine]

im makin idiot mistakes here gimme a minute

andddd i got it

thank you very much mr SWR

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[subtle briar]

hello i've been trying to find a pattern for this sequence but it just doesn't make any sense!! idk where to begin ;-;

[edgy willow]

try reading across

[subtle briar]

ok wait ill try this

i think the third sequence doesn't follow the same pattern as the 2 above

[edgy willow]

third sequence as in the bottom 3 honeycombs?

[subtle briar]

yes

[edgy willow]

how would you know that it doesn't follow

you'd have to fill in the bottom right part right

[subtle briar]

no 😦 i'm supposed to only pick from the given choices and none seems to match

here

[edgy willow]

are you asking like how do you add blanks?

you sort of have to guess the number system you're using, it's almost like binary if you know that

[subtle briar]

what i did was this

[edgy willow]

ok ngl I can justify both A and D so uhhhh

[shrewd ridge]

whether you read vertically or horizontally it's the same thing seemingly

[subtle briar]

same to D

thank u sm!!!

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[shrewd ridge]

d makes some sense

[subtle briar]

i havent looked for A

but did u get it from reading vertically? @edgy willow

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[bronze plover]

Not sure how to start. I said force of gravity in x direction (parallel) - tension = ma. And that's about it lol

[bronze plover]

I seriously have no idea how to approach pulley with mass problems

[placid juniper]

suvat

get s with trigonometry

u is 0

v is the number in question

and u get acceleration from F=ma

[bronze plover]

what do u mean by "s" and "u"

[bright quarry]

so this is a conservation of energy question

alternatively you could use torque

and forces to find the acceleration

[bronze plover]

I thought about using energy but wasn't sure how the pulley would affect that

and then i wasn't sure how to find tension if i wanted to use torque lol

I was going to say the initial potential energy of the block + the rotational kinetic energy of the pulley = final kinetic energy of the block? That didn't seem right tho and I don't know the angular speed either

[bright quarry]

well it’s released from rest

so there’s only potential energy

and the pulleys potential energy doesn’t change

[bronze plover]

So i can literally just set mgh = 1/2 mv^2?

[bright quarry]

so if you used conservation you’d have Upulley+Ublock=Kblock+Upulley+Kpulley

[bronze plover]

oh lol

[bright quarry]

Upulley is on both sides

cancels

so since it’s a disk we assume I=1/2 MR^2

the rest should be trivial algebra

[umbral maple]

s is distance and u is initial velocity, not much point doing that method if you weren’t taught about it tho

[bronze plover]

Ok just to clarify as well, the initial potential energy of the block is being split between the final kinetic energy of the pulley and final KE of the block right

[bright quarry]

you mean kinematics?

[bronze plover]

And I can just use v = rw to get rid of angular speed as well?

[bright quarry]

mhm

yes

and the block isn’t rotating

[umbral maple]

yes

[bright quarry]

so it only has translational

[bright quarry]

not sure wym using kinematics you would need to use torque to incorporate any kinematic equations

[bronze plover]

We do have problems in the future where it it's a sphere rotating, so in that case I'd do initial PE of sphere = KE of pulley + KE of sphere + rotational KE of sphere

[bright quarry]

yes

but a block won’t rotate

so it’s rotational energy is zero

[bronze plover]

Ok awesome, that actually makes a lot of sense

Thank you so much!!

[bright quarry]

you’re welcome!

[bronze plover]

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[bronze plover]

Man i had such confidence but I got stuck again at part B lol

I said net torque is equal to moment of inertia * ang acceleration

Then said net toruqe is equal to radius * tension

but I dont know how I'm supposed to get tension

Could I set up a system with net force and net torque and use a = radius * angular acceleration?

I'm a little confused about v = rw and a = r*α.. I thought v and a corresponded to a spot on a rotating object, but they're also used as velocity and acceleration of the center of mass of the block?

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[bronze plover]

I'm here

[bright quarry]

hello

[bronze plover]

Hey man, sorry I don't know why I'm not understanding rotations. Our whole final is on rotations tho so I'm trying to figure them out lol

[bright quarry]

yes

[bright quarry]

oof

[bright quarry]

center of mass velocity is for rolling motion

the center of mass linear velocity for the fixed pulley is zero

it’s not moving

but yes

[bronze plover]

So does the equation v = v of CM = rw just mean that the velocity of the center of mass of the rolling object is the same as the velocity of the point where the "string" is coming off of the pulley

if that makes any sense lol

and same idea with acceleration?

[bright quarry]

so

for the pulley

v corresponds to the velocity of a point as it moves around the pulley at a distance r from the axis of rotation

this all comes from s=rtheta

from geometry

or the arc length equals the radius multiplied by the angle

then to derive a and v

we differentiate with respect to time

for both equations

[bronze plover]

So it is just a point and when you set the radius to being the radius of the disk, it becomes the velocity of the cm of the rolling object

That would make sense intuitively to me

[bright quarry]

yes

[bronze plover]

I might need to watch some remedial geometry videos

[bright quarry]

$100 dollars is a hefty price💀

[bright quarry]

just look up arc length geometry is you forgot what i was talking about

like length of a sector

[bronze plover]

it's worth it tho trust

[bright quarry]

sure

you should have mgsintheta-T=ma

TR=Ialpha

and you can solve this

[bronze plover]

I did that as we were talking and ended up with a of cm = 1/3 g sin(theta)

Does that seem right? that it's only dependent on theta?

I'm prone to algebra mistakes

[bright quarry]

so R cancels

T=(1/2)(ma)

mgsintheta=3/2ma

a=2/3 gsintheta

or wait

[bronze plover]

I can send a picture of my work, I just figured you wouldn't feel like checking it, and hopefully the prof posts the key in a few days

[bright quarry]

3/2 ma

so how’d you get 1/3

did you forget the 1/2

[bronze plover]

im looking lol idk

[bright quarry]

in inertia

[bronze plover]

idk I checked it earlier and check it again but I get pretty blind when I check my work

[bright quarry]

show

[bronze plover]

I'm more worried about being able to set these up and understand them

[bright quarry]

oh wait

i messed up i think

the m isn’t the same

yea block is M/4

my fault

yea nice job

looks good

[bronze plover]

All good, thank you lots for helping me set it up. It's feels simple after I do haha

[bright quarry]

i’d hope so

[bronze plover]

Onto part c... something tells me I'll be back here soon

[bright quarry]

LOL

you can do it

believe in yourself

[trim jolt]

@bronze plover Has your question been resolved?

[bronze plover]

alright I may just be stuck on some algebra, or I may have set it up completely wrong

I'm not sure

Basically I set the initial PE of block = the total energy at y(c) = the KE of the block at the bottom

but I got stuck with two variables, y(c) and v at y(c)

[bright quarry]

hello

back again

[bronze plover]

I live in this channel now

[bright quarry]

oh wow this is a fun question

so interesting

love physics

[bronze plover]

I panicked when i read it, i had to read it like 3 times before i could understand it lol

[bright quarry]

so we need to look at the two situations of course

before and after the string is cut

[bright quarry]

well hold on

you’re thinking right but

what’s the kinetic energy of the block at yc?

…

[bronze plover]

uh

[bright quarry]

nonzero

ITS MOVING!!!

does this make sense

it has velocity when the string is cut

[bronze plover]

I thought I accounted for that in my equation

[bright quarry]

so the total energy is the potential + kinetic

[bronze plover]

Wait wait wiatwaitiatiw

[bright quarry]

i didn’t look at your work

[bronze plover]

lol

[bright quarry]

let me read it but based on what you said

[bronze plover]

I said that right before the string is cut, energy is equal to KE of the pulley + PE of the block + KE of the block

all at y(c)

[bright quarry]

oh lol

then yes

💀

i thought you did PE block at yc=K block at bottom

but yes what you did is correct then let me read your work

[bronze plover]

I have been applying your earlier lessons on energy from part a

Attempting to lol

[bright quarry]

wait vyc is for the block

[bronze plover]

yeah

I said it was the velocity of the block right before the string is cut

[bright quarry]

you forgot omega squared

2nd line

and 3rd line when you plugged in

[bronze plover]

well

i guess I'd rather it be an algebra mistake than a conceptual misunderstanding

[bright quarry]

yea

you’re good at physics

you’ll do fine

[bronze plover]

I appreciate you looking through my equations

[bright quarry]

usually people need much more help

and i have to explain for 30 min

or longer

[bronze plover]

I've spent like 3 hours today trying to teach myself rotations before I finally decided to go on discord lol

[bright quarry]

you do it without even telling you what to do

[bright quarry]

physics enthusiast

incline problems are always fun

[bronze plover]

More like I had an existential crisis when every single problem on the practice final involved some kind of rotational motion lol

[bright quarry]

i like these rotation problems

[bronze plover]

You're the opposite of my Ta, he said he avoided them at all costs haha

[bright quarry]

lol rotation is so fundamental to physics

how is he the Ta💀

[bronze plover]

He's actually really good, honestly learned more from him in discussion than the prof, I unfortunately had class during his office hours though

[bright quarry]

never learned from a teacher in school tbh

i always just learn on my own through books and the internet

[bronze plover]

Yeah I was told that this is my introduction to classes that are a lot of self-teaching, as this is my first physics class

[bright quarry]

yea you need to self study for stem

you can’t just go to class and that’s it

[bronze plover]

Yeahhhhhh I'm realizing that now lol

[bright quarry]

are you a math major?

[bronze plover]

But it's really cool, I've never thought about all the conservation laws and stuff, like it's crazy

[bright quarry]

or some sort of life science

[bronze plover]

I'm undeclared

[bright quarry]

so you’re taking physics 1

[bronze plover]

Yeah very first intro physics class and im struggling lol

[bright quarry]

conservation of momentum is newton’s third law

good video

[bright quarry]

you didn’t take physics in high school?

[bronze plover]

Not really, I did an online thing which covered the main topics in a super basic way

questions with two steps max

[bright quarry]

seems like intro physics

algebra based

almost no problems are more than 3 or 4 steps

most only take 2

[bronze plover]

Yeah it was also online so not to out myself or anything but...

[bright quarry]

cheated

despicable

[bronze plover]

I would never admit to that

I was not motivated back then lol

[bright quarry]

happens

[bronze plover]

But my brother is a physics major and he's always telling me about cool stuff he's learning which is kinda inspiring

[bright quarry]

man of taste

[bronze plover]

this guy is physics, electrical engineering, and philosophy minor

[bright quarry]

tell him to teach you Lagrangian mechanics

[bronze plover]

its crazy

[bright quarry]

or hamiltonian mechanics

he will be impressed

[bronze plover]

I told him ab having trouble with rotations and he told me to learn lagrangian mechanics in like 4 days

[bright quarry]

💀

[bronze plover]

before my final

[bright quarry]

see

😹😹😹

[bronze plover]

Yeah he said he almost failed his intro classical mechanics class but lagrangian is much easier

So I got hope for the future

[bright quarry]

principle of least action

real fun

[bright quarry]

i’d recommend taylor’s classical mechanics

[bronze plover]

As a supplementary textbook?

Or just to read in bed by candlelight

[bright quarry]

oh not for intro lol

it’s like 3rd year undergrad level

[bronze plover]

oh upper div stuff is a long ways away for me

I gotta not fail intro physics first

[bright quarry]

i can’t imagine you do

[bronze plover]

I appreciate all your help though, thank you so much

Super super helpful

[bright quarry]

mhm you’re welcome

good luck on your final

[bronze plover]

I might be back on here later (or really soon) bc ima start another problem set lol

thank you

[bright quarry]

ok if i’m still on i’ll help

it is 1020 though

[bronze plover]

Yeah no worries, you've already helped me a ton, I feel bad for taking up so much time

[bright quarry]

i wouldn’t do it if it wasn’t fun

volunteering

[bronze plover]

You physics people man

Such intellectuals

[bright quarry]

💀

dm me if you need help and i’m not in a help channel or ping me

[bronze plover]

I appreciate that! I'll try not to spam you haha

Nah bc I'm already looking at the next problem set... not looking good so far

[bright quarry]

you’ll be fine

just rotation

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[bronze plover]

Seems simple, pretty sure I'm missing something. I set initial PE of block = rot KE + final PE but im not even sure what rot KE would be since wouldn't it change instantaneously

Then I said all the rotational energy goes back into potential energy but that doesnt seem right

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[fluid compass]

The question asks me which equation is the solution of the value of x of point P and Q.
2x^2 +3x -4 = 0
3x^2 +2x -4 = 0
2x^2 -3x -4 = 0
3x^2 -2x -4 = 0
I absolutely have no idea how to solve for this.

[split chasm]

consider something simpler,
would you be able to determine where the lines
y = x + 3 and y = -x = 5
intersect algebraically , and if so how would you approach it. (just how, i don't really care about the final solution)

[fluid compass]

They both would have to same value of y?

[split chasm]

and then

[fluid compass]

Set the two equations to be equal to each other?

and slowly arrange them step by step

[split chasm]

yeh

you could view it in two ways here, either as substitution of y into the other equation,
or transitive property of equality

substitution is more relevant for what you have

[fluid compass]

Oh.... I understand it now

I absolutely forgot about substitution

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[fluid compass]

.reopen

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✅

[fluid compass]

How do I turn this to available so that other people can use it?

[exotic pine]

use .close, it will automatically be usable to other people after some time

[fluid compass]

.close

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[sterile egret]

Do you know range of
acosx±bsinx?

[merry zodiac]

Ohh yes

[sterile egret]

Use that

[merry zodiac]

But here we need to find solutions?

[-√k^2+9<=y<=√k^2+9] =k+1

[-k-1-√k^2+9<=y<=-k-1√k^2+9]

@sterile egret

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[soft badge]

i havent started but i think its something to do with quadratic model

[frozen plover]

drawing a graph of the function on the left would prove to be useful

then you need only find the interval for which y = k intersects the graph at 4 points

[soft badge]

how would i graph y=k

[worldly bloom]

oh

I see

My bad

[frozen plover]

its a straight line

parallel to the x axis

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[nova raptor]

for question b, what is the best way of finding P from here? i know that the magnitude of the lines leading from it to A and B must be equal to AC and BC (which i found to be sqrt29), but i'm honestly unsure what to do next.

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@nova raptor Has your question been resolved?

[vagrant prism]

notice that you got BC from AB and AC in part a

notice that should be symmetric around AB

so you should invert vector AC and do the same thing

[nova raptor]

hmm i see

i'll give that a try and see how i go

thank you

[wraith hinge]

anyone know how

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[next narwhal]

does anyone get this?

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[next narwhal]

<@&286206848099549185>

[dark herald]

can you type?

[next narwhal]

hold on

Drawn below is a sketch of a bus intersection. To ease the traffic situation, new roads have to be built. The city engineer suggested constructing a road passing to the Gingerbread building. The building is already abandoned and is schedule to demolition. If the road is to be parallel to the side of a triangular history Park that is not a long of a along the road, how long will the road be?

sorry for my handwriting lol

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[edgy willow]

it's hard to tell what's happening in the picture 😢

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[strange delta]

Find the arc length of the following
curves on the given interval.

[uncut aspen]

what's the formula for arc length?

[strange delta]

I've gotten to-

But that seems like an insane integral for this class

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@strange delta Has your question been resolved?

[strange delta]

<@&286206848099549185>

[uncut aspen]

are you sure you are not allowed to use a calculator for this class?

for homework on arc length, we were allowed to use calculators

[strange delta]

we're allowed to use calculators

[uncut aspen]

your calculator can do integrals, right?

[strange delta]

you can't use graphing calculators that do derivatives/integrals

ti-83/84 series only

[uncut aspen]

those can do integrals

[strange delta]

where? Found it, for definite integrals at least

either way, these have to be done by hand for credit

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@strange delta Has your question been resolved?

[radiant star]

turn it into a single fraction

[strange delta]

I shall try

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[wraith hinge]

hi! im a bit confused. this book, when talking about the history of sets it says "a major stumbling block was how to use sets to define an ordered pair because the definition of a set is unaffected by the order in which the elements are listed." I dont quite get this, why would we need to use sets to define an ordered pair? why cant we just define an ordered pair as a pair with distinct elements where order has significance? Please help me understand the reasoning behind this. btw the answer proposed to this was {{a},{a,b}}

[grim sparrow]

hold on

[scarlet escarp]

because the classical building blocks for mathematics are sets

[grim sparrow]

you've posted this in 3 places already lmao

[wraith hinge]

oh okay ty!

[scarlet escarp]

it's like if i asked you to define what 1 is

[wraith hinge]

sorry i rlly wanna sleep

[scarlet escarp]

you can't just say "well it's just 1"

the building blocks you can use are sets, so you would define it using sets

[wraith hinge]

aah i see, so its kind of a way of proving it??

[scarlet escarp]

no just defining it

[wraith hinge]

oh

[scarlet escarp]

what you can prove though

[wraith hinge]

yeah idk..

[scarlet escarp]

is that (a,b) =/= (b,a) using this construction

which is exactly what we want

[wraith hinge]

okay thats the perf answer i dont know why i thought i caught my textbook in a error

thank you!

[shrewd ridge]

slothcucumber.png

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[civic thorn]

How do I solve $7\cdot\cos(\frac{\pi}{2}x)=3$ over 0<=x<2\pi$

[solid kiln]

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

[civic thorn]

not quite sure why it says its wrong everthing there is right lmao

anyway im not sure how to solve this, im pretty sure its gonna involve some sort of identity but im not sure how to solve it

nvm im a dumbass haha

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[civic thorn]

How do I turn $\frac{\sin(x)}{\sec(x)-\cos(x)}$ into a single trig function with no fractions

[solid kiln]

guy

[civic thorn]

so far what ive done is $\frac{\sin(x)}{\frac{1}{\cos(x)}-cos(x)}$ which i havent been able to do muchw ith

[solid kiln]

guy

[scarlet escarp]

try multiplying the top and bottom by something

[civic thorn]

like sine?

[scarlet escarp]

like it, but not that

[round mango]

go ahead, simplify the denominator by cross multipltying or otherwise

[civic thorn]

ok im not sure then, i was thinking sine because it would make it a sine^2

[civic thorn]

ok

ill try that

[scarlet escarp]

well what other thing would be good to multiply by?

[civic thorn]

cosine is my only other guess, but im not sure wy

why

[round mango]

a general tip, never stop your own chain of thought always keep moving your hands when doing math, youll always arrive somewhere

[civic thorn]

okay will do

[round mango]

good

continue with your knowledge of fractions

[round mango]

perfect

you almost got it

[civic thorn]

OHHH i think i get it

one sec

[round mango]

nice

[civic thorn]

awesome

tysm for the help

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[vapid tiger]

Helep

[vapid tiger]

Why p suddenly becomes multiply

Btw p equals to lg5

[fallen zealot]

what is q

[vapid tiger]

q=log2 base 3

[fallen zealot]

so which part you dont understand exactly

[cyan zinc]

what's the point of any of this

[vapid tiger]

Express in terms of p and q

[vapid tiger]

Wdym

[cyan zinc]

like what's the goal here

[fallen zealot]

.

[vapid tiger]

How does the p turn to multiply in 3rd step

I'm lost after it substituted the p

[fallen zealot]

it was always a multiply ?

a/b = a * 1/b

[vapid tiger]

Huh how

[fallen zealot]

https://ibb.co/N300sBg

[vapid tiger]

Ah mann didn't see thsy

[fallen zealot]

np 🙂

[vapid tiger]

FAYYUM

SANKYU

.close

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[vapid tiger]

I'm still lost

Why is there 1 - p there suddenly

.reopen

[trim jolt]

✅

[vapid tiger]

Log is so annoying

It's so time consuming

I don't have time

I got physics and chem to cover dammit

Someone helep plssss

[vernal warren]

u get this
$$log_{3}5 - plog_{3} 5 = pq$$
here we can factor out $log_{3}5$ , and we then get
$$log_{3}5 (1 - p) = pq$$

[solid kiln]

JustToPro

[vapid tiger]

Bro how do you see this

Tell me how you can see that

Dang it

I'm still a long way away

[vernal warren]

practice ig

[vapid tiger]

How much practicr

[vernal warren]

i mean i saw pq was independent and the rest of term was missing

then i just saw what could have happened to the other term

[vapid tiger]

I think it's cuz my brain is fried. Have been studying phys chem and math back to back for 5 hours

[vapid tiger]

Such a simple explanation for a complex idea 😭😭

Ok tq

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[foggy dragon]

can someone help me do 1a?

its a very simple question but ive expanded it like 4 times now and i keep getting it wrong

im just curious if its just incorrect

you just have to sub in 3i + 2

and expan

can someone try it and tell me if they get 0?

oh nbmind i see what i did wrong....

fuck thats such a stupid ass mistake

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[vapid tiger]

Helep

[vapid tiger]

Why they divide

I mena what made you think that way

I wanna understand the thought process

No answer sheet in exams

Bruh why is maths like this

[lilac cloud]

probably so you can cancel out the a factor?

[vapid tiger]

Ok so in simultaneous situations like this

I can choose to divide multiply minus or plus?

In order to cancel out a factor?

Is that correct?

Breh I'll just use substitution method lah like this

[analog shale]

you're right, though the presented solution is efficient enough

[sacred sapphire]

you follow the operation of the thing you want to cancel

like in there the a is multiplied with something else

to cancel it you divide

if for example it was added with something like, a + something

you use minus

[vapid tiger]

Damn thanks for that

Mighty help

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[lofty elk]

i dont have any issue solving this, but i do have an issue with why k=8 is rejected

[lofty elk]

the way this is done for my syllabus is to apply the transformation to a point, and map that point onto another point on the line

https://www.quora.com/What-is-the-set-of-values-of-k-for-which-the-transformation-represented-by-begin-bmatrix-2-1-k-4-end-bmatrix-has-two-distinct-invariant-lines-through-the-origin-Why-is-k-not-equal-to-8-part-of-the-answer

[lofty elk]

this is a resource i found

so my questions is: why is k=8 rejected? every point on the line should be mapped to the origin, which is part of the line

[sacred sapphire]

2x + y is the same as 8x + 4y

they're not distinct lines

[lofty elk]

ohhhhhh

[scarlet escarp]

2x + y is not a line on its own

[lofty elk]

btw do you guys have a matrix calculator resource where i can play with this stuff

might be easier for me to visualise

actually nvm i think i got it

tysm guys ❤️

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[little oar]

Hello, I have the following problem: We have a point A outside of a circle with origo O and radius r, a point B inside this circle. We know vectors AB and AO, now we extend AB such that it intersects the circle, we call this point C, we are looking for the OC vector.

I was aiming to find the angle between OC and the x-axis because then I could just apply the rotation on a x-axis unit vector and scale it with r.

I think I got the COA angle, but I just can't see how to get the desired angle from it.

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@little oar Has your question been resolved?

[little oar]

I guess I can get the angle between -AO vector (OA) and the x-axis and then on some cases I need to add it to beta and sometimes subtract it...

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[little oar]

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[vagrant prism]

!status

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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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vagrant prism]

also is this the full problem?

no other context?

hm

i think this is an FTC question

fundamental theorem of calculus

also interesting obervation - if x = 3, then x^2(x+2) = 45

the integral becomes $F(x^2 (x+2)) - F(0) = x$

[solid kiln]

artemetra

[vagrant prism]

then differentiate both sides w.r.t. x

@sterile compass

derivative of a constant is 0

F(0) is a constant

huh?

F' = f

i think you did something wrong

$f(x^2 (x+2)) = \frac{1}{3x^2 + 4x}$

[solid kiln]

artemetra

[vagrant prism]

then use the fact that x=3 makes x^2(x+2) = 45, which is what you are looking for

so $f(45) = \frac{1}{3(3)^2 + 4\cdot3}$

[solid kiln]

artemetra

[vagrant prism]

seems like so

yes

because derivative of a constant is always 0

F(0) does not depend on the value of x

so it is a constant

no problem, if you are done type ".close"

uhh

numerically seems to converge to 2/3 but i am still thinking about how to prove it

i wonder whether this is any useful

b = 1 obv

do you need to find the limit or determine whether it is convergent or not

the latter is easy

$\frac{(2-1)(3-1)\cdots (n-1)\cdot (2^2 + 2 + 1)(3^2 + 3 + 1)\cdots (n^2 + n + 1)}{(2+1)(3+1)\cdots(n+1)\cdot (2^2 - 2 + 1)(3^2 - 3 + 1)\cdots(n^2 - n + 1)}$

[solid kiln]

artemetra

[vagrant prism]

ooooh

$\frac{(2-1)(3-1)\cancel{(4-1)}\cancel{\cdots (n-1)}\cdot (2^2 + 2 + 1)(3^2 + 3 + 1)\cdots (n^2 + n + 1)}{\cancel{(2+1)(3+1)}\cdots\cancel{(n-1)}(n+1)\cdot (2^2 - 2 + 1)(3^2 - 3 + 1)\cdots(n^2 - n + 1)}$

[solid kiln]

artemetra

[vagrant prism]

so $=2\frac{(2^2 + 2 + 1)(3^2 + 3 + 1)\cdots (n^2 + n + 1)}{(2^2 - 2 + 1)(3^2 - 3 + 1)\cdots(n^2 - n + 1)}$

[solid kiln]

artemetra

[vagrant prism]

n^2 ± n + 1 looks like a geometric sequence sum

so $n^2 + n + 1 = \frac{n^3 - 1}{n -1}$ and $n^2 - n + 1 = \frac{(-n)^3 - 1}{(-n) -1} = \frac{-n^3 - 1}{-n -1} = -\frac{n^3 + 1}{n + 1}$

[solid kiln]

artemetra

[vagrant prism]

so $=2 \frac{\frac{2^3 - 1}{2 - 1}\frac{3^3 - 1}{3 - 1}\cdots \frac{n^3 - 1}{n -1}}{(-1)^n \frac{2^3 + 1}{2 + 1}\frac{3^3 + 1}{3 + 1}\cdots \frac{n^3 + 1}{n +1}}$

[solid kiln]

artemetra

[vagrant prism]

okay the (-1)^n in the denominator is not nice

i am not sure where to take this further but i see some possible cancellations

hopefully some other helper will be able to help

i gtg

[turbid gazelle]

Can someone explain this

[fathom oasis]

Yo you

[turbid gazelle]

I get this part now

Its because when you diff F(x^2(x+2)) it becomes (3x^2+4x)f(x^2(x+2))

Ye

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[old spindle]

help with finding the b and c values of this graph

[old spindle]

i thought the b value would be 4 but it comes as wrong

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[old spindle]

<@&286206848099549185>

[old spindle]

<@&286206848099549185>

[azure glacier]

b = 1/2
a = 4
c = 11π/4

[old spindle]

are you sure about c it comes back as incorrect

[azure glacier]

throughly

[old spindle]

then what would this be

the value of a wolud be 3

but the rest im not sure

[azure glacier]

No, I think (a) should be 4 and (c) should be 1
As for (b), I can say that it becomes π/2

[old spindle]

all of it comes incorrect 😭

the max of the graph is 5 and lowest is -1

so isnt a just 6/2 = 3

yeah it is that comes correct

[azure glacier]

The maximum value of the blue curve is 5 and the minimum value is -3, the amplitude is half the difference: ( 5-(-3))/2
=8/2=4

[old spindle]

how is the minimum value -3 when it only goes down to y = -1 around x=-2

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[merry zodiac]

Maximum value of cos theta+cos 2 theta

[tribal jetty]

can you get rid of cos(2theta) ?

[prime lynx]

differentiate with respect to theta?

when d/dx(cos theta+cos 2 theta) = 0 {it is a turning point]

[marsh forum]

use $cos(2x)=2cos^2(x)-1$

[solid kiln]

Why am. I here

[marsh forum]

then compare it to the equation of a parabola

[marsh forum]

,w minima of cos(x)+cos(2x)

[solid kiln]

Results provided by WolframAlpha

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

Requested by physicsrocks

[marsh forum]

looks right to me

[merry zodiac]

Thank you

[marsh forum]

no problem

[merry zodiac]

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[crimson rover]

How do you calculate lambda in poisson distribution?

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.reopen

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[sterile musk]

This is a bit of a poor question, in what context? In general the parameter lambda is equal to the mean number of successes over a certain time period.

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[brittle pier]

Could someone help me get started on solving these types of problems?

[edgy willow]

there are equations but they all come from the fact that if you add distributions, their variance adds as well

variance is o^2 so with 100 students you'd have variance 100o^2, then you'd have to take the square root to get the standard deviation you want

[brittle pier]

so it would be sqrt (100 * 0.2)

?

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@brittle pier Has your question been resolved?

[edgy willow]

oh the 0.2 is outside the square root

sqrt(100o^2)=10o=10*0.2

[brittle pier]

how did you know to use variance?

[edgy willow]

you just have to know that changing the variance is easier and convert to standard deviation at the end

it's from how the variance formula is just adding a bunch of things but the standard deviation formula has a big square root

like for a distribution the "natural" parameters are mean and variance

[brittle pier]

So in this problem, they are basically giving us the average and standard deviation. So from that, you took the variance of that and used the 100 students as a constant?

[edgy willow]

yea 100 is how many distributions you're adding up

usually called n in all the formulas

[brittle pier]

Ohhh I see

[brittle pier]

So if i wanted to find the average of the new 100 student distribution, I just multiply the old average (8oz) by that constant then right?

[edgy willow]

right

[brittle pier]

wow

i must not be able to read because I thought this problem was a lot harder lol

[edgy willow]

haha

[brittle pier]

ty so much !!

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[worldly monolith]

How do I find this when the mean and x are the same

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[edgy willow]

usually you'd adjust the standard deviation by o/sqrt(n) and do z-scores, here since the mean is the same it's z=0

but like clearly if you have a bell curve centered at x then the chance something is greater than the middle is 1/2 right?

[worldly monolith]

That’s what I figured but I didn’t think it was gonna be that simple 😭

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[civic thorn]

Why do we get multiple answers from this?

[scarlet escarp]

well do you see why
cos(x) = 0
has multiple answers?

[wraith hinge]

cos is a periodic function

[civic thorn]

like i already found one answer, but im not sure why or how there are more

[civic thorn]

oh yeah

so will there always be multiple answers for a cos function?

[wraith hinge]

arccos is one-to-one so you aint getting another solution with it

[civic thorn]

then how do i find the others?

[wraith hinge]

reference angles are your friends

[civic thorn]

how do i use them to find the other answers tho?

im looking at my teachers work, and mi not sure why he did this

[wraith hinge]

well its because cos(theta) = x in the unit circle

so

you have an angle in the first quadrant that is positive

you want to find where else 3/7 occurs in-between the interval 0 and 2pi

the second and third quadrants are out rhe window because x is negative there

[civic thorn]

oh so basically all we're looking for is where 3/7 occurs in the circle

[wraith hinge]

yes

[civic thorn]

ok that makes more sense

thank you

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[civic thorn]

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✅

[civic thorn]

wait one more thing

why is he subtracting the version of arccos(3/7) before its multiplied by 2/pi

should he do 2pi-(arccos(3/7)*2/pi)

instead of 2pi-arccos(3/7)

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[cosmic plinth]

how can I count the number of gumballs in this jar?

[placid juniper]

😭

[delicate belfry]

You’ll probably have to find the mass of the glass jar, and an average mass of a gum, and get an average answer (besides counting).

[pearl tendon]

+1 to Good's answer.. but just so we don't leave any stone unturned... do you have any additional information? Size of the gumballs? Size of the jar? Volume? diameter? anything?

[cosmic plinth]

nope I was just given this image and told to count the number of gumballs

[pearl tendon]

meh.. old county fair game... kinda lame, imho

imagine you're looking at a slice, vertically, through the jar

That's your diameter

[cosmic plinth]

so something like the shell method?

[pearl tendon]

The jar is relatively cylindrical, so you could take that number and, using "gumball" as your unit of measure, get the height of the cylinder, and then calculate volume

[pearl tendon]

Not familiar with that terminology

[cosmic plinth]

okay can try

[cosmic plinth]

It's a method used to find volumes. Usually tought in calc 2

[pearl tendon]

At the end, it boils down to a guessing game, and your instructor likely just wants to get you to try and think logically

[cosmic plinth]

yea I'll try

thanks!

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[wild dew]

Can anyone please help me with no. 12? Im sort off confused on what to do or even where to start

[pulsar plaza]

you have two points B and C

and you want to find the equation of the line passing between

so do you know what equation to apply here?

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[wild dew]

No im second guessing atp😭😭

[pulsar plaza]

equation of a line lol

[wild dew]

OHH

Okok i think i get it now thank you!!

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[summer ferry]

basic question but is the inflection point the singular point?

[quaint aurora]

dy/dx f(x) = 0, with d^2y/dx^2 < 0, is a maximum, the double derivative would be positive if it was a minimum. those are inflection points. you also need to check the bounds of the domain, because those might be maximums or minimums too, but are not contained in this test.

[summer ferry]

I understand the inflection point is the point where it changes concavity

but im just a bit confused if it is the singular point?

or is singular point a different thing

[quaint aurora]

the singular point is where dx/dy = 0 and the double deriviative is also exactly zero

[summer ferry]

so its when f'(x) = 0 and f''(x)= 0 at the same point?

[quaint aurora]

yeah

[summer ferry]

Okay wait sorry just to clarify

so if it's f'(x) = 0 and f''(x) = 0 at x=2

then the x=2 is the singular point?

[quaint aurora]

yes

[summer ferry]

ok and last dumb question haha

so if f'x = 0 and f''x = smth else at x=2

then that wouldnt be a singular point right

obv question but just making sure haha

[quaint aurora]

no it wouldn't be a singular point, but it would be a local max or min

[summer ferry]

ohhh so singular refers to the absolute max/min

[quaint aurora]

im not sure about that

i just know that the definition is that the double derivative and the derivative are both zero

[summer ferry]

okok no problem, ill just double check later

thank you though

[quaint aurora]

np

[summer ferry]

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[amber crescent]

Pls help on probability problem algebra 2

[stable ore]

well

how many ways are there to choose 2 democrats

and how many ways can you choose 2 people

[amber crescent]

So 22C2 and 10C2

You divide them?

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[lapis burrow]

I'M giga stuck on simplifying this

[lapis burrow]

I'm dead

[tepid hamlet]

how did you end up with stuff ^4?

[lusty delta]

by doing correct work

[lusty delta]

notice u can factor a common term out from this expression inside the square root

[lapis burrow]

raising the problem to 2nd power

[lapis burrow]

9?

[lusty delta]

yes

but there is more

[tepid hamlet]

oh right

[lapis burrow]

sin^2 and cos^2?

[lusty delta]

yes

so factor out those 3 things and rewrite ur expression accordingly,

[lapis burrow]

oh the sin and cos cancel out

on the inside?

[lusty delta]

can u show what u have

[lapis burrow]

9cosx^2sinx^2(cosx^2+sinx^2

[lusty delta]

btw

(-sint) squared is positive (sint)^2

so the minus sign goes away

[lapis burrow]

my bad your right

[lusty delta]

v nice

[lapis burrow]

so the inside should cancel out?

[lusty delta]

sorta ye

[lapis burrow]

or justbe 1

[lusty delta]

since u know (cost)^2 + (sint)^2 = 1

[lapis burrow]

yeah

thank you very much mate

[lusty delta]

welc 🎉

[lapis burrow]

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[latent axle]

71

[lusty delta]

do u have any thoughts on this question so far

[latent axle]

Not really

[lusty delta]

so let's disregard some terms for now

do u know what the limit as x approaches infinity of (- 3^x) / (4^x) would be

@latent axle

[latent axle]

Uh

Inf/inf?

[lusty delta]

is correct but i'm asking do u know what the limit actually evaluates to

[latent axle]

No

[lusty delta]

r u familiar with such equality

[latent axle]

Yes

[lusty delta]

applying it here, what do u get

[latent axle]

(-3/4)^x

[lusty delta]

the negative sign is on the outside not inside

but yes

[lusty delta]

and as x goes to infinity what does this -(3/4)^x tend to

[latent axle]

-inf

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[wheat bronze]

i need help

[wheat bronze]

i dont understand anything, teacher just handed this without intro or lesson.

<@&286206848099549185>

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[foggy wharf]

A rectangular parallelepiped has a side length of 5 cm, a diagonal cross-sectional area of 205 cm, and a base area of 360 cm. Find sides

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[foggy wharf]

<@&286206848099549185>

<@&286206848099549185>

i can find the answer but i dont like method which im using

<@&286206848099549185>

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[foggy wharf]

<@&286206848099549185>

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[foggy wharf]

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[foggy wharf]

<@&286206848099549185>

A rectangular parallelepiped has a side length of 5 cm, a diagonal cross-sectional area of 205 cm², and a base area of 360 cm². Find the lengths of its sides

[earnest granite]

hi

I am letting you know that I am working on it

[foggy wharf]

oh ty

i can solve it but my method is very bad

[earnest granite]

what answers did you find ?

[foggy wharf]

answer should be 40 and 9

[earnest granite]

hmmmm

[foggy wharf]

i solved it using biquadratic equation but

numbers are very high

so i used calculator

[earnest granite]

I found a way

[foggy wharf]

good

how

[foggy wharf]

i also tried this

but

numbers are very high and i dont like this

[earnest granite]

hmmmmm

I can't think of anything simpler

[foggy wharf]

and how to take sqrt form big numbers without calculator if u know any techniques

[earnest granite]

I tend to just gradually break up a large number until I get something okish

[foggy wharf]

7 digit number?

[earnest granite]

could you send the number

?

[foggy wharf]

and D=2307361

this middle thing is nothing

[earnest granite]

yeaaa.........

[foggy wharf]