help-10

[sweet edge]

what have u got so far

try to get y on one side of the equation

I-

[umbral orbit]

did u try squaring both sides?

[sweet edge]

it'll be easier if u move y over btw

[obtuse pebble]

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[whole spruce]

Focus on number 10

[timber fox]

any thoughts so far?

[whole spruce]

Nevermind

.close

[obtuse pebble]

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[rustic niche]

Why isn't this correct

[rustic niche]

Because the 3y + 3y should be 6y
Instead of 3y^2

Maybe someone can explain this?

[glossy basalt]

3y+3y=6y

[rustic niche]

Yeah I do get that

[glossy basalt]

3y means there are 3 y's

3 y's add to another 3 y's will become a total of 6 y's

however for y², it means there are y y's, which is a total different thing

[rustic niche]

Okay, so if I do thing like=

-p • -p that does result in p^2

It's just the operator decides it

[glossy basalt]

you can say that

[rustic niche]

If i do 10p - 4p i get 6p

[glossy basalt]

yep

[rustic niche]

Yeah I do get it now. Just a silly mistake

Thank you for help!

[glossy basalt]

no problem

[rustic niche]

.close

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[lone bolt]

why isnt it c

[lone bolt]

a makes sense too

[paper epoch]

anyone have an easy way of solving this?

[lone bolt]

enji

i have my own channel right now

please delete

am i getting trolled

kinda working tbh

[sweet edge]

the value can exist but the limit may not

an example would be that it's a jump discontinuity

[lone bolt]

gotcha

ty

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[lone bolt]

why is this answer wrong? i graphed it and theres a jump

[lost tree]

Can u show ur graph

[lone bolt]

1 moment

[sweet edge]

u can take the two sided limits and see they don't match therefore it's discontinuous

[lost tree]

Is that 1/(4x)

U have entered in ur calculator

[lone bolt]

1/4x-7

[sweet edge]

oh wait, both limits are actually 8

that is continuous

[lone bolt]

let me guess

[lost tree]

Well does it know u mean (1/4)*x and not 1/(4x)

[lone bolt]

i typed it wrong

[sweet edge]

yeah lol

[lost tree]

Probably

Use parentheses ig

[lone bolt]

ugh

[sweet edge]

u can also just do 7+ x/4

if that helps

[lone bolt]

you always have to becareful

ur never safe

with paranthesis

[sweet edge]

haha fr

[royal basin]

idt that's the issue.

[royal basin]

there's a typo in your first piece here

the -7 should be +7

[lone bolt]

first i thought if u () the denominator, ur good, then sometimes u have to () the num too, now u have to separate it lol (1/4)x-7

[lost tree]

Oh yeah lol

[lone bolt]

so i typed it right then

minus the negative

part

[lost tree]

Yeah apart from the negative ig it seems ur calculator understands that 1/4x is (1/4)*x

[lone bolt]

1/ 4x is not the same as (1/4) x?

[sweet edge]

if ur calculator is like mine then it should be

but safer to have () prob

[lost tree]

Coz the part of the graph on Yr calculator also looks like a straight line

[sweet edge]

scale issue

[lone bolt]

genius

[lost tree]

Wdym

[lone bolt]

is a straight line

i zoomed out

two lines

[sweet edge]

the y axis scale is a lot zoomed out

[sweet edge]

ye

[lone bolt]

our calculators are overpowered

oh true

thanks guys

math wizards

❤️

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[halcyon heron]

How do i find the total enclosed area through integration. I am confused as of what to do and it seems overwhelming considering there are 7 functions

[obtuse pebble]

Please don't occupy multiple help channels.

[sweet edge]

can u shade in the part u need

total enclosed area like including eyes and stuff?

[halcyon heron]

ignore the eyes and stuff

like that area

[sweet edge]

so include the mouth?

but not eyes

[halcyon heron]

find the area assuming it is one enclosed area

so yes including the eyes and mouth

[obtuse pebble]

@halcyon heron Has your question been resolved?

[sweet edge]

go to the other channel

#help-21｜아리스킨충1

close this one

[obtuse pebble]

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[dense gulch]

Hi I'm looking for a little bit of a geometric interpretation. I'm looking for the euler characteristic of this torus. I know that the euler characteristic must equal zero so there must be 4 faces but what actually are the faces could someone point out please

[limpid widget]

oh

[dense gulch]

Would it be like splitting it into curved quarters?

For each face

[obtuse pebble]

@dense gulch Has your question been resolved?

[obtuse pebble]

@dense gulch Has your question been resolved?

[obtuse pebble]

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[torn cape]

how do i find derivative of this functin on R ?

[torn cape]

ik functiin is differentiable if both left and right limits exist but

[runic void]

Have you tried taking derivative of $x^2 \sin(\frac{1}{x})$?

[warm shale]

.doc

[torn cape]

ummm yes that woud be

2x sin 1/x-cos 1/x

[runic void]

If you verified that f is continuous, then now you can make a piecewise derivative function putting each of the derivatives

[torn cape]

WAIT NO it s not

but not importnt,

[torn cape]

ummm AHH so

how do i check if its contionuos

RIGHT

by finding a limit maube

hiw do i find this

[runic void]

@torn cape

We don’t want to evaluate that, we just wanted to prove LHD and RHD are equal at x=0, differentiability will imply continuity

R.H.D= $\lim_{h\to_0 }\frac{f(x+h)-f(x)}{h}$

[warm shale]

.doc

[runic void]

L.H.D=$\lim_{h\to_0 }\frac{f(x-h)-f(x)}{-h}$

[warm shale]

.doc

[runic void]

@torn cape Do you understand these definitions ?

You can think the difference quotient as finding the rise over run, for left hand derivative we move from f(x) to something left of f(x) ie f(x-h)

Here at x=0

R.H.D= $\lim_{h\to_0 }\frac{f(0+h)-f(0)}{h}$

[warm shale]

.doc

[runic void]

values of f(0) and f(0+h) can be obtained from the piecewise definition

[torn cape]

sorta

[torn cape]

wait let me try

[runic void]

Sure tyt

[torn cape]

nvm i cant

[torn cape]

how,

[obtuse pebble]

@torn cape Has your question been resolved?

[obtuse pebble]

@torn cape Has your question been resolved?

[obtuse pebble]

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[tame lotus]

How do you do this guys

The topic is functions and graphing

I dont understand how can we convert the inequality into that formula and graph it

<@&286206848099549185>

YO GUYS HELP

<@&286206848099549185>

[oak bear]

Ah ok

The shape of all parabola are the same, they just differ from scale, and where they are from the graph

Let's say a > 0, then it's just an upright parabola.

k describes where the parabola is in terms of y axis

[obtuse pebble]

@tame lotus Has your question been resolved?

[oak bear]

@tame lotus u still there?

sry bout that got to do something

so u finished?

[obtuse pebble]

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[lime cargo]

Can someone explain the algebraic steps to get from the top to the bottom.

[lime cargo]

I think the y_0 will cancel

and taking the ln of everything will cancel the e

but that would leave me with ln(1/2) = -4t_1/2

[daring rock]

-ln(1/2) = ln(2)

[lime cargo]

oh wow, that simple

ty

[daring rock]

np 👍

[lime cargo]

.close

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[ruby flame]

Solve using the quadratic formula

[trail cloak]

,rcw

[ruby flame]

I still don't understand where I went wrong

[copper shell]

its supposed to be (-3)^2 not -3^2

[trail cloak]

Your equation should equal 0

[high lily]

you didn't first rearrange the equation to general form

[trail cloak]

ax^2 + bx + c = 0

You had 3x^2 -3x = 2

You had to move the 2 to the other side

[high lily]

in addition other sign and notation issues

like fraction line of inappropriate length
and incorrect value for -b

[ruby flame]

Wait why do i need to move it

As shouldn't it be c

then i use it for my quadratic equation

[high lily]

the $c$ in the quadratic formula is the same $c$ in
$$ax^2 + bx + c= 0$$
$\red{\text{NOT}} \ ax^2 + bx = c$

[warm shale]

ℝamonov

[obtuse pebble]

@ruby flame Has your question been resolved?

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[timid silo]

Got this expression.

I am not a native english, but the first exercise I assume is to find the floor function of this, floor function was the integer part of any decimal number right?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

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[keen crane]

Hey, helping a friend in high school solve a logarithm problem. Photomath says there's a rule where $\log{3^4}{81} = \frac{1}{4} * \log{3}{81}$. Where does the rule that it becomes 1/4 come from? I've never seen it before nor was I taught it.

[warm shale]

CoreByte

[keen crane]

log_3 of 81, in this case.

It's essentially this rule, but I can't find any reason as to why this works. I'm familiar with the power rule, though, but that applies to the power of n (in the photos case).

[timid silo]

think of it in terms of the power

$log_{a}{b}=c \iff a^c = b$

[warm shale]

Jester

[keen crane]

I understand that, but I can't wrap my head around where the value would be on the right side if it was $a^d$ on the left, for example

[warm shale]

CoreByte

[timid silo]

Let $c = log_{a^m}n$ then $(a^m)^c = n$ and $a^{mc}=n$ from here $log_{a} n = mc$ i.e. $\frac{1}{m}log_{a} n$

[warm shale]

Jester

[keen crane]

ohhhhh

thank you very much, that makes a lot of sense

.close

[obtuse pebble]

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[fathom flicker]

I am wondering less about this specific question ^ and more about in general.

For a homogeneous second order linear ODE, are the solutions damped when you have real roots? oscillatory with complex roots? and undamped with repeated roots?

or what is the case, for each root case

for the characteristic polynomial

[dark stirrup]

Oscillatory with imaginary roots for sure if I remember my odes

[fathom flicker]

that's my current impression aswell

[dark stirrup]

I'm sure you know euler formula

[fathom flicker]

mhm

[dark stirrup]

Also, real roots will not always give you damped

Think of just e^x

[fathom flicker]

yeah real roots give like ce^blah + c2 e^blah

[dark stirrup]

It can be a solution to second order odes and is not damped

[fathom flicker]

is it

overdamped when p(x) > q(x)

and underdamped when q(x) > p(x)

for the y'' + p(x)y' +q(x)y

format

[dark stirrup]

Hmm for non constants, I do not know or remember the criteria

[fathom flicker]

for constants

my bad

[dark stirrup]

I have a work meeting, so I'll jump back on this later if you don't get help

[fathom flicker]

okay

Summarizing my questions here for whoever sees this:

for a second order ODE, y''+Ay'+By=0 where A and B are constants

are there oscillatory solutions when the roots to the char. polynomial is imaginary? are there damped solutions when the roots are real but not equal? If that last question is true, is it overdamped when A>B and underdamped when B>A? and finally, are there undamped solutions when the roots are real and equal?

<@&286206848099549185>

[dark stirrup]

Do you know how to solve the general ODE y''+Ay'+By=0?

[fathom flicker]

yes

I know what the solutions look like aswell

[dark stirrup]

So you know it's like e^(quadratic equation) yeah?

[fathom flicker]

well depending on the A and B the form changes no?

it depends on the roots

[dark stirrup]

Yeah, but I was just making sure you knew it was along these lines

[fathom flicker]

yes basically

[dark stirrup]

Okay cool. So the question then becomes what's the form for oscillatory, overdamped, and underdamped (and I guess critically damped if you're curious)

[fathom flicker]

yes

critically damped is r1=r2

I am 99% sure

and oscillatory when r=a+-bi

[dark stirrup]

I do not think this is right. I think you need pure imaginary for oscillatory

[fathom flicker]

yeah like r1=a+bi and r2=a-bi

[dark stirrup]

Oscillatory should be ${c_1}\sin{t}+{c_2}\cos{t}$, right?

[warm shale]

SWR

[fathom flicker]

I thought it was

[dark stirrup]

Yeah it should be

(btw sorry but I'm working at the same time, so slow replies)

[fathom flicker]

,, y=e^{at}(c_1\cos{bt}+c_2\sin{bt})

[dark stirrup]

But I am here

[warm shale]

AustinU

[dark stirrup]

And I'm only helping you

[fathom flicker]

thanks for your help, no worries

I appreciate it

I thought ^ that was the oscillatory solution form

[dark stirrup]

Let me grab my diffyQ book

It's been forever since I reviewed the wordings

[fathom flicker]

okay

Chartbit are you also interested in this lol

[unreal musk]

Like I wanna know how they word it now

[fathom flicker]

Here is what I have in my notes

but I didn't put a note for the oscillatory case

and I didn't put any reasoning behind why I named what I named whatever I named it

if that makes sense, that is why I am trying to confirm lol

[unreal musk]

Fair fairs! I’m over here watching to learn too haha

Feel like a lot of the time I know the working but not always the terms used because I’ve either never used them/they’re different haha

[fathom flicker]

yeah I feel like where I am learning it from uses a lot of niche terminology

this server is so helpful to me since I am just trying to learn stuff online right now

[dark stirrup]

Okay my book calls this simple harmonic motion or free undamped motion

[fathom flicker]

^ I have that in my notes somewhere too

it is a type of oscillation

for the form

[dark stirrup]

If a<0 here, then this is underdamped

[fathom flicker]

,, y''+w_0^{2}y=0

[warm shale]

AustinU

[fathom flicker]

even with complex roots

okay

so why is this also underdamped

,,y=e^{-3t}+e^{-t}

[warm shale]

AustinU

[fathom flicker]

is it just if there is a negative exponent on the exponential

regardless of the sine cosine waves being involved

[dark stirrup]

My book called this overdamped

[fathom flicker]

mine calls it underdamped :(

[fathom flicker]

also, mine calls when the a is negative in the oscillatory equation earlier that you said made it underdamped, mine says it is overdamped

...

[dark stirrup]

yours sound

[fathom flicker]

Okay, but that is in regards to the simple harmonic motion equation

mine is just in general form

so we don't even have a omega nought

I have the same notation as you for the simple harmonic equation btw

when P < W0 it is damped and P=0 is undamped

Maybe I probably should just drop this.. I am starting to fail to see the point

[dark stirrup]

From wikipedia:

[dark stirrup]

I believe this should definitely be called overdamped

[dark stirrup]

I don't wanna push you if it's getting exhausting

I need to do another work call anyway, so let's take a minute

[fathom flicker]

Well it's not, I just don't understand because yes that is clearly saying the opposite of what my notes say

[dark stirrup]

Could you send me pics of your book/notes on the definition(s) if you want?

[fathom flicker]

here were what I have

I didn't have real strict definitions

https://www.desmos.com/calculator/mn1rcqb5ps

and even when I graph what I have in my book , the underdamped example (that you and wikipedia say is overdamped) behaves the exact same as what my notes say is the overdamped

so perhaps just my notes are incorrect for the underdamped case? and then that begs the question why is it overdamped here ^ and also when there is imaginary roots like in my other example

or might one of those be incorrect aswell

[fathom flicker]

sounds good

[dark stirrup]

I see. I have a strong suspicion these are wrong. But I'm just one person and I don't want to mislead. Perhaps let's resolve this first in #odes-and-pdes then we can continue here

[fathom flicker]

I also feel like they are probably wrong now

bye channel

.close

[obtuse pebble]

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[flat sphinx]

I was wondering if anyone could explain the first step of simplifying the two equations

[timid silo]

r=R+h?

[flat sphinx]

I am a little confused why the denominator of the equation on the right turns into r+h

[timid silo]

because r = R+h

[flat sphinx]

herp derp

thanks

.close

[obtuse pebble]

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[prisma lintel]

.

[timid silo]

so ive been non stop testing different values for k and i cant figure out how the professors got whatever their answer is

here is the formula im using for the midpoint error

here are my values

so far ive tried

k being the actual maximum value of x on the interval which is f''(1)

k=19.2207559654

then i thought they rounded it so i put in k=20

then thought hey, maybe they just want me to plug in the largest value of sin/cos in the second derivative

so i tried 30

and now im just pissed off

.close

[obtuse pebble]

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[timid silo]

why wouldnt it be 1:2

[timid silo]

why do you think it's 1:2?

can you show your thought process/work?

well, i thought the 2nd one would be doubled

but now i see my mistake 😭

i just dont know what other answer i'd put for that considering 24 is just 6 more than 18

the 1st one would be 18 x 2 = 36 2nd one would be 24 x 2 = 48

im rlly confused 😭

hello?

[tardy epoch]

Perimeter scales as the side lengths of similar polygons

So if you double the side length, you double the perimeter

What do you multiply 18 by to get 24?

[timid silo]

1.5

right

no nvm

idk

.close

[obtuse pebble]

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[astral helm]

How do I get the partial fraction expansion of this function?

[dark stirrup]

If its a perfect square, I don't believe it can be decomposed

Is this for integrals?

[astral helm]

laplace

[dark stirrup]

ah

[astral helm]

i dont know how to get to 3 and -1

[spiral knot]

just add and subtract 3 from the numerator

[astral helm]

like -s+3-3?

[spiral knot]

yes

or if you wanna the hard way

$\frac{-s}{(s+3)^2} = \frac{A}{s+3} + \frac{B}{(s+3)^2}$

[warm shale]

Mehdi_Moulati

[spiral knot]

and try to find A and B

[astral helm]

I did something like that, but couldnt find a and b

I need to multiply both sides for (s+3)² right?

[spiral knot]

yes

[spiral knot]

what did you do ?

[astral helm]

then -s = A(s+3) + B

[spiral knot]

yes

correct

what's next ?

[astral helm]

if s= -3, B will be 6

[spiral knot]

nope

if s=-3 , B will be 3

[astral helm]

oh

i dont know how i got this 6

let me try to continue

[spiral knot]

of course

[astral helm]

As + 3A + 3 = -s

i cant continue

[spiral knot]

replace s with 1 and find A

for example

[astral helm]

in this equation?

[spiral knot]

or just compare the two equations

[spiral knot]

yes

any number

except for 0

[astral helm]

4A = -1-3

[spiral knot]

-3 not 3

[astral helm]

whoops

so A = -1

[spiral knot]

yes

[astral helm]

that makes sense

Sorry, im going crazy, i've studied all long the day and need to get some rest (test tomorrow😢 )

thanks for the help man

[spiral knot]

nahh , it's normal

the easy method

[spiral knot]

is this

,align
\frac{-s}{(s+3)²}
&= \frac{-s {\color{cyan}- 3 + 3}}{(s+3)²} \
&= \frac{-s-3}{(s+3)²} + \frac{3}{(s+3)²}\
&= \frac{-1}{s+3}+ \frac{3}{(s+3)²}

[warm shale]

Mehdi_Moulati

[astral helm]

why the -s-3 is equal to -1?

oh no

like

how did you transform from the second to the third line?

[spiral knot]

$\frac{-s-3}{(s+3)²} = \frac{-(s+3)}{(s+3)(s+3)} = \frac{-1}{s+3}$

[warm shale]

Mehdi_Moulati

[astral helm]

oooooh

makes sense

thanks, you helped me a lot

[spiral knot]

anytime bro

[astral helm]

.close

[obtuse pebble]

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[strange yoke]

Here Im integrating by parts

I've done it twice and now my integral at the very end matches the problem that I started with

I think I need to set that whole line equal to the original problem but i dont see how i could do it exactly

[fierce lagoon]

If it helps, let $I = \int e^{2x}\cos(5x)\dd x$

[warm shale]

Umbraleviathan

[strange yoke]

So I is like the original problem

[fierce lagoon]

Mainly for visual purposes because it's easier to spot an "I" in a sea of numbers

Yeah

[strange yoke]

ButI think my problem here is that theres a 5/2 fraction that multiplies a big thing of parenthesis

[fierce lagoon]

The original integral pops up again so replace that one with I

Then you're just isolating I

[strange yoke]

Ok i try

[fierce lagoon]

You just distribute

[strange yoke]

Wait

[fierce lagoon]

You distribute it normally

[strange yoke]

I dont know what normally is

Does this look good so far

I made it z instead of I

[fierce lagoon]

Yeah

[strange yoke]

I think this is good?

[fierce lagoon]

seems a bit off

Probably an algebra mistake

i gtg soon but the denomintaor should 2^2 + 5^2

[strange yoke]

I dont understand what 2^2 + 5^2 is? Why are things being squared and which denominator is it?

[fierce lagoon]

of the antiderivative

ive done enough problems like these to know what the denominator of the antiderivative should be

it should all be one fraction

whose denominator is 2^2 + 5^2

[obtuse pebble]

@strange yoke Has your question been resolved?

[strange yoke]

I don't understand how all of the variables and everything inside just becomes 2^2 + 5^2

I think we're talking about different things i just don't get it

I tried it again with a similar problem and im still getting it wrong

[obtuse pebble]

Channel closed

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[strange yoke]

.reopen

[obtuse pebble]

✅

[fierce lagoon]

This looks right

41 = 4^2 + 5^2

That's how I check these, is the denominator

,w integrate e^(ax) cos(bx)

[warm shale]

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Requested by Umbraleviathan#3607

[strange yoke]

So its the constant in the e exponent

And the constant inside the trig

Square the two and add them

[obtuse pebble]

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[lilac acorn]

Is the function y = 3(7) to the power of x, positive, negative, increasing or decreasing

[fierce lagoon]

If you need to, graph it out

[lilac acorn]

Yes I already did

Is it positive and increasing

[fierce lagoon]

Mmhm

[vast wren]

yeah

[lilac acorn]

Oh okay thank you

.close

[obtuse pebble]

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[timid silo]

Can someone check my answers for these (1-3)

[timid silo]

These are my answers

I realized I repeated a page on that pdf btw*

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[amber cypress]

I am doing math homework and my teacher was not able to teach long enough and we didnt cover any of tonights homework, right now I am trying to find the max or minumum value of y = 2x^2 - 10x + 13

[fallow cove]

Find the derivative of it

[fallow cove]

can you do that

[amber cypress]

Idk, prob not because ive never heard that word

[fallow cove]

oh so you grade 8

[amber cypress]

yeah

but this is algebra class

[fallow cove]

oh thought you were learning calc

[amber cypress]

oh no, 1 more year

[fallow cove]

oh ok

anyways

to find the minimum or maximum value of y we have to find the minimum value of x first

[brisk matrix]

do you know about vertex form?

[amber cypress]

no

[fallow cove]

the guy’s grade 8

[amber cypress]

I feel like im missing out on something

[brisk matrix]

have you heard of the vertex of a parabola?

[amber cypress]

yes\

like the peak?

[fallow cove]

Yeah

[brisk matrix]

have you ever seen something like $y = (x-h)^2 + k$

[warm shale]

maximo

[amber cypress]

no

not that

[brisk matrix]

ok

[amber cypress]

getting to that tmrw

[brisk matrix]

you called the vertex the "peak" which is good intuition for what you should be doing

[fallow cove]

For the so called maximum or minimum value of y

or x I kinda forgot

yeah for y and x

ok so your equation for y is y=-2x^2 - 10x + 13

[amber cypress]

positive 2x

[fallow cove]

ok

[amber cypress]

but yeah

[fallow cove]

so your first step for it is to complete the square

you know how to do that right?

[amber cypress]

no. what do you mean by square

[brisk matrix]

are you allowed to graph these @amber cypress

[amber cypress]

yeah, probably

[brisk matrix]

as in, with a graphing calculator?

[amber cypress]

yes

im in alg 1

Oh waitm I just graphed it, I get it noe

so it concaves up

[fallow cove]

yes

[amber cypress]

having a minumum

that makes so much sence

so it would be the y value?

[brisk matrix]

yes

look back at what the question asked

[amber cypress]

okay so its just like 0.5, im going to check anyways but I think its that

so it would be like the lowest value divided by X at that time?

?

???

hello?

all right then

[fallow cove]

sry for the wait, but the highest or lowest point you have would be the maximum or minimum value for x and y

the coordinayptes would be the answer to x and y

[amber cypress]

its okay, how do I find that on my calc, its not working

[fallow cove]

you have a graphing calculator right?

[amber cypress]

yep

84 plus ce

[fallow cove]

you type in the equation

[amber cypress]

done

[fallow cove]

Ok

so you will the find the maximum point right?

[amber cypress]

yeah

I tried 2md trace and min but it didnt work, unless im dumber then I think I am

[fallow cove]

So do you have the maximum point?

[amber cypress]

min? or max?

I can check for either just wanna make sure im doing the right thing]

[fallow cove]

The coefficient for.x^2 is positive right?

[amber cypress]

?

[fallow cove]

the first erm is 2x^2 right?

[amber cypress]

yeah

[fallow cove]

so it should be a maximum value

[amber cypress]

ooh

shoot okay

it says y = 13

[fallow cove]

for minimum value?

[amber cypress]

max

[fallow cove]

Ok

then that would be your answer

[amber cypress]

okay, thanks for all the help!

?solved

wait

.solved

ooh

.close

[obtuse pebble]

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[timid silo]

Can someone tell me if the probability for the teens only booth is 4/36 or 8/36

[timid silo]

It says 13, 14, 15, or 16 can be created in any order, so I'm saying 4/36 because if we roll a 1 or a 3, 4, 5, 6, we have a 1/6 and 4/6 chance in each scenario, hence 4/36

But my overthinking self is saying: "what if we roll 3, 4, 5, or 6, then 1?, so isn't that another 4/36 here"? Then again order doesn't matter :/ I'm honestly confused. Please help.

[obtuse pebble]

@timid silo Has your question been resolved?

[frosty spoke]

8/36

[timid silo]

I talked about it with a friend and he said 4/36

Because (1,3) or (3,1) is the same in this context

Since order doesn't matter

Like I can see how that works

[frosty spoke]

that's not exactly convincing reasoning

does your friend mean to say that (1, 3) and (3, 1) are the same outcome?

[timid silo]

Ok so if you let A=Event that you roll a one, it's probability is 1/6

If you let B=Event that you roll 3, 4, 5, 6, it's probability is 4/6

If A happens first then B, you can still score a point

If B happens first, then A, you can still score a point

Because order doesn't matter. It doesn't matter if A happens, then B (or vice versa)

So 1/6 x 4/6=4/36

[frosty spoke]

what is A, the event that you roll a one on any of the two dice?

[timid silo]

A is the event you roll a 1

And it's

[frosty spoke]

what's your space of outcomes again?

[timid silo]

When one of the dice is a one

(1, 1)
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(2, 5)
(2, 6)
(3, 1)
(3, 2)
(3, 3)
(3, 4)
(3, 5)
(3, 6)
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(4, 5)
(4, 6)
(5, 1)
(5, 2)
(5, 3)
(5, 4)
(5, 5)
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)

[timid silo]

Pasted them

[frosty spoke]

so A is the event that the first die comes out as a 1?

[timid silo]

Ah it says "roll a number cube 2 times"

[frosty spoke]

also, I recognize that the problem says roll a die twice

but that's basically the same as rolling two dice at the same time

I'm only saying two dice because maybe it makes a little more intuitive sense

[timid silo]

Ok but if you roll a 1 first, then a 3, or a 3 first, then a 1, you get a point

So I guess (1,3) and (3,1) are two possible outcomes

[frosty spoke]

they are

[timid silo]

I see

ugh

[frosty spoke]

maybe we should make the problem a little bit simpler

what if instead of just 3, 4, 5, 6, we allow 1,2,3,4,5,6

actually no that makes it tougher LOL

nevermind

[timid silo]

I think I understand what you're saying now

I have to fix my homework answers

so brb

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

please help me out wiht understanding how to solve this?

[obtuse pebble]

@timid silo Has your question been resolved?

[frosty spoke]

imagine that there are 10000 people. how many people test positive? of those how many have the virus?

[timid silo]

85 percent?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[wanton dagger]

How can I do this?

[fast cliff]

find how many 110 minutes are in 11 hours

[wanton dagger]

Ok

Then what do i do?

[wanton dagger]

6

[fast cliff]

every 110 minutes it cuts in half

so if it cuts in half 6 times

how do you multiply a number by itself 6 times

[wanton dagger]

Don’t we divide the atoms 6 times?

[fast cliff]

yea

wait

not

[wanton dagger]

Then what

[fast cliff]

divide by 2 6 times

[wanton dagger]

Wait

Ok

Wait divide 2 by what?

Hmm

@fast cliff

[royal basin]

every half-life, the amount of your radioactive substance is cut in half

so after six halflives you would multiply your original amount by 6 copies of (1/2)

[wanton dagger]

I see

Does that mean I divide 10000000 by 1/2 6 times recurringly?

[quaint finch]

you divide it by 2, six times continously

you can make those chain diagrams to help you

[wanton dagger]

Ok thx Ann, got it

.close

[obtuse pebble]

Channel closed

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[royal basin]

.reopen

[obtuse pebble]

✅

[royal basin]

what "chain diagrams", what "recurringly", what the hell are y'all on about

this is just 1,000,000 * (1/2)^6 for chrissakes

[quaint finch]

Haven't u ppl ever seen this

[fast cliff]

the correct way to do it is

initial * (1/2)^h/t

i just made up that formula so idk if those variables are right but

your methods dont work if 110 doesnt fit perfectly into whatever time

[royal basin]

overcomplicated honestly

[quaint finch]

It's just another way of writing that

[fast cliff]

yea but doesnt work with non integer exponenets

[quaint finch]

wdym

[fast cliff]

10000000*(1/2)^5.5

which is why you use exponents when working with half life

[quaint finch]

so what exactly do the variables mean in ur formula

[fast cliff]

uh

initial = starting value

h = half life

t = total time

[quaint finch]

what are u making it equal to

[fast cliff]

end quantity ig?

[frank hamlet]

t/h

[fast cliff]

ya

[quaint finch]

alright, cool

[obtuse pebble]

@wanton dagger Has your question been resolved?

[obtuse pebble]

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[crude ice]

how do i solve this qn

[crude ice]

how do i show a)

[obtuse pebble]

@crude ice Has your question been resolved?

[sharp pecan]

solve it normally by using perimeter and area formulas of both

[crude ice]

i understand x^2 but how did they get area for circle isnt it just pi r^2

[sharp pecan]

first of all, if you want to keep the channel press the x

just solve like you would normally

perimeter and total area equations in a system, solve for area

so 500=perimeter of square + perimeter of circle

and area = area of sqaure + area of circle

if you get the same equation you have proven it

[obtuse pebble]

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[elfin abyss]

if the sum of a series (un)=1, is there an N for which if n>N, un=o(1/n^2)

[elfin abyss]

so in latex if $$\sum_{n=0}^{\infty} u_n =1$$ and u_n positive, is there an N for which for all n >= N, $$u_n =o(\frac{1}{n^2})$$

[warm shale]

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

[elfin abyss]

i dont understand why its messed up

but i hope its understandable

[obtuse pebble]

@elfin abyss Has your question been resolved?

[rigid agate]

If the series over u_n converges, then, in particular, u_n is bounded, so u_n = O(1).

I think that's all one can say without additional information about u_n.

[obtuse pebble]

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[timid silo]

math help can we solve the quadratic equation thru quadratic formula and middle term break?

[timid silo]

like i know we can solve thru quadratic formula what about middle term break will it work too

[lost tree]

Yes u can use either way to solve

[timid silo]

oh thanks

[lost tree]

I'm assuming by middle term break u mean

x² + 5x + 6 = x² + 3x + 2x + 6

That kind of thing

[timid silo]

middle term break would only work if the equation is factorable though

[lost tree]

But sometimes the factors are not nice

[west sierra]

We(India)call middle term break as middle term factorisation

[lost tree]

The quadratic formula also would only work if it is factorable

[timid silo]

oh yes i got it

[lost tree]

When it is not factorable the quadratic formula leads you to square roots of negative nunbers

[timid silo]

when factors are not nice, the middle term break wont work

then quadratic formula would

[west sierra]

To see if it is factorable or not you have to find D(b^2-4ac) is greater than or equal to 0 or not

[timid silo]

thank you i guess the misconception got cleared

.close

[obtuse pebble]

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[ruby path]

$\int \frac{1 + 2x^{2022}}{x + x^{2023}} \dd{x}$

[warm shale]

NEONPerseus

[ruby path]

Honestly it's probably very easy but I just can't get it

Tried splitting it up and that worked well until I hit $\int \frac{\dd{x}}{1 + x^{2022}}$

[warm shale]

NEONPerseus

[ruby path]

The 2 seems to be a hint

$\int \frac{1 + x^{2022}}{x(1 + x^{2022})} + \frac{x^{2022}}{x + x^{2023}} \dd{x}$

[warm shale]

NEONPerseus

[ruby path]

Did some funky stuff like this

[fierce lagoon]

It's a u-sub I think

A horrid one

[ruby path]

[fierce lagoon]

Wait no

Shitttt

[robust sleet]

Just u sub denom

[ruby path]

$\int \frac{1}{x} + \frac{x^{2021}}{1 + x^{2022}} \dd{x}$

[warm shale]

NEONPerseus

[ruby path]

Oh bruh

[fierce lagoon]

LOL

[ruby path]

As you can see I've made a

massive fool

out of mysel

[fierce lagoon]

Gaming moment

[ruby path]

and shall not integrate

again

💀

thanks for whoever cared to look though

.coose

.close

[obtuse pebble]

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[obtuse pebble]

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[proper ivy]

The system of components shown in Figure 2-15
below functions as long as components 1 and 2 both function or components 3 and 4 both function. Each of the
four components functions with probability 0.9 independently of the others. Find the probability that the system
functions.```

[proper ivy]

that is the question

and this is the answer

i dont understand why we subtract 0.9^4

shouldnt it be 2((0.9)^2 * (0.1)^2) + 4(0.9)^3 * 0.1 + (0.9)^4

probability that 2 work and 2 dont work + probability that 3 work and 1 doesnt work + probability that all 4 work

[tender urchin]

I think P[AuB] = p[A] + p[B] - P[AnB]

So u subtract 0.9^4 as thst is P[AnB]

[proper ivy]

yeah i know the formula

but i dont get why it applies here

[tender urchin]

Wdym

[proper ivy]

i mean whats wrong with what i said

[tender urchin]

I think ur method should work but

Not sure

[proper ivy]

no i dont think so

[tender urchin]

Cuz that's all the possible outcomes

Hmm

[proper ivy]

.close

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[crude ice]

how do i find acceleration after 2 seconds

[west sierra]

is a=4sin(pi*t/2)

[crude ice]

yeah

[west sierra]

Why don't you put t=2 to find acceleration after 2 seconds

[crude ice]

im concerned about the statement displacement at t=0 is 0

[west sierra]

What does that have to do with finding a after 2 seconds though

Acceleration is only dependent on time, not position/displacement

[high lily]

some of the info is irrelevant to part i)
its relevant in later parts

e.g if a=2, b=5
and you're asked what is a+2?
you don't need to concern yourself/ care about b = 5 to answer this

[crude ice]

i put in 2 for t and got 0.219

i think its wrong

[crude ice]

yeah

[west sierra]

Isn't sinpi=0?

Can you show me how you 0.219 after putting t=2

[crude ice]

i put the values in the calculator

[west sierra]

Show me

[crude ice]

4sin(pi(2)/2)

[west sierra]

Why do you put double bracket in (2)

[west sierra]

2pi/2 = pi

Sin pi=0

4 * 0=0

[crude ice]

why doesnt this work tho why did it show 0.219

[west sierra]

Is your calculator working good?

[oak bear]

I think u didn't set it to radical

It must be in degrees

[crude ice]

i have it in degree

yeah

nvm

when do i have to use degree and radians>

[west sierra]

Even if it's in degree/radian 2 and 2 should get cancelled

Why does calculator give that answer

[oak bear]

No because, even if it is cancelled it would spit out pi

which is 3.14 something something

It might think it is 3.14 degrees instead of 180 degrees

[west sierra]

I calculated sin(3.14) value doesn't matcj

[oak bear]

Yeah cause u still have it in radians

[west sierra]

Oh

So it's 3.14 degree

[obtuse pebble]

@crude ice Has your question been resolved?

[obtuse pebble]

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[fathom wolf]

Hello

[fathom wolf]

,

I am sixth

and my question is

Find the value of x so that the given four numbers are in proportion..

x, 4, 15 and 30

What is the value of x??

<@&286206848099549185>

Umm no thankyou

.close

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[median tangle]

How do i solve this?

[west imp]

yoo

isnt this the same question?

[median tangle]

it was wrong

[west imp]

oh

wait lemme look at it again

[sage geode]

Note that 2x + 28 + 2(3x - 70) = 360

[west imp]

yea

but why do we need to make it 360

[sage geode]

Because the angle measure of the arc ADB is 2x + 28

[median tangle]

should it also be 2(2x+28)

shouldnt

[sage geode]

And the angle measure of the arc ACB is 2(3x - 70)

They should add up to 360

[west imp]

is the answer coming 59 degrees?

[sage geode]

2x + 28 is the central angle, why multiply by 2?

[west imp]

kuroko what is the answer?

[median tangle]

ok so you dont multipy the central angle thanks got it

[median tangle]

idk yet

[west imp]

ok kuroko have you studied the theorem ( angle subtended at the centre is double the angle subtended in any given part of the circle)

if you use the ur angle is would be x + 14

and ABCD is a cyclic quad

cyclic quad is when opposite sides have a sum of 180 degrees

[median tangle]

x=31

[west imp]

alr

[median tangle]

thats what i got

[west imp]

thats what ur getting?

idk cause im getting smth else

[median tangle]

2x+28 +2(3x-70)=360
8x-112=360
8x=248
x=31

oh shit i messed up

[west imp]

yea

the answer is 59

x= 59

[median tangle]

ya its 59

[west imp]

so there ya go by 2 methods

[median tangle]

thanks

.close

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[merry vault]

$\dfrac{1-\dfrac{1}{n^2}}{\dfrac{1}{n}}$

[warm shale]

madmike

[merry vault]

What is this for n -> infty?

Is it just 1 or am I missing something?

1/n -> 0

1/n^2 -> 0

1 / 0 =

[royal basin]

have you been able to simplify your fraction?

[merry vault]

yes actually let me just post the whole thing

this was just an example

[gloomy pine]

Infinity

[merry vault]

$\dfrac{1 - \dfrac{2}{n^2} + \dfrac{e^{-n}}{n^2}}{\dfrac{1}{n} + \dfrac{1}{n^2}} - \dfrac{1 + \dfrac{2}{n^3}}{\dfrac{1}{n} + \dfrac{1}{n^3}}$

[warm shale]

madmike

[merry vault]

n -> infty

[timid silo]

it is dying to be simplified

[merry vault]

really

hmmm

[timid silo]

anyways for this, multiply numerator and denominator by n

[merry vault]

so this would be infty then?

because 1n - 1/n

1/n -> 0

hmmmmmmm

[merry vault]

what should I do here

[timid silo]

yeah

it would diverge to infinity

[merry vault]

factor out 1/n^2 in the first frac and 1/n^3 in the second?

[gloomy pine]

Why dont you work it out slowly

[timid silo]

honestly, just begin by simplifying everything

[gloomy pine]

Unify the denominators

[timid silo]

we can definitely go from there

[merry vault]

I don't see how

[gloomy pine]

I would hit the ratio on the left with n^2 and thé ratio on the right with n^3

[merry vault]

that's basically undoing my work

lel

the original problem is

$\dfrac{n^2-2+e^{-n}}{n+1} - \dfrac{n^3 + 2}{n^2 + 1}$

[warm shale]

madmike

[merry vault]

I thought I would factor out n^2 on the left frac

and n^3 on the right frac

[gloomy pine]

Why dont you unify denominators and sée what you get

[merry vault]

what do you mean unify

I don't rly know how because of the + 1 there

[gloomy pine]

Put it all in one fraction together

[merry vault]

is it rly possible?

[gloomy pine]

Multiply thé ratio on the right by the denominator of thé left and vice versa

Then add thé two fractions together

[merry vault]

ahhhhh right I forgot this trick

$\dfrac{-n^3 - n^2 + 2n^2 e^{-n} -2n + e^{-n} - 4}{n^3 + n^2 + n + 1}$

[warm shale]

madmike

[merry vault]

I have this now

Would you factor out n^3 now?

in numerator and denominator ?

[gloomy pine]

Exactly

[merry vault]

and now? 😄

$\dfrac{-1 - \dfrac{1}{n} + \dfrac{2e^{-n}}{n} - \dfrac{2}{n^2} + \dfrac{e^{-n}}{n^3} - \dfrac{4}{n^3}}{1 + \dfrac{1}{n} + \dfrac{1}{n^2} + \dfrac{1}{n^3}}$